I gave a presentation for a seminar class yesterday on Fourier analysis, and introduced the sawtooth function as a counterexample, for a function whose Fourier series is not termwise differentiable. This example was motivated by my claim that physicists run around willy-nilly, and differentiate term-by-term without regard for convergence.

So, in short, if $f(x)=x$ for $0\le x\le 2\pi$ and we extend by periodicity, then $f$ has Fourier series $\sum_{n\neq 0, n\in\mathbb{Z}}^\infty -\frac{1}{in} e^{inx}$, which converges pointwise (conditionally). If we differentiate termwise, we get the series $-\sum_{n\neq 0} e^{inx}$, which never converges, and yet we should have gotten $1$ off of $2\pi\mathbb{Z}$.

My point: just after my presentation, a friend with a greater inclination to physics actually showed a "physicist's justification" of the above series actually equaling $1$. I.e. if we pretend that the geometric sum $\sum_{n=0}^\infty r^n$ actually converges for $|r|=1, r\neq 1$ then we can write the above series as $-\sum_{n=1}^\infty e^{-inx}+e^{inx}=-\frac{e^{-ix}}{1-e^{-ix}}-\frac{e^{ix}}{1-e^{ix}}$, which actually equals $1$ for $x\notin 2\pi\mathbb{Z}$!

My question is: is there a formal justification for this manipulation? I have in mind one involving the power series ring $\mathbb{C}[[x]]$, so that we can formally ignore convergence issues. The fact that this method, which is so simple, actually gives the right answer has convinced me that there must be some truth to it. Thanks for any advice or references you may be able to provide!

  • $\begingroup$ en.wikipedia.org/wiki/… $\endgroup$ Dec 12 '14 at 1:49
  • $\begingroup$ Does the series converges uniformly ?. $\endgroup$ Dec 12 '14 at 1:52
  • $\begingroup$ @QiaochuYuan So this class was actually based on Friedlander & Joshi, so I'm comfortable with distributions, and I actually believe that my series converges in $D'(\mathbb{R})$ to $1-2\pi\sum_{n=-\infty}^\infty \delta(x-2\pi n)$. Maybe to be more clear, I think I'm looking for a method that justifies using the geometric series manipulation, which I don't think distributions quite do. And uniform convergence implies pointwise convergence, so no. $\endgroup$
    – user59193
    Dec 12 '14 at 2:05

This is a valid manipulation of formal Laurent series $\mathbb{C}[[e^{ix},e^{-ix}]]$. Note that formal Laurent series don't form a ring because some pairs of elements can't be multiplied, but we indeed have that $\sum_{k=0}^{\infty}{e^{ikx}}$ is the multiplicative inverse of $1-e^{ix}$ and similarly for negative powers.

  • $\begingroup$ This isn't a ring? I might be wrong, but I thought this should just be $\mathbb{C}[x,y]/(xy-1)$, which in my head is a ring. $\endgroup$
    – user59193
    Dec 12 '14 at 2:12
  • $\begingroup$ @forgetfulfunctor Substitution doesn't work for infinite sums. That's a polynomial property. Try squaring $\sum_{k=-\infty}^{\infty}{x^k}$ and you'll see it doesn't work. $\endgroup$ Dec 12 '14 at 3:31
  • $\begingroup$ But in $\mathbb{C}[[x,y]]/(xy-1)$ like, that polynomial is in the same equivalence class, i.e. is the same, as $1+\sum_{k=1}^\infty x^k+y^k$, which you can square nicely, right? And every Laurent series can be put like that, uniquely, can't they? $\endgroup$
    – user59193
    Dec 12 '14 at 9:30
  • $\begingroup$ *, that series. Sorry, it says comments can only be edited for 5 minutes. $\endgroup$
    – user59193
    Dec 12 '14 at 10:02
  • 1
    $\begingroup$ @forgetfulfunctor You don't even really need to go to something that advanced, I shouldn't think. For example, the two simplifications really do happen in each separate power series ring. In ${\mathbb{C}}[[e^{ix}]]$ we have $\sum_{k=1}^{\infty}{e^{ikx}}=\frac{e^{ix}}{1-e^{ix}}$. $\endgroup$ Dec 14 '14 at 14:45

Your Answer

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