Solving a sum of series of exponential function with a sum of series of cosine function inside

Here is the equation I'm going to solve.

where epsilon is a variable.

I wrote the following code.

 r = 1:(2^h)-1;
v = 1:h-1
T = exp(x*h/1000)+sum(exp((x/1000)*(cos(2*pi*r/(2^h))+cos(2*pi*r*(2^v)/(2^h)))));

I can run the code when h=2; however, when h becomes greater than or equal to 3, an error that input must be a scalar and a square matrix shows up.

Could anyone please help me on this error?

[Cheung Ka Ho]

After I changed the code into 2.^v, another main problem comes out, which is inner matrix dimensions must agree. I just want to calculate the summation. How can I rewrite the code to ignore the dimensions of r and 2.^v?

[Torsten]

Is "epsilon" in your equation a given function ?

Is "e" in your equation the usual Euler-number ?

A closing round parenthesis is missing in your picture.

What does e{...} mean ? Does it mean exp(...) ?

Please clarify.

Best wishes

Torsten.

[Cheung Ka Ho]

epsilon is only a variable while e is the usual Euler-number. I mistakenly type it as e{...}. It should be e^{...}. Thank you for your help.

Answer 1

[Torsten]

h=...;
epsilon=...;
r=1:2^h-1;
v=1:h-1;
result=sum(exp(epsilon*sum(cos(pi/2^(h-1)*(2.^v).'*r))));

Best wishes

Torsten.

[Cheung Ka Ho]

Thank you!

Answer 2

[Matthew Taliaferro]

You cannot raise things to a power unless they are scalar or square (like the warning said). If you want to square each element, the notation is a little different.

h = 1:10
h_square = h.^2    % as opposed to h^2, which won't work

You also cannot divide something element by element unless it is a scalar.

   r = 1:10; h = 1:10;
   r_over_h = r./(h.^2);    % as opposed to r/(h^2), which won't do what you think it does

[Cheung Ka Ho]

Thank you for your help. It solves the above error. However, another problem that inner matrix dimensions must agree comes out. How can I resolve this error?

Answer 3

[mohammed alzubaidy]