HP-15C
Owner’s Handbook
HP Part Number: 00015-90001
Edition 2.4, Sep 2011
Legal Notice
This manual and any examples contained herein are provided “as is”
and are subject to change without notice. Hewlett-Packard Company
makes no warranty of any kind with regard to this manual, including,
but not limited to, the implied warranties of merchantability noninfringement and fitness for a particular purpose. In this regard, HP
shall not be liable for technical or editorial errors or omissions
contained in the manual.
Hewlett-Packard Company shall not be liable for any errors or
incidental or consequential damages in connection with the furnishing,
performance, or use of this manual or the examples contained herein.
Copyright © 2011 Hewlett-Packard Development Company, LP.
Reproduction, adaptation, or translation of this manual is prohibited
without prior written permission of Hewlett-Packard Company, except
as allowed under the copyright laws.
Hewlett-Packard Company
Palo Alto, CA
94304
USA
Introduction
Congratulations! Whether you are new to HP calculators or an experienced
user, you will find the HP-15C a powerful and valuable calculating tool.
The HP-15C provides:
 448 bytes of program memory (one or two bytes per instruction) and
sophisticated programming capability, including conditional and
unconditional branching, subroutines, flags, and editing.
 Four advanced mathematics capabilities: complex number calculations,
matrix calculations, solving for roots, and numerical integration.
 Direct and indirect storage in up to 67 registers.
This handbook is written for you, regardless of your level of expertise.
The beginning part covers all the basic functions of the HP-15C and how to
use them. The second part covers programming and is broken down into
three subsections – The Mechanics, Examples, and Further Information – in
order to make it easy for users with varying backgrounds to find the
information they need. The last part describes the four advanced
mathematics capabilities.
Before starting these sections, you may want to gain some operating and
programming experience on the HP-15C by working through the
introductory material, The HP-15C: A Problem Solver, on page 12.
The various appendices describe additional details of calculator operation,
as well as warranty and service information. The Function Summary and
Index and the Programming Summary and Index at the back of this manual
can be used for quick reference to each function key and as a handy page
reference to more comprehensive information inside the manual.
Also available from Hewlett-Packard is the HP-15C Advanced Functions
Handbook, which provides applications and technical descriptions for the
root-solving, integration, complex number, and matrix functions.
Note: You certainly do not need to read every part of the manual
before delving into the HP-15C Advanced Functions if you are
already familiar with HP calculators. The use of _ and f
requires a knowledge of HP-15C programming.
3
Contents
The HP-15C: A Problem Solver ....................................
A Quick Look at v .................................................
Manual Solutions ............................................................
Programmed Solutions .....................................................
12
12
13
14
Part I: HP-15C Fundamentals ................................
17
Section 1: Getting Started ..........................................
18
18
18
18
19
19
19
20
21
22
22
22
Power On and Off ..........................................................
Keyboard Operation .......................................................
Primary and Alternate Functions .....................................
Prefix Keys ..................................................................
Changing Signs ...........................................................
Keying in Exponents .....................................................
The "CLEAR" Keys ........................................................
Display Clearing: ` and − ...................................
Calculations ...................................................................
One-Number Functions .................................................
Two-Number Functions and v ...............................
Section 2: Numeric Functions .....................................
Pi ..................................................................................
Number Alteration Functions ............................................
One-Number Functions ....................................................
General Functions ........................................................
Trigonometric Operations ..............................................
Time and Angle Conversions .........................................
Degrees/Radians Conversions .......................................
Logarithmic Functions ...................................................
Hyperbolic Functions ....................................................
Two-Number Functions ....................................................
The Power Function ......................................................
Percentages .................................................................
Polar and Rectangular Coordinate Conversions ...............
24
24
24
25
25
26
26
27
28
28
29
29
29
30
Section 3: The Automatic Memory Stack, LAST X, and
Data Storage ........................................................
32
4
Contents
5
The Automatic Memory Stack and Stack Manipulation ........
Stack Manipulation Functions ........................................
The LAST X Register and K .......................................
Calculator Functions and the Stack .................................
Order of Entry and the v Key ...............................
Nested Calculations .....................................................
Arithmetic Calculations With Constants ...........................
Storage Register Operations ............................................
Storing and Recalling Numbers .....................................
Clearing Data Storage Registers ....................................
Storage and Recall Arithmetic ........................................
Overflow and Underflow ..............................................
Problems ........................................................................
32
33
35
36
37
38
39
42
42
43
43
45
45
Section 4: Statistics Functions .....................................
47
47
48
49
52
53
53
54
55
57
Probability Calculations ...................................................
Random Number Generator .............................................
Accumulating Statistics .....................................................
Correcting Accumulated Statistics ...................................
Mean ..........................................................................
Standard Deviation .......................................................
Linear Regression .........................................................
Linear Estimation and Correlation Coefficient ...................
Other Applications .......................................................
Section 5: The Display and Continuous Memory ...........
Display Control ..............................................................
Fixed Decimal Display ..................................................
Scientific Notation Display ............................................
Engineering Notation Display ........................................
Mantissa Display .........................................................
Round-Off Error ............................................................
Special Displays .............................................................
Annunciators ...............................................................
Digit Separators ...........................................................
Error Display ...............................................................
Overflow and Underflow ..............................................
Low-Power Indication ....................................................
Continuous Memory ........................................................
Status .........................................................................
58
58
58
59
59
60
60
60
60
61
61
61
62
62
62
6
Contents
Resetting Continuous Memory ........................................
63
Part II: HP-15C Programming ...............................
65
Section 6: Programming Basics ..................................
66
66
66
66
68
68
69
70
74
74
74
75
77
78
78
79
79
80
81
The Mechanics ...............................................................
Creating a Program .....................................................
Loading a Program ......................................................
Intermediate Program Stops ...........................................
Running a Program .......................................................
How to Enter Data ........................................................
Program Memory .........................................................
Further Information ..........................................................
Program Instructions .....................................................
Instruction Coding ........................................................
Memory Configuration ..................................................
Program Boundaries .....................................................
Unexpected Program Stops ...........................................
Abbreviated Key Sequences ..........................................
User Mode ..................................................................
Polynomial Expressions and Horner's Method ..................
Nonprogrammable Functions .........................................
Problems ........................................................................
Section 7: Program Editing ........................................
The Mechanics ...............................................................
Moving to a Line in Program Memory .............................
Deleting Program Lines .................................................
Inserting Program Lines .................................................
Examples .......................................................................
Further Information ..........................................................
Single-Step Operations .................................................
Line Position ................................................................
Insertions and Deletions ................................................
Initializing Calculator Status ..........................................
Problems ........................................................................
Section 8: Program Branching and Controls .................
The Mechanics ...............................................................
Branching ...................................................................
Conditional Tests ..........................................................
82
82
82
83
83
83
85
85
86
87
87
87
90
90
90
91
Contents
7
Flags ..........................................................................
Examples .......................................................................
Example: Branching and Looping ...................................
Example: Flags ............................................................
Further Information ..........................................................
GoTo ..........................................................................
Looping ......................................................................
Conditional Branching ..................................................
Flags ..........................................................................
The System Flags: Flags 8 and 9 ....................................
92
93
93
95
97
97
98
98
98
99
Section 9: Subroutines ...............................................
101
101
101
102
102
105
105
105
The Mechanics ...............................................................
GoTo Subroutine and Return ..........................................
Subroutine Limits ..........................................................
Examples .......................................................................
Further Information ..........................................................
The Subroutine Return ...................................................
Nested Subroutines ......................................................
Section 10: The Index Register and Loop Control ...........
The V and % Keys ....................................................
Direct Versus Indirect Data Storage With
The Index Register .....................................................
Indirect Program Control With the Index Register .............
Program Loop Control ...................................................
The Mechanics ...............................................................
Index Register Storage and Recall ..................................
Index Register Arithmetic ...............................................
Exchanging the X-Register .............................................
Indirect Branching With V .........................................
Indirect Flag Control With V ......................................
Indirect Display Format Control With V .......................
Loop Control with Counters: I and e ..................
Examples .......................................................................
Examples: Register Operations .......................................
Example: Loop Control With s .................................
Example: Display Format Control ....................................
Further Information ..........................................................
Index Register Contents .................................................
106
106
106
107
107
107
107
108
108
108
109
109
109
111
111
112
114
115
115
8
Contents
I and e ..........................................................
Indirect Display Control ...........................................
116
116
Part III: HP-15C Advanced Functions ....................
119
Section 11: Calculating With Complex Numbers ..........
120
120
120
121
121
121
124
124
124
125
128
129
130
130
131
131
131
132
133
133
135
137
The Complex Stack and Complex Mode ............................
Creating the Complex Stack ..........................................
Deactivating Complex Mode .........................................
Complex Numbers and the Stack ......................................
Entering Complex Numbers ...........................................
Stack Lift in Complex Mode ...........................................
Manipulating the Real and Imaginary Stacks ..................
Changing Signs ..........................................................
Clearing a Complex Number .......................................
Entering a Real Number ...............................................
Entering a Pure Imaginary Number ...............................
Storing and Recalling Complex Numbers .......................
Operations With Complex Numbers ................................
One-Number Functions ................................................
Two-Number Functions .................................................
Stack Manipulation Functions .......................................
Conditional Tests .........................................................
Complex Results from Real Numbers ..............................
Polar and Rectangular Coordinate Conversions .................
Problems .......................................................................
For Further Information ...................................................
Section 12: Calculating With Matrices ........................
Matrix Dimensions .........................................................
Dimensioning a Matrix .................................................
Displaying Matrix Dimensions .......................................
Changing Matrix Dimensions ........................................
Storing and Recalling Matrix Elements ..............................
Storing and Recalling All Elements in Order ...................
Checking and Changing Matrix Elements Individually .....
Storing a Number in All Elements of a Matrix .................
Matrix Operations .........................................................
Matrix Descriptors .......................................................
The Result Matrix .........................................................
138
140
141
142
142
143
143
145
147
147
147
148
Contents
Copying a Matrix .......................................................
One-Matrix Operations ................................................
Scalar Operations .......................................................
Arithmetic Operations ..................................................
Matrix Multiplication ...................................................
Solving the Equation AX = B ..........................................
Calculating the Residual ...............................................
Using Matrices in LU Form ............................................
Calculations With Complex Matrices ...............................
Storing the Elements of a Complex Matrix ......................
The Complex Transformations Between ZP and Z .............
Inverting a Complex Matrix ..........................................
Multiplying Complex Matrices ......................................
Solving the Complex Equation AX = B ............................
Miscellaneous Operations Involving Matrices .....................
Using a Matrix Element With Register Operations ............
Using Matrix Descriptors in the Index Register .................
Conditional Tests on Matrix Descriptors ..........................
Stack Operation for Matrix Calculations ............................
Using Matrix Operations in a Program ..............................
Summary of Matrix Functions ...........................................
For Further Information ....................................................
149
149
151
153
154
156
159
160
160
161
164
165
166
168
173
173
173
174
174
176
177
179
Section 13: Finding the Roots of an Equation ................
180
180
186
188
192
193
193
193
Using _ .................................................................
When No Root Is Found ..................................................
Choosing Initial Estimates ................................................
Using _ in a Program ..............................................
Restriction on the Use of _ .......................................
Memory Requirements .....................................................
For Further Information ....................................................
Section 14: Numerical Integration ..............................
Using f.......................................................................
Accuracy of f .............................................................
Using f in a Program ..................................................
Memory Requirements .....................................................
For Further Information ....................................................
194
194
200
203
204
204
9
10
Contents
Appendix A: Error Conditions ....................................
205
Appendix B: Stack Lift and the LAST X Register ...............
209
209
209
210
210
211
212
Digit Entry Termination ....................................................
Stack Lift ........................................................................
Disabling Operations ...................................................
Enabling Operations ....................................................
Neutral Operations ......................................................
LAST X Register ...............................................................
Appendix C: Memory Allocation ................................
The Memory Space .........................................................
Registers .....................................................................
Memory Status (W) ..................................................
Memory Reallocation ......................................................
The m % Function ................................................
Restrictions on Reallocation ...........................................
Program Memory ............................................................
Automatic Program Memory Reallocation ........................
Two-Byte Program Instructions .......................................
Memory Requirements for the Advanced Functions .............
Appendix D: A Detailed Look at _ ......................
How _ Works .......................................................
Accuracy of the Root .......................................................
Interpreting Results ..........................................................
Finding Several Roots ......................................................
Limiting the Estimation Time ..............................................
Counting Iterations .......................................................
Specifying a Tolerance .................................................
For Advanced Information ................................................
Appendix E: A Detailed Look at f ...........................
How f Works .............................................................
Accuracy, Uncertainty, and Calculation Time .....................
Uncertainty and the Display Format ...................................
Conditions That Could Cause Incorrect Results ....................
Conditions That Prolong Calculation Time ..........................
Obtaining the Current Approximation to an Integral ...........
For Advanced Information ................................................
213
213
213
215
215
215
216
217
217
218
218
220
220
222
226
233
238
239
239
239
240
240
241
245
249
254
257
258
Contents
Appendix F: Batteries .............................................
Low-Power Indication .......................................................
Installing New Batteries ................................................
Verifying Proper Operation (Self-Tests) ...............................
Function Summary and Index .....................................
11
259
259
259
261
Complex Functions ..........................................................
Conversions ...................................................................
Digit Entry ......................................................................
Display Control ..............................................................
Hyperbolic Functions .......................................................
Index Register Control .....................................................
Logarithmic and Exponential Functions ..............................
Mathematics ..................................................................
Matrix Functions .............................................................
Number Alteration ..........................................................
Percentage .....................................................................
Prefix Keys .....................................................................
Probability .....................................................................
Stack Manipulation .........................................................
Statistics ........................................................................
Storage .........................................................................
Trigonometry ..................................................................
262
262
262
262
263
263
263
263
264
264
265
266
266
266
266
267
267
268
Programming Summary and Index ..............................
269
Subject Index ...........................................................
271
The HP-15C:
A Problem Solver
The HP-15C Advanced Programmable Scientific Calculator is a powerful
problem solver, convenient to carry and easy to hold. Its continuous
memory retains data and program instructions indefinitely until you choose
to reset it. Though sophisticated, it requires no prior programming
experience or knowledge of programming languages to use it.
The new HP-15C is a modern re-release of the original HP-15C introduced
in 1982. While the battery life of the new version is now estimated to be 1
year for normal use, the calculator is now at least 150 times faster than the
original. The low-power indicator gives you plenty of warning before the
calculator stops functioning.
The HP-15C also conserves power by automatically shutting its display off
if it is left inactive for a few minutes. But don't worry about losing data –
any information contained in the HP-15C is saved by Continuous Memory.
A Quick Look at v
Your Hewlett-Packard calculator uses a unique operating logic, represented
by the v key, that differs from the logic in most other calculators.
You will find that using v makes nested and complicated
calculations easier and faster to work out. Let's get acquainted with how this
works.
For example, let's look at the arithmetic functions. First we have to get the
numbers into the machine. Is your calculator on? If not, press =. Is the
display cleared? To display all zeros, you can press | ` that is, press
|, then −.* To perform arithmetic, key in the first number, press v
to separate the first number from the second, then key in the second number
and press +, -, * or ÷. The result appears immediately after you
press any numerical function key.
*
If you have not used an HP calculator before, you will notice that most keys have three labels. To use the
primary function – the one printed in white on top of the key – just press that key. For those printed in gold
or blue, press the gold ´ key or the blue | key first.
12
The HP-15C: A Problem Solver
13
The display format used in this handbook is • 4 (the decimal point is
―fixed‖ to show four decimal places) unless otherwise mentioned. If your
calculator does not show four decimal places, you may want to press
´• 4 to match the displays in the examples.
Manual Solutions
Run through the following two-number calculations. It is not necessary to
clear the calculator between problems. If you enter a digit incorrectly, press
− to undo the mistake, then key in the correct number.
To Compute
9-6=3
9 × 6 = 54
9 ÷ 6 = 1.5
6
9 = 531,441
Keystrokes
9v69v6*
9v6÷
9v6Y
Display
3.0000
54.0000
1.5000
531,441.0000
Notice that in the four examples:
 Both numbers are in the calculator before you press the function key.
 v is used only to separate two numbers that are keyed in one
after the other.
 Pressing a numeric function key, in this case - * ÷ or Y,
executes the function immediately and displays the result.
To see the close relationship between manual and programmed problem
solving, let's first calculate the solution to a problem manually, that is, from
the keyboard. Then we'll use a program to calculate the solution to the same
problem with different data.
14
The HP-15C: A Problem Solver
The time an object takes to fall to the ground (ignoring air friction) is given
by the formula
t
2h
,
g
where t = time in seconds,
h = height in meters,
g = the acceleration due to gravity,
9.8 m/s2.
Example: Compute the time taken by a
stone falling from the top of the Eiffel
Tower (300.51 meters high) to the earth.
Keystrokes
300.51 v
2*
9.8 ÷
¤
Display
300.5100
601.0200
61.3286
7.8313
Enter h.
Calculates 2h.
(2h) /g.
Falling time, seconds.
Programmed Solutions
Suppose you wanted to calculate falling times from various heights. The
easiest way is to write a program to cover all the constant parts of a
calculation and provide for entry of variable data.
Writing the Program. The program is similar to the keystroke sequence
you used above. A label is useful to define the beginning of a program, and
a return is useful to mark the end of a program. Also, the program must
accommodate the entry of new data.
Loading the Program. You can load a program for the above problem by
pressing the following keys in sequence. (The display shows information
which you can ignore for now, though it will be useful later.)
The HP-15C: A Problem Solver
Keystrokes
15
Display
|¥
000-
´ CLEAR M
000-
´bA
001-42,21,11
2
*
9
002003004-
.
8
÷
¤
|n
005006007008009-
|¥
7.8313
Sets HP-15C to Program
mode. (PRGM
annunciator on.)
Clears program memory.
(This step is optional
here.)
Label "A" defines the
beginning of the
program.
2
20
9
48
8
10
11
43 32
The same keys you
pressed to solve the
problem manually.
―Return‖ defines the end
of the program.
Switches to Run mode.
(No PRGM
annunciator.)
Running the Program. Enter the following information to run the
program.
Keystrokes
300.51
Display
300.51
´A
7.8313
1050 ´A
14.6385
Height of the Eiffel Tower.
Falling time you calculated
earlier.
The time (seconds) for a stone
to reach the ground after release
from a blimp 1050 m high.
16
The HP-15C: A Problem Solver
With this program loaded, you can quickly calculate the time of descent of
an object from different heights. Simply key in the height and press
´A. Find the time of descent for objects released from heights of
100 m, 2 m, 275 m, and 2,000 m.
The answers are: 4.5175 s; 0.6389 s; 7.4915 s; and 20.2031 s.
That program was relatively easy. You will see many more aspects and
details of programming in part II. For now, turn the page to take an in-depth
look at some of the calculator's important operating basics.
Part l
HP-15C
Fundamentals
Section 1
Getting Started
Power On and Off
The = key turns the HP-15C on and off.* To conserve power, the
calculator automatically turns itself off after a few minutes of inactivity.
Keyboard Operation
Primary and Alternate Functions
Most keys on your HP-15C perform one primary and two alternate, shifted
functions. The primary function of any key is indicated by the character(s)
on the face of the key. The alternate functions are indicated by the gold
characters printed above the key and the blue characters printed on the
lower face of the key.
 To select the primary function printed on
the face of a key, press only that key. For
example: ÷.

To select the alternate function printed in
gold or blue, press the like-colored prefix
key (´ or |) followed by the function
key. For example: ´ _; |
£.
Throughout this handbook, we will observe certain conventions in referring
to alternate functions. References to the function itself will appear as just the
key name in a box, such as ―the W function.‖ References to the use of
the key will include the prefix key, such as ―press | W.‖ References
to the four gold functions printed under the bracket labeled ―CLEAR‖ will
be preceded by the word ―CLEAR‖, such as "the CLEAR Q function,‖
or ―press ´ CLEAR M.‖
*
Note that the = key is lower than the other keys to help prevent its being pressed inadvertently.
18
Section 1: Getting Started
Notice that when you press the ´ or |
prefix key, an f or g annunciator appears
and remains in the display until a function
key is pressed to complete the sequence.
19
0.0000
f
Prefix Keys
A prefix key is any key which must precede another key to complete the
key sequence for a function. Certain functions require two parts: a prefix
key and a digit or other key. For your reference, the prefix keys are:
" ^ •
m ´ |
G
P
f
I
>
l
i
F
s ?
H
b
<
_ X
t
O
T
If you make a mistake while keying in a prefix for a function, press ´
CLEAR u to cancel the error. The CLEAR u key is also used
to show the mantissa of a displayed number, so all 10 digits of the number
in the display will appear for a moment after the u key is pressed.
Changing Signs
Pressing “ (change sign) will change the sign (positive or negative) of
any displayed number. To key in a negative number, press “ after its
digits have been keyed in.
Keying in Exponents
‛ (enter exponent) is used when keying in a number with an exponent.
First key in the mantissa, then press ‛ and key in the exponent.
For a negative exponent press “ after keying in the exponent.* For
example, to key in Planck's constant (6.6262×10-34 Joule-seconds) and
multiply it by 50:
*
“ may also be pressed after ‛ and before the exponent, with the same result (unlike the mantissa,
where digit entry must precede “).
20
Section 1: Getting Started
Keystrokes
6.6262
‛
3
4
“
v
50 *
Display
6.6262
6.6262
00
6.6262
6.6262
6.6262
6.6262
3.3131
03
34
-34
-34
-32
The 00 prompts you to
key in the exponent.
(6.6262×103).
(6.6262×1034).
(6.6262×10-34).
Enters number.
Joule-seconds.
Note: Decimal digits from the mantissa that spill into the exponent
field will disappear from the display when you press ―, but will be
retained internally.
To prevent a misleading display pattern, ‛ will not operate with a
number having more than seven digits to the left of the radix mark (decimal
point), nor with a mantissa smaller than 0.000001. To key in such a number,
use a form having a greater exponent value (whether positive or negative).
For example, 123456789.8×1023 can be keyed in as 1234567.898×1025;
0.00000025×10-15 can be keyed in as 2.5×10-22.
The “CLEAR” Keys
Clearing means to replace a number with zero. The clearing operations in
the HP-15C are (the table is continued on the next page):
Clearing Sequence
|`
−
In Run mode:
In Program mode:
´ CLEAR ∑
Effect
Clears display (X-register).
Clears last digit or entire display.
Deletes current instruction.
Clears statistics storage registers, display,
and the memory stack (described in
section 3).
Section 1: Getting Started
Clearing Sequence
´ CLEAR M
In Run mode:
In Program mode:
´ CLEAR Q
´ CLEAR u*
21
Effect
Repositions program memory to line 000.
Deletes all program memory.
Clears all data storage registers.
Clears any prefix from a partially entered
key sequence.
* Also temporarily displays the mantissa.
Display Clearing: ` and −
The HP-15C has two types of display clearing operations: ` (clear X)
and − (back arrow).
In Run mode:

` clears the display to zero.

− deletes only the last digit in the display if digit entry has not
been terminated by v or most other functions. You can then
key in a new digit or digits to replace the one(s) deleted. If digit entry
has been terminated, then − acts like `.
Keystrokes
12345
−
9
¤
−
Display
12,345
1,234
12,349
111.1261
0.0000
Digit entry not terminated.
Clears only the last digit.
Terminates digit entry.
Clears all digits to zero.
In Program mode:
 ` is programmable: it is stored as a programmed instruction,
and will not delete the currently displayed instruction.
 − is not programmable, so it can be used for program correction.
Pressing − will delete the entire instruction currently displayed.
22
Section 1: Getting Started
Calculations
One-Number Functions
A one-number function performs an operation using only the number in the
display. To use any one-number function, press the function key after the
number has been placed in the display.
Keystrokes
45
|o
Display
45
1.6532
Two-Number Functions and v
A two-number function must have two numbers present in the calculator
before executing the function. +, -, * and ÷ are examples of
two-number functions.
Terminating Digit Entry. When keying in two numbers to perform an
operation, the calculator needs a signal that digit entry is terminated for the
first number. This is done by pressing v to separate the two numbers.
If, on the other hand, one of the numbers is already in the calculator as the
result of a previous operation, you do not need to use the v key. All
functions except the digit entry keys themselves* have the effect of
terminating digit entry.
Notice that, regardless of the number, a decimal point always appears and a
set number of decimal places are displayed when you terminate digit entry
(as by pressing v).
Chain Calculations. In the following calculations, notice that:
 The v key is used only for separating the sequential entry of
two numbers.
 The operator is keyed in only after both operands are in the calculator.
 The result of any operation may itself become an operand. Such
intermediate results are stored and retrieved on a last-in, first-out
basis. New digits keyed in following an operation are treated as a new
number.
*
The digit keys, +, “, ‛, and −.
Section 1: Getting Started
23
Example: Calculate (9 + 17  4) ÷ 4.
Keystrokes
9v
17 +
44÷
Display
9.0000
26.0000
22.0000
5.5000
Digit entry terminated.
(9 + 17).
(9 + 17 – 4).
(9 + 17 – 4) ÷ 4.
Even more complicated problems are solved in the same manner-using
automatic storage and retrieval of intermediate results. It is easiest to work
from the inside of parentheses outwards, just as you would with calculations
on paper.
Example: Calculate (6 + 7) × (9  3)
Keystrokes
6v
Display
6.0000
7+
9v
13.0000
9.0000
3*
6.0000
78.0000
First solve for the
intermediate result of (6 + 7).
Then solve for the
intermediate result of (9  3).
Then multiply the
intermediate results together
(13 and 6) for the final
answer.
Try your hand at the following problems. Each time you press v or a
function key in a calculation, the preceding number is saved for the next
operation.
(16 × 38) – (13 × 11) = 465.0000
4 × (17 – 12) ÷ (10 – 5) = 4.0000
232 – (13 × 9) + 1/7 = 412.1429
[(5.4  0.8)  (12.5  0.72 )]  0.5998
Section 2
Numeric Functions
This section discusses the numeric functions of the HP-15C (excluding
statistics and advanced functions). The nonnumeric functions are discussed
separately (digit entry in section 1, stack manipulation in section 3, and
display control in section 5).
The numeric functions of the HP-15C are used in the same way whether
executed from the keyboard or in a program. Some of the functions (such as
a) are, in fact, primarily of interest for programming.
Remember that the numeric functions, like all functions except digit entry
functions, automatically terminate digit entry. This means a numeric
function does not need to be preceded or followed by v.
Pi
Pressing | $ places the first 10 digits of π into the calculator. $
does not need to be separated from other numbers by v.
Number Alteration Functions
The number alteration functions act upon the number in the display
(X-register).
Integer Portion. Pressing | ‘ replaces the number in the display
with the nearest integer of lesser or equal magnitude.
Fractional Portion. Pressing ´ q replaces the number in the display
with its fractional part (that is, the difference between the number and its
integer part).
Rounding. Pressing | & rounds all 10 internally held digits of the
mantissa of the displayed value to the number of digits specified by the
current •, i, or ^ display format.
Absolute Value. Pressing | a yields the absolute value of the
number in the display.
24
Section 2: Numeric Functions
Keystrokes
Display
123.0000
123.4567 |‘
-123.0000
|K “ |‘
|K ´q
1.23456789 “
|&
´ CLEAR u
(release)
|a
25
Reversing the sign does
not alter digits.
-0.4567
-1.2346
1234600000 Temporarily displays all
-1.2346
digits in the mantissa.
1.2346
One-Number Functions
One-number math functions in the HP-15C operate only upon the number in
the display (X-register).
General Functions
Reciprocal. Pressing ∕ calculates the reciprocal of the number in the
display.
Factorial and Gamma. Pressing ´ ! calculates the factorial of the
displayed value, where x is an integer 0≤x≤69.
You can also use ! to calculate the Gamma function, Γ(x), used in
advanced mathematics and statistics. Pressing ´ ! calculates Γ(x + 1),
so you must subtract 1 from your initial operand to get Γ(x). For the Gamma
function, x is not restricted to nonnegative integers.
Square Root. Pressing ¤ calculates the positive square root of the
number in the display.
Squaring. Pressing | x calculates the square of the number in the
display.
Keystrokes
25 ∕
8´!
3.9 ¤
12.3 | x
Display
0.0400
40,320.0000
1.9748
151.2900
Calculates 8! or Γ(9).
26
Section 2: Numeric Functions
Trigonometric Operations
Trigonometric Modes. The trigonometric functions operate in the
trigonometric mode you select. Specifying a trigonometric mode does not
convert any number already in the calculator to that mode; it merely tells
the calculator what unit of measure (degrees, radians, or grads) to assign a
number for a trigonometric function.
Pressing | D sets Degrees mode. No annunciator appears in the
display. Degrees are in decimal, not minutes-seconds form.
Pressing | R sets Radians mode. The RAD annunciator appears in
the display. In Complex mode, all functions (except : and ;) assume
values are in radians, regardless of the trigonometric annunciator displayed.
Pressing | g sets Grads mode. The GRAD annunciator appears in
the display.
Continuous Memory will maintain the last trigonometric mode selected. At
"power up" (initial condition or when Continuous Memory is reset), the
calculator is in Degrees mode,
Trigonometric Functions. Given x in the display (X-register):
Pressing
[
|,
\
|{
]
|/
Calculates
sine of x
arc sine of x
cosine of x
arc cosine of x
tangent of x
arc tangent of x
Before executing a trigonometric function, be sure that the calculator is set
to the desired trigonometric mode (Degrees, Radians, or Grads).
Time and Angle Conversions
Numbers representing time (hours) or angles (degrees) can be converted by
the HP-15C between a decimal-fraction and a minutes-seconds format:
Section 2: Numeric Functions
27
Hours.Decimal Hours
Hours.Minutes Seconds Decimal Seconds
(H.h)
(H.MMSSs)
Degrees.Decimal Hours
Degrees.Minutes Seconds Decimal Seconds
(D.d)
(D.MMSSs)
Hours/Degrees-Minutes-Seconds Conversion. Pressing ´ h
converts the number in the display from a decimal hours/degrees format to
an hours/degree-minutes-seconds-decimal seconds format.
For example, press ´ h to convert
1.2 3 4 5
1.1404
seconds
minutes
hours
to
hours
Press ´ u to display the value to all possible decimal places:
1140420000
to the hundred-thousandth of a second.
Decimal Hours (or Degrees) Conversion. Pressing | À converts the
number in the display from an hours/degrees-minutes-seconds-decimal
seconds format to a decimal hours/degrees format.
Degrees/Radians Conversions
The d and r functions are used to convert angles to degrees
or radians (D.dR.r). The degrees must be expressed as decimal numbers,
and not in a minutes-seconds format.
Keystrokes
Display
0.7069
40.5 ´ r
40.5000
|d
Radians.
40.5 degrees (decimal fraction).
28
Section 2: Numeric Functions
Logarithmic Functions
Natural Logarithm. Pressing |Z calculates the natural logarithm of
the number in the display; that is, the logarithm to the base e.
Natural Antilogarithm. Pressing ' calculates the natural antilogarithm
of the number in the display; that is, raises e to the power of that number.
Common Logarithm. Pressing | o calculates the common
logarithm of the number in the display; that is, the logarithm to the base 10.
Common Antilogarithm. Pressing @ calculates the common
antilogarithm of the number in the display; that is, raises 10 to the power of
that number.
Keystrokes
45 |Z
Display
3.8067
Natural log of 45.
3.4012 '
30.0001
Natural antilog of 3.4012.
12.4578 | o
3.1354 @
1.0954
1,365.8405
Common log of 12.4578.
Common antilog of
3.1354.
Hyperbolic Functions
Given x in the display (X-register):
Pressing
´P[
|H[
´P\
|H\
´P]
Calculates
hyperbolic sine of x
inverse hyperbolic sine of x
hyperbolic cosine of x
inverse hyperbolic cosine of x
hyperbolic tangent of x
|H]
inverse hyperbolic tangent of x
Section 2: Numeric Functions
29
Two-Number Functions
The HP-15C performs two-number math functions using two values entered
sequentially into the display. If you are keying in both numbers, remember
that they must be separated by v or any other function – like |
‘ or ∕ – that terminates digit entry.
For a two-number function, the first value entered is considered the y-value
because it is placed into the Y-register for memory storage. The second
value entered is considered the x-value because it remains in the display,
which is the X-register.
The arithmetic operators, +, -, *, and ÷, are the four basic twonumber functions. Others are given below.
The Power Function
Pressing Y calculates the value of y raised to the x power. The base
number, y, is keyed in before the exponent, x.
To Calculate
1.4
2
2-1.4
(-2)3
3
2 or 2
1/3
Keystrokes
2 v 1.4 Y
2 v 1.4 “ Y
2“v3Y
2v3∕Y
Display
2.6390
0.3789
-8.0000
1.2599
Percentages
The percentage functions, k and ∆, preserve the value of the original
base number along with the result of the percentage calculation. As shown
in the example below, this allows you to carry out subsequent calculations
using the base number and the result without re-entering the base number.
Percent. The k function calculates the specified percentage of a base
number.
30
Section 2: Numeric Functions
For example, to find the sales tax at 3% and total cost of a $15.76 item:
Keystrokes
15.76 v
3 |k
+
Display
15.7600
0.4728
16.2328
Enters the base number (the price).
Calculates 3% of $15.76 (the tax).
Total cost of item ($15.76 + $0.47).
Percent Difference. The ∆ function calculates the percent difference
between two numbers. The result expresses the relative increase (a positive
result) or decrease (a negative result) of the second number entered
compared to the first number entered.
For example, suppose the $15.76 item only cost $14.12 last year. What is
the percent difference in last year’s price relative to this year’s?
Keystrokes
15.76 v
Display
15.7600
14.12 |∆
-10.4061
This year's price (our base number)
Last year's price was 10.41% less
than this year's price.
Polar and Rectangular Coordinate Conversions
The : and ; functions are provided in the
HP-15C for conversions between polar
coordinates and rectangular coordinates. The
angle θ is assumed to be in the mode, whether
degrees (in a decimal format, not a minutesseconds format), radians, or grads. θ is
measured as shown in the illustration at right.
Polar Conversion. Pressing |:
(polar) converts a set of rectangular coordinates (x, y) to polar coordinates
(magnitude r, angle θ). The y-value must be entered first, the x-value
second. Upon executing |: r will appear in the display. Press ®
(X exchange Y) to bring θ out of the Y-register and into the display (Xregister). θ will be returned as a value between -180° and 180°, between -π
and π radians, or between -200 and 200 grads.
Section 2: Numeric Functions
31
Rectangular Conversion. Pressing ´; (rectangular) converts a set of
polar coordinates (magnitude r angle θ) into rectangular coordinates (x, y). θ
must be entered first then r. Upon executing ´;, x will be displayed
first; press ® to display y.
Keystrokes
|D
Display
Set to Degrees mode (no annunciator).
5v
10
|:
®
5.0000
10
11.1803
26.5651
30 v
12
´;
®
30.0000
12
10.3923
6.0000
y-value.
x-value.
r.
θ; rectangular coordinates converted to
polar coordinates.
θ.
r.
x-value.
y-value. Polar coordinates converted to
rectangular coordinates.
Section 3
The Automatic Memory Stack,
LAST X, and Data Storage
The Automatic Memory Stack
and Stack Manipulation
HP operating logic is based on a mathematical logic known as ―Polish
Notation,‖ developed by the noted Polish logician Jan Łukasiewicz
(Wookashye'veech) (1878-1956). Conventional algebraic notation places the
algebraic operators between the relevant numbers or variables when
evaluating algebraic expressions. Łukasiewicz’s notation specifies the
operators before the variables. For optimal efficiency of calculator use, HP
applied the convention of specifying (entering) the operators after
specifying (entering) the variable(s). Hence the term "Reverse Polish
Notation" (RPN).
The HP-15C uses RPN to solve complicated calculations in a
straightforward manner, without parentheses or punctuation. It does so by
automatically retaining and returning intermediate results. This system is
implemented through the automatic memory stack and the v key,
minimizing total keystrokes.
The Automatic
Memory Stack Registers
T 0.0000
Z 0.0000
Y 0.0000
X 0.0000
Always displayed
When the HP-15C is in Run mode (no PRGM annunciator displayed), the
number that appears in the display is the number in the X-register.
32
Section 3: The Memory Stack, LAST X, and Data Storage
33
Any number that is keyed in or results from the execution of a numeric
function is placed into the display (X-register). This action will cause
numbers already in the stack to lift, remain in the same register, or drop,
depending upon both the immediately preceding and the current operation.
Numbers in the stack are stored on a last-in, first-out basis. The three stacks
drawn below illustrate the three types of stack movement. Assume x, y, z,
and t represent any numbers which may be in the stack.
Stack Lift
No Stack Lift or Drop
lost
T
t
z
T
t
t
Z
z
y
Z
z
z
Y
y
x
Y
y
y
X
x
π
X
x
x
Keys:
|$
¤
Stack Drop
T
t
t
Z
z
t
Y
y
z
X
x
x+y
Keys:
+
Notice the number in the T-register remains there when the stack drops,
allowing this number to be used repetitively as an arithmetic constant.
Stack Manipulation Functions
v. Pressing v separates two numbers keyed in one after the
other. It does so by lifting the stack and copying the number in the display
(X-register) into the Y-register. The next number entered then writes over
the value in the X-register; there is no stack lift. The example below shows
what happens as the stack is filled with the numbers 1, 2, 3, 4. (The
34
Section 3: The Memory Stack, LAST X, and Data Storage
shading indicates that the contents of that register will be written over
when the next number is keyed in or recalled.)
T t
Z z
Y y
X x
Keys:
lost
lost
z
y
x
1
y
x
1
1
v
1
lost
y
x
1
2
x
1
2
2
v
2
lost
T
Z
Y
X
x
1
2
2
Keys:
x
1
2
3
1
2
3
3
v
3
1
2
3
4
4
) (roll down), ( (roll up), and ® (X exchange Y). ) and (
roll the contents of the stack registers up or down one register (one value
moves between the X- and the T-register). No values are lost. ®
exchanges the numbers in the X- and Y-registers. If the stack were loaded
with the sequence 1, 2, 3, 4, the following shifts would result from
pressing )) and ®.
T
Z
Y
X
Keys:
1
2
3
4
4
1
2
3
)
1
2
3
4
|
(
1
2
4
3
®
Section 3: The Memory Stack, LAST X, and Data Storage
35
The LAST X Register and K
The LAST X register, a separate memory register, preserves the value that
was last in the display before execution of a numeric operation.* Pressing
|K (LAST X) places a copy of the contents of the LAST X register
into the display (X-register). For example:
lost
T
t
t
z
Z
z
z
y
Y
y
y
16
X
4
LAST X:
16
|x
Keys:
/
4
|K
4
4
The K feature saves you from having to re-enter numbers you want to
use again (as shown under Arithmetic Calculations With Constants, page
39). It can also assist you in error recovery, such as executing the wrong
function or keying in the wrong number.
For example, suppose you mistakenly entered the wrong divisor in a chain
calculation:
Keystrokes
287 v
12.9 +
|K
*
Display
287.0000
22.2481
12.9000
Oops! The wrong divisor.
Retrieves from LAST X the last
entry to the X-register (the
incorrect divisor) before +
was executed.
Unless that operation was ’, S, or L, which don’t use or preserve the value in the display (Xregister), but instead calculate from data in the statistics storage registers (R2 to R7). For a complete list of
operations which save x in LAST X, refer to appendix B.
36
Section 3: The Memory Stack, LAST X, and Data Storage
Keystrokes
*
Display
287.0000
13.9 +
20.6475
Reverses the function that
produced the wrong answer.
The correct answer.
Calculator Functions and the Stack
When you want to key in two numbers, one after the other, you press
v between entries of the numbers. However, when you want to key
in a number immediately following any function (including manipulations
like )), you do not need to use v. Why? Executing most HP-15C
functions has this additional effect:
• The automatic memory stack is lift-enabled that is, the stack will lift
automatically when the next number is keyed or recalled into the
display.
• Digit entry is terminated, so the next number starts a new entry.
lost
T
Z
Y
X
Keys:
t
z
y
t
z
y
4
2
¤
z
y
z
z
y
2
5
5
7
+
There are four functions – v, `, z, and w – that disable stack
lift.* They do not provide for the lifting of the stack when the next number is
keyed in or recalled. Following the execution of one of these functions, a new
number will simple write over the currently displayed number instead of causing
the stack to lift. (Although the stack lifts when v is pressed, it will not lift
when the next number is keyed in or recalled. The operation of v
illustrated on page 34 shows how v thus disables the stack.) In most
cases, the above effects will come so naturally that you won’t even think
about them.
*
− will also disable the stack lift if digit entry is terminated, making − clear the entire display like
`. Otherwise, it is neutral. For a further discussion of the stack, refer to appendix B.
Section 3: The Memory Stack, LAST X, and Data Storage
37
lost
T
Z
Y
X
z
z
y
z
z
y
z
z
y
z
z
y
7
0
6
y
|`
Keys:
6
Y
6
Order of Entry and the v Key
An important aspect of two-number functions is the positioning of the
numbers in the stack. To execute an arithmetic function, the numbers should
be positioned in the stack in the same way that you would vertically
position them on paper. For example:
98
-15
98
+15
98
x15
98
15
As you can see, the first (or top) number would be in the Y-register, while
the second (or bottom) number would be in the X-register. When the
mathematics operation is performed, the stack drops, leaving the result in
the X-register. Here is how a subtraction operation is executed in the
calculator:
T
Z
Y
X
Keys:
t
z
y
x
lost
lost
z
y
x
y
x
y
x
98
98
98
15
98
98
v
15
y
y
x
83
-
The same number positioning would be used to add 15 to 98, multiply 98 by
15, or divide 98 by 15.
38
Section 3: The Memory Stack, LAST X, and Data Storage
Nested Calculations
The automatic stack lift and stack drop make it possible to do nested
calculations without using parentheses or storing intermediate results. A
nested calculation is solved simply as a series of one- and two-number
operations.
Almost every nested calculation you are likely to encounter can be done
using just the four stack registers. It is usually wisest to begin the
calculation at the innermost number or pair of parentheses and work
outward (as you would for a manual calculation). Otherwise, you may need
to place an intermediate result into a storage register. For example, consider
the calculation of
3 [4 + 5 (6 + 7)] :
Keystrokes
6v7+
Display
13.0000
5*
65.0000
4+
69.0000
3*
207.0000
Intermediate result of
(6 + 7).
Intermediate result of
5 (6 + 7).
Intermediate result of
[4 + 5 (6 + 7)].
Final result:
3 [4 + 5 (6 + 7)].
The following sequence illustrates the stack manipulation in this example.
The stack automatically drops after each two-number calculation, and then
lifts when a new number is keyed in. (For simplicity, throughout the rest of
this handbook we will not show arrows between the stacks.)
T
t
z
y
y
y
Z
z
y
x
x
y
Y
y
x
6
6
x
X
x
6
6
7
13
Keys:
6
v
7
+
Section 3: The Memory Stack, LAST X, and Data Storage
T
y
y
y
y
Z
y
x
y
x
Y
x
13
x
65
X
13
5
65
4
Keys:
*
5
4
T
y
y
y
y
Z
x
y
x
y
Y
65
x
69
x
X
4
69
3
207
Keys:
+
39
3
*
Arithmetic Calculations With Constants
There are three ways (without using a storage register) to manipulate the
memory stack to perform repeated calculations with a constant:
1.
Use the LAST X register.
2.
Load the stack with a constant and operate upon different
numbers. (Clear the X-register every time you want to change
the number operated upon)
3.
Load the stack with a constant and operate upon an
accumulating number. (Do not change the number in the Xregister.)
LAST X. Use your constant in the X-register (that is, enter it second) so
that it always will be saved in the LAST X register. Pressing |K will
retrieve the constant and place it into the X-register (the display). This can
be done repeatedly.
40
Section 3: The Memory Stack, LAST X, and Data Storage
Example: Two close stellar neighbors of Earth
are Rigel Centaurus (4.3 light-years away) and
Sirius (8.7 light-years away). Use the speed of
light, c (3.0×108 meters/second, or 9.5×1015
meters/year), to figure the distances to these
stars in meters. (The stack diagrams show only
one decimal place.)
T
t
z
y
y
Z
z
y
x
x
4.3
4.3
Y
y
x
X
Keys:
LAST X:
x
4.3
v
4.3
4.3
9.5 15
9.5 ‛ 15
/
/
/
/
T
y
y
y
x
Z
x
y
x
4.1 16
4.3
x
4.1 16
8.7
9.5 15
4.1 16
8.7
Y
X
Keys:
LAST X:
*
8.7
/
9.5 15
T
x
x
Z
4.1 16
x
Y
8.7
4.1 16
X
9.5 15
8.3 16
*
Keys:
LAST X:
9.5 15
9.5 15
9.5 15
|K
9.5 15
9.5 15
(Rigel Centaurus is
4.1×1016 meters away.)
(Sirius is 8.3×1016
meters away.)
Section 3: The Memory Stack, LAST X, and Data Storage
41
Loading the Stack with a Constant. Because the number in the T-register
is replicated when the stack drops, this number can be used as a constant in
arithmetic operations.
T
c
c
Z
c
c
Y
c
c
X
x
cx
Keys:
New constant
generation.
Drops to interact
with X-register.
*
Fill the stack with a constant by keying it into the display and pressing
v three times. Key in your initial argument and perform the
arithmetic operation. The stack will drop, a copy of the constant will "fall"
into the Y-register, and a new copy of the constant will be generated in the
T-register.
If the variables change (as in the preceding example), be sure and clear the
display before entering the new variable. This disables the stack so that the
arithmetic result will be written over and only the constant will occupy the
rest of the stack.
If you do not have different arguments, that is, the operation will be
performed upon a cumulative number, then do not clear the display—simply
repeat the arithmetic operation.
Example: A bacteriologist tests a certain strain
of microorganisms whose population typically
increases by 15% each day (a growth factor of
1.15). If she starts with a sample culture of
1000, what will be the bacteria population at
the end of each day for four consecutive days?
Keystrokes
1.15
vv
v
1000
Display
1.15
1.1500
1,000
Growth factor.
Filling the stack.
Initial culture size.
42
Section 3: The Memory Stack, LAST X, and Data Storage
Keystrokes
*
*
*
*
Display
1,150.0000
1,322.5000
1,520.8750
1,749.0063
Population at the end of day 1.
Day 2.
Day 3.
Day 4.
Storage Register Operations
When numbers are stored or recalled, they are copied between the display
(X-register) and the data storage registers. At ―power-up‖ (initial turn-on or
Continuous Memory reset) the HP-15C has 21 directly accessible storage
registers: R0 through R9, R.0 through R.9, and the Index register (RI) (see the
diagram of the registers on the inside back cover). Six registers, R2 to R7,
are also used for statistics calculations.
The number of available data storage registers can be increased or
decreased. The m function, which is used to reallocate registers in
calculator memory, is discussed in appendix C, Memory Allocation. The
lowest-numbered registers are the last to be deallocated from data storage,
therefore it is wisest to store data in the lowest-numbered registers
available.
Storing and Recalling Numbers
O (store). When followed by a storage register address (0 through 9 or
.0 through .9*), this function copies a number from the display (X-register)
into the specified data storage register. It will replace any existing contents
of that register.
l (recall). Similarly, you can recall data from a particular register into
the display by pressing l followed by the register address. This brings
a copy of the desired data into the display; the contents of the storage
register remain unaltered.
X (X exchange). Followed by 0 through .9,* this function exchanges the
contents of the X-register and the addressed data storage register. This is
useful to view storage registers without disturbing the stack.
*
All storage register operations can also be performed with the Index register (using V or %), which is
covered in section 10, and with matrices, section 12.
Section 3: The Memory Stack, LAST X, and Data Storage
43
The above are stack lift-enabling operations, so the number remaining in the
X-register can be used for subsequent calculations. If you address a
nonexistent register, the display will show Error 3.
Example: Springtime is coming and you want to keep track of 24 crocuses
planted in your garden. Store the number of crocuses blooming the first day
and add to this the number of new blooms the second day.
Keystrokes
3O0
Display
3.0000
Stores the number of first-day
blooms in R0.
Turn the calculator off. Next day, turn it back on again.
l0
3.0000
5+
8.0000
Recalls the number of crocuses that
bloomed yesterday.
Adds today's new blooms to get the
total blooming crocuses.
Clearing Data Storage Registers
Pressing ´ CLEAR Q (clear registers) clears the contents of all data
storage registers to zero. (It does not affect the stack or the LAST X
register.) To clear a single data storage register, store zero in that register.
Resetting Continuous Memory clears all registers and the stack.
Storage and Recall Arithmetic
Storage Arithmetic. Suppose you not only wanted to store a number, but
perform arithmetic with it and store the result in the same register. You can
do this directly – without using l – by using the following procedure.
1.
2.
3.
4.
Have your second operand (besides the one in storage) in the display
(as the result of a calculation, a recall, or keying in).
Press O.
Press +, -, *, or ÷.
Key in the register address (0 to 9, .0 to .9). (The Index register,
discussed in section 10, can also be used.)
44
Section 3: The Memory Stack, LAST X, and Data Storage
The number in the register is determined as follows:
For storage arithmetic,
new contents
of register
R0
=
r


×

old contents
of register
T
t
Z
R0
number in
display
T
t
z
Z
z
Y
y
Y
y
X
x
X
x
r-x
O-0
Keys:
Recall Arithmetic. Recall arithmetic allows you to perform arithmetic with
the displayed value and a stored value without lifting the stack, that is,
without losing any values from the Y-, Z, and T-registers. The keystroke
sequence is the same as for storage arithmetic using l in place of
O.
For recall arithmetic,
new display
R0
r
=
old display
T
t
Z


×

T
t
z
Z
z
Y
y
Y
y
X
x
X
x-r
Keys:
R0
contents of
register
l-0
r
Section 3: The Memory Stack, LAST X, and Data Storage
45
Example: Keep a running count of your newly blooming crocuses for two
more days.
Keystrokes
8O0
Display
8.0000
4O+0
4.0000
3O+0
24 l - 0
3.0000
9.0000
l0
15.0000
Places the total number of blooms as of
day 2 in R0.
Day 3: adds four new blooms to those
already blooming.
Day 4: adds three new blooms.
Subtracts total number of blooms
summed in R0(15) from the total
number of plants (24); 9 crocuses have
not bloomed.
(The number in R0 does not change.)
Overflow and Underflow
If an attempted storage or recall arithmetic operation would result in
overflow in a data storage register, the value in the affected register will be
replaced with ±9.999999999×1099 and the display will blink. To stop the
blinking (clear the overflow condition), press − or = or | " 9.
In case of underflow, the value in the register will be replaced with zero (no
display blinking). Overflow and underflow are discussed further on page
61.
Problems
1.
Calculate the value of x in the following equation.
x
8.33 (4  5.2)  [(8.33  7.46) 0.32]
4.3 (3.15  2.75)  (1.71) (2.01)
Answer: 4.5728.
A possible keystroke solution is:
4 v 5.2 - 8.33 * | K 7.46 - 0.32 * ÷ 3.15
v 2.75 - 4.3 * 1.71 v 2.01 * - ÷ ¤
46
Section 3: The Memory Stack, LAST X, and Data Storage
2.
Use arithmetic with constants to calculate the remaining
balance of a $1000 loan after six payments of $100 each and
an interest rate of 1% (0.01) per payment period.
Procedure: Load the stack with (1 + i), where i = interest rate,
and key in the initial loan balance. Use the following formula
to find the new balance after each payment.
New Balance = ((Old Balance)×(1 + i)) - Payment
The first part of the key sequence would be:
1.01 vvv 1000
For each payment, execute:
* 100 Balance after six payments: $446.32.
3.
Store 100 in R5. Then:
1. Divide the contents of R5 by 25.
2. Subtract 2 from the contents of R5.
3. Multiply the contents of R5 by 0.75.
4. Add 1.75 to the contents of R5.
5. Recall the contents of R5.
Answer: 3.2500.
Section 4
Statistics Functions
A word about the statistics functions: their use is based on an understanding
of memory stack operation (Section 3). You will find that order of entry is
important for most statistics calculations.
Probability Calculations
The input for permutation and combination calculations is restricted to
nonnegative integers. Enter the y-value before the x-value. These functions,
like the arithmetic operators, cause the stack to drop as the result is placed
in the X-register.
Permutations. Pressing ´p calculates the number of possible
different arrangements of y different items taken in quantities of x items at a
time. No item occurs more than once in an arrangement, and different
orders of the same x items in an arrangement are counted separately. The
formula is
Py , x 
y!
( y  x)!
Combinations. Pressing |c calculates the number of possible sets
of y different items taken in quantities of x items at a time. No item occurs
more than once in a set, and different orders of the same x items in a set are
not counted separately. The formula is
C y, x 
y!
x!( y  x)!
Examples: How many different arrangements are possible of five pictures
which can be hung on the wall three at a time?
Keystrokes
5v3
´p
Display
3
60.0000
Five (y) pictures put up three (x) at a
time.
Sixty different arrangement possible.
47
48
Section 4: Statistics Functions
How many different four-card hands can be dealt from a deck of 52 cards?
Keystrokes
52 v 4
|c
Display
4
Fifty-two (y) cards dealt four
(x) at a time.
270,725.0000
Number of different hands
possible.
The maximum size of x or y is 9,999,999,999.
Random Number Generator
Pressing ´# (random number) will generate a random number
(part of a uniformly distributed pseudo-random number sequence) in the
range 0 ≤ r <1.*
At initial power-up (including reset of Continuous Memory), the HP-15C
random number generator will use zero as a ―seed‖ to initiate a random
number sequence. Any time you generate a random number, that number
becomes the seed for the next random number. You can initiate a different
random number sequence by storing a new seed for the random number
generator. (Repetition of a random number seed will produce repetition of
the random number sequence.)
O´# will store the X-register number (0 ≤ r < 1) as a new seed
for the random number generator. (A value for r outside this range will be
converted to fit within the range.)
l´# will recall to the display the current random number seed.
Keystrokes
.5764
O´
#
´#
´#
−
*
Display
0.5764
0.5764
0.3422
0.2809
0.0000
Stores 0.5764 as random number seed.
(The ´ keystroke may be omitted.)
Random number sequence initiated by the
above seed.
Passes the spectral test (D. Knuth, The Art of Computer Programming. Vol. 2. Seminumerical Algorithms,
Third Edition, 1998).
Section 4: Statistics Functions
Keystrokes
l´
#
Display
0.2809
49
Recall last random number generated,
which is the new seed. (The ´ may be
omitted.)
Accumulating Statistics
The HP-15C performs one- and two-variable statistical calculations. The
data is first entered into the Y- and X-registers. Then the z function
automatically calculates and stores statistics of the data in storage registers
R2 through R7. These registers are therefore referred to as the statistics
registers.
Before beginning to accumulate statistics for a new set of data, press
´ CLEAR ∑ to clear the statistics registers and stack. (If you have
reallocated registers in memory and any of the statistics registers no longer
exist, Error 3 will be displayed when you try to use CLEAR ∑, z, or
w Appendix C explains how to reallocate memory.)
In one-variable statistical calculations, enter each data point (x-value) by
keying in x and then press z.
In two-variable statistical calculations, enter each data pair (the x- and yvalues) as follows:
1.
2.
3.
4.
Key y into the display first.
Press v. The displayed y-value is copied into the Y-register.
Key x into the display.
Press z. The current number of accumulated data points, n, will be
displayed. The x-value is saved in the LAST X register and y remains
in the Y-register. z disable stack lift, so the stack will not lift
when the next number is keyed in.
50
Section 4: Statistics Functions
In some cases involving x or y data values that differ by a relatively small
amount, the calculator cannot compute s, r, linear regression, or ŷ, and will
display Error 2. This will not happen, however, if you normalize the data by
keying in only the difference between each value and the mean or
approximate mean of the values. This difference must be added back to the
calculations of x, ŷ, and the y-intercept (L). For example, if your x-values
were 665999, 666000, and 666001, you should enter the data as -1, 0, and 1;
then add 666000 back to the relevant results.
The statistics of the data are compiled as follows:
Register
R2
n
R3
Σx
Contents
Number of data points accumulated (n also
appears in the X-register).
Summation of x-values.
2
R4
Σx
R5
Σy
Summation of y-values.
R6
Σy2
Summation of squares of y-values.
R7
Σxy
Summation of products of x- and y-values.
Summation of squares of x-values.
You can recall any of the accumulated statistics to the display (X-register)
by pressing l and the number of the data storage register containing the
desired statistic. If you press l z, Σy and Σx will be copied
simultaneously from R3 and R5 respectively, into the X-register and the Yregister, respectively. (The sequence l z lifts the stack twice if stack
lift is enabled, once if not, and then enables stack lift.)
Example: Agronomist Silas Farmer has
developed a new variety of high-yield rice,
and has measured the plant's yield as a
function of fertilization. Use the z function
to accumulate the data below to find the values
for Σx, Σx2 Σy, Σy2, and Σxy for nitrogen
fertilizer application (x) versus grain yield (y).
Section 4: Statistics Functions
X
Y
NITROGEN APPLIED
0.00
(kg per hectare *), x
GRAIN YIELD
4.63
(metric tons per
hectare), y
51
20.00
40.00
60.00
80.00
4.78
6.61
7.21
7.78
*A hectare equals 2.47 acres.
Keystrokes
´ CLEAR ∑
Display
0.0000
´•2
0.00
4.63 v
4.63
0z
1.00
4.78 v
4.78
20 z
2.00
6.61v
6.16
40 z
3.00
7.21 v
7.21
60 z
4.00
7.78 v
7.78
80 z
l3
5.00
l4
12.000.00
l5
31.01
l6
200.49
l7
1,415.00
200.00
Clears statistical storage
registers (R2 through R7 and
the stack).
Limits display to two decimal
places, like the data.
First data point.
Second data point.
Third data point.
Fourth data point.
Fifth data point.
Sum of x-values, Σx (kg of
nitrogen).
Sum
of squares of x-values,
Σx2.
Sum of y-values, Σy (grain
yield).
Sum
of squares of y-values,
Σy2.
Sum of products of x- and
y-values, Σxy.
52
Section 4: Statistics Functions
Correcting Accumulated Statistics
If you discover that you have entered data incorrectly, the accumulated
statistics can be easily corrected. Even if only one value of an (x, y) data
pair is incorrect, you must delete and re-enter both values.
1.
2.
3.
4.
Key the incorrect data pair into the Y- and X-register.
Press |w to delete the incorrect data.
Key in the correct values for x and y.
Press z.
Alternatively, if the incorrect data point or pair is the most recent one
entered and z has been pressed, you can press |K |w to
remove the incorrect data.*
Example: After keying in the preceding data. Farmer realizes he misread a
smeared figure in his lab book. The second y-value should have been 5.78
instead of 4.78. Correct the data input.
Keystrokes Display
4.78
4.78
v
4.00
20 |w
5.78
5.78
v
5.00
20 z
Keys in the data pair we want to replace
and deletes the accompanying statistics.
The n-value drops to four.
Keys in and accumulates the replacement
data pair.
The n -value is back to five.
We will use these statistics in the rest of the examples in this section.
*
Note that these methods of data deletion will not delete any rounding errors that may have been generated
in the statistics registers. This difference will not be serious unless the erroneous pair has a magnitude that
is enormous compared with the correct pair, in such a case, it would be wise to start over!
Section 4: Statistics Functions
53
Mean
The ’ function computes the arithmetic mean (average) of the x-and yvalues using the formulas shown in appendix A and the statistics
accumulated in the relevant registers. When you press |’ the contents
of the stack lift (two registers if stack lift is enabled, one if not); the mean of
x (x) is copied into the X-register as the mean of y (y) is copied
simultaneously into the Y-register. Press ® to view y.
Example: From the corrected statistics data we have already entered and
accumulated, calculate the average fertilizer application, x. and average
grain yield y, for the entire range.
Keystrokes
|’
®
Display
40.00
6.40
Average kg of nitrogen, x, for all cases.
Average tons of rice, y, for all cases.
Standard Deviation
Pressing |S computes the standard deviation of the accumulated
statistics data. The formulas used to compute sx, the standard deviation of
the accumulated x-values, and sy, the standard deviation of the accumulated
y-values, are given in appendix A.
This function gives an estimate of the population standard deviation from
the sample data, and is therefore termed the sample standard deviation.*
When you press |S, the contents of the stack registers are lifted (twice
if stack lift is enabled, once if not); sx is placed into the X-register and sy is
placed into the Y-register. Press ® to view sy.
*
When your data constitutes not just a sample of a population but all of the population, the standard
deviation of the data is the true population standard deviation (denoted ). The formula for the true
population standard deviation differs by a factor of
(n  1) / n from the formula used for the S
function. The difference between the values is small for large n, and for most applications can be ignored.
But if you want to calculate the exact value of the population standard deviation for an entire population,
you can easily do so: simply add, using z, the mean (x) of the data to the data before pressing |S.
The result will be the population standard deviation. (If you subsequently correct any of your accumulated
data values, remember to delete the first mean value and add the corrected one.)
54
Section 4: Statistics Functions
Example: Calculate the standard deviation about the mean calculated
above.
Keystrokes
|S
Display
31.62
1.24
®
Standard deviation about the mean nitrogen
application, x.
Standard deviation about the mean grain
yield, y.
Linear Regression
Linear regression is a statistical method for finding a straight line that best
fits a set of two or more data pairs, thus providing a relationship between
two or more data pairs, thus providing a relationship between two variables.
By the method of least squares, ´L will calculate the slope, A, and yintercept, B, of the linear equation:
y=Ax+B
1.
Accumulate the statistics of your data using the z key.
2.
Press ´L. The y-intercept, B, appears in the display (Xregister). The slope, A, is copied simultaneously into the Yregister.
Press ® to view A. (As is the case with the functions ’
and S, L causes the stack to lift two registers if it's
enabled, one if not).
3.
T
t
y
y
Z
z
x
y
Y
y
A
slope
B
y-intercept
X
x
B
y-intercept
A
slope
Keys:
´L
®
The slope and y-intercept of the least squares line of the accumulated data
are calculated using the equations shown in appendix A.
Section 4: Statistics Functions
55
Example: Find the y-intercept and slope of the linear approximation of the
data and compare to the plotted data on the graph below.
Keystrokes
´L
®
Display
4.86
0.04
y-intercept of the line.
Slope of the line.
Linear Estimation and Correlation Coefficient
When you press ´j the linear estimate, ŷ, is placed in the X-register
and the correlation coefficient, r, is placed in the Y-register. To display r,
press ®.
56
Section 4: Statistics Functions
Linear Estimation. With the statistics accumulated, an estimated value for
y, denoted ŷ, can be calculated by keying in a proposed value for x and
pressing ´j.
An Estimated value for x (denoted xˆ ) can be calculated as follows:
1.
2.
3.
Press ´L.
Key in the known y-value.
Press ® - ® ÷.
Correlation Coefficient. Both linear regression and linear estimation
presume that the relationship between the x and y data values can be
approximated by a linear function. The correlation coefficient, r, is a
determination of how closely your data fit a straight line. The range is -1  r
 1, with -1 representing a perfectly negative correlation and +1
representing a perfectly positive correlation.
Note that if you do not key in a value for x before executing ´j, the
number previously in the X-register will be used (usually yielding a
meaningless value for ŷ).
Example: What if 70 kg of nitrogen fertilizer were applied to the rice field?
Predict the grain yield based on Farmer’s accumulated statistics. Because
the correlation coefficient is automatically included in the calculation, you
can view how closely the data fit a straight line by pressing ® after the
y prediction appears in the display.
Section 4: Statistics Functions
Keystrokes
70 ´j
®
Display
7.56
0.99
57
Predicted grain yield in tons/hectare.
The original data closely approximates a
straight line.
Other Applications
Interpolation. Linear interpolation of tabular values, such as in
thermodynamics and statistics tables, can be carried out very simply on the
HP-15C by using the j function. This is because linear interpolation is
linear estimation: two consecutive tabular values are assumed to form two
points on a line, and the unknown intermediate value is assumed to fall on
that same line.
Vector Arithmetic. The statistical accumulation functions can be used to
perform vector addition and subtraction. Polar vector coordinates must be
converted to rectangular coordinates upon entry (θ, v, r ;, z).
The results are recalled from R3 (Σx) and R5 (Σy) (using l z) and
converted back to polar coordinates, if necessary. Remember that for polar
coordinates the angle is between -180° and 180° (or -π and π radians, or 200 and 200 grads). To convert to a positive angle, add 360 (or 2π or 400)
to the angle.
For the second vector entered, the final keystroke will be either z or
w, depending on whether the two vectors should be added or subtracted.
Section 5
The Display
and Continuous Memory
Display Control
The HP-15C has three display formats – •, i, and ^ – that
use a given number (0 through 9) to specify display format. The illustration
below shows how the number 123,456 would be displayed specified to four
places in each possible mode.
´•4
:
123,456.0000
´i4
:
1.2346
05
´^4
:
123.46
03
Owing to Continuous Memory, any change you make in the display format
will be preserved until Continuous Memory is reset.
The current display format takes effect when digit entry is terminated; until
then, all digits you key in (up to 10) are displayed.
Fixed Decimal Display
• (fixed decimal) format displays a figure with the number of decimal
places you specify (up to nine, depending on the size of the integer portion.)
Exponents will be displayed if the number is too small or too large for the
display. At ―power-up,‖ the HP-15C is in • 4 format. The key sequence
is ´• n.
Keystrokes
123.4567895
´• 4
´• 6
´• 4
Display
123.4567895
123.4568
123.456790
123.4568
58
Display is rounded to six decimal
places. (Ten places are stored
internally.)
Usual • 4 display.
Section 5: The Display and Continuous Memory
59
Scientific Notation Display
i (scientific) format displays a number in scientific notation. The
sequence ´i n specifies the number of decimal places to be shown.
Up to six decimal places can be shown since the exponent display takes
three spaces. The display will be rounded to the specified number of
decimal places; however, if you specify more decimal places than the six
places the display can hold (that is, i 7, 8, or 9), rounding will occur in
the undisplayed seventh, eighth, or ninth decimal place.*
With the previous number still in the display:
Keystrokes
´i 6
Display
1.234568
02
´i 8
1.234567
02
Rounds to and shows six
decimal places.
Rounds to eight decimal places,
but displays only six.
Engineering Notation Display
^ (engineering) format displays numbers in an engineering notation
format in a manner similar to i, except:
 In engineering notation, the first significant digit is always present in
the display. The number you key in after ´^ specifies the
number of additional digits to which you want to round the display.
 Engineering notation shows all exponents in multiples of three.
Keystrokes
.012345
´^
1
´^ 3
10 *
Display
0.012345
12.
´• 4
0.1235
*
12.35
123.5
-03
-03
-03
Rounds to the first digit after
the leading digit.
Decimal shifts to maintain
multiple of three in exponent.
Usual • 4 format.
Therefore, the display shows no distinction among i. 7, 8, and 9 unless the number rounded up is a 9,
which carries a 1 over into the next higher decimal place .
60
Section 5: The Display and Continuous Memory
Mantissa Display
Regardless of the display format, the HP-15C always internally holds each
number as a 10-digit mantissa and a two-digit exponent of 10. For example,
π is always represented internally as 3.141592654×1000, regardless of what
is in the display.
When you want to view the full 10-digit mantissa of a number in the Xregister, press ´ CLEAR u. To keep the mantissa in the display,
hold the u key down.
Keystrokes
Display
3.1416
|$
´ CLEAR
u (hold)
3141592654
Round-Off Error
As mentioned earlier, the HP-15C holds every value to 10 digits internally.
It also rounds the final result of every calculation to the 10th digit. Because
the calculator can provide only a finite approximation for numbers such as 
or 2/3 (0.666…), a small error due to rounding can occur. This error can be
increased in lengthy calculations, but usually is insignificant. To accurately
assess this effect for a given calculation requires numerical analysis beyond
our scope and space here! Refer to the HP-15C Advanced Functions
Handbook for a more detailed discussion.
Special Displays
Annunciators
The HP-15C display contains eight annunciators that indicate the status of
the calculator for various operations. The meaning and use of these
annunciators is discussed on the following pages:
*
USER
f and g
RAD and GRAD
C
PRGM
Low-power indication, page 62.
User mode, pages 79 and 144.
Prefixes for alternate functions, pages 18-19.
Trigonometric modes, page 26.
Complex mode, page 121.
Program mode, page 66.
Section 5: The Display and Continuous Memory
61
Digit Separators
The HP-15C is set at power-up so that it separates integral and fractional
portions of a number with a period (a decimal point), and separates groups
of three digits in the integer portion with a comma. You can reverse this
setting to conform to the numerical convention used in many countries. To
do so, turn off the calculator. Press and hold =, press and hold .,
release =, then release . (= / .). (Repeating this sequence will
set the calculator to the previous display convention.)
Keystrokes
12345.67
=/.
=/.
Display
12,345.67
12.345.6700
12,345.6700
Error Display
If you attempt an improper operation—such as division by zero—an error
message (Error followed by a digit) will appear in the display. For a
complete listing of error messages and their causes, refer to appendix A.
To clear the Error display and restore the calculator to its prior condition,
press any key. You can then resume normal operation.
Overflow and Underflow
Overflow. When the result of a calculation in any register is a number with
a magnitude greater than 9.999999999×1099, ± 9.999999999×1099 is placed
in the affected register and the overflow flag, flag 9, is set.* Flag 9 causes
the display to blink. When overflow occurs in a running program, execution
continues until completion of the program, and then the display blinks.
The blinking can be stopped and flag 9 cleared by pressing −, = or
|" 9.
Underflow. If the result of a calculation in any register is a number with a
magnitude less than 1.000000000×10-99, that number will be replaced by
zero. Underflow does not have any other effect.
*
Recall that display does not include the last three digits of the mantissa.
62
Section 5: The Display and Continuous Memory
Low-Power Indication
When a flashing asterisk, which indicates
low battery power, appears in the lower
left-hand side of the display, there is no
reason to panic. You still have plenty of
calculator time remaining: at least 10
minutes if you continuously run programs,
and at least an hour if you do calculations
manually. Refer to appendix F (page 259)
for information on replacing the batteries.
0.0000
*
Continuous Memory
Status
The Continuous Memory feature of the HP-15C retains the following in the
calculator, even when the display is turned off:







All numeric data stored in the calculator.
All programs stored in the calculator.
Position of the calculator in program memory.
Display mode and setting.
Trigonometric mode (Degrees, Radians, or Grads).
Any pending subroutine returns.
Flag settings (except flag 9, which clears when the display is
manually turned off).
 User mode setting.
 Complex mode setting.
When the HP-15C is turned on, it always ―wakes up‖ in Run mode. If the
calculator is turned off, Continuous Memory will be preserved for a short
period while the batteries are removed. Data and programs are preserved
longer than other aspects of calculator status. Refer to appendix F for
instructions on changing batteries.
Section 5: The Display and Continuous Memory
63
Resetting Continuous Memory
If at any time you want to reset (entirely clear) the HP-15C Continuous
Memory:
1. Turn the calculator off.
2. Press and hold the = key, then press and hold the key.
3. Release the = key, then the - key. (This convention
is represented as = / -.)
When Continuous Memory is reset, Pr Error (power error) will be
displayed. Press any key to clear the display.
Note: Continuous Memory can inadvertently be interrupted and
reset if the calculator is dropped or otherwise traumatized.
Part ll
HP-15C
Programming
Section 6
Programming Basics
The next five sections are dedicated to explaining aspects of programming
the HP-15C. Each of these programming sections will first discuss basic
techniques (The Mechanics), then give examples for the implementation of
these techniques (Examples), and lastly discuss finer points of operation in
greater detail (Further Information). Read only as far as you need to support
your use of the HP-15C.
The Mechanics
Creating a Program
Programming the HP-15C is an easy matter, based simply on recording the
keystroke sequence used when calculating manually. (This is called
―keystroke programming‖.) To create a program out of a series of
calculation steps requires two extra manipulations: deciding where and how
to enter your data; and loading and storing the program. In addition,
programs can be instructed to make decisions and perform iterations
through conditional and unconditional branching.
As we step through the fundamentals of programming, we'll rework the
falling object program illustrated in the Problem Solver (page 14).
Loading a Program
Program Mode. Press | ¥ (program/run) to set the calculator to
Program mode (PRGM annunciator on). Functions are stored and not
executed when keys are pressed in Program mode.
Keystrokes
|¥
Display
000-
Switches to Program mode;
PRGM annunciator and line
number (000) displayed.
66
Section 6: Programming Basics
67
Location in Program Memory. Program memory – and therefore the
calculator's position in program memory – is demarcated by line numbers.
Line 000 marks the beginning of program memory and cannot be used to
store an instruction. The first line that contains an instruction is line 001.
Program lines other than 000 do not exist until instructions are written for
them.
You can start a program at any existent line (designated nnn), but it is
simplest and safest to start an independent program (as opposed to a
subroutine) at the beginning of program memory. As you write, any existing
program lines will be preserved and ―bumped‖ down in program memory.
Press t “ 000 (in Program or Run mode) to move to line 000
without recording the t statement. In Run mode, ´ CLEAR M
will also reset the calculator to line 000- without clearing program memory.
Alternatively, you can clear program memory, which will erase all
programs in memory and position you to line 000. To do so, press ´
CLEAR M in Program mode.
Program Begin. A label instruction – ´b followed by a letter
(A through E) or number (0 through 9 or .0 through .9) – is used to
define the beginning of a program or routine. The use of labels allows you
to quickly select and run one particular program or routine out of several.
Keystrokes
´ CLEAR
M
´bA
Display
000-
Clears program memory and
sets to line 000 (start of
program memory).
001-42,21,11
Recording a Program. Any key pressed—operator or constant—will be
recorded in memory as a programmed instruction.*
*
Except the nonprogrammable functions, which are listed on page 80.
68
Section 6: Programming Basics
Keystrokes
2
*
9
.
8
÷
¤
Display
002003004005006007008-
2
20
9
48
8
10
11
Given h in the X-register,
lines 002 to 008 calculate
2h
.
9.8
Program End. There are three possible endings for a program:
 | n (return) will end a program, return to line 000, and halt.
 ¦ will stop a program without moving to line 000.
 The end of program memory contains an automatic n.
Keystrokes
|n
Display
009-
43 32
Optional if this is the last
program in memory.
Intermediate Program Stops
Use ´ © (pause) as a program instruction to momentarily stop a
program and display an intermediate result. (Use more than one © for a
longer pause.)
Use a ¦ (run/stop) instruction to stop the program indefinitely. The
program will remain positioned at that line. You can resume program
execution (from that line) by pressing ¦ during Run mode, that is, from
the keyboard.
Running a Program
Run Mode. Switch back to Run mode when you are done programming:
| ¥. Program execution must take place in Run mode.
Section 6: Programming Basics
Keystrokes
|¥
69
Display
Run mode; no PRGM annunciator
displayed. (The display will depend
on any previous result.)
The position in program memory does not change when modes are
switched. Should the calculator be shut off, it always ―wakes up‖ in
Run mode.
Executing a Program. In Run mode, press ´ letter label or G digit
(or letter) label. This addresses a program and starts its execution. The
display will flash running.
Keystrokes
300.51
´A
Display
300.51
7.8313
Key a value for h into the X-register.
The result of executing program
―A‖. (The number of seconds it
takes an object dropped from 300.51
meters high to hit the ground.)
Restarting a Program. Press ¦ to continue execution of a program
that was stopped with a ¦ instruction.
User Mode. User mode is an optional condition to save keystrokes when
executing letter-named programs. Pressing ´ U will interchange the
´-shifted and primary functions of the A through E keys. You can
then execute a program using just one keystroke (skipping the ´ or
G).
How to Enter Data
Every program must take into account how and when data will be supplied.
This can be done in Run mode before running the program or during an
interruption in the program.
1.
Prior entry. If a variable value will be used in the first line of the
program, enter it into the X-register before starting the program. If it
will be used later, you can store it (with O) into a storage
register, and recall it (with a programmed l) within the
program.
70
Section 6: Programming Basics
This is the method used above, where h was placed in the X-register
before running the program. No v instruction is necessary
because program execution (here: ´A) both terminates digit
entry and enables the stack lift. The above program then multiplied
the contents of the X-register (h) by 2.
The presence of the stack even makes it possible to load more than
one variable prior to running a program. Keeping in mind how the
stack moves with subsequent calculations and how the stack can be
manipulated (as with ®), it is possible to write a program to use
variables which have been keyed into the X-, Y-, Z-, and T-registers.
2.
Direct entry. Enter the data as needed as the program runs. Write a
¦ (run/stop) instruction into the program where needed so the
program will stop execution. Enter your data, then press ¦ to
restart the program.
Do not key variable data into the program itself. Any values that will vary
should be entered anew with each program execution.
Program Memory
At power-up (Continuous Memory reset), the HP-15C offers 322 bytes of
program memory and 21 storage registers. Most program steps
(instructions) use one byte, but some use two. The distribution of memory
capacity can be altered, as explained in appendix C. The maximum
attainable program memory is 448 bytes (with the permanent storage
registers—RI, R0, and R1 — remaining); maximum number of storage
registers is 67 (with no program memory).
Example. Mother's Kitchen, a canning
company, wants to package a ready-toeat spaghetti mix containing three
different cylindrical cans: one of
spaghetti sauce, one of grated cheese,
and one of meatballs. Mother's needs to
calculate the base areas, total surface
areas, and volumes of the three different
cans. It would also like to know, per
package, the total base area, surface
area, and volume.
Section 6: Programming Basics
71
The program to calculate this information uses these formulas and data:
base area = r2.
volume = base area × height = r2h.
surface area = 2 base areas + side area = 2r2 + 2rh.
Radius, r
2.5cm
4.0
4.5
Height, h
Base Area
Volume
Surface Area
?
?
?
?
?
?
?
?
?
?
?
?
8.0 cm
10.5
4.0
TOTALS
Method:
1. Enter an r value into the calculator and save it for other calculations.
Calculate the base area (r2), store it for later use, and add the base
area to a register which will hold the sum of all base areas.
2. Enter h and calculate the volume (r2h). Add it to a register to hold
the sum of all volumes.
3. Recall r. Divide the volume by r and multiply by 2 to yield the side
area. Recall the base area, multiply by 2, and add to the side area to
yield the surface areas. Sum the surface areas in a register.
Do not enter the actual data while writing the program – just provide for
their entry. These values will vary and so will be entered before and/or
during each program run.
Key in the following program to solve the above problem. The display
shows line numbers and keycodes (the row and column location of a key),
which will be explained under Further Information.
Keystrokes
|¥
´ CLEAR M
Display
000000-
Sets calculator to Program
mode (PRGM displayed).
Clears program memory. Starts
at line 000.
72
Section 6: Programming Basics
Keystrokes
´bA
Display
001-42,21,11
O0
002-
44
0
|x
003-
43 11
|$
*
O4
O+1
00443 26
00520
00644 4
007-44,40, 1
¦
008-
31
*
009-
20
´©
010-
42 31
O+2
011-44,40, 2
l0
÷
2
*
l4
012013014015016-
2
*
017018-
45
0
10
2
20
45 4
2
20
Assigns this program the label
―A‖.
Stores the contents of X-register
into R0. r must be in the Xregister before running the
program.
Squares the contents of the Xregister (which will be r).
r2, the BASE AREA of a can.
Stores the BASE AREA in R4.
Keeps a sum of all BASE
AREAS in R1.
Stops to display BASE AREA
and allow entry of the h value.
Multiplies h by the BASE
AREA, giving VOLUME.
Pauses briefly to display
VOLUME.
Keeps a sum of all can
VOLUMES in R2.
Recalls r.
Divides VOLUME by r.
2 rh, the SIDE AREA of a can.
Recalls the BASE AREA of the
can.
Multiplies base area by two (for
top and bottom).
Section 6: Programming Basics
Keystrokes
+
Display
019–
40
O+3
020–44,40, 3
|n
021–
43 32
73
SIDE AREA + BASE AREA
= SURFACE AREA.
Keeps a sum of all SURFACE
AREAS in R3.
Ends the program and returns
program memory to line 000.
Now, let's run the program:
Keystrokes
|¥
Display
´ CLEAR Q
2.5
´A
(or: G A)
2.5
19.6350
8
8
¦
157.0796
164.9336
4
50.2655
10.5
527.7876
364.4247
4
¦
10.5
¦
4.5
¦
4.5
63.6173
Sets calculator to Run mode.
(PRGM cleared.)
Clears all storage registers. The
display does not change.
Enter r of the first can.
Starts program A. BASE AREA
of first can.
(running flashes during
execution.)
Enter h of first can. Then restart
program.
VOLUME of first can.
SURFACE AREA of first can.
Enter r of the second can.
BASE AREA of second can.
Enter h of second can.
VOLUME of second can.
SURFACE AREA of
second can.
Enter r of the third can.
BASE AREA of third can.
74
Section 6: Programming Basics
Keystrokes
4
¦
l1
l2
l3
Display
4
254.4690
240.3318
133.5177
939.3362
769.6902
Enter h of third can.
VOLUME of third can.
SURFACE AREA of third can.
Sum of BASE AREAS.
Sum of VOLUMES.
Sum of SURFACE AREAS.
The preceding program illustrates the basic techniques of programming. It
also shows how data can be manipulated in Program and Run modes by
entering, storing, and recalling data (input and output) using v,
O, l, storage register arithmetic, and programmed stops.
Further Information
Program Instructions
Each digit, decimal point, and function key is considered an instruction
and is stored in one line of program memory. An instruction may include
prefixes (such as ´, O, t and b) and still occupy only one
line. Most instructions require one byte of program memory; however, some
require two. For a complete list of two-byte instructions, refer to
Appendix C.
Instruction Coding
Each key on the HP-15C keyboard – except for the digit keys 0 through
9 – is identified in Program mode by a two-digit ―keycode‖ that
corresponds to the key's position on the keyboard.
Instruction
O+1
´eV
Code
006-44,40, 1
XXX-42, 5,25
Sixth program line.
e is just ―5‖.
The first digit of a keycode refers to the row (1 to 4 from top to bottom),
and the second digit refers to the column (1, 2, 9, 0 from left to right).
Exception: the keycode for a digit key is simply that digit.
Section 6: Programming Basics
75
Keycode 25: second row, fifth key.
Memory Configuration
Understanding memory configuration is not essential to your use of the
HP-15C. It is essential, however, for obtaining maximum efficiency in
memory and programming use. The more you program, the more useful this
knowledge will be. Memory configuration and allocation is thoroughly
explained in appendix C, Memory Allocation.
Should you ever get an Error 10, you have run up against limitations of the
HP-15C memory. If you learn how to reallocate memory, you can greatly
increase your ability to store information in the HP-15C.
The HP-15C memory consists of 67 registers (R0 to R65 and the Index
register) divided between data storage and programming/advanced function
capability. The initial configuration is:
 46 registers for both programming and the advanced functions
(_, f, the imaginary stack, and > functions). At seven
bytes of memory per register, this is worth 322 program bytes if no
memory is dedicated to advanced functions.
 21 registers for data storage (R0 to R9, R.0 to R.9, and the Index
register).
76
Section 6: Programming Basics
Initial Memory Configuration
Memory is reallocated by telling the calculator which data storage register
shall be the highest data register; all other registers are left for programming
and advanced functions.
Keystrokes
*
60 ´ m %
*
Display
60.0000
R60 and below allocated to data
storage; five (R61 to R65) remain
for programming.
The optional omission of the ´ keystroke after another prefix key is explained on page 78, Abbreviated
Key Sequences.
Section 6: Programming Basics
Keystrokes
1´m%
Display
1.0000
19 ´ m%
19.0000
lm%
19.0000
77
R1 and R0 allocated for data
storage; R2 to R65 available for
programming and advanced
functions.
Original allocation: R19 (R.9) and
below for data storage; R20, to
R65 for programming and
advanced functions.*
Displays the current highest data
register.
The m and W (memory status) functions are described in detail in
appendix C.
Keep in mind that an error message will result (given the above memory
configuration) if
1. You try to address a register higher than R19 (R.9), which initially is
the highest register allocated to data storage (Error 3).
2. You have 322 occupied program bytes and try to load more program
lines (Error 4).
3. You try to run an advanced function with insufficient available
memory (Error 10).
Program Boundaries
End. Not every program needs to end with a n or ¦ instruction. If
you are at the end of occupied program memory, there is an automatic
n instruction, so you do not need to enter one. This can save you one
line of memory. On the other hand, a program can ―end‖ by simply
transferring execution to another routine using t (section 7).
Labels. Labels in a program (or subroutine) are markers telling the
calculator where to begin execution. Following an ´ label or G label
instruction, the calculator will search downward in program memory for the
*
For memory allocation and indirect addressing, registers R .0 through R.9 are referred to as R10 through R19.
78
Section 6: Programming Basics
corresponding label. If need be, the search will wrap around at the end of
program memory and continue at line 000. When it encounters an
appropriate label, the search stops and execution begins.
If a label is encountered as part of a running program, it has no effect, that
is, execution simply continues. Therefore, you can label a subordinate
routine within a program (more on subroutines in section 9).
Since the calculator searches in only one direction from its present position,
it is possible (though not advisable) to use duplicate program labels.
Execution will begin at the first appropriately labeled line encountered.
If an ´ A entry starts the search
for ―A‖ here,
it then proceeds downward through
memory, wraps around to line 000,
and stops at label ―A‖. Execution then
starts and continues (ignoring any
other labels) until a halt instruction.
000´bA
´b3
¦
(stop)
end of memory
Unexpected Program Stops
Pressing Any Key. Pressing any key will halt program execution. It will
not halt in the middle of an operation. This instruction will be completed
before the program stops.
Error Stops. Program execution is immediately halted when the calculator
attempts an improper operation that results in an Error display.
To see the line number and keycode of the error-causing instruction (the
line at which the program stopped), press any one key to remove the Error
message, then switch to Program mode.
If the display is flashing when a program stops, an overflow condition exists
(page 61). Press − =, or | " 9 to stop the blinking.
Abbreviated Key Sequences
In certain cases, an ´ prefix you might expect to include in a key
sequence is not needed. The rule for using an abbreviated key sequence is:
the ´ prefix key is unnecessary after any other prefix key. (Page 19
contains a list of prefix keys.)
Section 6: Programming Basics
79
For example, ´b´A becomes ´bA, ´m´%
becomes ´m%, and O´# becomes O#.
The removal of the ´ is not ambiguous because the ´-shifted function
is the only logical one in these cases. The keycodes for such instructions do
not include the extraneous ´ even if you do key it in.
User Mode
User mode is a convenience to save keystrokes when addressing (calling
up) programs for execution. Pressing ´U will exchange the primary
functions and ´-shifted functions of the A through E keys only. In
User mode (USER annunciator displayed):
´ shift
Primary
| shift
A
¤
2
x
B
'
LN
C
D
@ y
LOG 
E
∕

Press | U again to deactivate User mode.
Polynomial Expressions and Horner's Method
Some expressions, such as polynomials, use the same variable several times
for their solution. For example, the expression
f(x) = Ax4 + Bx3 + Cx2 + Dx + E
uses the variable x four different times. A program to solve such an
equation could repeatedly recall a stored copy of x from a storage register.
A shorter programming method, however, would be to use a stack which
has been filled with the constant (refer to Loading the Stack with a
Constant, page 41).
Horner's Method is a useful means of rearranging polynomial expressions to
cut calculation steps and calculation time. It is especially expedient in
_ and f, two rather long-running functions that use subroutines.
This method involves rewriting a polynomial expression in a nested fashion
to eliminate exponents greater than 1:
Ax4 + Bx3 + Cx2 + Dx + E
(Ax3 + Bx2 + Cx + D)x + E
((Ax2 + Bx + C)x + D)x + E
(((Ax + B)x + C)x + D)x + E
80
Section 6: Programming Basics
Example: Write a program for 5x4 + 2x3 as (((5x + 2)x)x)x, then evaluate
for x = 7
Keystrokes
|¥
´bB
5
*
2
+
*
*
*
|n
Display
000-
Assumes position in memory
is line 000. If it is not, clear
program memory.
001-42,21,12
0025
00320
0042
00540
00620
00720
00820
00943 32
5x.
5x + 2.
(5x + 2)x.
(5x + 2)x2.
(5x + 2)x3.
Returns to Run mode, Prior
result remains in display.
|¥
7vv
v
7.0000
´B
12,691.0000
Loads the stack (X-, Y-, Z-,
and T-registers) with 7.
Nonprogrammable Functions
When the calculator is in Program mode, almost every function on the
keyboard can be recorded as an instruction in program memory. The
following functions cannot be stored as instructions in program memory.
´ CLEAR u
´ CLEAR M
´%
´U
|‚
|W
|¥
t “ nnn
Â
−
=/.
=/-
Section 6: Programming Basics
81
Problems
1.
The village of Sonance has installed a 12-o'clock whistle in the
firehouse steeple. The sound level at the firehouse door, 3.2 meters
from the whistle, is 138 decibels. Write a program to find the sound
level at various distances from the whistle.
Use the equation L = L0 – 20 log (r/r0), where: L0 is the known sound
level (138 db) at a point near the source,
r0 is the distance of that point from the source (3.2 m), L is the
unknown sound level at a second point, and r is the distance of the
second point from the source in meters.
What is the sound level at 3 km from the source (r = 3 km)?
A possible keystroke sequence is:
|¥ ´bC 3.2 ÷ |o 20 * “ 138
+ |n |¥ taking 15 program lines and 15 bytes of
memory. This problem can be solved in a more general way by
removing the specific values 3.2 and 138 from the program, and
instead recalling the L0 and r0 values from storage registers; or by
removing 3.2 and 138 and loading L0, r, and r0 into the stack before
execution: L0 v r v r0.
(Answer: for r = 3 km, L = 78.5606 db.)
2.
A "typical large" tomato weighs about 200 grams, of which about
188 g (94%) are water. A tomato grower is trying to produce
tomatoes of lower percentage water. Write a program to calculate the
percent change in water content of a given tomato compared to the
typical tomato. Use a programmed stop to enter the water weight of
the new tomato.
What is the percent change in water content for a 230 g tomato of
which 205 g are water?
A possible keystroke sequence is:
´bÁ .94 v ¦ (enter water weight of new tomato)
v ¦ (enter total weight of new tomato) ÷ |∆
|n taking 11 program lines and 11 bytes of memory.
(Answer: for the 230 g tomato above, the percent change in percent
water weight is -5.1804%.)
Section 7
Program Editing
There are many reasons to modify a program after you've already stored it:
you might want to add or delete an instruction (like O, ©, or
¦), or you might even find some errors! The HP-15C is equipped with
several editing features to make this process as easy as possible.
The Mechanics
Making a program modification of any kind involves two steps: moving to
the proper line (the location of the needed change) and making the
deletion(s) and/or insertion(s).
Moving to a Line in Program Memory
The Go To (t) Instruction. The sequence t “ nnn will move
program memory to line number nnn, whether pressed in Run mode or
Program mode (PRGM displayed). This is not a programmable sequence; it
is for manually finding a specific position in program memory. The line
number must be a three-digit number satisfying 000 ≤ nnn ≤ 448.
The Single Step (Â) Instruction. To move only one line at a time
forward through program memory, press  (single step). This function
is not programmable.
In Program mode: Â will move the memory position forward one line
and display that instruction. The instruction is not executed. If you hold the
key down, the calculator will continuously scroll through the lines in
program memory.
In Run mode: Â will display the current program line while the key is
held down. When the key is released, the current instruction is executed, the
result displayed, and the calculator steps forward to the next program line to
be executed.
82
Section 7: Program Editing
83
The Back Step (‚) Instruction. To move one line backwards in
program memory, press ‚ (back step) in Program or Run mode. This
function is not programmable. ‚ will scroll (with the key held down) in
Program mode. Program instructions are not executed.
Deleting Program Lines
Deletions of program instructions are made with − (back arrow) in
Program mode. Move to the line you want to delete, then press −. Any
remaining following lines will be renumbered to stay in sequence.
Pressing − in Run mode does not affect program memory, but is used for
display clearing. (Refer to page 21.)
Inserting Program Lines
Additions to a program are made by moving to the line preceding the point
of insertion. Any instruction you key in will be added following the line
currently in the display. To alter an instruction, first delete it, then add the
new version.
Examples
Let's refer back to the can volume program on page 71 in section 6 and
make a few changes in the instructions. (The can program as listed below is
assumed to be in memory starting on line 001.)
Deletions: If we don't need the summed base area, volume, and surface area
values, we can delete the storage register additions (lines 007, 011, and
020).
Changes: To eliminate the need to stop the program to enter the height
value (h), change the ¦ instruction to a l 1 instruction (because of
the above deletions, R1 is no longer being used) and store h in R1 before
running the program. To clean things up, let's also alter O 4 (line 006)
to O 2 and l 4 (old line 016) to l 2, since we are no longer
using R2 and R3.
The editing process is diagrammed on the next page.
84
Section 7: Program Editing
Let's start at the end of the program and work backwards. In this way, deletions
will not change the line numbers of the preceding lines in the program.
Keystrokes
|¥
t “ 020
(or use Â)
Display
000020-44,40, 3
Program mode. (Assumes
position is at line 000.)
Moves position to line 020
(instruction O + 3.)
Section 7: Program Editing
Keystrokes
Display
−
019| ‚ (hold)
016-
45
−
l2
t “ 011
(or hold ‚)
−
| ‚ (hold)
01520
01645 2
011-44,40, 2
42 31
31
−
l1
|‚
−
007-44,40,
00845
007-44,40,
00644
−
O2
005006-
010008-
44
40
4
1
1
1
4
20
2
85
Line 020 deleted.
The next line to edit is line
016 (l 4).
Line 016 deleted.
Line 016 changed to l 2.
Moves to line 011 (O+
2).
Line 011 deleted.
Stop! (Single-stepping
backwards to line 008:
¦.)
¦ deleted.
Line 008 changed to l 1.
Back-step to line 007.
Line 007 (O+ 1)
deleted.
Line 006 (O 4) deleted.
Changed to O 2.
The replacement of a line proceeds like this:
Further Information
Single-Step Operations
Single-Step Program Execution. If you want to check the contents of a
program or the location of an instruction, you can single step through the
program in Program mode. If, on the other hand, running the program
produces an error, or you suspect that a portion of the program is faulty,
86
Section 7: Program Editing
you can check the program by executing it stepwise. This is done by
pressing  in Run mode.
Keystrokes
Display
|¥
´ CLEAR Q
tA
8O1
2.5
 (hold)
(release)
Â
Â
Â
Â
8.0000
2.5
00142,21,11
2.5000
002
44
2.5000
0
003
43 11
6.2500
004
43 26
3.1416
005
19.6350
20
Run mode.
Clear storage registers.
Move to first line of program
A.
Store a can height.
Enter a can radius.
Keycode for line 001 (label).
Result of executing line 001.
t 0.
Result.
| x.
Result.
| $.
Result.
*
Result: the base area of the can.
Wrapping. Â will not move program position into ―unoccupied‖
program territory. Instead, the calculator will ―wrap around‖ to line 000. (In
Run mode, Â will perform any instructions at the end of program
memory, such as n, t or G.)
Line Position
Recall that the calculator's position in program memory does not change
when it is shut off or Program/Run modes are changed. Upon returning to
Program mode, the calculator line position will be where you left it. (If you
executed a program ending with n, the position will be at line 000.)
Therefore, if the calculator is left on and shuts itself off, you need only turn
it on and switch to Program mode (the calculator always "wakes up" in Run
mode) to be back where you were.
Section 7: Program Editing
87
Insertions and Deletions
After an insertion, the display will show the instruction you just added.
After a deletion, the display will show the line prior to the deleted (now
nonexistent) one.
If all space available in memory is occupied, the calculator will not accept
any program instruction insertions and Error 4 will be displayed.
Initializing Calculator Status
The contents of storage registers and the status of calculator settings will
affect a program if the program uses those registers or depends on a certain
status setting. If the current status is incorrect for the program being run,
you will get incorrect results. Therefore, it is wise to clear registers and set
relevant modes either just prior to running a program or within the program
itself. A self-initializing program is more mistake-proof—but it also uses
more program lines.
Calculator-initializing functions are: ´ CLEAR ∑, ´ CLEAR
M, ´ CLEAR Q, | D, | R, | g, |
F, and | ".
Problems
It is good programming technique to avoid using identical program labels.
(This shouldn't be hard, since the HP-15C provides 25 different labels.) To
ensure against duplication of labels, you can clear program memory first.
1. The following program is used by the manager of a savings and loan
company to compute the future values of savings accounts according
to the formula FV = PV (l + i)n, where FV is future value, PV is
present value, i is the periodic interest rate, and n is the number of
periods. Enter PV first (into the Y-register) and n second (into the Xregister) before executing the program. Given is an annual interest
rate of 7.5% (so i = 0.075).
88
Section 7: Program Editing
Keystrokes
´b.1
´ •2
1
.
0
7
5
®
y
*
|n
Display
001-42,21,.1
002-42, 7, 2
0031
00448
0050
0067
0075
00834
00914
01020
01143 32
Interest.
(1 + i)n
PV (1 + i)n
Load the program and find the future value of $1,000 invested for 5
years; of $2,300 invested for 4 years. Remember to use G to run
a program with a digit label. (Answers: $1,435.63; $3,071.58.)
Alter the program to make the annual interest rate 8.0%.
Using the edited program, find the future value of $500 invested for
4 years; of $2,000 invested for 10 years. (Answers: $680.24;
$4,317.85.)
2.
Create a program to calculate the length of a chord ℓ subtended by an
angle  (in degrees) on a circle of radius r, according to the equation
θ
ℓ=2r sin .
2
Find ℓ when θ = 30° and r = 25.
(Answer: 12.9410. A possible program is: ´bA |D
´•4 2 * ® 2 ÷ [ * |n). (Assumes
 in Y-register and r in X-register when program is run.)
Section 7: Program Editing
89
Make any necessary modifications in the program to also find and display s,
the length of the circular arc cut by θ (in radians), according to the equation
s = r θ.
Complete the following table:
θ
45°
90°
270°
r
50
100
100
ℓ
?
?
?
s
?
?
?
(Answers: 38.2683 and 39.2699; 141.4214 and 157.0796; 141.4214 and
471.2389.
A possible new sequence is:
´bA |D ´•4 O0 2* ® O1 2÷
[ * ´© ´© l0 l1 ´r *
|n).
Section 8
Program Branching
and Controls
Although the instructions in a program are normally executed sequentially,
it is often desirable to transfer execution to a part of the program other than
the next line. Branching in the HP-15C may be simple, or it may depend on
a certain condition. By branching to a previous line, it is possible to execute
part of a program more than once – a process called looping.
The Mechanics
Branching
The Go To (t) Instruction. Simple branching – that is, unconditional
branching – is carried out with the instruction t label. In a running
program, t will transfer execution to the next appropriately labeled
program or routine (not to a line number).
The calculator searches forward in memory, wrapping around through line
000 if necessary, and resumes execution at the first line containing the
proper label.
Looping. If a t instruction specifies a label at a lower-numbered line
(that is, a prior line), the series of instructions between the t and the
label will be executed repeatedly – possibly indefinitely. The continuation
90
Section 8: Program Branching and Controls
91
of this loop can be controlled by a conditional branch, an ¦ instruction
(written into the loop), or simply by pressing any key during execution
(which stops the program).
Conditional Tests
Another way to alter the sequence of program execution is by a conditional
test, a true/false test which compares the number in the X-register either to
zero or to the number in the Y-register. The HP-15C provides 12 different
tests, two explicit on the keyboard and 10 others accessible using |
T n.*
1. Direct: | £ and | ~ .
2. Indirect: | T n.
n
0
1
2
3
4
*
Test
x≠0
x>0
x<0
x≥0
x≤0
n
5
6
7
8
9
Test
x=y
x≠y
x>y
x<y
x≥y
Four of the conditional tests can also be used for complex values, as explained in section 11 on page 132.
92
Section 8: Program Branching and Controls
Following a conditional test, program execution follows the "Do if True"
Rule: it proceeds sequentially if the condition is true, and it skips one
instruction if the condition is false. A t instruction is often placed right
after a conditional test, making it a conditional branch; that is, the t
branch is executed only if the test condition is met.
Flags
Another conditional test for programming is a flag test. A flag is a status
indicator that is either set (= true) or clear (= false). Again, execution
follows the "Do if True" Rule: it proceeds sequentially if the flag is set, and
skips one line if the flag is clear.
The HP-15C has eight user flags, numbered 0 to 7, and two system flags,
numbered 8 (Complex mode) and 9 (overflow condition). The system flags
are discussed later in this section. All flags can be set, cleared, and tested as
follows:
 | F n will set flag number n (0 to 9).
 | " n will clear flag number n.
 | ? n will check if flag n is set.
A flag n that has been set remains set until it is cleared either by a "
n instruction or by clearing (resetting) Continuous Memory.
Section 8: Program Branching and Controls
93
Examples
Example: Branching and Looping
A radiobiology lab wants to predict the
diminishing radioactivity of a test amount of
131
I, a radioisotope. Write a program to figure
the radioactivity at 3-day intervals until a
given limit is reached. The formula for Nt, the
amount of radioisotope remaining after t days,
is
Nt = No (2-t/k),
where k = 8 days, the half-life of 131I, and N0 is the initial amount.
The following program uses a loop to calculate the number of millicuries
(mci) of isotope theoretically remaining at 3-day intervals of decay.
Included is a conditional test to check the result and end the program when
radioactivity has fallen to a given value (a limit).
The program assumes t1 – the first day of measurement – is stored in R0, N0
– the initial amount of isotope – is stored in R1, and the limit value for
radioactivity is stored in R2.
Keystrokes
|¥
´ CLEAR M
´bA
Display
000000001-42,21,11
l0
002-
45
0
´©
8
÷
“
2
®
Y
003004005006007008009-
42 31
8
10
16
2
34
14
Program mode.
(Optional.)
Each loop returns to this
line.
Recalls current t which
changes with each loop.
Pauses to display t.
k
–t/k.
2–t/k.
94
Section 8: Program Branching and Controls
Keystrokes
Display
l*1
010-45,20, 1
´©
l2
|T9
01142 31
01245 2
013-43,30, 9
|n
3
O+ 0
tA
01443 32
0153
016-44,40, 0
01722 11
Recall multiplication with the
contents of R1 (N0), yielding Nt,
the mci of 131I remaining after t
days
Pauses to display Nt.
Recalls limit value to X-register.
x ≥ y ? Tests whether limit value
(in X) meets or exceeds Nt
(in Y).
If so, program ends.
If not, program continues.
Adds 3 days to t in R0.
Go to ―A‖ and repeat execution
to find a new Nt from a new t.
Notice that without lines 012 to 014, the loop would run indefinitely (until
stopped from the keyboard).
Let's run the program, using t1 = 2 days, N0 = 100 mci, and a limit value of
half of N0 (50 mci).
Keystrokes
|¥
2O0
100 O 1
50 O 2
´A
Display
2.0000
100.0000
50.0000
2.0000
84.0896
5.0000
64.8420
8.0000
50.0000
50.0000
Run mode (display will vary).
t1.
N0.
Limit value for Nt.
t1.
N1.
t2.
N2.
t3.
N3.
Nt limit; program ends.
Section 8: Program Branching and Controls
95
Example: Flags
Calculations on debts or investments can be calculated in two ways: for
payments made in advance (at the beginning of a given period) and for
payments made in arrears (at the end of a given period). If you write a
program to calculate the value (or ―present value‖) of a debt or investment
with periodic interest and periodic payments, you can use a flag as a status
indicator to tell the program whether to assume payments are made in
advance or payments are made in arrears.
Suppose you are planning the payment of your child's future college tuition.
You expect the cost to be about $3,000/year or about $250/month. If you
wanted to withdraw the monthly payments from a bank account yielding
6% per year, compounded monthly (which equals 0.5% per month), how
much must you deposit in the account at the start of the college years to
fund monthly payments for the next 4 years?
The formula is
1  (1  i)  n 
V P 
 (1  i)
i


if payments are to be made
each month in advance,
and the formula is
1  (1  i)  n 
V P 

i


if payments are to be made
each month in arrears.
V is the total value of the deposit you must make in the account;
P is the size of the periodic payment you will draw from the account;
i is the periodic interest rate (here: ―periodic‖ means monthly, since interest
is compounded monthly); and
n is the number of compounding periods (months).
The following program allows for either payment mode. It assumes that,
before the program is run, P is in the Z-register, n is in the Y-register, and i
is in the X-register.
96
Section 8: Program Branching and Controls
Keystrokes
|¥
´ bB
Display
000001-42,21,12
|"0
002-43, 5, 0
t1
´bE
00322 1
004-42,21,15
|F0
005-43, 4, 0
´b1
O1
1
+
®
“
y
“
1
+
l÷1
006-42,21, 1
00744 1
0081
00940
01034
01116
01214
01316
0141
01540
016-45,10, 1
*
|?0
|n
01720
018-43, 6, 0
01943 32
l1
1
+
*
020021022023-
45
|n
024-
43 32
1
1
40
20
Program mode.
Start at "B" if payments to be
made at the beginning.
Flag 0 clear (false); indicates
advance payments.
Go to main routine.
Start at "E" if payments to be
made at the end.
Flag 0 set (true); indicates
payment in arrears.
Routine 1 (main routine).
Stores i (from X-register).
(1+i).
Puts n in X; (l + i) in Y.
– n.
(1 + i)-n.
– (1 + i)-n.
1 – (1 + i)-n.
Recall division with R1 (i) to
get [l– (l + i)-n]/i.
Multiplies quantity by P.
Flag 0 set?
End of calculation if flag 0 set
(for payments in arrears).
Recalls i.
(1 + i).
Multiplies quantity by final
term.
End of calculation if flag 0
clear.
Section 8: Program Branching and Controls
97
Now run the program to find the total amount needed in an account from
which you want to take $250/month for 48 months. Enter the periodic
interest rate as a decimal fraction, that is, 0.005 per month. First find the
sum needed if payments will be made at the beginning of the month
(payments in advance), then calculate the sum needed if payments will be
made at the end of the month (in arrears).
Keystrokes
|¥
250 v
48 v
.005
´B
(Repeat stack entries.)
´E
Display
Set to Run mode.
250.0000
Monthly payment.
48.0000
Payment periods (4 years × 12
months).
0.005
Monthly interest rate as a
decimal fraction.
10,698.3049 Deposit necessary for
payments to be made in
advance.
10,645.0795 Deposit necessary for payment to
be made in arrears. (The
difference between this deposit
and the tuition cost ($12,000)
represents interest earned on the
deposit!)
Further Information
Go to
In contrast to the nonprogrammable sequence t “ nnn, the
programmable sequence t label cannot be used to branch to a line
number, but only to program label (a line containing ´ b label).*
Execution continues from the point of the new label, and does not return to
the original routine unless given another t instruction.
t label can also be used in Run mode (that is, from the keyboard) to
move to a labeled position in program memory. No execution occurs.
*
It is possible to branch under program control to a particular line number by using indirect addressing,
discussed in section 10.
98
Section 8: Program Branching and Controls
Looping
Looping is an application of branching which uses a t instruction to
repeat a portion of the program. A loop can continue indefinitely, or may be
conditional. A loop is frequently used to repeat a calculation with different
variables. At the same time, a counter, which increments with each loop,
may be included to keep track of loop iterations. This counter can then be
checked with a conditional test to determine when to exit the loop. (This is
shown in the example on page 112.)
Conditional Branching
There are two general applications for conditional branching. One is to
control loops, as explained above. A conditional test can check for either a
certain calculated value or a certain loop count.
The other major use is to test for options and pursue one. For example, if a
salesperson made a variable commission depending on the amount of sale,
you could write a program which takes the amount of sale, compares it to a
test value, and then calculates a specific commission depending on whether
the sale is less than or greater than the test value.
Tests. A conditional test takes what is in the X-register (“x”) and compares
it either to zero (such as ~) or to “y”, that is, what is in the Y-register
(such as £). For an x:y comparison, therefore, you must have the x- and
y-values juxtaposed in the X- and Y-registers. This might require that you
store a test value and then recall it (bringing it into the X-register). Or, the
value might be in the stack and be moved, as necessary, using ®, ),
or (.
Tests With Complex Numbers and Matrix Descriptors. Four of the
conditional tests also work with complex numbers and matrix descriptors:
~, T 0 (x≠ 0), T 5 (x = y), and T 6 (x≠ y). Refer to
sections 11 and 12 for more information.
Flags
As a conditional test can be used to pick an option by comparing
two numbers in a program, a flag can be used to pick an option externally.
Usually, a flag is set or cleared first thing in a program by choosing a
different starting point (using different labels) depending on the condition
or mode you want (refer to the example on page 95).
Section 8: Program Branching and Controls
99
In this way, a program can accommodate two different modes of input, such
as degrees and radians, and make the correct calculation for the mode
chosen. You set a flag if a conversion needs to be made, for instance, and
clear it if no conversion is needed.
Suppose you had an equation requiring temperature input in degrees Kelvin,
although sometimes your data might be in degrees Celsius. You could use a
program with a flag to allow either a Kelvin or Celsius input. In part, such a
program might include:
´ bC
|"7
t1
´bÁ
|F7
´b1
|?7
t2
2
7
3
+
´b2
Start program at ―C‖ for degrees Celsius.
Flag 7 cleared (=false).
Start program at ―D‖ for degrees Kelvin.
Flag 7 set (=true).
(Assuming temperature in X-register.)
Checks for flag 7 (checks for Celsius or Kelvin
input).
If set (Kelvin input), goes to a later routine, skipping
the next few instructions.
If cleared (Celsius input), adds 273 to the
value in the X-register, since °K = °C + 273.
Calculation continues for both modes.
⋮
The System Flags: Flags 8 and 9
Flag 8. Setting flag 8 will activate Complex mode (described in section 11),
turning on the C annunciator. If another method is used to activate Complex
mode, flag 8 will automatically be set. Complex mode is deactivated only
by clearing flag 8; flag 8 is cleared in the same manner as the other flags.
100
Section 8: Program Branching and Controls
Flag 9. An overflow condition (described on page 61) automatically sets
flag 9. Flag 9 causes the display to blink or, if a program is running, waits
until execution is complete and then starts blinking the display.
Flag 9 may be cleared in three ways:

Press | " 9 (the common procedure for clearing flags).

Press −. This will only clear flag 9 and stop the blinking—it will
not clear the display.

Turn the calculator off. (Flag 9 is not cleared if the calculator turns
itself off.)
If you set flag 9 manually (F 9), it causes the display to blink irrespective
of the overflow status of the calculator. As usual, a program will run to
completion before the display starts blinking. Therefore, flag 9 can be used
as a programming tool to provide a visual signal for a selected condition.
Section 9
Subroutines
When the same set of instructions needs to be used at more than one point
in a program, memory space can be conserved by storing those instructions
as a single subroutine.
The Mechanics
Go To Subroutine and Return
The G (go to subroutine) instruction is executed in the same way as the
t branch, with one major difference: it establishes a pending return
condition. G label, like t label,* transfers program execution to the
line with the corresponding label (A to E, 0 to 9 or .0 to .9). However,
execution then continues until the first subsequent n instruction is
encountered – at which point execution transfers back to the instruction
immediately following the last G instruction, and continues on from
there.
Subroutine Execution
´bA
´b.1
G.1
|n
|n
END
RETURN
Execution transfers to line 000
and halts.
*
Execution transfers back to
original routine after
G.1
A G or t instruction followed by a letter label is an abbreviated key sequence (no ´
necessary). Abbreviated key sequences are explained on page 78.
101
102
Section 9: Subroutines
Subroutine Limits
A subroutine can call up another subroutine, and that subroutine can call up
yet another subroutine. This ―subroutine nesting‖—the execution of a
subroutine within a subroutine—is limited to stack of subroutines seven
levels deep (this does not count the main program level). The operation of
nested subroutines is as shown below:
Main Program
bA
b1
G1
b2
b3
b4
G3
G2
n
End
n
G4
n
n
n
Examples
Example: Write a program to
calculate the slope of the secant line
joining points (x1, y1) and (x2, y2) on
the graph shown, where y = x2 - sin x
(given x in radians).
The secant slope is:
2
y 2  y1
( x  sin x2 )  ( x12  sin x1 )
, or 2
x 2  x1
x 2  x1
The solution requires that the equation for y be evaluated twice—once for y1
and once for y2, given the data input for x1 and x2. Since the same
calculation must be made for different values, it will save program space to
call a subroutine to calculate y.
The following program assumes that x1 has been entered into the Y-register
and x2 into the X-register.
Section 9: Subroutines
MAIN PROGRAM
|¥
´ CLEAR M
000001- ´ b 9
002- | R
003- O 0
004- ®
005- O - 0
006- G .3
103
(Not programmable.)
Start main program.
Radians mode.
Stores x2 in R0.
Brings x1 into X; x2 into Y.
(x2 - x1) in R0.
Transfer to subroutine ―.3‖ with x1.
Return from subroutine ―.3‖.
007- “
008- ®
009- G .3
- y1.
Brings x2 into X-register.
Transfer to subroutine with x2.
Return from subroutine ―.3‖.
010- +
011- l ÷ 0
012- | n
SUBROUTINE
013- ´ b .3
014- | x
015- | K
016- [
017- 018- | n
y2 - y1.
Recalls (x2 – x1) from R0 and
calculates (y2 - y1)/(x2 - x1).
Program end (return to line 000).
Start subroutine .3.
x2.
Recall x.
sin x.
x 2 – sin x, which equals y.
Return to origin in main program.
Calculate the slope for the following values of x1 and x2: 0.52, 1.25; -1, 1;
0.81, 0.98. Remember to use G 9 (rather than ´ 9) when addressing a
routine with a digit label.
Answers: 1.1507; -0.8415; 1.1652.
104
Section 9: Subroutines
Example: Nesting. The following subroutine, labeled ―.4‖, calculates the
value of the expression x 2  y 2  z 2  t 2 as part of a larger calculation in a
larger program. The subroutine calls upon another subroutine (a nested
subroutine), labeled ―.5‖, to do the repetitive squaring.
The program is executed after placing the variables t, z, y, and x into the T-,
Z-, Y-, and X-registers.
Keystrokes
´ b.4
|x
G.5
G.5

G.5

¤

Start of main
subroutine.
x2.
Calculates y2 and
x2 + y2.
Calculates z2 and
2
2
2
X +y +z .
Calculates t2 and
x2 + y2 + z2 + t2.
2
2
2 2
x  y  z t
|n
End of main subroutine;
returns to main program.
´ b.5
Start of nested
subroutine.
®
|x
+
|n
Calculates a square and
adds it to current sum of squares.
End of nested sub-routine; returns
to main subroutine.
If you run the subroutine (with its nested subroutine) alone using x = 4.3,
y = 7.9, z = 1.3, and t = 8.0, the answer you get upon pressing G.4 is
12.1074.
Section 9: Subroutines
105
Further Information
The Subroutine Return
The pending return condition means that the n instruction occurring
subsequent to a G instruction causes a return to the line following the
G rather than a return to line 000. This is what makes a subroutine
useful and reuseable in different parts of a program: it will always return
execution to where it branched from, even as that point changes. The only
difference between using a G branch and a t branch is the transfer
of execution after a n.
Nested Subroutines
If you attempt to call a subroutine that is nested more than seven levels
deep, the calculator will halt and display Error 5 when it encounters the
G instruction at the eighth level.
Note that there is no limitation (other than memory size) on the number of
nonnested subroutines or sets of nested subroutines that you may use.
Section 10
The Index Register
and Loop Control
The Index register (RI) is a powerful tool in advanced programming of the
HP-15C. In addition to storage and recall of data the Index register can use
an index number to:

Count and control loops.

Indirectly address storage registers, including those beyond R.9
(R19).

Indirectly branch to program line numbers, as well as to labels.

Indirectly control the display format.

Indirectly control flag operations.
The V and % Keys
Direct Versus Indirect Data Storage With the Index Register
The Index register is a data storage register that can be used directly, with
V, or indirectly, with %.* The difference is important to note:
V
The V function uses the
number itself in the Index
register.
*
%
The % function uses the absolute
value of the integer portion of the
number in the Index register to
address another data storage
register. This is called indirect
addressing.
Note that the matrix functions and complex functions use the V and % keys also, but for different
purposes. Refer to sections 11 and 12 for their usage.
106
Section 10: The Index Register and Loop Control
107
Indirect Program Control With the Index Register
The V key is used for all forms of indirect program control other than
indirect register addressing. Hence, V (not %) is used for indirect
program branching, indirect display format control, and indirect flag
control.
Program Loop Control
Program loop counting and control can be carried out in the HP-15C by any
storage register: R0 through R9, R.0 through R.9, or the Index register (V).
Loop control can also be carried out indirectly with %.
The Mechanics
Both V and % can be used in abbreviated key sequences, omitting the
preceding ´ prefix (as explained on page 78).
Index Register Storage and Recall
Direct. O V and l V. Storage and recall between the Xregister and the Index register operate in the same manner as with other data
storage registers (page 42).
Indirect. O (or l) % stores into (or recalls from) the data storage
register whose number is addressed by the integer portion of the value (0 to
65) in the Index register. See the table below and on the next page.
Indirect Addressing
If RI contains:
± 0
⋮
9
10
11
⋮
19
20
% will address:
R0
⋮
R9
R.0
R.1
⋮
R.9
R20
t V or G V will
transfer to:*
´b0
⋮
´b9
"
" .0
"
" .1
⋮
´ b .9
"
" A
*For RI  0 only.
(Continued on next page.)
108
Section 10: The Index Register and Loop Control
Indirect Addressing
If RI contains:
21
22
23
24
⋮
65
% will address:
R21
R22
R23
R24
⋮
R65
t V or GV will
transfer to:*
´bB
" " C
" " Á
" " E
—
—
*For RI  0 only.
Index Register Arithmetic
Direct. O or l { + , -, *, ÷ } V. Storage or recall
arithmetic operates with the Index register in the same manner as upon
other data storage registers (page 43).
Indirect. O or l { + , -, *, ÷ } % carries out storage
or recall arithmetic with the contents of the data storage register addressed
by the integer portion of the number (0 to 65) in the Index register. See the
above table.
Exchanging the X-Register
Direct. ´ X V exchanges contents between the X-register and the
Index register. (Works the same as X n does with registers 0 through .9.)
Indirect. ´ X % exchanges contents between the X-register and the
data storage register addressed by the number (0 to 65) in the Index register.
See the above table.
Indirect Branching With V
The V key—but not the % key—can be used for indirect branching
(tV) and subroutine calls (GV). (Only the integer portion of
the number in RI is used.) (% is only used for indirect addressing of
storage registers).
Section 10: The Index Register and Loop Control
109
To Labels. If the RI value is positive, t V and G V will
transfer execution to the label which corresponds to the number in the Index
register (see the above table).
For instance, if the Index register contains 20.00500, then a tV
instruction will transfer program execution to ´b A. See the chart
on page 107.
To Line numbers. If the RI value is negative, tV causes branching
to that line number (using the absolute value of the integer portion of the
value in RI).
For instance, if RI contains –20.00500, then a tV instruction will
transfer program execution to program line 020.
Indirect Flag Control With V
F V, " V, or ? V will set, clear, or test the flag (0 to 9)
specified in RI (by the magnitude of the integer portion).
Indirect Display Format Control With V
´ • V, ´ i V, and ´ ^ V will format the
display in their customary manner (refer to pages 58–59), using the number
in RI (integer part only) for n, which must be from 0 to 9.*
Loop Control With Counters: I and e
The I (increment and skip if greater than) and e (decrement and
skip if less than or equal to) functions control loop execution by referencing
and altering a loop control number in a given register. Program execution
(skipping a line or not) then depends on that number.
The key sequence is ´ { I, e } register number. This number is
0 to 9, .0 to .9, V ,or %.
The Loop Control Number. The format of the loop control number is:
nnnnn.xxxyy, where
*
Except when using f (section 14)
±nnnnn
xxx
yy
is the current counter value,
is the test (goal) value, and
Is the increment of decrement value
110
Section 10: The Index Register and Loop Control
For example, the number 0.05002 in a storage register represents:
nnnnn x x x y y
0.0 5 0 0 2
Start count at zero.
Count by twos.
Count up to 50.
I and e Operation. Each time a program encounters I or
e it increments or decrements nnnnn (the integer portion of the loop
control number), thereby keeping count of the loop iterations. It compares
nnnnn to xxx, the prescribed test value, and exits the loop by skipping the
next line if the loop counter (nnnnn) is either greater than (I) or less
than or equal to (e) the test value (xxx). The amount that nnnnn is
incremented or decremented is specified by yy.
With these functions (as opposed to the other conditional tests), the rule is
―Skip if True‖.
False (nnnnn  xxx)
True (nnnnn > xxx)
instruction
´IV
loop
t. 1
instruction
exit loop
For I: given nnnnn.xxxyy, increment nnnnn to nnnnn + yy, compare
it to xxx, and skip the next program line if the new value satisfies nnnnn >
xxx. This allows you to exit a loop at this point when nnnnn becomes
greater than xxx.
Section 10: The Index Register and Loop Control
111
True (nnnnn  xxx)
False (nnnnn > xxx)
instruction
´sV
t. 1
loop
Instruction
exit loop
For e: given nnnnn.xxxyy, decrement nnnnn to nnnnn - yy, compare
it to xxx, and skip the next program line if the new value satisfies nnnnn ≤
xxx. This allows you to exit a loop at this point when nnnnn becomes less
than or equal to xxx.
For example, loop iterations will alter these control numbers as follows:
Iterations
Operation
I
0
0.00602
1
2.00602
2
4.00602
3
6.00602
e
6.00002
4.00002
2.00002
0.00002
(skip next
line)
4
8.00602
(skip next
line)
Examples
Examples: Register Operations
Storing and Recalling
Keystrokes
´ CLEAR Q
12.3456
OV
7¤
O%
lV
Display
Clears all storage registers.
12.3456
12.3456 Stores in RI.
2.6458
2.6458
Storage in R.2 by indirect addressing
(RI = 12.3456).
12.3456 Recalls contents of RI.
112
Section 10: The Index Register and Loop Control
Keystrokes
l%
´ X .2
Display
2.6458
2.6458
Indirectly recalls contents of R.2.
Check: same contents recalled by
directly addressing R.2.
Exchanging the X-Register
Keystrokes
´XV
Display
12.3456
lV
´ X%
2.6458
0.0000
l%
´X2
2.6458
2.6458
Exchanges contents of RI and Xregister.
Present contents of RI.
Exchanges contents of R2 (which is
zero) with X.
Check: directly address R2.
Storage Register Arithmetic
Keystrokes
Display
10.0000
10 O + V
12.6458
lV
3.1416
|$O÷
%
0.8422
l%
0.8422
´ X.2
Adds 10 to RI.
New contents of RI (= old + 10).
Divides contents of R.2 by .
New contents of R.2.
Check: directly address R.2.
Example: Loop Control with e
Remember the program in section 8 which used a loop to calculate
radioactive decay? (Refer to page 93.) This program used a test condition (x
≥ y?) to exit the loop when the calculated result passed the given limit (50).
As we've seen in this section, there's another way to control loop execution:
through a stored loop counter that is monitored by the I or e
function.
Section 10: The Index Register and Loop Control
113
Here is a revision of the original radioisotope decay program. This
time, we will limit the program to three executions of the loop rather
than setting a specific limit value. This example uses e with a
loop control number in R2 of
3.0 0 0 0 1.
initial loop counter
decrement value
test (goal) value
Make the following changes to the program (assuming it is in memory). A
loop counter will be stored in R2 and a line number in the Index register.
Keystrokes
|¥
t“013
Display
000013-43,30, 9
−−
´e 2
01142 31
012-42, 5, 2
tV
013-
22 25
Program mode.
The second of the two loop
test condition lines.
Delete lines 013 and 012.
Add your loop counter
function (counter stored in
R2).
Go to given line number
(015).
Now when the loop counter (stored in R2) has reached zero, it will skip line
013 and go on to 014, the n instruction, thereby ending the program. If
the loop counter has not yet decreased to zero, execution continues with line
013. This branches to line 015 and continues the program and the looping.
To run the program, put t1 (day 1) in R0, N0 (initial isotope batch) in R1 the
loop counter in R2, and the line number for branching in the Index register.
Keystrokes
|¥
2O0
100 O 1
3.000001 O 2
Display
2.00000
100.0000
3.0000
Run mode.
t1.
N0.
Loop counter. (This
instruction could also be
programmed.)
114
Section 10: The Index Register and Loop Control
Keystrokes
Display
15 “ O -15.0000
V´A
2.0000
84.0896
5.0000
64.8420
8.0000
50.0000
50.0000
Branch line number.
Running program loop
= 3.
counter
Loop counter = 2.
Loop counter = 1.
Loop counter = 0; program ends.
Example: Display Format Control
The following program pauses and displays an example of • display
format for each possible decimal place. It utilizes a loop containing a s
instruction to automatically change the number of decimal places.
Keystrokes
|¥
´CLEAR M
´bB
9
OV
´b0
´•V
lV
´©
´eV
t0
|T1
t0
|n
nnnnn = 9. Therefore, xxx = 0 and by default yy
= 1 (yy cannot be zero).
Displays current value of nnnnn.
Value in RI is decremented and tested. Skip a line
if nnnnn  test value.
Continue loop if nnnnn > test value (0).
Tests whether value in display is greater than 0, so
loop will continue when nnnnn has reached 0 but
display still only shows 1.0.
Section 10: The Index Register and Loop Control
115
To display fixed point notation for all possible decimal places on the
HP-15C:
Keystrokes
|¥
´B
Display
Run mode.
9.000000000
8.00000000
7.0000000
6.000000
5.00000
4.0000
3.000
2.00
1.0
0.
0.
Display at ´©instruction.
Display when program halts.
Further Information
Index Register Contents
Any value stored in the Index register can be referenced in three different
ways:
 Using V like any other storage register. The value in R I can be
manipulated as it is: stored, recalled, exchanged, added to, etc.
 Using V as a control number. The absolute value of the integer
portion in RI is a separate entity from the fractional portion. For indirect
branching, flag control, and display format control with V, only this
portion is used. For loop control, the fractional portion is also used, but
separately from the integer portion.*
 Using % as a reference to the contents of another storage register.
The % key uses the indirect addressing system shown in the tables on
pages 107 and 108. (In turn, the contents of that second register may be
used as a loop control number, in the fashion described above.)
*
This is also true for the value in any storage register used for indirect loop control.
116
Section 10: The Index Register and Loop Control
I and e
For the purpose of loop control, the integer portion (the counter value) of
the stored control number can be up to five digits long (nnnnn.xxxyy). The
counter value (nnnnn) is zero if not specified otherwise.
xxx, in the decimal portion of the control number, must be specified as a
three-digit number. (For example, ―5‖ must be ―005‖.) xxx is zero if not
specified otherwise. Whenever I or e is encountered, nnnnn is
compared internally to xxx, which represents the end level for incrementing
or decrementing.
yy must be specified as a two-digit number. yy cannot be zero, so if left (or
specified) as 00, the value for yy defaults to 1. The value nnnnn is altered
by the amount of yy each time the loop runs through I or e. Both
yy and xxx are reference values, which do not change with loop execution.
Indirect Display Control
While you can use the Index register to format the display manually (that is,
from the keyboard), this function is most commonly used in programming.
This capability is especially valuable for the f function, for which
accuracy can be stipulated by specifying the number of digits to be
displayed (as described in section 14).
There are, as usual, certain display limitations to keep in mind. Recall that
any display format function merely alters the number of decimal places to
which the display is rounded. In its memory, the calculator always retains a
number in scientific notation as a 10-digit mantissa with a two-digit
exponent.
The integer portion of the number in the Index register specifies the number
of decimal places to which the display is rounded. A number less than zero
defaults to zero (zero decimal places displayed in • format), while a
number greater than 9 defaults to 9 (9 decimal places displayed in •).*
* Note that in i and ^ format modes, the maximum display is a seven-digit mantissa
with a two-digit exponent. However, a format number greater than six (and less than or equal
to nine) will alter the decimal place at which rounding occurs. (Refer to page 58-59.)
Section 10: The Index Register and Loop Control
117
An exception is in the case of f where the display format number in R I
may range from -6 to +9. (This is discussed in appendix E on page 247.) A
number less than zero will not affect the display format, but will affect
accuracy with this function.
118
Part lll
HP-15C
Advanced Functions
Section 11
Calculating With
Complex Numbers
The HP-15C enables you to calculate with complex numbers, that is,
numbers of the form
a + ib,
where
a is the real part of the complex number,
b is the imaginary part of the complex number, and
i 
1.
As you will see, the beauty of calculating with the HP-15C in Complex
mode is that once the complex numbers are keyed in, most operations are
executed in the same manner as with real numbers.
The Complex Stack and Complex Mode
Calculations with complex numbers are
performed using a complex stack composed
of two parallel four-register stacks (and two
LAST X registers). One of these parallel
stacks – referred to as the real stack –
contains the real parts of complex numbers
used in calculations. (This is the same stack
used in ordinary calculations.) The other
stack – referred to as the imaginary stack –
contains the imaginary parts of complex
numbers used in calculations.
Creating the Complex Stack
The imaginary stack is created (by converting five storage registers as
described in appendix C) when you activate Complex mode; it does not
exist when the calculator is not in Complex mode.
120
Section 11: Calculating With Complex Numbers
121
Complex mode is activated
1) automatically, when executing ´ V or ´ }; or
2) by setting flag 8, the Complex mode flag (|F 8).
When the calculator is in Complex mode, the C annunciator in the display
is lit. This tells you that flag 8 is set and the complex stack exists. In or out
of Complex mode, the number appearing in the display is the number in the
real X-register.
Note: In Complex mode (signified by the C annunciator), the HP15C performs all trigonometric functions using radians. The
trigonometric mode annunciator in the display (RAD, GRAD, or
blank for Degrees) applies to two functions only: ; and :
(as explained later in this section).
Deactivating Complex Mode
Since Complex mode requires the allocation of five registers from memory,
you will have more memory available for programming and other advanced
functions if you deactivate Complex mode when you are working solely
with real numbers.
To deactivate Complex mode, clear flag 8 (keystroke sequence: | "
8). The C annunciator will disappear.
Complex mode is also deactivated when Continuous Memory is reset (as
described on page 63). In any case, deactivating Complex mode dissolves
the imaginary stack, and all imaginary numbers there are lost.
Complex Numbers and the Stack
Entering Complex Numbers
To enter a number with real and imaginary parts;
1.
2.
3.
4.
Key the real part of the number into the display.
Press v
Key the imaginary part of the number into the display.
Press ´ V. (If not already in Complex mode, this creates the
imaginary stack and displays the C annunciator.)
122
Section 11: Calculating With Complex Numbers
Example: Add 2 + 3i and 4 + 5i. (The operations are illustrated in the stack
diagrams following the keystroke listing.)
Keystrokes
´•4
2v
Display
2.0000
3
3
´V
2.0000
4v
4.0000
5
5
´V
4.0000
+
´ % (hold)
(release)
6.0000
8.0000
6.0000
Keys real part of first number
into (real) Y-register.
Keys imaginary part of first
number into (real)
X-register.
Creates imaginary stack; moves
the 3 into the imaginary Xregister, and drops the 2 into the
real X-register.
Keys real part of second number
into (real) Y-register.
Keys imaginary part of second
number into (real) X-register.
Copies 5 from real
X-register into imaginary
X-register, copies 4 from real Yregister into real X-register, and
drops stack.
Real part of sum.
Displays imaginary part
of sum while the % key is held.
(This also terminates digit entry.)
The operation of the real and imaginary stacks during this process is
illustrated below. (Assume that the stack registers have been loaded already
with the numbers shown as the result of previous calculations). Note that
the imaginary stack, which is shown below at the right of the real stack, is
not created until ´ V is pressed. (Recall also that the shading of the
stack indicates that those contents will be written over when the next
number is keyed in or recalled.)
Section 11: Calculating With Complex Numbers
Re
Im
Re
Im
Re
Im
Re
Im
123
Re
Im
T
9
8
7
7
7
0
Z
8
7
6
6
7
0
Y
7
6
2
2
6
0
X
6
2
3
2
Keys:
2
3
v
2
´V
3
The execution of ´ V causes the entire stack to drop, the T contents to
duplicate, and the real X contents to move to the imaginary X-register.
When the second complex number is entered, the stacks operate as shown
below. Note that v lifts both stacks.
Re
Im
Re
Im
Re
Im
Re
Im
T
7
0
7
0
6
0
6
0
Z
7
0
6
0
2
3
2
3
Y
6
0
2
3
4
0
4
0
X
2
3
4
0
4
0
5
0
Keys:
v
4
5
Re
Im
Re
Im
Re
Im
T
6
0
6
0
6
0
Z
2
3
6
0
6
0
Y
4
0
2
3
6
0
X
5
0
4
5
6
8
Keys:
´V
+
A second method of entering complex numbers is to enter the imaginary
part first, then use } and −. This method is illustrated under
Entering Complex Numbers With −, page 127.
124
Section 11: Calculating With Complex Numbers
Stack Lift in Complex Mode
Stack lift operates on the imaginary stack as it does on the real stack (the
real stack behaves identically in and out of Complex mode). The same
functions that enable, disable, or are neutral to lifting of the real stack will
enable, disable, or be neutral to lifting of the imaginary stack. (These
processes are explained in detail in section 3 and appendix B.)
In addition, every nonneutral function, except − and ` causes the
clearing of the imaginary X-register when the next number is entered. That
is, these functions cause a zero to be placed in the imaginary X-register
when the next number is keyed in or recalled. Refer to the stack diagrams
above for illustrations. This feature allows you to execute calculator
operations using the same key sequences you use outside of Complex
mode.*
Manipulating the Real and Imaginary Stacks
} (real exchange imaginary). Pressing ´ } will exchange
the contents of the real and imaginary X-registers, thereby converting the
imaginary part of the number into the real part and vice-versa. The Y-, Z-,
and T-registers are not affected. Press ´ } twice restore a number
to its original form.
} also activates Complex mode if it is not already activated.
Temporary Display of the Imaginary X-Register. Press ´ % to
momentarily display the imaginary part of the number in the X-register
without actually switching the real and imaginary parts. Hold the key down
to maintain the display.
Changing Signs
In Complex mode, the “ function affects only the number in the real Xregister – the imaginary X-register does not change. This enables you to
change the sign of the real or imaginary part without affecting the other. To
key in a negative real or imaginary part, change the sign of that part as you
enter it.
If you want to find the additive inverse of a complex number already in the
X-register, however, you cannot simply press “ as you would outside
*
Except for the : and ; functions, as explained in this section (page 133).
Section 11: Calculating With Complex Numbers
125
of Complex mode. Instead, you can do either of the following:
 Multiply by -1.
 If you don't want to disturb the rest of the stack, press “ ´
} “ ´ }.
To find the negative of only one part of a complex number in the X-register:
 Press “ to negate the real part only.
 Press ´ } “ ´ } to negate the imaginary
part only, forming the complex conjugate.
Clearing a Complex Number
Inevitably you will need to clear a complex number. You can clear only one
part at a time, but you can then write over both parts (since − and `
disable the stack).
Clearing the Real X-Register. Pressing − (or | `) with the
calculator in Complex mode clears only the number in the real X-register; it
does not clear the number in the imaginary X-register.
Example: Change 6 + 8i to 7 + 8i and subtract it from the previous entry.
(Use ´ } or ´ % to view the imaginary part in X.) Assume a,
b, c and d represent parts of complex numbers.
Re
Im
Re
Im
Re
Im
Re
Im
T
a
b
a
b
a
b
a
b
Z
c
d
c
d
c
d
a
b
Y
6
0
6
0
6
0
c
d
X
6
8
0
8
7
8
-1
-8
Keys:
−
7
- (or other
operation)
Since clearing disables the stack (as explained above), the next number you
enter will replace the cleared value. If you want to replace the real part with
zero, after clearing use v or any other function to terminate digit
entry (otherwise the next number you enter will write over the zero); the
imaginary part will remain unchanged. You can then continue with any
calculator function.
126
Section 11: Calculating With Complex Numbers
Clearing the Imaginary X-Register. To clear the number in the imaginary
X-register, press ´ }, then press −. Press ´ } again to
return the zero, or any new number keyed in, to the imaginary X-register.
Example: Replace -1 -8i by -1 + 5i.
Re
Im
Re
Im
Re
Im
Re
Im
Re
Im
T
a
b
a
b
a
b
a
b
a
b
Z
c
d
c
d
c
d
c
d
c
d
Y
e
f
e
f
e
f
e
f
e
f
X
-1
-8
-8
-1
0
-1
5
-1
-1
5
Keys:
}
−
}
(continue with
any operation)
5
Clearing the Real and Imaginary X-Registers. If you want to clear or
replace both the real and imaginary parts of the number in the X-register,
simply press −, which will disable the stack, and enter your new number.
(Enter zeros if you want the X-register to contain zeros.) Alternatively, if
the new number will be purely real (including 0 + 0i), you can quickly clear
or replace the old, complex number by pressing ) followed by zero or
the new, real number.
Example: Replace -1 + 5i with 4 + 7i.
Re
Im
Re
Im
Re
Im
Re
Im
Re
Im
T
a
b
a
b
c
d
c
d
c
d
Z
c
d
c
d
e
f
e
f
c
d
Y
e
f
e
f
4
5
4
5
e
f
X
-1
5
0
5
4
5
7
0
4
7
Keys:
−
4v
7
´V
(continue with
any operation)
Section 11: Calculating With Complex Numbers
127
Entering Complex Numbers with −. The clearing functions − and
` can also be used with } as an alternative method of entering
(and clearing) complex numbers. Using this method, you can enter a
complex number using only the X-register, without affecting the rest of the
stack. (This is possible because − and ` disable stack lift.)
Executing } will also create an imaginary stack if one is not already
present.
Example: Enter 9 + 8i without moving the stack and then find its square.
Keystrokes
(−)
Display
(0.0000)
8
´}
8
7.0000
−
0.0000
9
9
|x
17.0000
144.0000
17.0000
´ % (hold)
(release)
Prevents stack lift when the
next digit (8) is keyed in.
Omit this step if you'd rather
save what's in X and lose
what's in T.
Enter imaginary part first.
Displays real part; Complex
mode activated.
Disables stack. (Otherwise, it
would lift following }.)
Enters real part (digit entry not
terminated).
Real part.
Imaginary part.
T
Re
a
Im
b
Re
a
Im
b
Re
a
Im
b
Re
a
Im
b
Z
c
d
c
d
c
d
c
d
Y
e
f
e
f
e
f
e
f
X
4
7
0
7
8
7
7
8
Keys:
−
8
´}
128
Section 11: Calculating With Complex Numbers
Re
Im
Re
Im
Re
Im
Re
Im
T
a
b
a
b
a
b
a
b
Z
c
d
c
d
c
d
c
d
Y
e
f
e
f
e
f
e
f
X
7
8
0
8
9
8
17
144
−
Keys:
|x
9
Entering a Real Number
You have already seen two ways of entering a complex number. There is a
shorter way to enter a real number: simply key it (or recall it) into the
display just as you would if the calculator were not in Complex mode. As
you do so, a zero will be placed in the imaginary X-register (as long as the
previous operation was not − or `, as explained on page 124).
The operation of the real and imaginary stacks during this process is
illustrated below. (Assume the last key pressed was not − or ` and
the contents remain from the previous example.)
Re
Im
Re
Im
Re
Im
T
a
b
c
d
e
f
Z
c
d
e
f
17
144
Y
e
f
17
144
4
0
X
17
144
4
0
4
0
Keys:
4
v
(Followed by
another number.)
Section 11: Calculating With Complex Numbers
129
Entering a Pure Imaginary Number
There is a shortcut for entering a pure imaginary number into the X-register
when you are already in Complex mode: key in the (imaginary) number and
press ´ }
Example: Enter 0 + 10i (assuming the last function executed was not −
or `.
Keystrokes
10
Display
10
Keys 10 into the displayed
real X-register and zero into
the imaginary X-register.
Exchanges numbers in real
and imaginary X-registers.
Display again shows that the
number in the real Xregister is zero — as it
should be for a pure
imaginary number.
0.0000
´}
The operation of the real and imaginary stacks during this process is
illustrated below. (Assume the stack registers contain the numbers resulting
from the preceding examples.)
T
Re
e
Im
f
Re
e
Im
f
Re
e
Im
f
Z
17
144
17
144
17
144
Y
4
0
4
0
4
0
X
4
0
10
0
0
10
Keys:
10
´}
(Continue with
any operation.)
Note that pressing ´ } simply exchanges the numbers in the real
and imaginary X-registers and not those in the remaining stack registers.
130
Section 11: Calculating With Complex Numbers
Storing and Recalling Complex Numbers
The O and l functions act on the real X-register only; therefore,
the imaginary part of a complex number must be stored or recalled
separately. The keystrokes to do this can be entered as part of a program
and executed automatically.*
To store a + ib from the complex X-register to R1 and R2, you can use the
sequence
O 1 ´} O 2
You can follow this by ´ } to return the stack to its original
condition if desired. To recall a + ib from R1 and R2 you can use the
sequence
l1
l2
´V
If you wish to avoid disturbing the rest of the stack, you can recall the
number using the sequence
l2
´} − l1
(In Program mode, use | ` instead of −.)
Operations With Complex Numbers
Almost all functions performed on real numbers will yield the same answer
whether executed in or out of Complex mode,† assuming the result is also
real. In other words, Complex mode does not restrict your ability to
calculate with real numbers.
Any functions not mentioned below or in the rest of this section
(Calculating With Complex Numbers) ignore the imaginary stack.
* You can use the HP-15C matrix function, described in section 12, to make storing and
recalling complex numbers more convenient. By dimensioning a matrix to be n×2, n
complex numbers can be stored as rows of the matrix. (This technique is demonstrated in
the HP-15C Advanced Functions Handbook, section 3, under Applications.)
† The exceptions are : and ;, which operate differently in Complex mode in order to
facilitate converting complex numbers to polar form (page 133).
Section 11: Calculating With Complex Numbers
131
One-Number Functions
The following functions operate on both the real and imaginary parts of the
number in the X-register, and place the real and imaginary parts of the
answer back into those registers.
¤xNo∕@'a:;
All trigonometric and hyperbolic functions and their inverses also belong to
this group.*
The a function gives the magnitude of the number in the X-registers
(the square root of the sum of the squares of the real and imaginary parts);
the imaginary part of the magnitude is zero.
: converts to polar form and ; converts to rectangular form,
as described later in this section (page 133).
For the trigonometric functions, the calculator considers numbers in the real
and imaginary X-registers to be expressed in radians—regardless of the
current trigonometric mode. To calculate trigonometric functions for values
given in degrees, use r to convert those values to radians before
executing the trigonometric function.
Two-Number Functions
The following functions operate on both the real and imaginary parts of the
numbers in the X- and Y-registers, and place the real and imaginary parts of
the answer into the X-registers. Both stacks drop, just as the ordinary stack
drops after a two-number function not in Complex mode.
+-*÷y
Stack Manipulation Functions
When the calculator is in Complex mode, the following functions
simultaneously manipulate both the real and imaginary stacks in the same
way as they manipulate the ordinary stack when the calculator is not in
Complex mode. The ® function. for instance, will exchange both the
real and imaginary parts of the numbers in the X- and Y-registers.
®)(vK
* Refer to the HP-15C Advanced Functions Handbook for definitions of complex
trigonometric functions and further information about doing calculations in Complex mode.
132
Section 11: Calculating With Complex Numbers
Conditional Tests
For programming, the four conditional tests below will work in the complex
sense: ~ and T 0 compare the complex number in the (real and
imaginary) X-registers to 0 + 0i, while T 5 and T 6 compare the
complex numbers in the (real and imaginary) X- and Y-registers. All other
conditional tests besides those listed below ignore the imaginary stack.
~ T 0 (x ≠ 0) T 5 (x = y) T 6 (x ≠ y)
Example: Complex Arithmetic. The characteristic impedance of a ladder
network is given by an equation of the form
Z0 
A
B
,
where A and B are complex numbers. Find Z0 for the hypothetical values
A = 1.2 + 4.7i and B = 2.7 + 3.2i.
Keystrokes
1.2 v 4.7 ´ V
Display
1.2000
2.7 v 3.2 ´ V
2.7000
÷
¤
1.0428
1.0491
´ % (hold)
0.2406
(release)
1.0491
Enters A into real and
imaginary X-registers.
Enters B into real and
imaginary X-registers,
moving A into real and
imaginary Y-registers.
Calculates A/B.
Calculates Z0 and
displays real part.
Displays imaginary part
of Z0 while % is held
down.
Again displays real part
of Z0.
Section 11: Calculating With Complex Numbers
133
Complex Results from Real Numbers
In the preceding examples, the entry of complex numbers had ensured the
(automatic) activation of Complex mode. There will be times, however,
when you will need Complex mode to perform certain operations on real
numbers, such as  5 . (Without Complex mode, such as operation would
result in an Error 0 – improper math function.) To activate Complex mode
at any time and without disturbing the stack contents, set flag 8 before
executing the function in question.*
Example: The arc sine (sin-1) of 2.404 normally would result in an Error 0.
Assuming 2.404 in the X-register, the complex value arc sin 2.404 can be
calculated as follows:
Keystrokes
|F8
|,
Display
´ % (hold)
-1.5239
(release)
1.5708
1.5708
Activates Complex Mode.
Real part of
arc sin 2.404.
Imaginary part of
arc sin 2.404.
Display shows real part
again when % is
released.
Polar and Rectangular Coordinate Conversions
In many applications, complex numbers are represented in polar form,
sometimes using phasor notation. However, the HP-15C assumes that any
complex numbers are in rectangular form. Therefore, any numbers in polar
or phasor form must be converted to rectangular form before performing a
function in Complex mode.
*
Pressing ´ } twice will accomplish the same thing. The sequence ´ V is not used because
it would combine any numbers, in the real X-. and Y-registers into a single complex number.
134
Section 11: Calculating With Complex Numbers
r (cos θ + i sin θ) = reiθ
(polar)
a + ib =
rθ
(phasor)
; and : can be used to interconvert the rectangular and polar forms of
a complex number. They operate in Complex mode as follows:
´
;
converts the polar (or phasor) form of a complex number to its
rectangular form by replacing the magnitude r in the real Xregister with a, and replacing the angle θ in the imaginary Xregister with b.
|
:
converts the rectangular coordinates of a complex number to the
polar (or phasor) form by replacing the real part a in the real Xregister with r, and replacing the imaginary part b in the
imaginary X-register with θ.
These are the only functions in Complex mode that are affected by the
current trigonometric mode setting. That is, the angular units for θ must
correspond to the trigonometric mode indicated by the annunciator
(or absence thereof).
Section 11: Calculating With Complex Numbers
135
Example: Find the sum 2(cos 65° + i sin 65°) + 3(cos 40° + i sin 40°) and
express the result in polar form, (In phasor form, evaluate 2  65° +
3  40°.)
Keystrokes
|D
Display
Sets Degrees mode for any polarrectangular conversions.
2v
65 ´ V
2.0000
2.0000
´;
0.8452
3v
40 ´ V
´;
3.0000
3.0000
2.2981
+
|:
3.1434
4.8863
´% (hold)
(release)
49.9612
4.8863
C annunciator displayed;
Complex mode activated.
Converts polar to rectangular
form; real part (a) displayed.
Converts polar to rectangular
form; real part (a) displayed.
Converts rectangular to polar
form; r displayed.
θ (in degrees).
Problems
By working through the following problems, you will see that calculating
with complex numbers on the HP-15C is as easy as calculating with real
numbers. In fact, once your numbers are entered, most mathematical
operations will use exactly the same keystrokes. Try it and see!
1. Evaluate:
2i ( 8  6i )
3
(4  2 5 i ) (2  4 5i )
136
Section 11: Calculating With Complex Numbers
Keystrokes
2´}
8“v
6´V
3Y
*
4v
5¤
2“*
´V
Display
0.0000
-8.0000
-8.0000
352.0000
-1.872.0000
4.0000
2.2361
-4.4721
4.0000
-295.4551
÷
2i. Display shows real part.
-8 + 6i.
(-8 + 6i)3.
2 i (-8 + 6i)3.
2 5 .
4  2 5i .
2i(-8  6i )3
4 - 2 5i
2v5¤
4“*
.
2.2361
-8.9443
´V
2.0000
÷
´%
9.3982
-35.1344
9.3982
2  4 5i .
Real part of result.
Answer: 9.3982 -35.1344i.
2. Write a program to evaluate the function ω 
2z  1
5z  3
for different
values of z. ( ω represents a linear fractional transformation, a class of
conformal mappings.) Evaluate ω for z = l+2i.
(Answer: 0.3902 + 0.0122i. One possible keystroke sequence is: ´
bAvv2*1+®5*3+÷
¦ ´ } | n.)
3. Try your hand at a complex polynomial and rework the example on
page 80. You can use the same program to evaluate P(z) = 5z4 + 2z3,
where z is some complex number.
Load the stack with z = 7 + 0i and see if you get the same answer as
before. (Answer: 12,691.0000 + 0.0000i.)
Now run the program for z = 1 + i. (Answer -24.0000 + 4.0000i.)
Section 11: Calculating With Complex Numbers
137
For Further Information
The HP-15C Advanced Functions Handbook presents more detailed and
technical aspects of using complex numbers in various functions with the
HP-15C. Applications are included. The topics include:
 Accuracy considerations.
 Principal branches of multi-valued functions.
 Complex contour integrals.
 Complex potentials.
 Storing and recalling complex numbers using a matrix.
 Calculating the nth roots of a complex number.
 Solving an equation for its complex roots.
 Using _ and f in Complex mode.
Section 12
Calculating With Matrices
The HP-15C enables you to perform matrix calculations, giving you the
capability to handle advanced problems with ease. The calculator can work
with up to five matrices, which are named A through E since they are
accessed using the corresponding A through E keys. The HP-15C lets
you specify the size of each matrix, store and recall the values of matrix
elements, and perform matrix operations – for matrices with real or
complex elements. (A summary of matrix functions is listed at the end of
this section.)
A common application of matrix calculations is solving a system of linear
equations. For example, consider the equations
3.8x1 + 7.2x2 = 16.5
1.3x1 - 0.9x2 = -22.1
for which you must determine the values of x1 and x2.
These equations can be expressed in matrix form as AX = B, where
A
3.8
1.3
7.2
 0.9

,
X
 x1 
 x  , and
 2
B
 16.5 .
 22.1
The following keystrokes show how easily you can solve this matrix
problem using your HP-15C. (The matrix operations used in this example
are explained in detail later in this section.)
First, dimension the two known matrices, A and B, and enter the values of
their elements, from left to right along each row from the first row to the
last. Also, designate matrix C as the matrix that you will use to store the
result of your matrix calculation (C = X).
138
Section 12: Calculating with Matrices
Keystrokes
|"8
2v´mA
´>1
´U
3.8 O A
7.2 O A
1.3 O A
.9 “ O A
2v1´m
B
16.5 O B
22.1 “ O B
´<C
139
Display
Deactivates Complex
mode.
2.0000
Dimensions matrix A
to be 2×2.
2.0000
Prepares for automatic
entry of matrix
elements in User mode.
2.0000
(Turns on the USER
annunciator.)
A
1,1 Denotes matrix A, row
1, column 1. (A display
like this appears
momentarily as you
enter each element and
remains as long as you
hold the letter key.)
3.8000
Stores a11.
7.2000
Stores a12.
1.3000
Stores a21.
-0.9000
Stores a22.
1.0000
Dimensions matrix B to
be 2×l.
16.5000
Stores b11.
-22.1000
Stores b21.
-22.1000
Designates matrix C
for storing the result.
Using matrix notation, the solution of the matrix equation AX = B is
X = A-1B
where A–1 is the inverse of matrix A. You can perform this operation by
entering the ―descriptors‖ for matrices B and A into the Y- and X-registers
and then pressing ÷. (A descriptor shows the name and dimensions of a
matrix.) Note that if A and B were numbers, you could calculate the answer
in a similar manner.
140
Section 12: Calculating with Matrices
Keystrokes
Display
b
2
l>B
l>A
A
2
running
÷
C
2
1 Enters descriptor for B, the 2×1
constant matrix.
2 Enters descriptor for A, the 2×2
coefficient matrix, into the Xregister, moving the descriptor
for B into the Y-register.
Temporary display while A-1B is
being calculated and stored in
matrix C.
1 Descriptor for the result matrix,
C, a 2×1 matrix.
Now recall the elements of matrix C – the solution to the matrix equation.
(Also remove the calculator from User mode and clear all matrices.)
Keystrokes
lC
lC
´U
´>0
Display
C
1,1
-11.2887
8.2496
8.2496
8.2496
Denotes matrix C, row 1, column
1.
Value of c11 (x1).
Value of c21 (x2).
Deactivates User mode.
Clears all matrices.
The solution to the system of equations is x1 = -11.2887 and x2 = 8.2496.
Note: The description of matrix calculations in this section
presumes that you are already familiar with matrix theory and
matrix algebra.
Matrix Dimensions
Up to 64 matrix elements can be stored in memory. You can use all
64 elements in one matrix or distribute them among up to five matrices.
Section 12: Calculating with Matrices
141
Matrix inversion, for example, can be performed on an 8×8 matrix with real
elements (or on a 4×4 matrix with complex elements, as described later*).
To conserve memory, all matrices are initially dimensioned as 0×0. When a
matrix is dimensioned or redimensioned, the proper number of registers is
automatically allocated in memory. You may have to increase the number
of registers allocated to matrix memory before dimensioning a matrix or
before performing certain matrix operations. Appendix C describes how
memory is organized, how to determine the number of registers currently
available for storing matrix elements, and how to increase or decrease that
number.
Dimensioning a Matrix
To dimension a matrix to have y rows and x columns, place those numbers
in the Y- and X-registers, respectively, and then execute ´ m
followed by the letter key specifying the matrix:
1.
Key the number of rows (y) into
the display, then press v
to lift it into the Y-register.
2.
Key the number of columns (x)
into the X-register.
3.
Press ´ m followed by a
letter key, A through E,
that specifies the name of the
matrix.†
Y
number of
rows
X
number of
columns
*
The matrix functions described in this section operate on real matrices only. (In Complex mode, the
imaginary stack is ignored during matrix operation.) However, the HP-15C has four matrix functions that
enable you to calculate using real representations of complex matrices, as described on pages 160-173.
†
You don't need to press ´ before the letter key. (Refer to Abbreviated Key Sequences on page 78.)
142
Section 12: Calculating with Matrices
Example: Dimension matrix A to be a 2×3 matrix.
Keystrokes
2v
Display
2.0000
3
3
´mA
3.0000
Keys number of rows into
Y-register.
Keys number of columns into Xregister.
Dimensions matrix A to be 2×3.
Displaying Matrix Dimensions
There are two ways you can display the dimensions of a matrix:

Press l > followed by the letter key specifying the
matrix. The calculator displays the name of the matrix at the left,
and the number of rows followed by the number of columns at the
right.

Press l m followed by the letter key specifying the
matrix. The calculator places the number of rows in the Y-register
and the number of columns in the X-register.
Keystrokes
l>B
lmA
®
Display
b
0
3.0000
2.0000
0 Matrix B has 0 rows and 0
columns, since it has not been
dimensioned otherwise.
Number of columns in A.
Number of rows in A.
Changing Matrix Dimensions
Values of matrix elements are stored in memory in order from left to right
along each row, from the first row to the last. If you redimension a matrix to
a smaller size, the required values are reassigned according to the new
dimensions and the extra values are lost. For example, if the 2×3 matrix
shown at the left below is redimensioned to 2×2, then
Section 12: Calculating with Matrices
143
If you redimension a matrix to a larger size, elements with the value 0 are
added at the end as required by the new dimensions. For example, if the
same 2×3 matrix is re dimensioned, to 2×4, then
When you have finished calculating with matrices, you'll probably want to
redimension all five matrices to 0×0, so that the registers used for storing
their elements will be available for program lines or for other advanced
functions. You can redimension all five matrices to 0×0 at one time by
pressing ´ > 0. (You can dimension a single matrix to 0×0 by
pressing 0 ´ m {A through E}.)
Storing and Recalling Matrix Elements
The HP-15C provides two ways of storing and recalling values of matrix
elements. The first method allows you to progress through all of the
elements in order. The second method allows you to access elements
individually.
Storing and Recalling All Elements in Order
The HP-15C normally uses storage registers R0 and
R1 to indicate the row and column numbers of a
matrix element. If the calculator is in User mode,
the row and column numbers are automatically
incremented as you store or recall each matrix
element, from left to right along each row from the
first row to the last.
To set the row and column numbers in R0 and R1 to row 1, column 1,
press ´ > 1.
144
Section 12: Calculating with Matrices
To store or recall sequential elements of a matrix:
1.
Be sure the matrix is properly dimensioned.
2.
Press ´ >1. This stores 1 in both storage registers R0 and
R1, so that elements will be accessed starting at row 1, column 1.
3.
Activate User mode by pressing ´ U. With the calculator in
User mode, after each element is stored or recalled the row number
in R0 or the column number in R1 is automatically incremented by 1,
as shown in the example following.
4.
If you are storing elements, key in the value of the element to be
stored in row 1, column 1.
5.
Press O or l followed by the letter key specifying the
matrix.
6.
Repeat steps 4 and 5 for all elements of the matrix. The row and
column numbers are incremented according to the dimensions of the
matrix you specify.
While the letter key specifying the matrix is held down after O or
l is pressed, the calculator displays the name of the matrix followed by
the row and column numbers of the element whose value is being stored or
recalled. If the letter key is held down for longer than about 3 seconds, the
calculator displays null, doesn't store or recall the element value, and
doesn't increment the row and column numbers. (Also, the stack registers
aren't changed.)
After the last element of the matrix has been accessed, the row and column
numbers both return to 1.
Example: Store the values shown below in the elements of the matrix A
dimensioned above. (Be sure matrix A is dimensioned to 2×3.)
 a11 a12
a21 a22
A
a13  1 2 3


a23  4 5 6
Section 12: Calculating with Matrices
Keystrokes
´>1
Display
´U
1OA
A
2OA
3OA
4OA
5OA
6OA
lA
1.0000
2.0000
3.0000
4.0000
5.0000
6.0000
A
1,1
lA
lA
lA
lA
lA
´U
145
1,1
1.0000
2.0000
3.0000
4.0000
5.0000
6.0000
6.0000
Sets beginning row and column
numbers in R0 and R1 to 1.
(Display shows the previous
result.)
Activates User mode.
Row 1, column 1 of A.
(Displayed momentarily while
A key held down.)
Value of a11.
Value of a12.
Value of a13.
Value of a21.
Value of a22.
Value of a23.
Recalls element in row 1,
column l. (R0 and R1 were reset
in preceding step.)
Value of a11.
Value of a12.
Value of a13.
Value of a21.
Value of a22.
Value of a23.
Deactivates User mode.
Checking and Changing Matrix Elements Individually
The calculator provides two ways to check (recall) and change (store) the
value of a particular matrix element. The first method uses storage registers
R0 and R1 in the same way as described above – except that the row and
column numbers aren't automatically changed when User mode is
deactivated. The second method uses the stack to define the row and
column numbers.
146
Section 12: Calculating with Matrices
Using R0 and R1. To access a particular matrix element, store its row
number in R0 and its column number in R1. These numbers won't change
automatically (unless the calculator is in User mode).

To recall the element value (after storing the row and column
numbers), press l followed by the letter key specifying the
matrix.

To store a value in that element (after storing the row and column
numbers), place the value in the X-register and press O
followed by the letter key specifying the matrix.
Example: Store the value 9 as the element in row 2, column 3 of matrix A
from the previous example.
Keystrokes
2O0
3O1
9
OA
Display
2.0000
3.0000
9
A
2,3
9.0000
Stores row number in R0.
Stores column number in R1.
Keys the new element value into
the X-register.
Row 2, column 3 of A.
Value of a23.
Using the Stack. You can use the stack registers to specify a particular matrix
element. This eliminates the need to change the numbers in R0 and R1.

To recall an element value, enter the row number and column
number into the stack (in that order). Then press l |
followed by the letter key specifying the matrix. The element value
is placed in the X-register. (The row and column numbers are lost
from the stack.)

To store an element value, first enter the value into the stack
followed by the row number and column number. Then press
O | followed by the letter key specifying the matrix. (The
row and column numbers are lost from the stack; the element value
is returned to the X-register.)
Note that these are the only operations in which the blue | key precedes
a gold letter key.
Section 12: Calculating with Matrices
147
Example: Recall the element in row 2, column 1 of matrix A from the
previous example. Use the stack registers.
Keystrokes
2v1
l|A
Display
1
4.0000
Enters row number into Yregister and column number into
X-register.
Value of a21.
Storing a Number in All Elements of a Matrix
To store a number in all elements of a matrix, simply key that number into
the display, then press O> followed by the letter key specifying
the matrix.
Matrix Operations
In many ways, matrix operations are like numeric calculations. Numeric
calculations require you to specify the numbers to be used; often you define
a register for storing the result. Similarly, matrix calculations require you to
specify one or two matrices that you want to use. A matrix descriptor is
used to specify a particular matrix. For many calculations, you also must
specify a matrix for storing the result. This is the result matrix.
Because matrix operations usually require many individual calculations, the
calculator flashes the running display during most matrix operations.
Matrix Descriptors
Earlier in this section you saw that when you press l > followed
by a letter key specifying a matrix, the name of the matrix appears at the left
of the display and the number of rows followed by the number of columns
appears at the right. The matrix name is called the descriptor of the matrix.
Matrix descriptors can be moved among the stack and data storage registers
just like a number – that is, using O, l, v, etc. Whenever a
matrix descriptor is displayed in the X-register, the current dimensions of
that matrix are shown with it.
You use matrix descriptors to indicate which matrices are used in each
matrix operation. The matrix operations discussed in the rest of this section
148
Section 12: Calculating with Matrices
operate on the matrices whose descriptors are placed in the X-register and
(for some operations) the Y-register.
Two matrix operations – calculating a determinant and solving the matrix
equation AX = B – involve calculating an LU decomposition (also known
as an LU factorization) of the matrix specified in the X-register.* A matrix
that is an LU decomposition is signified by two dashes following the matrix
name in the display of its descriptor. (Refer to page 160 for using a matrix
in LU form.)
The Result Matrix
For many operations discussed in this section, you need to define the matrix
in which the result of the operation should be stored. This matrix is called
the result matrix.
Other matrix operations do not use or affect the result matrix. (This is noted
in the descriptions of these operations.) Such an operation either replaces
the original matrix with the result of the operation (if the result is a matrix,
such as a transpose) or returns a number to the X-register (if the result is a
number, such as a row norm).
Before you perform an operation that uses the result matrix, you must
designate the result matrix. Do this by pressing ´ < followed by
the letter key specifying the matrix, (If the descriptor of the intended result
matrix is already in the X-register, you can press O< instead.)
The designated matrix remains the result matrix until another is designated.†
To display the descriptor of the result matrix, press l <.
When you perform an operation that affects the result matrix, the matrix is
automatically redimensioned to the proper size. If this redimensioning
would require more additional elements than there are available in matrix
memory (a maximum of 64 for all five matrices), then the operation can't be
performed. This restriction can often be overcome by designating the result
matrix to be one of the matrices being operated on. (However, there are
certain operations for which the result matrix can not be the same one as
either of the matrices being operated on – this is noted in the description of
these operations.)
*
†
The LU decomposition of a matrix A is another matrix in which is encoded a lower-triangular matrix, L,
and an upper-triangular matrix, U, whose product LU equals matrix A (possibly with same rows
interchanged). The HP-15C Advanced Functions Handbook discusses LU decomposition in detail.
Matrix A is automatically designated as the result matrix whenever Continuous Memory is reset.
Section 12: Calculating with Matrices
149
While the key used for any matrix operation that stores a result in the result
matrix is held down, the descriptor of the result matrix is displayed. If the
key is released within about 3 seconds, the operation is performed, and the
descriptor of the result matrix is placed in the X-register. If the key is held
down longer, the operation is not performed and the calculator displays
null.
Copying a Matrix
To copy the elements of a matrix into the corresponding elements of
another matrix, use the O > sequence:
1.
Press l > followed by the letter key specifying the
matrix to be copied. This enters the descriptor of the matrix into
the display.
2.
Press O> followed by the letter key specifying the
matrix to be copied into.
If the matrix specified after l does not have the same dimensions as the
matrix specified after O, the second matrix is redimensioned to agree
with the first. The matrix specified after O need not already be
dimensioned.
Example: Copy matrix A from the previous example into matrix B.
Keystrokes
l>
A
O>
B
l>
B
Display
A
2
3
A
2
3
b
2
3
Displays descriptor of
matrix to be copied.
Redimensions matrix B and
copies A into B.
Displays descriptor of new
matrix B.
One-Matrix Operations
The following table shows functions that operate on only the matrix
specified in the X-register. Operations involving a single matrix plus a
number in another stack register are described under Scalar Operations
(page 151).
150
Section 12: Calculating with Matrices
One-Matrix Operations:
Sign Change, Inverse, Transpose, Norms, Determinant
Keystroke(s)
Result in
X-register
“
No change.
∕
(´∕ in
User Mode)
´> 4
Descriptor of
result matrix.
´> 7
´> 8
´> 9
Descriptor of
transpose.
Row norm of
specified
matrix.*
Frobenius or
Euclidean norm
of specified
matrix. †
Determinant of
specified
matrix.
Effect on Matrix
Effect on Result
Specified in
Matrix
X-register
Changes sign of
all elements.
None. ‡
None. ‡
Inverse of
specified matrix.
§
Replaced by
transpose.
None.
None. ‡
None.
None.
None.‡
LU decomposition of specified
matrix.§
None.
*
The row norm is the largest sum of the absolute values of the elements in
each row of the specified matrix.
†
The Frobenius of Euclidean norm is the square root of the sum of the
squares of all elements in the specified matrix.
‡
Unless the result matrix is the same matrix specified in the X-register.
§
If the specified matrix is a singular matrix (that is, one that doesn’t have an
inverse), then the HP-15C modifies the LU form by an amount that is
usually small compared to round-off error. For ∕, the calculated inverse
is the inverse of a matrix close to the original, singular matrix. (Refer to the
HP-15C Advanced Functions Handbook for further information.)
Section 12: Calculating with Matrices
151
Example: Calculate the transpose of matrix B. Matrix B was set in
preceding examples to
1 2 3
B
.
 4 5 9
Keystrokes
l>B
´>4
Display
b
2
b
3
3
2
Displays descriptor of
2×3 matrix B.
Descriptor of 3×2
transpose.
Matrix B (which you can view using l B in User mode) is now
1 4 


B  2 5.
3 9
Scalar Operations
Scalar operations perform arithmetic operations between a scalar (that is, a
number) and each element of a matrix. The scalar and the descriptor of the
matrix must be placed in the X- and Y-registers – in either order. (Note that
the register position will affect the outcome of the - and ÷ functions.)
The resulting values are stored in the corresponding elements of the result
matrix.
The possible operations are shown in the following table.
152
Section 12: Calculating with Matrices
Operation
+
*
-
÷
Elements of Result Matrix*
Matrix in Y-Register
Scalar in Y-Register
Scalar in X-Register
Matrix in X-Register
Adds scalar value to each matrix element.
Multiplies each matrix element by scalar value.
Subtracts scalar value
Subtracts each matrix
from each matrix
element from scalar value.
element.
Divides each matrix
Calculates inverse of matrix
element by scalar value. and multiplies each element
by scalar value.
* Result matrix may be the specified matrix.
Example: Calculate the matrix B = 2A. then subtract 1 from every element
in B. From before, use
1 2 3
A
.
 4 5 9
Keystrokes
´<B
l> A
2*
Display
A
b
2
2
3
3
Designates matrix B as result
matrix.
Displays descriptor of matrix A.
Redimensions matrix B to the
same dimensions as A, multiplies
the elements of A by 2, stores
those values in the corresponding
elements of B, and displays the
descriptor of the result matrix.
Section 12: Calculating with Matrices
Keystrokes
1-
Display
b
2
3
153
Subtracts 1 from the elements of
matrix B and stores those values in the
same elements of B.
The result (which you can view using lB in User mode) is
1
B
7
3 5
.
9 17
Arithmetic Operations
With matrix descriptors in both the X- and Y-registers, pressing + or
- calculates the sum or difference of the matrices.
Pressing
Calculates*
+
Y+X
Y-X
* Result is stored in result matrix.
Result matrix may be X or Y
Example: Calculate C = B - A, where A and B are defined in the previous
example.
A
Keystrokes
´< C
l> B
l> A
1 2 3
4 5 9
and B 
1 3 5 
7 9 17.
Display
b
2
A
2
Designates C as result matrix.
3 Recalls descriptor of matrix B.
(This step can be skipped if
descriptor is already in X-register.)
3 Recalls descriptor of matrix A into
X-register, moving descriptor of
matrix B to Y-register.
154
Section 12: Calculating with Matrices
Keystrokes
-
Display
C
2
3
Calculates B - A and stores
values in redimensioned result
matrix C.
0 1 2 

3 4 8 
The result is C  
Matrix Multiplication
With matrix description in both the X- and Y-registers, you can calculate
three different matrix products. The table below shows the results of the
three functions for a matrix X specified in the X-register and a matrix Y
specified in the Y-register. The matrix X-1 is the inverse of X, and the
matrix YT is the transpose of Y.
Pressing
*
´>5
÷
Calculates*
YX
Y TX
X-1Y
* Result is stored in result matrix. For ÷, the
result matrix can be Y but not X. For the others,
the result matrix must be other than X or Y.
Note: When you use the ÷ function to evaluate the expression
A-1B, you must enter the matrix descriptors in the order B, A rather
than in the order that they appear in the expression.*
The value stored in each element of the result matrix is determined
according to the usual rules of matrix multiplication.
For > 5, the matrix specified in the Y-register isn't changed by this
operation, even though its transpose is used. The result is identical to that
obtained using > 4 (transpose) and *.
*
This is the same order you would use if you were entering b and a for evaluating a-1b = b/a
Section 12: Calculating with Matrices
155
For ÷, the matrix specified in the X-register is replaced by its LU
decomposition. The ÷ function calculates X–1Y using a more direct
method than does ∕ and *, giving the result faster and with improved
accuracy.
Example: Using matrices A and B from the previous example, calculate
C = AT B.
A
Keystrokes
l>
A
l>
B
´<
C
´> 5
1 2 3
 4 5 9


Display
A
and B 
 1 3 5
7 9 17


2
3
Recalls descriptor for matrix A.
b
2
3
b
2
3
C
3
3
Recalls descriptor for matrix B
into X-register, moving matrix
A descriptor into Y-register.
Designates matrix C as result
matrix.
Calculates AT B and stores
result in matrix C, which is
redimensioned to 3×3.
The result, matrix C, is
29 39 73


C  37 51 95 .
66 90 168
156
Section 12: Calculating with Matrices
Solving the Equation AX = B
The ÷ function is useful for solving
matrix equations of the form AX = B,
where A is the coefficient matrix, B is
the constant matrix, and X is the
solution matrix. The descriptor of the
constant matrix B should be entered in
the Y-register and the descriptor of the
coefficient matrix A should be entered
in the X-register Pressing ÷ then
calculates the solution X=A-1B.*
Y
constant matrix
X
coefficient
matrix
Remember that the ÷ function replaces the coefficient matrix by its LU
decomposition and that this matrix must not be specified as the result
matrix. Furthermore, using ÷ rather than ∕ and * gives a solution
faster and with improved accuracy.
At the beginning of this section, you found the solution for a system of
linear equations in which the constant matrix and the solution matrix each
had one column. The following example illustrates that you can use the HP15C to find solutions for more than one set of constants—that is, for a
constant matrix and solution matrix with more than one column.
Example: Looking at his receipts for his
last three deliveries of cabbage and
broccoli, Silas Farmer sees the following
summary.
* If A is a singular matrix (that is, one that doesn’t have an inverse), then the HP-15C modifies the LU form
of A by an amount that is usually small compared to round-off error. The calculated solution corresponds
to that for a nonsingular coefficient matrix close to the original, singular matrix.
Section 12: Calculating with Matrices
1
Total Weight (kg)
Total Value
274
$120.32
Week
2
233
$112.96
157
3
331
$151.36
Silas knows that he received $0.24 per kilogram for his cabbage and $0.86
per kilogram for his broccoli. Use matrix operations to determine the
weights of cabbage and broccoli he delivered each week.
Solution: Each week's delivery represents two linear equations (one for
weight and one for value) with two unknown variables (the weights of
cabbage and broccoli). All three weeks can be handled simultaneously using
the matrix equation
1  d11 d12 d13 
1


 
0.24
0.86 d 21 d 22 d 23 

or
274
233
331

= 

120.32 112.96 151.36
AD = B
where the first row of matrix D is the weights of cabbage for the three
weeks and the second row is the weights of broccoli.
Keystrokes
Display
2.0000 Dimensions A as 2×2 matrix.
2
v´mA
2.0000 Sets row and column numbers in R0
´> 1
and R1 to 1.
2.0000 Activates User mode.
´U
1.0000 Stores a11.
1 OA
1.0000 Stores a12.
OA
0.2400 Stores a21.
.24 OA
0.8600 Stores a22.
.86 OA
3.0000 Dimensions B as 2×3 matrix.
2v3
´mB
158
Section 12: Calculating with Matrices
Keystrokes
274 OB
233 OB
331 OB
120.32 OB
112.96 OB
151.36 OB
´< Á
Display
274.0000
233.0000
331.0000
120.3200
112.9600
151.3600
151.3600
l> B
b2
3
l> A
A2
2
÷
d2
3
lÁ
186.0000
lÁ
141.0000
lÁ
lÁ
lÁ
lÁ
´U
215.0000
88.0000
92.0000
116.0000
116.0000
*
Stores b11.*
Stores b12.
Stores b13.
Stores b21.
Stores b22.
Stores b23.
Designates matrix D as result
matrix.
Recalls descriptor of constant
matrix.
Recalls descriptor of coefficient
matrix A into X-register, moving
descriptor of constant matrix B
into Y-register.
Calculates A-1B and stores result
in matrix D.
Recalls d11, the weight of cabbage
for the first week.
Recalls d12 the weight of cabbage
for the second week.
Recalls d13.
Recalls d21.
Recalls d22.
Recalls d23.
Deactivates User mode.
Note that you did not need to press ´> 1 before beginning to store the elements of matrix B. This
is because after you stored the last element of matrix A, the row and column numbers in R0 and R1 were
automatically reset to 1.
Section 12: Calculating with Matrices
159
Silas' deliveries were:
Cabbage (kg)
Broccoli (kg)
1
186
88
Week
2
141
92
3
215
116
Calculating the Residual
The HP-15C enables you to calculate the residual, that is, the matrix
Residual = R–YX
where R is the result matrix and X and Y are the matrices specified in the
X- and Y-registers.
This capability is useful, for example, in doing iterative refinement on the
solution of a system of equations and for linear regression problems. For
example, if C is a possible solution for AX = B, then B – AC indicates how
well this solution satisfies the equation. (Refer to the HP-15C Advanced
Functions Handbook for information about iterative refinement and linear
regression.)
The residual function (> 6) uses the current contents of the result
matrix and the matrices specified in the X- and Y-registers to calculate the
residual defined above. The residual is stored in the result matrix, replacing
the original result matrix. A matrix specified in the X- or Y-register can not
be the result matrix.
Using > 6 rather than * and - gives a result with improved
accuracy, particularly if the residual is small compared to the matrices being
subtracted.
To calculate the residual:
1.
Enter the descriptor of the Y matrix into the Y-register.
2.
Enter the descriptor of the X matrix into the X-register.
3.
Designate the R matrix as the result matrix.
4.
Press ´> 6. The residual replaces the original result
matrix (R). The descriptor of the result matrix is placed in the Xregister.
160
Section 12: Calculating with Matrices
Using Matrices in LU Form
As noted earlier, two matrix operations (calculating a determinant and
solving the matrix equation (AX = B) create an LU decomposition of the
matrix specified in the X-register. The descriptor of such a matrix has two
dashes following the matrix name. A matrix in LU form has elements that
differ from the elements of the original matrix.
However, the descriptor for a matrix in LU form can be used in place of the
descriptor for the original matrix for operations involving the inverse of the
matrix and for the determinant operation. That is, either the original matrix
or its LU decomposition can be used for these operations:
∕
÷ for the matrix in the X-register
>9
For these three functions, using the LU form of the matrix to be inverted
gives a result that is identical to that using the original matrix.
As an example, if you solved the matrix equation AX = B, matrix A would
be changed to its LU form. If you wanted to change the B matrix and solve
the equation again, you could do so without changing the A matrix – the LU
matrix will give the correct solution.
For all other matrix operations, a matrix that is an LU decomposition is not
recognized as representing its original matrix. Instead, the elements of the
LU matrix are used just as they appear in matrix memory and the result is
not the result you would obtain using the original matrix.
Calculations With Complex Matrices
The HP-15C enables you to perform matrix multiplication and matrix
inversion with complex matrices (that is, matrices whose elements are
complex numbers) and to solve systems of complex equations (that is,
equations whose coefficients and variables are complex).
However, the HP-15C stores and operates on only real matrices. The
capability of doing calculations with complex matrices is completely
independent of the capability of doing calculations with complex numbers
described in the preceding section. You don’t need to activate Complex
mode for calculations with complex matrices.
Section 12: Calculating with Matrices
161
Instead, calculations with complex matrices are performed by using real
matrices derived from the original complex matrices – in a manner to be
described below – and performing certain transformations in addition to the
regular matrix operations. These transformations are performed by four
calculator functions. This section will describe how to do these calculations.
(There are more examples of calculations with complex matrices in the
HP-15C Advanced Functions Handbook.)
Storing the Elements of a Complex Matrix
Consider an m×n complex matrix Z = X + iY, where X and Y are real
m×n matrices. This matrix can be represented in the calculator as a
2m×n ―partitioned‖ matrix:
X } Real Part
ZP   
Y } Imaginary Part
The superscript P signifies that the complex matrix is represented by a
partitioned matrix.
All of the elements of ZP are real numbers – those in the upper half
represent the elements of the real part (matrix X), those in the lower half
represent the elements of the imaginary part (matrix Y). The elements of ZP
are stored in one of the five matrices (A, for example) in the usual manner,
as described earlier in this section.
For example, if Z = X + iY, where
x 
x
y
X   11 12  and Y   11
x
x
21
22


 y21
y12 
,
y22 
then Z can be represented in the calculator by
 x11

x
 X
A  Z P      21

y11
Y
 

y
 21
x12 

x22 
.
y12 

y 22 
162
Section 12: Calculating with Matrices
Suppose you need to do a calculation with a complex matrix that is not
written as the sum of a real matrix and an imaginary matrix – as was the
matrix Z in the example above – but rather written with an entire complex
number in each element, such as
 x  iy11 x12  iy12 
Z   11
.
 x21  iy21 x22  iy22 
This matrix can be represented in the calculator by a real matrix that looks
very similar – one that is derived simply by ignoring the i and the + sign.
The 2 × 2 matrix Z shown above, for example, can be represented in the
calculator in ―complex‖ form by the 2 × 4 matrix.
AZ
C

 x11
x
 21
y11
y 21
x12
x 22
y12 
.
y 22 
The superscript C signifies that the complex matrix is represented in a
"complex-like" form.
Although a complex matrix can be initially represented in the calculator by
a matrix of the form shown for ZC, the transformations used for multiplying
and inverting a complex matrix presume that the matrix is represented by a
matrix of the form shown for ZP. The HP-15C provides two transformations
that convert the representation of a complex matrix between ZC and ZP:
Pressing
´p
|c
Transforms
ZC
ZP
Into
ZP
ZC
To do either of these transformations, recall the descriptor of ZC or ZP into
the display, then press the keys shown above. The transformation is done to
the specified matrix; the result matrix is not affected.
Section 12: Calculating with Matrices
163
Example: Store the complex matrix
4  3i 7  2i 
Z

1  5i 3  8i 
in the form ZC, since it is written in a form that shows ZC. Then transform
ZC into the form ZP.
You can do this by storing the elements of ZC in matrix A and then using
the p function, where
c
AZ 
Keystrokes
´> 0
2v4
´mA
´> 1
´U
4 OA
3 OA
7 OA
2 “ OA
1 OA
5 OA
3 OA
8 OA
´U
l> A
´p
 4 3 7  2
.
1 5 3
8

Display
4.0000
4.0000
4.0000
4.0000
3.0000
7.0000
-2.0000
1.0000
5.0000
3.0000
8.0000
8 0000
A
2
4
A
4
2
Clears all matrices.
Dimensions matrix A to be
2×4.
Sets beginning row and
column numbers in R0 and
R1 to 1.
Activates User mode.
Stores a11.
Stores a12.
Stores a13.
Stores a14.
Stores a21.
Stores a22.
Stores a23.
Stores a24.
Deactivates User mode.
Display descriptor of
matrix A.
Transforms ZC into ZP and
redimensions matrix A.
164
Section 12: Calculating with Matrices
Matrix A now represents the complex matrix Z in ZP form:
7
4


1
3
A  ZP  
.
 3  2


8
 5
}
}
Real P art
Imaginary P art
The Complex Transformations Between ZP and Z
An additional transformation must be done when you want to calculate the
product of two complex matrices, and still another when you want to
calculate the inverse of a complex matrix. These transformations convert
between the ZP representation of an m×n complex matrix and a 2m×2n
partitioned matrix of the following form:
 X Y 
Z
.
X
Y
The matrix created by the > 2 transformation has twice as many
elements as ZP.
For example, the matrices below show how
is related to ZP.
 1 6
~  1 6 4 5
ZP  
Z

 4 5
  4 5 1  6
The transformations that convert the representation of a complex matrix
between ZP and are shown in the following table.
Pressing
´>2
´>3
Transforms
ZP
Into
ZP
To do either of these transformations, recall the descriptor of ZP or into
the display, then press the keys shown above. The transformation is done to
the specified matrix; the result matrix is not affected.
Section 12: Calculating with Matrices
165
Inverting a Complex Matrix
You can calculate the inverse of a complex matrix by using the fact that
( )-1 = ( -1).
To calculate inverse, Z-1, of a complex matrix Z:
1. Store the elements of Z in memory, in the form either of ZP or of ZC
2. Recall the descriptor of the matrix representing Z into the display.
3. If the elements of Z were entered in the form ZC, press ´p to
transform ZC into ZP
4. Press ´ > 2 to transform ZP into .
5. Designate a matrix as the result matrix. It may be the same as the
matrix in which is stored.
6. Press ∕. This calculates ( )-1, which is equal to ( -1). The values
of these matrix elements are stored in the result matrix, and the
descriptor of the result matrix is placed in the X-register.
7. Press ´ > 3 to transform (
-1
) into (Z-1)P.
8. If you want the inverse in the form (Z-1)C, press | c
You can derive the complex elements of Z-1 by recalling the elements of ZP
or ZC and then combining them as described earlier.
Example: Calculate the inverse of the complex matrix Z from the previous
example.
7
4


1
3
.
A  ZP  
 3  2


8
 5
Keystrokes
l>A
´>2
Display
A 4
A 4
2
4
Recalls descriptor of matrix A.
Transforms ZP into and
redimensions matrix A.
166
Section 12: Calculating with Matrices
Keystrokes
´<
B
∕
´> 3
Display
A
4
4
b
4
4
b
4
2
Designates B as the result
matrix.
Calculates ( )-1 = ( -1) and
places the result in matrix B.
Transforms ( -1) into
( -1)P.
The representation of Z-1 in partitioned form is contained in matrix B.
 0.0254 0.2420


 0.0122  0.1017
B
 0.2829  0.0022


 0.1691  0.1315
}
}
Real P art
Imaginary P art
Multiplying Complex Matrices
The product of two complex matrices can be calculated by using the fact
P
that (YX)P =
.
To calculate YX, where Y and X are complex matrices:
1. Store the elements of Y and X in memory, in the form either of
ZP or ZC.
2. Recall the descriptor of the matrix representing Y into the
display.
3. If the elements of Y were entered in the form of YC, press
´p to transform YC into YP.
4. Press ´> 2 to transform YP into .
5. Recall the descriptor of the matrix representing X into the
display.
6. If the elements of X were entered in the form XC, press
´p to transform XC into XP.
7. Designate the result matrix; it must not be the same matrix as
either of the other two.
Section 12: Calculating with Matrices
167
8. Press * to calculate XP = (YX)P. The values of these matrix
elements are placed in the result matrix, and the descriptor of
the result matrix is placed in the X-register.
9. If you want the product in the form (YX)C, press |c
Note that you don't transform XP into .
You can derive the complex elements of the matrix product YX by recalling
the elements of (XY)P or (YX)C and combining them according to the
conventions described earlier.
Example: Calculate the product ZZ-1, where Z is the complex matrix given
in the preceding example.
Since elements representing both matrices are already stored (
(Z-1)P in B), skip steps 1, 3, 4, and 6.
Keystrokes
l>A
l>B
Display
A
4
b
4
´<C
*
´U
lC
b
C
C
C
lC
lC
lC
lC
lC
lC
lC
´U
1.0000
–2.8500
–4.0000
1.0000
1.0000
3.8000
1.0000
–1.0500
–1.0500
4
4
4
1,1
in A and
4 Displays descriptor of matrix A.
2 Displays descriptor of matrix
B.
2 Designates C as result matrix.
2 Calculates (Z-1)P = (ZZ-1)P.
2 Activates User mode.
Matrix C, row 1, column 1.
(Displayed momentarily while
last key held down.)
Value of c11.
–10 Value of c12.
–11 Value of c21.
Value of c22.
–11 Value of c31.
–10 Value of c32.
–11 Value of c41.
–10 Value of c42.
–10 Deactivates User mode.
168
Section 12: Calculating with Matrices
Writing down the elements of C,
 1.0000

 4.0000 1011
C
 1.0000 1011

11
 1.0000 10
 2.8500 1010 

1.0000
  ZZ 1
3.8000 1010 

 1.0500 1010 


P
,
where the upper half of matrix C is the real part of ZZ-1 and the lower half
is the imaginary part. Therefore, by inspection of matrix C,
 1.0000
 2.85001010 
ZZ 1  

11
1.0000
 4.000010

1.00001011
3.80001011 
i 

11
 1.05001010 
1.000010
As expected,
1 0 0 0
ZZ -1  
i 

0 1  0 0 
Solving the Complex Equation AX = B
You can solve the complex matrix equation AX = B by finding X = A-1B.
Do this by calculating XP = (Ã)-1 BP.
To solve the equation AX = B, where A, X, and B are complex matrices:
1. Store the elements of A and B in memory, in the form either of ZP or
of ZC.
2. Recall the descriptor of the matrix representing B into the display.
3. If the elements of B were entered in the form BC, press ´p to
transform BC into BP.
Section 12: Calculating with Matrices
169
4. Recall the descriptor of the matrix representing A into the display.
5. If the elements of A were entered in the form of AC, press ´
p to transform AC into AP.
6. Press ´> 2 to transform AP into Ã.
7. Designate the result matrix; it must not be the same as the matrix
representing A.
8. Press ÷; this calculates XP. The values of these matrix elements
are placed in the result matrix, and the descriptor of the result matrix
is placed in the X-register.
9. If you want the solution in the form XC, press |c.
Note that you don't transform BP into .
You can derive the complex elements of the solution X by recalling the
elements of XP or XC and combining them according to the conventions
described earlier.
Example: Engineering student A. C. Dimmer wants to analyze the
electrical circuit shown below. The impedances of the components are
indicated in complex form. Determine the complex representation of the
currents I1 and I2.
This system can be represented by the complex matrix equation
200i   I1  5
10  200i

    
  200i (200  30)i   I 2  0
or
AX = B.
170
Section 12: Calculating with Matrices
In partitioned form,
0
 10
5 


 
0
0
 and B  0 ,
A
 200  200
0 


 

200
170


0
where the zero elements correspond to real and imaginary parts with zero
value.
Keystrokes
Display
4 v2´mA 2.0000
´> 1
´U
10 OA
0OA
OA
OA
200 OA
“OA
OA
170 OA
4 v 1´m
B
0 O>B
2.0000
2.0000
10.0000
0.0000
0.0000
0.0000
200.0000
–200.0000
–200.0000
170.0000
1.0000
0.0000
5v1v
1.0000
O|B
l> B
5.0000
b
4 1
l> A
A
4
2
Dimensions matrix A to be
4×2.
Set beginning row and column
numbers in R0 and R1 to 1.
Activates User mode.
Stores a11.
Stores a12.
Stores a21.
Stores a22.
Stores a31.
Stores a32.
Stores a41.
Stores a42.
Dimensions matrix B to be
4×1.
Stores value 0 in all elements
of B.
Specifies value 5 for row 1,
column 1.
Stores value 5 in b11.
Recalls descriptor for matrix
B.
Places descriptor for matrix A
into X-register, moving
descriptor for matrix B into Yregister.
Section 12: Calculating with Matrices
Keystrokes
´> 2
´< C
÷
|c
lC
lC
lC
lC
´U
´> 0
Display
A
A
C
C
0.0372
0.1311
0.0437
0.1543
0.1543
0.1543
4
4
4
4
4
2
1
2
171
Transforms AP into Ã.
Designates matrix C as
result matrix.
Calculates XP and stores
in C.
Transforms XP into XC.
Recalls c11.
Recalls c12.
Recalls c21.
Recalls c22.
Deactivates User mode.
Redimensions all matrices
to 0×0.
The currents, represented by the complex matrix X, can be derived from C
 I  0.0372 0.1311i 
X   1  

 I 2  0.0437 0.1543i 
Solving the matrix equation in the preceding example required 24 registers
of matrix memory – 16 for the 4×4 matrix A (which was originally entered
as a 4×2 matrix representing a 2×2 complex matrix), and four each for the
matrices B and C (each representing a 2×1 complex matrix). (However, you
would have used four fewer registers if the result matrix were matrix B.)
Note that since X and B are not restricted to be vectors (that is, singlecolumn matrices), X and B could have required more memory.
The HP-15C contains sufficient memory to solve, using the method
described above, the complex matrix equation AX = B with X and B having
up to six columns if A is 2×2, or up to two columns if A is 3×3.* (The
allowable number of columns doubles if the constant matrix B is used as the
result matrix.) If X and B have more columns, or if A is 4×4, you can solve
the equation using the alternate method below. This method differs from the
preceding one in that it involves separate inversion and multiplication
operations and fewer registers.
*
If all available memory space is dimensioned to the common pool (W: 1 64 0-0). Refer to appendix C,
Memory Allocation.
172
Section 12: Calculating with Matrices
1. Store the elements of A in memory, in the form either of AP or of
AC.
2. Recall the descriptor of the matrix representing A into the display.
3. If the elements of A were entered in the form AC, press ´ p
to transform AC into AP.
4. Press ´> 2 to transform AP into Ã.
5. Press O< to designate the matrix representing A as the
result matrix.
6. Press ∕ to calculate (Ã)-1.
7. Redimension A to have half the number of rows as indicated in the
display of its descriptor after the preceding step.
8. Store the elements of B in memory, in the form either of BP or
of BC.
9. Recall the descriptor of the matrix representing A into the display.
10. Recall the descriptor of the matrix representing B into the display.
11. If the elements of B were entered in the form BC, press ´p to
transform BC into BP.
12. Press ´> 2 to transform BP into
13. Designate the result matrix; it must not be the same matrix as either
of the other two.
14. Press *.
15. Press ´> 4 to transpose the result matrix.
16. Press ´> 2.
17. Redimension the result matrix to have half the number of rows as
indicated in the display of its descriptor after the preceding step.
18. Press l< to recall the descriptor of the result matrix.
19. Press ´> 4 to calculate XP.
20. If you want the solution in the form XC, press |c
Section 12: Calculating with Matrices
173
A problem using this procedure is given in the HP-15C Advanced Functions
Handbook under Solving a Large System of Complex Equations.
Miscellaneous Operations Involving Matrices
Using a Matrix Element With Register Operations
If a letter key specifying a matrix is pressed after any of the following
function keys, the operation is performed using the matrix element specified
by the row and column numbers in R0 and R1, just as though it were a data
storage register.
*
*
O
l
O{+, -, *, ÷} l{+, -, *,
÷}
e
I
X
Using Matrix Descriptors in the Index Register
In certain applications, you may want to perform a programmed sequence
of matrix operations using any of the matrices A through E. In this
situation, the matrix operations can refer to whatever matrix descriptor is
stored in the index register (RI).
If the Index register contains a matrix descriptor:


*
Pressing % after any of the functions listed above performs the
operations using the element specified by R0 and R1 and the matrix
specified in RI.
Pressing % after O| or l| performs the operation
using the element specified by the row and column numbers in the
Y- and X-registers and the matrix specified in RI.
Also, in User mode the row and column numbers in R 0 and R1 are incremented according to the
dimensions of the specified matrix.
174
Section 12: Calculating with Matrices

Pressing ´mV dimensions the matrix specified in RI
according to the dimensions in the X- and Y-registers.

Pressing lmV recalls to the X- and Y-registers the
dimensions of the matrix specified in RI.

Pressing GV or tV has the same result as pressing
G or t followed by the letter of the matrix specified in R I.
(This is not actually a matrix operation – only the letter in the
matrix descriptor is used.)
Conditional Tests on Matrix Descriptors
Four conditional tests – ~, T 0 (x≠ 0), T 5 (x = y), and T
6 (x≠y) – can be performed with matrix descriptors in the X- and Yregisters, Conditional tests can be used to control program execution, as
described in section 8.
If a matrix descriptor is in the X-register, the result of ~ will be false
and the result of T 0 will be true (regardless of the element values in
the matrix.)
If matrix descriptors are in the X- and Y-registers when T 5 or T 6
conditional test is performed, x and y are equal if the same descriptor is in
the X- and Y-registers, and not equal otherwise. The comparison is made
between the descriptors themselves, not between the elements of the
specified matrices.
Other conditional tests can't be used with matrix descriptors.
Stack Operation for Matrix Calculations
During matrix calculations, the contents of the stack registers shift much
like they do during numeric calculations.
For some matrix calculations, the result is stored in the result matrix. The
arguments – one or two descriptors or numbers in the X-register or the Xand Y-registers – are combined by the operation, and the descriptor of the
result matrix is placed in the X-register. (The argument from the X-register
is placed in the LAST X register.)
Section 12: Calculating with Matrices
175
Several matrix functions operate on the matrix specified in the X-register
only and store the result in the same matrix. For these operations the
contents of the stack (including the LAST X register) are not moved –
although the display changes to show the new dimensions if necessary.
For the > 7, > 8, and > 9 functions, the matrix
descriptor specified in the X-register is placed in the LAST X register and
the norm or (for > 9) the determinant is placed in the X-register. The
Y-, Z-, and T-registers aren't changed.
When you recall descriptors or matrix elements into the X-register (with the
stack enabled), other descriptors and numbers already in the stack move up
in the stack – and the contents of the T-register are lost. (The LAST X
register is not changed.) When you store descriptors or matrix elements, the
stack (and the LAST X register) isn't changed.
In contrast to the operation described above, the O| and l|
functions do not affect the LAST X register and operate as shown on the
next page.
176
Section 12: Calculating with Matrices
Using Matrix Operations in a Program
If the calculator is in User mode during program entry when you enter a
O or l{A through E, %} instruction to store or recall a
matrix element, a u replaces the dash usually displayed after the line
number. When this line is executed in a running program, it operates as
though the calculator were in User mode. That is, the row and column
numbers in R0 and R1 are automatically incremented according to the
dimensions of the specified matrix. This allows you to access elements
sequentially. (The USER annunciator has no effect during program
execution.)
In addition, when the last element is accessed by the ―User‖ O or l
instruction – when R0 and R1 are returned to 1 – program execution skips
the next line. This is useful for programming a loop that stores or recalls
each matrix element, then continues executing the program. For example,
the following sequence squares all elements of matrix D:
Section 12: Calculating with Matrices
177
The > 7 (row norm) and > 8 (Frobenius norm) functions also
operate as conditional branching instructions in a program. If the X-register
contains a matrix descriptor, these functions calculate the norm in the usual
manner, and program execution continues with the next program line. If the
X-register contains a number, program execution skips the next line. In both
cases, the original contents of the X-register are stored in the LAST X
register. This is useful for testing whether a matrix descriptor is in the Xregister during a program.
Summary of Matrix Functions
Keystroke(s)
Results
|c
Transforms ZP into ZC.
“
Changes sign of all elements in matrix specified in
X-register.
´m {A
through E, V}
Dimensions specified matrix.
´> 0
Dimensions all matrices to 0×0.
´> 1
Sets row and column numbers in R0 and R1 to 1.
´> 2
Transform ZP into .
´> 3
Transforms
´> 4
Calculate transpose of matrix specified in X-register.
´> 5
Multiplies transpose of matrix specified in Yregister with matrix specified in X-register. Stores in
into ZP.
178
Section 12: Calculating with Matrices
Keystroke(s)
Results
result matrix.
´> 6
Calculates residual in result matrix.
´> 7
Calculates row norm of matrix specified in Xregister.
´> 8
Calculates Frobenius or Euclidean norm of matrix
specified in X-register.
´> 9
Calculates determinant of matrix specified in Xregister, Place LU in result matrix.
´p
Transforms ZC into ZP.
l{A through
E, %}
Recalls value from specified matrix, using row and
column numbers in R0 and R1.
l|{A
through E, %}
Recalls value from specified matrix using row and
column numbers in Y- and X-registers.
lm {A
through E, %}
Recalls dimensions of specified matrix into X- and
Y-registers.
l> {A Displays descriptor of specified matrix.
through E}
l<
Displays descriptor of result matrix.
´<{A
through E}
Designates specified matrix as result matrix.
O{A through
E %}
Stores value from display into element of specified
matrix, using row and column numbers in R0 and R1.
O|{A
through E %}
Stores value from Z-register into element of
specified matrix, using row and column numbers in
Y- and X-registers.
O> {A If matrix descriptor is in display, copies all elements
through E}
of that matrix into corresponding elements of
specified matrix. If number is in display, stores that
value in all elements of specified matrix.
Section 12: Calculating with Matrices
179
Keystroke(s)
Results
O<
Designates matrix specified in X-register as result
matrix.
´U
Row and column numbers in R0 and R1 are
automatically incremented each time O or l
{A through E, %} is pressed.
∕
Inverts matrix specified in X-register. Stores in result
matrix. Use ´ ∕ if User mode is on.
+, -
If matrix descriptors specified in both X- and Yregisters, adds or subtracts corresponding elements of
matrices specified. If matrix descriptor specified in only
one of these registers, performs addition or subtraction
with all elements in specified matrix and scalar in other
register. Stores in result matrix.
*
If matrix descriptors specified in both X- and Yregisters, calculates product of specified matrices (as
YX). If matrix specified in only one of these registers,
multiplies all elements in specified matrix by scalar in
other register. Stores in result matrix.
÷
If matrix descriptors specified in both X- and Yregisters, multiplies inverse of matrix specified in Xregister with matrix specified in Y-register. If matrix
specified in only Y-register, divides all elements of
specified matrix by scalar in other register. If matrix
specified in only X-register, multiplies each element of
inverse of specified matrix by scalar in other register.
Stores in result matrix.
For Further Information
The HP-15C Advanced Functions Handbook presents more detailed and
technical aspects of the matrix functions in the HP-15C, including
applications. The topics include: least-squares calculations, solving
nonlinear equations, ill-conditioned and singular matrices, accuracy
considerations, iterative refinement, and creating the identity matrix.
Section 13
Finding the Roots
of an Equation
In many applications you need to solve equations of the form
f(x)=0.*
This means finding the values of x that
satisfy the equation. Each such value
of x is called a root of the equation f(x)
= 0 and a zero of the function f(x).
These roots (or zeros) that are real
numbers are called real roots (or real
zeros). For many problems the roots of
an equation can be determined
analytically
through
algebraic
manipulation; in many other instances,
this is not possible. Numerical
techniques can be used to estimate the
roots when analytical methods are not suitable. When you use the _
key on your HP-15C, you utilize an advanced numerical technique that lets
you effectively and conveniently find real roots for a wide range of
equations.†
Using _
In calculating roots, the _ operation repeatedly calls up and executes
a subroutine that you write for evaluating f(x).
*
Actually, any equation with one variable can be expressed in this form. For example, f(x) = a is equivalent
to f(x) – a = 0, and f(x) = g(x) is equivalent to f(x) – g(x) = 0.
†
The _ function does not use the imaginary stack. Refer to the HP-15C Advanced Functions
Handbook for information about complex roots.
180
Section 13: Finding the Roots of an Equation
181
The basic rules for using _ are:
1.
In Program mode, key in a subroutine that evaluates the function
f(x) that is to be equated to zero. This subroutine must begin with a
label instruction (´b label) and end up with a result for f(x) in
the X-register.
In Run mode:
2.
Key two initial estimates of the desired root, separated by v,
into the X- and Y-registers. These estimates merely indicate to the
calculator the approximate range of x in which it should initially
seek a root of f(x) = 0.
3.
Press ´ _ followed by the label of your subroutine. The
calculator then searches for the desired zero of your function and
displays the result. If the function that you are analyzing equals zero
at more than one value of x, the routine will stop when it finds any
one of those values. To find additional values, you can key in
different initial estimates and use _ again.
Immediately before _ addresses your subroutine it places a value of x
in the X-, Y-, Z-, and T-registers. This value is then used by your subroutine
to calculate f(x). Because the entire stack is filled with the x-value, this
number is continually available to your subroutine. (The use of this
technique is described on page 41).
Example: Use _ to find the values of x for which
f(x) = x2 –3x – 10 = 0.
Using Horner's method (refer to page 79), you can rewrite f(x) so that it is
programmed more efficiently:
f(x) = (x – 3)x – 10.
In Program mode, key in the following subroutine to evaluate f(x).
Keystrokes
|¥
´ CLEAR M
Display
000000-
Program mode.
Clear program memory.
182
Section 13: Finding the Roots of an Equation
Keystrokes
´b0
3
*
1
0
|n
Display
001–42,21, 0
002–
003–
004–
005–
006–
007–
008–
3
30
20
1
0
30
43 32
Begin with b instruction.
Subroutine assumes stack
loaded with x.
Calculate x – 3.
Calculate (x – 3)x.
Calculate (x – 3)x – 10.
In Run mode, key two initial estimates into the X- and Y-registers.
Try estimates of 0 and 10 to look for a positive root.
Keystrokes
|¥
0v
10
Display*
Run mode.
0.0000
10
Initial estimates.
You can now find the desired root by pressing ´_ 0. When you do
this, the calculator will not display the answer right away. The HP-15C uses
an iterative algorithm† to estimate the root. The algorithm analyzes your
function by sampling it many times, perhaps a dozen times or more. It does
this by repeatedly executing your subroutine. Finding a root will usually
require about 2 to 10 seconds; but sometimes the process will require even
more time.
Press ´_ 0 and sit back while your HP-15C exhibits one of its
powerful capabilities. The display flashes running while _ is
operating.
*
†
Press ´• 4 to obtain the displays shown here. The display setting does not influence the operation
of _.
An algorithm is a step-by-step procedure for solving a mathematical problem. An iterative algorithm is one
containing a portion that is executed a number of times in the process of solving the problem.
Section 13: Finding the Roots of an Equation
Keystrokes
´_ 0
Display
5.0000
183
The desired root.
After the routine finds and displays the root, you can ensure that the
displayed number is indeed a root of f(x) = 0 by checking the stack. You have
seen that the display (X-register) contains the desired root. The Y-register
contains a previous estimate of the root, which should be very close to the
displayed root. The Z-register contains the value of your function
evaluated at the displayed root.
Keystrokes
)
)
Display
5.0000
0.0000
A previous estimate of the
root.
Value of the function at the
root showing that f(x) = 0.
Quadratic equations, such as the one you are solving, can have two roots. If
you specify two new initial estimates, you can check for a second root. Try
estimates of 0 and -10 to look for a negative root.
Keystrokes
0v
10 “
´_0
)
)
Display
0.0000
–10
–2.0000
–2.0000
0.0000
Initial estimates.
The second root.
A previous estimate of the
root.
Value of f(x) at second root.
184
Section 13: Finding the Roots of an Equation
You have now found the two roots of f(x)
= 0. Note that this quadratic equation
could have been solved algebraically – and
you would have obtained the same roots
that you found using _.
G
G
r
The convenience and power of the _ key become more apparent
when you solve an equation for a root that cannot be determined
algebraically.
Example: Champion ridget hurler Chuck
Fahr throws a ridget with an upward
velocity of 50 meters/second. If the height
of the ridget is expressed as
h = 5000(1 – e–t/20) – 200t,
how long does it take for it to reach the
ground again? In this equation, h is the
height in meters and t is the time in seconds.
Solution: The desired solution is the positive value of t at which h = 0.
Use the following subroutine to calculate the height.
Keystrokes
|¥
´ bA
2
0
÷
Display
000–
001–42,21,11
002–
2
003–
004–
0
10
Begin with label.
Subroutine assumes t is
loaded in X-and Y-registers.
Section 13: Finding the Roots of an Equation
Keystrokes
“
'
“
1
+
5
0
0
0
*
®
Display
005–
006–
007–
008–
009–
010–
011–
012–
013–
014–
015–
2
0
0
*
|n
016–
017–
018–
019–
020–
021–
16
12
16
1
40
5
0
0
0
20
34
2
0
0
20
30
43 32
185
– t / 20.
– e– t / 20.
1 – e– t / 20.
5000 (1 – e– t / 20).
Brings another t-value
into X-register.
200t.
5000(1 – e– t / 20) – 200t.
Switch to Run mode, key in two initial estimates of the time (for example, 5
and 6 seconds) and execute _.
Keystrokes
|¥
5v
6
´_A
Display
Run mode.
5.0000
6
9.2843
Initial estimates.
The desired root.
Verify the root by reviewing the Y- and Z-registers.
Keystrokes
)
)
Display
9.2843
0.0000
A previous estimate of the root.
Value of the function at the root
showing that h = 0.
186
Section 13: Finding the Roots of an Equation
Fahr's ridget falls to the ground
9.2843 seconds after he hurls it—a
remarkable toss.
When No Root Is Found
You have seen how the _ key estimates and displays a root of an
equation of the form f(x) = 0. However, it is possible that an equation has no
real roots (that is, there is no real value of x for which the equality is true).
Of course, you would not expect the calculator to find a root in this case.
Instead, it displays Error 8.
Example: Consider the equation
|x| = – 1.
which has no solution since the absolute
value function is never negative. Express
this equation in the required form
|x| + 1 = 0
and attempt to use _ to find a
solution.
G
r
G
Keystrokes
|¥
´b 1
|a
1
+
|n
Display
000–
001–42,21, 1
002–
43 16
003–
1
004–
40
005–
43 32
Program mode.
Section 13: Finding the Roots of an Equation
187
Because the absolute-value function is minimum near an argument of zero,
specify the initial estimates in that region, for instance 1 and -1. Then
attempt to find a root.
Keystrokes
Display
Run mode.
|¥
1.0000
1v
Initial estimates.
–1
1“
Error 8
This display indicates that no
´_1
root was found.
0.0000
Clear error display.
−
As you can see, the HP-15C stopped seeking a root of f(x) = 0 when it
decided that none existed – at least not in the general range of x to which it
was initially directed. The Error 8 display does not indicate that an ―illegal‖
operation has been attempted; it merely states that no root was found where
_ presumed one might exist (based on your initial estimates).
If the HP-15C stops seeking a root and displays an error message, one of
these three types of conditions has occurred:



If repeated iterations all produce a constant nonzero value for the
specified function, execution stops with the display Error 8.
If numerous samples indicate that the magnitude of the function
appears to have a nonzero minimum value in the area being
searched, execution stops with the display Error 8.
If an improper argument is used in a mathematical operation as part
of your subroutine, execution stops with the display Error 0.
In the case of a constant function value, the routine can see no indication of
a tendency for the value to move toward zero. This can occur for a function
whose first 10 significant digits are constant (such as when its graph levels
off at a nonzero horizontal asymptote) or for a function with a relatively
broad, local ―flat‖ region in comparison to the range of x-values being tried.
In the case where the function's magnitude reaches a nonzero minimum, the
routine has logically pursued a sequence of samples for which the
magnitude has been getting smaller. However, it has not found a value of
x at which the function's graph touches or crosses the x-axis.
188
Section 13: Finding the Roots of an Equation
The final case points out a potential deficiency in the subroutine rather than
a limitation of the root-finding routine. Improper operations may sometimes
be avoided by specifying initial estimates that focus the search in a region
where such an outcome will not occur. However, the _ routine is very
aggressive and may sample the function over a wide range. It is a good
practice to have your subroutine test or adjust potentially improper
arguments prior to performing an operation (for instance, use a prior to
¤). Rescaling variables to avoid large numbers can also be helpful.
The success of the _ routine in locating a root depends primarily
upon the nature of the function it is analyzing and the initial estimates at
which it begins searching. The mere existence of a root does not ensure that
the casual use of the _ key will find it. If the function f(x) has a
nonzero horizontal asymptote or a local minimum of its magnitude, the
routine can be expected to find a root of f(x) = 0 only if the initial estimates
do not concentrate the search in one of these unproductive regions—and, of
course, if a root actually exists.
Choosing Initial Estimates
When you use _ to find the root of an equation, the two initial
estimates that you provide determine the values of the variable x at which
the routine begins its search. In general, the likelihood that you will find the
particular root you are seeking increases with the level of understanding that
you have about the function you are analyzing. Realistic, intelligent
estimates greatly facilitate the determination of a root.
The initial estimates that you use may be chosen in a number of ways:
If the variable x has a limited range in which it is conceptually meaningful
as a solution, it is reasonable to choose initial estimates within this range.
Frequently an equation that is applicable to a real problem has, in addition
to the desired solution, other roots that are physically meaningless. These
usually occur because the equation being analyzed is appropriate only
between certain limits of the variable. You should recognize this restriction
and interpret the results accordingly.
Section 13: Finding the Roots of an Equation
189
If you have some knowledge of the behavior of the function f(x) as it varies
with different values of x, you are in a position to specify initial estimates in
the general vicinity of a zero of the function. You can also avoid the more
troublesome ranges of x such as those producing a relatively constant
function value or a minimum of the function's magnitude.
Example: Using a rectangular piece
of sheet metal 4 decimeters by 8
decimeters, an open-top box having a
volume of 7.5 cubic decimeters is to
be formed. How should the metal be
folded? (A taller box is preferred to a
shorter one.)
Solution: You need to find the height
of the box (that is, the amount to be
folded up along each of the four sides)
that gives the specified volume. If x is
the height (or amount folded up), the
length of the box is (8 – 2x) and the width is (4 – 2x). The volume V is
given by
V = (8 – 2x)(4 – 2x) x.
By expanding the expression and then using Horner's method (page 79), this
equation can be rewritten as
V = 4 ((x – 6) x + 8) x.
To get V= 7.5, find the values of x for which
f(x) = 4 ((x – 6) x + 8) x – 7.5 = 0.
The following subroutine calculates f(x):
Keystrokes
|¥
´b 3
6
Display
000–
001–42,21, 3
002–
6
Program mode.
Label.
Assumes stack loaded with x.
190
Section 13: Finding the Roots of an Equation
Keystrokes
*
8
+
*
4
*
7
.
5
|n
Display
003–
004–
005–
005–
007–
008–
009–
010–
011–
012–
013–
014–
30
20
8
40
20
4
20
7
48
5
30
43 32
(x – 6) x.
((x – 6) x + 8) x.
4 ((x – 6) x + 8) x.
It seems reasonable that either a tall, narrow box or a short, flat box could
be formed having the desired volume. Because the taller box is preferred,
larger initial estimates of the height are reasonable. However, heights
greater than 2 decimeters are not physically possible (because the metal is
only 4 decimeters wide). Initial estimates of 1 and 2 decimeters are
therefore appropriate.
Find the desired height:
Keystrokes
|¥
1v
2
´_3
)
)
Display
Run mode.
1.0000
2
1.5000
1.5000
0.0000
Initial estimates.
The desired height.
Previous estimate.
f(x) at root.
Section 13: Finding the Roots of an Equation
191
By making the height 1.5 decimeters, a
5.0×1.0×1.5-decimeter box is specified.
If you ignore the upper limit on the
height and use initial estimates of 3 and
4 decimeters (still less than the width),
you will obtain a height of 4.2026
decimeters – a root that is physically
meaningless. If you use small initial
estimates such as 0 and 1 decimeter,
you will obtain a height of 0.2974
decimeter – producing an undesirably
short, flat box.
Graph of f(x)
As an aid for examining the behavior of a function, you can easily evaluate
the function at one or more values of x using your subroutine in program
memory. To do this, fill the stack with x. Execute the subroutine to calculate
the value of the function (press ´ letter label or G label.
The values you calculate can be plotted to give you a graph of the function.
This procedure is particularly useful for a function whose behavior you do
not know. A simple-looking function may have a graph with relatively
extreme variations that you might not anticipate. A root that occurs near a
localized variation may be hard to find unless you specify initial estimates
that are close to the root.
If you have no informed or intuitive concept of the nature of the function or
the location of the zero you are seeking, you can search for a solution using
trial-and-error. The success of finding a solution depends partially upon the
function itself. Trial-and-error is often – but not always – successful.
 If you specify two moderately large positive or negative estimates and
the function's graph does not have a horizontal asymptote, the routine
will seek a zero which might be the most positive or negative (unless
the function oscillates many times, as the trigonometric functions do).
 If you have already found a zero of the function, you can check for
another solution by specifying estimates that are relatively distant
from any known zeros.
192
Section 13: Finding the Roots of an Equation
 Many functions exhibit special behavior when their arguments
approach zero. You can check your function to determine values of x
for which any argument within your function becomes zero, and then
specify estimates at or near those values.
Although two different initial estimates are usually supplied when using
_, you can also use _ with the same estimate in both the X- and
Y-registers. If the two estimates are identical, a second estimate is generated
internally. If your single estimate is nonzero, the second estimate differs
from your estimate by one count in the seventh significant digit. If your
estimate is zero, 1×10-7 is used as the second estimate. Then the root-finding
procedure continues as it normally would with two estimates.
Using _ in a Program
You can use the _ operation as part of a program. Be sure that the
program provides initial estimates in the X- and Y-registers just prior to the
_ operation. The _ routine stops with a value of x in the
X-register and the corresponding function value in the Z-register. If the xvalue is a root, the program proceeds to the next line. If the x-value is not a
root, the next line is skipped. (Refer also to Interpreting Results on page 226
for a further explanation of roots.) Essentially, the _ instruction tests
whether the x-value is a root and then proceeds according to the ―Do if
True‖ rule. The program can then handle the case of not finding a root, such
as by choosing new initial estimates or changing a function parameter.
The use of _ as an instruction in a program utilizes one of the seven
pending returns in the calculator. Since the subroutine called by _
utilizes another return, there can be only five other pending returns.
Executed from the keyboard, on the other hand, _ itself does not
utilize one of the pending returns, so that six pending returns are available
for subroutines within the subroutine called by _. Remember that if
all seven pending returns have been utilized, a call to another subroutine
will result in a display of Error 5. (Refer to page 105.)
Section 13: Finding the Roots of an Equation
193
Restriction on the Use of _
The one restriction regarding the use of _ is that _ cannot be
used recursively. That is, you cannot use _ in a subroutine that is
called during the execution of _. If this situation occurs, execution
stops and Error 7 is displayed. It is possible, however, to use _ with
f thereby using the advanced capabilities of both of these keys.
Memory Requirements
_ requires five registers to operate. (Appendix C explains how they
are automatically allocated from memory.) If five unoccupied registers are
not available, _ will not run and Error 10 will be displayed.
A routine that combines _ and f requires 23 registers of space.
For Further Information
In appendix D, Advanced Use of _, additional techniques and
explanations for using _ are presented. These include:
 How _ works.
 Accuracy of the root.
 Interpreting results.
 Finding several roots.
 Limiting estimation time.
Section 14
Numerical Integration
Many problems in mathematics, science, and
engineering require calculating the definite
integral of a function. If the function is
denoted by f(x) and the interval of integration
is a to b, the integral can be expressed
mathematically as
b
I   a f ( x ) dx.
The quantity I can be interpreted
geometrically as the area of a region bounded by the graph of f(x), the
x-axis, and the limits x = a and x = b.*
When an integral is difficult or impossible to evaluate by analytical
methods, it can be calculated using numerical techniques. Usually, this can
be done only with a fairly complicated computer program. With your
HP-15C, however, you can easily do numerical integration using the f
(integrate) key.†
Using f
The basic rules for using f are:
1. In Program mode, key in a subroutine that evaluates the function f(x) that
you want to integrate. This subroutine must begin with a label
instruction (´b label) and end up with a value for f(x) in the Xregister.
*
Provided that f(x) is nonnegative throughout the interval of integration.
†
The f function does not use the imaginary stack. Refer to the HP-15C Advanced Functions Handbook
for information about using f in Complex mode.
194
Section 14: Numerical Integration
195
In Run mode:
2. Key the lower limit of integration (a) into the X-register, then press
v to lift it into the Y-register.
3. Key the upper limit of integration (b) in to the X-register.
4. Press ´ f followed by the label of your subroutine.
Example: Certain problems in physics and engineering require calculating
Bessel functions. The Bessel function of the first kind of order 0 can be
expressed as
J 0 ( x) 
1
π
π
 cos ( x sin θ) dθ .
0
Find
J 0 (1) 
1
π
π
 cos (sin θ) dθ .
0
In Program mode, key in the following subroutine to evaluate the function
f(θ) = cos (sin θ).
Keystrokes
Display
000–
|¥
´ CLEAR M 000–
001–42,21, 0
´b 0
[
\
|n
002–
003–
004–
23
24
43 32
Program mode.
Clear program memory.
Begin subroutine with a
b instruction.
Subroutine assumes a
value of θ is in X-register.
Calculate sin θ.
Calculate cos (sin θ).
Now, in Run mode key the lower limit of integration into the Y-register and
the upper limit into the X-register. For this particular problem, you also
need to specify Radians mode for the trigonometric functions.
196
Section 14: Numerical Integration
Keystrokes
Display
|¥
0v
0.0000
|$
3.1416
|R
3.1416
Run mode.
Key lower limit, 0, into Yregister.
Key upper limit, π, into Xregister.
Specify Radians mode for
trigonometric functions.
Now you are ready to press ´f 0 to calculate the integral. When you
do so, you'll find that – just as with _ – the calculator will not display
the result right away, as it does with other operations. The HP-15C
calculates integrals using a sophisticated iterative algorithm. Briefly, this
algorithm evaluates f(x), the function to be integrated, at many values of x
between the limits of integration. At each of these values, the calculator
evaluates the function by executing the subroutine you write for that
purpose. When the calculator must execute the subroutine many times – as
it does when you press f – you can't expect any answer right away. Most
integrals will require on the order of 2 to 10 seconds; but some integrals
will require even more. Later on we'll discuss how you can decrease the
time somewhat; but for now press ´f 0 and take a break (or read
ahead) while the HP-15C takes care of the drudgery for you.
Keystrokes
´f 0
Display
2.4040
 0π cos (sin θ) dθ .
In general, don't forget to multiply the value of the integral by whatever
constants, if any, are outside the integral. In this particular problem, we
need to multiply the integral by 1/ π to get J0 (1):
Keystrokes
|$
÷
Display
3.1416
0.7652
J0 (1).
Section 14: Numerical Integration
197
Before calling the subroutine you provide to evaluate f(x), the f
algorithm – just like the _ algorithm – places the value of x in the X-,
Y-, Z-, and T-registers. Because every stack register contains the x-value,
your subroutine can calculate with this number without having to recall it
from a storage register. The subroutines in the next two examples take
advantage of this feature. (A polynomial evaluation technique that assumes
the stack is filled with the value of x is discussed on page 79.)
Note: Since the calculator puts the value of x into all stack
registers, any numbers previously there will be replaced by x.
Therefore, if the stack contains intermediate results that you'll
need after you calculate an integral, store those numbers in
storage registers and recall them later.
Occasionally you may want to use the subroutine that you wrote
for the f operation to merely evaluate the function at some
value of x. If you do so with a function that gets x from the stack
more than once, be sure to fill the stack manually with the value
of x, by pressing vvv, before you execute the
subroutine.
Example: The Bessel function of the first kind of order 1 can be expressed
as
J1 ( x ) 
1
π

π
J1 (1) 
1
π

π
cos ( - x sin θ ) dθ.
0
Find
Key in the following
f(θ) = cos (θ - sin θ).
Keystrokes
|¥
´b1
cos ( - sin θ ) dθ.
0
subroutine
Display
000001-42,21,
1
that
evaluates
the
function
Program mode.
Begin subroutine with a label.
198
Section 14: Numerical Integration
Keystrokes
[
-
Display
002–
003–
23
30
004–
005–
24
43 32
\
|n
Calculate sin θ.
Since a value of θ will be
placed into the Y-register by
the f algorithm before it
executes this subroutine, the
- operation at this point
will calculate
(θ – sin θ).
Calculate cos (θ – sin θ).
In Run mode, key the limits of integration into the X- and Y-registers. Be
sure that the trigonometric mode is set to Radians, then press ´f 1 to
calculate the integral. Finally, multiply the integral by 1/π to calculate
J1 (1).
Keystrokes
|¥
0v
Display
0.0000
|$
3.1416
|R
3.1416
´f 1
1.3825
|$ ÷
0.4401
Run mode.
Key lower limit into
Y-register.
Key upper limit into
X-register.
(If not already in
Radians mode.)
J1 (1).
Example:
Certain
problems
in
communications theory (for example, pulse
transmission through idealized networks)
require calculating an integral (sometimes
called the sine integral) of the form
Si(t ) 

t sin(x)
0
x
dx .
Section 14: Numerical Integration
199
Find Si(2).
Key in the following subroutine to evaluate the function f(x) = (sin x) / x.*
Keystrokes
|¥
´ b .2
Display
000–
001–42,21, .2
[
®
002–
003–
23
34
÷
|n
004–
005–
10
43 32
Program mode.
Begin subroutine with a b
instruction.
Calculate sin x.
Since a value of x will be
placed in the Y-register by the
f algorithm before it
executes this subroutine, the
® operation at this point
will return x to the X-register
and move sin x to the Yregister.
Divide sin x by x.
Now key the limits of integration into the X- and Y-registers. In Radians
mode, press ´f .2 to calculate the integral.
Keystrokes
|¥
0v
*
Display
0.4401
0.0000
2
2
|R
2.0000
´f .2
1.6054
Run mode.
Key lower limit into Yregister.
Key upper limit, into Xregister.
(If not already in Radians
mode.)
Si(2).
If the calculator attempted to evaluate f(x) = (sin x)/x at x = 0, the lower limit of integration, it would
terminate with Error 0 in the display (signifying an attempt to divide by zero), and the integral could not
be calculated. However, the f algorithm normally does not evaluate functions at either limit of
integration, so the calculator can calculate the integral of a function that is undefined there. Only when the
endpoints of the interval of integration are extremely close together, or the number of sample points is
extremely large, does the algorithm evaluate the function at the limits of integration.
200
Section 14: Numerical Integration
Accuracy of f
The accuracy of the integral of any function depends on the accuracy of the
function itself. Therefore, the accuracy of an integral calculated using f
is limited by the accuracy of the function calculated by your subroutine. * To
specify the accuracy of the function, set the display format so that the
display shows no more than the number of digits that you consider accurate
in the function's values.† If you specify fewer digits, the calculator will
compute the integral more quickly;‡ but it will presume that the function is
accurate to only the number of digits specified in the display format. We'll
show you how you can determine the accuracy of the calculated integral
after we say another word about the display format.
You'll recall that the HP-15C provides three types of display formatting:
•, i, and ^. Which display format should be used is largely a
matter of convenience, since for many integrals you'll get about the same
results using any of them (provided that the number of digits is specified
correctly, considering the magnitude of the function). Because it's more
convenient to use i display format when calculating most integrals,
we'll use i when calculating integrals in subsequent examples.
Note: Remember that once you have set the display format, you
can change the number of digits appearing in the display by storing
a number in the Index register and then pressing ´ • V,
´ i V, or ´ ^ V, as described in section 10.
This capability is especially useful when f is executed as part
of a program.
*
†
‡
It is possible that integrals of functions with certain characteristics (such as spikes or very rapid
oscillations) might be calculated inaccurately. However, this possibility is very small. The general
characteristics of functions that could cause problems, as well as techniques for dealing with them, are
discussed in appendix E.
The accuracy of a calculated function depends on such considerations as the accuracy of empirical
constants in the function as well as round–off error in the calculations. These considerations are discussed
in more detail in the HP-15C Advanced Functions Handbook.
The reason for this is discussed in appendix E.
Section 14: Numerical Integration
201
Because the accuracy of any integral is limited by the accuracy of the
function (as indicated in the display format), the calculator cannot compute
the value of an integral exactly, but rather only approximates it. The
HP-15C places the uncertainty* of an integral's approximation in the Yregister at the same time it places the approximation in the X-register. To
determine the accuracy of an approximation, check its uncertainty by
pressing ®.
Example: With the display format set to i 2, calculate the integral in
the expression for J1(1) (from the example on page 197).
Keystrokes
0v
Display
0.0000
|$
3.1416
|R
´i2
´f1
3.1416
3.14
00
1.3
00
8
1.8
8
03
®
Key lower limit into
Y-register.
Key upper limit into
X-register.
(If not already in Radians mode.)
Set display format to i 2.
Integral approximated in i 2.
Uncertainty of i 2
approximation.
The integral is 1.38 ± 0.00188. Since the uncertainty would not affect the
approximation until its third decimal place, you can consider all the
displayed digits in this approximation to be accurate. In general, though, it
is difficult to anticipate how many digits in an approximation will be
unaffected by its uncertainty. This depends on the particular function being
integrated, the limits of integration, and the display format.
*
No algorithm for numerical integration can compute the exact difference between its approximation and
the actual integral. But the algorithm in the HP-15C estimates an ―upper bound‖ on this difference, which
is the uncertainty of the approximation. For example, if the integral Si (2) is 1.6054 ± 0.0001, the
approximation to the integral is 1.6054 and its uncertainty is 0.0001. This means that while we don't know
the exact difference between the actual integral and its approximation, we do know that it is highly
unlikely that the difference is bigger than 0.0001. (Note the first footnote on page 200.)
202
Section 14: Numerical Integration
If the uncertainty of an approximation is larger than what you choose to
tolerate, you can decrease it by specifying a greater number of digits in the
display format and repeating the approximation.*
Whenever you want to repeat an approximation, you don't need to key the
limits of integration back into the X- and Y-registers. After an integral is
calculated, not only are the approximation and its uncertainty placed in the
X- and Y-registers, but in addition the upper limit of integration is placed in
the Z-register, and the lower limit is placed in the T-register. To return the
limits to the X- and Y-registers for calculating an integral again, simply
press ) ).
Example: For the integral in the expression for J1(l), you want an answer
accurate to four decimal places instead of only two.
Keystrokes
´i4
))
Display
1.8826
3.1416
-03
00
´f 1
1.3825
00
®
1.7091
-05
Set display format to i 4.
Roll down stack until upper
limit appears in X-register.
Integral approximated in
i4.
Uncertainty of i
4 approximation.
The uncertainty indicates that this approximation is accurate to at least four
decimal places. Note that the uncertainty of the i 4 approximation is
about one-hundredth as large as the uncertainty of the i 2
approximation. In general, the uncertainty of any f approximation
decreases by about a factor of 10 for each additional digit specified in the
display format.
*
Provided that f(x) is still calculated accurately to the number of digits shown in the display.
Section 14: Numerical Integration
203
In the preceding example, the uncertainty indicated that the approximation
might be correct to only four decimal places. If we temporarily display all
10 digits of the approximation, however, and compare it to the actual value
of the integral (actually, an approximation known to be accurate to a
sufficient number of decimal places), we find that the approximation is
actually more accurate than its uncertainty indicates.
Keystrokes
®
´ CLEAR u
Display
1.382
00 Return approximation to
5
display.
1382459676 All 10 digits of
approximation.
The value of this integral, correct to eight decimal places, is 1.38245969. The
calculator's approximation is accurate to seven decimal places rather than
only four. In fact, since the uncertainty of an approximation is calculated
very conservatively, the calculator's approximation, in most cases will be
more accurate than its uncertainty indicates. However, normally there is no
way to determine just how accurate an approximation is.
For a more detailed look at the accuracy and uncertainty of f
approximations, refer to appendix E.
Using f in a Program
f can appear as an instruction in a program provided that the program is
not called (as a subroutine) by f itself. In other words, f cannot be
used recursively. Consequently, you cannot use f to calculate multiple
integrals; if you attempt to do so, the calculator will halt with Error 7 in the
display. However, f can appear as an instruction in a subroutine called
by _.
The use of f as an instruction in a program utilizes one of the seven
pending returns in the calculator. Since the subroutine called by f
utilizes another return, there can be only five other pending returns.
Executed from the keyboard, on the other hand, f itself does not utilize
one of the pending returns, so that six pending returns are available for
subroutines within the subroutine called by f Remember that if all seven
pending returns have been utilized, a call to another subroutine will result in
a display of Error 5. (Refer to page 105.)
204
Section 14: Numerical Integration
Memory Requirements
f requires 23 registers to operate. (Appendix C explains how they are
automatically allocated from memory.) If 23 unoccupied registers are not
available, f will not run and Error 10 will be displayed.
A routine that combines f and _ also requires 23 registers of
space.
For Further Information
This section has given you the information you need to use f with
confidence over a wide range of applications. In appendix E, more esoteric
aspects of f are discussed. These include:
 How f works.
 Accuracy, uncertainty, and calculation time.
 Uncertainty and the display format.
 Conditions that could cause incorrect results.
 Conditions that prolong calculation time.
 Obtaining the current approximation to an integral.
Appendix A
Error Conditions
If you attempt a calculation containing an improper operation – say division
by zero – the display will show Error and a number. To clear an error
message, press any one key. This also restores the display prior to the Error
display.
The HP-15C has the following error messages. (The description of Error 2
includes a list of statistical formulas used.)
Error 0: Improper Mathematics Operation
Illegal argument to math routine:
÷, where x = 0.
y, where:
 out of Complex mode, y < 0 and x is noninteger;
 out of Complex mode, y = 0 and x ≤ 0; or
 in Complex mode, y = 0 and Re(x) ≤ 0.
¤, where, out of Complex mode, x < 0.
∕, where x = 0.
o, where:
 out of Complex mode, x ≤ 0; or
 in Complex mode, x = 0.
Z, where:
 out of Complex mode, x ≤ 0; or
 in Complex mode, x = 0.
,, where, out of Complex mode, │x│> l.
{, where, out of Complex mode, │x│> l.
O ÷, where x = 0.
l ÷, where the contents of the addressed register = 0.
∆, where the value in the Y-register is 0.
H \, where, out of Complex mode, x< 1.
H ], where, out of Complex mode, │x│> 1.
c p, where:
205
206
Appendix A: Error Conditions




x or y is noninteger;
x < 0 or y < 0;
x > y;
x or y ≥ 1010.
Error 1: Improper Matrix Operation
Applying an operation other than a matrix operation to a matrix, that is,
attempting a nonmatrix operation while a matrix is in the relevant register
(whether the X- or Y-register or a storage register).
Error 2: Improper Statistics Operation
’
S
j
L
n=0
n≤1
n≤1
n≤1
Error 2 is also displayed if division by zero or the square root of a negative
number would be required during computation with any of the following
formulas:
x
yˆ 
y
M
s 
x
A
x
n
s 
y
n ( n  1)
P
M
B
N
n ( n  1)
y
n
r
M  y  P x
nM
M  y  Pn  x   x 
nM
where:
M = nΣx2 – (Σx)2
N = nΣy2 – (Σy)2
P = nΣxy – ΣxΣy
(A and B are the values returned by the operation
L, where y= Ax + B.)
P
M N
Appendix A: Error Conditions
207
Error 3: Improper Register Number or Matrix Element
Storage register named is nonexistent or matrix element indicated is
nonexistent.
Error 4: Improper Line Number or Label Call
Line number called for is currently unoccupied or nonexistent (>448); or
you have attempted to load a program line without available space; or the
label called does not exist; or User mode is on and you did not press ´
before ¤, ', @, y or ∕.
Error 5: Subroutine Level Too Deep
Subroutine nested more than seven deep.
Error 6: Improper Flag Number
Attempted a flag number >9.
Error 7: Recursive _ or f
A subroutine which is called by _ also contains a _ instruction;
a subroutine which is called by f also contains an f instruction.
Error 8: No Root
_ unable to find a root using given estimates.
Error 9: Service
Self-test discovered circuitry problem, or wrong key pressed during key
test.
Error 10: Insufficient Memory
There is not enough memory available to perform a given operation.
Error 11: Improper Matrix Argument
Inconsistent or improper matrix arguments for a given matrix operation:
208
Appendix A: Error Conditions
+ or -, where the dimensions are incompatible.
*, where:
 the dimensions are incompatible; or
 the result is one of the arguments.
∕, where the matrix is not square.
scalar/matrix ÷, where the matrix is not square.
÷, where:
 the matrix in the X-register is not square;
 the dimensions are incompatible; or
 the result is the matrix in the X-register.
> 2, where the input is a scalar; or the number of rows is odd.
> 3, where the input is a scalar; or the number of columns is odd.
> 4, where the input is scalar.
> 5, where:
 the input is a scalar;
  the dimensions are incompatible; or
  the result is one of the arguments.
> 6, where:
 the input is scalar;
 the dimensions are incompatible (including the result); or
 the result is one of the arguments.
> 9, where the matrix is not square.
l m V, where contents of RI are scalar.
m V, where contents of RI are scalar.
O <, where the input is scalar.
p, where the number of columns is odd.
c, where the number of rows is odd.
Pr Error (Power Error)
Continuous Memory interrupted and reset because of power failure.
Appendix B
Stack Lift and
the LAST X Register
The HP-15C calculator has been designed to operate in a natural manner.
As you have seen working through this handbook, most calculations do not
require you to think about the operation of the automatic memory stack.
There are occasions, however – especially as you delve into programming –
when you need to know the effect of a particular operation upon the stack.
The following explanation should help you.
Digit Entry Termination
Most operations on the calculator, whether executed as instructions in a
program or pressed from the keyboard, terminate digit entry. This means
that the calculator knows that any digits you key in after any of these
operations are part of a new number.
The only operations that do not terminate digit entry are the digit entry keys
themselves:
0 through 9 “
.
‛
−
Stack Lift
There are three types of operations on the calculator based on how they
affect stack lift. These are stack-disabling operations, stack-enabling
operations, and neutral operations.
When the calculator is in Complex mode, each operation affects both the
real and imaginary stacks. The stack lift effects are the same. In addition,
the number keyed into the display (real X-register) after any operation
except − or ` is accompanied by the placement of a zero in the
imaginary X-register.
209
210
Appendix B: Stack Lift and the LAST X Register
Disabling Operations
Stack Lift. There are four stack-disabling operations on the calculator.*
These operations disable the stack lift, so that a number keyed in after one
of these disabling operations writes over the current number in the
displayed X-register and the stack does not lift. These special disabling
operations are:
v
`
z
w
Imaginary X-Register. A zero is placed in the imaginary X-register when
the next number following v, z, or w is keyed or recalled into
the display (real X-register). However, the next number keyed in or recalled
after − or ` does not change the contents of the imaginary Xregister.
Enabling Operations
Stack Lift. Most of the operations on the keyboard, including one-and twonumber mathematical functions like x and *, are stack-enabling
operations. This means that a number keyed in after one of these operations
will lift the stack (because the stack has been ―enabled‖ to lift). Both the
real and imaginary stacks are affected. (Recall that a shaded X-register
means that its contents will be written over when the next number is keyed
in or recalled.)
T
t
z
y
y
Z
z
y
x
x
Y
y
x
4.0000
4.0000
X
x
Keys:
(Assumes
stack
enabled.)
*
4
Refer to footnote, page 36.
4.0000
v
4
Stack
lifts.
Stack
disabled.
3
3
No stack
lift.
Appendix B: Stack Lift and the LAST X Register
211
T
y
y
y
y
Z
x
x
x
x
Y
4.0000
53.1301
53.1301
53.1301
X
3
5.0000
0.0000
Keys:
|:
|`
Stack
Stack
enabled.
disabled.
7
7
No stack
lift.
Imaginary X-Register. All enabling functions provide for a zero to be
placed in the imaginary X-register when the next number is keyed or
recalled into the display.
Neutral Operations
Stack Lift. Some operations, like •, are neutral; that is, they do not
alter the previous status of the stack lift. Thus, if you disable the stack lift
by pressing v, then press ´ • n and key in a new number, that
number will write over the number in the X-register and the stack will not
lift. Similarly, if you have previously enabled the stack lift by executing,
say ¤, then execute a • instruction followed by a digit entry
sequence, the stack will lift.*
The following operations are neutral on the HP-15C:
•
i
^
D
R
g
t “ nnn
‚
Â
W
U
CLEAR u
CLEAR Q
CLEAR ∑
©
¦
¥
%†
Imaginary X-Register. The above operations are also neutral with respect
to clearing the imaginary X-register.
*
All digit entry functions are also neutral during digit entry. After digit entry termination, “ and ‛
are lift enabling, − is disabling.
†
That is, the ´ % sequence used to view the imaginary X-register.
212
Appendix B: Stack Lift and the LAST X Register
LAST X Register
The following operations save x in the LAST X register:
+
*
÷
a
q
‘
&
∕
!
¤
x
[
\
]
,
{
/
P[
P\
P]
H[
H\
H]
h
À
d
r
N
'
o
@
Y
*
Except when used as a matrix function.
†
f uses the LAST X register in a special way, as described in appendix E.
k
∆
:
;
*
p
c*
z
w
j
> 5 through 9
†
f
Appendix C
Memory Allocation
The Memory Space
Storage registers, program lines, and advanced function execution * all draw
on a common memory space in the HP-15C. The availability of memory for
a specific purpose depends on the current allocation of memory, as well as
on the total memory capacity of the calculator.
Registers
Memory space in the HP-15C is allocated on the basis of registers. This
space is partitioned into two pools, which strictly define how a register may
be used. There is always a combined total of 67 registers in these two pools.
*

The data storage pool contains registers which may be used only
for data storage. At power-up (Continuous Memory reset) this
equals 21 registers. This pool contains at least three registers at all
times: RI, R0, and R1.

The common pool contains uncommitted registers available for
allocation to programming, matrices, the imaginary stack, and
_ and f operation. At power-up there are 46 uncommitted
registers in the common pool.
The use of _, f, Complex mode, or matrices temporarily requires extra memory space, as
explained later in this appendix.
213
214
Appendix C: Memory Allocation
Total allocatable memory: 64 registers, numbered R2 through R65.
[(dd – 1) + uu + pp + (matrix elements) + (imaginary stack) + (_
and f)] = 64. For memory allocation and indirect addressing, data
registers R.0 through R.9 are referred to as R10 through R19.
Appendix C: Memory Allocation
215
Memory Status (W)
To view the current memory configuration of the calculator, press |
W (memory), holding W to retain the display.* The display will be
four numbers,
dd
uu pp-b
where:
dd = the number of the highest-numbered register in the data storage
pool (making the total number of data registers dd + 2 because of R0
and RI);
uu = the number of uncommitted registers in the common pool;
pp = the number of registers containing program instructions; and
b = the number of bytes left before uu is decremented (to supply seven
more bytes of program memory) and pp is incremented.
The initial status of the HP-15C at power-up is:
19
46
0-0
The movable boundary between the data storage and common pools is
always between Rdd and Rdd + 1.
Memory Reallocation
There are 67 registers in memory, worth seven bytes each. Sixty-four of
these registers (R2 to R65) are interconvertible between the data storage and
common pools.
The m % Function
If you should require more common space (as for programming) or more
data storage space (but not both simultaneously!), you can make the
†
necessary register reallocation using m %. The procedure is:
*
MEM is nonprogrammable.
†
m (dimension) is so called because it is also used (with A through E or V) to dimension
matrices. Above, however, it is used (with %) to ―dimension‖ the size of the data storage pool.
216
Appendix C: Memory Allocation
1. Place dd, the number of the highest data storage register you want
allocated, into the display. 1dd65. The number of registers in the
uncommitted pool (and therefore potentially available for
programming) will be (65 – dd).
2. Press ´ m %.
There are two ways to review your allocation:

Press lm % to recall into the stack the number of the
highest-allocated data storage register, dd. (Programmable.)

Press | W (as explained above) to view a more complete
memory status (dd uu pp-b).
Keystrokes
Display
(assuming a cleared program memory)*
R1, R0, and RI
1 ´ m % 1.0000
allocated for data storage. Sixty| W (hold) 1 64 0-0
four registers are uncommitted;
none contain program instructions.
19.0000
R19 (R.9) is the highest-numbered
19 ´ m
data storage register. Forty-six
%
registers left in the common pool.
l m % 19.0000
Restrictions on Reallocation
Continuous Memory will maintain the configuration you allocate until a
new m % is executed or Continuous Memory is reset. If you try to
allocate a number less than 1, dd = 1. If you try to allocate a number greater
than 65, Error 10 results.
*
If program memory is not cleared, the number of uncommitted registers (uu) is less owing to allocation of
registers to program memory (pp). Therefore, pp would be >0 and b would vary.
Appendix C: Memory Allocation
217
When converting registers, note that:
 You can convert registers from the common pool only if they are
uncommitted. If, for example, you try to convert registers which
contain program instructions, you will get an Error 10 (insufficient
memory).
 You can convert occupied registers from the data storage pool,
causing a loss of stored data. An Error 3 results if you try to
address a "lost" – that is, nonexistent – register. Therefore, it is
good practice to store data in the lowest-numbered registers first,
as these are the last to be converted.
Program Memory
As mentioned before, each register consists of seven bytes of memory.
Program instructions use one or two bytes of memory. Most program lines
use one byte; those using two bytes are listed on page 218.
The maximum programming capacity of the HP-15C is 448 program bytes
(64 convertible registers at seven bytes per register). At power-up, memory
can hold up to 322 program bytes (46 allocated registers at seven bytes
per register).
Automatic Program Memory Reallocation
Within the common register pool, program memory will automatically
expand as needed. One uncommitted register at a time, starting with the
highest-numbered register available, will be allocated to seven bytes of
program memory.
Conversion of Uncommitted Registers to Program Memory
218
Appendix C: Memory Allocation
Your very first program instruction will commit R65 (all seven bytes) from
an uncommitted register to a program register. Your eighth program
instruction commits R64, and so on, until the boundary of the common pool
is encountered. Registers from the data storage pool (at power-up, this is R19
and below) are not available for program memory without reallocating
registers using m %.
Two-Byte Program Instructions
The following instructions are the only ones which require two bytes of
calculator memory. (All others require only one byte.)
´ b . label
t . label
| " (n or V)
|F (n or V)
| ? (n or V)
´ • (n or V)
´ i (n or V)
´ ^ (n or V)
´_
´f
´ > {0 to 9}
´ X {2 to 9, .0 to .9}
´ e {2 to 9, .0 to .9}
´ I {2 to 9, .0 to .9}
O {+, -, *, ÷}
l {+, -, *, ÷}
O > {A to E}
O {A to E, %} in User
mode
l {A to E, %} in User
mode
O|%
l|%
Memory Requirements for the Advanced Functions
The four advanced functions require temporary register space from the
common register pool.
Function
_
f
Complex Stack
Matrices
Registers Needed
5
23
23 if executed
together
5
1 per matrix element
Appendix C: Memory Allocation
219
For _ and f, allocation and deallocation of the required register
space takes place automatically.* Memory is thereby allocated only for the
duration of these operations.
Space for the imaginary stack is allocated whenever ´ V, ´
}, or | F 8 is pressed. The imaginary stack is deallocated when
" 8 is executed.
Space for matrix registers is not allocated until you dimension it (using
m). Reallocation takes place when you redimension a matrix. >
0 dimensions all matrices to 0  0.
*
If you should interrupt a _ or f routine in progress by pressing a key, you could deallocate its
registers by pressing |n or ´ CLEAR M in Run mode.
Appendix D
A Detailed Look at _
Section 13, Finding the Roots of an Equation, includes the basic
information needed for the effective use of the _ algorithm. This
appendix presents more advanced, supplemental considerations regarding
_.
How _ Works
You will be able to use _ most effectively by having a basic
understanding of how the algorithm works.
In the process of searching for a zero of the
specified function, the algorithm uses the
value of the function at two or three
previous estimates to approximate the shape
of the function’s graph. The algorithm uses
this shape to intelligently ―predict‖ a new
estimate where the graph might cross the xaxis. The function subroutine is then
executed, computing the value of the
function at the new estimate. This procedure is performed repeatedly by the
_ algorithm.
If any two estimates yield function values
with opposite signs, the algorithm presumes
that the function's graph must cross the xaxis in at least one place in the interval
between these estimates. The interval is
systematically narrowed until a root of the
equation is found.
A root is successfully found either if the
computed function value is equal to zero or
if two estimates, differing by one unit in their last significant digit, give
function values having opposite signs. In this case, execution stops and the
estimate is displayed.
220
Appendix D: A Detailed Look at _
221
As discussed in section 13, page 186, the occurrence of other situations in
the iteration process indicates the apparent absence of a function zero. The
reason is that there is no way to logically predict a new estimate that is
likely to have a function value closer to zero. In such cases, Error 8 is
displayed.
You should note that the initial estimates you provide are used to begin the
"prediction" process. By permitting more accurate predictions than might
otherwise occur, properly chosen estimates greatly facilitate the
determination of the root you seek.
The _ algorithm will always find a root provided one exists
(within the overflow bounds), if any one of four conditions are met:
 Any two estimates have function
values with opposite signs.
 The function is monotonic, meaning
that f(x) either always decreases or
else always increases as x is
increased.
222
Appendix D: A Detailed Look at _
 The function's graph is either
convex everywhere or concave
everywhere.
 The only local minima and
maxima of the function's graph
occur singly between adjacent
zeros of the function.
In addition, it is assumed that the _ algorithm will not be interrupted
by an improper operation.
Accuracy of the Root
When you use the _ key to find a root of an equation, the root is
found accurately. The displayed root either gives a calculated function value
(f(x)) exactly equal to zero or else is a 10-digit number virtually adjacent to
the place where the function's graph crosses the x-axis. Any such root has
an accuracy within two or three units in the 10th significant digit.
In most situations the calculated root is an accurate estimate of the
theoretical (infinitely precise) root of the equation. However, certain
conditions can cause the finite accuracy of the calculator to give a result that
appears to be inconsistent with your theoretical expectation.
Appendix D: A Detailed Look at _
223
If a calculation has a result whose magnitude is smaller than
1.000000000×10-99, the result is set equal to zero. This effect is referred to
as ―underflow.‖ If the subroutine that calculates your function encounters
underflow for a range of x and if this affects the value of the function, then a
root in this range may be expected to have some inaccuracy. For example,
the equation
x4 = 0
has a root at x = 0. Because of underflow, _ produces a root of
1.5060
-25 (for initial estimates of 1 and 2). As another example,
consider the equation
l / x2 = 0
whose root is infinite in value. Because of underflow, _ gives a root
of 3.1707
49 (for initial estimates of 10 and 20). In each of these
examples, the algorithm has found a value of x for which the calculated
function value equals zero. By understanding the effect of underflow, you
can readily interpret results such as these.
The accuracy of a computed value sometimes can be adversely affected by
―round-off‖ error, by which an infinitely precise number is rounded to 10
significant digits. If your subroutine requires extra precision to properly
calculate the function for a range of x, the result obtained by _ may
be inaccurate. For example, the equation
| x2 – 5 | = 0
has a root at x = 5 . Because no 10-digit number exactly equals 5 , the
result of using _ is Error 8 (for any initial estimates) because the
function never equals zero nor changes sign. On the other hand, the
equation
[(|x| + 1) + 1015]2 = 1030
has no roots because the left side of the equation is always greater than the
right side. However, because of round-off in the calculation of
f(x) = [(|x| + 1) + 1015]2 - 1030,
224
Appendix D: A Detailed Look at _
the root 1.0000 is found for initial estimates of 1 and 2. By recognizing
situations in which round-off error may influence the operation of _,
you can evaluate the results accordingly and perhaps rewrite the function to
reduce the effects of round-off.
In a variety of practical applications, the parameters in an equation – or
perhaps the equation itself – are merely approximations. Physical
parameters have an inherent accuracy (or inaccuracy). Mathematical
representations of physical processes are only models of those processes,
accurate only to the extent that the underlying assumptions are true. An
awareness of these and other inaccuracies can be used to your advantage.
By structuring your subroutine to return a function value of zero when the
calculated value is negligible for practical purposes, you can usually save
considerable time in finding a root with _ – particularly for cases that
would normally take a long time.
Example: Ridget hurlers such as Chuck Fahr can throw a ridget to heights
of 105 meters and more. In fact, Fahr’s hurls usually reach a height of
107 meters. How long does it take for his remarkable toss, described on
page 184 in section 13, to reach 107 meters?
Solution: The desired solution is the value of t at which h = 107. Enter the
subroutine from page 184 that calculates the height of the ridget. This
subroutine can be used in a new function subroutine to calculate
f(t) = h(t) – 107.
The following subroutine calculates f(t):
Keystrokes
|¥
´b B
GA
1
0
7
|n
Display
000–
001–42,21,12
002–
32 11
003–
1
004–
0
005–
7
006–
30
007–
43 32
Program mode.
Begin with new label.
Calculates h(t).
Calculates h(t) – 107.
Appendix D: A Detailed Look at _
225
In order to find the first time at which the height is 107 meters, use initial
estimates of 0 and 1 second and execute _ using B.
Keystrokes
|¥
0v
1
´_
B
)
)
Display
Run mode.
0.0000
1
4.1718
4.1718
0.0000
Initial estimates.
The desired root.
A previous estimate of the root.
Value of f(t) at root.
It takes 4.1718 seconds for the ridget to reach a height of exactly 107
meters. (It takes approximately two seconds to find this solution.)
However, suppose you assume that the function h(t) is accurate only to the
nearest whole meter. You can now change your subroutine to give f(t) = 0
whenever the calculated magnitude of f(t) is less than 0.5 meter. Change
your subroutine as follows:
Keystrokes
|¥
t “ 006
|a
.
5
|T7
|`
|T0
|K
Display
000006–
30
007–
43 16
008–
48
009–
5
010–43,30, 7
011–
43 35
012–43,30, 0
013–
43 36
Program mode.
Line before n instruction.
Magnitude of f(t).
Accuracy
Test for x > y and return
zero if accuracy >
magnitude (0.5 > | f(t) | ).
Test for x ≠ 0 and restore
f(t) if value is nonzero.
226
Appendix D: A Detailed Look at _
Execute _ again:
Keystrokes
|¥
0v
1
´vB
)
)
Display
Run mode.
0.0000
1
4.0681
4.0681
0.0000
Initial estimates.
The desired root.
A previous estimate of the
root.
Value of modified f(t) at root.
After 4.0681 seconds, the ridget is at a height of 107 ± 0.5 meters. This
solution, although different from the previous answer, is correct considering
the uncertainty of the height equation. (And this solution is found in just
under half the time of the earlier solution.)
Interpreting Results
The numbers that _ places in the X-, Y-, and Z-registers help you
evaluate the results of the search for a root of your equation. * Even when no
root is found, the results are still significant.
When _ finds a root of the specified
equation, the root and function values are
placed in the X- and Z-registers. A function
value of zero is the expected result.
However, a nonzero function value is also
acceptable because it indicates that the
function's graph apparently crosses the xaxis within an infinitesimal distance from
the calculated root. In most such cases, the
function value will be relatively close to
zero.
*
The number in the T-register is the same number that was left in the Y-register by the final execution of
your function subroutine. Generally, this number is not of interest.
Appendix D: A Detailed Look at _
227
Special consideration is required for a different
type of situation in which _ finds a root
with a nonzero function value. If your
function's graph has a discontinuity that
crosses the x-axis, _ specifies as a root
an x-value adjacent to the discontinuity. This is
reasonable because a large change in the
function value between two adjacent values of
x might be the result of a very rapid,
continuous transition. Because this cannot be
resolved by the algorithm, the root is displayed
for you to interpret.
A function may have a pole, where its
magnitude approaches infinity. If the function
value changes sign at a pole, the corresponding
value of x looks like a possible root of your
equation, just as it would for any other
discontinuity crossing the x-axis. However, for
such functions, the function value placed into
the Z-register when that root is found will be
relatively large. If the pole occurs at a value of
x that is exactly represented with 10 digits, the
subroutine may try that value and halt prematurely with an error indication.
In this case, the _ operation will not be completed. Of course, this
may be avoided by the prudent use of a conditional statement in your
subroutine.
Example: In her analysis of the stresses in a
structural component, design consultant Lucy
I. Beame has determined that the shear stress
can be expressed as
3
2

3x  45x  350
Q

1000

for 0  x  10
for 10  x  14
where Q is the shear stress in newtons per
square meter and x is the distance from one end in meters. Write a
subroutine to compute the shear stress for any value of x. Use _ to
find the location of zero shear stress.
228
Appendix D: A Detailed Look at _
Solution: The equation for the shear stress for x between 0 and 10 is more
efficiently programmed after rewriting it using Horner's method:
Q = (3x–45)x2 + 350
Keystrokes
|¥
´b2
1
0
|£
t9
|`
3
*
4
5
*
*
3
5
0
+
|n
´b9
‛
3
|n
for 0 < x < 10.
Display
000–
001–42,21, 2
002–
1
003–
0
004–
43 10
005–
22 9
006–
43 35
007–
3
008–
20
009–
4
010–
5
011–
30
012–
20
013–
20
014–
3
015–
5
016–
0
017–
40
013–
43 32
019–42,21, 9
020–
26
021–
3
022–
43 32
Program mode.
Test for x range.
Branch for x ≥ 10.
3x.
(3x – 45).
(3x – 45)x2.
(3x – 45)x 2 + 350.
End subroutine.
Subroutine for x ≥ 10.
103=1000.
End subroutine.
Execute _ using initial estimates of 7 and 14 to start at the outer end
of the beam and search for a point of zero shear stress.
Appendix D: A Detailed Look at _
Keystrokes
|¥
7v
14
´_ 2
))
229
Display
Run mode.
7.0000
14
10.0000
1,000.0000
Initial estimates.
Possible root.
Stress not zero.
The large stress value at the root points out that the _ routine has
found a discontinuity. This is a place on the beam where the stress quickly
changes from negative to positive. Start at the other end of the beam
(estimates of 0 and 7) and use _ again.
Keystrokes
0v
7
´_2
))
Display
0.0000
7
3.1358
2.0000
Initial estimates.
-07
Possible root.
Negligible stress.
Beame's beam has zero shear stress at
approximately 3.1358 meters and an
abrupt change of stress at 10.0000 meters.
When no root is found and Error 8 is displayed, you can press − or any
one key to clear the display and observe the estimate at which the function
was closest to zero. By also reviewing the numbers in the Y- and Zregisters, you can often determine the nature of the function near the root
estimate and use this information constructively.
230
Appendix D: A Detailed Look at _
If the algorithm terminates its search near a
local minimum of the function's magnitude,
clear the Error 8 display and observe the
numbers in the X-, Y-, and Z-registers by
rolling down the stack. If the value of the
function saved in the Z-register is relatively
close to zero, it is possible that a root of
your equation has been found – the number
returned in the X-register may be a 10-digit
number very close to a theoretical root. You
can explore this potential minimum further by rolling the stack until the
returned estimates are back in the X- and Y-registers and then executing
_ again using these numbers as initial estimates. If an actual
minimum has been found, Error 8 will again be displayed and the number
in the X-register will be approximately the same as before, but possibly
closer to the actual location of the minimum.
Of course, you may deliberately use _ to find the location of a local
minimum of the function's magnitude. However, in this case you must be
careful to confine the search in the region of the minimum. Remember,
_ tries hard to find a zero of the function.
If the algorithm stops searching and
displays Error 8 because it is working on a
horizontal asymptote (when the value of
the function is essentially constant for a
large range of x), the estimates in X- and
Y-registers usually are significantly
different from each other. The number in
the Z-register is the value of the potential
asymptote. If you execute _ again
using as initial estimates the numbers that
were returned in the X- and Y-registers, a
horizontal asymptote may again cause Error 8, but with numbers in the Xand Y-registers that will differ from the previous numbers. The value of the
function in the Z-register would then be about the same as that obtained
previously.
Appendix D: A Detailed Look at _
231
If Error 8 is displayed as a result of a search that
is concentrated in a local ―flat‖ region of the
function, the estimates in the X- and Y-registers
will be relatively close together or extremely
small. Execute _ again using for initial
estimates the numbers from the X- and Yregisters (or perhaps two numbers somewhat
further apart). If the magnitude of the function is
neither a minimum nor constant, the algorithm
will eventually expand its search and find a
more significant result.
Example: Investigate the behavior of the function
f ( x)  3  e
 x / 10
 2e x
2  x
e
as evaluated in the following subroutine.
Keystrokes
|¥
´ b .0
|a
“
Display
000–
001–42,21,.0
002–
43 16
003–
16
'
®
004–
005–
12
34
|x
*
'
2
006–
007–
008–
009–
43 11
20
12
2
*
“
®
|a
“
1
0
010–
011–
012–
013–
014–
015–
016–
20
16
34
43 16
16
1
0
Program mode.
e
x
.
Bring x-value into X-register.
x 2e
x
.
2 x
 2e x e
.
Bring x-value into X-register.
232
Appendix D: A Detailed Look at _
Keystrokes
÷
'
Display
017–
018–
10
12
+
3
019–
020–
40
3
+
|n
021–
022–
40
43 32
 x / 10.
e
 x /10
3e
 2e
 x /10
x 2e
 2e
x
.
x 2e
x
.
Use _ with the following single initial estimates: 10, 1, and 10-20.
Keystrokes
|¥
10 v
´ _ .0
−
)
)
| (| (
´ _.0
−
))
Display
10.0000
Error 8
455.335
48,026,721.85
1.0000
455.4335
Error 8
48,026,721.85
1.0000
1v
´ _.0
−
)
)
| (| (
´ _.0
−
))
1.0000
Error 8
2.1213
2.1471
0.3788
2.1213
Error 8
2.1213
0.3788
‛ “ 20
v
1.0000
Run mode.
Single estimate.
Best x-value.
Previous value.
Function value.
Restore the stack.
Another x-value
Same function value
(an asymptote).
Single estimate.
Best x-value.
Previous value.
Function value.
Restore the stack.
–20
Same x-value.
Same function value
(a minimum).
Single estimate.
Appendix D: A Detailed Look at _
Keystrokes
´ _.0
−
)
)
| (| (
´ _ .0
−
)
)
Display
Error 8
1.0000
1.1250
2.0000
1.0000
Error 8
1.1250
1.5626
2.0000
–20
–20
–20
–20
–16
233
Best x-value.
Previous value.
Function value.
Restore the stack.
Another x-value.
Previous value.
Same function value.
In each of the three cases, _ initially
searched for a root in a direction suggested by
the graph around the initial estimate. Using
10 as the initial estimate, _ found the
horizontal asymptote (value of 1.0000).
Using 1 as the initial estimate, a minimum of
0.3788 at x = 2.1213 was found. Using 10 –20
as the initial estimate, the function was
essentially constant (at a value of 2.0000) for
the small range of x that was sampled.
Finding Several Roots
Many equations that you encounter have more than one root. For this
reason, you will find it helpful to understand some techniques for finding
several roots of an equation.
The simplest method for finding several roots is to direct the root search in
different ranges of x where roots may exist. Your initial estimates specify
the range that is initially searched. This method was used for all examples
in section 13. You can often find the roots of an equation in this manner.
Another method is known as deflation. Deflation is a method by which roots are
"eliminated" from an equation. This involves modifying the equation so that the
first roots found are no longer roots, but the rest of the roots remain roots.
If a function f(x) has a value of zero at x = a, then the new function
f(x)/(x – a) will not approach zero in this region (if a is a simple root of
f(x) = 0). You can use this information to eliminate a known root. Simply
234
Appendix D: A Detailed Look at _
add a few program lines at the end of your function subroutine. These lines
should subtract the known root (to 10 significant digits) from the x-value
and divide this difference into the function value. In many cases the root
will be a simple one, and the new function will direct _ away from
the known root. On the other hand, the root may be a multiple root. A
multiple root is one that appears to be present repeatedly, in the following
sense: at such a root, not only does the graph of f(x) cross the x-axis, but its
slope (and perhaps the next few higher-order derivatives) also equals zero.
If the known root of your equation is a multiple root, the root is not
eliminated by merely dividing by the factor described above. For example,
the equation
f(x) = x(x – a)3 = 0
has a multiple root at x = a (with a multiplicity of 3). This root is not
eliminated by dividing f(x) by (x – a). But it can be eliminated by dividing
by (x – a)3.
Example: Use deflation to help find the roots of
60x4 – 944x3 + 3003x2 + 6171x – 2890 = 0.
Using Horner's method, this equation can be rewritten in the form
(((60x – 944)x + 3003)x + 6171)x – 2890 = 0.
Program a subroutine that evaluates the polynomial.
Keystrokes
|¥
´ CLEAR
M
´b2
6
0
*
9
4
4
Display
000000001-42,21, 2
002–
6
003–
0
004–
20
005–
9
006–
4
007–
4
Program mode.
Appendix D: A Detailed Look at _
Keystrokes
*
3
0
0
3
+
*
6
1
7
1
+
*
2
8
9
0
|n
Display
008–
009–
010–
011–
012–
013–
014–
015–
016–
017–
018–
019–
020–
021–
022–
023–
024–
025–
026–
027–
235
30
20
3
0
0
3
40
20
6
1
7
1
40
20
2
8
9
0
30
43 32
In Run mode, key in two large, negative initial estimates (such as -10 and
-20) and use _ to find the most negative root.
Keystrokes
|¥
10 “ v
20 “
´_2
O0
))
Display
Run mode.
–10.0000
–20
–1.6667
–1.6667
4.0000 –06
Initial estimates.
First root.
Stores root for deflation.
Function value near zero.
236
Appendix D: A Detailed Look at _
Return to Program mode and add instructions to your subroutine to
eliminate the root just found.
Keystrokes
|¥
|‚|
‚
®
l0
÷
Display
000026–
027–
028–
029–
030–
30
34
45 0
30
10
Program mode.
Line before n.
Brings x into X-register.
Divides by (x – a), where
a is known root.
Now use the same initial estimates to find the next root.
Keystrokes
|¥
10 “ v
20 “
´_2
O1
))
Display
4.0000 -06
–10.0000
–20
0.4000
0.4000
0.0000
Run mode.
Same initial estimates.
Second root.
Stores root for deflation.
Deflated function value.
Now modify your subroutine to eliminate the second root.
Keystrokes
|¥
|‚|
‚
®
l1
÷
Display
000030–
10
031–
032–
033–
034–
34
45 1
30
10
Program mode.
Line before n.
Brings x into X-register.
Deflation for second root.
Appendix D: A Detailed Look at _
237
Again, use the same initial estimates to find the next root.
Keystrokes
|¥
10 “ v
20 “
´_2
O2
))
Display
0.0000
–10.0000
–20
8.4999
8.4999
–1.0929 –07
Run mode.
Same initial estimates.
Third root.
Stores root for deflation.
Deflated function value near
zero.
Now change your subroutine to eliminate the third root.
Keystrokes
|¥
|‚|
‚
®
l2
÷
Display
000–
034–
035–
036037–
038–
10
45
34
2
30
10
Program mode.
Line before n.
Brings x into X-register.
Deflation for third root.
Find the fourth root.
Keystrokes
|¥
10 “ v
20 “
´_2
O3
))
Display
–1.0929 –07
–10.0000
–20
8.5001
8.5001
–0.0009
Same initial estimates.
Fourth root.
Stores root for reference.
Deflated function value
near zero.
238
Appendix D: A Detailed Look at _
Using the same initial estimates each
time, you have found four roots for this
equation involving a fourth-degree
polynomial. However, the last two
roots are quite close to each other and
are actually one root (with a
multiplicity of 2). That is why the root
was not eliminated when you tried
deflation once at this root. (Round-off
error causes the original function to
have small positive and negative values
for values of x between 8.4999 and
8.5001; for x = 8.5 the function is
exactly zero.)
In general, you will not know in advance the multiplicity of the root you are
trying to eliminate. If, after you have attempted to eliminate a root, _
finds that same root again, you can proceed in a number of ways:

Use different initial estimates with the deflated function in an
attempt to search for a different root.

Use deflation again in an attempt to eliminate a multiple root. If
you do not know the multiplicity of the root, you may need to
repeat this a number of times.

Examine the behavior of the deflated function at x-values near the
known root. If the function's calculated values cross the x-axis
smoothly, either another root or a greater multiplicity is indicated.

Analyze the original function and its derivatives algebraically. It
may be possible to determine its behavior for x-values near the
known root. (A Taylor series representation, for example, may
indicate the multiplicity of a root.)
Limiting the Estimation Time
Occasionally, you may desire to limit the time used by _ to find a
root. You can use two possible techniques to do this – counting iterations
and specifying a tolerance.
Appendix D: A Detailed Look at _
239
Counting Iterations
While searching for a root, _ typically samples your function at least
a dozen times. Occasionally, _ may need to sample it one hundred
times or more. (However, _ will always stop by itself.) Because your
function subroutine is executed once for each estimate that is tried, it can
count and limit the number of iterations. An easy way to do this is with an
I instruction to accumulate the number of iterations in the Index
register (or other storage register).
If you store an appropriate number in the register before using _, your
subroutine can interrupt the _ algorithm when the limit is exceeded.
Specifying a Tolerance
You can shorten the time required to find a root by specifying a tolerable
inaccuracy for your function. Your subroutine should return a function
value of zero if the calculated function value is less than the specified
tolerance. This tolerance that you specify should correspond to a value that
is negligible for practical purposes or should correspond to the accuracy of
the computation. This technique eliminates the time required to define the
estimate more accurately than is justify by the problem. The example on
page 224 uses this method.)
For Advanced Information
In the HP-15C Advanced Functions Handbook, additional, advanced
techniques and applications for using _ are presented. These topics
include:
 Using _ with polynomials.
 Solving a system of equations.
 Finding local extremes of a function.
 Using _ for financial problems.
 Using _ in Complex mode.
 Solving an equation for its complex roots.
Appendix E
A Detailed Look at f
Section 14, Numerical Integration, presented the basic information you need
to use f This appendix discusses more intricate aspects of f that are
of interest if you use f often.
How f Works
The f algorithm calculates the integral of a function f(x) by computing a
weighted average of the function's values at many values of x (known as
sample points) within the interval of integration. The accuracy of the result
of any such sampling process depends on the number of sample points
considered: generally, the more sample points, the greater the accuracy. If
f(x) could be evaluated at an infinite number of sample points, the algorithm
could – neglecting the limitation imposed by the inaccuracy in the
calculated function f(x) – provide an exact answer.
Evaluating the function at an infinite number of sample points would take a
very long time (namely, forever). However, this is not necessary, since the
maximum accuracy of the calculated integral is limited by the accuracy of
the calculated function values. Using only a finite number of sample points,
the algorithm can calculate an integral that is as accurate as is justified
considering the inherent uncertainty in f(x).
The f algorithm at first considers only a few sample points, yielding
relatively inaccurate approximations. If these approximations are not yet as
accurate as the accuracy of f(x) would permit, the algorithm is iterated (that
is, repeated) with a larger number of sample points. These iterations
continue, using about twice as many sample points each time, until the
resulting approximation is as accurate as is justified considering the
inherent uncertainty in f(x).
240
Appendix E: A Detailed Look at f
241
The uncertainty of the final approximation is a number derived from the
display format, which specifies the uncertainty for the function.* At the end
of each iteration, the algorithm compares the approximation calculated
during that iteration with the approximations calculated during two previous
iterations. If the difference between any of these three approximations and
the other two is less than the uncertainty tolerable in the final
approximation, the algorithm terminates, placing the current approximation
in the X-register and its uncertainty in the Y-register.
It is extremely unlikely that the errors in each of three successive
approximations – that is, the differences between the actual integral and the
approximations – would all be larger than the disparity among the
approximations themselves. Consequently, the error in the final
approximation will be less than its uncertainty. † Although we can't know the
error in the final approximation, the error is extremely unlikely to exceed
the displayed uncertainty of the approximation. In other words, the
uncertainty estimate in the Y-register is an almost certain ―upper bound‖ on
the difference between the approximation and the actual integral.
Accuracy, Uncertainty, and Calculation Time
The accuracy of an f approximation does not always change when you
increase by just one the number of digits specified in the display format,
though the uncertainty will decrease. Similarly, the time required to
calculate an integral sometimes changes when you change the display
format, but sometimes does not.
Example: The Bessel function of the first kind, of order four, can be
expressed as
J 4 ( x) 
*
†
1

 0 cos4θ  x sinθ dθ
π
The relationship between the display format, the uncertainly in the function, and the uncertainty in the
approximation to its integral are discussed later in this appendix.
Provided that f(x) does not vary rapidly, a consideration that will be discussed in more detail later in this
appendix.
242
Appendix E: A Detailed Look at f
Calculate the integral in the expression for J4 (1),

 cos(4  sin )d
0
First, switch to Program mode and key in a subroutine that evaluates the
function f(θ) = cos (4θ – sin θ).
Keystrokes
|¥
´ CLEAR M
´b0
4
*
®
[
\
|n
Display
000000001–42,21,
002–
003–
004–
005–
006–
007–
008–
43
Program mode.
0
4
20
34
23
30
24
32
Now, switch to Run mode and key the limits of integration into the X- and
Y-registers. Be sure the trigonometric mode is set to Radians, and set the
display format to i 2. Finally, press ´ f0 to calculate the integral.
Keystrokes
|¥
0v
|$
|R
´i2
´f0
®
Display
0.0000
3.1416
3.1416
3.14
7.79
1.45
00
-03
-03
Run mode.
Keys lower limit into Y-register.
Keys upper limit into X-register.
Sets the trigonometric mode to
Radians.
Sets display format to i 2.
Integral approximated in i 2.
Uncertainty of i 2
approximation.
Appendix E: A Detailed Look at f
243
The uncertainty indicates that the displayed digits of the approximation
might not include any digits that could be considered accurate. Actually,
this approximation is more accurate than its uncertainty indicates.
Keystrokes
®
´ CLEAR u
(hold)
Display
7.79
-03
7785820888
Return
display.
approximation
to
All 10 digits of i 2
approximation.
The actual value of this integral, correct to five significant digits, is
7.7805×10-3. Therefore, the error in this approximation is about
(7.7858  7.7805)×10-3 = 5.3×10-6. This error is considerably less than the
uncertainty, 1.45×10-3 The uncertainty is only an upper bound on the error
in the approximation; the actual error will generally be smaller.
Now calculate the integral in i 3 and compare the accuracy of the
resulting approximation to that of the i 2 approximation.
Keystrokes
´i3
Display
7.786
–03
))
3.142
00
´f0
7.786
–03
®
1.448
–04
®
7.786
–03
Changes display format
to i 3.
Rolls down stack until
upper limit appears in Xregister.
Integral approximated in
i3
Uncertainty of i 3
approximation.
Returns approximation to
display.
´ CLEAR u
(hold)
7785820888
All 10 digits of i
3 approximation.
244
Appendix E: A Detailed Look at f
All 10 digits of the approximations in i 2 and i 3 are identical: the
accuracy of the approximation in i 3 is no better than the accuracy in
i 2 despite the fact that the uncertainty in i 3 is less than the
uncertainty in i 2. Why is this? Remember that the accuracy of any
approximation depends primarily on the number of sample points at which
the function f(x) has been evaluated. The f algorithm is iterated with
increasing numbers of sample points until the disparity among three
successive approximations is less than the uncertainty derived from the
display format. After a particular iteration, the disparity among the
approximations may already be so much less than the uncertainty that it
would still be less if the uncertainty were decreased by a factor of 10. In
such cases, if you decreased the uncertainty by specifying one more digit in
the display format, the algorithm would not have to consider additional
sample points, and the resulting approximation would be identical to the
approximation calculated with the larger uncertainty.
If you calculated the two preceding approximations on your calculator, you
may have noticed that it did not take any longer to calculate the integral in
i 3 than in i 2. This is because the time to calculate the integral of
a given function depends on the number of sample points at which the
function must be evaluated to achieve an approximation of acceptable
accuracy. For the i 3 approximation, the algorithm did not have to
consider more sample points than it did in i 2, so it did not take any
longer to calculate the integral.
Often, however, increasing the number of digits in the display format will
require evaluating the function at additional sample points, so that
calculating the integral will take more time. Now calculate the same integral
in i 4.
Keystrokes
´i4
Display
7.7858
))
3.1416
´f0
7.7807
–03 i 4 display.
00 Rolls down stack until upper
limit appears in X-register.
–03 Integral approximated in i 4.
Appendix E: A Detailed Look at f
245
This approximation took about twice as long as the approximation in i
3 or i 2. In this case, the algorithm had to evaluate the function at
about twice as many sample points as before in order to achieve an
approximation of acceptable accuracy. Note, however, that you received a
reward for your patience: the accuracy of this approximation is better, by
almost two digits, than the accuracy of the approximation calculated using
half the number of sample points.
The preceding examples show that repeating the approximation of an
integral in a different display format sometimes will give you a more
accurate answer, but sometimes it will not. Whether or not the accuracy is
changed depends on the particular function, and generally can be
determined only by trying it.
Furthermore, if you do get a more accurate answer, it will come at the cost
of about double the calculation time. This unavoidable trade-off between
accuracy and calculation time is important to keep in mind if you are
considering decreasing the uncertainty in hopes of obtaining a more
accurate answer.
The time required to calculate the integral of a given function depends not
only on the number of digits specified in the display format, but also, to a
certain extent on the limits of integration. When the calculation of an
integral requires an excessive amount of time, the width of the interval of
integration (that is, the difference of the limits) may be too large compared
with certain features of the function being integrated. For most problems,
however, you need not be concerned about the effects of the limits of
integration on the calculation time. These conditions, as well as techniques
for dealing with such situations, will be discussed later in this appendix.
Uncertainty and the Display Format
Because of round-off error, the subroutine you write for evaluating f(x)
cannot calculate f(x) exactly, but rather calculates
fˆ ( x)  f ( x)  1 ( x),
where δ1 (x) is the uncertainty of f(x) caused by round-off error. If f(x)
relates to a physical situation, then the function you would like to integrate
is not f(x) but rather
246
Appendix E: A Detailed Look at f
F ( x)  f ( x)  δ2 ( x) ,
where δ2(x) is the uncertainty associated with f(x) that is caused by the
approximation to the actual physical situation.
Since f ( x)  fˆ ( x)  δ1 ( x) , the function you want to integrate is
F ( x)  fˆ ( x)  δ1 ( x)  δ2 ( x)
F ( x)  fˆ ( x)  δ( x) ,
or
where δ(x) is the net uncertainty associated with f(x).
Therefore, the integral you want is
b
b
a
a
 F ( x) dx   [ fˆ ( x)  δ( x)]dx


b

b
fˆ ( x) dx   ( x) dx
a
a
 I 
b
where I is the approximation to
 F ( x) dx
and ∆ is the uncertainty
a
associated with the approximation. The f algorithm places the number I
in the X-register and the number ∆ in the Y-register.
The uncertainty δ(x) of fˆ ( x) , the function calculated by your subroutine, is
determined as follows. Suppose you consider three significant digits of the
function's values to be accurate, so you set the display format to i 2.
The display would then show only the accurate digits in the mantissa of a
function's values: for example, 1.23
–04.
Since the display format rounds the number in the X-register to the
number displayed, this implies that the uncertainty in the function's values
is ± 0.005×10–4 = ± 0.5×10–2×10–4 = ± 0.5×10-6. Thus, setting the display
Appendix E: A Detailed Look at f
247
format to i n or ^ n, where n is an integer,* implies that the
uncertainty in the function’s values is
δ(x)  0.5  10 n  10m( x )
 0.5 10 n m( x)
In this formula, n is the number of digits specified in the display format and
m(x) is the exponent of the function's value at x that would appear if the
value were displayed in i display format.
The uncertainty is proportional to the factor 10m(x), which represents the
magnitude of the function's value at x. Therefore, i and ^ display
formats imply an uncertainty in the function that is relative to the function's
magnitude.
Similarly, if a function value is display in • n, the rounding of the
display implies that the uncertainty in the function's values is
δ(x)  0.5 10n.
Since this uncertainty is independent of the function's magnitude, •
display format implies an uncertainty that is absolute.
Each time the f algorithm samples the function at a value of x, it also
derives a sample of δ(x), the uncertainty of the function's value at x. This is
calculated using the number of digits n currently specified in the display
format and (if the display format is set to i or ^) the magnitude
m(x) of the function's value at x. The number Δ, the uncertainty of the
approximation to the desired integral, is the integral δ (x):
*
Although i 8 or 9 generally results in the same display as i 7, it will result in a smaller
uncertainty of a calculated integral. (The same is true for the ^ format.) A negative value for n (which
can be set by using the Index register) will also affect the uncertainty of an f calculation. The minimum
value for n that will affect uncertainty is -6. A number in RI less than -6 will be interpreted as -6.
248
Appendix E: A Detailed Look at f
Δ
b
 δ(x) dx
a


b
[0.5 10 n  m( x) ] dx .
a
This integral is calculated using the samples of δ(x) in roughly the same
ways that the approximation to the integral of the function is calculated
using the samples of fˆ ( x) .
Because Δ is proportional to the factor 10 -n, the uncertainty of an
approximation changes by about a factor of 10 for each digit specified in the
display format. This will generally not be exact in i or ^ display
format, however, because changing the number of digits specified may
require that the function be evaluated at different sample points, so that
δ(x) ~ 10m(x) would have different values.
Note that when an integral is approximated in • display format, m(x) =
0 and so the calculated uncertainty in the approximation turns out to be
Δ = 0.5×10-n (b – a).
Normally you do not have to determine precisely the uncertainty in the
function. (To do so would frequently require a very complicated analysis.)
Generally, it's more convenient to use i or ^ display format if the
uncertainty in the function's values can be more easily estimated as a
relative uncertainty. On the other hand, it’s more convenient to use •
display format if the uncertainty in the function’s values can be more easily
estimated as an absolute uncertainly. • display format may be
inappropriate to use (leading to peculiar results) when you are integrating a
function whose magnitude and uncertainty have extremely small values
within the interval of integration. Likewise, i display format may be
inappropriate to use (also leading to peculiar results) if the magnitude of the
function becomes much smaller than its uncertainty. If the results of
calculating an integral seem strange, It may be more appropriate to calculate
the integral in the alternate display format.
Appendix E: A Detailed Look at f
249
Conditions That Could Cause Incorrect Results
Although the f algorithm in the HP-15C is one of the best available, in
certain situations it – like nearly all algorithms for numerical integration –
might give you an incorrect answer. The possibility of this occurring is
extremely remote. The f algorithm has been designed to give accurate
results with almost any smooth function. Only for functions that exhibit
extremely erratic behavior is there any substantial risk of obtaining an
inaccurate answer. Such functions rarely occur in problems related to actual
physical situations; when they do, they usually can be recognized and dealt
with in a straightforward manner.
As discussed on page 240, the f algorithm samples the function f(x) at
various values of x within the interval of integration. By calculating a
weighted average of the function's values at the sample points, the
algorithm approximates the integral of f(x).
Unfortunately, since all that the algorithm knows about f(x) are its values at
the sample points, it cannot distinguish between f(x) and any other function
that agrees with f(x) at all the sample points. This situation is depicted in the
illustration on the next page, which shows (over a portion of the interval of
integration) three of the infinitely many functions whose graphs include the
finitely many sample points.
250
Appendix E: A Detailed Look at f
With this number of sample points, the algorithm will calculate the same
approximation for the integral of any of the functions shown. The actual
integrals of the functions shown with solid lines are about the same, so the
approximation will be fairly accurate if f(x) is one of these functions.
However, the actual integral of the function shown with a dashed line is
quite different from those of the others, so the current approximation will be
rather inaccurate if f(x) is this function.
The f algorithm comes to know the general behavior of the function by
sampling the function at more and more points. If a fluctuation of the
function in one region is not unlike the behavior over the rest of the interval
of integration, at some iteration the algorithm will likely detect the
fluctuation. When this happens, the number of sample points is increased
until successive iterations yield approximations that take into account the
presence of the most rapid, but characteristic, fluctuations.
For example, consider the approximation of


0
xe  x dx.
Appendix E: A Detailed Look at f
251
Since you’re evaluating this integral numerically, you might think (naively
in this case, as you'll see) that you should represent the upper limit of
99
integration by 10 – which is virtually the largest number you can key into
the calculator. Try it and see what happens.
Key in a subroutine that evaluates the function f(x) = xe-x
Keystrokes
|¥
´b1
“
'
*
|n
Display
000001-42,21, 1
0021
6
00312
00420
00543 32
Program mode.
Set the calculator to Run mode. Then set the display format to i 3 and
key the limits of integration into the X- and Y-registers.
Keystrokes
|¥
Display
Run mode.
Sets display format to i 3.
´i 3
0v
0.000
00
‛ 99
1
99
´f1
0.000
00
Keys lower limit into Yregister.
Keys upper limit into Xregister.
Approximation of integral.
The answer returned by the calculator is clearly incorrect, since the actual
integral of f(x) = xe-x from 0 to  is exactly 1. But the problem is not that
you represented  by 1099 since the actual integral of this function from 0 to
1099 is very close to 1. The reason you got an incorrect answer becomes
apparent if you look at the graph of f(x) over the interval of integration:
252
Appendix E: A Detailed Look at f
The graph is a spike very close to the origin. (Actually, to illustrate f(x) the
width of the spike has been considerably exaggerated. Shown in actual scale
over the interval of integration, the spike would be indistinguishable from
the vertical axis of the graph.) Because no sample point happened to
discover the spike, the algorithm assumed that f(x) was identically equal to
zero throughout the interval of integration. Even if you increased the
number of sample points by calculating the integral in i 9, none of the
additional sample points would discover the spike when this particular
function is integrated over this particular interval. (Better approaches to
problems such as this are mentioned at the end of the next topic, Conditions
That Prolong Calculation Time.)
You've seen how the f algorithm can give you an incorrect answer when
f(x) has a fluctuation somewhere that is very uncharacteristic of the
behavior of the function elsewhere. Fortunately, functions exhibiting such
aberrations are unusual enough that you are unlikely to have to integrate
one unknowingly.
Functions that could lead to incorrect results can be identified in simple
terms by how rapidly it and its low-order derivatives vary across the
interval of integration. Basically, the more rapid the variation in the
function or its derivatives, and the lower the order of such rapidly varying
derivatives, the less quickly will the f algorithm terminate, and the less
reliable will the resulting approximation be.
Appendix E: A Detailed Look at f
253
Note that the rapidity of variation in the function (or its low-order
derivatives) must be determined with respect to the width of the interval of
integration. With a given number of sample points, a function f(x) that has
three fluctuations can be better characterized by its samples when these
variations are spread out over most of the interval of integration than if they
are confined to only a small fraction of the interval. (These two situations
are shown in the next two illustrations.) Considering the variations or
fluctuations as a type of oscillation in the function, the criterion of interest
is the ratio of the period of the oscillations to the width of the interval of
integration: the larger this ratio, the more quickly the algorithm will
terminate, and the more reliable will be the resulting approximation.
254
Appendix E: A Detailed Look at f
In many cases you will be familiar enough with the function you want to
integrate that you’ll know whether the function has any quick wiggles
relative to the interval of integration. If you're not familiar with the
function, and you have reason to suspect that it may cause problems, you
can quickly plot a few points by evaluating the function using the
subroutine you wrote for that purpose.
If for any reason, after obtaining an approximation to an integral, you have
reason to suspect its validity, there's a very simple procedure you can use to
verify it: subdivide the interval of integration into two or more adjacent
subintervals, integrate the function over each subinterval, then add the
resulting approximations. This causes the function to be sampled at a brand
new set of sample points, thereby more likely revealing any previously
hidden spikes. If the initial approximation was valid, it will equal the sum of
the approximations over the subintervals.
Conditions That Prolong Calculation Time
In the preceding example (page 251), you saw that the algorithm gave an
incorrect answer because it never detected the spike in the function. This
happened because the variation in the function was too quick relative to the
width of the interval of integration. If the width of the interval were smaller,
you would get the correct answer; but it would take a very long time if the
interval were still too wide.
For certain integrals such as the one in that example, calculating the integral
may be unduly prolonged because the width of the interval of integration is
too large relative to certain features of the functions being integrated.
Consider an integral where the interval of integration is wide enough to
require excessive calculation time but not so wide that it would be
calculated incorrectly. Note that because f(x) = xe-x approaches zero very
quickly as x approaches , the contribution to the integral of the function at
large values of x is negligible. Therefore, you can evaluate the integral by
replacing , the upper limit of integration, by a number not so large as 10 99,
say 103.
Appendix E: A Detailed Look at f
Keystrokes
0v
Display
0.000
‛3
1
´f1
®
1.000
1.824
00
03
00
-04
255
Keys lower limit into
Y-register.
Keys upper limit into
X-register.
Approximation to integral.
Uncertainty of
approximation.
This is the correct answer, but it took almost 60 seconds. To understand
why, compare the graph of the function over the interval of integration,
which looks about identical to that shown on page 252, to the graph of the
function between x = 0 and x = 10.
By comparing the two graphs, you can see that the function is "interesting"
only at small values of x. At greater values of x, the function is
"uninteresting," since it decreases smoothly and gradually in a very
predictable manner.
As discussed earlier, the f algorithm will sample the function with
higher densities of sample points until the disparity between successive
approximations becomes sufficiently small. In other words, the algorithm
samples the function at increasing numbers of sample points until it has
sufficient information about the function to provide an approximation that
changes insignificantly when further samples are considered.
256
Appendix E: A Detailed Look at f
If the interval of integration were (0, 10) so that the algorithm needed to
sample the function only at values where it was interesting but relatively
smooth, the sample points after the first few iterations would contribute no
new information about the behavior of the function. Therefore, only a few
iterations would be necessary before the disparity between successive
approximations became sufficiently small that the algorithm could
terminate with an approximation of a given accuracy.
On the other hand, if the interval of integration were more like the one
shown in the graph on page 252, most of the sample points would capture
the function in the region where its slope is not varying much. The few
sample points at small values of x would find that values of the function
changed appreciably from one iteration to the next. Consequently the
function would have to be evaluated at additional sample points before the
disparity between successive approximations would become sufficiently
small.
In order for the integral to be approximated with the same accuracy over
the larger interval as over the smaller interval, the density of the sample
points must be the same in the region where the function is interesting. To
achieve the same density of sample points, the total number of sample
points required over the larger interval is much greater than the number
required over the smaller interval. Consequently, several more iterations are
required over the larger interval to achieve an approximation with the same
accuracy, and therefore calculating the integral requires considerably more
time.
Because the calculation time depends on how soon a certain density of
sample points is achieved in the region where the function is interesting, the
calculation of the integral of any function will be prolonged if the interval
of integration includes mostly regions where the function is not interesting.
Fortunately, if you must calculate such an integral, you can modify the
problem so that the calculation time is considerably reduced. Two such
techniques are subdividing the interval of integration and transformation of
variables. These methods enable you to change the function or the limits of
integration so that the integrand is better behaved over the interval(s) of
integration. (These techniques are described in the HP-15C Advanced
Functions Handbook.)
Appendix E: A Detailed Look at f
257
Obtaining the Current Approximation
to an Integral
When the calculation of an integral is requiring more time than you care to
wait, you may want to stop and display the current approximation. You can
obtain the current approximation, but not its uncertainty.
Pressing ¦ while the HP-15C is calculating an integral halts the
calculation, just as it halts the execution of a running program. When you
do so, the calculator stops at the current program line in the subroutine you
wrote for evaluating the function, and displays the result of executing the
preceding program line. Note that after you halt the calculation, the current
approximation to the integral is not the number in the X-register nor the
number in any other stack register. Just as with any program, pressing
¦ again starts the calculation from the program line at which it was
stopped.
The f algorithm updates the current approximation and stores it in the
LAST X register after evaluating the function at each new sample point. To
obtain the current approximation, therefore, simply halt the calculator,
single-step if necessary through your function subroutine until the calculator
has finished evaluating the function and updating the current
approximation. Then recall the contents of the LAST X register, which are
updated when the n instruction in the function subroutine is executed.
While the calculator is updating the current approximation, the display is
blank and does not show running. (While the calculator is executing your
function subroutine, running is displayed.) Therefore, you might avoid
having to single-step through your subroutine by halting the calculator at a
moment when the display is blank.
In summary, to obtain the current approximation to an integral, follow the
steps below.
1.
Press ¦ to halt the calculator, preferably while the display is
blank.
2.
When the calculator halts, switch to Program mode to check the
current program line.

If that line contains the subroutine label, return to Run
mode and view the LAST X register (step 3).
258
Appendix E: A Detailed Look at f

If any other program line is displayed, return to Run mode
and single-step (Â) through the program until you
reach a n instruction (keycode 43 32) or line 000 (if
there is no n). (Be sure to hold the  key down
long enough to view the program line numbers and
keycodes.)
Press | K to view the current approximation. If you want to
continue calculating the final approximation, press − +
¦. This refills the stack with the current x-value and restarts
the calculator.
3.
For Advanced Information
The HP-15C Advanced Functions Handbook explores more esoteric aspects
of f and its applications. These topics include:

Accuracy of the function to be integrated.

Shortening calculation time.

Calculating difficult integrals.

Using f in Complex mode.
Appendix F
Batteries
Batteries
The HP-15C is shipped with two 3 Volt CR2032 Lithium batteries. Battery
life depends on how the calculator is used. If the calculator is being used to
perform operations other than running programs, it uses much less power.
Low-Power Indication
A battery symbol () shown in the upper-left corner of the display when
the calculator is on signifies that the available battery power is running low.
When the battery symbol begins flashing, replace the battery as soon as
possible to avoid losing data.
Use only a fresh battery. Do not use rechargeable batteries.
Warning
There is the danger of explosion if the battery is
incorrectly replaced. Replace only with the same or
equivalent type recommended by the manufacturer.
Dispose of used batteries according to the manufacturer’s
instructions. Do not mutilate, puncture, or dispose of
batteries in fire. The batteries can burst or explode,
releasing hazardous chemicals. Replacement battery is a
Lithium 3V Coin Type CR2032.
Installing New Batteries
To prevent memory loss, never remove two old batteries at the same time.
Be sure to remove and replace the batteries one at a time.
259
260
Appendix F: Batteries
To install new batteries, use the following procedure:
1. With the calculator turned off, slide the battery cover off.
2. Remove the old battery.
3. Insert a new CR2032 lithium battery, making sure that the positive
sign (+) is facing outward.
4. Remove and insert the other battery as in steps 2 through 3. Make sure
that the positive sign (+) on each battery is facing outward.
5. Replace the battery cover.
Note: Be careful not to press any keys while the battery is
out of the calculator. If you do so, the contents of Continuous
Memory may be lost and keyboard control may be lost (that
is, the calculator may not respond to keystrokes).
6. Press = to turn on the power. If for any reason Continuous Memory
has been reset (that is, if its contents have been lost), the display will
show Pr Error. Pressing any key will clear this message.
Appendix F: Batteries
261
Verifying Proper Operation (Self-Tests)
If it appears that the calculator will not turn on or otherwise is not operating
properly, use the following procedures to access the test system;
1) Turn the calculator off.
2) Press and HOLD the | and v keys (keep both keys held
down for the next step).
3) Press the = key (while both | and v keys are held down
from Step 2 above).
4) Release the = key.
5) Release the | and v keys.
You will be presented with a main test screen that displays the following:
1.L 2.C 3.H

Press 1 to perform the LCD test (all LCD segments will be turned on).
Press any key to exit

Press 2 to perform the checksum test and see the copyright messages.
Press any key to go from one screen to the next until you return to the
main test screen.

Press 3 to perform the keyboard test. You then need to press EVERY
key on the keyboard until all the keys have been pressed at least once
(the screen will progressively turn off). You can press the keys in any
order and any number of times. Once all the keys have been pressed
and the screen is clear, press on any key to return to the test screen.
Press = to exit the test system. This will also turn the calculator off.
If the calculator detects an error at any point, it will display an error
message.
If you still experience difficulty, write or telephone Hewlett-Packard at an
address or phone number listed on the web at: www.hp.com/support.
Function Summary and Index
= Turns the
calculator's display on
and off (page 18). It is
also used in resetting
Continuous Memory
(page 63), changing the
digit separator (page
61), and in various tests
of the calculator's
operation (pages 261).
Complex
Functions
} Real exchange
imaginary. Activates
Complex mode
(establishing an
imaginary stack) and
exchanges the real and
imaginary X-registers
(page 124).
V Used to enter
complex numbers.
Activates Complex
mode (establishing an
imaginary stack)
(page 121). Also used
with m to indirectly
dimension matrices
(page 174). (For Index
register functions, refer
to Index Register
Control keys,
page 263.)% Displays
the contents of the
imaginary X-register
while the key is held
(page 124).
F 8 Sets flag 8,
which activates
Complex mode
(page 121).
" 8 Clears flag 8,
deactivating Complex
mode (page 121).
Conversions
; Converts polar
magnitude r and angle
θ in X- and Y-registers
respectively to
rectangular x- and ycoordinates (page 31).
For operation in
Complex mode, refer to
page 134.
h Converts
decimal hours (or
degrees) to hours,
minutes, seconds (or
degrees, minutes,
seconds) (page 27).
À Converts hours,
minutes, seconds (or
degrees, minutes,
seconds) to decimal
hours (or degrees) (page
27).
r Converts
degrees to radians
(page 27).
d Converts
radians to degrees (page
27).
Digit Entry
v Enters a copy
of number in X-register
: Converts x, y
(display) into Y-register;
rectangular coordinates used to separate multiple
placed in X- and Ynumber entries (pages
registers respectively to 22, 37).
polar magnitude r and
angle θ (page 30). For “ Change sign of
operation in Complex
number or exponent of
mode, refer to page 134. 10 in display (pages 19,
124).
262
263
Function Summary and Index
‛ Enter exponent;
next digits keyed in are
exponents of 10
(page 19).
0 through 9 digit
keys (page 22).
. Decimal point
(page 22)
Display Control
• Selects fixed
point display mode
(page 58).
i Selects scientific
notation display mode
(page 59).
^ Selects
engineering notation
display mode (page
59).
Mantissa. Pressing
´ CLEAR u
displays all 10 digits of
the number in the Xregister as long as the
u key is held
down (page 60). It also
clears any partial key
sequences
(page 19).
Hyperbolic
Functions
P[
P\
P] Compute
hyperbolic sine,
hyperbolic cosine, or
hyperbolic tangent,
respectively (page 28).
% Indirect
operations. Used to
address another storage
register through RI for
purposes of storage,
recall, storage,
arithmetic, and program
loop control (page
107). Also used with
m to allocate
storage registers (page
215).
H [, H
\, H ]
Compute inverse
Logarithmic and
hyperbolic sine, inverse
Exponential
hyperbolic cosine, or
Functions
inverse hyperbolic
tangent, respectively
N Computes natural
(page 28).
logarithm (page 28).
Index Register
Control
' Natural
antilogarithm. Raises e
to power of number in
V Index register (RI). display (X-register)
(page 28).
Storage register for:
indirect program
execution – branching
with t and G,
looping with I and
s – indirect flag
control, and indirect
display format control
(page 107). Also used
to enter complex
numbers and activate
Complex mode (page
121).
o Computes
common logarithm
(base 10) (page 28).
@ Common
antilogarithm. Raises
10 to power of number
in display (X-register)
(page 28).
Y Raises number in
Y-register to power of
264
Function Summary and Index
number in display (Xregister) (enter y, then
x). Causes the stack to
drop (page 29).
Mathematics
-+-÷
Arithmetic operators;
cause the stack to drop
(page 29).
¤ Computes square
root x (page 25).
f Integrate.
Computes the definite
integral of f(x), with the
expression f(x) defined
by the user in a labeled
subroutine (page 194).
Matrix Functions
m Dimensions a
matrix of a given name
{A to E, V}
(page 141).
} Stores or recalls
matrix elements using
the row and column
numbers in the Y- and
X-registers (page 146).
O and
l> { A
to E } Stores or
recalls matrices for the
specified matrix (pages
142, 147).
O and l
< Stores or
< Designates the recalls descriptor of the
matrix into which the
x Computes
result matrix (page
result of certain matrix 148).
the square of x
operations is placed
(page 25).
(page 148).
l m {A
! Calculates the
through E, V}
U User mode. Row Recalls the dimensions
factorial (n!) of x or
Gamma function (Γ) of and column numbers in of the given matrix into
R0 and R1 are
(1 + x) (page 25).
the Y- (row) and Xautomatically
(column) registers
incremented each time (page 142).
∕ Computes
O or l {A
reciprocal (page 25).
(For matrix use, refer to to E, %} is pressed ∕ Inverts the matrix
Matrix Functions, page (page 144).
whose descriptor is
264.)
displayed and places
O and l { A the result in the
to E, %} Stores or specified result matrix.
$ Places value of π
recalls matrix elements The descriptor of the
in display (page 24).
using the row and
result matrix is then
_ Solves for real column numbers in R0 displayed (page 150).
and R1 (pages 144,
root of a function f(x),
with the expression for 146).
+ - * Adds,
subtracts, or multiplies
f(x) defined by the user
the corresponding
in a labeled subroutine O | and l
| {A to E, % elements of two
(page 180).
Function Summary and Index
matrices or of one
matrix and a scalar.
Stores in result matrix
(page 152-155).
complex transform.
to ZP (page164).
> {0 through 9}
Matrix operations.
> 9 Calculates
determinant of matrix
specified in X-register
(also does LU
decomposition of the
matrix) (page 150).
265
"partitioned form" (Z P)
(page 162).
> 4 Transpose X ~ T 0 T 5
to XT (page 150).
T 6 Conditional
÷ For two matrices,
tests for matrix
multiplies inverse of
descriptors in the X- or
> 5 Transpose
matrix in X by matrix
X- and Y-registers.
multiply: Y and X to
in Y. For only one
~ and T 0 (x ≠
YTX (page 154).
matrix, if matrix in Y,
0) test the quantity in
divides all elements of > 6 Calculates
the X-register for zero.
matrix by scalar in X; if residuals in result
Matrix descriptors are
matrix in X, multiplies matrix (page 159).
considered nonzero.
each element of inverse
T 5 (x = y) and
of matrix by the scalar > 7 Calculates
T
in Y. Stores in result
6 (x ≠ y) test if the
row norm of matrix
matrix (pages 152descriptors in X and Y
specified in X-register
155).
are the same. The result
(page 150).
affects program
execution: skip (one
“ changes sign of
> 8 Calculates
line) if false (page
all elements in matrix
Frobenius norm of
174).
specified in X-register
matrix specified in X(page 150).
register (page 150).
> 0 Dimensions
all matrices to 0×0
(page 143).
Number Alteration
a Yields absolute
value of number in
display (page 24).
q Leaves only
fractional portion of
> 1 Sets row and c Transforms
number in display
column numbers in R0 matrix stored in
and R1 to 1 (page 143). "partitioned form" (Z P) (X-register) by
to "complex form" (ZC) truncating integer
portion (page 24).
> 2 Complex
(page 162).
transform: ZP to
‘ Leaves only
(page 164).
p Transforms
integer portion of
matrix stored in
C
> 3 inverse
"complex form" (Z ) to number in display (X-
266
Function Summary and Index
register) by truncating
the blue function
fractional portion (page printed below that key
24).
(page 18).
different items taken x
at a time, and causes the
stack to drop (page 47).
(For matrix use, refer to
Matrix Functions keys,
& Rounds mantissa For other prefix keys,
refer to Display Control page 264.)
of entire (10-digit)
number in X-register to keys (page 263),
match display format
Storage keys (page
Stack
(page 24).
267), and the
Programming Summary
Manipulation
and Index (page 269).
Percentage
k Percent. Computes
x% (value in display) of
number in the Yregister (page 29).
Unlike most twonumber functions, k
does not drop the stack.
CLEAR u
Cancels any prefix
keystrokes and partially
entered instructions
such as ´ i
(page 19). Also
displays the complete
10-digit mantissa of the
number in the display
(page 60).
∆ Percent difference.
Computes percent of
change between number
Probability
in Y-register and
number in display
c Combination.
(page 30). Does not
drop the stack.
Computes the number
of possible sets of y
different items taken x
Prefix Keys
at a time, and causes the
stack to drop (page 47).
´ Pressed before a
(For matrix use, refer to
function key to select
Matrix Functions keys,
the gold function
page 264.)
printed above that key
(page 18).
p Permutation.
Computes the number
| Pressed before a
of possible different
function key to select
arrangements of y
® Exchanges
contents of X- and Ystack registers (page
34).
X X-register
exchange. Exchanges
contents of X-register
with those of any other
named storage register.
Used with V , %,
digit, or . digit
address (page 42).
} Real exchange
imaginary. Exchanges
the contents of the real
and imaginary Xregisters and activates
Complex mode (page
124).
) Rolls down
contents of stack (page
34).
( Rolls up contents
of stack (page 34).
Function Summary and Index
` Clears contents
of display (X-register)
to zero (page 21).
− In Run mode:
removes the last digit
keyed in, or clears the
display (if digit entry
has been terminated)
(page21).
Statistics
z Accumulates
numbers from X- and
Y-registers into storage
registers R2 through R7
(page 49).
w Removes numbers
in X- and Y-registers
from storage registers
R2 through R7 for
correcting z
accumulations (page
52).
and correlation
coefficient. Computes
estimated value of y (ŷ)
for a given value of x
by least squares method
and places result in Xregister. Computes the
correlation coefficient,
r, of the accumulated
data and places result in
Y-register (page 55).
Storage
O Store. Stores a
copy of a number into
the storage register
specified {0 to 9, .0 to
.9, V, %} (page
42). Also used for
storage register
arithmetic: new register
contents = old register
contents { +, -,
L Linear Regression. *, ÷ } display
Computes the y(page 44).
intercept and slope for
the linear function best
l Recall. Recalls a
approximating the
copy of the number
accumulated data. The from the storage
value of the y-intercept
register specified {0 to
is placed in the X9, .0 to .9, V, % }
register; the value of the
(page 42). Also used
slope is placed in the Yfor storage register
register (page 54).
arithmetic: new display
# Random
number. Yields a
pseudorandom number
as generated from a
’ Computes mean of seed stored using O
# (page 48).
x- and y-values
accumulated by z
(page 53).
CLEAR ∑ Clears
contents of the statistics
S Computes sample registers (R2 to R7)
(page 49).
standard deviations of
x- and y-values
accumulated by z
(page 53).
j Linear estimate
267
= old display {+,
- *, ÷}
register contents (page
44).
CLEAR Q Clears
contents of all storage
registers to zero
(page 43).
K Recalls into
the display the number
present before the
previous operation
(page 35).
268
Function Summary and Index
Trigonometry
D Sets decimal
Degrees mode for
trigonometric
functions—indicated by
absence of GRAD or
RAD annunciator (page
26). Not operative for
complex trigonometry.
R Sets Radians
mode for trigonometric
functions—indicated by
RAD annunciator (page
26).
[, \, ]
Compute sine, cosine,
or tangent, respectively,
of number in display
(X-register) (page 26).
g Sets Grads mode
for trigonometric
functions—indicated by
GRAD annunciator
(page 26) Not operative
for complex
trigonometry.
, , {, /
Compute arc sine, arc
cosine, or arc tangent,
respectively, of number
in display (X-register)
(page 26).
Programming Summary and Index
¥ Program/Run
mode. Sets the
calculator to Program
mode (PRGM
annunciator on) or Run
mode (PRGM
annunciator cleared)
(page 66).
W Displays current
status of calculator
memory (number of
registers dedicated to
data storage, the
common pool, and
program memory)
(page 215).
W Displays current
status of calculator
memory (number of
registers dedicated to
data storage, the
common pool, and
program memory)
(page 215).
− Back arrow. In
Program mode, deletes
displayed instruction
from program memory.
All subsequent
instructions are moved
up (page 83).
b Label. Used with
the label designations
below to denote the
start of a program
routine (page 67).
a label designator
(listed above) or V to
transfer the position of
the calculator to the
ABCÁE 0 designated label. If it is
1 2 3 4 5 6 7 8 9 .0 .1 .2 a program instruction,
.3 .4 .5 .6 .7 .8 .9 Label program execution
continues. If it is not a
designations. When
program instruction,
preceded by b,
define the beginning of only the position
change occurs (page
a program routine
90). If a negative
(page 67). Also used
number is stored in RI,
(without b) to
t V will effect a
initiate execution of a
transfer to a line
specific routine
number (page 109).
(page 69).
U Activates and
deactivates User mode,
which exchanges the
primary (white) and
gold alternate functions
(A through E) of
the top left five
functions (page 69).
User mode also affects
the matrix use of O
or l {A
throughE , %}
User mode
automatically
increments R0 (row
number) or R1 (column
number) for storage or
recall of matrix
elements (page 144).
t Go to. Used with
269
t “ nnn Go to
line number. Positions
calculator to the
existing line number
specified by nnn. Not
programmable (page
82).
G Go to subroutine.
Used with a label
designator (listed
above) or start the
execution of a given,
labeled routine. Can be
used both in a program
and from the keyboard
(in Run mode). A
n instruction
transfers execution back
to the first line
270
Programming Summary and Index
following the G
(page 101).
calculator to return to
line 000 and halt
execution (if running)
(page 68). If in a
‚ Back step.
Moves calculator back subroutine, merely
returns to line after
one or more lines in
program memory. (Also G (page 101).
scrolls in Program
mode.) Displays line
F Set flag (= true).
number and contents of Sets designated flag (0
previous program line
to 9). Flags 0 through 7
(page 83).
are user flags, flag 8
signifies Complex
mode, and flag 9
 Single step. In
Program mode: moves signifies an overflow
calculator forward one condition (page 92).
or more lines in
" Clear flag (=
program memory. In
Run mode: displays and false). Clears
designated flag (0 to 9)
executes the current
program line, then steps (page 92).
to next line to be
? Is flag set? Tests
executed (page 82).
for designated flag. If
set, program execution
© Pause. Halts
continues; If cleared,
program execution for
program execution
about 1 second to
skips one line before
display contents of Xcontinuing (page 92).
register, then resumes
execution (page 68).
£ ~ T {0
through 9} Conditional
¦ Run/Stop.
tests. Each test
Begins program
compares value in Xexecution from current register against 0 or
line number in program value in Y-register as
memory. Stops
indicated. If true,
execution if program is calculator executes
running (page 68).
instruction in next line
of program memory. If
n Return. Causes
false, calculator skips
one line in program
memory before
resuming execution
(page 91). ~ and
T 0, 5, and 6 are
also valid for complex
numbers and matrix
descriptors (pages 132.
174).
T0x≠0
T1x>0
T2x<0
T3x≥0
T4x≤0
T5x=y
T6x≠y
T7x>y
T8x<y
T9x≥y
s Decrement and
skip if equal to or less
than. Decrements
counter value in given
register as stipulated.
Skips one program line
if new counter value is
equal to or less than
specified test value
(page 109).
I Increment and
skip if greater than.
Increments counter
value in given register
as stipulated. Skips one
program line if new
counter value is greater
than specified test value
(page 109).
Subject Index
Page numbers in bold type indicate primary references; page numbers in
regular type indicate secondary references.
A ___________________________________________
Abbreviated key sequences, 78
Absolute value (a ), 24
Allocating memory, 42, 213-219
Altering program lines, 83
Annunciators,
complex, 121
list of, 60
PRGM, 32, 66
trigonometric, 26
Antilogarithms, common and natural, 28
Arithmetic operation, 29, 37
Asymptotes, horizontal, 230
Automatic incrementing of row and column numbers, 143
B ___________________________________________
Back-stepping (‚), 83
Bacterial population example, 41
Battery life, 259
Battery replacement, 260, 259-260
Bessel functions, 195, 197
Branching,
conditional, 91, 98, 177, 192
indirect, 108-109, 112-114, 115
simple, 90
C ___________________________________________
C annunciator, 99, 121
Can volume and area example, 70-74
Chain calculations, 22-23, 38
Changing signs, 19
in Complex mode, 124-125
271
272
Subject Index
in matrices, 177
“, 19
Clearing
blinking in display, 100
complex numbers, 125-127
display, 21
memory, 63
operations, 20-21
overflow condition, 45, 61
prefix keys, 19
statistics registers, 49
Coefficient matrix, 156
Combinations function (c), 47
Common pool, 213
Complex arithmetic example, 132
Complex conjugate, forming, 125
Complex matrix,
inverting, 162, 164, 165
multiplying, 162, 164, 166
storing elements, 161
transforming, 162, 164
Complex mode, 120-121
activating, 99, 120-121, 133
deactivating, 121
mathematics functions in, 131
stack lift in, 124
Complex numbers,
clearing, 125-127
converting polar and rectangular forms, 133-135
entering, 121, 127, 128-129
storing and recalling, 130
Conditionals, indirect, 109-111, 112, 116
Conditional tests, 91, 98, 192
in Complex mode, 132
with matrix descriptors, 174
Constant matrix, 156
Constants,
calculations with, 39-42
using in arithmetic calculations, 35, 39-42
Subject Index
Continuous Memory,
duration of, 62
resetting (clearing), 63
what it retains, 43, 48, 58, 61, 62
Conventions, handbook, 18
Conversions,
degrees and radians, 27
polar and rectangular coordinates, 30-31
time and angle, 26-27
Correcting accumulated statistics data, 52
Correlation coefficient, find the (j), 55-56
\,{, 26
Counters in program loops, 98, 112-114
Crocus example, 43
Cumulative calculations, 41
D ___________________________________________
Data storage, 42
Data storage pool, 213-214
Debt payment example, 95
Decimal point, 22
Decimal point display, 61
Deflation, 233, 234, 237
D, 26
Determinant, 150
Digit entry, 22
in Complex mode, 121, 125, 127, 128-129
termination, 22, 36, 209
Digit separator display, 61
m, 76-77, 215-217
Disabling stack lift, 36
Display (See also X-register),
blinking, 100
clearing, 21
error messages, 61
full mantissa, 60
in Complex mode, 121
Display format, 58-59, 61
effect on ´ 200, 241, 244, 245-249
Do if True rule, 92, 192
273
274
Subject Index
s 109-111, 112, 116
E ___________________________________________
‛, 19
Electrical circuit example, 169-171
Enabling stack lift, 36
^, 59
Engineering notation, 59
v, 12, 33-34, 36
effect on digit entry, 22, 29
effect on stack movement, 37, 41
Entering data for statistical analysis, 49
Error
conditions, 205-208
display, 61
stops, 78
Errors,
with f, 203-204
with _, 187, 192, 193
Euclidean norm (See Frobenius norm)
Exchanging the real and imaginary stacks, 124
Exponential function (See Power function)
Exponents, 19, 20
F____________________________________________
´, 18
Factorial function (!), 25
Falling stone example, 14
•, 58
Fixed decimal notation, 58
Flag tests, 92, 98
Flag 8, 99
Flag 9, 100
Format, handbook, 2, 18
Fractional portion (q), 24
Frobenius norm, 150, 177
Functions, nonprogrammable, 80
Functions, one-number, 22, 25
Functions, primary and alternate, 18
Functions, two-number, 22, 29
Subject Index
G ___________________________________________
|, 18
Gamma function (!), 25
g, 26
G, 101
t, 90, 97, 98
t “, 82
H ___________________________________________
Horner's Method, 79, 181 Hyperbolic
functions, 28
I ____________________________________________
Imaginary stack,
clearing the, 124
creation of, 121-123, 133
display of, 124
stack lift of, 124
Index register
arithmetic, 108, 112
display format control, 109, 114, 115, 116
exchange with X-register, 108, 112
flag control, 109, 115
loop control, 107, 109-111
storage and recall, 107, 111, 115
Indirect addressing, 106-108, 115
Initialization, 87
Instructions, 74
Integer portion (‘), 24
Integrate function (f), 194-204
accuracy of, 200-203, 240, 241-245
algorithm for, 196, 240-241, 249-251, 255-256
display format with, 245-249
execution time for, 196, 200, 244, 245, 254-256
memory usage, 204
obtaining an approximation for, 257-258
problems with erratic functions, 249-254
programmed, 203-204
recursive use of, 203
subroutines for, 194-195
275
276
Subject Index
uncertainty in, 202-203, 240-244, 245-249
Interchanging functions (See User mode)
Interference, radio and television, 271
Intermediate results, 22, 38
Interpolation, using j, 57
I, 109-111, 116
Iterations using I and s, 111
K ___________________________________________
Keycodes, 74-75
Keying in
chain calculations, 22
exponents, 19-20
one-number functions, 22
two-number functions, 22, 29
L ____________________________________________
Labels, 67, 77, 90, 97
LAST X register, 35
in matrix functions, 174-176
operations saved by, 212
putting constants in, 39-40
to correct statistics data, 52
Linear equations, solving with matrices, 138, 156
Linear estimation (j), 55-56
Linear regression (L), 54
Loading the stack with constants, 39, 41
Logarithmic functions, common and natural, 28
Loop control number, 109, 116
Looping, 90, 98
Low-power indication, 62, 259
LU decomposition, 148, 155, 156, 160
Łukasiewicz, Jan, 32
M ___________________________________________
Mantissa, displaying full 10 digits, 60
Matrix
complex, 160-163
copying, 149
descriptors, 139, 147, 160, in RI, 173-174
Subject Index
dimensioning, 140, 142, 142, 174
dimensions, displaying, 142, 147
equation, complex, 168
memory, 140, 171
name (See Matrix descriptors)
partitioned, 161, 164
Matrix elements,
accessing individually, 145-147
displaying, 144
storing and recalling, 143-144, 147, 149, 176
Matrix functions,
using RI, 173-174
using registers, 173
arithmetic, 153
conditional, 177
inverse, 150, 154
multiplication, 154
one-matrix, 149-151
programmed, 176-177
reciprocal, 150
residual, 159
row norm, 150, 177
summary, 177-179
transpose, 150, 151, 154
Mean (’). 53
W, 215
Memory
allocation, 76, 215-217
availability, 75-77, 213, 215
configuration, initial, 75-76
distribution, 75, 213-214
limitations, 75, 77, 217
requirements for advanced functions, 218-219
requirements for programming, 218
stack (See Stack)
status display, 215
registers in, 213-215
Metal box dimensions example, 189-191
Minima, finding with _, 230
Modes, trigonometric, 26
277
278
Subject Index
Multiple roots, 234
N ___________________________________________
Negative numbers, 19
in Complex mode, 124-125
Nested calculations, 38
Neutral operations, 211
Nonprogrammable functions, 80
Normalizing statistics data, 50
null display, 144, 149
Numerical integration, 194-204
O ___________________________________________
=,
and off, 18
to reset Continuous Memory, 63
to set decimal point display, 61
Overflow condition, 45, 61, 100
P____________________________________________
¥, 66, 68
Pause (©), 68
Percent difference (∆), 29
Percentage functions, 29-30
Permutations function (p), 47
Phasor notation, 133
Pi, 24
Polar coordinates, 30, in Complex mode, 133-135
Power function (y), 29
Prefix keys, 19
PRGM annunciator, 66, 82
Program
control, indirect, 107, 109-111
data entry techniques, 69-70
end, 68, 77
entering, 66-68
labels, 67, 77
loading, 66
loop counters, 109, 112-114, 116
mode, 66, 68, 86
Subject Index
position, changing, 82, 86
running, 68-69
starting, 69
stops, 68, 78
Program execution, 69
after G, 101
after t, 97
after overflow, 100
after test, 92
from or through labels, 78-79
Program lines (instructions), 67, 74
deleting, 83, 86
inserting, 83, 86
Program memory, 67, 70, 75, 217-219
automatic real location, 217-218
clearing, 67
moving in, 67
Q ___________________________________________
Quadratic equation, solving, 181
R ___________________________________________
R0 and R1, using to access matrix elements, 143, 146, 176
R, 26
Radioisotope example, 93-94
Random number generator (#), 48
Random number storage and recall, 48
Recall arithmetic, 44
Recalling accumulated statistics data, 50
Recalling numbers (l), 42, 44, with matrices, 144, 149, 176
Reciprocal (∕), 25, with matrix, 150
Rectangular coordinates, 31, in Complex mode, 133-135
Registers, converting, 215-217
Reset Continuous Memory, 63
Residual, 159
Result matrix, 147, 148, 150, 152
Return (n), 68, 77
Returns, pending, 101, 105, 192, 204
Reverse Polish Notation, 32
} 124, 127
279
280
Subject Index
Rice yield example, 50-56
Ridget hurling example, 184-186, 224-226
Roll down, 34
Roll up, 34
Roots, eliminating, 233, 234, 237
Roots, meaningless, 188, 191
Rounding (&), 24
Rounding in the display, 59
Round-off errors, 52, 60, with _, 223, 237
Row norm, 150, 177
Run/Stop (¦), 68, 91
running display, 69, 147, 182
S ___________________________________________
Scalar operations, 151-153
i, 58
Scientific notation, 58
Scrolling, 82
Secant line calculation example, 102
Self-tests, 261
Service information, 267-270
Shear stress example, 227-228
[, ,, 26
Sine integral example, 198-199
Single-stepping (Â), 82, 85
Skip if True rule, 110
Slope, finding the, 54
_, 180-181
accuracy, 222-226, specifying, 238
algorithm, 182, 187-188, 220-222, 230-231
conditions necessary for, 221-222
constant function value with, 187, 189
execution time, 238
illegal math routine with, 187-188
initial estimates with, 181, 188-192, 221, 233, 237
memory usage, 193
nonzero minimum of function with, 187
programmed, 192
recursive use of, 193
restrictions on, 193
Subject Index
using as a conditional test, 192
using functions with discontinuities, 227
using functions with poles, 227
using functions with several roots, 233-238
with no root, 186-188, 192, 229
Square root (¤), 25
Squaring (x), 25
Stack
contents, with f, 197, 202
drop, 33, 38
lift, 33, 36, 38, 44, 209-211
manipulation functions, 33-34, in Complex mode, 131
imaginary, 120-125
used to access matrix elements, 146-147
Stack-disabling operations, 210
Stack-enabling operations, 210-211
Stack movement, 32, 33-37
in matrix functions, 174-176
with _, 181
Standard deviation (S), 53, sample vs. population, 53
Star example, 40
Statistics, accumulation of data (z), 49
Statistics, correction of accumulated data (z), 52
Statistics functions,
combinations, 47
correlation coefficient, 55
linear estimation, 55
linear regression, 54
mean, 53
permutations, 47
probability, 47
standard deviation, 53
Statistics registers, 49-50
Status indicators, 60
Storage and recall (O, l), 42, 43, 44
complex numbers, 130
direct (with V), 106, 107
indirect, 106-107, 111
matrices, 144, 149, 176
matrix elements, 143-144, 147, 149
281
282
Subject Index
Storage arithmetic, 43
Storage registers, 42
allocation, 42, 215-217
arithmetic, 43
clearing, 43
statistics, 42, 49
Subroutine
levels, 102, 105
limits, 102, 105
nesting example, 103
returns, 101, 105
using with _, 180-181, 192
System flags, 92, 99
T____________________________________________
T-register, 32, 33
in matrix functions, 174-176
with f, 202
] /, 26
T, 91
Tracing, 82
Transpose, 150, 151, 154
Trigonometric modes in Complex mode, 121, 134
Trigonometric operations, 26
U ___________________________________________
u display, 176
Uncommitted registers, 213, 215, 217
Underflow,
in any register, 61
storage register arithmetic, 45
with _, 223
User flags, 92
User mode, 69, 79, with matrices, 143, 176
V ___________________________________________
Vector arithmetic, using statistics functions, 57
W __________________________________________
Wrapping, 86, 90
Subject Index
X ___________________________________________
X exchange (X), 42
X exchange Y (®), 34
X-register, 32, 35, 37, 42, 60, 209-210
imaginary, 210, 211
in matrix functions, 141, 156, 175-176
with f, 202
with _, 181, 183, 102, 226
Y ___________________________________________
y-intercept, finding, 54
Y-register, 32, 37
in matrix functions, 141,156, 175-176
with f, 202
_, 181, 183, 192, 226
Z ___________________________________________
Z-register, 32
in matrix functions, 174-176
with f, 202
with _,181, 183, 192, 226
283
Product Regulatory &
Environment Information
Federal Communications Commission Notice
This equipment has been tested and found to comply with the limits for
a Class B digital device, pursuant to Part 15 of the FCC Rules. These
limits are designed to provide reasonable protection against harmful
interference in a residential installation. This equipment generates,
uses, and can radiate radio frequency energy and, if not installed and
used in accordance with the instructions, may cause harmful
interference to radio communications. However, there is no guarantee
that interference will not occur in a particular installation. If this
equipment does cause harmful interference to radio or television
reception, which can be determined by turning the equipment off and
on, the user is encouraged to try to correct the interference by one or
more of the following measures:

Reorient or relocate the receiving antenna.

Increase the separation between the equipment and the receiver.


Connect the equipment into an outlet on a circuit different from
that to which the receiver is connected.
Consult the dealer or an experienced radio or television
technician for help.
Modifications
The FCC requires the user to be notified that any changes or
modifications made to this device that are not expressly approved by
Hewlett-Packard Company may void the user’s authority to operate the
equipment.
284
Declaration of Conformity for Products Marked with FCC
Logo, United States Only
This device complies with Part 15 of the FCC Rules. Operation is
subject to the following two conditions: (1) this device may not cause
harmful interference, and (2) this device must accept any interference
received, including interference that may cause undesired operation.
If you have questions about the product that are not related to this declaration,
write to
Hewlett-Packard Company
P. O. Box 692000, Mail Stop 530113
Houston, TX 77269-2000
For questions regarding this FCC declaration, write to
Hewlett-Packard Company
P. O. Box 692000, Mail Stop 510101
Houston, TX 77269-2000
or call HP at 281-514-3333
To identify your product, refer to the part, series, or model number located on
the product.
Canadian Notice
This Class B digital apparatus meets all requirements of the Canadian
Interference-Causing Equipment Regulations.
Avis Canadien
Cet appareil numérique de la classe B respecte toutes les exigences du
Règlement sur le matériel brouilleur du Canada.
European Union Regulatory Notice
Products bearing the CE marking comply with the following EU
Directives:
• Low Voltage Directive 2006/95/EC
• EMC Directive 2004/108/EC
• Ecodesign Directive 2009/125/EC, where applicable
CE compliance of this product is valid if powered with the correct CEmarked AC adapter provided by HP.
Compliance with these directives implies conformity to applicable
harmonized European standards (European Norms) that are listed in
the EU Declaration of Conformity issued by HP for this product or
product family and available (in English only) either within the product
documentation or at the following web site: www.hp.eu/certificates
(type the product number in the search field).
The compliance is indicated by one of the following conformity
markings placed on the product:
For non-telecommunications products and
for EU harmonized telecommunications
products, such as Bluetooth® within power
class below 10mW.
xxxx*
For EU non-harmonized telecommunications
products (If applicable, a 4-digit notified
body number is inserted between CE and !).
Please refer to the regulatory label provided on the product.
The point of contact for regulatory matters is:
Hewlett-Packard GmbH, Dept./MS: HQ-TRE, Herrenberger Strasse
140, 71034 Boeblingen, GERMANY.
286
Japanese Notice
Korean Notice
Disposal of Waste Equipment by Users in Private
Household in the European Union
This symbol on the product or on its packaging
indicates that this product must not be disposed of
with your other household waste. Instead, it is your
responsibility to dispose of your waste equipment by
handing it over to a designated collection point for
the recycling of waste electrical and electronic
equipment. The separate collection and recycling of
your waste equipment at the time of disposal will
help to conserve natural resources and ensure that it
is recycled in a manner that protects human health
and the environment. For more information about
where you can drop off your waste equipment for
recycling, please contact your local city office, your
household waste disposal service or the shop where
you purchased the product.
Chemical Substances
HP is committed to providing our customers with information about the
chemical substances in our products as needed to comply with legal
requirements such as REACH (Regulation EC No 1907/2006 of the
European Parliament and the Council). A chemical information report
for this product can be found at: www.hp.com/go/reach.
Perchlorate Material - special handling may apply
This calculator's Memory Backup battery may contain perchlorate and
may require special handling when recycled or disposed in
California.
288
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