I studied fourier series as an undergrad and grad. student in EE but did not fully grasp the concepts. Now that I am involved in medical imaging (MRI) understanding the basics of fourier series and transforms is very important and I am frustrated at my level of understanding. For example, a problem in a MRI physics book asks:

Prove the following fourier series identity: $$\sum_{n=-\infty}^\infty \exp(2i\pi n a) = \sum_{m=-\infty}^\infty \delta(a-m)$$. The author gives a hint: "consider integrations over small intervals that either include or exclude the region where the argument of one of the delta functions vanishes".

My approach was to write out the left side of the equation for n = -5 to 5. I end up with the value 11. something's wrong for sure. Any help is greatly appreciated!

Thank you -Dave

  • 1
    $\begingroup$ Strictly speaking, that is not a Fourier series. Did you ever read about distribution theory? $\endgroup$ – AD. Jan 6 '11 at 13:07
  • $\begingroup$ Can you please give us the book title and the author? $\endgroup$ – AD. Jan 6 '11 at 13:08
  • $\begingroup$ How did you get an $11$ when both sides of the equation depend on the variable $a$? $\endgroup$ – Mariano Suárez-Álvarez Jan 6 '11 at 13:22
  • $\begingroup$ The hint suggests you integrate both sides from $p-\epsilon$ to $p+\epsilon$ for small $\epsilon$ and 1)a particular integer $p$, then for a non-integer $p$ (so no integers are in the interval of integration). I'll do the RHS for non-integer $p$-it is zero. Do you see why? Do you see why this might help? $\endgroup$ – Ross Millikan Jan 6 '11 at 14:31

What you have described is, strictly speaking, an identity of distributions and not functions, as the $\delta$-"function" is a distribution. So the identity means that if you "evaluate" both sides with respect to a Schwarz function (i.e. one which decreases faster than any polynomial asymptotically, and so do all its derivatives), you will obtain the same answer. By definition, "evaluation" means multiplying by a Schwarz function and integrating.

Fix a Schwarz function $\phi$. Then the left side (I am assuming that there is a small typo in what you have written, namely that you have omitted an $n$ in the exponent) is $$\sum_n \int_{\mathbb{R}} f(x) e^{2 \pi i nx} dx$$ while the right hand side is $$\sum_n f(n)$$ because, by definition, multiplying a function by the Dirac function and integrating just picks out the function's value at zero.

So the identity you ask for is really the Poisson summation formula $$\sum_n \hat{f}(n) = \sum f(n).$$ (Here $\hat{f}$ denotes the Fourier transform.)

The proof of the Poisson summation formula is to consider the auxiliary function $g(x) = \sum_{n} f(x + n x)$. This is $1$-periodic and smooth because $f$ is Schwarz, so we can write by the theory of Fourier series, $$g(0) = \sum_n e^{2\pi i n 0} \left( \int_0^1 g(x) e^{2\pi i nx} dx \right),$$ which in turn yields the summation formula: $g(0)$ is the sum $\sum_n f(n)$, while one can check by expanding out the definition of $g$ and using the definition of the Fourier transform that the right-hand-side is the sum $\sum_n \hat{f}(n)$.


First off, there's a typo in the left hand side.. the $n$th term of the series is given by ${\mathbb\exp (2\pi i n a)}$.

What you're being asked to prove is equivalent to the Poisson summation formula. Since Akhil Mathew already proved this as an answer, I'll just try to give some intuition.

You have to be careful when saying ${\mathbb\sum_{n = -\infty}^{\infty}\exp (2\pi i n a) = \lim_{m \rightarrow \infty} \sum_{n = -m}^{m} \exp (2\pi i n a)}$ since the partial sums only converge as distributions and the terms don't go to zero for most $a$. Similarly one has the distributional limit $$ \sum_{n = -\infty}^{\infty} \exp (2\pi i n a) = \lim_{r \rightarrow 1-} \sum_{n = -\infty}^{\infty} r^{|n|}\exp (2\pi i n a)$$ The right-hand sum can be written as the sum two geometric series which can be explicitly summed to $P_r(2\pi a)$, where $P_r(\theta)$ is the famous Poisson kernel. (see http://en.wikipedia.org/wiki/Poisson_kernel). The explicit formula is $$P_r(2\pi a) = {1 - r^2 \over 1 - 2r\cos(2\pi a) + r^2}$$ Note that $P_r(2\pi a)$ has period $1$. It can be shown that for $r$ near $1$, $P_r(2\pi a)$ is the sum of bump functions surrounding each integer, and as $r$ goes to $1$ from below, this sum of bump functions converges (in the sense of distributions) to the sum of delta functions at all integers, which is exactly your right-hand side.

Incidentally, if you tried to do the analogous thing with the distributional limit $\lim_{m \rightarrow \infty} \sum_{n = -m}^{m} \exp (2\pi i n a)$ in place of the above, you'd end out dealing with the Dirichlet kernel instead of the Poisson kernel. One can still show analogous convergence, but it's trickier. This is related to how showing pointwise convergence of Fourier series tends to be harder than showing Abel convergence.


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