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Consider the expression given by $$ \large e^{-\Large\sum_{i=-k}^k(k-|i|)x_i} $$ Is there a way of simplifying this expression?

For example, provided $\{x_i\}$ is bounded and "smooth" enough ($|x_i-x_{i\pm 1}|< \epsilon$, $\forall \epsilon>0$), can we estimate such an expression based, for example on the local mean of $\{x_i\}$ for each $k$, i.e, $\frac{1}{2k+1}\sum_{i=-k}^k x_i$?

MathematicianByMistake
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sam wolfe
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    Well, $x_i$ appears with coefficient $k-|i|$ so you can get one copy of the whole sum on $[-k+1,k-1]$ then a copy of the sum on $[-k+2,k-2]$ etc. – Ian Mar 13 '23 at 11:12
  • @Ian great to see you again! The motivation for this question actually comes from a previous question of mine, which is a follow-up to a question that you answered. If you could look at that one, perhaps you have a bit more insight. That is actually the goal and I wonder what can be said about those variables. Thanks! – sam wolfe Mar 13 '23 at 11:23
  • One idea is to rewrite it as $k(k+1) \sum \frac{(1-|i|/k)x_i}{1+1/k} \frac{1}{k}$ and interpret it for large $k$ as an integral of $(1-|i|/k)x_i$ (let $t=i/k$ and then you just have to assume $x_i$ asymptotically behaves as some $f(t)$). – Ian Mar 13 '23 at 13:55
  • That's an interesting approach. Regarding the linked question on the heterogenous activation process, however, I still think the small valued $k$ terms are important, as these seem to contribute the most to the expectation. Could we somehow transform the sum over $k$ to an integral as well? In practice, all I need is a suitable way of estimating the values $x_i$ given smooth data of the expectation to, essentially, transform the system, even if only locally. I am trying to recover the activation rates from timing data. – sam wolfe Mar 13 '23 at 14:02
  • On the outer sum the integral approximation seems like a non-starter really, except in the $f_i/v \ll 1$ limit. – Ian Mar 13 '23 at 14:16
  • I understand. It seems, in general, quite untractable, but what gives me a bit of hope of finding a local and data-dependent inverse transform is the fact that numerical simulations with a fitting inverse power law on the local mean of the activation rates give me relatively close timing values to the expectation data (though far from ideal). And this is simply on top of the approximation from the constant activation rate case, which you answered (see this neat approximation). – sam wolfe Mar 13 '23 at 14:27
  • Do you have an example of a nontrivial $f_i$ that you tried to reconstruct and more or less succeeded? – Ian Mar 13 '23 at 15:09
  • Let us continue this discussion in chat. – sam wolfe Mar 13 '23 at 15:15

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