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Question-72671




Question Number 72671 by aliesam last updated on 31/Oct/19
Commented by kaivan.ahmadi last updated on 31/Oct/19
a)  ∣x_(n+1) −x_n ∣≤c∣x_n −x_(n−1) ∣≤c^2 ∣x_(n−1) −x_(n−2) ∣≤...≤c^n ∣x_1 −x_0 ∣  b)  ∣x_(n+1) −x_0 ∣=∣(x_(n+1) −x_n )+(x_n −x_(n−1) )+...+(x_1 −x_0 )∣≤  ∣x_(n+1) −x_n ∣+∣x_n −x_(n−1) ∣+...+∣x_1 −x_0 ∣≤  c^n ∣x_1 −x_0 ∣+c^(n−1) ∣x_1 −x_0 ∣+...+∣x_1 −x_0 ∣=  (c^n +c^(n−1) +...+1)∣x_1 −x_0 ∣  c)  x_(n+1) −x_k =(x_(n+1) −x_n )+(x_n −x_(n−1) )+...+(x_(k+1) −x_k )⇒  c  is clear.
$$\left.{a}\right) \\ $$$$\mid{x}_{{n}+\mathrm{1}} −{x}_{{n}} \mid\leqslant{c}\mid{x}_{{n}} −{x}_{{n}−\mathrm{1}} \mid\leqslant{c}^{\mathrm{2}} \mid{x}_{{n}−\mathrm{1}} −{x}_{{n}−\mathrm{2}} \mid\leqslant…\leqslant{c}^{{n}} \mid{x}_{\mathrm{1}} −{x}_{\mathrm{0}} \mid \\ $$$$\left.{b}\right) \\ $$$$\mid{x}_{{n}+\mathrm{1}} −{x}_{\mathrm{0}} \mid=\mid\left({x}_{{n}+\mathrm{1}} −{x}_{{n}} \right)+\left({x}_{{n}} −{x}_{{n}−\mathrm{1}} \right)+…+\left({x}_{\mathrm{1}} −{x}_{\mathrm{0}} \right)\mid\leqslant \\ $$$$\mid{x}_{{n}+\mathrm{1}} −{x}_{{n}} \mid+\mid{x}_{{n}} −{x}_{{n}−\mathrm{1}} \mid+…+\mid{x}_{\mathrm{1}} −{x}_{\mathrm{0}} \mid\leqslant \\ $$$${c}^{{n}} \mid{x}_{\mathrm{1}} −{x}_{\mathrm{0}} \mid+{c}^{{n}−\mathrm{1}} \mid{x}_{\mathrm{1}} −{x}_{\mathrm{0}} \mid+…+\mid{x}_{\mathrm{1}} −{x}_{\mathrm{0}} \mid= \\ $$$$\left({c}^{{n}} +{c}^{{n}−\mathrm{1}} +…+\mathrm{1}\right)\mid{x}_{\mathrm{1}} −{x}_{\mathrm{0}} \mid \\ $$$$\left.{c}\right) \\ $$$${x}_{{n}+\mathrm{1}} −{x}_{{k}} =\left({x}_{{n}+\mathrm{1}} −{x}_{{n}} \right)+\left({x}_{{n}} −{x}_{{n}−\mathrm{1}} \right)+…+\left({x}_{{k}+\mathrm{1}} −{x}_{{k}} \right)\Rightarrow \\ $$$${c}\:\:{is}\:{clear}. \\ $$

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