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Author: Tinku Tara

Une-fonction-P-est-dite-quasi-polynomiale-s-il-existe-pour-k-N-k-1-fonction-periodique-c-i-i-0-k-de-Z-dans-R-telles-que-P-n-k-1-n-c-i-n-n-i-1-Montrez-que-l-ensemble-des-fonction-

Question Number 219267 by zetamaths last updated on 21/Apr/25 $${Une}\:{fonction}\:{P}\:{est}\:{dite}\:{quasi}\:{polynomiale}\:{s}'{il}\:{existe}\:\left({pour}\:{k}\in\mathbb{N}\:\right)\:{k}+\mathrm{1}\:{fonction}\:{periodique}\left({c}_{{i}} \right)_{{i}\in\left[\mid\mathrm{0};{k}\mid\right]} {de}\:\mathbb{Z}\:{dans}\:\mathbb{R} \\ $$$$\:{telles}\:{que}\:{P}\left({n}\right)=\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{c}_{{i}} \left({n}\right){n}^{{i}} \\ $$$$\left(\mathrm{1}\right)\:{Montrez}\:{que}\:{l}'{ensemble}\:{des}\:{fonction}\:{quasi}\:{polynomiale}\:{forme}\:{un}\:\mathbb{R}−{ev}\left({real}\:{space}\:{vector}\right). \\ $$$$\left(\mathrm{2}\right){Montrez}\:{que}\:{si}\:{P},{Q}:\mathbb{Z}\rightarrow\mathbb{R}\:{sont}\:{desfonction}\:{quasi}\:{polynomiale}\:{tel}\:{que}\:{P}\left({n}\right)={Q}\left({n}\right)\:\forall{n}\in\mathbb{N}\:{alors}\:{P}={Q} \\ $$ Answered by…

Question-219135

Question Number 219135 by Rojarani last updated on 20/Apr/25 Answered by Frix last updated on 20/Apr/25 $${a}^{\frac{\mathrm{1}}{\mathrm{3}}} +{b}^{\frac{\mathrm{1}}{\mathrm{3}}} ={c}^{\frac{\mathrm{1}}{\mathrm{3}}} \\ $$$${a}+{b}+\mathrm{3}{a}^{\frac{\mathrm{1}}{\mathrm{3}}} {b}^{\frac{\mathrm{1}}{\mathrm{3}}} \underset{={c}^{\frac{\mathrm{1}}{\mathrm{3}}} } {\underbrace{\left({a}^{\frac{\mathrm{1}}{\mathrm{3}}}…

x-1-x-2-1-x-3-dx-

Question Number 219193 by fantastic last updated on 20/Apr/25 $$\int\sqrt{\frac{{x}+\mathrm{1}}{{x}+\mathrm{2}}}\:.\frac{\mathrm{1}}{{x}+\mathrm{3}}\:{dx}=? \\ $$ Answered by aleks041103 last updated on 20/Apr/25 $$\frac{{x}+\mathrm{1}}{{x}+\mathrm{2}}={u}=\mathrm{1}−\frac{\mathrm{1}}{{x}+\mathrm{2}} \\ $$$$\Rightarrow{x}=\frac{\mathrm{1}}{\mathrm{1}−{u}}−\mathrm{2}=\frac{\mathrm{2}{u}−\mathrm{1}}{\mathrm{1}−{u}} \\ $$$$\Rightarrow{dx}=\frac{\mathrm{2}\left(\mathrm{1}−{u}\right)−\left(−\mathrm{1}\right)\left(\mathrm{2}{u}−\mathrm{1}\right)}{\left(\mathrm{1}−{u}\right)^{\mathrm{2}} }{du}=…

Question-219185

Question Number 219185 by fantastic last updated on 20/Apr/25 Answered by MrGaster last updated on 20/Apr/25 $$\bigtriangledown\centerdot\left(\overset{\rightarrow} {{F}}×\overset{\rightarrow} {{G}}\right)=\partial_{{i}} \left(\epsilon_{{ijk}} {F}_{{j}} {G}_{{k}} \right) \\ $$$$=\epsilon_{{ijk}}…

Question-219181

Question Number 219181 by hardmath last updated on 20/Apr/25 Answered by devdutt last updated on 20/Apr/25 $$\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{{k}^{\mathrm{2}} }{\mathrm{2}{k}^{\mathrm{2}} −\mathrm{2}{nk}+{n}^{\mathrm{2}} }\:=\:\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}\frac{{k}^{\mathrm{2}} }{{k}^{\mathrm{2}}…

Question-219232

Question Number 219232 by Spillover last updated on 20/Apr/25 Answered by A5T last updated on 21/Apr/25 Commented by A5T last updated on 21/Apr/25 $$\mathrm{AD}=\mathrm{bsin}\theta\:;\:\mathrm{AE}=\mathrm{bcos}\theta\:;\:\mathrm{AC}=\mathrm{asin}\theta\:;\:\mathrm{AB}=\mathrm{acos}\theta \\…