Question Number 45594 by maxmathsup by imad last updated on 14/Oct/18 $${calculate}\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\:\frac{\mathrm{1}}{\left(\mathrm{4}{n}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} } \\ $$ Commented by maxmathsup by imad last updated…
Question Number 45517 by maxmathsup by imad last updated on 13/Oct/18 $${find}\:{S}\left({x}\right)=\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\frac{{x}^{{n}} }{{n}^{\mathrm{2}} }\:\:{with}\:\mid{x}\mid<\mathrm{1}\:\:. \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 45518 by maxmathsup by imad last updated on 14/Oct/18 $${calculate}\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{{cos}\left({n}\theta\right)}{{n}^{\mathrm{2}} }\:\:{and}\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\frac{{sin}\left({n}\theta\right)}{{n}^{\mathrm{2}} } \\ $$ Commented by maxmathsup by imad…
Question Number 176501 by infinityaction last updated on 20/Sep/22 $$\:\mathrm{find}\:\mathrm{the}\:\mathrm{range}\:\mathrm{of}\:\mathrm{x}+\mathrm{y}\:\mathrm{such}\:\mathrm{that} \\ $$$$\:\left({x}−\mathrm{2}\right)^{\mathrm{2}} +\:\left({y}−\mathrm{4}\right)^{\mathrm{2}} \:=\:\mathrm{49} \\ $$ Commented by cortano1 last updated on 20/Sep/22 $$\:\mathrm{let}\:\mathrm{x}+\mathrm{y}=\mathrm{k}\:\Rightarrow\mathrm{x}+\mathrm{y}−\mathrm{k}=\mathrm{0} \\…
Question Number 110954 by mathmax by abdo last updated on 01/Sep/20 $$\mathrm{verify}\:\mathrm{the}\:\mathrm{formulae} \\ $$$$\sum_{\mathrm{n}=−\infty} ^{+\infty} \:\frac{\mathrm{1}}{\left(\mathrm{na}\:+\mathrm{1}\right)^{\mathrm{p}} }\:=−\frac{\pi}{\mathrm{a}^{\mathrm{n}} }\:\mathrm{lim}_{\mathrm{z}\rightarrow−\frac{\mathrm{1}}{\mathrm{a}}} \:\:\:\frac{\mathrm{1}}{\left(\mathrm{p}−\mathrm{1}\right)!}\left\{\mathrm{cotan}\left(\pi\mathrm{z}\right)\right\}^{\left(\mathrm{p}−\mathrm{1}\right)} \\ $$$$\left.\mathrm{inthis}\:\mathrm{case}\:\:\mathrm{1}\right)\:\:\mathrm{a}\:=\mathrm{1}\:\mathrm{and}\:\mathrm{p}=\mathrm{2} \\ $$$$\left.\mathrm{2}\right)\:\mathrm{a}=\mathrm{2}\:\:\:\mathrm{and}\:\mathrm{p}=\mathrm{2} \\ $$$$\left.\mathrm{3}\right)\mathrm{a}=\mathrm{2}\:\mathrm{and}\:\mathrm{p}=\mathrm{3}…
Question Number 110948 by bemath last updated on 01/Sep/20 $$\:\:\:\:\:\sqrt{\mathrm{bemath}} \\ $$$$\mathrm{If}\:\mathrm{each}\:\mathrm{point}\:\mathrm{on}\:\mathrm{the}\:\mathrm{line}\:\mathrm{3x}+\mathrm{4y}=\mathrm{2} \\ $$$$\mathrm{is}\:\mathrm{transformed}\:\mathrm{by}\:\mathrm{matrix}\:\mathrm{M}=\begin{pmatrix}{\mathrm{2}\:\:\:\mathrm{0}}\\{\mathrm{0}\:\:\:\mathrm{1}}\end{pmatrix}\:,\:\mathrm{the} \\ $$$$\mathrm{image}\:\mathrm{is}\:\mathrm{a}\:\mathrm{line}\:\_\_\_ \\ $$ Commented by bemath last updated on 01/Sep/20…
Question Number 176336 by mnjuly1970 last updated on 16/Sep/22 $$ \\ $$$$\:{in}\:{A}\overset{\Delta} {{B}}_{\:} {C}\::\:\:\frac{{b}−{c}}{{h}_{{a}} \:}\:={k}\:, \\ $$$$\:\:\:\:\:\:\:\:\:\:{and}\:\:\hat {{A}}\:{is}\:{given}. \\ $$$$\:\:\:\:\:\:\:\:\hat {{B}}\:,\:\hat {{C}}\:=?\:\: \\ $$ Answered…
Question Number 45240 by maxmathsup by imad last updated on 10/Oct/18 $${calculate}\:\sum_{{n}=\mathrm{2}} ^{\infty} \:\:\frac{\mathrm{2}{n}+\mathrm{1}}{{n}^{\mathrm{4}} −{n}^{\mathrm{2}} } \\ $$ Commented by maxmathsup by imad last updated…
Question Number 45237 by maxmathsup by imad last updated on 10/Oct/18 $${calculate}\:\sum_{{n}=\mathrm{2}} ^{\infty} \:\frac{\mathrm{1}}{{n}^{\mathrm{3}} −{n}} \\ $$ Commented by maxmathsup by imad last updated on…
Question Number 110669 by Aina Samuel Temidayo last updated on 30/Aug/20 $$\mathrm{Given}\:\mathrm{that}\:\mathrm{f}\left(\mathrm{x}\right)=\left(\mathrm{3}+\mathrm{2x}\right)^{\mathrm{3}} \left(\mathrm{4}−\mathrm{x}\right)^{\mathrm{4}} \:\mathrm{on} \\ $$$$\mathrm{the}\:\mathrm{interval}\:−\frac{\mathrm{3}}{\mathrm{2}}<\mathrm{x}<\mathrm{4}.\:\mathrm{Find}\:\mathrm{the} \\ $$$$\left(\mathrm{a}\right)\:\mathrm{Maximum}\:\mathrm{value}\:\mathrm{of}\:\mathrm{f}\left(\mathrm{x}\right) \\ $$$$\left(\mathrm{b}\right)\:\mathrm{The}\:\mathrm{value}\:\mathrm{of}\:\mathrm{x}\:\mathrm{that}\:\mathrm{gives}\:\mathrm{the} \\ $$$$\mathrm{maximum}\:\mathrm{in}\:\left(\mathrm{a}\right) \\ $$ Commented…