How prove that x_1=x_(2000) implies x_2 ne = x_(1999), where x_(n+2)=(x_n x_(n+1)+5x_n^4)/(x_n-x_(n+1)) ?

inurbandojoa

inurbandojoa

Answered question

2022-11-17

How prove that x 1 = x 2000 implies x 2 x 1999 , where x n + 2 = x n x n + 1 + 5 x n 4 x n x n + 1 ?

Answer & Explanation

Samsonitew7b

Samsonitew7b

Beginner2022-11-18Added 15 answers

First note that by the definition, x n x n + 1 for all n 1. Subtract x n + 1 from both sides of our recurrence relation to end up with
x n + 2 x n + 1 = 5 x n 4 + x n + 1 2 x n x n + 1 .
So, for any n 1 , ( x n + 2 x n + 1 ) ( x n + 1 x n ) < 0. Now, if x 2 > x 1 , then x 2 > x 3 and x 4 > x 3 so iterating this process leads to x 2 > x 1 = x 2000 > x 1999 . In the other case, namely x 2 < x 1 , we have x 2 < x 3 and x 4 < x 3 etc. Finally, x 1999 > x 2000 = x 1 > x 2 . In both cases, x 2 x 1999 .

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