Brad,

I assume $x$ is an integer. Set $w=\exp (2i\pi /x)$. Now try expanding:

$1+e^w+e^{w^2}+\dots +e^{w^{x-1}}$

Note that it is:

$\sum _{m=0}^{x-1}{e}^{w^m}=\sum _{m=0}^{x-1}\sum _{n=0}^{\infty }{w}^{mn}/n!=\sum _{n=0}^{\infty }\sum _{m=0}^{x-1}{w}^{mn}/n!=\sum _{n=0}^{\infty }1/n!\times [1+w^n+w^{2n}+\dots +w^{(x-1)n}]$

The term in square brackets is obviously $x$ if $w^n=1$, i.e. if $n$ is a multiple of $x$; otherwise it's 0 (using theformula for a geometric progression). So the expression is:

$\sum _{n=0}^{\infty }1/n!\times x$

where the sum is now only over values of $n$ which are multiples of $x$. Therefore it works out as $x(1/0!+1/x!+1/(2x)!+\dots )$ which is $x$ times your original sum. So your original sum is:

$1/x\times \sum _{m=0}^{x-1}{e}^{w^m}$ (*)

We can write this in terms of trig functions, using the fact that $w^m=\exp (2\pi im/x)=\cos (2\pi m/x)+i\sin (2\pi m/x)$ we get your sum to be:

$1/x\times \sum _{m=0}^{x-1}{e}^{\cos (2\pi m/x)+i\sin (2\pi m/x)}$

$=1/x\times \sum _{m=0}^{x-1}{e}^{\cos (2\pi m/x)}(\cos (\sin (2\pi m/x))+i\sin (\sin (2\pi m/x)))$

But since the answer is real, the imaginary part must vanish.

[Which means that we have proven, as a bonus, that

$\sum _{m=0}^{x-1}{e}^{\cos (2\pi m/x)}\sin (\sin (2\pi m/x))=0$]

So your sum works out as:

$1/x\times \sum _{m=0}^{x-1}{e}^{\cos (2\pi m/x)}\cos (\sin (2\pi m/x))$