# Prove that $F_{2n}(x)=0$ has exactly one root in the interval $x\in(0,1),$ and this root $\to 0$ when $n \to \infty.$

Define $$f_1(x)=x\\f_2(x)=x^x\\\vdots\\f_{n+1}(x)=x^{f_n(x)}$$

Let $F_n(x)=f_n^{'}(x).$ Hence $$F_1(x)=1\\F_2(x)=x^x(1+\log(x))\\\vdots$$

Prove that $F_{2n}(x)=0$ has exactly one root in the interval $x\in(0,1),$ and this root $\to 0$ when $n \to \infty.$

Here are the images of $f_{2n}(x)$ for $n=1,2,\ldots,10.$ • @Jaycob Coleman How do you get this? I think $F_3(x)=0$ has no real roots. – Next Aug 10 '13 at 14:20
• @Hecke Do you have a source for this problem? – user1337 Aug 11 '13 at 22:55
• Since $f_n$ converges only for $x \in [e^{-e},e^{1/e}]$ I would expect that the root of $F_n$ tends to $e^{-e}$ as $n \to \infty$. – Antonio Vargas Nov 10 '15 at 11:50

## 1 Answer

I would suggest to study formally $f_n(x)$ :

• look at the values in 0 and 1, and the variations in between
• study $f_n$ recursively
• At this point prove that there is 1 root only $r_n$
• prove that $r_n$ decreases
• show that the limit cannot be > 0.
• This would be more appropriate as a comment I think. – Antonio Vargas Nov 10 '15 at 11:53