# Approximating a sum of exponentials

1. Sep 12, 2010

### mooshasta

I came across the following statement:

$$\sum_n p(n)e^{-in\theta} \approx exp[-i\theta \langle n\rangle - \theta^2 \langle ( \delta n)^2 \rangle / 2]$$

where $\theta$ is small, $\sum_n p(n) = 1$, $\langle n \rangle = \sum_n p(n)n$, and $\langle ( \delta n)^2 \rangle = \sum_n p(n)(n-\langle n \rangle)^2$.

I am pretty stumped trying to figure out how this asymptotic expression is derived. I tried writing out the exponents as sums to no avail. I can see that $(-i\theta)^2 = -\theta^2$ but I am pretty confused regarding the presence of $\langle n \rangle$ and $\langle (\delta n)^2 \rangle$ in the exponential. Any suggestions are greatly appreciated!

2. Sep 14, 2010

### trambolin

I did not spend much time on it but if "exp" means the exponential, then this looks like a small angle approximation.

3. Sep 14, 2010

### Mute

Look up "Cumulant expansion". Alas, wikipedia has no page for this, but it is the method one would use to arrive at this approximation (and get better approximations). If you expand the exponential factor in the sum to second order and do the averages, the result looks just like the series for the right hand side of the equation to first order.

A related result that holds exactly: If $\theta$ is Gaussian distributed, then

$$\langle \exp(i\theta) \rangle = \exp(-\langle \theta^2 \rangle)$$