Question Number 205607 by SANOGO last updated on 25/Mar/24 $${soit}\:{H}\:{espace}\:{de}\:{Hilbert} \\ $$$${montrez}\:{que}: \\ $$$${d}\left({x},\left\{{a}\right\}\right)^{\lfloor} \:=\:\frac{\mid<{x},{a}>\mid}{\mid\mid{a}\mid\mid} \\ $$ Answered by Berbere last updated on 25/Mar/24 $$\frac{{a}}{\mid{a}\mid},{a}_{\mathrm{2}}…
Question Number 205516 by MathedUp last updated on 23/Mar/24 $$\underset{{h}=\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{\boldsymbol{\zeta}\left(\mathrm{2}{h}\right)−\mathrm{1}}{{h}}\:=\:…..? \\ $$$$\underset{{h}=\mathrm{1}} {\overset{\infty} {\sum}}\:\left(\boldsymbol{\zeta}\left(\mathrm{2}{h}+\mathrm{1}\right)−\mathrm{1}\right)=……? \\ $$$$\mathrm{pls}\:\mathrm{help}\:\mathrm{me} \\ $$ Terms of Service Privacy Policy…
Question Number 205487 by MathedUp last updated on 22/Mar/24 $${f}\left({t}\right)={t}\centerdot\int_{\mathrm{0}} ^{\infty} \:{e}^{−{st}+\mathrm{tanh}\left({t}\right)} \mathrm{d}{t} \\ $$$$\underset{{t}\rightarrow\mathrm{0}} {\mathrm{lim}}{f}\left({t}\right)=??? \\ $$$$\mathrm{g}\left({t}\right)=\frac{\mathrm{1}}{{t}}{f}\left({t}\right) \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\left\{\mathrm{g}\left({n}\right)−\underset{{h}=\mathrm{1}} {\overset{{n}} {\sum}}\:\frac{\mathrm{1}}{{h}}\right\}=??? \\ $$…
Question Number 205451 by MrGHK last updated on 21/Mar/24 $$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\int_{\mathrm{0}} ^{{x}} \frac{{dt}}{{e}^{\mathrm{2}{t}} {t}} \\ $$ Answered by Berbere last updated on 21/Mar/24 $$\left.{x}\overset{{f}} {\rightarrow}{e}^{−\mathrm{2}{x}}…
Question Number 205446 by 073 last updated on 21/Mar/24 Answered by namphamduc last updated on 21/Mar/24 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 205472 by SANOGO last updated on 21/Mar/24 Answered by TheHoneyCat last updated on 01/Apr/24 $$\mathrm{This}\:\mathrm{will}\:\mathrm{be}\:\mathrm{true}\:\mathrm{if}\:\left(\mathrm{and}\:\mathrm{only}\:\mathrm{if}\right)\:\mathrm{the}\:{A}_{{n}} \:\mathrm{are}\:\mathrm{all} \\ $$$$\mathrm{disjoint}. \\ $$$$\mathrm{If}\:\mathrm{they}\:\mathrm{are}\:\mathrm{not},\:\mathrm{take}\:{x}\in{A}_{{i}} \cap{A}_{{k}} \:\left(\mathrm{with}\:{k}\neq{i}\right) \\…
Question Number 205396 by MathedUp last updated on 20/Mar/24 $$\mathrm{Can}'\mathrm{t}\:\mathrm{understand}\:\mathrm{why} \\ $$$$\int_{\mathrm{0}} ^{\infty} \left[{e}_{\:} ^{\mathrm{sin}\left({z}\right)} −{e}^{\mathrm{cos}\left({z}\right)} \right]\mathrm{d}{z}\:\mathrm{isn}'{t}\:\frac{\mathrm{1}}{\mathrm{2}}\pi\left(\boldsymbol{\mathrm{I}}_{\mathrm{0}} ^{\:} \left(\mathrm{1}\right)−\boldsymbol{\mathrm{L}}_{\mathrm{0}} \left(\mathrm{1}\right)\right)\approx\mathrm{3}.\mathrm{104}… \\ $$$$\mathrm{I}\:\mathrm{Just}\:\mathrm{guessing}\:{f}\left(\mathrm{0}\right)=\:\frac{\mathrm{1}}{\mathrm{2}}\pi\boldsymbol{\mathrm{I}}_{\mathrm{0}} \left(\mathrm{1}\right)−\boldsymbol{\mathrm{L}}_{\mathrm{0}} \left(\mathrm{1}\right) \\…
Question Number 205371 by 073 last updated on 19/Mar/24 Commented by lepuissantcedricjunior last updated on 19/Mar/24 $$\boldsymbol{\mathrm{I}}=\int_{\mathrm{0}} ^{\mathrm{2}} \boldsymbol{\mathrm{log}}\left(\boldsymbol{\mathrm{x}}^{\mathrm{3}} +\mathrm{8}\right)\boldsymbol{\mathrm{dx}} \\ $$$$\boldsymbol{\mathrm{posons}}\:\begin{cases}{\boldsymbol{\mathrm{u}}=\boldsymbol{\mathrm{log}}\left(\boldsymbol{\mathrm{x}}^{\mathrm{3}} +\mathrm{8}\right)}\\{\boldsymbol{\mathrm{v}}'=\mathrm{1}}\end{cases}\Leftrightarrow\begin{cases}{\boldsymbol{\mathrm{u}}'=\frac{\mathrm{3}\boldsymbol{{x}}^{\mathrm{2}} }{\left(\boldsymbol{{x}}^{\mathrm{3}} +\mathrm{8}\right)\boldsymbol{\mathrm{ln}}\mathrm{10}}}\\{\boldsymbol{\mathrm{v}}=\boldsymbol{\mathrm{x}}}\end{cases}…
Question Number 205360 by Davidtim last updated on 18/Mar/24 Answered by A5T last updated on 18/Mar/24 Commented by A5T last updated on 19/Mar/24 $${FE}={EG};{EB}=\sqrt{\mathrm{4}^{\mathrm{2}} +\left(\mathrm{3}−{FE}\right)^{\mathrm{2}}…
Question Number 205323 by liuxinnan last updated on 16/Mar/24 $${If}\:\:\:\:{log}_{\sqrt{\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }{\mathrm{2}}}} \frac{{a}+{b}}{{b}}\geqslant{log}_{\sqrt{{ab}}} \frac{\mathrm{2}}{\frac{\mathrm{1}}{{a}}+\frac{\mathrm{1}}{{b}}} \\ $$$${when}\:{a}>\mathrm{1}\:{b}>\mathrm{1} \\ $$ Commented by liuxinnan last updated on 17/Mar/24…