Question Number 30407 by abdo imad last updated on 22/Feb/18 $${let}\:{give}\:{s}\left({x}\right)=\:\sum_{{n}=\mathrm{1}} ^{\infty} {nx}^{{n}} \:\:{and}\:{w}\left({x}\right)=\sum_{{n}=\mathrm{1}} ^{\infty} \frac{\mathrm{1}}{{n}}{x}^{{n}−\mathrm{1}} \:\:{for}\mid{x}\mid<\mathrm{1} \\ $$$${find}\:{s}\left({x}\right).{w}\left({x}\right)\:{at}\:{form}\:{of}\:{series} \\ $$$$\left.\mathrm{2}\right)\:{find}\:{s}\left({x}\right).{w}\left({x}\right)\:{at}\:{form}\:{of}\:{function}. \\ $$ Terms of…
Question Number 95941 by mathmax by abdo last updated on 28/May/20 $$\mathrm{prove}\:\mathrm{that}\:\frac{\mathrm{1}}{\mathrm{sinx}}\:=\sum_{\mathrm{n}=−\infty} ^{+\infty} \:\frac{\left(−\mathrm{1}\right)^{\mathrm{n}} }{\mathrm{x}+\mathrm{n}\pi} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 161442 by nadovic last updated on 18/Dec/21 $$\mathrm{Use}\:\mathrm{the}\:\mathrm{binomial}\:\mathrm{theorem}\:\mathrm{to}\:\mathrm{write} \\ $$$$\mathrm{the}\:\mathrm{first}\:\mathrm{four}\:\mathrm{terms}\:\mathrm{of}\:\mathrm{the}\:\mathrm{expansion} \\ $$$$\mathrm{of}\:\sqrt{\mathrm{2}+\mathrm{3}{x}−{x}^{\mathrm{2}} } \\ $$ Answered by mathmax by abdo last updated on…
Question Number 161387 by savitar last updated on 17/Dec/21 $$\:{sin}\sqrt{\mathrm{1}+\pi^{\mathrm{2}} {n}^{\mathrm{2}} }\:\sim\:\frac{\left(−\mathrm{1}\right)^{{n}} }{\mathrm{2}\pi{n}}\:\:? \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 161391 by cortano last updated on 17/Dec/21 $$\:\:{f}^{\:\mathrm{3}} \left({x}\right)+{x}^{\mathrm{2}} \:{f}\left({x}\right)=\mathrm{2}{x}^{\mathrm{3}} +\mathrm{4}{x}^{\mathrm{2}} +\mathrm{3}{x}+\mathrm{1} \\ $$$$\:\forall{x}\in\mathbb{R}\: \\ $$ Commented by Rasheed.Sindhi last updated on 17/Dec/21…
Question Number 95839 by mathmax by abdo last updated on 28/May/20 $$\mathrm{solve}\:\mathrm{y}^{\left(\mathrm{3}\right)} −\mathrm{2y}^{\left(\mathrm{2}\right)} \:+\mathrm{3y}\:\:−\mathrm{2y}\:=\mathrm{sinx} \\ $$ Commented by john santu last updated on 28/May/20 $$\mathrm{homogenous}\:\mathrm{solution}…
Question Number 95838 by mathmax by abdo last updated on 28/May/20 $$\mathrm{determine}\:\mathrm{L}\left(\mathrm{f}^{\left(\mathrm{3}\right)} \left(\mathrm{x}\right)\:\:\mathrm{with}\:\mathrm{L}\:\mathrm{is}\:\mathrm{laplace}\:\mathrm{transform}\right. \\ $$ Answered by mathmax by abdo last updated on 28/May/20 $$\mathrm{L}\left(\mathrm{f}^{\left(\mathrm{3}\right)}…
Question Number 95837 by mathmax by abdo last updated on 28/May/20 $$\mathrm{let}\:\mathrm{p}\left(\mathrm{x}\right)=\left(\mathrm{1}+\mathrm{ix}+\mathrm{x}^{\mathrm{2}} \right)^{\mathrm{n}} −\left(\mathrm{1}−\mathrm{ix}\:+\mathrm{x}^{\mathrm{2}} \right)^{\mathrm{n}} \\ $$$$\left.\mathrm{1}\right)\:\mathrm{determine}\:\mathrm{roots}\:\mathrm{of}\:\mathrm{p}\left(\mathrm{x}\right) \\ $$$$\left.\mathrm{2}\right)\:\mathrm{find}\:\mathrm{p}\left(\mathrm{x}\right)\:\mathrm{at}\:\mathrm{form}\:\Sigma\:\mathrm{a}_{\mathrm{i}} \:\mathrm{x}^{\mathrm{i}} \\ $$$$\left.\mathrm{3}\right)\mathrm{ddtermne}\:\mathrm{p}\left(\mathrm{x}\right)\:\mathrm{at}\:\mathrm{form}\:\mathrm{arctan} \\ $$$$\left.\mathrm{4}\right)\:\mathrm{factorize}\:\mathrm{p}\left(\mathrm{x}\right)\:\mathrm{inside}\:\mathrm{C}\left[\mathrm{x}\right] \\…
Question Number 95786 by abdomathmax last updated on 27/May/20 $$\mathrm{let}\:\mathrm{f}\left(\mathrm{x}\right)\:=\frac{\mathrm{1}}{\mathrm{x}}\mathrm{ln}\left(\mathrm{1}+\mathrm{2x}\right) \\ $$$$\left.\mathrm{1}\right)\:\mathrm{calculate}\:\mathrm{f}^{\left(\mathrm{n}\right)} \left(\mathrm{x}\right)\mathrm{and}\:\mathrm{f}^{\left(\mathrm{n}\right)} \left(\mathrm{1}\right) \\ $$$$\left.\mathrm{2}\right)\mathrm{developp}\:\mathrm{f}\:\mathrm{at}\:\mathrm{integr}\:\mathrm{serie}\:\mathrm{at}\:\mathrm{x}_{\mathrm{0}} =\mathrm{1} \\ $$$$\left.\mathrm{3}\right)\mathrm{developp}\:\mathrm{f}\:\mathrm{at}\:\mathrm{integr}\:\mathrm{serie}\:\:\mathrm{at}\:\mathrm{x}_{\mathrm{0}} =\mathrm{0} \\ $$ Commented by Rio…
Question Number 95784 by abdomathmax last updated on 27/May/20 $$\mathrm{calculate}\:\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{H}_{\mathrm{n}} }{\mathrm{n}^{\mathrm{2}} } \\ $$$$\mathrm{H}_{\mathrm{n}} =\sum_{\mathrm{k}=\mathrm{1}} ^{\mathrm{n}} \:\frac{\mathrm{1}}{\mathrm{k}} \\ $$ Terms of Service Privacy…