Another short one. This one bugged me for quite a while until I realized what the answer was a few years ago. The Sampling Theorem states that (under the right conditions) $\displaystyle x(t) = \sum_{n=-\infty}^{\infty} x(nT) \;\textrm{sinc}\left(\frac{t - nT}{T}\right)$

where $\displaystyle \textrm{sinc}(x) = \frac{\sin(\pi x)}{\pi x}$

(the normalized sinc function). The problem is this: Where does the sinc function come from? (In a philosophical sense. It plops out of the proof sure enough, but that’s not what I mean). Fourier theory is full of (trigonometric) polynomials (i.e. sine/cosine waves when you’re dealing with real-valued signals), so where does the factor of x in the denominator suddenly come from?

The answer is a nice identity discovered by Euler: $\displaystyle \textrm{sinc}(x) = \prod_{n=1}^{\infty} \left(1 - \frac{x^2}{n^2}\right)$

With some straightforward algebraic manipulations (ignoring convergence issues for now) you get: $\displaystyle \textrm{sinc}(x) = \prod_{n=1}^{\infty} \left(1 - \frac{x}{n}\right) \left(1 + \frac{x}{n}\right)$ $\displaystyle = \prod_{\substack{n \in \mathbb{Z} \\ n \ne 0}} \left(1 - \frac{x}{n}\right) = \prod_{\substack{n \in \mathbb{Z} \\ n \ne 0 }} \frac{n - x}{n}$

Compare this with the formula for Lagrange basis polynomials: $\displaystyle L_{i;n}(x) = \prod_{\substack{0 \le j \le n \\ j \ne i}} \frac{x_j - x}{x_j - x_i}$

in other words, the sinc function is the limiting case of Lagrange polynomials for an infinite number of equidistant control points. Which is pretty neat :)

From → Maths

One Comment
1. 