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Question Number 219267 by zetamaths last updated on 21/Apr/25
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)=Σ_(k=1) ^n c_i (n)n^i (1) Montrez que l′ensemble des fonction quasi polynomiale forme un R−ev(real space vector). (2)Montrez que si P,Q:Z→R sont desfonction quasi polynomiale tel que P(n)=Q(n) ∀n∈N alors P=Q
$${Une}\:{fonction}\:{P}\:{est}\:{dite}\:{quasi}\:{polynomiale}\:{s}'{il}\:{existe}\:\left({pour}\:{k}\in\mathbb{N}\:\right)\:{k}+\mathrm{1}\:{fonction}\:{periodique}\left({c}_{{i}} \right)_{{i}\in\left[\mid\mathrm{0};{k}\mid\right]} {de}\:\mathbb{Z}\:{dans}\:\mathbb{R} \\ $$$$\:{telles}\:{que}\:{P}\left({n}\right)=\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{c}_{{i}} \left({n}\right){n}^{{i}} \\ $$$$\left(\mathrm{1}\right)\:{Montrez}\:{que}\:{l}'{ensemble}\:{des}\:{fonction}\:{quasi}\:{polynomiale}\:{forme}\:{un}\:\mathbb{R}−{ev}\left({real}\:{space}\:{vector}\right). \\ $$$$\left(\mathrm{2}\right){Montrez}\:{que}\:{si}\:{P},{Q}:\mathbb{Z}\rightarrow\mathbb{R}\:{sont}\:{desfonction}\:{quasi}\:{polynomiale}\:{tel}\:{que}\:{P}\left({n}\right)={Q}\left({n}\right)\:\forall{n}\in\mathbb{N}\:{alors}\:{P}={Q} \\ $$
Answered by MrGaster last updated on 22/Apr/25
(1): ∀P,Q∈V,∃k,m∈N,P(n)=Σ_(i=0) ^k c_i (n)n^i ,Q(n)=Σ_(j=0) ^m d_j (n)n^j let t=max{k,m}⇒P(n)=Σ_(i=0) ^t c′_i (n)n^i Q(n)=Σ_(i=0) ^t d_i ^′ (n)n^i (zero fill) P+Q(n)=Σ_(i=0) ^t (c′_i (n)+d_i ^′ (n))n^i ∈V αP(n)=Σ_(i=0) ^t (αc_i ^′ (n))n^i ∈V 0(n)=Σ_(i=0) ^t 0(n)n^i ∈V ∴Vis a R-vector sqace (2): Let P(n)=Σ_(i=0) ^k c_i (n)n^i ,Q(n)=Σ_(i=0) ^k d_i (n)n^i ,P(n)=Q(n) ∀n∈N Assume there exists a maximum i such that c_i (n)≠d_i (n) As n→∞,P(n)−Q(n)∼(c_k (n)−d_k (n))n^k ⇒c_(k ) (n)=d_k (n) ∀n∈N By the property of periodic functions,c_k =d_k Recursively apply to i=k,…,0,⇒c_i =d_i ∴P=Q
$$\left(\mathrm{1}\right): \\ $$$$\forall{P},{Q}\in\mathcal{V},\exists{k},{m}\in\mathbb{N},{P}\left({n}\right)=\underset{{i}=\mathrm{0}} {\overset{{k}} {\sum}}{c}_{{i}} \left({n}\right){n}^{{i}} ,{Q}\left({n}\right)=\underset{{j}=\mathrm{0}} {\overset{{m}} {\sum}}{d}_{{j}} \left({n}\right){n}^{{j}} \\ $$$$\mathrm{let}\:{t}=\mathrm{max}\left\{{k},{m}\right\}\Rightarrow{P}\left({n}\right)=\underset{{i}=\mathrm{0}} {\overset{{t}} {\sum}}{c}'_{{i}} \left({n}\right){n}^{{i}} {Q}\left({n}\right)=\underset{{i}=\mathrm{0}} {\overset{{t}} {\sum}}{d}_{{i}} ^{'} \left({n}\right){n}^{{i}} \left(\mathrm{zero}\:\mathrm{fill}\right) \\ $$$${P}+{Q}\left({n}\right)=\underset{{i}=\mathrm{0}} {\overset{{t}} {\sum}}\left({c}'_{{i}} \left({n}\right)+{d}_{{i}} ^{'} \left({n}\right)\right){n}^{{i}} \in\mathcal{V} \\ $$$$\alpha{P}\left({n}\right)=\underset{{i}=\mathrm{0}} {\overset{{t}} {\sum}}\left(\alpha{c}_{{i}} ^{'} \left({n}\right)\right){n}^{{i}} \in\mathcal{V} \\ $$$$\mathrm{0}\left({n}\right)=\underset{{i}=\mathrm{0}} {\overset{{t}} {\sum}}\mathrm{0}\left({n}\right){n}^{{i}} \in\mathcal{V} \\ $$$$\therefore\mathcal{V}\mathrm{is}\:\mathrm{a}\:\mathbb{R}-\mathrm{vector}\:\mathrm{sqace} \\ $$$$\left(\mathrm{2}\right): \\ $$$$\mathrm{Let}\:{P}\left({n}\right)=\underset{{i}=\mathrm{0}} {\overset{{k}} {\sum}}{c}_{{i}} \left({n}\right){n}^{{i}} ,{Q}\left({n}\right)=\underset{{i}=\mathrm{0}} {\overset{{k}} {\sum}}{d}_{{i}} \left({n}\right){n}^{{i}} ,{P}\left({n}\right)={Q}\left({n}\right)\:\forall{n}\in\mathbb{N} \\ $$$$\mathrm{Assume}\:\mathrm{there}\:\mathrm{exists}\:\mathrm{a}\:\mathrm{maximum}\:{i}\:\mathrm{such}\:\mathrm{that}\:{c}_{{i}} \left({n}\right)\neq{d}_{{i}} \left({n}\right) \\ $$$${As}\:{n}\rightarrow\infty,{P}\left({n}\right)−{Q}\left({n}\right)\sim\left({c}_{{k}} \left({n}\right)−{d}_{{k}} \left({n}\right)\right){n}^{{k}} \\ $$$$\Rightarrow{c}_{{k}\:} \left({n}\right)={d}_{{k}} \left({n}\right)\:\forall{n}\in\mathbb{N} \\ $$$$\mathrm{By}\:\mathrm{the}\:\mathrm{property}\:\mathrm{of}\:\mathrm{periodic}\:\mathrm{functions},{c}_{{k}} ={d}_{{k}} \\ $$$$\mathrm{Recursively}\:\mathrm{apply}\:\mathrm{to}\:{i}={k},\ldots,\mathrm{0},\Rightarrow{c}_{{i}} ={d}_{{i}} \\ $$$$\therefore{P}={Q} \\ $$

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