Question Number 196914 by SANOGO last updated on 03/Sep/23 Answered by witcher3 last updated on 03/Sep/23 $$−\frac{\mathrm{1}}{\mathrm{n}+\mathrm{1}}>−\frac{\mathrm{1}}{\mathrm{n}} \\ $$$$\mathrm{I}_{\mathrm{n}} =\left[−\frac{\mathrm{1}}{\mathrm{n}},\mathrm{1}\right] \\ $$$$\mathrm{I}_{\mathrm{n}+\mathrm{1}} \:\:\subseteq\mathrm{I}_{\mathrm{n}} \:\:\mathrm{suite}\:\mathrm{decroissante}\:\mathrm{minore} \\…
Question Number 196915 by sonukgindia last updated on 03/Sep/23 Commented by mr W last updated on 23/Sep/23 $${see}\:{Q}\mathrm{197578} \\ $$ Terms of Service Privacy Policy…
Question Number 196913 by ERLY last updated on 02/Sep/23 $${soit}\:\left\{_{{r}_{{n}+\mathrm{1}} ={r}_{{n}} /\left(\mathrm{2}+{r}_{{n}} ^{\mathrm{2}} \right)} ^{{r}_{\mathrm{0}} =\mathrm{1}} \right. \\ $$$${demontrer}\:{sans}\:{recurrence}\:{que}\:{r}_{{n}} >\mathrm{0} \\ $$$${demontrer}\:{par}\:{recurrence}\:{que}\:{r}_{{n}+\mathrm{1}} \leq\frac{\mathrm{1}}{\mathrm{2}}{r}_{{n}} \\ $$$${demontrer}\:{sans}\:{recurrence}\:{que}\:{r}_{{n}}…
Question Number 196910 by ERLY last updated on 02/Sep/23 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 196848 by sonukgindia last updated on 01/Sep/23 Commented by Frix last updated on 01/Sep/23 $$\mathrm{The}\:\mathrm{answer}\:\mathrm{is}\:\mathrm{42}\:\left(\mathrm{or}\:\mathrm{else}\:\mathrm{prove}\:\mathrm{it}'\mathrm{s}\:\boldsymbol{{not}}\:\mathrm{42}\right) \\ $$ Answered by AST last updated on…
Question Number 196852 by SANOGO last updated on 01/Sep/23 Answered by Mathspace last updated on 02/Sep/23 $$\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{t}^{{a}−\mathrm{1}} }{\mathrm{1}+{t}^{{b}} }{dt}\:=\int_{\mathrm{0}} ^{\mathrm{1}} {t}^{{a}−\mathrm{1}} \sum_{{n}=\mathrm{0}} ^{\infty}…
Question Number 196850 by sonukgindia last updated on 01/Sep/23 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 196846 by mokys last updated on 01/Sep/23 $$\mathrm{2}{xy}''\:+\:\left(\mathrm{1}−\mathrm{4}{x}\right){y}'\:+\:\left(\mathrm{2}{x}−\mathrm{1}\right){y}\:=\:{y} \\ $$ Answered by witcher3 last updated on 04/Sep/23 $$\mathrm{2xy}''+\left(\mathrm{1}−\mathrm{4x}\right)\mathrm{y}'+\left(\mathrm{2x}−\mathrm{1}\right)\mathrm{y}=\mathrm{0} \\ $$$$\mathrm{y}\left(\mathrm{x}\right)=\mathrm{e}^{\mathrm{x}} ..\mathrm{solution} \\ $$$$\mathrm{y}=\mathrm{ze}^{\mathrm{x}}…
Question Number 196843 by sonukgindia last updated on 01/Sep/23 Answered by qaz last updated on 02/Sep/23 $${xyy}''={yy}'+{x}−{x}\left({y}'\right)^{\mathrm{2}} \\ $$$$\Rightarrow{x}\left({yy}'\right)'={yy}'+{x} \\ $$$${yy}'={e}^{\int\frac{{dx}}{{x}}} \left({C}_{\mathrm{1}} +\int{e}^{−\int\frac{{dx}}{{x}}} {dx}\right)={C}_{\mathrm{1}} {x}+{xlnx}…
Question Number 196828 by ERLY last updated on 01/Sep/23 Answered by Skabetix last updated on 01/Sep/23 $$\left.\mathrm{2}.{a}\right)\:{U}_{{n}+\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{1}}{{U}_{{n}} } \\ $$$${Comme}\:{U}_{{n}} >\mathrm{0}\rightarrow{U}_{{n}} \geqslant\frac{\mathrm{1}}{{U}_{{n}} }\rightarrow{U}_{{n}+\mathrm{1}} \leqslant\frac{\mathrm{1}}{\mathrm{2}}{U}_{{n}}…