# given-z-lim-n-n-z-z-1-z-n-n-z-proof-that-i-z-1-z-z-ii-wiertrass-definition-of-gamma-function-1-z-ze-z-m-1-1-z-m-e-z-m-is-euler-macheroni-con

Question Number 2891 by 123456 last updated on 29/Nov/15
$$\mathrm{given} \\$$$$\Gamma\left({z}\right)=\underset{{n}\rightarrow+\infty} {\mathrm{lim}}\:\frac{{n}!}{{z}\left({z}+\mathrm{1}\right)\centerdot\centerdot\centerdot\left({z}+{n}\right)}{n}^{{z}} \\$$$$\mathrm{proof}\:\mathrm{that} \\$$$$\left.{i}\right) \\$$$$\Gamma\left({z}+\mathrm{1}\right)={z}\Gamma\left({z}\right) \\$$$$\left.{ii}\right)\:\mathrm{wiertrass}\:\mathrm{definition}\:\mathrm{of}\:\mathrm{gamma}\:\mathrm{function} \\$$$$\frac{\mathrm{1}}{\Gamma\left({z}\right)}={ze}^{{z}\gamma} \underset{{m}=\mathrm{1}} {\overset{+\infty} {\prod}}\:\left(\mathrm{1}+\frac{{z}}{{m}}\right){e}^{−\frac{{z}}{{m}}} \\$$$$\gamma\:\mathrm{is}\:\mathrm{euler}\:\mathrm{macheroni}\:\mathrm{constant}\:\left(\mathrm{Q2818},\mathrm{Q2783}\right) \\$$
Commented by Filup last updated on 30/Nov/15
$${z}\left({z}+\mathrm{1}\right)\left({z}+\mathrm{2}\right)\centerdot\centerdot\centerdot\left({z}+{n}\right)=\frac{\left({z}+{n}\right)!}{\left({z}−\mathrm{1}\right)!} \\$$$$\therefore\Gamma\left({z}\right)=\underset{{n}\rightarrow+\infty} {\mathrm{lim}}\frac{{n}!\centerdot\left({z}−\mathrm{1}\right)!}{\left({z}+{n}\right)!}\centerdot{n}^{{z}} \\$$
Answered by Filup last updated on 30/Nov/15
$$\Gamma\left({z}\right)=\left({z}−\mathrm{1}\right)\left({z}−\mathrm{2}\right)\centerdot\centerdot\centerdot×\mathrm{3}×\mathrm{2}×\mathrm{1} \\$$$$\Gamma\left({z}+\mathrm{1}\right)={z}\left({z}−\mathrm{1}\right)\left({z}−\mathrm{2}\right)\centerdot\centerdot\centerdot×\mathrm{3}×\mathrm{2}×\mathrm{1} \\$$$$\Gamma\left({z}+\mathrm{1}\right)={z}\Gamma\left({z}\right) \\$$$$\\$$$$\mathrm{I}'\mathrm{m}\:\mathrm{sure}\:\mathrm{you}\:\mathrm{wanted}\:\mathrm{a}\:\mathrm{proof}\:\mathrm{in}\:\mathrm{terms}\:\mathrm{of} \\$$$$\mathrm{the}\:\mathrm{limit}\:\mathrm{you}\:\mathrm{gave}\:\mathrm{above},\:\mathrm{but}\:\mathrm{I}\:\mathrm{am} \\$$$$\mathrm{unsure}\:\mathrm{as}\:\mathrm{to}\:\mathrm{how}\:\mathrm{to}\:\mathrm{achieve}\:\mathrm{the}\:\mathrm{result} \\$$$$\mathrm{you}\:\mathrm{most}\:\mathrm{likely}\:\mathrm{desired}.\:\mathrm{Maybe}\:\mathrm{you}\:\mathrm{can} \\$$$$\mathrm{further}\:\mathrm{prove}\:\mathrm{by}\:\mathrm{applying}\:\mathrm{the}\:\mathrm{limit}? \\$$$$\\$$$$\mathrm{I}\:\mathrm{shall}\:\mathrm{update}\:\mathrm{this}\:\mathrm{post}\:\mathrm{if}\:\mathrm{I}\:\mathrm{can}.\:\mathrm{I}\:{might} \\$$$$\mathrm{have}\:\mathrm{an}\:\mathrm{idea}\:\mathrm{to}\:\mathrm{prove}\:\left({i}\right)\:\mathrm{in}\:\mathrm{terms} \\$$$$\mathrm{of}\:\mathrm{the}\:\mathrm{limit} \\$$
Commented by 123456 last updated on 30/Nov/15
$$\mathrm{the}\:\mathrm{limit}\:\mathrm{above}\:\mathrm{is}\:\mathrm{the}\:\mathrm{most}\:\mathrm{general}\:\mathrm{definition} \\$$$$\mathrm{possible}\:\mathrm{for}\:\Gamma\:\left(\mathrm{analitic}\:\mathrm{continuation}\:\mathrm{of}\:\Gamma\right) \\$$$$\mathrm{its}\:\mathrm{extend}\:\Gamma\:\mathrm{to}\:\mathrm{almost}\:\mathrm{all}\:\mathbb{C}\left(\mathrm{with}\:\mathrm{exeption}\right. \\$$$$\left.\mathrm{of}\:\mathrm{negatives}\:\mathrm{number}\:\mathrm{and}\:\mathrm{0}\right) \\$$$$\mathrm{however}\:\mathrm{its}\:\mathrm{is}\:\mathrm{so}\:\mathrm{hard}\:\mathrm{to}\:\mathrm{use} \\$$
Commented by Filup last updated on 30/Nov/15
$$\mathrm{I}\:\mathrm{am}\:\mathrm{trying}\:\mathrm{a}\:\mathrm{method}\:\mathrm{of}\:\mathrm{proof}\:\mathrm{now}. \\$$$$\mathrm{If}\:\mathrm{it}\:\mathrm{leads}\:\mathrm{nowhere},\:\mathrm{i}'\mathrm{ll}\:\mathrm{delete}\:\mathrm{my}\:\mathrm{2nd}\:\mathrm{post} \\$$
Answered by Filup last updated on 30/Nov/15
$${Proof}\:{with}\:{limit} \\$$$$\Gamma\left({z}\right)=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\frac{{n}!}{{z}\left({z}+\mathrm{1}\right)\centerdot\centerdot\centerdot\left({z}+{n}\right)}{n}^{{z}} \\$$$$\therefore\Gamma\left({z}\right)=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\frac{{n}!\left({z}−\mathrm{1}\right)!}{\left({z}+{n}\right)!}{n}^{{z}} \\$$$$\\$$$$\mathrm{Show}\:\mathrm{that}\:\Gamma\left({z}\right)=\left({z}−\mathrm{1}\right)\Gamma\left({z}−\mathrm{1}\right) \\$$$$\therefore\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\frac{{n}!\left({z}−\mathrm{1}\right)!}{\left({z}+{n}\right)!}{n}^{{z}} =\left({z}−\mathrm{1}\right)\left(\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\frac{{n}!\left({z}−\mathrm{2}\right)!}{\left({z}+{n}−\mathrm{1}\right)!}{n}^{{z}−\mathrm{1}} \right) \\$$$$\mathrm{Let}\:\mathrm{us}\:\mathrm{assume}\:\Gamma\left({z}\right)=\left({z}−\mathrm{1}\right)\Gamma\left({z}−\mathrm{1}\right).\:\mathrm{If}\:\mathrm{this}\:\mathrm{is} \\$$$$\mathrm{true}\:\mathrm{then}\:\mathrm{LHS}=\mathrm{RHS} \\$$$$\\$$$$\Gamma\left({x}\right)=\left({x}−\mathrm{1}\right)! \\$$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\frac{{n}!\centerdot\Gamma\left({z}\right)}{\Gamma\left({z}+{n}+\mathrm{1}\right)}{n}^{{z}} =\left({z}−\mathrm{1}\right)\left(\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\frac{{n}!\centerdot\Gamma\left({z}−\mathrm{1}\right)}{\Gamma\left({z}+{n}\right)}{n}^{{z}−\mathrm{1}} \right) \\$$$$\therefore\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\frac{{n}!\centerdot\Gamma\left({z}\right)}{\Gamma\left({z}+{n}+\mathrm{1}\right)}{n}^{{z}} =\left({z}−\mathrm{1}\right)\left(\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\frac{{n}!\centerdot\frac{\Gamma\left({z}\right)}{{z}−\mathrm{1}}}{\frac{\Gamma\left({z}+{n}+\mathrm{1}\right)}{{z}−\mathrm{1}}}{n}^{{z}−\mathrm{1}} \right) \\$$$$\therefore\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\frac{{n}!\centerdot\Gamma\left({z}\right)}{\Gamma\left({z}+{n}+\mathrm{1}\right)}{n}^{{z}} =\left({z}−\mathrm{1}\right)\left(\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\left(\frac{{n}!\centerdot\Gamma\left({z}\right)\centerdot\left({z}−\mathrm{1}\right)}{\left({z}−\mathrm{1}\right)\centerdot\Gamma\left({z}+{n}+\mathrm{1}\right)}\right){n}^{{z}−\mathrm{1}} \right) \\$$$$\therefore\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\frac{{n}!\centerdot\Gamma\left({z}\right)}{\Gamma\left({z}+{n}+\mathrm{1}\right)}{n}^{{z}} =\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\frac{\left({z}−\mathrm{1}\right)\centerdot{n}!\centerdot\Gamma\left({z}\right)}{\Gamma\left({z}+{n}+\mathrm{1}\right)}{n}^{{z}−\mathrm{1}} \\$$$$\Gamma\left({z}+{n}+\mathrm{1}\right)=\left({z}+{n}\right)! \\$$$$\therefore\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\frac{{n}!\centerdot\Gamma\left({z}\right)}{\left({z}+{n}\right)!}{n}^{{z}} =\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\frac{{n}!\centerdot\Gamma\left({z}\right)^{\mathrm{2}} }{\left({z}+{n}\right)!\centerdot\Gamma\left({z}−\mathrm{1}\right)}{n}^{{z}−\mathrm{1}} \\$$$$\therefore\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\frac{{n}!}{\left({z}+{n}\right)!}{n}^{{z}} =\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\frac{{n}!\centerdot\Gamma\left({z}\right)}{\left({z}+{n}\right)!\centerdot\Gamma\left({z}−\mathrm{1}\right)}{n}^{{z}−\mathrm{1}} \\$$$$\therefore\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\frac{{n}!}{\left({z}+{n}\right)!}{n}^{{z}} =\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\frac{{n}!\centerdot\left({z}−\mathrm{1}\right)}{\left({z}+{n}\right)!}{n}^{{z}−\mathrm{1}} \\$$$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{{x}!}{\left({x}+{a}\right)!}=\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:\:\left(?\right) \\$$$$\therefore\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:{n}^{{z}} =\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\left({z}−\mathrm{1}\right){n}^{{z}−\mathrm{1}} \\$$$$\mathrm{TRUE} \\$$$$\therefore\Gamma\left({z}+\mathrm{1}\right)={z}\Gamma\left({z}\right) \\$$$$\mathrm{please}\:\mathrm{let}\:\mathrm{me}\:\mathrm{know}\:\mathrm{of}\:\mathrm{any}\:\mathrm{errors} \\$$
Commented by 123456 last updated on 30/Nov/15
$$\Gamma\left({z}+\mathrm{1}\right)={z}\Gamma\left({z}\right) \\$$$$\Gamma\left({z}\right)=\left({z}−\mathrm{1}\right)\Gamma\left({z}−\mathrm{1}\right) \\$$
Commented by Filup last updated on 30/Nov/15
$$\mathrm{How}\:\mathrm{should}\:\mathrm{I}\:\mathrm{apply}\:\mathrm{this}? \\$$
Commented by 123456 last updated on 30/Nov/15
$$\mathrm{you}\:\mathrm{have}\:\mathrm{a}\:\mathrm{typo}\:\mathrm{in}\:\mathrm{the}\:\mathrm{5th}\:\mathrm{line} \\$$
Commented by Filup last updated on 30/Nov/15
$$\mathrm{T}{hanks}!\:{I}'{ll}\:{re}\:{read}\:{everything}! \\$$$$\mathrm{I}\:\mathrm{believe}\:\mathrm{I}\:\mathrm{have}\:\mathrm{fixed}\:\mathrm{everything} \\$$
Commented by 123456 last updated on 30/Nov/15
$$\mathrm{hint}:\:\frac{{x}}{{x}}=\mathrm{1},\underset{{n}\rightarrow\infty} {\mathrm{lim}}\frac{{n}+{a}}{{n}+{b}}=\mathrm{1} \\$$$$\mathrm{try}\:\mathrm{to}\:\mathrm{write}\:\Gamma\left({z}\right)\:\mathrm{and}\:\Gamma\left({z}+\mathrm{1}\right) \\$$$$\mathrm{and}\:\mathrm{compare}\:\mathrm{each}\:\mathrm{other}\:\mathrm{to}\:\mathrm{see}\:\mathrm{what} \\$$$$\mathrm{you}\:\mathrm{need}\:\mathrm{to}\:\mathrm{manipulate}\::\mathrm{D} \\$$$$\mathrm{ps}:\mathrm{if}\:\mathrm{you}\:\mathrm{want}\:\mathrm{i}\:\mathrm{can}\:\mathrm{post}\:\mathrm{the}\:\mathrm{proof}\:\mathrm{for}\:\mathrm{this} \\$$$$\mathrm{part},\:\mathrm{but}\:\mathrm{try}\:\mathrm{to}\:\mathrm{do}\:\mathrm{it} \\$$
Commented by Filup last updated on 30/Nov/15
$$\mathrm{I}'\mathrm{ll}\:\mathrm{let}\:\mathrm{you}\:\mathrm{know}\:\mathrm{if}\:\mathrm{i}\:\mathrm{can}'\mathrm{t}\:\mathrm{do}\:\mathrm{it}! \\$$
Commented by Filup last updated on 30/Nov/15
$$\mathrm{Am}\:\mathrm{I}\:\mathrm{able}\:\mathrm{to}\:\mathrm{remove}\:\mathrm{like}\:\mathrm{terms}\:\mathrm{within} \\$$$$\mathrm{limits}?\:\mathrm{e}.\mathrm{g}.: \\$$$$\underset{{x}\rightarrow{n}} {\mathrm{lim}}\:{f}\left({x}\right)=\underset{{x}\rightarrow{n}} {\mathrm{lim}}\:{g}\left({x}\right) \\$$$$\Rightarrow\underset{{x}\rightarrow{n}} {\mathrm{lim}}\:\frac{{f}\left({x}\right)}{{g}\left({x}\right)}=\mathrm{1} \\$$
Commented by Filup last updated on 30/Nov/15
$$\mathrm{is}\:\mathrm{this}\:\mathrm{correct}? \\$$
Answered by Yozzi last updated on 30/Nov/15
$$\left({i}\right)\:{We}\:{are}\:{given} \\$$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\Gamma\left({z}\right)=\underset{{n}\rightarrow+\infty} {\mathrm{lim}}\frac{{n}!{n}^{{z}} }{\underset{{r}=\mathrm{0}} {\overset{{n}} {\prod}}\left({z}+{r}\right)}.\: \\$$$${Here}\:{I}\:{assume}\:{z}\:{is}\:{such}\:{that}\:{the}\: \\$$$${function}\:\Gamma\left({z}\right)\:{is}\:{well}\:{defined}.\:{A}\: \\$$$${possible}\:{constriction}\:{on}\:{the}\:{domain} \\$$$${for}\:\Gamma\:{is}\:{z}>\mathrm{0}.\:{Therefore}\:{we}\:{have} \\$$$$\Gamma\left({z}+\mathrm{1}\right)=\underset{{n}\rightarrow+\infty} {\mathrm{lim}}\frac{{n}!{n}^{{z}+\mathrm{1}} }{\left({z}+\mathrm{1}\right)\left({z}+\mathrm{2}\right)…\left({z}+{n}\right)\left({z}+{n}+\mathrm{1}\right)} \\$$$$\Gamma\left({z}+\mathrm{1}\right)=\underset{{n}\rightarrow+\infty} {\mathrm{lim}}\frac{{n}!{n}^{{z}+\mathrm{1}} {z}}{\left\{\underset{{r}=\mathrm{0}} {\overset{{n}} {\prod}}\left({z}+{r}\right)\right\}\left({z}+\mathrm{1}+{n}\right)} \\$$$$\Gamma\left({z}+\mathrm{1}\right)=\underset{{n}\rightarrow+\infty} {\mathrm{lim}}\left(\frac{{n}!{n}^{{z}} }{\underset{{r}=\mathrm{0}} {\overset{{n}} {\prod}}\left({z}+{r}\right)}\right)×\frac{{zn}}{{z}+\mathrm{1}+{n}} \\$$$${Since}\:\underset{{x}\rightarrow+\infty} {\mathrm{lim}}{f}\left({x}\right)×{g}\left({x}\right)=\left(\underset{{x}\rightarrow+\infty} {\mathrm{lim}}{f}\left({x}\right)\right)\left(\underset{{x}\rightarrow+\infty} {\mathrm{lim}}{g}\left({x}\right)\right)\:{in} \\$$$${general},\:{we}\:{obtain} \\$$$$\Gamma\left({z}+\mathrm{1}\right)=\Gamma\left({z}\right)×\underset{{n}\rightarrow+\infty} {\mathrm{lim}}\frac{{zn}}{{z}+\mathrm{1}+{n}}. \\$$$${But},\:\frac{{zn}}{{z}+\mathrm{1}+{n}}=\frac{{z}}{\frac{{z}}{{n}}+\frac{\mathrm{1}}{{n}}+\mathrm{1}}. \\$$$$\Rightarrow\Gamma\left({z}+\mathrm{1}\right)=\Gamma\left({z}\right)\underset{{n}\rightarrow+\infty} {\mathrm{lim}}\frac{{z}}{\frac{{z}}{{n}}+\frac{\mathrm{1}}{{n}}+\mathrm{1}} \\$$$$\Gamma\left({z}+\mathrm{1}\right)=\Gamma\left({z}\right)×\frac{{z}}{\frac{{z}}{\infty}+\frac{\mathrm{1}}{\infty}+\mathrm{1}} \\$$$$\Gamma\left({z}+\mathrm{1}\right)=\Gamma\left({z}\right)×\frac{{z}}{\mathrm{0}+\mathrm{0}+\mathrm{1}} \\$$$$\Gamma\left({z}+\mathrm{1}\right)={z}\Gamma\left({z}\right).\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\Box \\$$$$\\$$$${Remarks}: \\$$$$\left(\mathrm{1}\right)\:{If}\:{it}\:{is}\:{required}\:{to}\:{calculate}\:{the}\: \\$$$${value}\:{of}\:\Gamma\left({x}\right)\:{for}\:{x}<\mathrm{0}\:{we}\:{use}\:{the} \\$$$${recurrence}\:{relation}\:\Gamma\left({x}\right)=\frac{\mathrm{1}}{{x}}\Gamma\left({x}+\mathrm{1}\right) \\$$$${until}\:{the}\:{part}\:\Gamma\left({x}+\mathrm{1}\right)\:{reduces}\:{fully}\:{to}\: \\$$$${some}\:{non}−{negative}\:{form}. \\$$$${E}.{g}\:\:\Gamma\left(\frac{−\mathrm{5}}{\mathrm{2}}\right)=\frac{−\mathrm{2}}{\mathrm{5}}\Gamma\left(\frac{−\mathrm{3}}{\mathrm{2}}\right)=\frac{−\mathrm{2}}{\mathrm{5}}×\frac{−\mathrm{2}}{\mathrm{3}}\Gamma\left(\frac{−\mathrm{1}}{\mathrm{2}}\right) \\$$$$\Gamma\left(−\frac{\mathrm{5}}{\mathrm{2}}\right)=\frac{\mathrm{4}}{\mathrm{15}}×\frac{−\mathrm{2}}{\mathrm{1}}\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\right) \\$$$${Since}\:\Gamma\left(\mathrm{1}/\mathrm{2}\right)=\sqrt{\pi}, \\$$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\Gamma\left(\frac{−\mathrm{5}}{\mathrm{2}}\right)=\frac{−\mathrm{8}}{\mathrm{15}}\sqrt{\pi}. \\$$$$\left(\mathrm{2}\right)\:{For}\:\mathrm{0}<{x}<\mathrm{1}\:{we}\:{have}\: \\$$$$\Gamma\left({x}\right)\Gamma\left(\mathrm{1}−{x}\right)=\frac{\pi}{{sin}\pi{z}}. \\$$$${So},\:{if}\:{x}=\mathrm{1}/\mathrm{2}\:{we}\:{derive}\:{the}\:{result} \\$$$$\Gamma\left(\mathrm{0}.\mathrm{5}\right)=\sqrt{\pi}. \\$$$$\Gamma\left(\mathrm{0}.\mathrm{5}\right)\Gamma\left(\mathrm{1}−\mathrm{0}.\mathrm{5}\right)=\frac{\pi}{{sin}\mathrm{0}.\mathrm{5}\pi} \\$$$$\left(\Gamma\left(\mathrm{0}.\mathrm{5}\right)\right)^{\mathrm{2}} =\frac{\pi}{\mathrm{1}}\Rightarrow\Gamma\left(\mathrm{0}.\mathrm{5}\right)=\pm\sqrt{\pi} \\$$$${However},\:{for}\:{x}>\mathrm{0}\:{the}\:{limit}\:{definition} \\$$$${for}\:\Gamma\left({x}\right)=\underset{{n}\rightarrow+\infty} {\mathrm{lim}}\frac{{n}!{n}^{{x}} }{\underset{{r}=\mathrm{0}} {\overset{{n}} {\prod}}\left({x}+{r}\right)}\:{indicates}\:{that} \\$$$$\Gamma\left({x}\right)\:{is}\:{non}−{negative}. \\$$$$\Rightarrow\Gamma\left(\mathrm{0}.\mathrm{5}\right)\neq−\sqrt{\pi}\Rightarrow\Gamma\left(\mathrm{0}.\mathrm{5}\right)=\sqrt{\pi}\:.\:\:\:\: \\$$$$\\$$$$\\$$
Commented by 123456 last updated on 02/Dec/15
$$\mathrm{the}\:\mathrm{limit}\:\mathrm{form}\:\mathrm{of}\:\Gamma\:\mathrm{function}\:\mathrm{also}\:\mathrm{tell} \\$$$$\mathrm{you}\:\mathrm{that}\:\mathrm{its}\:\mathrm{have}\:\mathrm{simple}\:\mathrm{pole}\:\mathrm{at} \\$$$${z}\in\left\{…,−\mathrm{5},−\mathrm{4},−\mathrm{3},−\mathrm{2},−\mathrm{1},\mathrm{0}\right\} \\$$