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Question Number 73974 by Raxreedoroid last updated on 17/Nov/19

$$\mathrm{Prove}\mathrm{this}\mathrm{equation}$$$$\frac{2^{x-1}(x-\frac{1}{2})!}{\frac{1}{2}!}=\frac{(2x-1)!}{2^{x-1}(x-1)!}$$

Answered by mind is power last updated on 17/Nov/19

$$\mathrm{\Gamma }\left(2\mathrm{x}\right)=(2\mathrm{x}-1)!$$$$\beta (\mathrm{x},\mathrm{x})=\frac{\mathrm{\Gamma }\left(\mathrm{x}\right).\mathrm{\Gamma }\left(\mathrm{x}\right)}{\mathrm{\Gamma }\left(2\mathrm{x}\right)}$$$$\beta (\mathrm{x},\mathrm{x})=2{\int }_0^{\frac{\pi }{2}}{\cos }^{2\mathrm{x}-1}\left(\mathrm{t}\right).{\sin }^{2\mathrm{x}-1}\left(\mathrm{t}\right)\mathrm{dt}$$$$=2{\int }_0^{\frac{\pi }{2}}.{\left(\sin \left(\mathrm{t}\right)\cos \left(\mathrm{t}\right)\right)}^{2\mathrm{x}-1}.\mathrm{dt}$$$$=2{\int }_0^{\frac{\pi }{2}}{\left(\frac{\sin \left(2\mathrm{t}\right)}{2}\right)}^{2\mathrm{x}-1}\mathrm{dt}$$$$=\frac{1}{2^{2\mathrm{x}-2}}.{\int }_0^{\frac{\pi }{2}}.{\sin }^{2\mathrm{x}-1}\left(2\mathrm{t}\right)\mathrm{dt}$$$$\mathrm{u}=2\mathrm{t}\Rightarrow \beta (\mathrm{x},\mathrm{x})=\frac{1}{2^{2\mathrm{x}-2}}.{\int }_0^{\pi }{\sin }^{2\mathrm{x}-1}\left(\mathrm{u}\right)\mathrm{du}=\frac{1}{2^{2\mathrm{x}-2}}{\int }_0^{\frac{\pi }{2}}{\sin }^{2\mathrm{x}-1}\left(\mathrm{u}\right)\mathrm{du}+\frac{1}{2^{2\mathrm{x}-2}}{\int }_{\frac{\pi }{2}}^{\pi }{\sin }^{2\mathrm{x}-1}\left(\mathrm{u}\right)\mathrm{du}$$$${\int }_0^{\frac{\pi }{2}}{\sin }^{2\mathrm{x}-1}\left(\mathrm{u}\right)\mathrm{du}=\frac{1}{2}\beta (\frac{1}{2},\mathrm{x})$$$${\int }_{\frac{\pi }{2}}^{\pi }{\sin }^{2\mathrm{x}-1}\left(\mathrm{u}\right)\mathrm{du}={\int }_0^{\frac{\pi }{2}}{\sin }^{2\mathrm{x}-1}(\mathrm{u}+\frac{\pi }{2})\mathrm{d}\left(\mathrm{u}\right)=\frac{1}{2}\beta (\mathrm{x},\frac{1}{2})=\frac{1}{2}\beta (\frac{1}{2},\mathrm{x})$$$$\Rightarrow {\int }_0^{\pi }{\sin }^{2\mathrm{x}-1}\left(\mathrm{t}\right)\mathrm{dt}=\beta (\mathrm{x},\frac{1}{2})$$$$\Rightarrow \beta (\mathrm{x},\mathrm{x})=\frac{1}{2^{2\mathrm{x}-2}}.\beta (\mathrm{x},\frac{1}{2})$$$$\iff \frac{\mathrm{\Gamma }\left(\mathrm{x}\right).\mathrm{\Gamma }\left(\mathrm{x}\right)}{\mathrm{\Gamma }\left(2\mathrm{x}\right)}=\frac{1}{2^{2\mathrm{x}-2}}.\frac{\mathrm{\Gamma }\left(\mathrm{x}\right).\mathrm{\Gamma }\left(\frac{1}{2}\right)}{\mathrm{\Gamma }(\mathrm{x}+\frac{1}{2})}\iff \frac{\mathrm{\Gamma }\left(2\mathrm{x}\right)}{\mathrm{\Gamma }\left(\mathrm{x}\right)2^{\mathrm{x}-1}}=\frac{\mathrm{\Gamma }(\mathrm{x}+\frac{1}{2})}{\mathrm{\Gamma }\left(\frac{1}{2}\right)}.2^{\mathrm{x}-1}$$$$\mathrm{\Gamma }\left(\mathrm{x}\right)=(\mathrm{x}-1)!,\mathrm{\Gamma }(\mathrm{x}+\frac{1}{2})=(\mathrm{x}-\frac{1}{2})!,\mathrm{\Gamma }\left(2\mathrm{x}\right)=(2\mathrm{x}-1)!$$$$\iff \frac{2^{\mathrm{x}-1}.(\mathrm{x}-\frac{1}{2})!}{\left(\frac{1}{2}\right)!}=\frac{(2\mathrm{x}-1)!}{(\mathrm{x}-1)!.2^{\mathrm{x}-1}}$$

Commented by Raxreedoroid last updated on 17/Nov/19

$$\mathrm{My}\mathrm{proof}\mathrm{is}\mathrm{the}\mathrm{following}$$$$\frac{(2n-1)!}{2^{n-1}(n-1)!}=1,3,15,105,945...$$$$\underset{k=1}{\prod ^{n-1}}2k+1=1\cdot 3\cdot 5\cdot 7\cdot 9...(2n+1)$$$$\mathrm{let}\mathrm{u}=2k+1$$$$n=1,\underset{k=1}{\prod ^0}u=1$$$$n=2,\underset{k=1}{\prod ^1}u=1\cdot 3=3$$$$n=3,\underset{k=1}{\prod ^2}u=1\cdot 3\cdot 5=15$$$$n=4,\underset{k=1}{\prod ^3}u=1\cdot 3\cdot 5\cdot 7=105$$$$n=5,\underset{k=1}{\prod ^4}u=945$$$${\mathrm{It}}^{\prime }\mathrm{s}\mathrm{the}\mathrm{same}$$$$\underset{k=1}{\prod ^{n-1}}2k+1=\underset{k=1}{\prod ^{n-1}}2(k+\frac{1}{2})=\left(\underset{k=1}{\prod ^{n-1}}2\right)(\underset{k=1}{\prod ^{n-1}}k+\frac{1}{2})$$$$\underset{k=1}{\prod ^{n-1}}2=2^{n-1}$$$$\underset{k=1}{\prod ^{n-1}}x+\frac{1}{2}=\underset{k=1+\frac{1}{2}}{\prod ^{n-1+\frac{1}{2}}}x=\frac{(n-\frac{1}{2})!}{(1\frac{1}{2}-1)!}=\frac{(n-\frac{1}{2})!}{\left(\frac{1}{2}\right)!}$$$$\underset{k=1}{\prod ^{n-1}}2x+1=\frac{2^{x-1}(x-\frac{1}{2})!}{\left(\frac{1}{2}\right)!}$$

Commented by mind is power last updated on 17/Nov/19

$$\mathrm{yeah}\mathrm{nice}\mathrm{this}\mathrm{for}\mathrm{x}\in \mathbb{N}$$

Commented by Raxreedoroid last updated on 17/Nov/19

$$\mathrm{I}\mathrm{am}\mathrm{surprised}\mathrm{becuase}\mathrm{I}\mathrm{proved}\mathrm{this}$$$$\mathrm{in}\mathrm{a}\mathrm{simple}\mathrm{way}\mathrm{even}\mathrm{though}\mathrm{I}{\mathrm{can}}^{\prime }\mathrm{t}\mathrm{understand}$$$$\mathrm{most}\mathrm{of}\mathrm{this}$$

Commented by mind is power last updated on 17/Nov/19

$$\mathrm{if}\mathrm{x}\in \mathbb{N}\mathrm{or}\mathrm{x}\in \mathbb{C}?$$

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