Let $ (x_1,x_2,\cdots)$ be a sequence of positive numbers such that $ (8x_2 - 7x_1)x_1^7 = 8$ and \[ x_{k + 1}x_{k - 1} - x_k^2 = \frac {x_{k - 1}^8 - x_k^8}{x_k^7x_{k - 1}^7} \text{ for }k = 2,3,\ldots \] Determine real number $ a$ such that if $ x_1 > a$, then the sequence is monotonically decreasing, and if $ 0 < x_1 < a$, then the sequence is not monotonic.
Problem
Source: China Girls Math Olympiad 2008, Problem 6
Tags: algebra unsolved, algebra
20.05.2009 12:33
orl wrote: Let $ (x_1,x_2,\cdots)$ be a sequence of positive numbers such that $ (8x_2 - 7x_1)x_1^7 = 8$ and \[ x_{k + 1}x_{k - 1} - x_k^2 = \frac {x_{k - 1}^8 - x_k^8}{x_k^7x_{k - 1}^7} \text{ for }k = 2,3,\ldots \] Determine real number $ a$ such that if $ x_1 > a$, then the sequence is monotonically decreasing, and if $ 0 < x_1 < a$, then the sequence is not monotonic. We have $ x_{k + 1} = \frac{x_k^2}{x_{k-1}} + \frac {x_{k - 1}^8 - x_k^8}{x_k^7x_{k - 1}^8}$ and so $ 8x_{k + 1}x_k^7 = 8\frac{x_k^9}{x_{k-1}} + 8\frac {x_{k - 1}^8 - x_k^8}{x_{k - 1}^8}$ $ 8x_{k + 1}x_k^7 -7x_k^8 = 8\frac{x_k^9}{x_{k-1}} -7 x_k^8 + 8\frac {x_{k - 1}^8 - x_k^8}{x_{k - 1}^8}$ $ \implies$ $ 8x_{k + 1}x_k^7 -7x_k^8 = \frac{x_k^8}{x_{k-1}}(8x_k - 7x_{k-1}) + 8-8\frac { x_k^8}{x_{k - 1}^8}$ $ \implies$ $ (8x_{k + 1} -7x_k)x_k^7 = 8 + \frac{x_k^8}{x_{k-1}^8}((8x_k - 7x_{k-1})x_{k-1}^7 -8)$ $ \implies$ since $ (8x_2 - 7x_1)x_1^7 = 8$ : $ (8x_{k + 1} -7x_k)x_k^7 = 8$ And so $ \boxed{x_{k+1}=\frac{7}{8}x_k+\frac{1}{x_k^7}\text{ for }k>=1}$ Let $ f(x)=\frac{7}{8}x+\frac{1}{x^7}$ defined in $ \mathbb R^{+*}$. It's easy to see that : $ f(x) > 8^{\frac{1}{8}}$ $ \forall x\neq 8^{\frac{1}{8}}$ $ f(x) < x$ $ \forall x>8^{\frac{1}{8}}$ $ f(x) > x$ $ \forall x<8^{\frac{1}{8}}$ $ f(8^{\frac{1}{8}})=8^{\frac{1}{8}}$ So : If $ x_1=8^{\frac{1}{8}}$, $ x_n=x_1$ is a constant sequence. If $ x_1 > 8^{\frac{1}{8}}$ $ x_n$ is a strictly decreasing sequence (always $ >8^{\frac{1}{8}}$) If $ x_1 < 8^{\frac{1}{8}}$ $ x_2 > x_1$ and $ x_n$ becomes strictly decreasing from there and so $ x_n$ is not monotonic. And so the answer is $ \boxed{a=8^{\frac{1}{8}}}$
13.01.2023 15:31
orl wrote: Let $ (x_1,x_2,\cdots)$ be a sequence of positive numbers such that $ (8x_2 - 7x_1)x_1^7 = 8$ and \[ x_{k + 1}x_{k - 1} - x_k^2 = \frac {x_{k - 1}^8 - x_k^8}{x_k^7x_{k - 1}^7} \text{ for }k = 2,3,\ldots \] Determine real number $ a$ such that if $ x_1 > a$, then the sequence is monotonically decreasing, and if $ 0 < x_1 < a$, then the sequence is not monotonic. $x_{k+1}x_{k-1}-x_k^2=\dfrac{x_{k-1}^8-x_k^8}{x_k^7x_{k-1}^7}\Rightarrow \dfrac{x_{k+1}}{x_k}-\dfrac{x_k}{x_{k-1}}=\dfrac{1}{x_k^8}-\dfrac{1}{x_{k-1}^8}\Rightarrow \dfrac{x_{k+1}}{x_k}-\dfrac{1}{x_k^8}=\dfrac{x_k}{x_{k-1}}-\dfrac{1}{x_{k-1}^8}=\dfrac{x_2}{x_1}-\dfrac{1}{x_1^8}=\dfrac{7}{8}$ $\therefore x_{k+1}=x_k-\dfrac{1}{x_k^7}$. Let $f(x)=\dfrac{7}{8}x-\dfrac{1}{x^7}$, then $f(a)=a\Rightarrow a=8^{\frac{1}{8}}$, and the problem becomes easy.