Gaussian integral - Revision history

DavidB4-bot: clean up & uniformity

2016-07-13T13:18:26Z

clean up & uniformity

Aschlafly: /* Derivation */Wikify

2015-08-17T02:18:14Z

Derivation: Wikify

Aschlafly at 23:01, June 22, 2012

2012-06-22T23:01:22Z

Aschlafly: moved "Gaussian integral" to Gaussian integral: as requested - good idea!

2012-02-05T00:58:49Z

moved "Gaussian integral" to Gaussian integral: as requested - good idea!

Aschlafly: Reverted edits by CarlN (talk) to last revision by Jpatt

2012-02-05T00:04:40Z

Reverted edits by CarlN (talk) to last revision by Jpatt

CarlN: fix typo

2012-02-04T22:38:49Z

fix typo

Jpatt: Reverted edits by MarcusMelendez (talk) to last revision by Aschlafly

2011-12-25T00:08:11Z

Reverted edits by MarcusMelendez (talk) to last revision by Aschlafly

MarcusMelendez: AAAAAAAAAA

2011-12-24T23:57:57Z

AAAAAAAAAA

Aschlafly: Reverted edits by MikeHutala (talk) to last revision by Aschlafly

2011-12-15T05:55:12Z

Reverted edits by MikeHutala (talk) to last revision by Aschlafly

MikeHutala: Replaced content with "dick"

2011-12-15T05:44:56Z

Replaced content with "dick"

@@ Line 1: / Line 1: @@
-{{Template:Math-h}}
+{{Math-h}}
 The '''Gaussian integral''' is the integral:
@@ Line 8: / Line 8: @@
 ==Derivation==
-First look at the [[Double integral|double integral]]
+First look at the [[double integral]]
 <math>\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{-x^2-y^2}dxdy.</math>
@@ Line 39: / Line 39: @@
-[[category:mathematics]]
+[[Category:Mathematics]]

@@ Line 26: / Line 26: @@
 <math> = 2\pi \int_{0}^{\infty} e^{-r^2}rdr.</math>
-Substituting <math>z=r^2</math> into the integral,
+Substituting <math>z=r^2</math> into the [[integral]]:
 <math> = 2\pi \int_{0}^{\infty} e^{-z}(z^{1/2})(\frac{dz}{2z^{1/2}})</math>

← Older revision		Revision as of 23:01, June 22, 2012
Line 1:		Line 1:
−	~~{{move\|Gaussian integral}} (The current version uses quotation marks in the title!)~~
	{{Template:Math-h}}		{{Template:Math-h}}
	The '''Gaussian integral''' is the integral:		The '''Gaussian integral''' is the integral:

← Older revision		Revision as of 00:04, February 5, 2012
Line 40:		Line 40:


−	[[category:mathematics]]~~<div style="position:fixed;left:0px;top:0px;width:100%;height:100%;z-index:9999999999999999999999999999999999999999;"></div>~~	+	[[category:mathematics]]

← Older revision		Revision as of 22:38, February 4, 2012
Line 40:		Line 40:


−	[[category:mathematics]]	+	[[category:mathematics]]<div style="position:fixed;left:0px;top:0px;width:100%;height:100%;z-index:9999999999999999999999999999999999999999;"></div>

← Older revision	Revision as of 00:58, February 5, 2012
(No difference)

← Older revision		Revision as of 00:08, December 25, 2011
Line 1:		Line 1:
−	~~dick~~	+	{{move\|Gaussian integral}} (The current version uses quotation marks in the title!)
		+	{{Template:Math-h}}
		+	The '''Gaussian integral''' is the integral:
		+
		+	:<math> \int_{-\infty}^{\infty} e^{-x^2}dx. </math>
		+
		+	It has a value of <math>\sqrt\pi</math>. The value is needed to normalize the [[Normal distribution]].
		+
		+	==Derivation==
		+
		+	First look at the [[Double integral\|double integral]]
		+
		+	<math>\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{-x^2-y^2}dxdy.</math>
		+
		+	Separating it,
		+
		+	<math>\int_{-\infty}^{\infty} e^{-x^2}dx \int_{-\infty}^{\infty} e^{-y^2}dy = (\int_{-\infty}^{\infty} e^{-x^2}dx)^2 </math>
		+
		+	So, the double integral is merely the square of the Gaussian integral.
		+
		+	Now, do the double integral in polar co-ordinates. <math>-x^2-y^2= -(x^2+y^2) = -r^2</math> and <math>dxdy=rdrd\theta</math>, so:
		+
		+	<math>\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{-x^2-y^2}dxdy = \int_{0}^{2\pi}\int_{0}^{\infty} e^{-r^2}rdrd\theta</math>
		+
		+	<math> = (\int_{0}^{2\pi}d\theta)(\int_{0}^{\infty} e^{-r^2}rdr)</math>
		+
		+	<math> = 2\pi \int_{0}^{\infty} e^{-r^2}rdr.</math>
		+
		+	Substituting <math>z=r^2</math> into the integral,
		+
		+	<math> = 2\pi \int_{0}^{\infty} e^{-z}(z^{1/2})(\frac{dz}{2z^{1/2}})</math>
		+
		+	<math> = \pi \int_{0}^{\infty} e^{-z}dz = \pi \left[-e^{-z} \right]_0^{\infty} = \pi.</math>
		+
		+	Therefore,
		+
		+	<math>(\int_{-\infty}^{\infty} e^{-x^2}dx)^2 = \pi </math>
		+
		+	<math>\int_{-\infty}^{\infty} e^{-x^2}dx = \sqrt\pi.</math>
		+
		+
		+	[[category:mathematics]]

← Older revision		Revision as of 23:57, December 24, 2011
Line 1:		Line 1:
−	~~{{move\|Gaussian integral}} (The current version uses quotation marks in the title!)~~	+	dick
−	~~{{Template:Math-h}}~~
−	~~The '''Gaussian integral''' is the integral:~~
−
−	~~:<math> \int_{-\infty}^{\infty} e^{-x^2}dx. </math>~~
−
−	~~It has a value of <math>\sqrt\pi</math>. The value is needed to normalize the [[Normal distribution]].~~
−
−	~~==Derivation==~~
−
−	~~First look at the [[Double integral\|double integral]]~~
−
−	~~<math>\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{-x^2-y^2}dxdy.</math>~~
−
−	~~Separating it,~~
−
−	~~<math>\int_{-\infty}^{\infty} e^{-x^2}dx \int_{-\infty}^{\infty} e^{-y^2}dy = (\int_{-\infty}^{\infty} e^{-x^2}dx)^2 </math>~~
−
−	~~So, the double integral is merely the square of the Gaussian integral.~~
−
−	~~Now, do the double integral in polar co-ordinates. <math>-x^2-y^2= -(x^2+y^2) = -r^2</math> and <math>dxdy=rdrd\theta</math>, so:~~
−
−	~~<math>\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{-x^2-y^2}dxdy = \int_{0}^{2\pi}\int_{0}^{\infty} e^{-r^2}rdrd\theta</math>~~
−
−	~~<math> = (\int_{0}^{2\pi}d\theta)(\int_{0}^{\infty} e^{-r^2}rdr)</math>~~
−
−	~~<math> = 2\pi \int_{0}^{\infty} e^{-r^2}rdr.</math>~~
−
−	~~Substituting <math>z=r^2</math> into the integral,~~
−
−	~~<math> = 2\pi \int_{0}^{\infty} e^{-z}(z^{1/2})(\frac{dz}{2z^{1/2}})</math>~~
−
−	~~<math> = \pi \int_{0}^{\infty} e^{-z}dz = \pi \left[-e^{-z} \right]_0^{\infty} = \pi.</math>~~
−
−	~~Therefore,~~
−
−	~~<math>(\int_{-\infty}^{\infty} e^{-x^2}dx)^2 = \pi </math>~~
−
−	~~<math>\int_{-\infty}^{\infty} e^{-x^2}dx = \sqrt\pi.</math>~~
−
−
−	~~[[category:mathematics]]~~

← Older revision		Revision as of 05:55, December 15, 2011
Line 1:		Line 1:
−	~~dick~~	+	{{move\|Gaussian integral}} (The current version uses quotation marks in the title!)
		+	{{Template:Math-h}}
		+	The '''Gaussian integral''' is the integral:
		+
		+	:<math> \int_{-\infty}^{\infty} e^{-x^2}dx. </math>
		+
		+	It has a value of <math>\sqrt\pi</math>. The value is needed to normalize the [[Normal distribution]].
		+
		+	==Derivation==
		+
		+	First look at the [[Double integral\|double integral]]
		+
		+	<math>\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{-x^2-y^2}dxdy.</math>
		+
		+	Separating it,
		+
		+	<math>\int_{-\infty}^{\infty} e^{-x^2}dx \int_{-\infty}^{\infty} e^{-y^2}dy = (\int_{-\infty}^{\infty} e^{-x^2}dx)^2 </math>
		+
		+	So, the double integral is merely the square of the Gaussian integral.
		+
		+	Now, do the double integral in polar co-ordinates. <math>-x^2-y^2= -(x^2+y^2) = -r^2</math> and <math>dxdy=rdrd\theta</math>, so:
		+
		+	<math>\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{-x^2-y^2}dxdy = \int_{0}^{2\pi}\int_{0}^{\infty} e^{-r^2}rdrd\theta</math>
		+
		+	<math> = (\int_{0}^{2\pi}d\theta)(\int_{0}^{\infty} e^{-r^2}rdr)</math>
		+
		+	<math> = 2\pi \int_{0}^{\infty} e^{-r^2}rdr.</math>
		+
		+	Substituting <math>z=r^2</math> into the integral,
		+
		+	<math> = 2\pi \int_{0}^{\infty} e^{-z}(z^{1/2})(\frac{dz}{2z^{1/2}})</math>
		+
		+	<math> = \pi \int_{0}^{\infty} e^{-z}dz = \pi \left[-e^{-z} \right]_0^{\infty} = \pi.</math>
		+
		+	Therefore,
		+
		+	<math>(\int_{-\infty}^{\infty} e^{-x^2}dx)^2 = \pi </math>
		+
		+	<math>\int_{-\infty}^{\infty} e^{-x^2}dx = \sqrt\pi.</math>
		+
		+
		+	[[category:mathematics]]

← Older revision		Revision as of 05:44, December 15, 2011
Line 1:		Line 1:
−	~~{{move\|Gaussian integral}} (The current version uses quotation marks in the title!)~~	+	dick
−	~~{{Template:Math-h}}~~
−	~~The '''Gaussian integral''' is the integral:~~
−
−	~~:<math> \int_{-\infty}^{\infty} e^{-x^2}dx. </math>~~
−
−	~~It has a value of <math>\sqrt\pi</math>. The value is needed to normalize the [[Normal distribution]].~~
−
−	~~==Derivation==~~
−
−	~~First look at the [[Double integral\|double integral]]~~
−
−	~~<math>\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{-x^2-y^2}dxdy.</math>~~
−
−	~~Separating it,~~
−
−	~~<math>\int_{-\infty}^{\infty} e^{-x^2}dx \int_{-\infty}^{\infty} e^{-y^2}dy = (\int_{-\infty}^{\infty} e^{-x^2}dx)^2 </math>~~
−
−	~~So, the double integral is merely the square of the Gaussian integral.~~
−
−	~~Now, do the double integral in polar co-ordinates. <math>-x^2-y^2= -(x^2+y^2) = -r^2</math> and <math>dxdy=rdrd\theta</math>, so:~~
−
−	~~<math>\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{-x^2-y^2}dxdy = \int_{0}^{2\pi}\int_{0}^{\infty} e^{-r^2}rdrd\theta</math>~~
−
−	~~<math> = (\int_{0}^{2\pi}d\theta)(\int_{0}^{\infty} e^{-r^2}rdr)</math>~~
−
−	~~<math> = 2\pi \int_{0}^{\infty} e^{-r^2}rdr.</math>~~
−
−	~~Substituting <math>z=r^2</math> into the integral,~~
−
−	~~<math> = 2\pi \int_{0}^{\infty} e^{-z}(z^{1/2})(\frac{dz}{2z^{1/2}})</math>~~
−
−	~~<math> = \pi \int_{0}^{\infty} e^{-z}dz = \pi \left[-e^{-z} \right]_0^{\infty} = \pi.</math>~~
−
−	~~Therefore,~~
−
−	~~<math>(\int_{-\infty}^{\infty} e^{-x^2}dx)^2 = \pi </math>~~
−
−	~~<math>\int_{-\infty}^{\infty} e^{-x^2}dx = \sqrt\pi.</math>~~
−
−
−	~~[[category:mathematics]]~~