Sinc-squared function - Revision history

Vipul: /* Definition */

2011-12-19T18:56:41Z

Definition

← Older revision		Revision as of 18:56, 19 December 2011
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	<math>\operatorname{sinc}^2 x := \left\lbrace \begin{array}{rl} 1, & x = 0 \\ \frac{\sin^2x}{x^2}, & x \ne 0 \\\end{array}\right.</math>		<math>\operatorname{sinc}^2 x := \left\lbrace \begin{array}{rl} 1, & x = 0 \\ \frac{\sin^2x}{x^2}, & x \ne 0 \\\end{array}\right.</math>

			==Key data==

			{\| class="sortable" border="1"
			! Item !! Value
			\|-
			\| default [[domain]] \|\| all [[real number]]s, i.e., all of <math>\R</math>
			\|-
			\| [[range]] \|\| the [[closed interval]] <math>[0,1]</math>, i.e., the set <math>\{ y \mid 0 \le y \le 1 \}</math><br>[[absolute maximum value]]: 1, [[absolute minimum value]]: 0
			\|-
			\| [[period]] \|\| none; the function is not periodic
			\|-
			\| [[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^2 x}{x^2} = 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>.
			\|-
			\| [[local maximum value]]s and points of attainment \|\| The local maximum values occur at all points where <math>\tan^2x = x^2</math>. There is an anomalous local maximum at <math>x = 0</math> with value 1. All other local maxima occur at points close to odd multiples of <math>\pi/2</math> (''except'' <math>\pm \pi/2</math> themselves, though not exactly at those points.
			\|-
			\| [[local minimum value]]s and points of attainment \|\| The local minimum values occur at all points where <math>\sin x = 0</math> ''except'' the point <math>x = 0</math> itself. These are precisely the nonzero integer multiples of <math>\pi</math>. The values are all equal to zero.
			\|-
			\| [[derivative]] \|\| {{fillin}}
			\|-
			\| [[second derivative]] \|\| {{fillin}}
			\|-
			\| [[antiderivative]] \|\| <math>\operatorname{Si}(2x) - \frac{\sin^2x}{x} + C</math>. See [[#First antiderivative]]. Here <math>\operatorname{Si}</math> denotes the [[sine integral]], an antiderivative for the [[sinc function]].
			\|-
			\| [[power series]] and [[Taylor series]]\|\| The power series about 0 (which is also the Taylor series) is {{fillin}}
			\|}

	==Graph==		==Graph==

Vipul: /* First antiderivative */

2011-12-19T18:17:53Z

First antiderivative

← Older revision		Revision as of 18:17, 19 December 2011
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	Denote by <math>\operatorname{Si}(x)</math> the function <math>\int_0^x \operatorname{sinc} t \, dt = \int_0^x (\sin t)/t \, dt</math>. We integrate <math>\operatorname{sinc}^2</math> and obtain the following answer in terms of <math>\operatorname{Si}</math>:		Denote by <math>\operatorname{Si}(x)</math> the function <math>\int_0^x \operatorname{sinc} t \, dt = \int_0^x (\sin t)/t \, dt</math>. We integrate <math>\operatorname{sinc}^2</math> and obtain the following answer in terms of <math>\operatorname{Si}</math>:

	<math>\int_0^x \operatorname{sinc}^2t \, dt = \operatorname{Si}(2x) - \frac{\sin^2x}{x} + C</math>		<math>\int_0^x \operatorname{sinc}^2t \, dt = \operatorname{Si}(2x) - \frac{\sin^2x}{x}</math>

			For the indefinite integral, we can put a <math>+C</math> at the end..

	We do this using [[integration by parts]]:		We do this using [[integration by parts]]:

Vipul: /* Integration */

2011-12-19T18:15:48Z

Integration

← Older revision		Revision as of 18:15, 19 December 2011
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	The limit expression is zero because <math>\sin^2</math> has a zero of order 2 at zero. For the integration expression, set <math>u = 2t</math> and get <math>\int_0^{2x} \frac{\sin(u)}{u} \, du = \operatorname{Si}(2x)</math>. Plugging back in, we get the desired answer.		The limit expression is zero because <math>\sin^2</math> has a zero of order 2 at zero. For the integration expression, set <math>u = 2t</math> and get <math>\int_0^{2x} \frac{\sin(u)}{u} \, du = \operatorname{Si}(2x)</math>. Plugging back in, we get the desired answer.

			===Improper definite integral===

			We know that:

			<math>\int_0^\infty \operatorname{sinc} t \, dt = \lim_{x \to \infty} \operatorname{Si}(x) = \frac{\pi}{2}</math>

			Using this, we obtain that:

			<math>\int_0^\infty \operatorname{sinc}^2 t \,dt = \lim_{x \to \infty} \operatorname{Si}(2x) - \lim_{x \to \infty} \frac{\sin^2x}{x} = \frac{\pi}{2} - 0 = \frac{\pi}{2}</math>

			Similarly:

			<math>\int_{-\infty}^0 \operatorname{sinc}^2 t \, dt = \frac{\pi}{2}</math>

			Overall:

			<math>\int_{-\infty}^\infty \operatorname{sinc}^2 t \, dt = \pi</math>

			Note that the improper integral value is the ''same'' for the sinc function and its square. Roughly speaking, the sinc function is bigger than its square when both are positive, but the sinc function also takes negative values while its square does not, and so these differences balance out in the overall integration.

Vipul: /* First antiderivative */

2011-12-19T18:10:38Z

First antiderivative

← Older revision		Revision as of 18:10, 19 December 2011
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	===First antiderivative===		===First antiderivative===

	Denote by <math>\operatorname{Si}(x)</math> the function <math>\int_0^x \sinc t \, dt = \int_0^x (\sin t)/t \, dt</math>. We integrate <math>\operatorname{sinc}^2</math> and obtain the following answer in terms of <math>\operatorname{Si}</math>:		Denote by <math>\operatorname{Si}(x)</math> the function <math>\int_0^x \operatorname{sinc} t \, dt = \int_0^x (\sin t)/t \, dt</math>. We integrate <math>\operatorname{sinc}^2</math> and obtain the following answer in terms of <math>\operatorname{Si}</math>:

	<math>\int_0^x \operatorname{sinc}^2t \, dt = \operatorname{Si}(2x) - \frac{\sin^2x}{x} + C</math>		<math>\int_0^x \operatorname{sinc}^2t \, dt = \operatorname{Si}(2x) - \frac{\sin^2x}{x} + C</math>

Vipul: /* Integration */

2011-12-19T18:10:21Z

Integration

← Older revision		Revision as of 18:10, 19 December 2011
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	===First antiderivative===		===First antiderivative===

	Denote by <math>\operatorname{Si}(x)</math> the function <math>\int_0^x \sinc t \, dt = \int_0^x (\sin t)/t \, dt</math>. We integrate <math>\sinc^2</math> and obtain the following answer in terms of <math>\operatorname{Si}</math>:		Denote by <math>\operatorname{Si}(x)</math> the function <math>\int_0^x \sinc t \, dt = \int_0^x (\sin t)/t \, dt</math>. We integrate <math>\operatorname{sinc}^2</math> and obtain the following answer in terms of <math>\operatorname{Si}</math>:

	<math>\int_0^x \sinc^2t \, dt = \operatorname{Si}(2x) - \frac{\sin^2x}{x} + C</math>		<math>\int_0^x \operatorname{sinc}^2t \, dt = \operatorname{Si}(2x) - \frac{\sin^2x}{x} + C</math>

	We do this using [[integration by parts]]:		We do this using [[integration by parts]]:

	<math>\int_0^x \sinc^2t \, dt = \int_0^x \frac{\sin^2t}{t^2} \, dt</math>		<math>\int_0^x \operatorname{sinc}^2t \, dt = \int_0^x \frac{\sin^2t}{t^2} \, dt</math>

	Take <math>1/t^2 \, dt</math> as the part to integrate. We get:		Take <math>1/t^2 \, dt</math> as the part to integrate. We get:

Vipul: /* Graph */

2011-12-19T18:09:29Z

Graph

← Older revision		Revision as of 18:09, 19 December 2011
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	[[File:Sincsquaredbasicnottoscale.png\|800px]]		[[File:Sincsquaredbasicnottoscale.png\|800px]]

			==Integration==

			===First antiderivative===

			Denote by <math>\operatorname{Si}(x)</math> the function <math>\int_0^x \sinc t \, dt = \int_0^x (\sin t)/t \, dt</math>. We integrate <math>\sinc^2</math> and obtain the following answer in terms of <math>\operatorname{Si}</math>:

			<math>\int_0^x \sinc^2t \, dt = \operatorname{Si}(2x) - \frac{\sin^2x}{x} + C</math>

			We do this using [[integration by parts]]:

			<math>\int_0^x \sinc^2t \, dt = \int_0^x \frac{\sin^2t}{t^2} \, dt</math>

			Take <math>1/t^2 \, dt</math> as the part to integrate. We get:

			<math>\left[\sin^2t\left(\frac{-1}{t}\right)\right]_0^x - \int_0^x \sin(2t) \frac{-1}{t} \, dt</math>

			This becomes:

			<math>\frac{-\sin^2x}{x} + \lim_{t \to 0} \frac{\sin^2t}{t} + \int_0^x \frac{\sin(2t)}{t} \, dt</math>

			The limit expression is zero because <math>\sin^2</math> has a zero of order 2 at zero. For the integration expression, set <math>u = 2t</math> and get <math>\int_0^{2x} \frac{\sin(u)}{u} \, du = \operatorname{Si}(2x)</math>. Plugging back in, we get the desired answer.

Vipul: /* Graph */

2011-09-04T12:28:03Z

Graph

← Older revision		Revision as of 12:28, 4 September 2011
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	The graph is a little unclear, here is an alternative version where different scalings are used for the <math>x</math>-axis and <math>y</math>-axis:		The graph is a little unclear, here is an alternative version where different scalings are used for the <math>x</math>-axis and <math>y</math>-axis:

	[[File:Sincsquaredbasicnottoscale\|800px]]		[[File:Sincsquaredbasicnottoscale.png\|800px]]

Vipul: /* Graph */

2011-09-04T12:24:49Z

Graph

← Older revision		Revision as of 12:24, 4 September 2011
Line 20:		Line 20:
	The graph is a little unclear, here is an alternative version where different scalings are used for the <math>x</math>-axis and <math>y</math>-axis:		The graph is a little unclear, here is an alternative version where different scalings are used for the <math>x</math>-axis and <math>y</math>-axis:

	[[File:Sincsquaredbasicnottoscale.800px]]		[[File:Sincsquaredbasicnottoscale\|800px]]

Vipul: /* Definition */

2011-09-04T12:24:35Z

Definition

← Older revision		Revision as of 12:24, 4 September 2011
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	Alternatively, it is given as:		Alternatively, it is given as:

	<math>\sinc^2 x := \left\lbrace \begin{array}{rl} 1, & x = 0 \\ \frac{\sin^2x}{x^2}, & x \ne 0 \\\end{array}\right.</math>		<math>\operatorname{sinc}^2 x := \left\lbrace \begin{array}{rl} 1, & x = 0 \\ \frac{\sin^2x}{x^2}, & x \ne 0 \\\end{array}\right.</math>

	==Graph==		==Graph==

Vipul: Created page with "{{particular function}} {{angular function radian convention}} ==Definition== This function is defined as the composite of t..."

2011-09-04T12:24:16Z

Created page with "{{particular function}} {{angular function radian convention}} ==Definition== This function is defined as the composite of t..."

New page

{{particular function}}
{{angular function radian convention}}

==Definition==

This function is defined as the [[defining ingredient::composite of two functions|composite]] of the [[defining ingredient::square function]] and the [[defining ingredient::sinc function]]. Explicitly, it is given as:

<math>x \mapsto (\operatorname{sinc} \ x)^2</math>

Alternatively, it is given as:

<math>\sinc^2 x := \left\lbrace \begin{array}{rl} 1, & x = 0 \\ \frac{\sin^2x}{x^2}, & x \ne 0 \\\end{array}\right.</math>

==Graph==

Here is a graph on the interval <math>[-3\pi,3\pi]</math>:

[[File:Sincsquaredbasic.png|600px]]

The graph is a little unclear, here is an alternative version where different scalings are used for the <math>x</math>-axis and <math>y</math>-axis:

[[File:Sincsquaredbasicnottoscale.800px]]