Question Number 30754 by abdo imad last updated on 25/Feb/18 $${prove}\:{that}\:{e}^{{x}} =\sum_{{k}=\mathrm{0}} ^{{n}} \:\frac{{x}^{{k}} }{{k}!}\:+\frac{{x}^{{n}+\mathrm{1}} }{{n}!}\:\int_{\mathrm{0}} ^{} \left(\mathrm{1}−{t}\right)^{{n}} \:{e}^{{tx}} {dt} \\ $$$$\left.\mathrm{2}\right)\:{prove}\:{that}\:\:{e}^{{x}} =\:\sum_{{k}=\mathrm{0}} ^{\infty\:} \:\:\frac{{x}^{{k}}…
Question Number 30752 by abdo imad last updated on 25/Feb/18 $${let}\:{f}\left({x}\right)=\:\frac{\mathrm{1}}{\mathrm{1}+{x}^{\mathrm{2}} }\:\:{calculate}\:{f}^{\left({n}\right)} \left({x}\right). \\ $$ Commented by abdo imad last updated on 28/Feb/18 $${we}\:{have}\:{f}\left({x}\right)=\:\frac{\mathrm{1}}{\left({x}+{i}\right)\left({x}−{i}\right)}=\frac{\mathrm{1}}{\mathrm{2}{i}}\left(\:\frac{\mathrm{1}}{{x}−{i}}\:−\frac{\mathrm{1}}{{x}+{i}}\right)\:\Rightarrow \\…
Question Number 30753 by abdo imad last updated on 25/Feb/18 $${let}\:{f}_{{n}} \left({x}\right)=\frac{\mathrm{1}}{{x}^{{n}+\mathrm{1}} }\:\left({e}^{{x}} \:\:−\sum_{{p}=\mathrm{0}} ^{{n}\:} \:\:\:\frac{{x}^{{p}} }{{p}!}\right) \\ $$$$\left.\mathrm{1}\right)\:{prove}\:{that}\:{f}^{\left({n}\right)} \left({x}\right)=\frac{{Q}_{{n}} \left({x}\right)\:{e}^{{x}} \:−{P}_{{n}} \left({x}\right)}{{x}^{\mathrm{2}{n}+\mathrm{1}} }\:{find}\:{the} \\…
Question Number 30751 by abdo imad last updated on 25/Feb/18 $${study}\:{and}\:{give}\:{the}\:{graph}\:{for}\: \\ $$$${f}\left({x}\right)=\frac{{x}^{\mathrm{2}} }{{x}−\mathrm{1}}\:{e}^{\frac{\mathrm{1}}{{x}}} \:\:. \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 30750 by abdo imad last updated on 25/Feb/18 $${f}\:{function}\:{C}^{\infty} \:/{f}^{'} =\mathrm{1}+{f}^{\mathrm{2}} \:\:{let}\:{take}\:{a}_{{k}} =\frac{{f}^{\left({k}\right)} \left(\mathrm{0}\right)}{{k}!}\: \\ $$$${prove}\:{that}\:{a}_{{n}+\mathrm{1}} =\:\frac{\mathrm{1}}{{n}+\mathrm{1}}\:\sum_{{k}=\mathrm{0}} ^{{n}} \:{a}_{{k}} \:\:{a}_{{n}−{k}} \\ $$ Terms…
Question Number 30749 by abdo imad last updated on 25/Feb/18 $${let}\:{f}\left({x}\right)={arcsinx}\:{with}\:{x}\in\left[\mathrm{0},\mathrm{1}\right] \\ $$$$\left.\mathrm{1}\right)\:{prove}\:{that}\:\left(\mathrm{1}−{x}^{\mathrm{2}} \right){f}^{''} \left({x}\right)\:−{xf}^{'} \left({x}\right)=\mathrm{0} \\ $$$$\left.\mathrm{2}\right){prove}\:{that}\:\left(\mathrm{1}−{x}^{\mathrm{2}} \right){f}^{\left({n}+\mathrm{2}\right)} \left({x}\right)=\left(\mathrm{2}{n}+\mathrm{1}\right){x}\:{f}^{\left({n}+\mathrm{1}\right)} \left({x}\right)\:+{n}^{\mathrm{2}} {f}^{\left({n}\right)} \left({x}\right) \\ $$$$\left.\mathrm{3}\right)\:{prove}\:{that}\:\:{f}^{\left({n}\right)}…
Question Number 30748 by abdo imad last updated on 25/Feb/18 $${let}\:{a}>\mathrm{0}\:{and}\:{b}>\mathrm{0}\:{find}\:{lim}_{{x}\rightarrow\mathrm{0}^{+} } \:\:\left(\:\:\frac{{a}^{{x}} \:+{b}^{{x}} }{\mathrm{2}}\right)^{\frac{\mathrm{1}}{{x}}} \:\:. \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 30747 by abdo imad last updated on 25/Feb/18 $${let}\:{f}\left({x}\right)=\sqrt{\mathrm{1}+{x}^{\mathrm{2}} } \\ $$$$\left.\mathrm{1}\right){find}\:{a}\:{d}.{e}.{wich}\:{verify}\:{f}\left({x}\right) \\ $$$$\left.\mathrm{2}\right)\:{prove}\:{that}\:\forall{x}\in{R}\:,\forall{n}\in{N} \\ $$$$\left(\mathrm{1}+{x}^{\mathrm{2}} \right){f}^{\left({n}+\mathrm{2}\right)} \left({x}\right)+\left(\mathrm{2}{n}+\mathrm{1}\right){x}\:{f}^{\left({n}+\mathrm{1}\right)} \left({x}\right)\:+\left({n}^{\mathrm{2}} −\mathrm{1}\right){f}^{\left({n}\right)} \left({x}\right)=\mathrm{0} \\ $$$$\left.\mathrm{3}\right)\:{prove}\:{that}\:\:{f}^{\left(\mathrm{2}{n}+\mathrm{1}\right)}…
Question Number 96211 by mathmax by abdo last updated on 30/May/20 $$\mathrm{solve}\:\mathrm{inside}\:\mathrm{C}\:\:\left(\mathrm{x}−\frac{\mathrm{1}}{\mathrm{x}}\right)^{\mathrm{3}} \:+\left(\mathrm{x}−\frac{\mathrm{1}}{\mathrm{x}}\right)^{\mathrm{2}} \:+\left(\mathrm{x}−\frac{\mathrm{1}}{\mathrm{x}}\right)+\mathrm{1}\:=\mathrm{0} \\ $$ Answered by mr W last updated on 30/May/20 $${let}\:{t}={x}−\frac{\mathrm{1}}{{x}}…
Question Number 96200 by mathmax by abdo last updated on 30/May/20 $$\mathrm{solve}\:\:\mathrm{y}^{''} \:+\mathrm{y}^{'} \:−\mathrm{2y}\:=\mathrm{xcosx}\:\:\mathrm{with}\:\mathrm{y}^{\left(\mathrm{2}\right)} \left(\mathrm{0}\right)=\mathrm{1}\:\mathrm{and}\:\mathrm{y}^{'} \left(\mathrm{0}\right)\:=−\mathrm{2} \\ $$ Answered by mathmax by abdo last updated…