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

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

$$\mathrm{a})\mathrm{wehave}{\mathrm{u}}_{\mathrm{n}+1}=\mathrm{f}\left({\mathrm{u}}_{\mathrm{n}}\right)\mathrm{with}\mathrm{f}\left(\mathrm{x}\right)=\mathrm{x}-{\mathrm{x}}^2\mathrm{f}\mathrm{is}\mathrm{contnue}$$$$\mathrm{the}\mathrm{fix}\mathrm{point}\mathrm{verify}\mathrm{x}=\mathrm{x}-{\mathrm{x}}^2\Rightarrow \mathrm{x}=0\Rightarrow {\lim }_{\mathrm{n}\to +\mathrm{\infty }}{\mathrm{u}}_{\mathrm{n}}=0$$$$\mathrm{b})\mathrm{we}\mathrm{have}{\mathrm{u}}_{\mathrm{n}}^2={\mathrm{u}}_{\mathrm{n}}-{\mathrm{u}}_{\mathrm{n}+1}\Rightarrow \sum _{\mathrm{k}=0}^{\mathrm{n}}{\mathrm{u}}_{\mathrm{k}}^2=\sum _{\mathrm{k}=0}^{\mathrm{n}}({\mathrm{u}}_{\mathrm{k}}-{\mathrm{u}}_{\mathrm{k}+1})$$$$={\mathrm{u}}_0-{\mathrm{u}}_1+{\mathrm{u}}_1-{\mathrm{u}}_2+....+{\mathrm{u}}_{\mathrm{n}}-{\mathrm{u}}_{\mathrm{n}+1}={\mathrm{u}}_0-{\mathrm{u}}_{\mathrm{n}+1}\Rightarrow$$$${\lim }_{\mathrm{n}\to +\mathrm{\infty }}\sum _{\mathrm{k}=0}^{\mathrm{n}}{\mathrm{u}}_{\mathrm{k}}^2={\lim }_{\mathrm{n}\to +\mathrm{\infty }}({\mathrm{u}}_{\mathrm{o}}-{\mathrm{u}}_{\mathrm{n}+1})={\mathrm{u}}_0$$

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

Merci monsieur😃

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