Question Number 68592 by Abdo msup. last updated on 14/Sep/19 $${find}\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\frac{\mathrm{1}}{{n}^{\mathrm{2}} \left({n}+\mathrm{1}\right)^{\mathrm{3}} } \\ $$ Commented by Abdo msup. last updated on 16/Sep/19…
Question Number 68593 by Abdo msup. last updated on 14/Sep/19 $${calculate}\:\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$ Commented by Abdo msup. last updated on 06/Oct/19…
Question Number 68466 by mathmax by abdo last updated on 11/Sep/19 $${let}\:{f}\left({x}\right)\:=\int_{{x}^{} } ^{{x}^{\mathrm{2}} −{x}} \:{arctan}\left({e}^{−{x}−{t}} \right){dt} \\ $$$${calculate}\:{f}^{'} \left({x}\right)\:\:\:{and}\:{f}^{'} \left(\mathrm{0}\right). \\ $$ Commented by…
Question Number 2891 by 123456 last updated on 29/Nov/15 $$\mathrm{given} \\ $$$$\Gamma\left({z}\right)=\underset{{n}\rightarrow+\infty} {\mathrm{lim}}\:\frac{{n}!}{{z}\left({z}+\mathrm{1}\right)\centerdot\centerdot\centerdot\left({z}+{n}\right)}{n}^{{z}} \\ $$$$\mathrm{proof}\:\mathrm{that} \\ $$$$\left.{i}\right) \\ $$$$\Gamma\left({z}+\mathrm{1}\right)={z}\Gamma\left({z}\right) \\ $$$$\left.{ii}\right)\:\mathrm{wiertrass}\:\mathrm{definition}\:\mathrm{of}\:\mathrm{gamma}\:\mathrm{function} \\ $$$$\frac{\mathrm{1}}{\Gamma\left({z}\right)}={ze}^{{z}\gamma} \underset{{m}=\mathrm{1}} {\overset{+\infty}…
Question Number 133893 by mathlove last updated on 25/Feb/21 $$\:\:{if}\:\:\:\:\:\:\:{f}\left({x}\right)=\frac{\mathrm{10}^{{x}} −\mathrm{10}^{−{x}} }{\mathrm{10}^{{x}} +\mathrm{10}^{−{x}} }\:\:\:\:\:\:{then}\:\:{f}^{−\mathrm{1}} \left({x}\right)=? \\ $$ Answered by rs4089 last updated on 25/Feb/21 $${y}=\frac{\mathrm{10}^{{x}}…
Question Number 2720 by prakash jain last updated on 25/Nov/15 $$\mathrm{Analytical}\:\mathrm{Continuation} \\ $$$$\mathrm{Sum}\:\mathrm{of}\:\mathrm{the}\:\mathrm{below}\:\mathrm{divergent}\:\mathrm{series}\:\mathrm{was} \\ $$$$\mathrm{shown}\:\mathrm{to}\:\mathrm{be}\:\mathrm{using}\:\mathrm{analytical}\:\mathrm{continuation}. \\ $$$$\underset{{i}=\mathrm{1}} {\overset{{n}} {\sum}}\mathrm{2}^{{i}−\mathrm{1}} =−\mathrm{1}\:\:\:\:\:\:…\left(\mathrm{A}\right) \\ $$$$\zeta\left(−\mathrm{1}\right)=\underset{{i}=\mathrm{0}} {\overset{\infty} {\sum}}{i}=−\frac{\mathrm{1}}{\mathrm{12}}\:\:\:\:\:\:…\left(\mathrm{B}\right) \\…
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Question Number 68244 by mathmax by abdo last updated on 07/Sep/19 $${find}\:{nature}\:{of}\:{the}\:{serie}\:\sum_{{n}=\mathrm{1}} ^{\infty} \:{arctan}\left({n}+\frac{\mathrm{1}}{{n}}\right) \\ $$ Commented by mathmax by abdo last updated on 23/Sep/19…
Question Number 68242 by mathmax by abdo last updated on 07/Sep/19 $${find}\:{lim}_{{x}\rightarrow\mathrm{0}} \:\:\:\frac{{cos}\left(\pi{x}^{{x}} \right)+\mathrm{1}}{{x}^{\mathrm{2}} } \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 68243 by mathmax by abdo last updated on 07/Sep/19 $${let}\:{f}\left({x}\right)\:={arctan}\left({ax}\:+\mathrm{1}\right)\:\:{with}\:{a}\:{real} \\ $$$$\left.\mathrm{1}\right)\:{calculate}\:{f}^{\left({n}\right)} \left({x}\right)\:{and}\:{f}^{\left({n}\right)} \left(\mathrm{0}\right) \\ $$$$\left.\mathrm{2}\right)\:{developp}\:{f}\:{at}\:{integr}\:{serie} \\ $$$$\left.\mathrm{3}\right)\:{calculate}\:\int_{−\infty} ^{+\infty} \:\frac{{f}\left({x}\right)}{{x}^{\mathrm{2}} \:+\mathrm{4}}{dx} \\ $$…