Question Number 33302 by abdo imad last updated on 14/Apr/18 $${developp}\:{at}\:{integr}\:{serie}\:{f}\left({x}\right)\:=\:\frac{\mathrm{1}}{\mathrm{1}+{x}\:−\mathrm{2}{x}^{\mathrm{3}} } \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 33300 by abdo imad last updated on 14/Apr/18 $${find}\:{the}\:{value}\:{of}\:\:\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\:\frac{\left(−\mathrm{1}\right)^{{n}} }{\mathrm{4}{n}\:+\mathrm{3}}\:. \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 33299 by abdo imad last updated on 14/Apr/18 $${find}\:{the}\:{sum}\:{of}\:\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{{n}^{\mathrm{2}} \:+\mathrm{1}}{{n}+\mathrm{1}}\:{x}^{{n}\:} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 33295 by abdo imad last updated on 14/Apr/18 $${let}\:\:{f}\left({x}\right)=\:\left({x}+\mathrm{1}^{} \right)^{\mathrm{2}{n}} \:\:{e}^{−{nx}} \:\:{with}\:{n}\:{integr} \\ $$$$\left.\mathrm{1}\right)\:\:{calculste}\:\:{f}^{\left({p}\right)} \left({x}\right)\:\:{and}\:{f}^{\left({n}\right)} \left({x}\right) \\ $$$$\left.\mathrm{2}\right)\:{find}\:{f}^{\left({p}\right)} \left(\mathrm{0}\right) \\ $$$$\left.\mathrm{3}\right)\:{developp}\:{f}\left({x}\right)\:{at}\:{integr}\:{serie}. \\ $$…
Question Number 33294 by abdo imad last updated on 14/Apr/18 $${developp}\:{f}\left({x}\right)=\:{xln}\left(\:\mathrm{1}+{e}^{−{x}} \right)\:\:{at}\:{inter}\:{srie} \\ $$$$\left.\mathrm{2}\right)\:{find}\:\:\int_{\mathrm{0}} ^{\mathrm{1}} \:{f}\left({x}\right){dx}\:. \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 33293 by abdo imad last updated on 14/Apr/18 $${developp}\:{f}\left({x}\right)\:=\:\frac{{e}^{{x}} }{{x}−\mathrm{1}}\:{at}\:{integr}\:{serie} \\ $$ Commented by abdo imad last updated on 19/Apr/18 $$−{f}\left({x}\right)\:=\:\frac{{e}^{{x}} }{\mathrm{1}−{x}}\:\:{so}\:{if}\:\mid{x}\mid<\mathrm{1}\:\:{we}\:{have}\: \\…
Question Number 33292 by abdo imad last updated on 14/Apr/18 $${developp}\:{f}\left({x}\right)={arctan}\left({x}+\mathrm{1}\right)\:{at}\:{integr}\:{serie} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 33287 by abdo imad last updated on 14/Apr/18 $${study}\:{the}\:{sequence}\:\:{u}_{{n}+\mathrm{1}} =\:\sqrt{\frac{\mathrm{1}\:+{u}_{{n}} }{\mathrm{2}}} \\ $$$${with}\:\:\:\:\mathrm{0}<{u}_{\mathrm{0}} <\mathrm{1}\:. \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 33285 by abdo imad last updated on 14/Apr/18 $${study}\:{the}\:{sequence}\:\:{u}_{{n}+\mathrm{1}} \:\:=\sqrt{{u}_{{n}} \:\:\:+\frac{\mathrm{1}}{{n}+\mathrm{1}}} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 33284 by abdo imad last updated on 14/Apr/18 $${study}\:{the}\:{sequence}\:\:{u}_{{n}+\mathrm{1}} \:\:=\:\frac{{u}_{{n}} \:−{ln}\left(\mathrm{1}+{u}_{{n}} \right)}{{u}_{{n}} ^{\mathrm{2}} } \\ $$$${with}\:{u}_{\mathrm{0}} >\mathrm{0}\:. \\ $$ Terms of Service Privacy…