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Question-59089




Question Number 59089 by Tawa1 last updated on 04/May/19
Answered by arcana last updated on 08/Jun/19
let m,n≥1,m>n  x_m −x_n =x_m −x_(m−1) +x_(m+1) −x_(m−2) +...+x_(n+1) −x_n   ∣x_m −x_n ∣≤∣x_m −x_(m−1) ∣+∣x_(m−1) −x_(m−2) ∣+...+∣x_(n+1) −x_n ∣  ≤cσ^(m−1) +cσ^(m−2) +...+cσ^(n+1) +cσ^n   ≤c+c+...+c≤c(m−1−n)
$${let}\:{m},{n}\geqslant\mathrm{1},{m}>{n} \\ $$$${x}_{{m}} −{x}_{{n}} ={x}_{{m}} −{x}_{{m}−\mathrm{1}} +{x}_{{m}+\mathrm{1}} −{x}_{{m}−\mathrm{2}} +…+{x}_{{n}+\mathrm{1}} −{x}_{{n}} \\ $$$$\mid{x}_{{m}} −{x}_{{n}} \mid\leqslant\mid{x}_{{m}} −{x}_{{m}−\mathrm{1}} \mid+\mid{x}_{{m}−\mathrm{1}} −{x}_{{m}−\mathrm{2}} \mid+…+\mid{x}_{{n}+\mathrm{1}} −{x}_{{n}} \mid \\ $$$$\leqslant{c}\sigma^{{m}−\mathrm{1}} +{c}\sigma^{{m}−\mathrm{2}} +…+{c}\sigma^{{n}+\mathrm{1}} +{c}\sigma^{{n}} \\ $$$$\leqslant{c}+{c}+…+{c}\leqslant\mathrm{c}\left({m}−\mathrm{1}−{n}\right) \\ $$

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