Difference between revisions of "Gamma function"
(Created page with 'The '''Gamma funciton''' is defined as <math>\Gamma(z) = \int_{0}^{\infty}t^{z-1}e^{-t}dt</math> Using integration by parts, <math>\Gamma(z) = \int_{…') |
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| + | {{Math-h}} | ||
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The '''Gamma funciton''' is defined as | The '''Gamma funciton''' is defined as | ||
<math>\Gamma(z) = \int_{0}^{\infty}t^{z-1}e^{-t}dt</math> | <math>\Gamma(z) = \int_{0}^{\infty}t^{z-1}e^{-t}dt</math> | ||
| − | Using [[ | + | == Relations and values == |
| + | |||
| + | ===Factorial=== | ||
| + | Using [[integration by parts]], | ||
<math>\Gamma(z) = \int_{0}^{\infty}t^{z-1}d(e^{-t}dt)</math> | <math>\Gamma(z) = \int_{0}^{\infty}t^{z-1}d(e^{-t}dt)</math> | ||
| − | :<math>= \left[t^{z-1}(-e^{-t}) \right]_0^{\infty}- \int_{0}^{\infty}(z-1)t^{(z-2)}(-e^{-t})dt</math> | + | :<math>= \left[t^{z-1}(-e^{-t}) \right]_0^{\infty}- \int_{0}^{\infty}(z-1)t^{(z-2)}(-e^{-t})dt.</math> |
| + | |||
| + | At <math>t=0</math>, <math>t^{z-1}(-e^{-t})</math> goes to 0. Using [[L'Hopital's rule]], it's easy to show that <math>\lim_{t\to \infty} \frac{t^{z-1}}{e^{t}} = 0</math>. | ||
| + | |||
| + | So, | ||
| + | |||
:<math>=(z-1) \int_{0}^{\infty}t^{(z-1)-1}e^{-t}dt</math> | :<math>=(z-1) \int_{0}^{\infty}t^{(z-1)-1}e^{-t}dt</math> | ||
| − | :<math>=(z-1) \Gamma(z-1)</math> | + | :<math>=(z-1) \Gamma(z-1).</math> |
Thus, | Thus, | ||
| − | <math> \Gamma (z) = (z-1) \Gamma(z-1)</math> | + | :<math> \Gamma (z) = (z-1) \Gamma(z-1).</math> |
| + | |||
| + | Using this, and the fact that <math>\Gamma(1)=1</math>, then we can get the [[factorial]] function. For <math>n</math> a positive integer, | ||
| + | |||
| + | :<math> \Gamma(n) = (n-1)!. </math> | ||
| + | |||
| + | ===z=1/2=== | ||
| + | For z=1/2, | ||
| + | |||
| + | :<math> \Gamma(1/2) = \int_{0}^{\infty}t^{-1/2}e^{-t}dt </math> | ||
| + | |||
| + | Substituting <math>x=t^{1/2}</math>, | ||
| + | |||
| + | :<math>=\int_{0}^{\infty} (\frac{1}{x}) e^{-x^2} (2xdx )</math> | ||
| + | :<math>=2 \int_{0}^{\infty} e^{-x^2}dx </math> | ||
| + | :<math>= \int_{-\infty}^{\infty} e^{-x^2}dx </math> | ||
| + | |||
| + | where in the last line we used the fact that <math>e^{-x^2}</math> is an even function. The integral is called the Gaussian integral and has a well known value of <math>\sqrt\pi</math>. | ||
| + | |||
| + | Thus, | ||
| + | |||
| + | :<math> \Gamma(1/2) = \sqrt\pi. </math> | ||
| + | |||
| + | |||
| + | |||
| + | [[Category:Mathematics]] | ||
Latest revision as of 17:46, July 18, 2016
| Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{d}{dx} \sin x=?\,} | This article/section deals with mathematical concepts appropriate for late high school or early college. |
The Gamma funciton is defined as
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Gamma(z) = \int_{0}^{\infty}t^{z-1}e^{-t}dt}
Relations and values
Factorial
Using integration by parts,
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Gamma(z) = \int_{0}^{\infty}t^{z-1}d(e^{-t}dt)}
- Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle = \left[t^{z-1}(-e^{-t}) \right]_0^{\infty}- \int_{0}^{\infty}(z-1)t^{(z-2)}(-e^{-t})dt.}
At Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle t=0} , Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle t^{z-1}(-e^{-t})} goes to 0. Using L'Hopital's rule, it's easy to show that Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lim_{t\to \infty} \frac{t^{z-1}}{e^{t}} = 0} .
So,
- Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle =(z-1) \int_{0}^{\infty}t^{(z-1)-1}e^{-t}dt}
- Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle =(z-1) \Gamma(z-1).}
Thus,
- Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Gamma (z) = (z-1) \Gamma(z-1).}
Using this, and the fact that Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Gamma(1)=1} , then we can get the factorial function. For Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle n} a positive integer,
- Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Gamma(n) = (n-1)!. }
z=1/2
For z=1/2,
- Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Gamma(1/2) = \int_{0}^{\infty}t^{-1/2}e^{-t}dt }
Substituting Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x=t^{1/2}} ,
- Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle =\int_{0}^{\infty} (\frac{1}{x}) e^{-x^2} (2xdx )}
- Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle =2 \int_{0}^{\infty} e^{-x^2}dx }
- Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle = \int_{-\infty}^{\infty} e^{-x^2}dx }
where in the last line we used the fact that Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle e^{-x^2}} is an even function. The integral is called the Gaussian integral and has a well known value of .
Thus,
- Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Gamma(1/2) = \sqrt\pi. }