# Changes

## Sinc function

, 18:30, 19 December 2011
First derivative
$\operatorname{sinc}'(x) = \left\lbrace \begin{array}{rl} 0, & x = 0 \\ \frac{x \cos x - \sin x}{x^2}, & x \ne 0 \\\end{array}\right.$

===Second derivative===

''Computation at $x = 0$'':

{{point differentation incorrect}}

Here, we need to compute the derivative using first principles, as a limit of a [[difference quotient]]:

$\operatorname{sinc}'' 0 := \lim_{x \to 0} \frac{\operatorname{sinc}' x - \operatorname{sinc}' 0}{x - 0} = \lim_{x \to 0} \frac{(x \cox x - \sin x)/x^2 - 0}{x} = \lim_{x \to 0} \frac{x \cos x - \sin x}{x^3}$

The latter limit can be computed in many ways. One way is to use the [[L'Hopital rule]]:

$\lim_{x \to 0} \frac{x \cos x - \sin x}{x^3} \stackrel{*}{=} \lim_{x \to 0} \frac{-x \sin x + \cos x - \cos x}{3x^2} = \lim_{x \to 0} \frac{-\sin x}{3x} \stackrel{*}{=} \lim_{x \to 0} \frac{-\cos x}{3} = \frac{-1}{3}$

Thus, $\operatorname{sinc}'' 0 = -1/3$.

The computation at other points can be done using the usual method for computing derivatives, i.e., with the [[quotient rule for differentiation]] combined with other rules. Overall, we get:

$\operatorname{sinc}'(x) = \left\lbrace \begin{array}{rl} \frac{-1}{3}, & x = 0 \\ \frac{(2 - x^2)\sin x - 2x\cos x}{x^3}, & x \ne 0 \\\end{array}\right.$
==Taylor series and power series==
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