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Question Number 167181 by mathocean1 last updated on 08/Mar/22

Given x, c ∈ R.  (u_n )_(n∈N ) : { ((u_0 =0)),((u_(n+1) =xsin(u_n )+c)) :}  1.Show that ∣u_(n+1) −u_n ∣≤∣c∣∣x^n ∣.  2.Show that : (∣x∣<1 et m≥n ⇒∣u_m −u_n ∣≤((∣c∣∣x^n ∣)/(1−∣x∣))  3.Deduct that u_n  is convergent and  calculate its limit.

$${Given}\:{x},\:{c}\:\in\:\mathbb{R}. \\ $$$$\left({u}_{{n}} \right)_{{n}\in\mathbb{N}\:} :\begin{cases}{{u}_{\mathrm{0}} =\mathrm{0}}\\{{u}_{{n}+\mathrm{1}} ={xsin}\left({u}_{{n}} \right)+{c}}\end{cases} \\ $$$$\mathrm{1}.{Show}\:{that}\:\mid{u}_{{n}+\mathrm{1}} −{u}_{{n}} \mid\leqslant\mid{c}\mid\mid{x}^{{n}} \mid. \\ $$$$\mathrm{2}.{Show}\:{that}\::\:\left(\mid{x}\mid<\mathrm{1}\:{et}\:{m}\geqslant{n}\:\Rightarrow\mid{u}_{{m}} −{u}_{{n}} \mid\leqslant\frac{\mid{c}\mid\mid{x}^{{n}} \mid}{\mathrm{1}−\mid{x}\mid}\right. \\ $$$$\mathrm{3}.{Deduct}\:{that}\:{u}_{{n}} \:{is}\:{convergent}\:{and} \\ $$$${calculate}\:{its}\:{limit}. \\ $$

Answered by mindispower last updated on 09/Mar/22

n=0  ∣U_1 −U_0 ∣=∣c∣ True  suppose ∀n ∣U_(n+1) −U_n ∣<∣c∣∣x^n ∣  we show That ∣u_(n+2) −u_(n+1) ∣<∣c∣∣x∣^(n+1)   ∀n∈N^∗   ∣U_(n+2) −U_(n+1) ∣=∣xsin(u_(n+1) )−xsin(u_n )∣  =∣x∣.∣sin(U_(n+1) )−sin(U_n )∣....(E)  ∀(a,b)∈R^2  We have  ∣sin(a)−sin(b)∣<∣a−b∣  (E)≤∣x∣.∣U_(n+1) −U_n ∣≤∣x∣.∣c∣.∣x∣^n =∣c∣∣x^(n+1)  true  (2) use m=n+k  and ∣U_m −U_n ∣=∣Σ_(k=n) ^(m−1) U_(k+1) −U_k ∣<Σ∣U_(k+1) −U_k ∣ and Quatiom 1

$${n}=\mathrm{0} \\ $$$$\mid{U}_{\mathrm{1}} −{U}_{\mathrm{0}} \mid=\mid{c}\mid\:{True} \\ $$$${suppose}\:\forall{n}\:\mid{U}_{{n}+\mathrm{1}} −{U}_{{n}} \mid<\mid{c}\mid\mid{x}^{{n}} \mid \\ $$$${we}\:{show}\:{That}\:\mid{u}_{{n}+\mathrm{2}} −{u}_{{n}+\mathrm{1}} \mid<\mid{c}\mid\mid{x}\mid^{{n}+\mathrm{1}} \\ $$$$\forall{n}\in\mathbb{N}^{\ast} \:\:\mid{U}_{{n}+\mathrm{2}} −{U}_{{n}+\mathrm{1}} \mid=\mid{xsin}\left({u}_{{n}+\mathrm{1}} \right)−{xsin}\left({u}_{{n}} \right)\mid \\ $$$$=\mid{x}\mid.\mid{sin}\left({U}_{{n}+\mathrm{1}} \right)−{sin}\left({U}_{{n}} \right)\mid....\left({E}\right) \\ $$$$\forall\left({a},{b}\right)\in\mathbb{R}^{\mathrm{2}} \:{We}\:{have} \\ $$$$\mid{sin}\left({a}\right)−{sin}\left({b}\right)\mid<\mid{a}−{b}\mid \\ $$$$\left({E}\right)\leqslant\mid{x}\mid.\mid{U}_{{n}+\mathrm{1}} −{U}_{{n}} \mid\leqslant\mid{x}\mid.\mid{c}\mid.\mid{x}\mid^{{n}} =\mid{c}\mid\mid{x}^{{n}+\mathrm{1}} \:{true} \\ $$$$\left(\mathrm{2}\right)\:{use}\:{m}={n}+{k} \\ $$$${and}\:\mid{U}_{{m}} −{U}_{{n}} \mid=\mid\underset{{k}={n}} {\overset{{m}−\mathrm{1}} {\sum}}{U}_{{k}+\mathrm{1}} −{U}_{{k}} \mid<\Sigma\mid{U}_{{k}+\mathrm{1}} −{U}_{{k}} \mid\:{and}\:{Quatiom}\:\mathrm{1} \\ $$$$ \\ $$$$ \\ $$

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