Sinc function - Revision history

Vipul: /* Higher derivative at zero */

2011-12-19T18:50:32Z

Higher derivative at zero

← Older revision		Revision as of 18:50, 19 December 2011
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	<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>		<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 ~~derivative~~ at zero===		===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:		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:

Vipul: /* Higher derivative at zero */

2011-12-19T18:49:58Z

Higher derivative at zero

← Older revision		Revision as of 18:49, 19 December 2011
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	! <math>n</math> !! 0 !! 1 !! 2 !! 3 !! 4 !! 5 !! 6 !! 7 !! 8 !! 9		! <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		\| <math>\operatorname{sinc}^{(n)}(0)</math> \|\| 1 \|\| 0 \|\| -1/3 \|\| 0 \|\| 1/5 \|\| 0 \|\| -1/7 \|\| 0 \|\| 1/9 \|\| 0
	\|}		\|}

Vipul: /* Higher derivative at zero */

2011-12-19T18:49:39Z

Higher derivative at zero

← Older revision		Revision as of 18:49, 19 December 2011
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	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.		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 ~~derivatives~~ (starting from the zeroth) are:		The first few derivative values at zero (starting from the zeroth) are:

	<math>1, 0, -1/3, 0, 1/5, 0, -1/7, 0, 1/9~~, \dots</math>~~		{\| 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
			\|}

	==Integration==		==Integration==

Vipul: /* Higher derivative at zero */

2011-12-19T18:48:24Z

Higher derivative at zero

← Older revision		Revision as of 18:48, 19 December 2011
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	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:		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>		<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:		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:

Vipul: /* Second derivative */

2011-12-19T18:47:57Z

Second derivative

← Older revision		Revision as of 18:47, 19 December 2011
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	<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>		<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 derivative 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 derivatives (starting from the zeroth) are:

			<math>1, 0, -1/3, 0, 1/5, 0, -1/7, 0, 1/9, \dots</math>

	==Integration==		==Integration==

Vipul: /* Integration */

2011-12-19T18:41:16Z

Integration

← Older revision		Revision as of 18:41, 19 December 2011
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	<math>\int_{-\infty}^\infty \operatorname{sinc} x \, dx = \pi</math>		<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==		==Taylor series and power series==

Vipul: /* Differentation */

2011-12-19T18:37:58Z

Differentation

← Older revision		Revision as of 18:37, 19 December 2011
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	[[File:Sincfunctionbasicnottoscale.png\|800px]]		[[File:Sincfunctionbasicnottoscale.png\|800px]]

	==~~Differentation~~==		==Differentiation==

	===First derivative===		===First derivative===

Vipul: /* Taylor series and power series */

2011-12-19T18:36:59Z

Taylor series and power series

← Older revision		Revision as of 18:36, 19 December 2011
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	<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>		<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>

			==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>

	==Taylor series and power series==		==Taylor series and power series==

Vipul: /* Second derivative */

2011-12-19T18:32:12Z

Second derivative

← Older revision		Revision as of 18:32, 19 December 2011
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	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:		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>		<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>

	==Taylor series and power series==		==Taylor series and power series==

Vipul: /* Second derivative */

2011-12-19T18:31:52Z

Second derivative

← Older revision		Revision as of 18:31, 19 December 2011
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	Here, we need to compute the derivative using first principles, as a limit of a [[difference quotient]]:		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 \~~cox~~ x - \sin x)/x^2 - 0}{x} = \lim_{x \to 0} \frac{x \cos x - \sin x}{x^3}</math>		<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]]:		The latter limit can be computed in many ways. One way is to use the [[L'Hopital rule]]: