Why does my anonymous function give array indices error?

3 次查看（过去 30 天）
Ann
Ann 2024-10-16
回答： Ann 2024-10-17
I am trying to make a function that inputs no. of terms and calculates a fourier series to that many terms. To do this I am using a for loop to update an anonymous function to add each term.
apprxf = @(x) a0/2;
for n = 1:terms
apprxf = @(x) apprxf(x) + a(x)*cos(n*x) + b(x)*sin(n*x);
end
When I try to plot the function "apprxf", or just input any non integer (such as apprxf(pi) ), I get the error message saying:
"Array indices must be positive integers or logical values." and then points to what is line 4 here.
Does Matlab see the anonymous function as an array instead? There is nothing in the workspace to indicate that there is one that's overriding it, but maybe there still is.
How can I fix this in a way that allows me to plot this function as a graph from 0 to 2pi.
  2 个评论
Torsten
Torsten 2024-10-16
编辑：Torsten 2024-10-16
Where do you define a and b as functions of x ?
And if you want to evaluate "apprxf" for x being an array, you'll probably need pointwise operations:
apprxf = @(x) apprxf(x) + a(x).*cos(n*x) + b(x).*sin(n*x);
instead of
apprxf = @(x) apprxf(x) + a(x)*cos(n*x) + b(x)*sin(n*x);
Walter Roberson
Walter Roberson 2024-10-17
apprxf = @(x) apprxf(x) + a(x)*cos(n*x) + b(x)*sin(n*x);
when the input x is pi then that would be apprxf(pi) + a(pi)*cos(n*pi) + b(pi)*sin(n*pi)
If a and b are arrays, then that would be an attempt to index array a at location pi

请先登录，再进行评论。

请先登录，再回答此问题。

采纳的回答

Ann
Ann 2024-10-17
The answer to this was in fact that I didn't check through my code properly. It should have been "apprxf = @(x) apprxf(x) + a(n)*cos(n*x) + b(n)*sin(n*x);" Instead of "apprxf = @(x) apprxf(x) + a(x)*cos(n*x) + b(x)*sin(n*x);" As the comments pointed out, inputting a non integer attempts to read from a non integer index in arrays a and b, when in reality those should have been using n.

更多回答（0 个）

类别

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

标签

产品


版本

R2023b

Community Treasure Hunt

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

Start Hunting!

Translated by