Question Number 41051 by turbo msup by abdo last updated on 01/Aug/18 $${find}\:{the}\:{value}\:{of}\:\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{2}{n}+\mathrm{3}}{{n}^{\mathrm{2}} \left({n}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$ Commented by math khazana by abdo…
Question Number 40984 by prof Abdo imad last updated on 30/Jul/18 $${let}\:{f}\left({x}\right)\:=\int_{\mathrm{0}} ^{\infty} \:\:{te}^{−{t}^{\mathrm{2}} } \:{arctan}\left({xt}\right){dt} \\ $$$$\left.\mathrm{1}\right)\:{find}\:{a}\:{simple}\:{form}\:{of}\:{f}\left({x}\right) \\ $$$$\left.\mathrm{2}\right)\:{calculate}\:\int_{\mathrm{0}} ^{\infty} \:{te}^{−{t}^{\mathrm{2}} } {arctantdt}\:{and} \\…
Question Number 40898 by abdo.msup.com last updated on 28/Jul/18 $${let}\:{u}_{{n}} =\sum_{{k}=\mathrm{1}} ^{{n}−\mathrm{1}} \:\:\frac{{n}−{k}}{{n}−{k}+\mathrm{1}} \\ $$$${find}\:{a}\:{equivalent}\:{of}\:{u}_{{n}} \left({n}\rightarrow+\infty\right) \\ $$ Commented by maxmathsup by imad last updated…
Question Number 40897 by abdo.msup.com last updated on 28/Jul/18 $${calculate}\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\frac{{n}}{\left({n}+\mathrm{1}\right)^{\mathrm{2}} \left({n}+\mathrm{2}\right)} \\ $$ Commented by math khazana by abdo last updated on 03/Aug/18…
Question Number 40895 by abdo.msup.com last updated on 28/Jul/18 $${prove}\:{that}\:{for}\:\mid{x}\mid<\mathrm{1} \\ $$$$\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+{x}}}\:=\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\frac{\left(−\mathrm{1}\right)^{{n}} \:{C}_{\mathrm{2}{n}} ^{{n}} }{\mathrm{4}^{{k}} }\:{x}^{\mathrm{2}{k}} \\ $$ Terms of Service Privacy Policy…
Question Number 40896 by abdo.msup.com last updated on 28/Jul/18 $${for}\:\mid{x}\mid<\mathrm{1}\:{prove}\:{that} \\ $$$$\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}\:=\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{{C}_{\mathrm{2}{n}} ^{{n}} }{\mathrm{4}^{{n}} }\:{x}^{\mathrm{2}{n}} \\ $$ Terms of Service Privacy Policy…
Question Number 40885 by prof Abdo imad last updated on 28/Jul/18 $${prove}\:{that} \\ $$$$\left.\mathrm{1}\right)\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{t}^{{p}} {ln}\left({t}\right)}{{t}−\mathrm{1}}{dt}\:=\frac{\pi^{\mathrm{2}} }{\mathrm{6}}\:−\sum_{{k}=\mathrm{1}} ^{{p}} \:\frac{\mathrm{1}}{{k}^{\mathrm{2}} } \\ $$$$\left.\mathrm{2}\right)\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{t}^{\mathrm{2}{p}}…
Question Number 171955 by infinityaction last updated on 22/Jun/22 $$\:\:\boldsymbol{{f}}\left(\boldsymbol{{x}}\right)+\:\boldsymbol{{f}}\left(\frac{\mathrm{1}}{\mathrm{1}−\boldsymbol{{x}}}\right)\:=\:\mathrm{1}+\frac{\mathrm{1}}{\boldsymbol{{x}}\left(\mathrm{1}−\boldsymbol{{x}}\right)} \\ $$$$\:\:\:\:\:\boldsymbol{{f}}\left(\boldsymbol{{x}}\right)\:=\:\:??\:\:\:\:\:\:\:\:\: \\ $$ Answered by thfchristopher last updated on 22/Jun/22 $$\mathrm{1}+\frac{\mathrm{1}}{{x}\left(\mathrm{1}−{x}\right)} \\ $$$$=\frac{{x}−{x}^{\mathrm{2}} +\mathrm{1}}{{x}\left(\mathrm{1}−{x}\right)}…
Question Number 40882 by prof Abdo imad last updated on 28/Jul/18 $$\left.\mathrm{1}\right){prove}\:{that}\:\forall{n}\geqslant\mathrm{2}\left({n}\:{inyegr}\right) \\ $$$${x}^{\mathrm{2}{n}} −\mathrm{1}=\left({x}−\mathrm{1}\right)\left({x}+\mathrm{1}\right)\prod_{{k}=\mathrm{1}} ^{{n}−\mathrm{1}} \left({x}^{\mathrm{2}} \:−\mathrm{2}{cos}\left(\frac{{k}\pi}{{n}}\right){x}+\mathrm{1}\right) \\ $$$$\left.\mathrm{2}\right){find}\:{the}\:{value}\:{of}\:\:\int_{\mathrm{0}} ^{\pi} {ln}\left({x}^{\mathrm{2}} −\mathrm{2}{xcost}\:+\mathrm{1}\right){dt} \\ $$…
Question Number 40880 by prof Abdo imad last updated on 28/Jul/18 $${prove}\:{that}\:\sum_{{k}={n}} ^{\infty} \:\frac{\mathrm{1}}{{k}^{\alpha} }\:\sim\:\:\frac{\mathrm{1}}{\left(\alpha−\mathrm{1}\right){n}^{\alpha−\mathrm{1}} }{with}\:\alpha>\mathrm{1} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com