Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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?

share|improve this question
    
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
add comment

1 Answer 1

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!

share|improve this answer
1  
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
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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