Newtonian interpolation polynomial that interpolates f twice (value and 1st derivative)
1 次查看(过去 30 天)
显示 更早的评论
Construct a Newtonian interpolation polynomial that interpolates f at all points x = −2, −1,0,1,2 twice (value and 1st derivative). What is the value of the interpolation polynomial at x = 1.5?
This is my code but I do not get the correct answer:
format long
function d = deljeneDif_P3_2xInterp(X,Y,DY)
M = zeros(4);
M(:,1) = [Y(1); Y(1); Y(2); Y(2)];
M(:,2) = [0; DY(1); (Y(2)-Y(1))/(X(2)-X(1)); DY(2)];
M(:,3) = [0; 0; (M(3,2)-M(2,2))/(X(2)-X(1)); (M(4,2)-M(3,2))/(X(2)-X(1))];
M(4,4) = (M(4,3)-M(3,3))/(X(2)-X(1));
d = diag(M);
end
a = 1./2.80;
f = @(x) (8.*a.^3)./(x.^2+4.*a.^2);
df = @(x) (-16.*a.^3.*x)./(x.^2+4.*a.^2).^2;
interpX = [-2 -1 0 1 2];
interpY = f(interpX);
interpDY = df(interpX);
d = deljeneDif_P3_2xInterp(interpX,interpY,interpDY);
vrednost_2 = polyval(d, 1.5);
0 个评论
回答(1 个)
Davide Masiello
2022-5-18
If the problem is the error appearing in your question, i.e.
Function definitions in a script must appear at the end of the file.
then just change the script in the following way and it works.
format long
a = 1./2.80;
f = @(x) (8.*a.^3)./(x.^2+4.*a.^2);
df = @(x) (-16.*a.^3.*x)./(x.^2+4.*a.^2).^2;
interpX = [-2 -1 0 1 2];
interpY = f(interpX);
interpDY = df(interpX);
d = deljeneDif_P3_2xInterp(interpX,interpY,interpDY);
vrednost_2 = polyval(d, 1.5)
function d = deljeneDif_P3_2xInterp(X,Y,DY)
M = zeros(4);
M(:,1) = [Y(1); Y(1); Y(2); Y(2)];
M(:,2) = [0; DY(1); (Y(2)-Y(1))/(X(2)-X(1)); DY(2)];
M(:,3) = [0; 0; (M(3,2)-M(2,2))/(X(2)-X(1)); (M(4,2)-M(3,2))/(X(2)-X(1))];
M(4,4) = (M(4,3)-M(3,3))/(X(2)-X(1));
d = diag(M);
end
另请参阅
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!