# Changes

## Sinc function

, 13:12, 4 September 2011
Differentation
$\lim_{x \to 0} \frac{\sin x - x}{x^2} \stackrel{*}{=} \lim_{x \to 0} \frac{\cos x - 1}{2x} \stackrel{*}{=} \lim_{x \to 0} \frac{-\sin x}{2} = \frac{-\sin 0}{2} = 0$

{{even function derivative at zero}}
''Computation at $x \ne 0$'': At any such point, we know that the function looks like $(\sin x)/x$ in an [[open interval]] about the point, so we can use the [[quotient rule for differentiation]], which states that:
$\frac{d}{dx}(\operatorname{sinc}\ x) = \frac{x \cos x - \sin x}{x^2}$

Combining the two computations, we get:

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

==Taylor series and power series==

===Computation of power series===

We use the power series for the [[sine function]] (see [[sine function#Computation of power series]]):

$\! \sin x := x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \dots = \sum_{k=0}^\infty \frac{(-1)^kx^{2k+1}}{(2k + 1)!}$

Dividing both sides by $x$ (valid when $x \ne 0$), we get:

$\! \frac{\sin x}{x} = 1 - \frac{x^2}{3!} + \frac{x^4}{5!} - \frac{x^6}{7!} + \dots = \sum_{k=0}^\infty \frac{(-1)^kx^{2k}}{(2k+1)!}$

We note that the power series ''also'' works at $x = 0$ (because $\operatorname{sinc} \ 0 = 1$), hence it works globally, and is the power series for the sinc function.
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