Error: Undefined operator '+' for input arguments of type 'function_handle'

Here is my code;

function z = joy(t,z)

joymin = @(x) 150 - x;

joyadd = @(z) z;

z = integral2(@(x,y) normpdf(x,100,10).*normpdf(y,200,10),t,inf,joymin + joyadd,inf);

end

When I try to run joy(100,10), I get an error saying Undefined operator '+' for input arguments of type 'function_handle'.

The problem comes from joyadd because the code runs smoothly if I remove it.

Answer 1

[Star Strider]

You need to create a new anonymous function that adds the functions:

@(x)joymin(x) + joyadd(x)

so your full integral2 call becomes:

z = integral2(@(x,y) normpdf(x,100,10).*normpdf(y,200,10),t,inf,@(x)joymin(x) + joyadd(x),inf);

Your code will at least not throw that error.

[Emre Yazicioglu]

I tried that and now it doesn't give an error, but I feel like variable z didn't affect outcome of the integral at all.

I get same results if I completely delete function and only add the constant.

    z = integral2(@(x,y) normpdf(x,100,10).*normpdf(y,200,10),t,inf,150,inf);

[Emre Yazicioglu]

Okay, I just changed your solution a bit and it works. Thank you!

 z = integral2(@(x,y) normpdf(x,100,10).*normpdf(y,200,10),t,inf,@(x)joymin(x) + joyadd(z),inf);

[Walter Roberson]

It is confusing that you are using z as an input argument and that your output argument is also z.

It is better to only use the same variable for input and output if you are "updating" the variable. For example,

   function x = addone(x)
     x = x + 1;

or

   function options = eval50(options)
      options.MaxFunEvals = 50;

Those kinds of uses are clear, but the way you used z is confusing.

[Star Strider]

@Emre Yazicioglu — My pleasure!

@Walter Roberson — Thank you. It was confusing for me as well.