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Question Number 108239 by Dwaipayan Shikari last updated on 15/Aug/20

$$\left(\frac{7}{2}\right)!=?$$$$$$

Answered by abdomsup last updated on 15/Aug/20

$$\mathrm{\Gamma }(x+n)=\mathrm{\Gamma }(x+n-1+1)$$$$=(x+n-1)\mathrm{\Gamma }(x+n-2+1)$$$$=(x+n-1)(x+n-2)\mathrm{\Gamma }(x+n-2)$$$$=(x+n-1)(x+n-2)(x+n-3)\mathrm{\Gamma }(x+n-3)$$$$\Rightarrow \left(\frac{7}{2}\right)!=\mathrm{\Gamma }(\frac{7}{2}+1)$$$$=\mathrm{\Gamma }(\frac{1}{2}+4)=(\frac{1}{2}+3)(\frac{1}{2}+2)(\frac{1}{2}+1)\mathrm{\Gamma }(\frac{1}{2}+1)$$$$=\frac{7}{2}.\frac{5}{2}.\frac{3}{2}.\frac{1}{2}\mathrm{\Gamma }\left(\frac{1}{2}\right)$$$$=\frac{35.3}{8}\mathrm{\Gamma }\left(\frac{1}{2}\right)=\frac{105}{8}\mathrm{\Gamma }\left(\frac{1}{2}\right)$$$$\mathrm{\Gamma }\left(x\right)={\int }_0^{\mathrm{\infty }}{t}^{x-1}{e}^{-t}dt\Rightarrow$$$$\mathrm{\Gamma }\left(\frac{1}{2}\right)={\int }_0^{\mathrm{\infty }}\frac{e^{-t}}{\sqrt{t}}dt{=}_{t=u^2}{\int }_0^{\mathrm{\infty }}\frac{e^{-u^2}}{u}\left(2u\right)du$$$$=2{\int }_0^{\mathrm{\infty }}{e}^{-u^2}du=2.\frac{\sqrt{\pi }}{2}=\sqrt{\pi }\Rightarrow$$$$\left(\frac{7}{2}\right)!=\frac{105\sqrt{\pi }}{8}$$

Commented by abdomsup last updated on 15/Aug/20

$$sorry\left(\frac{7}{2}\right)!=\frac{105\sqrt{\pi }}{16}$$

Commented by abdomsup last updated on 15/Aug/20

$$generalywehave$$$$\mathrm{\Gamma }(x+n)=(x+n-1)(x+n-2)$$$$....(x+n-k)\mathrm{\Gamma }(x+n-k)$$$$k=n\Rightarrow \mathrm{\Gamma }(x+n)=(x+n-1)(x+n-2)$$$$......(x+1)x\mathrm{\Gamma }\left(x\right)$$$$exn=4\Rightarrow$$$$\mathrm{\Gamma }(x+4)=(x+3)(x+2)(x+1)\mathrm{\Gamma }\left(x\right)$$$$x=\frac{1}{m}\Rightarrow \mathrm{\Gamma }(\frac{1}{m}+4)$$$$=\left(\frac{1}{m+3}\right)(\frac{1}{m}+2)(\frac{1}{m}+1)\frac{1}{m}\mathrm{\Gamma }\left(\frac{1}{m}\right)$$

Commented by abdomsup last updated on 15/Aug/20

$$erroroftypo$$$$\mathrm{\Gamma }(\frac{1}{m}+4)=(\frac{1}{m}+3)(\frac{1}{m}+2)(\frac{1}{m}+1)\frac{1}{m}\mathrm{\Gamma }\left(\frac{1}{m}\right)$$

Commented by mathmax by abdo last updated on 15/Aug/20

$$\mathrm{\Gamma }(\mathrm{x}+4)=(\mathrm{x}+3)(\mathrm{x}+2)(\mathrm{x}+1)\mathrm{x}\mathrm{\Gamma }\left(\mathrm{x}\right)$$

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