Question-127865

Question Number 127865 by Ar Brandon last updated on 02/Jan/21

Answered by mathmax by abdo last updated on 03/Jan/21

a)wehave u_(n+1 ) =f(u_n ) with f(x)=x−x^2 f is contnue the fix point verify x=x−x^2 ⇒x=0 ⇒lim_(n→+∞) u_n =0 b)we have u_n ^2 =u_n −u_(n+1) ⇒Σ_(k=0) ^n u_k ^2 =Σ_(k=0) ^n (u_k −u_(k+1) ) =u_0 −u_1 +u_1 −u_2 +....+u_n −u_(n+1) =u_0 −u_(n+1) ⇒ lim_(n→+∞) Σ_(k=0) ^n u_k ^2 =lim_(n→+∞) (u_o −u_(n+1) ) =u_0

$$\left.\mathrm{a}\right)\mathrm{wehave}\:\mathrm{u}_{\mathrm{n}+\mathrm{1}\:} =\mathrm{f}\left(\mathrm{u}_{\mathrm{n}} \right)\:\mathrm{with}\:\mathrm{f}\left(\mathrm{x}\right)=\mathrm{x}−\mathrm{x}^{\mathrm{2}} \:\:\mathrm{f}\:\mathrm{is}\:\mathrm{contnue}\: \\ $$$$\mathrm{the}\:\mathrm{fix}\:\mathrm{point}\:\:\mathrm{verify}\:\mathrm{x}=\mathrm{x}−\mathrm{x}^{\mathrm{2}} \:\Rightarrow\mathrm{x}=\mathrm{0}\:\Rightarrow\mathrm{lim}_{\mathrm{n}\rightarrow+\infty} \mathrm{u}_{\mathrm{n}} =\mathrm{0} \\ $$$$\left.\mathrm{b}\right)\mathrm{we}\:\mathrm{have}\:\mathrm{u}_{\mathrm{n}} ^{\mathrm{2}} \:=\mathrm{u}_{\mathrm{n}} −\mathrm{u}_{\mathrm{n}+\mathrm{1}} \:\Rightarrow\sum_{\mathrm{k}=\mathrm{0}} ^{\mathrm{n}} \:\mathrm{u}_{\mathrm{k}} ^{\mathrm{2}} \:=\sum_{\mathrm{k}=\mathrm{0}} ^{\mathrm{n}} \left(\mathrm{u}_{\mathrm{k}} −\mathrm{u}_{\mathrm{k}+\mathrm{1}} \right) \\ $$$$=\mathrm{u}_{\mathrm{0}} −\mathrm{u}_{\mathrm{1}} +\mathrm{u}_{\mathrm{1}} −\mathrm{u}_{\mathrm{2}} +….+\mathrm{u}_{\mathrm{n}} −\mathrm{u}_{\mathrm{n}+\mathrm{1}} =\mathrm{u}_{\mathrm{0}} −\mathrm{u}_{\mathrm{n}+\mathrm{1}} \:\Rightarrow \\ $$$$\mathrm{lim}_{\mathrm{n}\rightarrow+\infty} \sum_{\mathrm{k}=\mathrm{0}} ^{\mathrm{n}} \:\mathrm{u}_{\mathrm{k}} ^{\mathrm{2}} \:=\mathrm{lim}_{\mathrm{n}\rightarrow+\infty} \left(\mathrm{u}_{\mathrm{o}} −\mathrm{u}_{\mathrm{n}+\mathrm{1}} \right)\:=\mathrm{u}_{\mathrm{0}} \\ $$

Commented by Ar Brandon last updated on 03/Jan/21

Merci monsieur😃

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