The following error occurred converting from sym to double: Unable to convert expression into double array.

This part is included in a function called myf2(w,k):

function f= myf2(w,k)
i = sqrt(-1);
fm = @(x)  (1/sqrt(2*pi))* exp(-x.^2/2);
dfm = @(x) -x.*fm(x);
fun = @(x) dfm(x) / (x-w/k);
w0 = max(imag(w/k),0);
a = -10+w0*i;
b = 10+w0*i;
N = 100;
R = dfm(w/k);
x = linspace(a,b,N+1);
syms wn x 
wn = sym(zeros(1, N)); % <------- Something is wrong here 
f_int = int(fun, x, a, b); 
A = vpa(f_int);
f = k^2 - A - 2*pi*i*R;
return

This is where I call the function:

for ix = 1:Nx
    for iy = 1:Ny
        z = x(ix) + I*y(iy);
        tmp = myf2(z,k);
        f(ix,iy) = abs(tmp);
    end
end

After this I'm getting an error! Please help me out!

[Walter Roberson]

In order to test that, we need values for x, y, I, and k; and we need to know that it is safe to assume that Nx = length(x) and Ny = length(y).

Inside myf2 you confusingly use x several times in ways that do not appear to link back to the x being used in the call to myf2 .

x = linspace(a,b,N+1);
syms wn x 

The second of those two lines is equivalent to

wn = sym('wn');
x = sym('x');

So in context you have

x = linspace(a,b,N+1);
wn = sym('wn');
x = sym('x');

That has the effect of overwriting the numeric x you determined, with a symbolic x. If you intended to do that, then why bother calculating the numeric x ? It just confuses things.

[Walter Roberson]

In the test I did a moment ago with random inputs, the expression to be integrated over x had a denominator that became zero between the calculated a and b, leading to a singularity and so to an undefined integration result.

[Angela Mehta]

I see and understand my mistake but then how will I numerically calculate the integral(to avoid the issue of running into division by 0) and still getting an approximately close answer? Initially, I tried to use the following in myf2:

height = (b-a)/N;
s = 1/2* (fun(a)+fun(b));
A = height * s;

[Walter Roberson]

In some cases you could hope that the singularity is "removable" to permit a Cauchy Principle Value to be calculated. However in cases where the function goes continuously to infinity, the singularity has no hope of being removable unless it is the same sign of infinity on both sides of the singularity. The singularity that I encounter with random data with of the form (expression in x) / (linear expression in x) and that has infinities with different sign as the linear expression becomes 0, so the singularity cannot be removed.

In other words, in the case I encountered, no integration was possible. The "approximately close answer" was undefined.

If you post your revised code and post values for x, y, I, and k so that we can test, then we can have another look.

Answer 1

[David Hill]

Do you have to perform symbolically? Does not look like the integral exists. Why not numerically?

function f= myf2(w,k)
dfm =@(x) -x.*(1/sqrt(2*pi)).* exp(-x.^2/2);
fun =@(x) -x.*(1/sqrt(2*pi)).* exp(-x.^2/2)./ (x-w/k);
w0 = max(imag(w./k),0);
a = -10+w0*1i;
b = 10+w0*1i;
R = dfm(w./k);
A = integral(fun,a,b); 
f = k^2 - A - 2*pi*1i*R;
end

[Angela Mehta]

Hello, I tried to perform the integral numerically but wasn’t getting the exact solution(with negligible error). So I tried to do it symbolically and am getting the exact solutions but now am running into that error!

[David Hill]

What error? If your exact answer has 'int' out in front, then it has not really been integrated symbolically.

[Angela Mehta]

I am actually new to Matlab so I might be misplacing my words here and there so I am sorry! When I run the code, it's gives an error of not being able to convert from sym to double.