Question Number 2851 by 123456 last updated on 28/Nov/15 $${x}^{\mathrm{2}} ={x}+{x}+{x}+\centerdot\centerdot\centerdot+{x}\:\left({x}\:\mathrm{times}\right) \\ $$$$\mathrm{taking}\:\mathrm{derivate} \\ $$$$\mathrm{2}{x}=\mathrm{1}+\mathrm{1}+\mathrm{1}+\centerdot\centerdot\centerdot+\mathrm{1}\:\left({x}\:\mathrm{times}\right) \\ $$$$\mathrm{2}{x}={x}\:\left({x}\neq\mathrm{0}\right) \\ $$$$\mathrm{2}=\mathrm{1} \\ $$$$\mathrm{where}\:\mathrm{the}\:\mathrm{problem}? \\ $$ Commented by…
Question Number 133923 by bemath last updated on 25/Feb/21 $$\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mid\mathrm{sin}\:\mathrm{x}\mid}{\mathrm{x}^{\mathrm{2}} }\:=? \\ $$ Answered by EDWIN88 last updated on 25/Feb/21 $$\:\mathrm{x}^{\mathrm{2}} \:=\:\mid\mathrm{x}\mid^{\mathrm{2}} \:\Rightarrow\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mid\mathrm{sin}\:\mathrm{x}\mid}{\mid\mathrm{x}\mid}\:.\frac{\mathrm{1}}{\mid\mathrm{x}\mid}=\:\underset{{x}\rightarrow\mathrm{0}}…
Question Number 133916 by shaker last updated on 25/Feb/21 Answered by TheSupreme last updated on 25/Feb/21 $${S}_{{n}} =\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}\frac{{e}^{{kx}} +{e}^{−{kx}} }{\mathrm{2}}=\frac{\mathrm{1}}{\mathrm{2}}\left\{\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}\left({e}^{{x}} \right)^{{k}}…
Question Number 2845 by Rasheed Soomro last updated on 28/Nov/15 $$\mathcal{W}{hile}\:{you}\:{are}\:{in}\:{between}\:{the}\:{project} \\ $$$$\mathcal{I}\:{am}\:{trying}\:{to}\:{improve}\:{my}\:{digestiblity}\:{to} \\ $$$${digest}\:{the}\:{concept}\:{of}\:'{analytical}\:{continuation}'. \\ $$$$ \\ $$$${First}\:{we}\:{make}\:{aformula}\:{to}\:{sum}\:{n}\:{terms}\:{of}\:{a}\:{powe}\:{series}: \\ $$$$\frac{{x}^{{n}} −\mathrm{1}}{{x}−\mathrm{1}}=\mathrm{1}+{x}+{x}^{\mathrm{2}} +…+{x}^{{n}} \\ $$$${latter}\:{we}\:{change}\:{it}\:{for}\:\mid{x}\mid<\mathrm{1}\:{and}\:{n}\rightarrow\infty\:\left[{x}^{{n}}…
Question Number 133918 by mohammad17 last updated on 25/Feb/21 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 2843 by prakash jain last updated on 28/Nov/15 $$\mathrm{Prove}\:\mathrm{by}\:\mathrm{mathematical}\:\mathrm{induction} \\ $$$${n}\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)…\left({n}+{r}−\mathrm{1}\right)\:\mathrm{is}\:\mathrm{divible}\:\mathrm{by}\:{r}! \\ $$$$ \\ $$$${n},{r}\in\mathbb{N}. \\ $$ Commented by 123456 last updated on…
Question Number 2842 by Yozzi last updated on 28/Nov/15 $${Prove}\:{that}\:\pi=\mathrm{3}.\mathrm{14}…\:{is}\:{irrational}. \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 133915 by nherly last updated on 25/Feb/21 Commented by TheSupreme last updated on 25/Feb/21 $${nice}\:{photo} \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 133911 by liberty last updated on 25/Feb/21 $$\mathcal{A}=\int\:\frac{\mathrm{cos}\:\mathrm{x}}{\mathrm{sin}\:\mathrm{5x}+\mathrm{sin}\:\mathrm{x}}\:\mathrm{dx}\:? \\ $$ Answered by Ar Brandon last updated on 25/Feb/21 $$\mathcal{A}=\int\frac{\mathrm{cosx}}{\mathrm{sin5x}+\mathrm{sinx}}\mathrm{dx}=\int\frac{\mathrm{cosx}}{\mathrm{2sin3xcos2x}}\mathrm{dx} \\ $$$$\:\:\:\:=\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\mathrm{cosx}\:\mathrm{dx}}{\left(\mathrm{3sinx}−\mathrm{4sin}^{\mathrm{3}} \mathrm{x}\right)\left(\mathrm{1}−\mathrm{2sin}^{\mathrm{2}} \mathrm{x}\right)}…
Question Number 133910 by Ar Brandon last updated on 25/Feb/21 $$\:\:\:\:\mathcal{D}\acute {\mathrm{e}montrer}\:\mathrm{que}; \\ $$$$\underset{\mathrm{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{\mathrm{n}} }{\mathrm{1}+\mathrm{n}^{\mathrm{4}} }=\frac{\mathrm{1}}{\mathrm{2}}\left[\frac{\pi}{\:\sqrt{\mathrm{2}}}\:\frac{\mathrm{sin}\left(\frac{\pi}{\:\sqrt{\mathrm{2}}}\right)\mathrm{cosh}\left(\frac{\pi}{\:\sqrt{\mathrm{2}}}\right)+\mathrm{sinh}\left(\frac{\pi}{\:\sqrt{\mathrm{2}}}\right)\mathrm{cos}\left(\frac{\pi}{\:\sqrt{\mathrm{2}}}\right)}{\mathrm{sinh}^{\mathrm{2}} \left(\frac{\pi}{\:\sqrt{\mathrm{2}}}\right)+\mathrm{sin}^{\mathrm{2}} \left(\frac{\pi}{\:\sqrt{\mathrm{2}}}\right)}\right] \\ $$ Terms of Service…