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For which values of $x\ge 1$ does the expression $x^{x^{x^{x^{.^{.^{.}}}}}}$ make sense? To tackle this, define $f_1(x)=x$ and $f_{n+1}(x)=x^{f_n(x)}$ for $x \ge 1$ and $n\ge1$.

a) Show that $f_{n+1}(x) \ge f_n(x)$ for all $n\ge1$.

b) When $L(x) = \lim_{n\to\infty} f_n(x)$ exists, find an equation for $L(x)$. Use it to find an upper bound for $x$.

c) For these values of $x$, show by induction that $f_n(x)$ is bounded above by $e$ for all $n\ge1$. What can you conclude?

d) What happens for larger $x$?

I'm having trouble showing it is increasing for a).

I solved b) with the fact that $\lim_{n\to\infty}f_n(x)=L(x)=x^{L(x)}\implies x=L(x)^{\frac{1}{L(x)}}$

Maximizing on $L>0$: $$\frac{d}{dL}L^{1/L}=\frac{d}{dL}e^\frac{\ln(L)}{L}=\frac{d}{dL}\left(\frac{\ln(L)}{L}\right)e^\frac{\ln(L)}{L}$$ $$=\frac{1-\ln(L)}{L^2}L^{\frac{1}{L}}=L^{\frac{1}{L}-2}(1-\ln(L))$$ $$=-L^{\frac{1}{L}-2}(\ln(L)-1)=0\iff\ln(L)-1=0 \text{ so } L=e$$

Therefore $\max\{L^{1/L}\}$ happens when $L=e$ so $x\le e^{1/e}$ and I now have bounds for $x$

It then follows for c):

Since $f_n$ is increasing on $[1,e^{1/e}], \max\{f_n(x)\}=f_n(e^{1/e})$

Base case: $f_1(x)=x\le e^{1/e}<e$.

Assume $f_n(x)<e$, consider $x^{f_n(x)}\le x^e$ $\implies f_{n+1}(x)\le f_{n+1}(e^{1/e})<x^e<(e^{1/e})^e=e$ therefore by principle of mathematical induction, $f_n(x)<e$ for $x\in[1,e^{1/e}]$


marked as duplicate by Daniel W. Farlow, user147263, Adam Hughes, user149792, Peter Taylor Mar 13 '15 at 8:54

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  • 1
    $\begingroup$ for $1$ it makes sense $\endgroup$ – Jorge Fernández Hidalgo Mar 13 '15 at 2:39
  • $\begingroup$ In general we have $x^{f_n(x)}\ge f_n(x)$ rearranging we get that $f_n(x)\ln(x)\ge\ln(f_n(x))$ but I'm not sure if this helps $\endgroup$ – dd19 Mar 13 '15 at 2:45
  • $\begingroup$ See infinite tetration. $\endgroup$ – Lucian Mar 13 '15 at 3:03

(a) will follow if we can prove that $f(x,y)=x^y/y\ge 1$ for all $y\ge x>1$. We handle $x=1$ separately, which the emperor did already in the comments.

Considering this is a function of $y$ on $[x,+\infty)$, we see that $f(x,x)=x^{x-1}\ge x^0=1$ for all $x> 1$. We also have $$\frac{\partial }{\partial y}f(x,y)=\frac{x^y(y\ln x - 1)}{y^2}$$

We have $y\ln x \ge x \ln x >1 $ for $x>1$, so $\frac{\partial }{\partial y}f(x,y)>0$. Since this is an increasing function on $[x,+\infty)$, and is $\ge 1$ at the left endpoint, it must be $\ge 1$ on the whole interval.

  • $\begingroup$ Great answer! Is there a way of doing it without partial derivatives as we haven't covered that yet? $\endgroup$ – dd19 Mar 13 '15 at 3:01
  • $\begingroup$ Well, it's not really a partial derivative, it's a function of $y$ with $x$ fixed. You can think of it as $f_x(y)=x^y/y$. $\endgroup$ – vadim123 Mar 13 '15 at 3:02

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