5
votes
0answers
111 views

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 $f_1(x)=x,f_2(x)=x^x,\dots f_{n+1}(x)=x^{f_n(x)}~(n\geq 1).$ Let $F_n(x)=f_n^{'}(x).$ Hence $F_1(x)=1, F_2(x)=x^x(1+\log(x))\dots.$ Prove that $F_{2n}(x)=0$ has exactly one root in the ...
50
votes
4answers
3k views

Are the solutions of $x^{x^{x^{x^{\cdot^{{\cdot}^{\cdot}}}}}}=2$ correct?

Problem: Find $x$ in $$\large x^{x^{x^{x^{ \cdot^{{\cdot}^{\cdot}} }}}}=2$$ Trick: $x^{x^{x^{x^{\cdot^{{\cdot}^{\cdot}}}}}}=2$, so, $x^{(x^{x^{x^{\cdot^{{\cdot}^{\cdot}}}}})}=x^2=2$, and, ...