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# chapter 8 interpolation 1 - PowerPoint PPT Presentation

Chapter 8 Interpolation (1). Table of Contents. 8.1 Polynomial Interpolation 8.1.1 Lagrange Interpolation 8.1.2 Newton Interpolation 8.1.3 Difficulties with Polynomial Interpolation 8.2 Hermite Interpolation 8.3 Rational-Function Interpolation. Lagrange Interpolation Polynomials.

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Chapter 8Interpolation (1)

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• 8.1 Polynomial Interpolation

• 8.1.1 Lagrange Interpolation

• 8.1.2 Newton Interpolation

• 8.1.3 Difficulties with Polynomial Interpolation

• 8.2 Hermite Interpolation

• 8.3 Rational-Function Interpolation

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• Basic concept

• The Lagrange interpolating polynomial is the polynomial of degree n-1 that passes through the n points.

• Using given several point, we can find Lagrange interpolation polynomial.

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• The general form of the polynomial is

p(x) = L1y1 + L2y2 + … + Lnyn

where the given points are (x1,y1), ….. , (xn,yn).

• The equation of the line passing through two points (x1,y1) and (x2,y2) is

• The equation of the parabola passing through three points (x1,y1), (x2,y2), and (x3,y3) is

• http://math.fullerton.edu/mathews/n2003/lagrangepoly/LagrangePolyProof.pdf

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• Example

• Given points (x1,y1)=(-2,4), (x2,y2)=(0,2), (x3,y3)=(2,8)

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• We can represent the Lagrange polynomial with coefficient ck.

• p(x)=c1N1+c2N2+ … +cnNn

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• Higher order interpolation polynomials

x = [ -2 -1 0 1 2 3 4],

y = [ -15 0 3 0 -3 0 15]

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• In Lagrange interpolation polynomial, it always go through given points.

Think with equation below

• The Lagrange form of polynomial is convenient when the same abscissas may occur in different applications.

• It is less convenient than the Newton form when additional data points may be added to the problem.

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• Newton form of the equation of a straight line passing through two points (x1, y1) and (x2, y2) is

• Newton form of the equation of a parabola passing through three points (x1, y1), (x2, y2), and (x3, y3) is

• the general form of the polynomial passing through n points (x1, y1), …,(xn, yn) is

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• Substituting (x1, y1) into

• Substituting (x2, y2) into

• Substituting (x3, y3) into

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• Passing through the points (x1, y1)=(-2, 4), (x2, y2)=(0, 2), and (x3, y3)=(2, 8).

• The equations is

• Where the coefficients are

• thus

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• Passing through the points (x1, y1)=(-2, 4), (x2, y2)=(0, 2), and (x3, y3)=(2, 8).

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• We extend the previous example, adding the points (x4, y4) = (-1, -1) and (x5, y5) = (1, 1)

• Divided-difference

table becomes

(with new entries

shown in bold)

• Newton interpolation

polynomial is

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• Consider again the data from

Example 8.4 with Lagrange form.

x = [ -2 -1 0 1 2 3 4 ],

y = [ -15 0 3 0 -3 0 15]

• Do it again with Newton form.

That the polynomial is cubic is clear.

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• If the y values are modified slightly, the divided-difference table shows the small contribution from the higher degree terms:

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>> x = [-2 -1 0 1 2 3 4];

>> y = [-15 0 3 0 -3 0 15];

>> Newton_Coef(x,y);

d =

15 -6 1 0 0 0

3 -3 1 0 0 0

-3 0 1 0 0 0

-3 3 1 0 0 0

3 6 0 0 0 0

15 0 0 0 0 0

>>

function a = Newton_Coef(x, y)

n = length(x);

%Calculate coeffiecients of

Newton interpolating polynomial

a(1) = y(1);

for k=1 : n-1

d(k,1) = (y(k+1) - y(k))/(x(k+1) - x(k));

%1st divided diff

end

for j=2 : n-1

for k=1 : n-j

d(k,j) = (d(k+1,j-1) - d(k,j-1))/(x(k+j) - x(k));

%jth divided diff

end

end

d

for j=2:n

a(j) = d(1, j-1);

end

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>> a = [-15 15 -6 1 0 0 0];

>> x = [-2 -1 0 1 2 3 4];

>> t = [0 1 2];

>> Newton_Eval(t, x, a);

t =

0 1 2

p =

3 0 -3

function p = Newton_Eval(t,x,a)

% t : input value of the polynomial (x)

% x : x values of interpolating points

% a : answer of previous MATLAB function, i.e,

the coefficients of Newton polynomial.

n = length(x);

hold on;

for i =1 : length(t)

ddd(1) = 1; %Compute first term

c(1) = a(1);

for j=2 : n

ddd(j) = (t(i) - x(j-1)).*ddd(j-1);

% Compute jth term

c(j) = a(j).*ddd(j);

end;

p(i) = sum(c);

%plot(t(i),p(i)); grid on;

end

t

p

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• The data

• x = [ -2 -1.5 -1 -0.5 0 – 0.5 1 1.5 2]

• y = [ 0 0 0 0.87 1 0.87 0 0 0]

illustrate the difficulty with using higher order polynomials to interpolate a moderately large number of points.

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• The data

• x = [ 0.00 0.20 0.80 1.00 1.20 1.90 2.00 2.10 2.95 3.00]

• y = [ 0.01 0.22 0.76 1.03 1.18 1.94 2.01 2.08 2.90 2.95]

• Not well suited with noisy straight line.

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• The function

is a famous example of the fact that polynomial interpolation does not produce a good approximation for some functions and that using more function values (at evenly spaced x values) does not necessarily improve the situation.

• Example 1.

• x = [ -1 -0.5 0.0 0.5 1.0 ]

• y = [0.0385 0.1379 1.0000 0.1379 0.0385]

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• Example 2.

• x = [-1.000 -0.750 -0.500 -0.250 0.000 0.250 0.500 0.750 1.000 ]

• y = [0.0385 0.0664 0.138 0.3902 1.000 0.3902 0.138 0.0664 0.0385]

• The interpolation polynomial overshoots the true polynomial muchmore severely than the polynomial formed by using only five points.

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8.2Hermite Interpolation

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• Hermite interpolation allows us to find a ploynomial that matched both function value and some of the derivative values

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• As with lower order polynomial interpolation, trying to interpolate in humped and flat regions cause overshoots.

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8.3Rational-Function Interpolation

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• Why use rational-function interpolation?

• Some functions are not well approximated by polynomials.(runge-function)

• but are well approximated by rational functions, that is quotients of polynomials.

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• Bulirsch-Stoer algorithm

• The approach is recursive, based on tabulated data(in a manner similar to that for the Newton form of polynomial interpolation).

• Given a set of m+1 data points (x1,y1), … , (xm+1, ym+1), we seek an interpolation function of the form

<Bulirsch-Stoer algorithm general pattern>

The proof is in J.Stoer and R.Bulirsch, 'Introduction to Numerical Analysis'

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• Bulirsch-Stoer method for three data points

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Second stage

data

First stage

x1 y1

R1= y1

x2 y2

R2= y2

x3 y3

R3=y3

x4 y4

R4=y4

x5 y5

R5=y5

Bulirsch-Stoer algorithm(cont’d)

• Bulirsch-Stoer method for five data points

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Forth stage

Bulirsch-Stoer algorithm(cont’d)

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data points:

x = [-1 -0.5 0.0 0.5 1.0]

y = [0.0385 0.1379 1.0000 0.1379 0.0385]

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