# approximation of sum of gaussian-like function?

Let:

$g(u; x,s) = \dfrac{1}{s\sqrt{2\pi}} \exp\left(-\dfrac{1}{2} \left(\dfrac{x-u}{s}\right)^2\right)$

Where $x,s$ are parameters

I'm looking for a closed-form solution or approximation of:

$f_1(k) = \dfrac{\sum\limits^{r=+\infty}_{r=-\infty} rg(k+rD)}{\sum\limits^{r=+\infty}_{r=-\infty} g(k+rD)}$

and

$f_2(k) = \dfrac{\sum\limits^{r=+\infty}_{r=-\infty} r^2g(k+rD)}{\sum\limits^{r=+\infty}_{r=-\infty} g(k+rD)}$

Typically:

$D > 0$

$0 < s < D$

$\dfrac{-D}{2} < k < \dfrac{D}{2}$

$\dfrac{-D}{2} < x < \dfrac{D}{2}$

In practice, the summation converges very quickly, so $r$ is summed from at most -10 to 10.

Background: this resulted from working with wrapped gaussians, where I need to solve this equation for $k$:

$\sum\limits^{N}_{n=0}p(n)\left(x_n - k - D \frac{\sum\limits^{\infty}_{r=-\infty} rg(k+rD;x_n,s_n)}{\sum\limits^{\infty}_{r=-\infty}g(k+rD;x_n, s_n)} \right) = 0$

If the nastiness with last fraction of sums wasn't there, it'd be super easy.

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Why are you using quantities that look like moments of Gaussians assuming that the values are integers, even though the Gaussian distribution is continuous instead of integer-valued? –  user2566092 Sep 13 '13 at 20:45
This resulted from working with: en.wikipedia.org/wiki/Wrapped_normal_distribution –  proteneer Sep 13 '13 at 20:54

Since you are dealing with series instead of integrals, and $e^{-z^2}$ approaches $0$ very very quickly as $|z|$ increases, and given your ranges on $D,k,x$ in terms of $s$, I'm pretty sure you will find no significantly faster approximation other than to simply evaluate the series for $r \in \{-R,-R+1,\ldots,-1,0,1,\ldots,R-1,R\}$ for some small choice of $R$. The only question is how many terms in the series of $e^{-z^2}$ you need to get a good approximation of something like $r^2 g(k+rD)$ for a given choice of $r$, but it should be very few (like $10$). Let me know if you would like estimates of the rates of convergence of your summation ratio quantities in terms of $R$ and the number of terms you use to estimate $e^{-z^2}$, so you know how large to choose to choose these based on how close you want your estimated quantities to be to the true values. But I can already foresee, if you estimate $e^{-z^2}$ using say $10$ terms, and choose $R$ to be no more than $10$, you will already get very, very close answers to the true quantities.
I actually do this already, but I was hoping for some approximate expression for $f_1(k)$ and $f_2(k)$ that would be linear in $k$. eg. $f_1(k) = h(x,r,s,D) + ak$, or least have some nice expression for $k$ –  proteneer Sep 13 '13 at 20:59