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Question Number 9233 by geovane10math last updated on 24/Nov/16

$$\mathrm{The}\mathrm{gamma}\mathrm{function}$$$$\mathrm{\Gamma }\left(s\right)={\int }_0^{\mathrm{\infty }}{e}^{-x}{x}^{s-1}dx$$$$\mathrm{How}\mathrm{to}\mathrm{calculate}\mathrm{the}\mathrm{gamma}\mathrm{function}$$$$\mathrm{in}\mathrm{an}\mathrm{easy}\mathrm{and}\mathrm{not}\mathrm{time}-\mathrm{consuming}$$$$\mathrm{way}?$$$$$$

Answered by 123456 last updated on 25/Nov/16

$$\mathrm{\Gamma }(s+1)=s!$$$$\mathrm{you}\mathrm{can}\mathrm{use}\mathrm{stirling}\mathrm{aproxiation}\mathrm{for}$$$$\mathrm{large}s,\mathrm{also}\mathrm{any}\mathrm{good}\mathrm{aproximation}$$$$\mathrm{can}\mathrm{be}\mathrm{made}\mathrm{by}$$$$\mathrm{\Gamma }(s+1)=s\mathrm{\Gamma }\left(s\right)\iff \mathrm{\Gamma }\left(s\right)=\frac{\mathrm{\Gamma }(s+1)}{s}$$$$1:\mathrm{for}\mid s\mid >ϵ\left(ϵ\mathrm{large}\mathrm{enough}\right)$$$$\mathrm{get}\mathrm{aproximate}\mathrm{value}\mathrm{via}\mathrm{stirling}$$$$2:\mathrm{for}\mid s\mid ⩽ϵ$$$$\mathrm{get}\mathrm{value}\mathrm{via}\mathrm{\Gamma }\left(s\right)=\frac{\mathrm{\Gamma }(s+1)}{s}$$$$----------------$$$$\mathrm{not}\mathrm{sure}\mathrm{if}\mathrm{it}\mathrm{eficient}\mathrm{but}\mathrm{it}\mathrm{work}:v$$

Commented by 123456 last updated on 25/Nov/16

$$\mathrm{\Gamma }\left(s\right)=\frac{\mathrm{\Gamma }(s+1)}{s}$$$$\mathrm{\Gamma }(s+1)=\frac{\mathrm{\Gamma }(s+2)}{s+1}\Rightarrow \mathrm{\Gamma }\left(s\right)=\frac{\mathrm{\Gamma }(s+2)}{s(s+1)}$$$$...$$$$\mathrm{\Gamma }\left(s\right)=\frac{\mathrm{\Gamma }(s+a)}{s(s+1)...(s+a-1)}a\in \mathbb{N},a>0$$

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