MAT137Y1a | Manualzz

MAT137Y1a.doc
Lecture #1 – Tuesday, September 9, 2003
SETS
•
•
non-example: A = {paintings that are beautiful}
example: A = {natural numbers divisible by 4}
Notation
•
•
4 ∈ A – “4 is in A”
3 ∉ A – “4 is not in A”
•
B = {natural numbers divisible by 7}
A ∪ B = {natural numbers divisible by 4 or 7}
•
– “union of A and B”
= {4,7,8,12,14,...}
•
A ∩ B = {natural numbers divisible by 4 and 7} – “intersection of A and B”
FACTS ABOUT REAL NUMBERS
•
•
•
•
N = natural numbers = {1,2,3,4,...}
Z = integers = {...,−3,−2,−1,0,1,2,3,...}
Q = rational numbers = {1,2,3,4,...}
R = real numbers = all this and more (+ irrational numbers)
Geometrically
-5
•
-4
-3
-2
-1
0
1
2
3
4
any point on the line represents a real number
Intervals
•
•
•
[− 1,2] = {x ∈ R : −1 ≤ x ≤ 2}
(− 1,2) = {x ∈ R : −1 < x < 2}
[0, ∞ )
Ordering (Inequalities)
•
If a, b are real numbers, then exactly one of the following is true:
Important Properties
•
•
If a 0 , then ac < bc
If a bc
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ab
5
The Real Line
MAT137Y1a.doc
Distance/Absolute Values
a, if a ≥ 0
•
a = distance from a to 0 =
•
a − b = distance between a and b
− a, if a < 0
Triangle Inequality
•
•
a+b ≤ a + b
analogous to triangle theorem from geometry
a
b
c
c ≤ a+b c
•
Lecture #2 – Thursday, September 11, 2003
TRIANGLE INEQUALITY
•
a + b ≤ a + b , a, b ∈ R
Proof (A Proof By Cases)
1) If a, b both ≥ 0
• Then a + b ≥ 0
a+b = a+b = a + b
•
2) If a > 0, b < 0, a ≥ b
•
•
3) If
4)
Then a + b ≥ 0
a+b = a+b
a + b = a −b
•
Because b < 0 , so a + b < a − b
•
•
Then a + b ≥ 0
a+b = a+b = a + b
a > 0, b < 0, a 0
Where is it = 0? x = 1,4,−2
3)
(x − 1)
(x − 4 )
(x + 2 )
-2
x <δ
•
x−c <δ
-
x
0
c-
c
-
+
+
-
-
-
+
-
+
+
+
Product +
∴ x ∈ (− 2,1) ∪ (4, ∞ )
-
+
−δ < 0 < δ
x ∈ (− δ , δ )
– it means that
x
c+
– it means that
x −3 > 5
•
x−3 > 5
-2
3
8
x − 3 < −5
∴ (− ∞,−2) ∪ (8, ∞ )
COORDINATE GEOMETRY
Rectangular Coordinates
y-axis
3
2
1
4
-
Inequalities With Absolute Values
•
1
(3,1)
1 2 3 x-axis
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c −δ < x < c +δ
x ∈ (c − δ , c + δ )
MAT137Y1a.doc
Pythagorean Theorem
(x1, y1)
|y1 – y0|
(x0, y0)
|x1 – x0|
d=
x1 − x 0
2
+ y1 − y 0
2
Line Equations
•
y − y 0 = m(x − x 0 ) – “point-slope” form
y = mx + b – “y-intercept” form
ax + by + c = 0 – standard form
•
If
•
•
y = m1 x + b1
y = m 2 x + b2
,
•
Two lines are parallel if m1 = m 2
•
Two lines are perpendicular if m1 m 2 = −1
Conic Sections
•
•
ax 2 + by 2 + cx + dy + e = 0, a, b, c, d , e are constants
Get ellipses, hyperbolas, parabolas
•
(x − 2)2 + ( y − 3)2 = 9
( x − 2 ) 2 + ( y − 3) 2 = 3
3
(2,3)
3
3
FUNCTIONS
•
A rule (“black box”) that takes an input and produces a single output
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Examples
•
f ( x) = x 2
• Domain of f is the set of legitimate inputs – domain ( f ) = (− ∞, ∞ )
• Range of f is the set of legitimate outputs – range( f ) = [0, ∞ )
•
f ( x) = 9 − x 2
•
domain ( f ) = [− 3,3]
range( f ) = [0,3]
•
Piecewise Defined Functions
•
x 2 if x ≥ 0
− x if x < 0
f (x ) =
GRAPHS AND FUNCTIONS
•
graph ( f ) = {( x, y ) : x ∈ dom( f ), y = f (x )}
Example
f (x ) = x 2 + 1
domain ( f ) = (− ∞, ∞ )
range( f ) = [1, ∞ )
(0,1)
SOME BASIC FUNCTIONS
Polynomials
•
f (x ) = a n x n + a n −1 x n −1 + ... + a1 x + a 0
• ai are constants
•
domain ( f ) = (− ∞, ∞ )
Rational Functions
•
f (x ) =
•
P(x )
, where P, Q are polynomials
Q( x )
domain ( f ) = {x ∈ R : Q( x ) ≠ 0}
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TRIGONOMETRIC FUNCTIONS
(x,y)
1
θ
(1,0)
sin θ = y
cos θ = x
y
tan θ =
x
1
csc θ =
y
1
sec θ =
x
x
cot θ =
y
Example of a Trigonometric Identity
•
sin 2 θ + cos 2 θ = 1 , because x 2 + y 2 = 1
COMBINING FUNCTIONS
•
If f, g are functions
• “sum” – ( f + g )(x ) = f (x ) + g ( x )
•
•
•
( f − g )(x ) = f (x ) − g (x )
“product” – ( f ⋅ g )(x ) = f (x ) ⋅ g (x )
Warning: domain is domain ( f ) ∩ domain (g )
“difference” –
Example
•
Given f (x ) = 1 + x and f (x ) = 1 + x , g ( x ) = − x + 5
domain ( f ) = [− 1, ∞ )
•
domain ( g ) = (− ∞,5]
•
domain ( f + g ) = [− 1,5]
COMPOSITION OF FUNCTIONS (TWO “BLACK BOXES” IN SUCCESSION)
•
•
(f
g )(x ) = f (g ( x ))
Warning: f g ≠ g f
Examples
•
Given f (x ) = x + 3 and g ( x ) = x 2
•
•
•
•
(f
(g
g )(x ) = x 2 + 3
f )(x ) = (x + 3)2
domain ( f g ) = {x ∈ domain ( g ) : g (x ) ∈ domain ( f )}
Given f (x ) = x − 2 and g (x ) = x
•
domain ( f g ) = [4, ∞ )
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Lecture #3 – Tuesday, September 16, 2003
INTRODUCTION TO PROOFS: “PROOFS ARE YOUR FRIENDS”
Mathematical Rigor: To What End?
•
Moral: Need rigor to obtain a correct, elegant solution which doesn’t come from brute force
Proofs: To What End?
•
Informally, a proof is a guarantee that a claim is true
Definitions
•
•
•
Proposition – a sentence that is either true or false
Theorem – a proposition that is guaranteed by a proof
Proof – showing a theorem follows logically from the set of axioms – ex: B follows logically from A if B is
true in all possible worlds where A is true
Examples
1) The acceleration of a rigid body is proportional to force applied.
• False is real world
2) For all integers n, if n > 2 , there are no positive integers a, b, c such that a n + b n = c n
• True
3) For all non-negative integers n, n 2 + n + 41 is prime
• False – n = 41 would make it false (counterexample)
Notation
•
•
∀ – “for all” – universal quantifier
∃ – “there exist”, “for some” – existential quantifier
P
Q
False
False
True
True
False
True
False
True
¬Q
not Q
True
True
False
False
P∧Q
P and Q
False
False
False
True
P∨Q
P or Q
False
True
True
True
I: PROOF BY ENUMERATION
Example 1
•
•
•
Given: Roses are red and violets are blue.
Prove: Roses are red.
Proof: 4th line from the table ( P ∧ Q )
Example 2
•
•
Given: If it rains today, I’ll eat my hat. It rains today.
Prove: I’ll eat my hat.
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P Q
if P, then Q
True
True
False
True
P⇔Q
P iff Q
True
False
False
True
¬Q
¬P
True
True
False
True
MAT137Y1a.doc
•
Proof: 4th line from the table ( P
Q)
Lecture #4 – Thursday, September 18, 2003
II: PROOF BY CONTRAPOSITIVE (“INDIRECT PROOF”)
Example 1
•
If John is at work, then he’s logged in. ( P
•
If John is not logged in, then he’s not at work. ( ¬Q
•
•
Warning: Converse of P Q is Q
P
Moral: Proving contrapositive is logically equivalent to proving the original statement
Q)
¬P )
Example 2
Prove: For any integer n, if n2 is even, then n is even.
Proof:
• Reformulate: If n is odd, then n2 is odd.
1) If n is odd, then n = 2a + 1, a ∈ Z
(
)
2) If n = 2a + 1 , then n 2 = (2a + 1)2 = 4a 2 + 4a + 1 = 2 2a 2 + 2a + 1
3) Because a ∈ Z , 2a 2 + 2a ∈ Z
4) Because 2a 2 + 2a ∈ Z , n2 is odd
QED
Example of A Non-Proof
Prove: The “Theorem” 1 = −1
Proof:
(− 1)(− 1) =
−1 ⋅ −1 =
( −1)
•
1= 1 =
•
•
But a ⋅ b ≠ a ⋅ b for all a and b!
Moral: Justify each step.
2
= −1 QED
Other Classical Errors
•
•
Divide both sides of an equation by a variable – ax = bx a = b – what if x = 0 ?
Divide both sides of an inequality by a variable – ax < bx a < b – what if x ≤ 0 ?
III: PROOF BY CONTRADICTION
•
•
Want to show (WTS): P
Idea: Assume P is not true, then show that a contradiction occurs when combining it with the axioms.
Example
Prove:
2 is irrational.
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Proof:
1) Assume that
2 is rational. That means ∃a, b ∈ Z such that
2=
a
, where a, b have no common
b
factors.
a2
, so 2b 2 = a 2
b2
3) That means a2 is even. So by theorem above, a is even.
4) a = 2k for some k ∈ Z , so a 2 = 4k 2 .
2) Then 2 =
5) So 2b 2 = 4k 2 , so b 2 = 2k 2 .
6) So b2 is even by theorem above.
7) 2 divides both a and b.
8) Contradiction: a, b have no common factors. Thus,
•
2 is irrational.
lemma(s) – smaller theorems used to prove bigger theorems
IV: PROVE BY CASES
•
Warning: It only works if there are a finite number of cases.
Example
Prove: The Triangle Inequality a + b ≤ a + b , a, b ∈ R
•
•
•
Proof:
Case 1: a, b ≥ 0 …
Case 2: a, b < 0 …
Case 3: a ≥ 0, b < 0, a ≥ b …
•
Case 4: a ≥ 0, b < 0, a < b …
V: PROOF BY INDUCTION
•
•
What do you do when there are infinitely many cases?
Example: Universally quantified statements over N – i.e. “for all n ∈ N …”
Principle of Induction
•
If you can prove that: some property holds for n = 0 (“base case”) and if the property holds for n = k , it
holds for n = k + 1 (“inductive step”), then the property holds for all n ∈ N .
Example
Prove: ∀n ∈ N , n 3 − n is divisible by 3.
Definition: For all a, b ∈ Z , “a divides b” ( a | b ) iff ∃q ∈ Z, b = a ⋅ q
Proof:
• “Base Case”: If n = 0 , the 0 3 − 0 = 0 . 3 | 0 because 0 = 3 ⋅ 0
•
“Inductive Step”: Suppose it is true for n = k so 3 | k 3 − k and k 3 − k = 3q for some q ∈ Z .
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•
Now WTS: 3 | (k + 1)3 − (k + 1)
(k + 1)3 − (k + 1)
= k 3 + 3k 2 + 3k + 1 − k − 1
= k 3 − k + 3k 2 + 3k
(
= 3q + 3 k 2 + k
)
So, by definition of divisibility, 3 | (k + 1)3 − (k + 1)
QED
Non-Proof Example
“Theorem”: All iMacs are the same colour.
Prove: All n ∈ N , a set of iMacs of size n is monochromatic.
•
•
Proof:
“Base Case”: n = 1 , then every set of one iMacs is one colour
“Inductive Step”: Suppose true for n = k , WTS for n = k + 1
Let S = {i1 , i 2 ,...i k +1 }
S1 = {i1 , i 2 ,...i k } , S 2 = {i 2 , i 3 ,...i k +1 }
S1 ∩ S 2 = {i 2 ...i k }
•
Since S1, S2 are monochromatic and they intersect in S1 ∩ S 2 , so S is monochromatic. So all iMacs are
monochromatic.
QED
•
But the “Inductive Step” fails for n = 2
VI: STRONG INDUCTION
•
Same as Induction, except n = k is replaced with “if the property holds for n ≤ k …”
Example (Half of “Fundamental Theorem of Arithmetic”)
Prove: Any natural number n, for n > 1 , can be written as a product of primes
Proof (Attempt with Induction):
• “Base Case”: n = 2 . n is a prime.
• “Inductive Step”: Suppose that n = k , k can be written as a product of primes, WTS for n = k + 1
STUCK – primes that divide k don’t divide k + 1
•
•
Proof (Attempt with Strong Induction):
“Base Case”: n = 2 . n is a prime.
“Inductive Step”: Suppose that n ≤ k , k can be written as a product of primes, WTS for n = k + 1
• Case 1: n = k + 1 is prime.
• Case 2: n = k + 1 is not prime, so ∃a, b ∈ N,1 < a, b < k + 1 so that k + 1 = a ⋅ b . Because
a, b ≤ k , so a ⋅ b can be written as a product of primes. Therefore, k + 1 is a product of primes
• Therefore, by strong induction, it is true for all n ∈ N .
QED
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Lecture #5 – Tuesday, September 23, 2003
WHY CALCULUS?
•
Bascially, because the study of natural phenomena involves:
• rates of change
• non-linear phenomena – calculus allows for linear approximation
LIMITS
Basic Idea
1) A (precise) way to take successive approximations and get an exact answer
2) Start with things you can easily compute (ex: lengths of line segments, area or rectangles) and build up to
computations for more complicated objects
Pictures
1) Length of a (non-linear) curve:
2) Volume of a solid:
The Question
What happens to a function value f (x ) as x approaches a fixed number c?
Example
•
•
f (x ) = x − 1
c=2
The function value then approaches 1 – lim(x − 1) = 1
x →1
f(x)
f(x) = x - 1
y=2
x
c=2
Notation
lim f (x ) = L
x→c
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Definition (Attempt 1)
•
f (x ) gets closer to L as x gets close c
Example
•
f (x ) = x 2
f(x)
c = −2
f(x) = x2
L=4
x
c = -2
Technical Note
Doesn’t matter whether the function is defined at x = c or not – ask only about values of x close to c, but not
equal c
Example
•
•
x 2 −1
x +1
c = −1
f(x)
f (x ) =
f(x) = x2 – 1
x+1
x 2 −1
= −2
x → −1 x + 1
c = -1
lim
x
L = -2
Example
Why is something not the limit?
f (x ) = x
•
c =1
f(x)
f(x) = x
L=2
L=2
x
c=1
•
What went wrong?
f ( x ) − 2 > 0.5 if x − 1 < 0.5
Definition (Attempt 2)
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x
f (x ) − 2
1.5
1.25
1.1
0.499
0.249
0.099
MAT137Y1a.doc
•
lim f (x ) = L if no matter what distance ε (on y-axis) I pick, if I say
x→c
f ( x ) − L < ε , then you should be able
to find another distance δ (on x-axis) so that you can guarantee that if x − c < δ , then
y
f (x ) − L < ε
f(x)
L+ε
y=L
L-ε
x
c-δ x=c c+δ
Lecture #6 – Thursday, September 25, 2003
LIMITS (CONTINUED)
Definition (Attempt 3 and Final): Epsilon-Delta Definition of Limits
•
•
Let f (x ) be defined in some interval of the form (c − p, c ) ∪ (c, c + p ) for some p > 0
We say lim f (x ) = L if for all ε > 0 , there exist a δ > 0 such that if x − c < δ then
x→c
∀ε > 0, ∃δ > 0 such that x − c < δ
f (x ) − L < ε –
f (x ) − L < ε
Example
f (x ) = 2 x + 1
If ε is equal to…
ε =1
y
c =1
L=3
(2 x + 1) − 3 < 1
ε =
10
Any ε
y=1
2x − 2 < 1
2 x −1 < 1
x −1 <
1
2
1
c=1
Can choose δ to be…
1
δ=
2
1
δ=
20
ε
δ=
2
x
Technical Note
•
•
•
Warning: The limit of the function at x = c can be different from the function value f (c )
A limit exists exactly when both left hand side and right hand side limits exist and they are equal
1
Another way that limits might not exist – ex: f (x ) = , c = 0
x
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Definition
Left hand limit:
• Assume f is defined on some interval (c − p, c ) for p > 0
Then lim− f (x ) = L if ∀ε > 0, ∃δ > 0 such that c − δ < x < c
•
x →c
Right hand limit:
• Assume f is defined on some interval (c, c + p ) for p > 0
Then lim+ f (x ) = L if ∀ε > 0, ∃δ > 0 such that c < x < c + δ
•
x →c
f (x ) − L < ε
f (x ) − L < ε
Example
Prove (using ε-δ) that lim x 2 = 4
x→ 2
•
Let ε > 0
•
Want: x 2 − 4 < ε – our task is to find a δ so that x − 2 < δ then x 2 − 4 < ε holds
•
Algebra: x 2 − 4 = (x + 2 )(x − 2 )
•
Trick: From now on, assume we’ll take δ < 1
That means we can assume x − 2 < 1 so 1 < x < 3
•
That means
•
Choose δ = min
•
Check: If δ <
(
)
x+2 < x +2
x+2 <5
, so x 2 − 4 = x + 2 x − 2 < 5 x − 2
ε
,1
5
ε
ε
, then x 2 − 4 < 5 x − 2 < 5
<ε
5
5
Q.E.D.
Example
Prove: lim x = 2
x→ 4
•
Want: For ε > 0 ,
•
By the properties of the square root function, we already know δ < 4 – so 0 < x < 8
•
Algebra:
x−4 =
•
2<
x − 2 < ε for small enough x − 4
(x − 4 ) = (
x +2
x +2
)(
x −2
)
x −2
x + 2 because x > 0
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1
x−4
2
x −2 <
•
•
Choose: δ = min (4,2ε )
•
Check: If x − 4 < δ , then
x −2 <
1
1
x − 4 < (2ε ) = ε
2
2
Q.E.D
Example
1
Prove: lim
x→ 2
x
2
1
4
=
1
Want: For ε > 0 , want
•
We already know δ < 2 , 0 < x < 4
•
Algebra:
•
•
•
1
x
2
−
1
x2
−
x
2
−
1
< ε for x close enough to 2 (ie: x − 2 )
4
•
1 4 − x 2 (2 + x )(2 − x )
=
=
4
4x 2
4x 2
1
(2 + x )(2 − x ) = (2 + x )(2 − x ) = (2 + x ) x − 2
=
4
4x 2
4x 2
4x 2
Make additional requirement: δ < 1 , so 1 < x < 3
2+ x < 2+ x < 2+3< 5
1
4x 2
<
1
because x > 1
4
(2 + x )
<
5
4
•
Therefore,
•
So
•
Choose δ = min 1,
•
Check: If x − 2 < δ , then
1
x2
−
4x
2
1 5
< x−2
4 4
4
ε
5
1
x
2
−
1 5
< x−2 < ε
4 4
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Lecture #7 – Tuesday, September 30, 2003
COMPUTATIONS WITH LIMITS
Building Blocks
1)
2)
3)
lim x = c
x →c
lim
x →
0
=
x
c
lim k = k
x→c
Theorem 1
•
If lim f (x ) = L and lim g (x ) = M
x→c
x →c
•
lim ( f ( x ) + g (x )) = L + M
•
lim (αf ( x )) = α ⋅ L
•
lim ( f (x ) ⋅ g (x )) = L ⋅ M
x →c
x→c
x→c
Proof 1
Given ε > 0 , we want to find δ > 0 so that if x − c < δ then
•
Algebra:
f ( x ) + g (x ) − L − M = ( f ( x ) − L ) + (g (x ) − M ) ≤ f ( x ) − L + g (x ) − M
•
We know that lim f (x ) = L and lim g (x ) = M
•
If
•
f ( x ) + g ( x ) − (L + M ) < ε
x→c
x →c
ε
ε
and g (x ) − M < , then we’re in business!
2
2
ε
ε
ε
For
> 0 , we can find δ1 so that f ( x ) − L <
if x − c < δ1 , and we can find δ2 so that g (x ) − M <
2
2
2
if x − c < δ 2
f (x ) − L <
•
To make sure both are true, we pick δ = min (δ1 , δ 2 )
•
Check: If x − c < δ ≤ δ1 , δ 2 , then
f ( x ) + g ( x ) − (L + M ) ≤ f ( x ) − L + g ( x ) − M <
Q.E.D.
ε ε
+ =ε
2 2
Corollaries (theorems that follows (almost) immediately)
1)
lim ( f ( x ) − g ( x )) = L − M
2)
lim (α 1 ⋅ f 1 (x ) + α 2 ⋅ f 2 ( x ) + ... + α n ⋅ f n ( x )) = α 1 L1 + α 2 L2 + ...α n L n if lim f i (x ) = Li
3)
lim ( f 1 ( x ) ⋅ f 2 ( x ) ⋅ ... ⋅ f n (x )) = L1 ⋅ L 2 ⋅ ... ⋅ L n
x →c
x→c
x→c
x→c
4) If P( x ) = polynomial = a n x n + a n −1 x n −1 + ... + a 0 , then lim P( x ) = P(c ) = a n (c )n + a n −1 (c )n −1 + ... + a 0
x→c
Example
•
lim 2 x 2 + 10 x = 28
x→ 2
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•
Now what happens if the polynomial is in the denominator?
Example
•
lim
x→2
1
2
2x + 1
=
1
9
Theorem 2
•
If lim g ( x ) = M , M ≠ 0 , then lim
x→c
x→c
1
1
=
g (x ) M
•
Proof: For ε > 0 we want δ > 0 so that
•
Algebra:
•
Choose δ1 so that g (x ) − M <
1
1
−
<ε
g (x ) M
M − g (x )
1
1
1
−
=
=
g (x ) − M
g (x ) M
g (x )M
g ( x )M
M
2
. If x is in the region, then
g (x ) − M
1
2
g (x ) − M <
= 2 g (x ) − M
2
g ( x )M
M
M
2
If I choose δ1 so that g (x ) − M <
•
M2
ε
2
1
1
1
−
<
g (x ) − M
2
g (x ) M
M
2
Check: If x − c < δ , then
•
<
1
M
⋅ε
2
M
2
2
=ε
2
Lecture #8 – Thursday, October 2, 2003
Examples
•
•
lim
1
2
=
2x + 1
1
lim 3 = −1
x → −1 x
x →3
1
19
Theorem 3
•
If lim f (x ) = L and lim g (x ) = M , M ≠ 0 , then lim
x→c
x→c
x→c
f (x ) L
=
g (x ) M
Page 17 of 33
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•
Proof:
lim
x→c
f (x )
1
= f (x ) ⋅
g (x )
g (x )
f (x )
1
= lim f ( x ) ⋅
g (x ) x →c
g (x )
= lim f ( x ) ⋅ lim
x →c
= L⋅
x →c
1
g (x )
1
L
=
M M
Example
•
•
2x 2 + x
lim
2
x −3
x→ 2
=
10
= 10
1
The hard case: What happens if the denominator approaches 0?
Example
•
lim
•
lim
x →1
1
DNE
x −1
x 2 −1
=2
x →1 x − 1
Theorem 4
•
If lim f (x ) = L, L ≠ 0 and lim g (x ) = 0 , then lim
x→c
•
x →c
Proof: By contradiction
•
Suppose lim
x→c
x→c
f (x )
=M
g (x )
lim f (x ) = lim
x→c
•
Then
f (x )
does not exist
g (x )
x →c
= lim
x →c
f (x )
⋅ g (x )
g (x )
f (x )
⋅ lim g ( x )
g ( x ) x →c
= M ⋅0 = 0
•
Contradiction because lim f (x ) ≠ 0 . So lim
x→c
x→c
f (x )
DNE.
g (x )
Moral
If you’re asked to do a limit computation
f (x )
, you should proceed in several steps:
g (x )
1) See if the denominator is equal to 0.
2) If NO, plug it in at x = c . Done.
If YES, check if numerator limit is 0.
3) If NO, then the limit does not exist. Done.
If YES, do algebra to simplify the expression and go back to Step 1.
Page 18 of 33
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Example
•
•
x−2
lim
x−2
x→4
lim
x→ 4
DNE (by Theorem 4)
x2 + 4
=2
x+2
CONTINUITY
Idea
•
A function is continuous if you can draw the graph “without lifting your pencil”
Example
(
c
)
Definition
•
•
Suppose f is defined on (c − p, c + p ) . The function f is continuous at c if:
•
The limit lim f (x ) exist
•
the limit lim f (x ) = f (c )
x→ c
x→c
How can a function not be continuous?
f (c )
• Case A: The limit exist, but
• Case B: The limit doesn’t exist
Example of Case A
•
c
Example of Case B – “Essential Discontinuity”
•
Different ways it could happen
Page 19 of 33
“removable discontinuity”
MAT137Y1a.doc
•
The one-side limits exist, but they don’t agree
c
•
One of the side limits exist, but only one
0, x ≤ 0
f (x ) =
1
sin
,x>0
x
•
Neither of the one sided limits exist
0, x ∈ Q
f (x ) =
1, x ∉ Q
Examples
•
Functions that are continuous:
• Polynomials: lim P( x ) = P(c )
x→c
•
Absolute value (by Building Blocks): lim x = c
•
Square root: lim x = c for c > 0
x→c
x →c
Theorem 1
•
If f and g are both continuous at c, then f + g , f − g , αf , f ⋅ g ,
f
if g (c ) ≠ 0 are continuous.
g
Example
•
h( x ) = x + 3 x +
1
is continuous at any x > 0
x +1
Another Definition For Continuity
•
f is continuous at c if ∀ε > 0 , ∃δ > 0 such that if x − c < δ then
f ( x ) − f (c ) < ε
Theorem 2
•
If g ( x ) is continuous at c and f is continuous at g (c ) , then f g is continuous at c
Page 20 of 33
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Example
g (x ) = x 2 + 1
g(x) = x2 + 1
f (x ) = x
f g = x 2 +1
g(x) = x2 + 1
g(c) = 2
f g(1)
c=1
•
g(1) = 2
Proof: Let ε > 0 . We want to find δ > 0 so that if x − c < δ then
f ( g (x )) − f (g (c )) < ε
•
•
Do one function at a time
Choose δ1 > 0 so that if t − g (c ) < δ1 , then
•
Choose δ > 0 so that if x − c < δ , then g (x ) − g (c ) < δ1 because g is continuous at c
•
g (c )
f (t ) − f (g (c )) < ε because f is continuous at
Check: If x − c < δ then g (x ) − g (c ) < δ1 then
f ( g (x )) − f (g (c )) < ε
Example
1
(continuous, c ≠ 0 )
x
•
If f (x) = x , g (x ) = x 2 + 1 , h( x ) =
•
k (x ) = h g f (x )
1
is also continuous for any c
=
2
x +1
Detecting Discontinuities
•
Definition: f is continuous from the left if lim− f (x ) = f (c ) ; f is continuous from the right if
lim+ f (x ) = f (c )
x →c
x →c
x=c
Another Way To Say A Function Is Continuous
•
f is continuous at x = c iff
•
lim− f ( x ) exists and is equal to f (c )
x→c
•
lim f (x ) exists and is equal to f (c )
x→c +
Page 21 of 33
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Example
Determine the discontinuities of the function:
2 x − 1 if x < 1
g ( x ) = 0 if x = 1
1
x2
•
•
if x > 1
Certainly the function is continuous at any point in (− ∞,1) or any point in (1, ∞ ) . So the only point in
question is c = 1 .
1
lim g (x ) = lim= 2 x − 1 = 1 and lim+ g (x ) = lim+ 2 = 1 which is not equal to g (1) = 0 . So this function is
x →1=
x →1
x →1
x →1 x
not continuous. c = 1 is removable discontinuity.
Last Definition (For Now) For Continuity Of Fuctions
•
Let (a, b ) be an interval where f is defined.
•
Definition: f is continuous on (a, b ) if f is continuous at every point c in
•
•
Let [a, b ] be an interval where f is defined.
Definition: f is continuous on [a, b ] if f is continuous on (a, b ) and it’s
continuous from the right at a and continuous from the left at b
(a, b )
a c
b
Lecture #9 – Tuesday, October 7, 2003
THEOREM “THE PINCHING THEOREM”
•
Let p > 0 . Suppose that for all x such that
•
If lim h( x ) = L = lim g (x ) , then lim f (x ) = L
0 < x − c 0 . Want:
•
•
•
c
c+p
f ( x ) − L < ε for x − c small enough.
Choose δ1 so that h(x ) − L < ε if x − c < δ1
Choose δ2 so that g (x ) − L < ε if x − c < δ 2
L − ε < h( x ) < L + ε
L − ε < g (x ) < L + ε
If we choose δ = min (δ1 , δ 2 ) , then L − ε < h( x ) ≤ f ( x ) ≤ g ( x ) < L + ε
Q.E.D.
Page 22 of 33
f (x ) − L < ε
MAT137Y1a.doc
APPLICATION: SINE AND COSINE
First A Figure
Q
P
1
x
tan x
sin x
cos x
O
B
A
1
•
First task: Compute lim sin x = 0 (1)
x→0
•
Proof: From the figure, for x small enough sin x < x
•
•
So in fact, − x < sin < x for x small
So by Pinching Theorem, lim (− x ) = 0 and lim (x ) = 0 , so lim sin x = 0
x →c
x→c
x→0
Q.E.D.
•
•
Because (by Pythagorean Theorem) sin 2 x + cos 2 x = 1 , cos x = 1 − sin 2 x for x small
So because of the theorem about composing continuous functions,
(
)
lim cos x = lim 1 − sin 2 x = lim 1 − sin 2 x = 1 (because lim sin x = 0 )
x→0
x →0
x→0
•
Therefore, lim cos x = 1 (2)
•
Now what about
x→0
x→0
lim sin x = sin c
x→c
lim cos x = cos c
sine and cosine are continuous
x→c
•
Proof of (3):
•
lim sin x is the same as lim sin (c + h )
x→c
h →0
•
Now, we can use the addition formulas: sin (c + h ) = sin c cosh + cos c sinh
•
h →0
lim sin (c + h ) = (sin c ) lim cosh + (cos c ) lim sinh
h →0
= sin c
h →0
Q.E.D.
Lecture #10 – Thursday, October 9, 2003
•
In fact, all the trigonometric functions are continuous (where defined): this follows from the theorem about
quotients of continuous functions being continuous.
Page 23 of 33
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TWO MORE IMPORTANT LIMITS
•
•
lim
x→0
sin x
1 − cos x
= 1 (1) and lim
= 0 (2)
x→0
x
x
Note: For both of these, if you “plug in”, you get “
0
” – need a subtler argument
0
Proof of (1)
•
Refer to Figure
•
Area of ∆OBP =
•
•
•
•
1
(sin x )(1) = sin x
2
2
x
Area of sector OAP =
2
tan x
Area of ∆OAQ =
2
sin x x tan x
< <
2
2
2
sin x
sin x 1
<1<
⋅
x
x cos x
sin x 1
1<
⋅
sin x
x cos x
Two inequalities:
< 1 and
sin x
x
cos x <
x
sin x < x <
sin x
cos x
sin x
< 1 – true for x “small” and x > 0 or x < 0
x
•
So cos x <
•
By Pinching Theorem, we can conclude that lim
sin x
= 1 because lim cos x = 1 and lim 1 = 1 .
x→0
x →0
x→0 x
Q.E.D.
Proof of (2)
•
•
•
1 − cos x 1 − cos x 1 + cos x 1 − cos 2 x
sin 2 x
sin x
sin x
=
⋅
=
=
=
⋅
x
x
1 + cos x x(1 + cos x ) x(1 + cos x )
x (1 + cos x )
1 − cos x
sin x
sin x
Therefore lim
= lim
⋅ lim
=0
x→0
x →0 x
x →0 (1 + cos x )
x
Q.E.D.
sin (ax )
1 − cos(ax )
= 0 and lim
=0
x→0
x→0
ax
ax
In fact, lim
Examples of Application
1)
2)
lim
x→0
sin (5 x )
sin (5 x ) 5
sin (5 x ) 5 5
sin (5 x ) 5
= lim
⋅ = lim
⋅ = lim
=
x →0
4x
4x
5 x →0 5 x
4 4 x →0 5 x
4
2x 2 + x
x(2 x + 1)
x
= lim
= lim
⋅ lim (2 x + 1) =
x → 0 sin x
x →0 sin x
x →0 sin x x →0
lim
1
x
x →0 sin x
lim
Page 24 of 33
⋅ lim (2 x + 1) = 1
x →0
MAT137Y1a.doc
x2
= lim
x → 0 sec x − 1
x →0
lim
3)
= lim
x 2 cos x(1 + cos x )
x2
x2
x 2 cos x
= lim
= lim
= lim
1
x →0 1 − cos x
x →0 1 − cos x
x →0 (1 − cos x )(1 + cos x )
−1
cos x
cos x
x 2 cos x + x 2 cos 2 x
2
x →0
= lim
x →0
x→0
sin x
x
sin x
= lim
2
cos x + lim
x →0
x
sin x
x2
2
sin x
2
⋅ cos x + lim
x →0
x2
2
sin x
⋅ cos 2 x
cos 2 x = 2
TWO VERY IMPORTANT THEOREMS ABOUT CONTINUOUS FUNCTIONS
•
•
•
Intermediate Value Theorem
Extreme Value Theorem
Understanding of the proof is going to require the Least Upper Bound Axioms
Idea For Intermediate Value Theorem (IVT)
•
A continuous function has a graph which is “an unbroken curve”
f(b)
f(b)
k
f(a)
f(a)
a
b
a
continuous
b
non-continuous
THEOREM: INTERMEDIATE VALUE THEOREM
•
If f is continuous on [a, b ] and k is any number between f (a ) and f (b ) , then there is a value c,
a < c 0 , then there is a solution c to f (c ) = 0
solution between a and b (by IVT)
a
b
Page 25 of 33
MAT137Y1a.doc
Example
f (x ) = cos
π
x − x 2 on [0,1]
2
•
Evaluate at x = 0 : cos
π
⋅ 0 − (0 )2 = 1 > 0
2
•
Evaluate at x = 1 : cos
π
⋅1 − (1)2 = −1 < 0
2
•
IVT says there is a solution to cos
•
Now evaluate at x =
•
Now IVT says there is root between
•
Can keep going and get an approximation to root – “bisection method”
π
x − x 2 = 0 between [0,1]
2
1
1
π
1
: f
= cos
−
2
4
2
2
2
≈ 0.45 > 0
1
and 1.
2
Application: Solving Inequalities
•
We were implicitly using IVT before!
•
Solve the inequality x( x + 2)( x − 3) > 0
• Find the zeros: 0, -2, 3
-2
•
0
3
If a, b are 2 zeros of P(x ) = x( x + 2)(x − 3) , then the IVT is saying that on (a, b ) , P(x ) has to be
all positive or negative – because if it’s positive f (a ) > 0, a ∈ (a, b ) , f (b ) < 0, a ∈ (a, b ) then
there is a zero between a and b; that’s a contradiction since you’ve already found all the zeros
BOUNDEDNESS AND EXTREME VALUES
•
•
Let f be a function, defined on an interval I
Definition: f is called bounded on I if there are constants k and K so that k < f ( x ) < K for all x ∈ I
Example
•
f (x ) = x 2 on [0,2]
4
2
Page 26 of 33
MAT137Y1a.doc
•
f (x ) = sin x on [0,2π]
1
2π
-1
•
Definition: If f is not bounded it is called unbounded (on I)
Example
•
1
on (0, ∞ )
g (x ) = x 2
0 at x = 0
•
•
•
g ( x ) is unbounded on (0, ∞ )
g ( x ) is bounded on [1, ∞ )
Note: A function can have a maximum value or a minimum value or both or neither on an interval I
Example
•
1
on (0, ∞ )
g (x ) = x 2
0 at x = 0
•
•
•
•
On [0, ∞ ) : unbounded, no maximum, has minimum
On [1, ∞ ) : bounded, has maximum, no minimum
On (1, ∞ ) : bounded, no maximum, no minimum
Note: This has nothing to do with continuity
Example
•
f (x ) =
1 if x ∈ Q
− 1 if x ∉ Q
is bounded, has maximum, has minimum on (− ∞, ∞ )
THEOREM: EXTREME VALUE THEOREM
•
If a function is continuous on a bounded, closed interval [a, b ] , then the function takes on both a maximum
value M and a minimum value m – i.e. m ≤ f ( x ) ≤ M on [a, b ] (m and M are called the “extreme
values”)
Page 27 of 33
MAT137Y1a.doc
Lecture #11 – Tuesday, October 14, 2003
•
Warning: Need all three of the assumptions
Counter-Examples
1) The interval must be bounded
f (x ) = x 2 on [0, ∞ )
2) The interval must be closed
f (x ) = x 2 on (0,1)
3) The function must be continuous
a
b
Idea of IVT
•
A continuous function “takes intervals to intervals”
Idea of IVT and EVT Together
•
A continuous function “takes bounded closed intervals to bounded closed intervals”
•
Proofs of IVT and EVT require an understanding of the “Least Upper Bound Axiom” of the real number
LEAST UPPER BOUNDS
•
Let S be a nonempty set of real numbers
Example
•
S1 = (− ∞,0 )
•
T1 = [0, ∞ )
•
T2 = [1, ∞ )
•
T1 = φ
0
1 2
n
, ,...,
,...
2 3
n +1
•
S2 =
•
S 3 = (1,2,3,.., n,...)
0
1
x
1
2
3
4
5
•
S4 =
•
Definition: A number M is an upper bound for S if x ≤ M for any x ∈ S
(x + 1)(x + 3)
:3≤ x ≤ 5
Page 28 of 33
MAT137Y1a.doc
•
•
•
•
Note: Not all sets have upper bounds – those with upper bounds in the example are S1, S2, S4
Definition: If S has an upper bound, we say S is bounded above
Now let’s think about the set of possible upper bounds for S (call it T)
What is the smallest possible upper bound? – in the Example, for S1, the smallest is 0; for S2, the smallest is
1
Definition
•
If S is nonempty set of real numbers that is bounded above, the least upper bound (lub) of S is an upper
bound that is less than or equal to any upper bound for S
LEAST UPPER BOUND AXIOM
•
Every nonempty set of real numbers that has an upper bound has a least upper bound
•
If you think this is “obvious”, it’s NOT TRUE, for example, the rational numbers!
Example
{
}
•
S = x < 2 , x is rational
•
Let T = the set of possible rational upper bound
•
for S = x > 2 : x is rational
Has no least “element”
{
0
2
}
Back To Real Numbers: Examples
1)
lub(− 4,−1) = −1
2)
1
1
1
lub − 1,− ,−
,...,− 3 ,... = 0
8 27
n
3)
lub x ∈ R : x 2 < 3 = 3
{
}
Lecture #12 – Thursday, October 16, 2003
Theorem
•
If M is the least upper bound of set S and ε > 0 is any positive number, then there is an element s in S such
that M − ε < s ≤ M
elements in S
M-ε
•
M
Idea: The elements in S get arbitrarily close to M
Proof
•
•
•
Let ε > 0 . Then since M is an upper bound, any element x of S satisfy x < M
So we only need to find s ∈ S so that M − ε < s
We’ll argue by contradiction. Suppose for all s ∈ S , s ≤ M − ε
• But that means M − ε is an upper bound for S
• But M − ε < M and M is supposed to be the lub: contradiction
Page 29 of 33
MAT137Y1a.doc
•
So there must be s, M − ε < s ≤ M
Example
S=
1 2
n
, ,...,
,...
2 3
n +1
999/1000
0
•
Take ε =
1/2
1
1
999
1000 9999
. Is it true that there is an element of S such that
< s < 1 ? Yes: s =
,
1000
1000
1001 10000
LOWER BOUND
Definition
•
Let S be a nonempty set of real numbers, then we say m is a lower bound of S if ∀x ∈ S , m ≤ x – if S has a
lower bound, then S is bounded below
•
If S is a nonempty set of real numbers bounded below, then the greatest lower bound (glb) is a lower bound
for S that is greater than or equal to any other lower bound for S
Theorem: Existence of Greatest Lower Bound
•
Every nonempty set of S of R bounded below has a greatest lower bound
Theorem
•
If m is the glb of S, ε > 0 , then ∃ an element s ∈ S such that m ≤ s < m + ε
Proofs
•
Use what you know for least upper bounds
PROOF OF INTERMEDIATE VALUE THEOREM
Lemma
•
Let f be continuous on [a, b ] . If f (a ) < 0 < f (b ) (or f (a ) > 0 > f (b ) ) then ∃ c, a < c < b such that
f (c ) = 0
Page 30 of 33
MAT137Y1a.doc
Picture
ξ
c
a
b
f(a)
Proof
•
•
•
•
•
Suppose f (a ) < 0 < f (b ) (The other situation can be proved similarly)
Let S = {ξ : f is negative on [a, ξ )}
By continuity of f, S is not empty
Because f (b ) > 0 , b is an upper bound for S. That means S has a least upper bound: call it c
Want: f (c ) = 0 (we’ll argue that it can’t be >0 and can’t be <0)
•
•
Suppose f (c ) > 0
•
By continuity of f at c, there is an interval (c − ε, c + ε ) where f is >0
•
Then because
c −
ε
> 0
, c−
ε
is also an upper bound for S, but c is the
2
least upper bound. Contradiction.
Suppose f (c ) > 0
•
By continuity of f at c, f is <0 on an interval (c − ε, c + ε )
ε
is in S, but c is an upper bound
2
ε
• Contradiction because c + > c
2
So f (c ) is neither <0 or >0, so f (c ) = 0
Q.E.D.
•
•
f
So c +
Theorem: IVT
•
If f is continuous on [a, b ] , and k is any value between f (a ) and f (b ) , then ∃ c, a < c 0 so g is continuous on [a, b ]
By the Lemma, there is a c, a < c < b so that g (c ) = 0
But then if g (c ) = 0 = f (c ) − k , then f (c ) = k
Q.E.D.
Page 31 of 33
MAT137Y1a.doc
PROOF OF EXTREME VALUE THEOREM
Lemma
•
If f is continuous on [a, b ] , then f is bounded on [a, b ]
Proof
•
•
Let S = {x ∈ R : x ∈ [a, b] and f is bounded on [a, x ]}
• S is bounded above by definition of S: b is an upper bound
• S is not empty: can take x = a
Let c = lub S . Want: c = b
• Because b is an upper bound and c is the least upper bound, then c ≤ b .
• We must rule out the possibility that c < b .
• Suppose c < b
• By continuity of f at c, f is bounded on some interval (c − ε, c + ε ) .
•
ε
. f is also bounded on
2
c−
ε
ε
,c+
2
2
ε
ε
is in S.
c+
2
2
• But c is an upper bound for S, so contradiction.
• So c = b .
Want: f is bounded on [a, b ] .
• Because f is bounded at b, there is an interval [b − δ, b ] where f is bounded.
is bounded on
•
a, c −
Because c is the lub, f is bounded on
a, c +
•
Because b is lub of S, f is bounded on
•
So f is bounded on [a, b ] .
a, b −
ε
.
2
Theorem: EVT
•
If f is continuous on [a, b ] , then f achieves
both a maximum M and minimum m value
on [a, b ] .
M
a
c
b
m
Proof
•
•
•
By Lemma, f is bounded.
Definition: M = lub{ f ( x ) : x ∈ [a, b]}
Want: ∃ c such that f (c ) = M .
•
Suppose not. Then f (x ) ≠ M for all x ∈ [a, b ] .
•
Then M − f (x ) ≠ 0 for all x ∈ [a, b ] (in fact M − f ( x ) > 0 for all x ∈ [a, b ] ).
•
Define g ( x ) =
•
Because M is lub, M − f (x ) can be made arbitrarily small, and g ( x ) can be made arbitrarily
large.
1
. So g is continuous on [a, b ] .
M − f (x )
Page 32 of 33
f
MAT137Y1a.doc
•
• So g is continuous on [a, b ] , but not bounded. This contradicts the Lemma.
Theforefore, there is a c such that f (c ) = M .
Q.E.D.
Page 33 of 33

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