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 $M$?


the decay time in your "simple case" is well approximated by the large-$M$ limit [*]


here is a plot of


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

[*] 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 $M\sigma\tau=2.84092\cdots$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.