17

What is the sum of a shifted sinc function:

$$g(y) \equiv \sum_{n=-\infty}^\infty \frac{\sin(\pi(n - y))}{\pi(n-y)} \, ?$$

DanielSank
  • 1,139
  • 7
  • 20

2 Answers2

20

We use the Poisson summation formula. Define $f(x) \equiv \sin(\pi x) / (\pi x)$. Then the sum we are trying to solve is $$g(y) = \sum_{n=-\infty}^\infty f(n-y) \, .$$ The Poisson summation formula converts the sum over values of $f$ to a sum over values of the Fourier transform of $f$.

Poisson summation

Note that $g(y)$ is periodic with period $1$. The Fourier series coefficients of $g$ are by definition \begin{align} g_\nu &= \int_0^1 dy \, g(y)e^{-i 2 \pi \nu y} \\ &= \int_0^1 dy \, \sum_{n=-\infty}^\infty f(n-y) e^{-i 2 \pi \nu y} \\ (\text{Let }x\equiv n-y) \qquad &= \sum_{n=-\infty}^\infty \int_{n-1}^n dx \, f(x) e^{-i 2 \pi \nu (n-x)} \\ &= \int_{-\infty}^\infty dx \, f(x) e ^{i 2 \pi \nu x} \\ &= \tilde{f}(-\nu) \, . \end{align} where $\tilde{f}$ is the Fourier transform of $f$.

By definition of the Fourier series, \begin{align} g(y) &= \sum_{\nu = -\infty}^\infty e^{i 2 \pi \nu y} g_\nu \\ \text{so} \qquad \sum_{n=-\infty}^\infty f(n-y) &= \sum_{\nu=-\infty}^\infty e^{-i 2 \pi \nu y} \tilde{f}(\nu) \end{align} which is the Poisson summation formula

Solution to the problem

Using the Poisson summation formula, we can write $$g(y) = \sum_{n=-\infty}^\infty f(n-y) = \sum_{\nu=-\infty}^\infty \tilde{f}(\nu) e^{-i 2 \pi \nu y} \, .$$ What is $\tilde{f}$? We can easily compute that the Fourier transform of the tophat function $$ T(x) = \left\{ \begin{array}{l} 1, \qquad -1/2 < x <1/2 \\ 0, \qquad \text{otherwise} \end{array} \right. $$ is $\tilde{T}(\nu)=\sin(\pi \nu) / (\pi \nu)$. By duality of the Fourier transform, that means that $\tilde{f}$ is the tophat function $T$. Therefore we have $$g(y) = \sum_{\nu=-\infty}^\infty T(\nu) e^{-i 2 \pi \nu y} = 1 \, .$$ This is a remarkable result: no matter how much you shift your sample points on a sinc function, the sum of those samples is constant.

DanielSank
  • 1,139
  • 7
  • 20
2

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \mrm{g}\pars{y} & \equiv \sum_{n = -\infty}^{\infty}{\sin\pars{\pi\bracks{n - y}} \over \pi\pars{n - y}} = {\sin\pars{\pi y} \over \pi y} + \sum_{n = 1}^{\infty}\bracks{% {\sin\pars{\pi\bracks{n - y}} \over \pi\pars{n - y}} + {\sin\pars{\pi\bracks{-n - y}} \over \pi\pars{-n - y}}} \\[5mm] & = -\,{\sin\pars{\pi y} \over \pi y} + \sum_{n = 0}^{\infty}\bracks{% {\sin\pars{\pi\bracks{n - y}} \over \pi\pars{n - y}} + {\sin\pars{\pi\bracks{n + y}} \over \pi\pars{n + y}}} \end{align}

Note that

$\ds{\bracks{{\sin\pars{\pi\bracks{z - y}} \over \pi\pars{z - y}} + {\sin\pars{\pi\bracks{z + y}} \over \pi\pars{z + y}}} \expo{-2\pi\,\verts{\Im\pars{z}}} \stackrel{\mrm{as}\ \Im\pars{z}\ \to\ \pm\infty}{\large\sim} \pm{\expo{-\pi\verts{\Im\pars{z}}} \over \pi\,\Im\pars{z}} \stackrel{\mrm{as}\ \Im\pars{z}\ \to\ \pm\infty}{\large\to}{\large 0}}$

such that the sum can be evaluated by means of the Abel-Plana Formula: \begin{align} \mrm{g}\pars{y} & = -\,{\sin\pars{\pi y} \over \pi y} + \int_{0}^{\infty}\bracks{% {\sin\pars{\pi\bracks{x - y}} \over \pi\pars{x - y}} + {\sin\pars{\pi\bracks{x + y}} \over \pi\pars{x + y}}}\dd x \\[2mm] & + {1 \over 2} \bracks{{\sin\pars{\pi\bracks{x - y}} \over \pi\pars{x - y}} + {\sin\pars{\pi\bracks{x + y}} \over \pi\pars{x + y}}}_{\ x\ =\ 0} \\[5mm] & = -\,{\sin\pars{\pi y} \over \pi y} + {1 \over 2}\int_{-\infty}^{\infty}\bracks{% {\sin\pars{\pi\bracks{x - y}} \over \pi\pars{x - y}} + {\sin\pars{\pi\bracks{x + y}} \over \pi\pars{x + y}}}\dd x \\[2mm] & + {\sin\pars{\pi y} \over \pi y} = {1 \over 2}\int_{-\infty}^{\infty}{\sin\pars{\pi x} \over \pi x}\,\dd x + {1 \over 2}\int_{-\infty}^{\infty}{\sin\pars{\pi x} \over \pi x}\,\dd x = \bbx{\Large 1} \end{align}

Felix Marin
  • 84,132
  • 10
  • 143
  • 185