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Question Number 9233 by geovane10math last updated on 24/Nov/16
The gamma function Γ(s) = ∫_0 ^∞ e^(−x) x^(s−1) dx How to calculate the gamma function in an easy and not time-consuming way?
$$\mathrm{The}\:\mathrm{gamma}\:\mathrm{function}\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\Gamma\left({s}\right)\:=\:\int_{\mathrm{0}} ^{\infty} {e}^{−{x}} {x}^{{s}−\mathrm{1}} \:{dx} \\ $$$$\mathrm{How}\:\mathrm{to}\:\mathrm{calculate}\:\mathrm{the}\:\mathrm{gamma}\:\mathrm{function} \\ $$$$\mathrm{in}\:\mathrm{an}\:\mathrm{easy}\:\mathrm{and}\:\mathrm{not}\:\mathrm{time}-\mathrm{consuming}\: \\ $$$$\mathrm{way}? \\ $$$$ \\ $$
Answered by 123456 last updated on 25/Nov/16
Γ(s+1)=s! you can use stirling aproxiation for large s, also any good aproximation can be made by Γ(s+1)=sΓ(s)⇔Γ(s)=((Γ(s+1))/s) 1: for ∣s∣>ε (ε large enough) get aproximate value via stirling 2: for ∣s∣≤ε get value via Γ(s)=((Γ(s+1))/s) −−−−−−−−−−−−−−−− not sure if it eficient but it work :v
$$\Gamma\left({s}+\mathrm{1}\right)={s}! \\ $$$$\mathrm{you}\:\mathrm{can}\:\mathrm{use}\:\mathrm{stirling}\:\mathrm{aproxiation}\:\mathrm{for} \\ $$$$\mathrm{large}\:{s},\:\mathrm{also}\:\mathrm{any}\:\mathrm{good}\:\mathrm{aproximation} \\ $$$$\mathrm{can}\:\mathrm{be}\:\mathrm{made}\:\mathrm{by} \\ $$$$\Gamma\left({s}+\mathrm{1}\right)={s}\Gamma\left({s}\right)\Leftrightarrow\Gamma\left({s}\right)=\frac{\Gamma\left({s}+\mathrm{1}\right)}{{s}} \\ $$$$\mathrm{1}:\:\mathrm{for}\:\mid{s}\mid>\epsilon\:\left(\epsilon\:\mathrm{large}\:\mathrm{enough}\right) \\ $$$$\mathrm{get}\:\mathrm{aproximate}\:\mathrm{value}\:\mathrm{via}\:\mathrm{stirling} \\ $$$$\mathrm{2}:\:\mathrm{for}\:\mid{s}\mid\leqslant\epsilon \\ $$$$\mathrm{get}\:\mathrm{value}\:\mathrm{via}\:\Gamma\left({s}\right)=\frac{\Gamma\left({s}+\mathrm{1}\right)}{{s}} \\ $$$$−−−−−−−−−−−−−−−− \\ $$$$\mathrm{not}\:\mathrm{sure}\:\mathrm{if}\:\mathrm{it}\:\mathrm{eficient}\:\mathrm{but}\:\mathrm{it}\:\mathrm{work}\::{v} \\ $$
Commented by 123456 last updated on 25/Nov/16
Γ(s)=((Γ(s+1))/s) Γ(s+1)=((Γ(s+2))/(s+1))⇒Γ(s)=((Γ(s+2))/(s(s+1))) ... Γ(s)=((Γ(s+a))/(s(s+1)...(s+a−1))) a∈N,a>0
$$\Gamma\left({s}\right)=\frac{\Gamma\left({s}+\mathrm{1}\right)}{{s}} \\ $$$$\Gamma\left({s}+\mathrm{1}\right)=\frac{\Gamma\left({s}+\mathrm{2}\right)}{{s}+\mathrm{1}}\Rightarrow\Gamma\left({s}\right)=\frac{\Gamma\left({s}+\mathrm{2}\right)}{{s}\left({s}+\mathrm{1}\right)} \\ $$$$… \\ $$$$\Gamma\left({s}\right)=\frac{\Gamma\left({s}+{a}\right)}{{s}\left({s}+\mathrm{1}\right)…\left({s}+{a}−\mathrm{1}\right)}\:\:{a}\in\mathbb{N},{a}>\mathrm{0} \\ $$

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