Why does my anonymous function give array indices error?

3 vues (au cours des 30 derniers jours)
Ann
Ann le 16 Oct 2024
Réponse apportée : Ann le 17 Oct 2024
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 commentaires
Torsten
Torsten le 16 Oct 2024
Modifié(e) : Torsten le 16 Oct 2024
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 le 17 Oct 2024
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

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Ann
Ann le 17 Oct 2024
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.

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