Question Number 28168 by abdo imad last updated on 21/Jan/18 $${find}\:{the}\:{nature}\:{of}\:\:\sum_{{n}=\mathrm{1}} ^{\infty} {arcos}\left(\mathrm{1}−\frac{\mathrm{1}}{{n}^{{a}} }\right){z}^{{n}} \:\:{with}\:{a}>\mathrm{1}. \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 28138 by abdo imad last updated on 21/Jan/18 $${studie}\:{and}?{give}\:{the}\:{graph}\:{for}\:{the}\:{function} \\ $$$${f}\left({x}\right)=\:{e}^{{x}} \:\:−{x}^{{e}} \:\:\:\:\:. \\ $$ Commented by abdo imad last updated on 26/Jan/18…
Question Number 93464 by i jagooll last updated on 13/May/20 $$\mathrm{If}\:\mathrm{f}\:\mathrm{a}\:\mathrm{function}\:\mathrm{such}\:\mathrm{that}\: \\ $$$$\mathrm{f}\left(\mathrm{a}\right).\mathrm{f}\left(\mathrm{b}\right)−\mathrm{f}\left(\mathrm{a}+\mathrm{b}\right)=\mathrm{a}+\mathrm{b}. \\ $$$$\mathrm{find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\mathrm{f}\left(\mathrm{2019}\right)\: \\ $$ Answered by prakash jain last updated on 13/May/20…
Question Number 93418 by abdomathmax last updated on 13/May/20 $${let}\:{A}=\begin{pmatrix}{\mathrm{1}\:\:\:\:\:\:\:\:\mathrm{2}\:\:\:\:\:\:\mathrm{3}}\\{\mathrm{3}\:\:\:\:\:\:\:\:\:\mathrm{2}\:\:\:\:\:\:\:\mathrm{1}}\end{pmatrix}\:\:\:\:\:\in{M}_{\mathrm{3}} \left({C}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{1}\:\:\:\:\:\:\:\:\mathrm{4}\:\:\:\:\:\:\:\mathrm{2}\:\right) \\ $$$$\left.\mathrm{1}\right)\:{find}\:{A}^{−\mathrm{1}} \:{if}\:{A}\:{inversible} \\ $$$$\left.\mathrm{2}\right)\:{calculate}\:{A}^{{n}} \\ $$$$\left.\mathrm{3}\right){find}\:{cosA}\:\:{and}\:{sinA} \\ $$$$\left.\mathrm{4}\right)\:{is}\:{cos}^{\mathrm{2}} \:{A}\:+{sin}^{\mathrm{2}} \:{A}\:={I}\:? \\…
Question Number 93415 by abdomathmax last updated on 13/May/20 $${let}\:\:{A}\:=\begin{pmatrix}{\mathrm{1}\:\:\:\:\:\:\:\:−\mathrm{1}}\\{\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{1}}\end{pmatrix} \\ $$$$\left.\mathrm{1}\right)\:{calculate}\:{A}^{−\mathrm{1}} \:{and}\:{A}^{−\mathrm{2}} \\ $$$$\left.\mathrm{2}\right)\:{calculate}\:{A}^{{n}} \\ $$$$\left.\mathrm{3}\right)\:{find}\:{e}^{{A}} \:{and}\:{e}^{−{A}} \\ $$ Commented by prakash jain last…
Question Number 93414 by abdomathmax last updated on 13/May/20 $${let}\:{p}\left({x}\right)=\frac{{x}^{{n}} \left(\mathrm{4}−\mathrm{2}{x}\right)^{{n}} }{{n}!} \\ $$$$\left.\mathrm{1}\right)\:{prove}\:{that}\:\:{p}^{\left({k}\right)} \left(\mathrm{0}\right)={p}^{\left({k}\right)} \left(\mathrm{2}\right)=\mathrm{0}\:{for}\:{all}\:{k}\in\left[\mathrm{1},{n}−\mathrm{1}\right] \\ $$$$\left.\mathrm{2}\right)\:\:{prove}\:{that}\:\:\forall{m}\in{N}\:\:\:\:{p}^{\left({m}\right)} \left(\mathrm{0}\right)\:{and}\:{p}^{\left({m}\right)} \left(\mathrm{2}\right)\:{are}\:{integrs} \\ $$ Terms of Service…
Question Number 27847 by NECx last updated on 15/Jan/18 $${Find}\:{the}\:{range}\:{of}\:{f}\left({x}\right)=\frac{{x}+\mathrm{5}}{{x}^{\mathrm{2}} −\mathrm{4}} \\ $$ Commented by NECx last updated on 16/Jan/18 $${please}\:{help} \\ $$ Answered by…
Question Number 93331 by john santu last updated on 12/May/20 $$\mathrm{If}\:\mathrm{f}\left(\mathrm{x}+\mathrm{y}\right)\:=\:\mathrm{f}\left(\mathrm{x}\right)+\mathrm{f}\left(\mathrm{y}\right)\:\mathrm{and} \\ $$$$\mathrm{f}\left(\mathrm{1}\right)=\mathrm{5}\:\mathrm{then}\:\mathrm{what}\:\mathrm{is}\:\mathrm{the}\:\mathrm{value} \\ $$$$\mathrm{of}\:\mathrm{f}\left(\mathrm{2020}\right)\:?\: \\ $$ Commented by mr W last updated on 12/May/20…
Question Number 27790 by abdo imad last updated on 14/Jan/18 $${let}\:{give}\:\:{f}\left({x}\right)=\:{x}^{{n}−\mathrm{1}} \:{ln}\left(\mathrm{1}+{x}\right)\:\:{with}\:{n}\:{fromN}^{\ast} \:\:\:{find}\:{f}^{\left({n}\right)} \left({x}\right) \\ $$ Commented by abdo imad last updated on 15/Jan/18 $$\:{by}\:{leibnitz}\:{formule}\:{f}^{\left({n}\right)}…
Question Number 158862 by mathlove last updated on 09/Nov/21 Answered by gsk2684 last updated on 09/Nov/21 $${f}\left({x}\right)+\mathrm{2}{f}\left(\frac{\mathrm{1}}{{x}}\right)={x}…\left(\mathrm{1}\right) \\ $$$${replace}\:{x}\:{by}\:\frac{\mathrm{1}}{{x}} \\ $$$${f}\left(\frac{\mathrm{1}}{{x}}\right)+\mathrm{2}{f}\left({x}\right)=\frac{\mathrm{1}}{{x}}\Rightarrow\frac{{x}−{f}\left({x}\right)}{\mathrm{2}}+\mathrm{2}{f}\left({x}\right)=\frac{\mathrm{1}}{{x}} \\ $$$$\frac{{x}−{f}\left({x}\right)+\mathrm{4}{f}\left({x}\right)}{\mathrm{2}}=\frac{\mathrm{1}}{{x}}\Rightarrow{x}+\mathrm{3}{f}\left({x}\right)=\frac{\mathrm{2}}{{x}} \\ $$$$\mathrm{3}{f}\left({x}\right)=\frac{\mathrm{2}}{{x}}−{x}\Rightarrow{f}\left({x}\right)=\frac{\mathrm{2}−{x}^{\mathrm{2}}…