This has to do with the Nyquist-Shannon sampling and reconstruction theorem and the so-called Whittaker–Shannon interpolation formula. I had previously asked an ancillary question about this here but this is about a specific nagging issue that seems to "periodically" crop up.

Let's begin with a periodic infinite sequence of real numbers, $a_n \in\mathbb{R}$, having period $N>0\in\mathbb{Z}$. That is:

$$ a_{n+N}=a_n \qquad \forall \ n\in\mathbb{Z}. $$

So there are only $N$ unique values of $a_n$.

Imagine these discrete (but ordered) values as equally spaced on the real number line and being interpolated (between integer $n$) as

$$f(x) = \sum_{n=-\infty}^{\infty} a_n \, \operatorname{sinc}(x-n),$$


$$ \operatorname{sinc}(u) \triangleq \begin{cases} \dfrac{\sin(\pi u)}{\pi u} & \text{if } u \ne 0, \\\;1 & \text{if } u = 0. \end{cases} $$

Clearly $f(x)$ is periodic with the same period $N$:

$$ f(x+N) = f(x) \qquad \forall \ x \in \mathbb{R}. $$

All terms are bandlimited to a maximum frequency of $\frac{1}{2}$, so the summation is bandlimited to the same bandlimit. And, in any case, we have

$$ f(x) \Big|_{x = n} = a_n, $$

so the reconstruction works out exactly at the sampling instances.

$$\begin{align} f(x) &= \sum_{n=-\infty}^{\infty} a_n \, \operatorname{sinc}(x-n) \\ &= \sum_{m=-\infty}^{\infty} \sum_{n=0}^{N-1} a_{(n+mN)} \, \operatorname{sinc}\big(x - (n+mN)\big) \\ &= \sum_{m=-\infty}^{\infty} \sum_{n=0}^{N-1} a_n \, \operatorname{sinc}\big(x - (n+mN)\big) \\ &= \sum_{n=0}^{N-1} \left(a_n \, \sum_{m=-\infty}^{\infty} \operatorname{sinc}\big(x - (n+mN)\big)\right). \\ \end{align}$$

Substituting $u \triangleq x-n$ gives

$$ f(x) = \sum_{n=0}^{N-1} a_n \, g(x-n), $$


$$ g(u) = \sum_{m=-\infty}^{\infty} \operatorname{sinc}(u-mN). $$

Clearly the continuous (and real) $g(u)$ is periodic with period $N$:

$$ g(u+N) = g(u) \qquad \forall u \in \mathbb{R}. $$

What is the closed-form expression for $g(u)$ in terms of $u$ and $N$?

For $N$ odd, we get the Dirichlet kernel:

$$ g(u) = \frac{\sin(\pi u)}{N \sin(\pi u/N)}. $$

I can then get that expression by an extension of the Discrete Fourier Transform (DFT) and relating it to the continuous Fourier series:

$$ \hat{a}_k \triangleq \frac{1}{\sqrt{N}} \sum\limits_{n=0}^{N-1} a_n \, e^{-i 2 \pi nk/N}, $$

$$ a_n = \frac{1}{\sqrt{N}} \sum\limits_{k=0}^{N-1} \hat{a}_k \, e^{+i 2 \pi nk/N}. $$

We know that both infinite sequences $a_n$ and $\hat{a}_k$ are periodic with period $N$.

Now, the continuous Fourier series for $f(x)$ is

$$ f(x) = \sum\limits_{k=-\infty}^{\infty} c_k \, e^{+i 2 \pi (k/N) x}, $$

and, because $f(x) \in \mathbb{R}$, we know we have conjugate symmetry

$$ c_{-k} = (c_k)^* \qquad \forall \ k \in \mathbb{Z}. $$

Being "bandlimited" means that

$$ c_k = 0 \qquad \forall \ |k| > \tfrac{N}{2}. $$

But when $N$ is even, what should $g(u)$ be? Now there is potentially a non-zero component to the DFT value at what we EEs call the "Nyquist frequency"; namely $\hat{a}_{N/2}$ exists and might not be zero.

The expression for $g(u)$ I get when $N$ is even is

$$ g(u) = \frac{\sin(\pi u)}{N \tan(\pi u/N)}. $$

But the question is: can it be, in the case that $N$ is even, that

$$ f(x) = \sum_{n=0}^{N-1} a_n \, g(x-n) + A \sin(\pi x),$$

where $A$ can be any real and finite number?

Do you math whiz-bangs know of a good way that I can say, for certain, that $A=0$?

So my most concise question is: for $N$ even and $a_n \in\mathbb{R}$ having period $N>0\in\mathbb{Z}$, namely

$$ a_{n+N}=a_n \qquad \forall \ n\in\mathbb{Z}, $$

is it true that

$$\sum_{n=-\infty}^{\infty} a_n \, \frac{\sin\big(\pi(x-n) \big)}{\pi(x-n)} = \sum_{n=0}^{N-1} a_n \frac{\sin\big(\pi (x-n)\big)}{N \tan\big(\pi (x-n)/N\big)} $$


Another way of looking at the question is this special case. Can anyone prove that

$$\sum_{n=-\infty}^{\infty} (-1)^n \, \frac{\sin\big(\pi(x-n) \big)}{\pi(x-n)} = \cos(\pi x) $$


  • $\begingroup$ i realize that. the Nyquist-Shannon sampling and reconstruction theorem is really built on the Poisson summation formula. but i can't quite get a grip on whether i can prove that $A=0$ when $N$ is even. $\endgroup$ – robert bristow-johnson Jul 8 '19 at 5:30
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    $\begingroup$ Please write a useful title... one that will help others with the same or similar problem find the answers given here. $\endgroup$ – David G. Stork Jul 8 '19 at 5:34
  • $\begingroup$ Bungo: since one of the sums is finite, you do not need absolute convergence to exchange the sums. Note, in the same way, that the convergence of these sums should be proved exactly like when proving that $\sum_{n \geq 1}{\operatorname{sinc}{n}}$ converges (and hold with the same meaning, aka $\sum_{m=-A}^B{\operatorname{sinc}(...)}$ has a limit when $A,B \rightarrow +\infty$. $\endgroup$ – Mindlack Jul 8 '19 at 8:08
  • $\begingroup$ no, @JohnBentin , the period for $\cos(\pi x)$ also is $2$. $\endgroup$ – robert bristow-johnson Jul 8 '19 at 23:52
  • $\begingroup$ okay @DavidG.Stork , is the new title sufficiently useful? $\endgroup$ – robert bristow-johnson Jul 9 '19 at 0:39

The final equation can be written as$$\sum_{n=1}^\infty\frac1{n^2-x^2}=\frac1{2x^2}-\frac\pi{2x}\cot\pi x\quad(x\in\Bbb R\setminus\Bbb Z),$$a proof of which can be found here.

  • $\begingroup$ By "final equation", do you mean this: $$\sum_{n=-\infty}^{\infty} (-1)^n \, \frac{\sin\big(\pi(x-n) \big)}{\pi(x-n)} = \cos(\pi x) \quad ?$$ Because I fail to see the intermediate steps and I fail to see how it gets us to to $\cos(\pi x)$. $\endgroup$ – robert bristow-johnson Jul 10 '19 at 18:39
  • $\begingroup$ @robertbristow-johnson Yes. To get the equation in my answer from your (final) equation, use the fact that $(-1)^n\sin(\pi x-n\pi)=\sin\pi x$, separate the term for $n=0$, namely $(\sin\pi x)/\pi x$, pair up the remaining terms using the identity$$\frac1{x-n}+\frac1{x+n}=\frac{2x}{x^2-n^2},$$and divide through by $2x\sin\pi x$. To prove that the equation is true is not so elementary, and I don't know of an easier proof than the one I cited. $\endgroup$ – John Bentin Jul 10 '19 at 19:51
  • $\begingroup$ but even so, how do you get from $\frac1{2x^2}-\frac\pi{2x}\cot(\pi x)$ to $\cos(\pi x)$? $\endgroup$ – robert bristow-johnson Jul 10 '19 at 19:55
  • $\begingroup$ @robertbristow-johnson After rearranging the equation, both sides of the equation change. Of course the expressions you give in your last comment do not equate. $\endgroup$ – John Bentin Jul 10 '19 at 20:02
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    $\begingroup$ grumble grumble. $\endgroup$ – robert bristow-johnson Jul 10 '19 at 20:38

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