Interpolation (27 pages)

Interpolation (27 pages)
Chapter 3
Interpolation
Interpolation is the process of defining a function that takes on specified values at
specified points. This chapter concentrates on two closely related interpolants: the
piecewise cubic spline and the shape-preserving piecewise cubic named “pchip.”
3.1
The Interpolating Polynomial
We all know that two points determine a straight line. More precisely, any two
points in the plane, (x1 , y1 ) and (x2 , y2 ), with x1 ̸= x2 , determine a unique firstdegree polynomial in x whose graph passes through the two points. There are
many different formulas for the polynomial, but they all lead to the same straight
line graph.
This generalizes to more than two points. Given n points in the plane,
(xk , yk ), k = 1, . . . , n, with distinct xk ’s, there is a unique polynomial in x of degree
less than n whose graph passes through the points. It is easiest to remember that n,
the number of data points, is also the number of coefficients, although some of the
leading coefficients might be zero, so the degree might actually be less than n − 1.
Again, there are many different formulas for the polynomial, but they all define the
same function.
This polynomial is called the interpolating polynomial because it exactly reproduces the given data:
P (xk ) = yk , k = 1, . . . , n.
Later, we examine other polynomials, of lower degree, that only approximate the
data. They are not interpolating polynomials.
The most compact representation of the interpolating polynomial is the Lagrange form


∑ ∏ x − xj

 yk .
P (x) =
xk − xj
k
j̸=k
September 16, 2013
1
2
Chapter 3. Interpolation
There are n terms in the sum and n − 1 terms in each product, so this expression
defines a polynomial of degree at most n − 1. If P (x) is evaluated at x = xk , all the
products except the kth are zero. Furthermore, the kth product is equal to one, so
the sum is equal to yk and the interpolation conditions are satisfied.
For example, consider the following data set.
x = 0:3;
y = [-5 -6
-1
16];
The command
disp([x;
y])
displays
0
-5
1
-6
2
-1
3
16
The Lagrangian form of the polynomial interpolating these data is
P (x) =
x(x − 2)(x − 3)
(x − 1)(x − 2)(x − 3)
(−5) +
(−6)
(−6)
(2)
x(x − 1)(x − 3)
x(x − 1)(x − 2)
+
(−1) +
(16).
(−2)
(6)
We can see that each term is of degree three, so the entire sum has degree at
most three. Because the leading term does not vanish, the degree is actually three.
Moreover, if we plug in x = 0, 1, 2, or 3, three of the terms vanish and the fourth
produces the corresponding value from the data set.
Polynomials are not usually represented in their Lagrangian form. More frequently, they are written as something like
x3 − 2x − 5.
The simple powers of x are called monomials, and this form of a polynomial is said
to be using the power form.
The coefficients of an interpolating polynomial using its power form,
P (x) = c1 xn−1 + c2 xn−2 + · · · + cn−1 x + cn ,
can, in principle, be computed by solving a system of simultaneous linear equations
 n−1
  c1   y1 
x1
x1n−2 · · · x1 1
c2   y2 
x2n−2 · · · x2 1  
 xn−1
=  . .
.. 
 2


···
··· ··· ··· 1
.   .. 
xn−1
xnn−2 · · · xn 1
n
cn
yn
The matrix V of this linear system is known as a Vandermonde matrix. Its
elements are
vk,j = xn−j
.
k
3.1. The Interpolating Polynomial
3
The columns of a Vandermonde matrix are sometimes written in the opposite order,
but polynomial coefficient vectors in Matlab always have the highest power first.
The Matlab function vander generates Vandermonde matrices. For our example data set,
V = vander(x)
generates
V =
0
1
8
27
0
1
4
9
0
1
2
3
1
1
1
1
Then
c = V\y’
computes the coefficients.
c =
1.0000
0.0000
-2.0000
-5.0000
In fact, the example data were generated from the polynomial x3 − 2x − 5.
Exercise 3.6 asks you to show that Vandermonde matrices are nonsingular if
the points xk are distinct. But Exercise 3.18 asks you to show that a Vandermonde
matrix can be very badly conditioned. Consequently, using the power form and
the Vandermonde matrix is a satisfactory technique for problems involving a few
well-spaced and well-scaled data points. But as a general-purpose approach, it is
dangerous.
In this chapter, we describe several Matlab functions that implement various
interpolation algorithms. All of them have the calling sequence
v = interp(x,y,u)
The first two input arguments, x and y, are vectors of the same length that define
the interpolating points. The third input argument, u, is a vector of points where
the function is to be evaluated. The output v is the same length as u and has
elements
v(k)=interp(x,y,u(k))
Our first such interpolation function, polyinterp, is based on the Lagrange
form. The code uses Matlab array operations to evaluate the polynomial at all
the components of u simultaneously.
4
Chapter 3.
Interpolation
function v = polyinterp(x,y,u)
n = length(x);
v = zeros(size(u));
for k = 1:n
w = ones(size(u));
for j = [1:k-1 k+1:n]
w = (u-x(j))./(x(k)-x(j)).*w;
end
v = v + w*y(k);
end
To illustrate polyinterp, create a vector of densely spaced evaluation points.
u = -.25:.01:3.25;
Then
v = polyinterp(x,y,u);
plot(x,y,’o’,u,v,’-’)
creates Figure 3.1.
25
20
15
10
5
0
−5
−10
−0.5
0
0.5
1
1.5
2
2.5
3
3.5
Figure 3.1. polyinterp.
The polyinterp function also works correctly with symbolic variables. For
example, create
symx = sym(’x’)
Then evaluate and display the symbolic form of the interpolating polynomial with
3.1. The Interpolating Polynomial
5
P = polyinterp(x,y,symx)
pretty(P)
which produces
P =
(x*(x - 1)*(x - 3))/2 + 5*(x/2 - 1)*(x/3 - 1)*(x - 1) +
(16*x*(x/2 - 1/2)*(x - 2))/3 - 6*x*(x/2 - 3/2)*(x - 2)
/ x
\
16 x | - - 1/2 | (x - 2)
x (x - 1) (x - 3)
/ x
\ / x
\
\ 2
/
----------------- + 5 | - - 1 | | - - 1 | (x - 1) + -----------------------2
\ 2
/ \ 3
/
3
/ x
\
- 6 x | - - 3/2 | (x - 2)
\ 2
/
This expression is a rearrangement of the Lagrange form of the interpolating polynomial. Simplifying the Lagrange form with
P = simplify(P)
changes P to the power form
P =
x^3 - 2*x - 5
Here is another example, with a data set that is used by the other methods
in this chapter.
x = 1:6;
y = [16 18 21 17 15 12];
disp([x; y])
u = .75:.05:6.25;
v = polyinterp(x,y,u);
plot(x,y,’o’,u,v,’r-’);
produces
1
16
2
18
3
21
4
17
5
15
6
12
and Figure 3.2.
Already in this example, with only six nicely spaced points, we can begin to
see the primary difficulty with full-degree polynomial interpolation. In between
the data points, especially in the first and last subintervals, the function shows
excessive variation. It overshoots the changes in the data values. As a result, fulldegree polynomial interpolation is hardly ever used for data and curve fitting. Its
primary application is in the derivation of other numerical methods.
6
Chapter 3. Interpolation
Full degree polynomial interpolation
22
20
18
16
14
12
10
1
2
3
4
5
6
Figure 3.2. Full-degree polynomial interpolation.
3.2
Piecewise Linear Interpolation
You can create a simple picture of the data set from the last section by plotting
the data twice, once with circles at the data points and once with straight lines
connecting the points. The following statements produce Figure 3.3.
x = 1:6;
y = [16 18 21 17 15 12];
plot(x,y,’o’,x,y,’-’);
To generate the lines, the Matlab graphics routines use piecewise linear interpolation. The algorithm sets the stage for more sophisticated algorithms. Three
quantities are involved. The interval index k must be determined so that
xk ≤ x < xk+1 .
The local variable, s, is given by
s = x − xk .
The first divided difference is
δk =
yk+1 − yk
.
xk+1 − xk
With these quantities in hand, the interpolant is
L(x) = yk + (x − xk )
yk+1 − yk
xk+1 − xk
3.2. Piecewise Linear Interpolation
7
Piecewise linear interpolation
22
20
18
16
14
12
10
1
2
3
4
5
6
Figure 3.3. Piecewise linear interpolation.
= yk + sδk .
This is clearly a linear function that passes through (xk , yk ) and (xk+1 , yk+1 ).
The points xk are sometimes called breakpoints or breaks. The piecewise linear
interpolant L(x) is a continuous function of x, but its first derivative, L′ (x), is not
continuous. The derivative has a constant value, δk , on each subinterval and jumps
at the breakpoints.
Piecewise linear interpolation is implemented in piecelin.m. The input u
can be a vector of points where the interpolant is to be evaluated, so the index k is
actually a vector of indices. Read this code carefully to see how k is computed.
function v = piecelin(x,y,u)
%PIECELIN Piecewise linear interpolation.
% v = piecelin(x,y,u) finds the piecewise linear L(x)
% with L(x(j)) = y(j) and returns v(k) = L(u(k)).
%
First divided difference
delta = diff(y)./diff(x);
%
Find subinterval indices k so that x(k) <= u < x(k+1)
n = length(x);
k = ones(size(u));
for j = 2:n-1
8
Chapter 3. Interpolation
k(x(j) <= u) = j;
end
%
Evaluate interpolant
s = u - x(k);
v = y(k) + s.*delta(k);
3.3
Piecewise Cubic Hermite Interpolation
Many of the most effective interpolation techniques are based on piecewise cubic
polynomials. Let hk denote the length of the kth subinterval:
hk = xk+1 − xk .
Then the first divided difference, δk , is given by
δk =
yk+1 − yk
.
hk
Let dk denote the slope of the interpolant at xk :
dk = P ′ (xk ).
For the piecewise linear interpolant, dk = δk−1 or δk , but this is not necessarily true
for higher order interpolants.
Consider the following function on the interval xk ≤ x ≤ xk+1 , expressed in
terms of local variables s = x − xk and h = hk :
P (x) =
3hs2 − 2s3
h3 − 3hs2 + 2s3
yk+1 +
yk
3
h
h3
s(s − h)2
s2 (s − h)
d
+
dk .
+
k+1
h2
h2
This is a cubic polynomial in s, and hence in x, that satisfies four interpolation
conditions, two on function values and two on the possibly unknown derivative
values:
P (xk ) = yk , P (xk+1 ) = yk+1 ,
P ′ (xk ) = dk , P ′ (xk+1 ) = dk+1 .
Functions that satisfy interpolation conditions on derivatives are known as Hermite
or osculatory interpolants, because of the higher order contact at the interpolation
sites. (“Osculari” means “to kiss” in Latin.)
If we happen to know both function values and first derivative values at a
set of data points, then piecewise cubic Hermite interpolation can reproduce those
data. But if we are not given the derivative values, we need to define the slopes
dk somehow. Of the many possible ways to do this, we will describe two, which
Matlab calls pchip and spline.
3.4. Shape-Preserving Piecewise Cubic
3.4
9
Shape-Preserving Piecewise Cubic
The acronym pchip abbreviates “piecewise cubic Hermite interpolating polynomial.” Although it is fun to say, the name does not specify which of the many
possible interpolants is actually being used. In fact, spline interpolants are also
piecewise cubic Hermite interpolating polynomials, but with different slopes. Our
particular pchip is a shape-preserving, “visually pleasing” interpolant that was introduced into Matlab fairly recently. It is based on an old Fortran program by
Fritsch and Carlson [2] that is described by Kahaner, Moler, and Nash [3]. Figure
3.4 shows how pchip interpolates our sample data.
Shape−preserving Hermite interpolation
22
20
18
16
14
12
10
1
2
3
4
5
6
Figure 3.4. Shape-preserving piecewise cubic Hermite interpolation.
The key idea is to determine the slopes dk so that the function values do not
overshoot the data values, at least locally. If δk and δk−1 have opposite signs or if
either of them is zero, then xk is a discrete local minimum or maximum, so we set
dk = 0.
This is illustrated in the first half of Figure 3.5. The lower solid line is the piecewise
linear interpolant. Its slopes on either side of the breakpoint in the center have
opposite signs. Consequently, the dashed line has slope zero. The curved line is
the shape-preserving interpolant, formed from two different cubics. The two cubics
interpolate the center value and their derivatives are both zero there. But there is
a jump in the second derivative at the breakpoint.
If δk and δk−1 have the same sign and the two intervals have the same length,
10
Chapter 3. Interpolation
then dk is taken to be the harmonic mean of the two discrete slopes:
)
(
1
1
1
1
=
+
.
dk
2 δk−1
δk
In other words, at the breakpoint, the reciprocal slope of the Hermite interpolant
is the average of the reciprocal slopes of the piecewise linear interpolant on either
side. This is shown in the other half of Figure 3.5. At the breakpoint, the reciprocal
slope of the piecewise linear interpolant changes from 1 to 5. The reciprocal slope
of the dashed line is 3, the average of 1 and 5. The shape-preserving interpolant is
formed from the 2 cubics that interpolate the center value and that have slope equal
to 1/3 there. Again, there is a jump in the second derivative at the breakpoint.
Figure 3.5. Slopes for pchip.
If δk and δk−1 have the same sign, but the two intervals have different lengths,
then dk is a weighted harmonic mean, with weights determined by the lengths of
the two intervals:
w1 + w2
w1
w2
=
+
,
dk
δk−1
δk
where
w1 = 2hk + hk−1 , w2 = hk + 2hk−1 .
This defines the pchip slopes at interior breakpoints, but the slopes d1 and dn
at either end of the data interval are determined by a slightly different, one-sided
analysis. The details are in pchiptx.m.
3.5
Cubic Spline
Our other piecewise cubic interpolating function is a cubic spline. The term “spline”
refers to an instrument used in drafting. It is a thin, flexible wooden or plastic tool
that is passed through given data points and defines a smooth curve in between. The
physical spline minimizes potential energy subject to the interpolation constraints.
The corresponding mathematical spline must have a continuous second derivative
3.5. Cubic Spline
11
and satisfy the same interpolation constraints. The breakpoints of a spline are also
referred to as its knots.
The world of splines extends far beyond the basic one-dimensional, cubic,
interpolatory spline we are describing here. There are multidimensional, high-order,
variable knot, and approximating splines. A valuable expository and reference text
for both the mathematics and the software is A Practical Guide to Splines by Carl
de Boor [1]. De Boor is also the author of the spline function and the Spline
Toolbox for Matlab.
Figure 3.6 shows how spline interpolates our sample data.
Spline interpolation
22
20
18
16
14
12
10
1
2
3
4
5
6
Figure 3.6. Cubic spline interpolation.
The first derivative P ′ (x) of our piecewise cubic function is defined by different
formulas on either side of a knot xk . Both formulas yield the same value dk at the
knots, so P ′ (x) is continuous.
On the kth subinterval, the second derivative is a linear function of s = x−xk :
P ′′ (x) =
(6h − 12s)δk + (6s − 2h)dk+1 + (6s − 4h)dk
.
h2
If x = xk , s = 0 and
P ′′ (xk +) =
6δk − 2dk+1 − 4dk
.
hk
The plus sign in xk + indicates that this is a one-sided derivative. If x = xk+1 ,
s = hk and
−6δk + 4dk+1 + 2dk
.
P ′′ (xk+1 −) =
hk
12
Chapter 3. Interpolation
On the (k − 1)st interval, P ′′ (x) is given by a similar formula involving δk−1 , dk ,
and dk−1 . At the knot xk ,
P ′′ (xk −) =
−6δk−1 + 4dk + 2dk−1
.
hk−1
Requiring P ′′ (x) to be continuous at x = xk means that
P ′′ (xk +) = P ′′ (xk −).
This leads to the condition
hk dk−1 + 2(hk−1 + hk )dk + hk−1 dk+1 = 3(hk δk−1 + hk−1 δk ).
If the knots are equally spaced, so that hk does not depend on k, this becomes
dk−1 + 4dk + dk+1 = 3δk−1 + 3δk .
Like our other interpolants, the slopes dk of a spline are closely related to the
differences δk . In the spline case, they are a kind of running average of the δk ’s.
The preceding approach can be applied at each interior knot xk , k = 2, . . . , n−
1, to give n − 2 equations involving the n unknowns dk . As with pchip, a different
approach must be used near the ends of the interval. One effective strategy is known
as “not-a-knot.” The idea is to use a single cubic on the first two subintervals,
x1 ≤ x ≤ x3 , and on the last two subintervals, xn−2 ≤ x ≤ xn . In effect, x2 and
xn−1 are not knots. If the knots are equally spaced with all hk = 1, this leads to
d1 + 2d2 =
5
1
δ1 + δ2
2
2
and
1
5
δn−2 + δn−1 .
2
2
The details if the spacing is not equal to one are in splinetx.m.
With the two end conditions included, we have n linear equations in n unknowns:
Ad = r.
2dn−1 + dn =
The vector of unknown slopes is


d1
 d2 

d=
 ...  .
dn
The coefficient matrix A is tridiagonal:
h
h2 + h1
2
h1
 h2 2(h1 + h2 )

h
2(h

3
2 + h3 )
A=
..

.



h2
..
.
hn−1
..
.
2(hn−2 + hn−1 )
hn−1 + hn−2
hn−2
hn−2



.



3.5. Cubic Spline
13
The right-hand side is




r = 3


h

r1
h2 δ1 + h1 δ2
h3 δ2 + h2 δ3
..
.
n−1 δn−2
+ hn−2 δn−1
rn



.



The two values r1 and rn are associated with the end conditions.
If the knots are equally spaced with all hk = 1, the coefficient matrix is quite
simple:


1 2
1 4 1





1 4
1

.
A=
.. .. ..

.
.
.



1
4 1
2
The right-hand side is

+ 16 δ2
+ δ2
δ2 + δ3
..
.
5
6 δ1
δ1
1








.
r = 3



 δ

n−2 + δn−1
1
5
6 δn−2 + 6 δn−1
In our textbook function, splinetx, the linear equations defining the slopes
are solved with the tridisolve function introduced in Chapter 2, Linear Equations.
In the spline functions distributed with Matlab and the Spline Toolbox, the slopes
are computed by the Matlab backslash operator
d = A\r
Because most of the elements of A are zero, it is appropriate to store A in a sparse
data structure. The backslash operator can then take advantage of the tridiagonal
structure and solve the linear equations in time and storage proportional to n, the
number of data points.
Figure 3.7 compares the spline interpolant, s(x), with the pchip interpolant,
p(x). The difference between the functions themselves is barely noticeable. The
first derivative of spline, s′ (x), is smooth, while the first derivative of pchip, p′ (x),
is continuous, but shows “kinks.” The spline second derivative s′′ (x) is continuous,
while the pchip second derivative p′′ (x) has jumps at the knots. Because both functions are piecewise cubics, their third derivatives, s′′′ (x) and p′′′ (x), are piecewise
constant. The fact that s′′′ (x) takes on the same values in the first two intervals
and the last two intervals reflects the “not-a-knot” spline end conditions.
14
Chapter 3. Interpolation
spline
pchip
21
21
18
18
15
15
12
12
2
4
6
8
5
5
0
0
−5
4
6
8
2
4
6
8
2
4
6
8
2
4
6
8
−5
2
4
6
8
20
20
0
0
−20
2
2
4
6
8
−20
30
30
0
0
−30
−30
2
4
6
8
Figure 3.7. The spline and pchip interpolants, and their first three derivatives.
3.6
pchiptx, splinetx
The M-files pchiptx and splinetx are both based on piecewise cubic Hermite
interpolation. On the kth interval, this is
P (x) =
3hs2 − 2s3
h3 − 3hs2 + 2s3
y
+
yk
k+1
h3
h3
s2 (s − h)
s(s − h)2
+
dk+1 +
dk ,
2
h
h2
where s = x − xk and h = hk . The two functions differ in the way they compute the
slopes, dk . Once the slopes have been computed, the interpolant can be efficiently
evaluated using the power form with the local variable s:
P (x) = yk + sdk + s2 ck + s3 bk ,
where the coefficients of the quadratic and cubic terms are
3δk − 2dk − dk+1
,
h
dk − 2δk + dk+1
.
bk =
h2
ck =
Here is the first portion of code for pchiptx. It calls an internal subfunction to
compute the slopes, then computes the other coefficients, finds a vector of interval
3.6. pchiptx, splinetx
15
indices, and evaluates the interpolant. After the preamble, this part of the code for
splinetx is the same.
function v = pchiptx(x,y,u)
%PCHIPTX Textbook piecewise cubic Hermite interpolation.
% v = pchiptx(x,y,u) finds the shape-preserving piecewise
% P(x), with P(x(j)) = y(j), and returns v(k) = P(u(k)).
%
% See PCHIP, SPLINETX.
% First derivatives
h = diff(x);
delta = diff(y)./h;
d = pchipslopes(h,delta);
% Piecewise polynomial coefficients
n = length(x);
c = (3*delta - 2*d(1:n-1) - d(2:n))./h;
b = (d(1:n-1) - 2*delta + d(2:n))./h.^2;
% Find subinterval indices k so that x(k) <= u < x(k+1)
k = ones(size(u));
for j = 2:n-1
k(x(j) <= u) = j;
end
% Evaluate interpolant
s = u - x(k);
v = y(k) + s.*(d(k) + s.*(c(k) + s.*b(k)));
The code for computing the pchip slopes uses the weighted harmonic mean
at interior breakpoints and a one-sided formula at the endpoints.
function d = pchipslopes(h,delta)
% PCHIPSLOPES Slopes for shape-preserving Hermite cubic
% pchipslopes(h,delta) computes d(k) = P’(x(k)).
%
%
%
%
%
%
Slopes at interior points
delta = diff(y)./diff(x).
d(k) = 0 if delta(k-1) and delta(k) have opposites
signs or either is zero.
d(k) = weighted harmonic mean of delta(k-1) and
delta(k) if they have the same sign.
16
Chapter 3. Interpolation
n = length(h)+1;
d = zeros(size(h));
k = find(sign(delta(1:n-2)).*sign(delta(2:n-1))>0)+1;
w1 = 2*h(k)+h(k-1);
w2 = h(k)+2*h(k-1);
d(k) = (w1+w2)./(w1./delta(k-1) + w2./delta(k));
%
Slopes at endpoints
d(1) = pchipend(h(1),h(2),delta(1),delta(2));
d(n) = pchipend(h(n-1),h(n-2),delta(n-1),delta(n-2));
function d = pchipend(h1,h2,del1,del2)
% Noncentered, shape-preserving, three-point formula.
d = ((2*h1+h2)*del1 - h1*del2)/(h1+h2);
if sign(d) ~= sign(del1)
d = 0;
elseif (sign(del1)~=sign(del2))&(abs(d)>abs(3*del1))
d = 3*del1;
end
The splinetx M-file computes the slopes by setting up and solving a tridiagonal system of simultaneous linear equations.
function d = splineslopes(h,delta);
% SPLINESLOPES Slopes for cubic spline interpolation.
% splineslopes(h,delta) computes d(k) = S’(x(k)).
% Uses not-a-knot end conditions.
%
Diagonals of tridiagonal system
n = length(h)+1;
a = zeros(size(h)); b = a; c = a; r = a;
a(1:n-2) = h(2:n-1);
a(n-1) = h(n-2)+h(n-1);
b(1) = h(2);
b(2:n-1) = 2*(h(2:n-1)+h(1:n-2));
b(n) = h(n-2);
c(1) = h(1)+h(2);
c(2:n-1) = h(1:n-2);
%
Right-hand side
r(1) = ((h(1)+2*c(1))*h(2)*delta(1)+ ...
h(1)^2*delta(2))/c(1);
r(2:n-1) = 3*(h(2:n-1).*delta(1:n-2)+ ...
3.7. interpgui
17
Piecewise linear interpolation
Full degree polynomial interpolation
Shape−preserving Hermite interpolation
Spline interpolation
Figure 3.8. Four interpolants.
h(1:n-2).*delta(2:n-1));
r(n) = (h(n-1)^2*delta(n-2)+ ...
(2*a(n-1)+h(n-1))*h(n-2)*delta(n-1))/a(n-1);
%
Solve tridiagonal linear system
d = tridisolve(a,b,c,r);
3.7
interpgui
Figure 3.8 illustrates the tradeoff between smoothness and a somewhat subjective
property that we might call local monotonicity or shape preservation.
The piecewise linear interpolant is at one extreme. It has hardly any smoothness. It is continuous, but there are jumps in its first derivative. On the other hand,
it preserves the local monotonicity of the data. It never overshoots the data and it
is increasing, decreasing, or constant on the same intervals as the data.
The full-degree polynomial interpolant is at the other extreme. It is infinitely
differentiable. But it often fails to preserve shape, particularly near the ends of the
interval.
The pchip and spline interpolants are in between these two extremes. The
spline is smoother than pchip. The spline has two continuous derivatives, while
pchip has only one. A discontinuous second derivative implies discontinuous cur-
18
Chapter 3. Interpolation
Interpolation
24
22
20
18
16
14
12
10
8
0
1
2
3
4
5
6
Figure 3.9. interpgui.
vature. The human eye can detect large jumps in curvature in graphs and in mechanical parts made by numerically controlled machine tools. On the other hand,
pchip is guaranteed to preserve shape, but the spline might not.
The M-file interpgui allows you to experiment with the four interpolants
discussed in this chapter:
• piecewise linear interpolant,
• full-degree interpolating polynomial,
• piecewise cubic spline,
• shape-preserving piecewise cubic.
The program can be initialized in several different ways:
• With no arguments, interpgui starts with 8 zeros.
• With a scalar argument, interpgui(n) starts with n zeros.
• With one vector argument, interpgui(y) starts with equally spaced x’s.
• With two arguments, interpgui(x,y) starts with a plot of y versus x.
After initialization, the interpolation points can be varied with the mouse. If x
has been specified, it remains fixed. Figure 3.9 is the initial plot generated by our
example data.
Exercises
19
Exercises
3.1. Reproduce Figure 3.8, with four subplots showing the four interpolants discussed in this chapter.
3.2. Tom and Ben are twin boys born on October 27, 2001. Here is a table of
their weights, in pounds and ounces, over their first few months.
%
W = [10
11
12
12
01
01
03
Date
27 2001
19 2001
03 2001
20 2001
09 2002
23 2002
06 2002
Tom
5 10
7 4
8 12
10 14
12 13
14 8
16 10
Ben
4 8
5 11
6 4
8 7
10 3
12 0
13 10];
You can use datenum to convert the date in the first three columns to a serial
date number measuring time in days.
t = datenum(W(:,[3 1 2]));
Make a plot of their weights versus time, with circles at the data points and
the pchip interpolating curve in between. Use datetick to relabel the time
axis. Include a title and a legend. The result should look something like
Figure 3.10.
3.3. (a) Interpolate these data by each of the four interpolants discussed in this
chapter: piecelin, polyinterp, splinetx, and pchiptx. Plot the results
for −1 ≤ x ≤ 1.
20
Chapter 3. Interpolation
Twins’ weights
18
16
14
Pounds
12
10
8
6
Tom
Ben
4
Nov01
Dec01
Jan02
Feb02
Mar02
Figure 3.10. Twins’ weights.
x
-1.00
-0.96
-0.65
0.10
0.40
1.00
y
-1.0000
-0.1512
0.3860
0.4802
0.8838
1.0000
(b) What are the values of each of the four interpolants at x = −0.3? Which
of these values do you prefer? Why?
(c) The data were actually generated from a low-degree polynomial with
integer coefficients. What is that polynomial?
3.4. Make a plot of your hand. Start with
figure(’position’,get(0,’screensize’))
axes(’position’,[0 0 1 1])
[x,y] = ginput;
Place your hand on the computer screen. Use the mouse to select a few
dozen points outlining your hand. Terminate the ginput with a carriage
return. You might find it easier to trace your hand on a piece of paper and
then put the paper on the computer screen. You should be able to see the
ginput cursor through the paper. (Save these data. We will refer to them in
Exercises
21
other exercises later in this book.)
Figure 3.11. A hand.
Now think of x and y as two functions of an independent variable that goes
from one to the number of points you collected. You can interpolate both
functions on a finer grid and plot the result with
n = length(x);
s = (1:n)’;
t = (1:.05:n)’;
u = splinetx(s,x,t);
v = splinetx(s,y,t);
clf reset
plot(x,y,’.’,u,v,’-’);
Do the same thing with pchiptx. Which do you prefer?
Figure 3.11 is the plot of my hand. Can you tell if it was done with splinetx
or pchiptx?
3.5. The previous exercise uses the data index number as the independent variable
for two-dimensional parametric interpolation. This exercise uses, instead, the
angle θ from polar coordinates. In order to do this, the data must be centered
so that they lie on a curve that is starlike with respect to the origin, that is,
every ray emanating from the origin meets the data only once. This means
that you must be able to find values x0 and y0 so that the Matlab statements
x = x - x0
22
Chapter 3. Interpolation
y = y - y0
theta = atan2(y,x)
r = sqrt(x.^2 + y.^2)
plot(theta,r)
produce a set of points that can be interpolated with a single-valued function,
r = r(θ). For the data obtained by sampling the outline of your hand, the
point (x0 , y0 ) is located near the base of your palm. See the small circle in
Figure 3.11. Furthermore, in order to use splinetx and pchiptx, it is also
necessary to order the data so that theta is monotonically increasing.
Choose a subsampling increment, delta, and let
t = (theta(1):delta:theta(end))’;
p = pchiptx(theta,r,t);
s = splinetx(theta,r,t);
Examine two plots:
plot(theta,r,’o’,t,[p s],’-’)
and
plot(x,y,’o’,p.*cos(t),p.*sin(t),’-’,...
s.*cos(t),s.*sin(t),’-’)
Compare this approach with the one used in the previous exercise. Which do
you prefer? Why?
3.6. This exercise requires the Symbolic Toolbox.
(a) What does vandal(n) compute and how does it compute it?
(b) Under what conditions on x is the matrix vander(x) nonsingular?
3.7. Prove that the interpolating polynomial is unique. That is, if P (x) and Q(x)
are two polynomials with degree less than n that agree at n distinct points,
then they agree at all points.
3.8. Give convincing arguments that each of the following descriptions defines
the same polynomial, the Chebyshev polynomial of degree five, T5 (x). Your
arguments can involve analytic proofs, symbolic computation, numeric computation, or all three. Two of the representations involve the golden ratio,
√
1+ 5
ϕ=
.
2
(a) Power form basis:
T5 (x) = 16x5 − 20x3 + 5x.
(b) Relation to trigonometric functions:
T5 (x) = cos (5 cos−1 x).
Exercises
23
(c) Horner representation:
T5 (x) = ((((16x + 0)x − 20)x + 0)x + 5)x + 0.
(d) Lagrange form:
x1 , x6 = ±1,
x2 , x5 = ±ϕ/2,
x3 , x4 = ±(ϕ − 1)/2,
yk = (−1)k , k = 1, . . . , 6,


∑ ∏ x − xj

 yk .
T5 (x) =
xk − xj
k
j̸=k
(e) Factored representation:
z1 , z5 = ±
√
√
(2 + ϕ)/4,
z2 , z4 = ± (3 − ϕ)/4,
z3 = 0,
5
∏
T5 (x) = 16 (x − zk ).
1
(f) Three-term recurrence:
T0 (x) = 1,
T1 (x) = x,
Tn (x) = 2xTn−1 (x) − Tn−2 (x) for n = 2, . . . , 5.
3.9. The M-file rungeinterp.m provides an experiment with a famous polynomial
interpolation problem due to Carl Runge. Let
f (x) =
1
,
1 + 25x2
and let Pn (x) denote the polynomial of degree n − 1 that interpolates f (x) at
n equally spaced points on the interval −1 ≤ x ≤ 1. Runge asked whether,
as n increases, Pn (x) converges to f (x). The answer is yes for some x, but
no for others.
(a) For what x does Pn (x) → f (x) as n → ∞?
(b) Change the distribution of the interpolation points so that they are not
equally spaced. How does this affect convergence? Can you find a distribution
so that Pn (x) → f (x) for all x in the interval?
3.10. We skipped from piecewise linear to piecewise cubic interpolation. How far
can you get with the development of piecewise quadratic interpolation?
24
Chapter 3. Interpolation
3.11. Modify splinetx and pchiptx so that, if called with two output arguments,
they produce both the value of the interpolant and its first derivative. That
is,
[v,vprime] = pchiptx(x,y,u)
and
[v,vprime] = splinetx(x,y,u)
compute P (u) and P ′ (u).
3.12. Modify splinetx and pchiptx so that, if called with only two input arguments,
they produce PP, the piecewise polynomial structure produced by the standard Matlab functions spline and pchip and used by ppval.
3.13. (a) Create two functions perpchip and perspline by modifying pchiptx
and splinetx to replace the one-sided and not-a-knot end conditions with
periodic boundary conditions. This requires that the given data have
yn = y1
and that the resulting interpolant be periodic. In other words, for all x,
P (x + ∆) = P (x),
where
∆ = xn − x1 .
The algorithms for both pchip and spline involve calculations with yk , hk ,
and δk to produce slopes dk . With the periodicity assumption, all of these
quantities become periodic functions, with period n − 1, of the subscript k.
In other words, for all k,
Exercises
25
yk
hk
δk
dk
= yk+n−1 ,
= hk+n−1 ,
= δk+n−1 ,
= dk+n−1 .
This makes it possible to use the same calculations at the endpoints that
are used at the interior points in the nonperiodic situation. The special case
code for the end conditions can be eliminated and the resulting M-files are
actually much simpler.
For example, the slopes dk for pchip with equally spaced points are given by
dk = 0 if sign(δk−1 ) ̸= sign(δk ),
(
)
1
1
1
1
=
+
if sign(δk−1 ) = sign(δk ).
dk
2 δk−1
δk
With periodicity, these formulas can also be used at the endpoints where
k = 1 and k = n because
δ0 = δn−1 and δn = δ1 .
For spline, the slopes satisfy a system of simultaneous linear equations for
k = 2, . . . , n − 1:
hk dk−1 + 2(hk−1 + hk )dk + hk−1 dk+1 = 3(hk δk−1 + hk−1 δk ).
With periodicity, this becomes
h1 dn−1 + 2(hn−1 + h1 )d1 + hn−1 d2 = 3(h1 δn−1 + hn−1 δ1 )
at k = 1 and
hn dn−1 + 2(hn−1 + h1 )dn + hn−1 d2 = 3(h1 δn−1 + hn−1 δ1 )
at k = n. The resulting matrix has two nonzero elements outside the tridiagonal structure. The additional nonzero elements in the first and last rows
are A1,n−1 = h1 and An,2 = hn−1 .
(b) Demonstrate that your new functions work correctly on
x = 0:pi/4:2*pi;
y = cos(x);
u = 0:pi/50:2*pi;
v = your_function(x,y,u);
plot(x,y,’o’,u,v,’-’)
(c) Once you have perchip and perspline, you can use the NCM M-file
interp2dgui to investigate closed-curve interpolation in two dimensions.
You should find that the periodic boundary conditions do a better job of
reproducing symmetries of closed curves in the plane.
26
Chapter 3. Interpolation
3.14. (a) Modify splinetx so that it forms the full tridiagonal matrix
A = diag(a,-1) + diag(b,0) + diag(c,1)
and uses backslash to compute the slopes.
(b) Monitor condest(A) as the spline knots are varied with interpgui. What
happens if two of the knots approach each other? Find a data set that makes
condest(A) large.
3.15. Modify pchiptx so that it uses a weighted average of the slopes instead of
the weighted harmonic mean.
3.16. (a) Consider
x = -1:1/3:1
interpgui(1-x.^2)
Which, if any, of the four interpolants linear, spline, pchip, and polynomial
are the same? Why?
(b) Same questions for
interpgui(1-x.^4)
3.17. Why does interpgui(4) show only three graphs, not four, no matter where
you move the points?
3.18. (a) If you want to interpolate census data on the interval 1900 ≤ t ≤ 2000
with a polynomial,
P (t) = c1 t10 + c2 t9 + · · · + c10 t + c11 ,
you might be tempted to use the Vandermonde matrix generated by
t = 1900:10:2000
V = vander(t)
Why is this a really bad idea?
(b) Investigate centering and scaling the independent variable. Plot some
data, pull down the Tools menu on the figure window, select Basic Fitting,
and find the check box about centering and scaling. What does this check
box do?
(c) Replace the variable t with
s=
t−µ
.
σ
This leads to a modified polynomial P̃ (s). How are its coefficients related to
those of P (t)? What happens to the Vandermonde matrix? What values of
µ and σ lead to a reasonably well conditioned Vandermonde matrix? One
possibility is
mu = mean(t)
sigma = std(t)
but are there better values?
Bibliography
[1] C. de Boor, A Practical Guide to Splines, Springer-Verlag, New York, 1978.
[2] F. N. Fritsch and R. E. Carlson, Monotone Piecewise Cubic Interpolation,
SIAM Journal on Numerical Analysis, 17 (1980), pp. 238-246.
[3] D. Kahaner, C. Moler, and S. Nash, Numerical Methods and Software,
Prentice–Hall, Englewood Cliffs, NJ, 1989.
27
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