Question Number 2796 by Rasheed Soomro last updated on 27/Nov/15 $${Prove}\:{that} \\ $$$$\left({i}\right)\:\zeta\left(\mathrm{2}\right)=\frac{\pi^{\mathrm{2}} }{\mathrm{6}}\:\:\:\:\:\:\:\:\:\:\:\:\:\left({ii}\right)\:\:\zeta\left(\mathrm{4}\right)=\frac{\pi^{\mathrm{4}} }{\mathrm{90}} \\ $$ Answered by prakash jain last updated on 28/Nov/15…
Question Number 2795 by Rasheed Soomro last updated on 27/Nov/15 $${Prove}\:{that} \\ $$$$\left(\mathrm{1}+{x}+{x}^{\mathrm{2}} +…\right)\left(\mathrm{1}+\mathrm{2}{x}+\mathrm{3}{x}^{\mathrm{2}} +…\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}.\mathrm{2}+\mathrm{2}.\mathrm{3}{x}+\mathrm{3}.\mathrm{4}{x}^{\mathrm{2}} +…\right) \\ $$ Answered by prakash jain last…
Question Number 68331 by Peculiar last updated on 08/Sep/19 $${Differentiate}\:{y}={ln}\:\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{3}{x}^{\mathrm{2}} \underset{} {\right)} \\ $$ Answered by MJS last updated on 08/Sep/19 $${y}={h}\left({g}\left({f}\left({x}\right)\right)\right) \\ $$$${y}'={h}'\left({g}\left({f}\left({x}\right)\right)\right)×{g}'\left({f}\left({x}\right)\right)×{f}'\left({x}\right)…
Question Number 68329 by 9102176137086 last updated on 08/Sep/19 $${y}=\mathrm{ln}\:\left({sinx}+{x}^{\mathrm{2}} \right) \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 2791 by Filup last updated on 27/Nov/15 $$\mathrm{Knowing}\:\mathrm{that}\:{e}=\underset{{i}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{i}!}, \\ $$$$\mathrm{Show}\:\mathrm{that}\:{e}\:\mathrm{is}\:\mathrm{finite}. \\ $$$$ \\ $$$$\mathrm{That}\:\mathrm{is},\:\mathrm{show}\:\mathrm{the}\:\mathrm{following}\:\mathrm{is}\:\mathrm{true}: \\ $$$${S}=\left\{\exists{x}\in\mathbb{R}:\mid{x}\mid<\infty,\:{e}={x}\right\} \\ $$$$\mathrm{Where}\:{S}\:\mathrm{is}\:\mathrm{the}\:\mathrm{solution} \\ $$ Commented…
Question Number 133860 by aurpeyz last updated on 24/Feb/21 $${an}\:{object}\:{placed}\:\mathrm{20}{cm}\:{from}\:{a}\:{converging} \\ $$$${lens}\:{forms}\:{a}\:{magnified}\:{clear}\:{image} \\ $$$${on}\:{a}\:{screen}.\:{when}\:{the}\:{lens}\:{is}\:{moved} \\ $$$$\mathrm{20}{cm}\:{towards}\:{the}\:{screen}.\:{a}\:{smaller} \\ $$$${clear}\:{image}\:{is}\:{formed}\:{on}\:{the}\:{screen}.\: \\ $$$${calculate}\:{the}\:{forcal}\:{length}\:{of}\:{the} \\ $$$${lens}. \\ $$$$\left({a}\right)\:\mathrm{1}.\mathrm{33} \\…
Question Number 68327 by 9102176137086 last updated on 08/Sep/19 $${y}=\left(\mathrm{1}−\mathrm{2}{x}^{−\mathrm{7}} \right)^{\mathrm{3}} \\ $$ Answered by $@ty@m123 last updated on 09/Sep/19 $${Please}\:{do}\:{not}\:{spoil}\:{the}\:{flow}\:{of}\:{this}\:{forum} \\ $$$${with}\:{a}\:{flood}\:{of}\:{similar}\:{questions}. \\ $$$${See}\:{solved}\:{examples}\:{in}\:{your}…
Question Number 133857 by mnjuly1970 last updated on 24/Feb/21 $$\:\:\:\:\:\:\:\:\:\:\:\:\:\:…..#{advanced}\:\:\:\:……………\:\:\:{calculus}#….. \\ $$$$\:\:\:\:{prove}\:\:{that}\::::\:\:\boldsymbol{\phi}=\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{{ln}^{\mathrm{2}} \left(\mathrm{1}−{x}\right)}{{x}}{dx}\overset{?} {=}\mathrm{2}\zeta\left(\mathrm{3}\right) \\ $$$$\:\:\:\:\:\:\:\:\overset{\mathrm{1}−{x}={t}} {=}\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{{ln}^{\mathrm{2}} \left({t}\right)}{\mathrm{1}−{t}}{dt}=\int_{\mathrm{0}} ^{\:\mathrm{1}} \underset{{n}=\mathrm{0}} {\overset{\infty}…
Question Number 133856 by mathocean1 last updated on 24/Feb/21 $$ \\ $$$$\begin{cases}{{U}_{\mathrm{0}} =\mathrm{1}}\\{{U}_{\mathrm{1}} =\mathrm{2}}\\{\:{U}_{{n}+\mathrm{2}} =\frac{\mathrm{3}}{\mathrm{2}}{U}_{{n}+\mathrm{1}} −\frac{\mathrm{1}}{\mathrm{2}}{U}_{{n}} }\end{cases} \\ $$$${Determinate}\:{the}\:{smallest}\:{integer} \\ $$$${n}_{\mathrm{0}} \:{such}\:{that}\:\forall\:{n}\geqslant{n}_{\mathrm{0}} \:{we}\:{have}\:\mid{U}_{{n}} −\mathrm{3}\mid\leqslant\mathrm{10}^{−\mathrm{4}} \\…
Question Number 133859 by mnjuly1970 last updated on 24/Feb/21 Answered by Ñï= last updated on 24/Feb/21 $$\underset{{n}=−\infty} {\overset{+\infty} {\sum}}\frac{\mathrm{1}}{\left({x}+{n}\pi\right)^{\mathrm{2}} } \\ $$$$=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left[\frac{\mathrm{1}}{\left({x}+{n}\pi\right)^{\mathrm{2}} }+\frac{\mathrm{1}}{\left({x}−{n}\pi\right)^{\mathrm{2}}…