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Solved! Get answer or ask a different Question 18631

MATLAB Problems:

1) For f (x) = e^(-|x|) on [-3, 3], you are to compare evenly-spaced interpolation with Chebyshev interpolation. In this exercise, you do not have to use any loops.

b) Modify your script from part (a) to make a new script file to do the following: Compute the n Chebyshev nodes, and call newtdd.m and nest.m the same way you did in part (a) to get an interpolating polynomial on x-values evenly spaced by 0.01 between -3 and 3.

Plot the new interpolating polynomial, and also plot the vector of the absolute errors between the interpolating polynomial and the exact function on these 0.01-spaced x-values for n = 10 and n = 30, where n is the number of nodes.

Use MATLAB to compute an estimate of the maximum of the absolute error in the interpolating polynomial for n = 30. c) Can you compute a theoretical error bound for parts (a) and (b) with the bounds discussed in class? Explain. On a related note, near what part of a function do you think large interpolation errors can occur, besides the endpoints?

(result from class that an upper bound on the absolute error in the unique interpolating polynomial on 2 nodes, x1 and x2, for any function f(x) is

((M2)/8) h^2, where h = x2 – x1 > 0, and M2 = max (x1< x < x2 ) |f ”(x)|.)

%Program 3.1 Newton Divided Difference Interpolation Method

%Computes coefficients of interpolating polynomial

%Input: x and y are vectors containing the x and y coordinates

%    of the n data points

%Output: coefficients c of interpolating polynomial in nested form

%Use with nest.m to evaluate interpolating polynomial

function c=newtdd(x,y,n)

for j=1:n

 v(j,1)=y(j);    % Fill in y column of Newton triangle

end

for i=2:n       % For column i,

 for j=1:n+1-i    % fill in column from top to bottom

  v(j,i)=(v(j+1,i-1)-v(j,i-1))/(x(j+i-1)-x(j));

 end

end

for i=1:n

 c(i)=v(1,i);    % Read along top of triangle

end          % for output coefficients

%Program 0.1 Nested multiplication

%Evaluates polynomial from nested form using Horner’s method

%Input: degree d of polynomial,

%    array of d+1 coefficients (constant term first),

%    x-coordinate x at which to evaluate, and

%    array of d base points b, if needed

%Output: value y of polynomial at x

function y=nest(d,c,x,b)

if nargin<4, b=zeros(d,1); end 

y=c(d+1);

for i=d:-1:1

 y = y.*(x-b(i))+c(i);

end

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