Revision as of 04:14, November 4, 2011

{{| }} ( !) -

<> \{-\}^{\} ^{-^}. </>

    <>\\</>.        [[ ]].

==

   [[ | ]]

<>\{-\}^{\} \{-\}^{\} ^{-^-^}.</>

<>\{-\}^{\} ^{-^} \{-\}^{\} ^{-^} = (\{-\}^{\} ^{-^})^ </>

, .

, -. <>-^-^= -(^+^) = -^</> <>=\</>, :

<>\{-\}^{\} \{-\}^{\} ^{-^-^} = \{}^{\}\{}^{\} ^{-^}\</>

<> = (\{}^{\}\)(\{}^{\} ^{-^})</>

<> = \ \{}^{\} ^{-^}.</>

<>=^</>   ,

<> = \ \{}^{\} ^{-}(^{/})(\{}{^{/}})</>

<> = \ \{}^{\} ^{-} = \ \[-^{-} \]^{\} = \.</>

,

<>(\{-\}^{\} ^{-^})^ = \ </>

<>\{-\}^{\} ^{-^} = \\.</>

[[:]]

@@ Line 1: / Line 1: @@
-{{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]]
+[[:]]

Difference between revisions of "Gaussian integral"

Revision as of 04:14, November 4, 2011

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