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Sinc function

4,967 bytes added, 18:50, 19 December 2011
Higher derivative at zero
| [[horizontal asymptote]]s || <math>y = 0</math>, i.e., the <math>x</math>-axis. This is because <math>\lim_{x \to \pm \infty} \frac{\sin x}{x} = 0</math>, which in turn can be deduced from the fact that the numerator is bounded while the magnitude of the denominator approaches <math>\infty</matH>.
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| [[local maximum value]]s and points of attainment || The local maxima occur at points <math>x</math> satisfying <math>\tan x = x</math> and <math>x \in [2n\pi,(2n + 1)\pi]</math> or <math>x \in [-(2n + 1)\pi,-2n\pi]</math> for <math>n</math> a positive integer.<br> There is an anomalous local maximum at <math>x = 0</math> with value 1. Apart from that, the other local maxima occur at points of the form <math>\pm(2n\pi + \alpha_n)</math> where <math>\alpha_n</math> is fairly close to <math>\pi/2</math> for all <math>n > 0</math>. The local maximum value at this point is slightly more than <math>1/(2n\pi + \pi/2)</math>.
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| [[local minimum value]]s and points of attainment || The local minima occur at points <math>x</math> satisfying <math>\tan x = -x</math> and <math>x \in [(2n - 1)\pi,2n\pi]</math> or <math>x \in [-2n\pi,-(2n - 1)\pi]</math> for <math>n</math> a positive integer.<br> The local minima occur at points of the form <math>\pm(2n\pi - \alpha_n)</math> where <math>\alpha_n</math> is fairly close to <math>\pi/2</math> for all <math>n > 0</math>. The local minimum value at this point is slightly less than <math>-1/(2n\pi - \pi/2)</math>.
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| [[point of inflection|points of inflection]] || {{fillin}}
[[File:Sincfunctionbasicnottoscale.png|800px]]
==DifferentationDifferentiation==
===First derivative===
''Computation at <math>x = 0</math>'':  {{point differentation incorrect}} 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>
<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 incorrecteven function derivative at zero}}
''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}(\operatorname{sinc}\ x) = \frac{x \cos x - \sin x}{x^2}</math>
 
Combining the two computations, we get:
 
<math>\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.</math>
 
===Second derivative===
 
''Computation at <math>x = 0</math>'':
 
{{point differentation incorrect}}
 
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{(x \cos x - \sin x)/x^2 - 0}{x} = \lim_{x \to 0} \frac{x \cos x - \sin x}{x^3}</math>
 
The latter limit can be computed in many ways. One way is to use the [[L'Hopital rule]]:
 
<math>\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}</math>
 
Thus, <math>\operatorname{sinc}'' 0 = -1/3</math>.
 
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:
 
<math>\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.</math>
 
===Higher derivatives 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:
 
<math>\operatorname{sinc} x = 1 - \frac{x^2}{3!} + \frac{x^4}{5!} - \dots = \sum_{k=0}^\infty \frac{(-1)^kx^{2k}}{(2k + 1)!}</math>
 
Since this is a power series, it is also a Taylor series. Thus, the coefficient of <math>x^{2k}</math> is <math>1/(2k)!</math> times <math>\operatorname{sinc}^{(2k)}(0)</math>. We thus get:
 
<math>\frac{1}{(2k + 1)!} = \frac{(-1)^k \operatorname{sinc}^{(2k)}(0)}{(2k)!}</math>
 
Rearranging and simplifying, we get:
 
<math>\operatorname{sinc}^{(2k)}(0) = \frac{(-1)^k}{2k + 1}</math>
 
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 derivative values at zero (starting from the zeroth) are:
 
{| class="sortable" border="1"
! <math>n</math> !! 0 !! 1 !! 2 !! 3 !! 4 !! 5 !! 6 !! 7 !! 8 !! 9
|-
| <math>\operatorname{sinc}^{(n)}(0)</math> || 1 || 0 || -1/3 || 0 || 1/5 || 0 || -1/7 || 0 || 1/9 || 0
|}
 
==Integration==
 
===First antiderivative===
 
These is no antiderivative for this among elementary functions. However, we can create a new function that serves as the antiderivative:
 
<math>\operatorname{Si}(x) := \int_0^x \operatorname{sinc} t \, dt</math>
 
The function <math>\operatorname{Si}</math> is termed the [[sine integral]].
 
===Integration of transformed versions of function===
 
We have:
 
<math>\int_0^x \operatorname{sinc}(mt) \, dt = \frac{1}{m} \operatorname{Si}(mx)</math>
 
In indefinite integral form:
 
<math>\int \operatorname{sinc}(mx + \varphi) \, dx = \frac{1}{m} \operatorname{Si}(mx + \varphi)</math>
 
===Improper definite integrals===
 
The following fact is true, but not easy to prove:
 
<math>\int_0^\infty \operatorname{sinc} x \, dx = \frac{\pi}{2}</math>
 
Because the function is an even function, this is equivalent to the following:
 
<math>\int_{-\infty}^0 \operatorname{sinc} x \, dx = \frac{\pi}{2}</math>
 
and
 
<math>\int_{-\infty}^\infty \operatorname{sinc} x \, dx = \pi</math>
 
===Second antiderivative===
 
The second antiderivative is:
 
<math>\int (\int \operatorname{sinc} x \, dx) \, dx = x \operatorname{Si}(x) + \cos x + C_1x + C_0</math>
 
To obtain this, we use [[integration by parts]] to integrate the function <math>\operatorname{Si}(x)</math>.
 
===Higher antiderivatives===
 
Higher antiderivatives of the sinc function can be computed in the same manner using [[integration by parts]]. Up to the arbitrary polynomial additive, the antiderivative is expressible as a polynomial linear combination of <math>\operatorname{Si}, \sin, \cos</math>.
 
==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]]):
 
<math>\! \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)!}</math>
 
Dividing both sides by <math>x</math> (valid when <math>x \ne 0</math>), we get:
 
<math>\! \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)!}</math>
 
We note that the power series ''also'' works at <math>x = 0</math> (because <math>\operatorname{sinc} \ 0 = 1</math>), hence it works globally, and is the power series for the sinc function.
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