Question Number 80863 by ~blr237~ last updated on 07/Feb/20 $$\:{Let}\:{W}\:{the}\:{lambert}\:{function}\:{defined}\:{as}\:{W}\left({xe}^{{x}} \right)={x}\:\:\:{x}\geqslant\mathrm{0} \\ $$$${Prove}\:{that}\:\:\:\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\:{W}\left(−{ulnu}\right)}{{u}}{du}=\frac{\zeta\left(\mathrm{2}\right)}{\mathrm{2}}\:\: \\ $$ Answered by Kamel Kamel last updated on 08/Feb/20…
Question Number 146366 by liberty last updated on 13/Jul/21 Answered by iloveisrael last updated on 13/Jul/21 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 80816 by ~blr237~ last updated on 06/Feb/20 $${let}\:\:\mathrm{0}<{a}<{b}\:\:{prove}\:{that} \\ $$$$\:{ln}\left(\mathrm{1}+\frac{{a}}{{b}}\right){ln}\left(\mathrm{1}+\frac{{b}}{{a}}\right)<\:\left({ln}\mathrm{2}\right)^{\mathrm{2}} \:\: \\ $$ Commented by ~blr237~ last updated on 06/Feb/20 $${Sir}\:\:,\:{i}\:{look}\:{like}\:\:{at}\:\left(\mathrm{3}−\mathrm{4}\right)\:{crossing}\:{you}\:{uze} \\ $$$${A}\leqslant{B}\:{and}\:{B}\geqslant{C}\:\Rightarrow\:{A}\leqslant{C}\:\:\:\:??\:…
Question Number 146300 by akolade last updated on 12/Jul/21 Answered by nimnim last updated on 12/Jul/21 $${cosB}=\frac{{a}^{\mathrm{2}} +{c}^{\mathrm{2}} −{b}^{\mathrm{2}} }{\mathrm{2}{ac}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\left({n}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} +\left({n}^{\mathrm{2}} −\mathrm{2}{n}\right)^{\mathrm{2}}…
Question Number 15179 by Joel577 last updated on 08/Jun/17 $$\mathrm{If}\:\:\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }\:+\:\mathrm{4}{y}\:=\:\mathrm{0}\: \\ $$$${y}\left(\mathrm{0}\right)\:=\:\mathrm{1}\:\mathrm{and}\:{y}\left(\frac{\pi}{\mathrm{6}}\right)\:=\:\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:+\:\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\mathrm{How}\:\mathrm{to}\:\mathrm{find}\:{y}\left({x}\right)\:? \\ $$ Answered by sma3l2996 last updated on 08/Jun/17…
Question Number 146164 by mnjuly1970 last updated on 11/Jul/21 $$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:{calulate}\::: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{S}\::\:=\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\:\mathrm{H}_{\frac{{n}}{\mathrm{2}}\:} }{\:\mathrm{2}^{\:{n}} }\:=? \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:…….{m}.{n}. \\ $$ Answered…
Question Number 146063 by mnjuly1970 last updated on 10/Jul/21 $$ \\ $$$$\:\:\:\:{if}\:\:{g}\left({x}\right)=\frac{{x}^{\:\mathrm{2}} −{x}}{\mathrm{2}{x}−\mathrm{1}}\:\:\:,\:{D}_{{g}} =\:\left[\mathrm{1}\:,\:\infty\right) \\ $$$$\:\:\:\:,\:{lim}_{{x}\rightarrow\infty} \frac{{g}^{\:−\mathrm{1}} \left({x}\right)}{{ax}\:+\:{b}}\:=\:{b}−{a}\:\:\left({a}\:<\mathrm{0}\:\right) \\ $$$$\:\:{then}\:{find}\:\:{the}\:{value}\:{of}\:{Max}\:\left({b}\:\right) \\ $$$$\:\: \\ $$$$\:\:{D}_{\:{g}} \:=\:{Domain}\:…
Question Number 146054 by mnjuly1970 last updated on 10/Jul/21 $$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\mathrm{I}\::=\:\int_{\mathrm{0}} ^{\:\:\infty} {e}^{\:−{x}} \:.\:\mathrm{J}_{\frac{\mathrm{1}}{\mathrm{2}}} \:\left({x}\:\right)\:{dx} \\ $$$$\:\:\:\:\:\:\:\:\:\:\mathrm{J}_{{v}\:} \:\left({x}\:\right)\:=\:{x}^{\:{v}} \:\underset{{n}=\mathrm{0}} {\overset{\:\infty} {\sum}}\frac{\left(−\:\mathrm{1}\:\right)^{\:{n}} \:{x}^{\:\mathrm{2}{n}} }{\mathrm{2}^{\:{n}\:+\:{v}} \:{n}\:!\:\Gamma\:\left(\:{n}\:+\:{v}\:+\mathrm{1}\:\right)}…
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Question Number 145701 by alcohol last updated on 07/Jul/21 Commented by MJS_new last updated on 07/Jul/21 $$\mathrm{1}. \\ $$$$\mathrm{square}−\mathrm{2}×\left(\mathrm{square}−\mathrm{quarter}\:\mathrm{circle}\right)= \\ $$$$=−\mathrm{square}+\mathrm{2}×\mathrm{quarter}\:\mathrm{circle}= \\ $$$$=\mathrm{half}\:\mathrm{circle}−\mathrm{square}=\mathrm{4}^{\mathrm{2}} \pi/\mathrm{2}−\mathrm{4}^{\mathrm{2}} =…