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Sinc function

, 12:53, 4 September 2011
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| [[derivative]] || $\operatorname{sinc}'x = \left\lbrace \begin{array}{rl} 0, & x = 0 \\ \frac{x \cos x - \sin x}{x^2} \\\end{array}\right.$
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| [[antiderivative]] || the [[sine integral]] (''this is defined as the antiderivative of the sinc function that takes the value 0 at 0'')
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==Graph==
[[File:Sincfunctionbasicnottoscale.png|800px]]

==Differentation==

===First derivative===

''Computation at $x = 0$'': Here, we need to compute the derivative using first principles, as a limit of a [[difference quotient]]:

$\operatorname{sinc}' 0 := \lim_{x \to 0} \frac{\operatorname{sinc} x - \operatorname{sinc} 0}{x - 0} = \lim_{x \to 0} \frac{(\sin x)/x - 1}{x} = \lim_{x \to 0} \frac{\sin x - x}{x^2}$

This limit can be computed in many ways. For instance, it can be computed using the [[L'Hopital rule]]:

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

{{point differentation incorrect}}

''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:

$\frac{d}{dx}\left[\frac{f(x)}{g(x)}\right] = \frac{g(x)f'(x) - f(x)g'(x)}{(g(x))^2}$

In this case, $f(x) = \sin x$ and $g(x) = x$, so we get:

$\frac{d}{dx}(\operatorname{sinc}\ x) = \frac{x \cos x - \sin x}{x^2}$
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