Question Number 29846 by abdo imad last updated on 12/Feb/18 $${give}\:{the}\:{developpement}\:\:{at}\:{integr}\:{series}\:{for} \\ $$$${f}\left({x}\right)=\frac{{ln}\left(\mathrm{1}+{x}\right)−{ln}\left(\mathrm{1}−{x}\right)}{{x}} \\ $$$$\left.\mathrm{2}\right){find}\:\:\:{lim}_{{x}\rightarrow\mathrm{0}} \:{f}\left({x}\right). \\ $$ Commented by maxmathsup by imad last updated…
Question Number 29844 by abdo imad last updated on 12/Feb/18 $${let}\:{give}\:{f}_{\alpha} \left({t}\right)={cos}\left(\alpha{t}\right)\:\:\mathrm{2}\pi\:{periodic}\:{with}\:{t}\:\in\left[−\pi,\pi\right]{and} \\ $$$$\alpha\in\:{R}−{Z} \\ $$$$\left.\mathrm{1}\right)\:{developp}\:{f}_{\alpha} \:\:{at}\:{fourier}\:{serie}\:{and}\:{prove}\:{that} \\ $$$${cotan}\left(\alpha\pi\right)=\:\frac{\mathrm{1}}{\alpha\pi}\:\:+\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\frac{\mathrm{2}\alpha}{\pi\left(\alpha^{\mathrm{2}} −{n}^{\mathrm{2}} \right)} \\ $$$$\left.\mathrm{2}\left.\right)\left.{let}\:{x}\in\right]\mathrm{0},\pi\left[\:{ant}\:{g}\left({t}\right)={cotant}\:−\frac{\mathrm{1}}{{t}}\:\:{if}\:{t}\in\right]\mathrm{0},{x}\right]{andg}\left(\mathrm{0}\right)=\mathrm{0}…
Question Number 29842 by abdo imad last updated on 12/Feb/18 $$\left.{prove}\:{that}\:\forall\:{x}\in\right]\mathrm{0},\mathrm{1}\left[\:\:\:\:\:\:\frac{\mathrm{1}}{\Gamma\left({x}\right).\Gamma\left(\mathrm{1}−{x}\right)}={x}\:\prod_{{n}=\mathrm{1}} ^{\infty} \left(\mathrm{1}−\frac{{x}^{\mathrm{2}} }{{n}^{\mathrm{2}} }\right).\right. \\ $$ Commented by abdo imad last updated on 12/Feb/18…
Question Number 29841 by abdo imad last updated on 12/Feb/18 $$\left.{prove}\:{that}\:\forall\:{x}\in\right]\mathrm{0},\mathrm{1}\left[\:\:\:\:\Gamma\left({x}\right).\Gamma\left(\mathrm{1}−{x}\right)=\:\frac{\pi}{{sin}\left(\pi{x}\right)}\right. \\ $$$$\left({compliments}\:{formula}\right). \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 29838 by abdo imad last updated on 12/Feb/18 $${find}\:\prod_{{n}=\mathrm{1}} ^{\infty} \left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{4}{n}^{\mathrm{2}} }\right). \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 29839 by abdo imad last updated on 12/Feb/18 $${find}\:\prod_{{n}=\mathrm{1}} ^{\infty} \:\left(\mathrm{1}−\frac{{x}^{\mathrm{2}} }{{n}^{\mathrm{2}} \pi^{\mathrm{2}} }\right). \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 29835 by abdo imad last updated on 12/Feb/18 $${let}\:{give}\:{f}\left({x}\right)=−{x}\:+\mathrm{2}\:+\frac{\sqrt{{x}+\mathrm{1}}}{{x}} \\ $$$$\left.\mathrm{1}\right)\:{study}\:{the}\:{variation}\:{of}\:{and}\:{give}\:{the}\:{graph}\:{C}_{{f}} \\ $$$$\left.\mathrm{2}\right){give}\:{the}\:{equation}\:{of}\:{tangent}\:{at}\:{C}_{{f}} \:{in}\:{point}\:{A}\left(\mathrm{1},{f}\left(\mathrm{1}\right)\right) \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 29831 by abdo imad last updated on 12/Feb/18 $$\left.{let}\:{give}\:{f}\left({x}\right)=\:\frac{\mathrm{1}}{\mathrm{1}+{x}^{\mathrm{2}} }\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{1}\right){prove}\:{that}\:\:\:{prove}\:{that} \\ $$$${f}^{\left({n}\right)} \left({x}\right)=\frac{{p}_{{n}} \left({x}\right)}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{{n}+\mathrm{1}} }\:{with}\:{p}_{{n}} {is}\:{a}\:{polynomial} \\ $$$$\left.\mathrm{2}\right)\:{prove}\:{that}\:{p}_{{n}+\mathrm{1}} \left({x}\right)=\left(\mathrm{1}+{x}^{\mathrm{2}} \right){p}_{{n}} ^{'} \left({x}\right)\:−\mathrm{2}\left({n}+\mathrm{1}\right){p}_{{n}}…
Question Number 95218 by mathmax by abdo last updated on 24/May/20 $$\mathrm{calculate}\:\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{\mathrm{n}} }{\mathrm{n}^{\mathrm{3}} \left(\mathrm{n}+\mathrm{1}\right)^{\mathrm{3}} } \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 95216 by mathmax by abdo last updated on 24/May/20 $$\mathrm{calculste}\:\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \:\:\:\frac{\mathrm{n}}{\left(\mathrm{2n}+\mathrm{1}\right)^{\mathrm{2}} \left(\mathrm{n}+\mathrm{3}\right)} \\ $$ Answered by abdomathmax last updated on 25/May/20 $$\mathrm{let}\:\mathrm{S}_{\mathrm{n}}…