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Author: Tinku Tara

x-x-e-xln-x-e-x-1-x-x-2-2-x-3-3-e-xln-x-1-xln-x-xln-x-2-2-e-xln-x-xln-x-0-0-xln-x-1-1-xln-x-2-2-x-x-e-xln-x-n-0

Question Number 6027 by FilupSmith last updated on 10/Jun/16 $${x}^{{x}} ={e}^{{x}\mathrm{ln}\:{x}} \\ $$$${e}^{{x}} =\mathrm{1}+{x}+\frac{{x}^{\mathrm{2}} }{\mathrm{2}!}+\frac{{x}^{\mathrm{3}} }{\mathrm{3}!}+… \\ $$$$\therefore\:{e}^{{x}\mathrm{ln}\:{x}} \:=\:\mathrm{1}+{x}\mathrm{ln}\left({x}\right)+\frac{\left({x}\mathrm{ln}\:{x}\right)^{\mathrm{2}} }{\mathrm{2}!}+… \\ $$$${e}^{{x}\mathrm{ln}\:{x}} \:=\:\frac{\left({x}\mathrm{ln}\:{x}\right)^{\mathrm{0}} }{\mathrm{0}!}+\frac{\left({x}\mathrm{ln}\:{x}\right)^{\mathrm{1}} }{\mathrm{1}!}+\frac{\left({x}\mathrm{ln}\:{x}\right)^{\mathrm{2}}…

advanced-calculus-please-evaluate-1-n-1-H-n-2n-1-2-2-0-1-xln-x-ln-1-x-1-x-dx-note-H-n-1-1-2-1-3-1-n-0-1-

Question Number 137093 by mnjuly1970 last updated on 29/Mar/21 $$\:\:\:\:\:\:\:\:\:…..{advanced}\:\:\:\:{calculus}…. \\ $$$$\:\:\:\:\:{please}\:\:{evaluate}::\:\: \\ $$$$\:\:\:\:\:\mathrm{1}:\:\underset{{n}=\mathrm{1}\:} {\overset{\infty} {\sum}}\frac{{H}_{{n}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{2}} }\:=? \\ $$$$\:\:\:\:\mathrm{2}:\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{{xln}\left({x}\right){ln}\left(\mathrm{1}−{x}\right)}{\mathrm{1}−{x}}{dx}=? \\ $$$$\:\:{note}:{H}_{{n}} =\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{3}}+…+\frac{\mathrm{1}}{{n}}=\int_{\mathrm{0}}…

Question-71550

Question Number 71550 by mhmd last updated on 17/Oct/19 Answered by MJS last updated on 17/Oct/19 $$\mathrm{it}'\:\mathrm{wrong},\:\mathrm{it}\:\mathrm{must}\:\mathrm{be} \\ $$$$\begin{pmatrix}{{n}+\mathrm{1}}\\{{k}}\end{pmatrix}\:=\begin{pmatrix}{{n}}\\{{k}}\end{pmatrix}\:+\begin{pmatrix}{{n}}\\{{k}−\mathrm{1}}\end{pmatrix} \\ $$$$\begin{pmatrix}{{n}}\\{{k}}\end{pmatrix}\:+\begin{pmatrix}{{n}}\\{{k}−\mathrm{1}}\end{pmatrix}\:=\frac{{n}!}{{k}!\left({n}−{k}\right)!}+\frac{{n}!}{\left({k}−\mathrm{1}\right)!\left({n}−{k}+\mathrm{1}\right)!}= \\ $$$$=\frac{{n}!\left({n}−{k}+\mathrm{1}\right)}{\left({k}−\mathrm{1}\right)!\left({n}−{k}+\mathrm{1}\right)!{k}}+\frac{{n}!{k}}{\left({k}−\mathrm{1}\right)!\left({n}−{k}+\mathrm{1}\right)!{k}}= \\ $$$$=\frac{{n}!\left({n}−{k}+\mathrm{1}\right)+{n}!{k}}{\left({k}−\mathrm{1}\right)!\left({n}−{k}+\mathrm{1}\right)!{k}}=\frac{{n}!\left({n}+\mathrm{1}\right)}{{k}!\left({n}+\mathrm{1}−{k}\right)!}=…