# Changes

## Sinc function

, 18:47, 19 December 2011
Second derivative
$\operatorname{sinc}''(x) = \left\lbrace \begin{array}{rl} \frac{-1}{3}, & x = 0 \\ \frac{(2 - x^2)\sin x - 2x\cos x}{x^3}, & x \ne 0 \\\end{array}\right.$

===Higher derivative at zero===

The limit procedure used for the first and second derivatives can be extended to the computation of higher derivatives at zero. However, this process is tedious. An alternative approach is to use the [[power series]] expansion:

$\operatorname{sinc} x = 1 - \frac{x^2}{3!} + \frac{x^4}{5!} - \dots = \sum_{k=0}^\infty \frac{(-1)^kx^{2k}}(2k + 1)!}$

Since this is a power series, it is also a Taylor series. Thus, the coefficient of $x^{2k}$ is $1/(2k)!$ times $\operatorname{sinc}^{(2k)}(0)$. We thus get:

$\frac{1}{(2k + 1)!} = \frac{(-1)^k \operatorname{sinc}^{(2k)}(0)}{(2k)!}$

Rearranging and simplifying, we get:

$\operatorname{sinc}^{(2k)}(0) = \frac{(-1)^k}{2k + 1}$

This gives an explicit expression for the even order derivatives. The odd order derivatives (first derivative, third derivative, etc.) are zero because the function is an even function.

The first few derivatives (starting from the zeroth) are:

$1, 0, -1/3, 0, 1/5, 0, -1/7, 0, 1/9, \dots$
==Integration==
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