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I have what looks like a Fourier series but I don't quite understand how (or if) it is possible to recover a function from this.


Any ideas? If changes to this series are necessary to make it a Fourier series from which an equation is recoverable then that also would be really interesting.

Thanks for the help, much appreciated.

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Note that $e^{i\pi/2}=i$, whence its cube is $-i$. Your series is essentially divergent. – Pedro Tamaroff Jul 23 '13 at 2:02
Fourier series are functions. What you wrote is a sum of constants, which is either a constant function, or (as in this case) not even that. – Andrés E. Caicedo Jul 23 '13 at 2:03
This series is just equal to $-i\sum_{k=1}^\infty{k}$. – Ataraxia Jul 23 '13 at 2:04
up vote 2 down vote accepted

You can factorise $e^{3i\pi/2}$ out of this expression, to get $e^{3i\pi/2}(1+2+3+4+5+\dots)$. So it's divergent (undefined, if you prefer).

It's not really a fourier series anyway: a fourier series is a function. Perhaps you're thinking of something like $f(x) = \dots + a_{-1}e^{-ix} + a_0 + a_1e^{ix} + a_2e^{2ix} + \dots$. Whenever you choose values for the $a_i$, though, you still have to make sure the function converges - that is, you can't take them all equal to $1$, because at $x = 0$ you get $f(0) = \dots + 1 + 1 + 1 + 1 + \dots$, which diverges.

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Fourier series do not have to converge. – tomasz Jul 23 '13 at 2:11
Strictly speaking a Fourier series isn't always a function. E.g. a Fourier series can be defined for a periodic integrable function, but the series need not converge (as tomasz points out). One has to add hypotheses, and decide what type of convergence one wants. E.g., the Fourier series of a continuous periodic function $f$ is uniformly Cesàro summable to $f$, although it might diverge in the ordinary sense. – Jonas Meyer Jul 23 '13 at 2:15
Noted. You're absolutely right, and it's sloppy of me to talk about "convergence" - I simply wasn't prepared to copy out half a textbook about when a fourier series is useful and when it's not. To the best of my knowledge (though correct me if I'm wrong), an arbitrary choice of coefficients $a_i$ is more often than not meaningless - I was simply trying to get across this idea. – Billy Jul 23 '13 at 2:38
One might say that this function is $-\frac{e^{3i\pi/2}}{12}$, since their are notions of summability that have $1+2+3+...=-\frac{1}{12}$. – Baby Dragon Jul 23 '13 at 5:09
For more info on @Baby Dragon's comment, see the question Why does $1+2+3+\dots = {-1\over 12}$?. It also came up in a meta thread. – Jonas Meyer Jul 23 '13 at 7:16

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