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I have an equation which is the sum of exponentials $V(t)=\sum_{pre}\epsilon(t-t_{pre})$ where $\epsilon(t)=\frac{\epsilon_0}{\tau_m}e^{\frac{-t}{\tau_m}}$. The terms in $pre$ are evenly spaced time intervals. Is there not some identity that relates this sum to a single exponential? I feel like it is something related to the fourier series, but I am not remembering.

Any clues?

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Not sure that I understand -- otherwise the question is entirely trivial. Are the $t_{pre}$ terms constants? If so, your definition implies $\epsilon(t-t_{pre}) = \exp(t_{pre}/\tau_m)\epsilon(t)$. Apply the distribution law. This can't be what you really mean, right? –  Jerry Gagelman Dec 5 '10 at 1:10
No, the $t_{pre}$ are a series of previous times, evenly spaced. So it is the sum of several decaying exponentials of a given rate. What I want to do is relate the rate at which the $t_{pre}$s occur to the value of $V(t)$ using a single exponential, without having to take the sum over each point in time of the previous events. –  kaleidomedallion Dec 5 '10 at 1:25
Pull out $\epsilon_0/\tau_m$. If $t_{pre}=0,1,2,\ldots$, then what remains is a sum of powers of $e$, and if I've calculated correctly, it sums to $\frac{e^{-\frac{t}{\tau_m}}}{e^{\frac{1}{\tau_m}}-1}$. –  Joseph O'Rourke Dec 5 '10 at 3:55
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1 Answer

up vote 5 down vote accepted

How's this:

Take the times $t_{pre}$ to be of the form $t_{pre} = T_{0} + iT$, where $i$ ranges over some set of integers of the form $\{0, 1, 2, . . . , N\}$, and we allow the possibility that $N$ is infinite and place no restriction on the sign of $T$, the interval size, though we take $T \ne\ 0$; since the case $T = 0$ boils down to there being only one value of $t_{pre}$, $T_{0}$, in which case $V(t)$ trivially becomes $(\epsilon_{0}/\tau)e^{-(t - T_{0})/\tau}$. (I'm dropping your subscripts to $\tau$ for convenience, i.e. to make typing slightly easier.) Then in general we have $V(t) = \sum_{i = 0}^{N}(\epsilon_{0}/\tau)e^{-(t-T_{0} - iT)/\tau}$; factoring out $(\epsilon_{0}/\tau)e^{-(t-T_{0})/\tau}$ yields $V(t) = (\epsilon_{0}/\tau)e^{-(t-T_{0})/\tau}\sum_{i = 0}^{N}e^{iT/\tau}$; in this form, $V(t)$ is a single exponential times a constant, as per your request. The sum can be cleaned up using standard formulae for geometric series (cf. http://en.wikipedia.org/wiki/Geometric_series), viz $\sum_{i = 0}^{N}e^{iT/\tau} = ((1 - e^{(N + 1)T/\tau})/(1- e^{T/\tau}))$, so finally $V(t) = (\epsilon_{0}/\tau)e^{-(t-T_{0})/\tau}((1 - e^{(N + 1)T/\tau})/(1- e^{T/\tau}))$. In the event $T > 0$, the sum diverges unless $N$ is finite; for $T < 0$, however, letting $N \to \infty$ we obtain $V(t) = (\epsilon_{0}/\tau)e^{-(t-T_{0})/\tau}(1/(1- e^{T/\tau}))$ or $V(t) = (\epsilon_{0}/\tau)(1/(1- e^{T/\tau}))e^{-(t-T_{0})/\tau}$.

It is interesting to note that, as $T < 0$ becomes smaller in magnitude, i.e. $T \to 0^{-}$, $V(t)$ explodes, as if it were the sum of an infinite number of identical terms $(\epsilon_{0}/\tau)e^{-(t-T_{0})/\tau}$; I leave a similar analysis when $T > 0$ to you.

I think you will find this stuff is generally covered under Laplace transforms, rather than Fourier series.

To the boys at http://tea.mathoverflow.net/discussion/784/question-being-bumped-to-the-front-page/: I did it again! Accidentally hit "Save" instead of "Preview"! Only lasted an hour this time. Whew! Gotta be more careful with those buttons!

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Wow, very thorough! I am trying to learn as much as I can, but very much starting out on all the math past differential equations. Much appreciated, truly! T is always greater than 1 in my application, so I won't run in to the situation you mention. Always good to keep in mind though. –  kaleidomedallion Dec 6 '10 at 8:10
Thanks a lot for your good words, and that phat check! –  drbobmeister Dec 6 '10 at 9:31
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