Interpolating Polynomials Question Code

6 Ansichten (letzte 30 Tage)
Kaylene Widdoes
Kaylene Widdoes am 9 Nov. 2015
Beantwortet: Walter Roberson am 10 Nov. 2015
I'm having some trouble getting this to run. Any suggestions are appreciated. This is what I have done so far and I am getting an error with Y = f(X)
% HW5 - Problem 2
% Chebyshev versus Newton
clear all
f = @(x) exp^(-(x^2)); % function
% Newton
X = -5:5; % equidistant points to compute the interpolating polynomial
Y = f(X); % corresponding y values
x = 0:0.01:10; % points to plot
y = f(x);
C = divdiff(X,Y); % computes the diagonal from the divided difference table
N = zeros(1,length(x));
for i=1:length(x)
N(i) = newtval(C,X,x(i)); % evaluates the newton polynomial at x(i)
end
subplot(1,2,1)
plot(X,Y,'o',x,y,'k',x,N,'r'),title('Newton Interpolation'),
legend('Data','Exact','Newton')
axis([0 10 -2 15])
% Chebyshev
N = length(X); % same number of points as in Newton;
t = zeros(1,N);
c = zeros(1,N);
for k=1:N
t(k) = cos( (2*k-1)*pi/(2*N) ); % Chebyshev points
c(k) = t(k)*(10-0)/2 + (10+0)/2; % Chebyshev points in [0,10]
end
Y = f(c);
C = divdiff(c,Y);
Ch = zeros(1,length(x));
for i=1:length(x)
Ch(i) = newtval(C,c,x(i));
end
subplot(1,2,2)
plot(c,Y,'o',x,y,'k',x,Ch,'r'),title('Chebyshev Interpolation'),
legend('Data','Exact','Chebyshev')
axis([0 10 -2 15])

Antworten (1)

Walter Roberson
Walter Roberson am 10 Nov. 2015
f = @(x) exp(-(x.^2)); % function

Kategorien

Mehr zu Interpolation finden Sie in Help Center und File Exchange

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by