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Sinc function

1,347 bytes added, 12:53, 4 September 2011
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| [[derivative]] || <math>\operatorname{sinc}'x = \left\lbrace \begin{array}{rl} 0, & x = 0 \\ \frac{x \cos x - \sin x}{x^2} \\\end{array}\right.</math>
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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 <math>x = 0</math>'': Here, we need to compute the derivative using first principles, as a limit of a [[difference quotient]]:
 
<math>\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}</math>
 
This limit can be computed in many ways. For instance, it can be computed using the [[L'Hopital rule]]:
 
<math>\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</math>
 
{{point differentation incorrect}}
 
''Computation at <math>x \ne 0</math>'': At any such point, we know that the function looks like <math>(\sin x)/x</math> in an [[open interval]] about the point, so we can use the [[quotient rule for differentiation]], which states that:
 
<math>\frac{d}{dx}\left[\frac{f(x)}{g(x)}\right] = \frac{g(x)f'(x) - f(x)g'(x)}{(g(x))^2}</math>
 
In this case, <math>f(x) = \sin x</math> and <math>g(x) = x</math>, so we get:
 
<math>\frac{d}{dx}(\operatorname{sinc}\ x) = \frac{x \cos x - \sin x}{x^2}</math>
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