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Define a sequence $(f_n)$ of rational functions by $f_1(x)=2,f_2(x)=x$ and

$$ f_{n}(x)=\frac{1}{f_{n-1}(x)}+\frac{1}{f_{n-2}(x)} $$

Let $I$ be the interval $I=[\frac{7}{6},\frac{3}{2}]$. Then

Fact. For $n\in [2,15]$, the minimum for $f_n$ on $I$ is attained on the boundary of $I$, i.e. at some $b_n\in \lbrace \frac{7}{6},\frac{3}{2} \rbrace$.

Can anyone prove or disprove my conjecture that this holds for any $n\geq 2$ ?

My thoughts : assuming the conjecture is true, denote the case where $b_n=\frac{7}{6}$ by $a$ and the case where $b_n=\frac{3}{2}$ by $b$, for $n=2$ to $n=30$, we have the following sequence :

$$abbabbaabaabaabbabbaabaabaabb$$

This sequence does not seem to follow any particular pattern.

Related : Proof of a limit for a recursively-defined sequence

Ewan Delanoy
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  • It looks like any $f_n$ is monotonous on interval $I=[\frac{7}{6},\frac{3}{2}]$. Could it be another conjecture (or part of yours ?) ? Thus your characterization by the bound $a$ or $b$ of $I$ in which the max is obtained could be expressed as the sign of $f 'n$ on I. A convincing fact in favour of the truth of the conjecture is as follows : consider differetiating the defining relationship : this gives $f '_n(x)= - f '{n-1}/ f_{n-1}^2 - f '{n-2}/ f{n-2}^2 $ ; out of that, one can conclude that if $f '{n-2}$ and $f '{n-1}$ have the same sign, $$f '_n$ will have an opposite sign [...] – Jean Marie Aug 21 '16 at 22:52
  • [ctd] This would account for the fact that behind a sequence $aa$ (resp $bb$) you have necessarily $b$ (resp $a$). But is it true that every $f_n$ is monotonous ?... – Jean Marie Aug 21 '16 at 22:53

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