Question Number 50385 by Abdo msup. last updated on 16/Dec/18 $${p}\:{is}\:{a}\:{polynom}\:{having}\:{n}\:{roots}\:{simples}\:{with}\:{x}_{{i}} \neq\overset{−} {+}\mathrm{1} \\ $$$${caculate}\:\sum_{{k}=\mathrm{1}} ^{{n}} \:\:\frac{\mathrm{1}}{\mathrm{1}−{x}_{{i}} }\:\:{and}\:\sum_{{k}=\mathrm{1}} ^{{n}} \:\:\frac{\mathrm{1}}{\mathrm{1}−{x}_{{i}} ^{\mathrm{2}} } \\ $$$$ \\…
Question Number 50381 by prof Abdo imad last updated on 16/Dec/18 $${calculate}\:{S}_{{n}} =\sum_{{p}=\mathrm{1}} ^{{n}} \:\:\frac{{p}}{\mathrm{1}+{p}^{\mathrm{2}} \:+{p}^{\mathrm{4}} } \\ $$$${and}\:{determine}\:{lim}_{{n}\rightarrow+\infty} \:{S}_{{n}} \\ $$ Terms of Service…
Question Number 50360 by prof Abdo imad last updated on 16/Dec/18 $${find}\:{radius}\:{of}\:{convergence}\:{for}\:{the}\:{serie} \\ $$$$\sum_{{n}=\mathrm{0}} ^{\infty} \left\{\left(\mathrm{1}+{i}\right)^{{n}} −\left(\mathrm{1}−{i}\right)^{{n}} \right\}{x}^{{n}} \\ $$ Terms of Service Privacy Policy…
Question Number 115885 by bemath last updated on 29/Sep/20 $${given}\:{f}\left(\frac{\mathrm{1}}{{x}}\right)+{f}\left(\mathrm{1}−{x}\right)={x}\:,\:{x}\neq\mathrm{0} \\ $$$${find}\:\mathrm{2}{f}\left({x}\right) \\ $$$$\left({a}\right)\:\frac{\mathrm{1}+{x}^{\mathrm{2}} +{x}^{\mathrm{3}} }{{x}^{\mathrm{2}} −{x}} \\ $$$$\left({b}\right)\:\frac{{x}^{\mathrm{2}} −\mathrm{1}−{x}^{\mathrm{3}} }{{x}^{\mathrm{2}} −{x}} \\ $$$$\left({c}\right)\:\frac{{x}^{\mathrm{2}} −{x}^{\mathrm{3}}…
Question Number 115854 by bemath last updated on 29/Sep/20 $${If}\:{f}\left(\mathrm{2}{x}\right)=\:{x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{1}\:,\:{what}\:{all} \\ $$$${values}\:{of}\:{t}\:{for}\:{which}\:{f}\left(\frac{{t}}{\mathrm{2}}\right)\:=\:−\frac{\mathrm{11}}{\mathrm{4}} \\ $$$${where}\:{f}\:{represents}\:{a}\:{function} \\ $$ Answered by bobhans last updated on 29/Sep/20 $$\:\Leftrightarrow\:{f}\left({x}\right)=\:\left(\frac{{x}}{\mathrm{2}}\right)^{\mathrm{2}}…
Question Number 50010 by Tawa1 last updated on 13/Dec/18 Commented by MJS last updated on 13/Dec/18 $$\mathrm{is}\:\mathrm{it}\:\mathrm{really}\:{x}\leqslant\mathrm{2}\:\mathrm{in}\:\mathrm{the}\:\mathrm{3}^{\mathrm{rd}} \:\mathrm{line}? \\ $$$$\mathrm{in}\:\mathrm{this}\:\mathrm{case}\:\mathrm{it}'\mathrm{s}\:\mathrm{not}\:\mathrm{defined}\:\mathrm{for}\:{x}>\mathrm{2}\:\mathrm{and}\:\mathrm{it}'\mathrm{s} \\ $$$${f}\left({x}\right)=\begin{cases}{−\infty<{x}<\mathrm{0}:\:\left(\mathrm{2}{x}−\mathrm{1}\right)\vee\mathrm{1}}\\{\:\:\:\:\:\:\:\mathrm{0}\leqslant{x}\leqslant\mathrm{2}:\:{x}^{\mathrm{2}} \vee\mathrm{1}}\end{cases} \\ $$…
Question Number 49983 by Tawa1 last updated on 12/Dec/18 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{domain}\:\mathrm{of}\:\mathrm{the}\:\mathrm{function}\:\mathrm{and}\:\mathrm{sketch}\:\mathrm{the}\:\mathrm{graph} \\ $$$$\:\:\:\:\mathrm{f}\left(\mathrm{x}\right)\:\:=\:\:\begin{cases}{\mathrm{2x}\:−\:\mathrm{1},\:\:\:\:\:\:\:\:\:\:\:\mathrm{if}\:\:\:\:\:\:\:\:\mathrm{x}\:<\:\mathrm{0}}\\{\mathrm{x}^{\mathrm{2}} \:,\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{if}\:\:\:\:\:\:\mathrm{0}\:\leqslant\:\mathrm{x}\:\leqslant\:\mathrm{2}}\\{\mathrm{1}\:,\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{if}\:\:\:\:\:\:\:\mathrm{x}\:\leqslant\:\mathrm{2}}\end{cases} \\ $$ Answered by kaivan.ahmadi last updated on 15/Dec/18 $$\mathrm{this}\:\mathrm{is}\:\mathrm{not}\:\mathrm{a}\:\mathrm{function}\:\mathrm{since}\:\mathrm{f}\left(\mathrm{0}.\mathrm{5}\right)=\mathrm{0}.\mathrm{25}\:\mathrm{for}\:\mathrm{f}\left(\mathrm{x}\right)=\mathrm{x}^{\mathrm{2}} \\ $$$$\mathrm{and}\:\mathrm{f}\left(\mathrm{0}.\mathrm{5}\right)=\mathrm{1}\:\mathrm{for}\:\mathrm{f}\left(\mathrm{x}\right)=\mathrm{1}.…
Question Number 49961 by maxmathsup by imad last updated on 12/Dec/18 $${find}\:{the}\:{sequence}\:\left({a}_{{n}} \right)\:{wich}\:{verify}\: \\ $$$$\left(\sum_{{n}=\mathrm{1}} ^{\infty} \:{x}^{{n}} \right)\left(\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\frac{\left(−{x}\right)^{{n}} }{{n}+\mathrm{1}}\right)=\sum_{{n}=\mathrm{0}} ^{\infty} \:{a}_{{n}} {x}^{{n}} \:\:{also}\:{find}\:{the}\:{radius}\:{of}\:{this}\:{serie}.…
Question Number 115489 by ruwedkabeh last updated on 26/Sep/20 $${find}\:{range}\:{for} \\ $$$$\mathrm{1}.\:{f}\left({x}\right)=\frac{\sqrt{{x}^{\mathrm{2}} −\mathrm{9}}}{{x}−\mathrm{1}} \\ $$$$\mathrm{2}.\:{f}\left({x}\right)=\mathrm{ln}\:\left(\sqrt{\mathrm{4}−\mathrm{9}{x}^{\mathrm{2}} }\right) \\ $$ Answered by PRITHWISH SEN 2 last updated…
Question Number 49952 by maxmathsup by imad last updated on 12/Dec/18 $$\left.\mathrm{1}\right)\:{simplify}\:{A}_{{n}} =\:\frac{\mathrm{1}}{\left(\mathrm{2}+{i}\sqrt{\mathrm{3}}\right)^{{n}} }\:+\frac{\mathrm{1}}{\left(\mathrm{2}−{i}\sqrt{\mathrm{3}}\right)^{{n}} } \\ $$$$\left.\mathrm{2}\right)\:{smplify}\:\:{B}_{{n}} =\frac{\mathrm{1}}{\left(\mathrm{2}+\sqrt{\mathrm{3}}\right)^{{n}} }\:+\frac{\mathrm{1}}{\left(\mathrm{2}−\sqrt{\mathrm{3}}\right)^{{n}} } \\ $$$${n}\:{integr}\:{natural}. \\ $$ Commented…