Get ready Part 16

Algebra Level 3

Let ( x n ) n ≥ 0 (x_n)_{n \ge 0} be a sequence of nonzero real numbers such that x n 2 − x n − 1 x n + 1 = 1 x^2_n-x_{n-1}x_{n+1}=1 for n = 1 , 2 , 3 , … . n=1,2,3,\dots. .Is there exists a real number a such that x n + 1 = a x n − x n − 1 x_{n+1} = ax_n - x_{n-1} for all n ≥ 1 n \ge 1 ?

No Yes

Gia Hoàng Phạm by Gia Hoàng Phạm , 19, Vietnam

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

2 solutions

Mark Hennings
Nov 20, 2018

Define a = x 2 + x 0 x 1 a = \frac{x_2+x_0}{x_1} , so that x 2 = a x 1 − x 0 x_2 = ax_1 - x_0 .

Suppose now that n ≥ 2 n \ge 2 and that x n = a x n − 1 − x n − 2 x_n = ax_{n-1} - x_{n-2} . Then x n + 1 = x n 2 − 1 x n − 1 = x n ( a x n − 1 − x n − 2 ) − 1 x n − 1 = a x n − x n x n − 2 + 1 x n − 1 = a x n − x n − 1 2 x n − 1 = a x n − x n − 1 \begin{aligned} x_{n+1} & = \; \frac{x_n^2-1}{x_{n-1}} \; = \; \frac{x_n\big(ax_{n-1} - x_{n-2}\big) - 1}{x_{n-1}} \; = \; ax_n - \frac{x_nx_{n-2} + 1}{x_{n-1}} \; = \; ax_n - \frac{x_{n-1}^2}{x_{n-1}} \; = \; ax_n - x_{n-1} \end{aligned} Thus the identity x n = a x n − 1 − x n − 2 n ≥ 2 x_n \; = \; ax_{n-1} - x_{n-2} \hspace{2cm} n \ge 2 follows by induction.

4 Helpful 0 Interesting 0 Brilliant 0 Confused

Using AM-GM inequality, the result becomes obvious

0 Helpful 0 Interesting 0 Brilliant 0 Confused

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...