Error in Hermite Polynomial

3 次查看（过去 30 天）
Mingxuan
Mingxuan 2023-8-20
编辑： Alan Stevens 2023-8-20
Hermite_Polynomial.m
function Q = Hermite_Polynomial(X, Y, YP)
n = length(X) - 1;
Q = zeros(2*n + 2, 2*n + 2);
for i = 0:n
z(2*i + 1) = X(i + 1);
z(2*i + 2) = X(i + 1);
Q(2*i + 1, 1) = Y(i + 1);
Q(2*i + 2, 1) = Y(i + 1);
Q(2*i + 2, 2) = YP(i + 1);
if i ~= 0
Q(2*i + 1, 2) = (Q(2*i + 1, 1) - Q(2*i - 1, 1)) / (z(2*i + 1) - z(2*i - 1));
end
end
for i = 2:2*n + 1
for j = 2:i
Q(i, j) = (Q(i, j - 1) - Q(i - 1, j - 1)) / (z(i) - z(i - j + 1));
end
end
result = Q(i,j);
end
main.m
clear all
X = [1, 2, 3];
Y = [1.105170918, 1.491824698, 2.459603111];
YP = [0.2210341836, 0.5967298792, 1.475761867];
x_interp = 1.25;
% Compute divided differences
Q = Hermite_Polynomial(X, Y, YP);
% Interpolate using Hermite polynomial
n = length(X) - 1;
result_H5 = Q(1, 1);
for i = 1:2*n+1
Q_ii = Q(i,i);
product = 1;
for j = 0:i-1
Q_ii = Q_ii * (x_interp - X(j));
end
result_H5 = result_H5 + Q_ii;
end
% Compute Hermite interpolation using H3(1.25)
result_H3 = Q(1, 1);
for i = 1:2*n-1
Q_ii = Q(i,i);
product = 1;
for j = 0:i-1
product = product * (x_interp - X(j));
end
result_H3 = result_H3 + Q_ii;
end
% Display results
disp(['H5(1.25) approximation: ', num2str(result_H5)]);
disp(['H3(1.25) approximation: ', num2str(result_H3)]);
The above is my code when I run it I get the error saying the following:
Array indices must be positive integers or logical values.
Error in main (line 15)
z_product = z_product * (x_interp - X(j);
Array indices must be positive integers or logical values.
Error in main (line 29)
product = product .* (x_interp - X(j));
Please help me debug the it

请先登录，再回答此问题。

回答（1 个）

Alan Stevens
Alan Stevens 2023-8-20
编辑：Alan Stevens 2023-8-20
In line
for j = 0:i-1
you have j starting at zero. Matlab's indices start at 1, so the following line should have X(j+1) not X(j).

类别

在 Help Center 和 File Exchange 中查找有关 Interpolation 的更多信息

标签

产品


版本

R2022b

Community Treasure Hunt

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

Start Hunting!

Translated by