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For a given $c$ define the power tower $p$ of $c$ as follows: $$p_0 = 1,~~p_i = c^{p_{i−1}} . $$ Find the largest $c$ such that this sequence is bounded.

Clearly $c=1$ works. So assumer $c>1.$ Note that for $c>1,$ $c^n$ is increasing as $n$ increases.

There are two cases:

  1. There exists $x\ge 1$ such that $c^x\le x$
  2. There does not exist $x\ge 1$ such that $c^x\le x$

Note that if there exists $x\ge 1$ such that $c^x\le x$ and hence we get the sequence $\{c^n\}$ to be bounded above by $x.$ Since $\{c^n\}$ is bounded above then so is $\{p_i\}.$

But I couldn't understand the following part.

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After this we have to find the maximum value of $c$ such that there exist $x$ such that $c^x\le x\implies x^{1/x}\le c$ or find maximum value of $c$ which is ( after differentiating) we get $e^{1/e}.$

My doubt is, I get that $c^x-x$ is increasing but it might be bounded above so how can we say that there exists $A$ such that $c^A>A+c-1$ ( we know that $c^x>x$ for all $x>1$).

Moreover, if $c^x-x$ is increasing, then in the interval $[1,A]$ wouldn't $c-1$ be the minimum?

And lastly, I did not understand how $p_i=c^{p_{i-1}}\ge p_{i-1}+m.$ But I understood that if $p_i=c^{p_{i-1}}\ge p_{i-1}+m$, then we must have $p_i\ge 1+m\cdot i.$

Any help or other solution is appreciated.

Raheel
  • 1,791

1 Answers1

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Explicitly writing few terms

$$1, c, c^c , c^{c^c}, c^{c^{c^c}}, \ldots$$

$c=1$ satisfies the boundedness condition. To find better solution let us assume $c>1$. Also, let $t_n$ is the $n$th term of the sequence.

Let the limit of the sequence is denoted by $x$, i.e. $$\lim_{n \to \infty} t_n = x$$ Therefore, $$c^x-x =0$$ because $c>0$.

We want to find the largest $c$ which satisfies this condition. From the equation above

$$ \frac{1}{c} \frac{dc}{d x} =\frac{\log(x) -1}{\log(x)^2}$$

Solving to $x=e$ and $c = e^{1/e}$

p.s.: Edited my completely wrong answer.

Lavneet
  • 283