How to use an anonymous function in a ''for''?

Hi everyone!!

I'm trying to write a general solution for a differential equation using a moment method (Galerkin, with polynomials as base functions). When I try to write the general solution por N polynomials, the anonymous function that I'm using, it doesnt update with every loop. I write my code:

N = 10; %number of elements;
l = zeros([N,N]);
g = zeros(N,1);
A = zeros (N,1);
f = @(x) 0;
for i=1:N 
    for k = 1:N;
    l(i,k) = i*k/(i+k+1);
    g(k,1) = (k*(8+3*k))/(2*(2+k)*(4+k));
    A = linsolve(l,g);    
    f = @(x) f(x) + A(i,1)*(x - x.^(i+1));
    end
    
end

Thanks!!

[Star Strider]

First, this will fail because it uses ‘recursion’:

f = @(x) f(x) + A(i,1)*(x - x.^(i+1));

Second, what is the argument to the function supposed to be? It cannot be ‘x’ (as you called it with ‘f(x)’) because ‘x’ does not exist.

[Migue Balma]

The function I want to create is:

Where the element A(i,1) has the coefficient I want. I use a for in that anonymous function because I want to have it regardless of the value of N. If N = 3, I should have 3 polynomials and so on.

Thank you!!

[Steven Lord]

Use the sum function and element-wise operations inside the anonymous function.

Answer 1

[Asad (Mehrzad) Khoddam]

if you don't need to use it inside loop. you can define f(x) after for loop:

N = 10; %number of elements;
l = zeros([N,N]);
g = zeros(N,1);
A = zeros (N,1);
for i=1:N 
    for k = 1:N;
    l(i,k) = i*k/(i+k+1);
    g(k,1) = (k*(8+3*k))/(2*(2+k)*(4+k));
    A = linsolve(l,g);    
    f = @(x) f(x) + A(i,1)*(x - x.^(i+1));
    end 
end
f = @(x) sum(A'.*(x-x.^(2:N+1)));

[Migue Balma]

Thank you for your answer!! But when I insert and array of values, for example, x1 = linspace (0,1). It gives me an error, that can't evaluate f(x1) to represent it in a plot, I don't know why.

Edit: I solve it using another ''for'' to evaluate the function:

x1 = linspace(0,1)
f1 = zeros(1,length(x1));
for j = 1: length (x1);
    f1(j) = f(x1(j));
end

Thank you so much!! You solved my question!