Use symsum with non-integer step?

Is it possible to use symsum with a non-integer step? If not, what is a summation work around for using non-integer steps in order to sum over a function?

Here is what I currently have. This is part of a double summation, but the second summation done has it's bounds set by the value of the first summation bounds so I had to use a for loop on that first summation. If you have any input on how to better do the double summation on this, feel free to share.

nn = 0:step:Nmax;
      for z = 1:length(nn)
          in = L(i)-abs(N(j)-nn(z));
          ll = L_le:step:abs(in);
          Ll = 0;
          func = sqrt((L(i) - Ll + 0.5).^2-(N(j) - nn(z) - 0.5).^2)/((L(i) - Ll + 0.5).*(N(j) - nn(z) - 0.5))...
                  - sqrt((L(i) - Ll + 0.5).^2-(N(j) - nn(z) + 0.5).^2)/((L(i) - Ll + 0.5).*(N(j) - nn(z) + 0.5));

          R = symsum(func,Ll,ll)*A(j,i)*B(j,i)*C(j,i)*Cp(j,i) + R;

Answer 1

[Walter Roberson]

No, the underlying symbolic engine only accepts integer steps.

Use integer indices but replace the variable with a function of the index variable that return the appropriate value. For example,

LL = @(IDX) L_e + (IDX-1) * step;