Question Number 163045 by mnjuly1970 last updated on 03/Jan/22 $$ \\ $$$$\:\:\:\:\:{prove}\:{that} \\ $$$$ \\ $$$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\:\left(\:\mathrm{2}{n}+\mathrm{1}\:\right)!!}{\left(\mathrm{2}{n}\:\right)!!}\:\frac{\mathrm{1}}{\mathrm{2}^{\:{n}} \left(\mathrm{2}{n}\:+\mathrm{1}\right)^{\:\mathrm{2}} }\:=\frac{\pi\sqrt{\mathrm{2}}}{\mathrm{4}}−\mathrm{1} \\ $$ Answered by qaz…
Question Number 31972 by abdo imad last updated on 17/Mar/18 $$\left.{solve}\:{inside}\:\right]−\mathrm{1},\mathrm{1}\left[\:{the}\:{d}.{e}.\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }\:{y}^{'} \:+{y}\:={e}^{−\mathrm{2}{x}} \:.\right. \\ $$ Commented by math khazana by abdo last updated on…
Question Number 31971 by abdo imad last updated on 17/Mar/18 $${let}\:{consider}\:{the}\:{d}.{e}.\:{x}\left({x}−\mathrm{1}\right){y}^{''} \:+\mathrm{3}{xy}^{'} \:+{y}\:=\mathrm{0} \\ $$$${find}\:{a}\:{solution}\:{at}\:{form}\:\Sigma{a}_{{n}} {x}^{{n}} \:\:. \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 163033 by mnjuly1970 last updated on 03/Jan/22 $$ \\ $$$$\:\:\:\:\Omega=\:\int_{\mathrm{0}} ^{\:\infty} \frac{\:{x}\:−\:{sin}\:\left({x}\:\right)}{{x}^{\:\mathrm{3}} }{dx} \\ $$$$−−−\:{solution}−−− \\ $$$$\:\:\:\:\:\Omega\overset{\mathscr{I}.\mathscr{B}.\mathscr{P}} {=}\:\left[\:\frac{−\mathrm{1}}{\mathrm{2}\:{x}^{\:\mathrm{2}} }\:\left({x}−{sin}\left({x}\right)\right)\right]_{\mathrm{0}} ^{\infty} +\frac{\mathrm{1}}{\mathrm{2}}\:\int_{\mathrm{0}} ^{\:\infty} \frac{\mathrm{1}−{cos}\:\left({x}\right)}{{x}^{\:\mathrm{2}}…
Question Number 97483 by Rio Michael last updated on 08/Jun/20 $$\mathrm{In}\:\mathrm{each}\:\mathrm{week}\:\mathrm{the}\:\mathrm{growth}\:\mathrm{of}\:\mathrm{a}\:\mathrm{plant}\:\mathrm{is}\:\mathrm{two}−\mathrm{thirds} \\ $$$$\mathrm{the}\:\mathrm{growth}\:\mathrm{of}\:\mathrm{the}\:\mathrm{previous}\:\mathrm{week}. \\ $$$$\mathrm{The}\:\mathrm{plant}\:\mathrm{grows}\:\mathrm{12}\:\mathrm{cm}\:\mathrm{in}\:\mathrm{the}\:\mathrm{first}\:\mathrm{week}. \\ $$$$\left(\mathrm{a}\right)\:\mathrm{Calculate}\:\mathrm{the}\:\mathrm{growth}\:\mathrm{of}\:\mathrm{the}\:\mathrm{plant}\:\mathrm{in}\: \\ $$$$\left(\mathrm{b}\right)\:\mathrm{the}\:\mathrm{limiting}\:\mathrm{height}\:\mathrm{of}\:\mathrm{the}\:\mathrm{pant} \\ $$ Commented by bobhans last…
Question Number 163002 by mnjuly1970 last updated on 03/Jan/22 $$ \\ $$$$\:\:\:{prove}\:{that} \\ $$$$ \\ $$$$\:{i}:\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\:\frac{\left(−\mathrm{1}\:\right)^{\:{n}} }{\left({n}\:+\frac{\mathrm{1}}{\mathrm{2}}\right){cosh}\left({n}+\frac{\mathrm{1}}{\mathrm{2}}\right)\pi}\:=\frac{\pi}{\mathrm{4}} \\ $$$$\:\:{ii}:\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\:{sin}\left(\:\pi\:{x}\:\right)}{{x}^{\:{x}} \left(\:\mathrm{1}−{x}\:\right)^{\:\mathrm{1}−{x}} }\:\frac{{dx}}{\mathrm{1}+{x}}\:=\frac{\pi}{\mathrm{4}}…
Question Number 97463 by john santu last updated on 08/Jun/20 Commented by bobhans last updated on 08/Jun/20 $$\frac{\mathrm{4}}{\mathrm{t}−\mathrm{4}}\:=\:\frac{\mathrm{R}}{\:\sqrt{\mathrm{R}^{\mathrm{2}} +\mathrm{t}^{\mathrm{2}} }}\:\Rightarrow\:\mathrm{16}\left(\mathrm{R}^{\mathrm{2}} +\mathrm{t}^{\mathrm{2}} \right)=\mathrm{R}^{\mathrm{2}} \left(\mathrm{t}^{\mathrm{2}} −\mathrm{8t}+\mathrm{16}\right) \\…
Question Number 97418 by Rio Michael last updated on 08/Jun/20 $$\mathrm{Verify}\:\mathrm{if}\:\mathrm{the}\:\mathrm{series}\: \\ $$$$\:\underset{{n}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{\mathrm{2}{n}\:+\:\mathrm{5}}{{n}^{\mathrm{2}} \:+\mathrm{3}{n}\:+\:\mathrm{2}}\:\mathrm{is}\:\mathrm{convergent}\:\mathrm{or}\:\mathrm{divergent}. \\ $$$$\mathrm{What}\:\mathrm{method}\:\mathrm{is}\:\mathrm{easier}? \\ $$ Commented by Tony Lin last…
Question Number 162939 by mnjuly1970 last updated on 02/Jan/22 $$ \\ $$$$\:\:{lim}_{\:{x}\rightarrow\:\mathrm{3}} \:\left(\:{a}\:\lfloor{x}\:\rfloor\:+\:\lfloor\:−{x}\rfloor\right).{tan}\left(\frac{\pi{x}}{\mathrm{2}}\:\right)=−\infty \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:{a}\:\in\:? \\ $$$$ \\ $$ Terms of Service Privacy…
Question Number 162924 by mnjuly1970 last updated on 02/Jan/22 $$\: \\ $$$$\:\boldsymbol{\phi}\:=\int_{\mathrm{0}} ^{\:\infty} \frac{\:{e}^{\:−{x}^{\:\mathrm{2}} } .\mathrm{ln}\left(\:{x}\:\right)}{\:\sqrt{{x}}}\:{dx}=\lambda\:\Gamma\left(\frac{\mathrm{1}}{\mathrm{4}}\right) \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\lambda=?\:\:\:\:\:\:\:\:\:\:\:\:\:\blacksquare \\ $$$$ \\ $$ Answered…