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Question Number 80613 by M±th+et£s last updated on 04/Feb/20

$$Q.find\frac{d}{dx}(x!)$$

Commented by Rio Michael last updated on 04/Feb/20

$$\mathrm{i}{\mathrm{don}}^{\prime }\mathrm{t}\mathrm{think}\mathrm{this}\mathrm{exist},\mathrm{from}\mathrm{my}\mathrm{understanding}$$$$x!=x(x-1)!\mathrm{has}\mathrm{a}\mathrm{domain}\mathrm{of}\mathrm{D}=x\in {\mathbb{N}}^{\ast }$$$$\mathrm{tbat}\mathrm{means}\mathrm{we}\mathrm{have}\mathrm{no}\mathrm{specific}\mathrm{interval}$$$$\mathrm{for}\mathrm{which}x!\mathrm{is}\mathrm{not}\mathrm{continous},\mathrm{A}\mathrm{non}-\mathrm{continous}\mathrm{function}$$$$\mathrm{is}\mathrm{non}-\mathrm{defferentiable}\mathrm{hence}\mathrm{i}\mathrm{think}$$$$\frac{d}{dx}(x!)\mathrm{cannot}\mathrm{be}\mathrm{calculated}.$$$$\mathrm{please}\mathrm{give}\mathrm{me}\mathrm{your}\mathrm{oppinion}$$$$$$

Commented by MJS last updated on 04/Feb/20

$$\frac{d}{dx}[x!]\mathrm{does}\mathrm{not}\mathrm{exist}\mathrm{because}x!\mathrm{is}\mathrm{only}$$$$\mathrm{defined}\mathrm{on}\mathbb{N}\Rightarrow {\mathrm{it}}^{\prime }\mathrm{s}\mathrm{not}\mathrm{continuous}$$$$\mathrm{\Gamma }\left(x\right)\mathrm{is}\mathrm{the}\mathrm{extension}\mathrm{of}x!\mathrm{but}\mathrm{both}\mathrm{are}\mathrm{not}$$$$\mathrm{the}\mathrm{same}$$$$\mathrm{so}\mathrm{if}\mathrm{you}\mathrm{ask}\mathrm{for}\frac{d}{dx}[x!]\mathrm{the}\mathrm{correct}\mathrm{answer}$$$$\mathrm{is}‘‘\mathrm{does}\mathrm{not}{\mathrm{exist}}^{″}$$$$\mathrm{mathematical}\mathrm{language}\mathrm{is}\mathrm{precise}\mathrm{language}$$$$$$$$x!=\mathrm{\Gamma }(x+1)\mathrm{is}\mathrm{true}\mathrm{\forall }x\in \mathbb{N}$$$$\mathrm{but}$$$$\mathrm{\Gamma }\left(x\right)\ne (x-1)!\mathrm{\forall }x\in \mathbb{R}\mathrm{\setminus }\mathbb{N}$$

Answered by M±th+et£s last updated on 04/Feb/20

$$\mathrm{\Gamma }\left(x\right)=(x-1)!={\int }_0^{\mathrm{\infty }}{t}^{x-1}{e}^{-t}dt$$$$\Rightarrow \frac{1}{\mathrm{\Gamma }\left(x\right)}={xe}^{\gamma x}\underset{k=1}{\prod ^{\mathrm{\infty }}}(1+\frac{x}{k})e^{\frac{-x}{k}}$$$$ln\left(\frac{1}{\mathrm{\Gamma }\left(x\right)}\right)=-ln\left(\mathrm{\Gamma }\left(x\right)\right)=lnx+\gamma x+\underset{k=1}{\sum ^{\mathrm{\infty }}}[ln(1+\frac{x}{k})-\frac{x}{k}]$$$$\frac{d}{dx}(-ln\left(\mathrm{\Gamma }\left(x\right)\right))=\frac{{\mathrm{\Gamma }}^{\prime }\left(x\right)}{\mathrm{\Gamma }\left(x\right)}=\frac{1}{x}+\gamma +\underset{k=1}{\sum ^{\mathrm{\infty }}}[\frac{1}{1+\frac{x}{k}}.\frac{1}{k}-\frac{1}{k}]$$$$\Rightarrow \frac{{\mathrm{\Gamma }}^{\prime }\left(x\right)}{\mathrm{\Gamma }\left(x\right)}=-\frac{1}{x}-\gamma +\underset{k=1}{\sum ^{\mathrm{\infty }}}[\frac{1}{k}-\frac{1}{k+x}]=\mathrm{\Psi }\left(x\right)$$$$\mathrm{\Psi }\left(x\right)=-\gamma +\underset{k=1}{\sum ^{\mathrm{\infty }}}(\frac{1}{k}-\frac{1}{k-1+x})$$$$\mathrm{\Psi }(x+1)=-\gamma +\underset{k=1}{\sum ^{\mathrm{\infty }}}[\frac{1}{k}-\frac{1}{k+x}]=\frac{{\mathrm{\Gamma }}^{\prime }(x+1)}{\mathrm{\Gamma }(x+1)}$$$${\mathrm{\Gamma }}^{\prime }(x+1)=\mathrm{\Psi }(x+1)\mathrm{\Gamma }(x+1)$$$$\frac{d}{dx}=\mathrm{\Gamma }(x+1)\mathrm{\Psi }(x+1)$$$$$$

Commented by M±th+et£s last updated on 04/Feb/20

$${\mathrm{\Psi }}^{\left(m\right)}\left(z\right)={(-1)}^{m+1}{\int }_0^{\mathrm{\infty }}\frac{t^m{e}^{-zt}}{1-e^{-t}}dt:m>0andRe\left(z\right)>0$$

Commented by mathmax by abdo last updated on 04/Feb/20

$$\mathrm{\Gamma }\left(x\right)=(x-1)!isonlyanotation...!$$

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