I am facing the following problem. I have a function which is defined through a discrete sum of Gaussians
$$F\_M(t) = 2\sum\limits\_{n=1}^{M}e^{-t^2 \sigma^2 n^2}\times \sum\limits\_{k=n}^{M}p\_k p\_{k-n} + \sum\limits\_{k=0}^{M}p\_k^2$$
where $\sigma$ is just a real parameter and $p\_k$ are weights
$$\sum\limits\_{k=0}^{M}p\_k = 1.$$
This functions decays from $F\_M(0)=1$ down to $F\_M(\infty)=\sum\limits\_{k=0}^{M}p\_k^2$. 


The question is: how the decay time $\tau$ defined through the condition $F\_M(\tau) = 1/2(1 - F\_{M}(\infty))$ (at half height) depends on $M$? 


Firstly, I tried to attack a particular case when $p\_k = 1/(M+1)$. Then
$$F\_M(t) = \frac{2}{(M+1)^2}\left[(M+1)\sum\limits\_{n=1}^{M}e^{-t^2 \sigma^2 n^2} - \sum\limits\_{n=1}^{M}n\ e^{-t^2 \sigma^2 n^2} \right] + \frac{1}{(M+1)}.$$
I didn't manage to obtain any answer even for this simple case. The size $M$ does not necessarily has to be very large, but it is also interesting what happens when $M \rightarrow \infty$. Any ideas how the decay time might depend on $M$?



Discrete summation of Gaussian functions. Decay time problem
the decay time in your "simple case" is well approximated by the large-$M$ limit [\*]


$$\lim\_{M\rightarrow\infty}M\sigma\tau=2.84$$


here is a plot of


$$f\_M(s)=\left.\frac{F\_M(t)}{1-F\_M(\infty)}\right|\_{t=s/(M\sigma)}$$


for $M=5,10,100$, that shows the half-time $s\approx 3$ is quite accurate already for not so large values of $M$


![](https://ilorentz.org/beenakker/MO/fMs.png)




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[\*] for the large-$M$ limit I replace the sum in the defition of $F\_M(t)$ by an integral and solve for the half-time,


$$\int\_0^M\frac{2(M-x)}{M^2}e^{-(\sigma \tau x)^2}\,dx=\frac{1}{2}$$


this gives $M\sigma\tau=2.84092\cdots$.


