Question Number 178635 by infinityaction last updated on 19/Oct/22 $$\mathrm{solution}\:\mathrm{set}\:\mathrm{of}\:\:\mathrm{log}_{\mathrm{x}^{\mathrm{2}\:\:\:} } \left(\frac{\mathrm{x}}{\mid\mathrm{x}\mid}−\mathrm{x}\right)\geqslant\mathrm{0} \\ $$ Commented by Frix last updated on 19/Oct/22 $${x}\leqslant−\mathrm{2}\vee\mathrm{0}<{x}<\mathrm{1} \\ $$ Commented…
Question Number 178624 by infinityaction last updated on 19/Oct/22 $$\:\:\:\:\:\:\:\boldsymbol{\mathrm{let}}\:\boldsymbol{\mathrm{f}}:\left[\mathrm{0},\mathrm{1}\right]\rightarrow\:\mathbb{R}\:\boldsymbol{\mathrm{be}}\:\boldsymbol{\mathrm{given}}\:\boldsymbol{\mathrm{by}} \\ $$$$\:\:\boldsymbol{\mathrm{f}}\left(\boldsymbol{\mathrm{x}}\right)\:=\:\:\frac{\left(\mathrm{1}+\boldsymbol{\mathrm{x}}^{\frac{\mathrm{1}}{\mathrm{3}}} \right)^{\mathrm{3}} +\left(\mathrm{1}−\boldsymbol{\mathrm{x}}^{\frac{\mathrm{1}}{\mathrm{3}}} \right)^{\mathrm{3}} }{\mathrm{8}\left(\mathrm{1}+\boldsymbol{\mathrm{x}}\right)}\:\:\:\boldsymbol{\mathrm{then}} \\ $$$$\:\:\boldsymbol{\mathrm{max}}\left\{\boldsymbol{\mathrm{f}}\left(\boldsymbol{\mathrm{x}}\right):\:\boldsymbol{\mathrm{x}}\in\left[\mathrm{0},\mathrm{1}\right]\right\}−\boldsymbol{\mathrm{min}}\left\{\boldsymbol{\mathrm{f}}\left(\boldsymbol{\mathrm{x}}\right):\boldsymbol{\mathrm{x}}\in\left[\mathrm{0},\mathrm{1}\right]\right\} \\ $$$$\mathrm{is} \\ $$ Answered by a.lgnaoui…
Question Number 112681 by bemath last updated on 09/Sep/20 $$\mathrm{solve}\:\mathrm{the}\:\mathrm{equation}\:\mathrm{of}\:\mathrm{function}\: \\ $$$$\left(\mathrm{f}\left(\mathrm{3x}\right)\right)^{\mathrm{2}} \:=\:\left(\mathrm{f}\left(\mathrm{2x}\right)\right)^{\mathrm{2}} +\left(\mathrm{f}\left(\mathrm{x}\right)\right)^{\mathrm{2}} \\ $$ Answered by bobhans last updated on 09/Sep/20 $$\:\left(\blacklozenge\right)\:\left(\mathrm{f}\left(\mathrm{3x}\right)\right)^{\mathrm{2}} \:=\:\left(\mathrm{f}\left(\mathrm{2x}\right)\right)^{\mathrm{2}}…
Question Number 47064 by maxmathsup by imad last updated on 04/Nov/18 $$\left.\mathrm{1}\right){calculate}\:\:{u}_{{n}} =\int_{\mathrm{0}} ^{\infty} \:\:\frac{{sin}\left({nx}\right)}{{sh}\left(\mathrm{2}{x}\right)}{dx}\:\:{with}\:\:{n}\:{integr}\:{natural} \\ $$$$\left.\mathrm{2}\right)\:{calculate}\:\sum_{{n}=\mathrm{0}} ^{\infty} \:{u}_{{n}} \:. \\ $$ Commented by maxmathsup…
Question Number 47063 by maxmathsup by imad last updated on 04/Nov/18 $$\left.\mathrm{1}\right){calculate}\:{u}_{{n}} =\:\int_{\mathrm{0}} ^{\infty} \:\:{e}^{−{nx}} {ln}\left(\mathrm{1}+{x}\right){dx}\:\:{with}\:{n}\:{integr}\:{natural} \\ $$$$\left.\mathrm{2}\right)\:{calculate}\:{lim}_{{n}\rightarrow+\infty} {u}_{{n}} \\ $$$$\left.\mathrm{3}\right)\:{find}\:\sum_{{n}=\mathrm{0}} ^{\infty} \:{u}_{{n}} \\ $$…
Question Number 46983 by maxmathsup by imad last updated on 03/Nov/18 $$\left({u}_{{n}} \right)\:{is}\:{a}\:{sequence}\:{wich}\:{verify}\:{u}_{{n}} ={n}\:{u}_{{n}−\mathrm{1}} \:−\lambda\:\:\left(\lambda\:{from}\:{R}\:{and}\:{n}\geqslant\mathrm{1}\right) \\ $$$${calculate}\:{u}_{{n}} {interm}\:{of}\:{n}\:{and}\:\lambda\:. \\ $$ Terms of Service Privacy Policy…
Question Number 178032 by TheHoneyCat last updated on 12/Oct/22 $$\bullet\:{D}=\left\{{z}\::\:\mid{z}\mid<\mathrm{1}\right\} \\ $$$$\bullet\:\mathscr{H}\:\left({A}\rightarrow{B}\right)\:\mathrm{denotes}\:\mathrm{the}\:\mathrm{set}\:\mathrm{of}\:\mathrm{holomorfic} \\ $$$$\mathrm{functions}\:\mathrm{from}\:{A}\:\mathrm{to}\:{B} \\ $$$$\bullet\:\mathrm{We}\:\mathrm{define}: \\ $$$${W}=\left\{{f}\in\mathscr{H}\:\left({D}\rightarrow\mathbb{R}\right)\::\:\mid\mid{f}\mid\mid_{{W}} <\infty\:\right\} \\ $$$$\mathrm{where}\:\:\mid\mid\:\centerdot\:\mid\mid_{{W}} \::\:\begin{cases}{{W}}&{\rightarrow}&{\mathbb{R}_{+} }\\{{f}}&{ }&{\underset{{n}=\mathrm{0}} {\overset{\infty}…
Question Number 46852 by maxmathsup by imad last updated on 01/Nov/18 $${find}\:{the}\:{value}\:{of}\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{{n}^{\mathrm{3}} }{\mathrm{3}^{{n}} } \\ $$ Commented by maxmathsup by imad last updated…
Question Number 46840 by maxmathsup by imad last updated on 01/Nov/18 $${find}\:{lim}_{{n}\rightarrow+\infty} \:\sum_{{k}=\mathrm{1}} ^{{n}} \:\:\frac{{n}^{\mathrm{2}\:} +{k}^{\mathrm{2}} }{{n}^{\mathrm{3}} \:+{k}^{\mathrm{3}} } \\ $$ Commented by maxmathsup by…
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