Let $a$ be a real number in the open interval $(0,1)$. Let $n\geq 2$ be a positive integer and let $f_n\colon\mathbb{R}\to\mathbb{R}$ be defined by $f_n(x) = x+\frac{x^2}{n}$. Show that \[\frac{a(1-a)n^2+2a^2n+a^3}{(1-a)^2n^2+a(2-a)n+a^2}<(f_n \circ\ \cdots\ \circ f_n)(a)<\frac{an+a^2}{(1-a)n+a}\] where there are $n$ functions in the composition.
Problem
Source: Romania TST 2014 Day 2 Problem 2
Tags: function, algebra unsolved, algebra
12.05.2015 13:27
Do you have any solution for this?
12.05.2015 13:42
I don't like even the look of it.... seems 'dangerous'.
12.05.2015 16:13
ComplexPhi wrote: Let $a$ be a real number in the open interval $(0 ,1)$, let $n$ be a positive integer and let $f_n\colon\mathbb{R}\to\mathbb{R}$, $f_n(x) = x + \frac{x^2}{n}$. Show that : \[\frac{a(1-a)n^2+2a^2n+a^3}{(1-a)^2n^2+a(2-a)n+a^2}<(f_n \circ \dots \circ f_n)(a)<\frac{an+a^2}{(1-a)n+a}\] where there are $n$ functions in the composition. Wrong. Choose as counter-example $n=1$ and $a=\frac 12$ and RHS becomes $\frac 34<\frac 34$, which is wrong
12.05.2015 16:43
This was just an extreme case and it happened to be an equality . I will change it to $n \geq 2$ . For $n=1$ the inequalities are : $\frac{a+a^2+a^3}{a^2+1}<a^2+a<a^2+a$ As pointed by pco the RHS is wrong and the LHS reduces to $a^4>0$ which is true. Now let's prove it for $ n \geq 2$ .
25.12.2016 19:45
We should consider the sequences a(1), a(2) ... a(n),such that a(i)=f(f(f(...(f(a))...)(there are i functions in this composition) and than (1/a(k))-(1/(a(k+1))=1/(a(k)+n) . with this information question become easy.
11.11.2022 05:59
this is quite old, but bumping this because I don't see how the substitution makes the question easy