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Category: Algebra

given-the-recursive-a-n-define-by-setting-a-1-0-1-a-n-1-a-n-1-a-n-n-1-prove-that-1-lim-n-na-n-1-2-b-n-n-1-na-n-is-a-incresing-sequence-and

Question Number 219414 by universe last updated on 24/Apr/25 $$\:\:\:\mathrm{given}\:\mathrm{the}\:\mathrm{recursive}\:\left\{\mathrm{a}_{\mathrm{n}} \right\}\:\mathrm{define}\:\mathrm{by}\:\mathrm{setting} \\ $$$$\:\:\mathrm{a}_{\mathrm{1}\:} \:\in\:\left(\mathrm{0},\mathrm{1}\right)\:\:\:,\:\:\:\:\mathrm{a}_{\mathrm{n}+\mathrm{1}} \:=\:\mathrm{a}_{\mathrm{n}} \left(\mathrm{1}−\mathrm{a}_{\mathrm{n}} \right)\:\:\:,\:\mathrm{n}\geqslant\mathrm{1} \\ $$$$\:\:\mathrm{prove}\:\mathrm{that}\:\:\left(\mathrm{1}\right)\:\:\:\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}}\:\mathrm{na}_{\mathrm{n}} =\:\mathrm{1} \\ $$$$\:\:\left(\mathrm{2}\right)\:\:\mathrm{b}_{\mathrm{n}} \:=\:\mathrm{n}\left(\mathrm{1}−\mathrm{na}_{\mathrm{n}} \right)\:\mathrm{is}\:\mathrm{a}\:\mathrm{incresing}\:\mathrm{sequence}…

given-g-x-x-2023-x-1-find-gogogogogog-2024-

Question Number 219404 by golsendro last updated on 24/Apr/25 $$\:\:\mathrm{given}\:\mathrm{g}\left(\mathrm{x}\right)=\:\frac{\mathrm{x}−\mathrm{2023}}{\mathrm{x}−\mathrm{1}} \\ $$$$\:\:\mathrm{find}\:\left(\mathrm{gogogogogog}\right)\left(\mathrm{2024}\right) \\ $$ Commented by kapoorshah last updated on 25/Apr/25 $${g}^{−\mathrm{1}} \left({x}\right)=\frac{{x}−\mathrm{2023}}{{x}−\mathrm{1}} \\ $$$$\:\:\:\:\:\left({g}\:{o}\:{g}\:{o}\:{g}\:{o}\:{g}\:{o}\:{g}\:{o}\:{g}\right)\left(\mathrm{2024}\right)…

Question-219281

Question Number 219281 by Rojarani last updated on 22/Apr/25 Answered by Ghisom last updated on 22/Apr/25 $${P}\left({x}\right)={x}^{\mathrm{3}} −\mathrm{3}{x}+\mathrm{1} \\ $$$${P}\:'\left({x}\right)=\mathrm{3}{x}^{\mathrm{2}} −\mathrm{3} \\ $$$${P}\:''\left({x}\right)=\mathrm{6}{x} \\ $$$$…

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…