How to use integral with an implicit function defined through a sum

Suppose we have a function define as follows:

fz=@(z) exp(-(z).^2/2);

g=@(y) sum ( v(1:n/2).*fz((y- v(n/2+1:n))));

here v is a vector of length n that is known and is used to define the fucntion g.

Now I want to integrated g from [-5,5] using command

integral(@(y) g(y), -5,5).

I, however, get an error that matrix dimensions must agree.

I get where the issue is. It occurs because I defined a function through a vector and pass another vector as input.

Is there a way of fixing this?

[David Hill]

What is the v function?

[Alex Dytso]

Right, I forgot to say what it is. It's a vector of length n that is known and is used to define a function g.

Answer 1

[Steven Lord]

By default, integral will call your function with an array and expects your function to return an array of the same size as the output. If you specify the name-value pair arguments 'ArrayValued', true in your integral call it will instead call your function with a scalar and expect an array as output.

v = 2:5;
y1 = integral(@(x) x.^v, 0, 1, 'ArrayValued', true)
y2 = integral(@(x) x.^v, 0, 1) % Throws an error