7
$\begingroup$

I want to prove the following power tower inequality:

$$ 3 \uparrow \uparrow 100 > 4 \uparrow \uparrow 99 $$

but I don't know how to do this. I think that induction will not work, because I think there will be an $N$ for which

$$ 3 \uparrow \uparrow N < 4 \uparrow \uparrow (N-1) $$

Could anyone help me in the right direction? Please don't answer a full solution, but I do need a hint.

$\endgroup$
2
$\begingroup$

Actually, $4 \uparrow \uparrow n < 3 \uparrow \uparrow (n+1)$ for all $n \ge 0$.

In fact, it can be shown that if $x$ and $y$ are such that $1 < x < y$, then $$y \uparrow \uparrow n \ \ < \ \ x \uparrow \uparrow (n+c)\ \ \quad(n \ge 0)$$ where $c$ is a postive integer depending only on $x$ and $y$.

Here's a proof-outline (adapted from this posting and followups):

  1. Suppose $1 < x < y$, and let $a_n = {\log_x}^n(y\uparrow\uparrow n)$ for $n \ge 0$. Consider the sequence $(a_n)$ as $n \to \infty.$
  2. Show that $(a_n)$ is strictly increasing, using properties of the logarithm.
  3. Show that $(a_n)$ has a limit, say $A$, using the Mean Value Theorem to prove extremely rapid convergence.
  4. Note that $y\uparrow\uparrow n = (x\uparrow)^n a_n < (x\uparrow)^n A \le (x\uparrow)^n (x\uparrow)^c 1$ = $x\uparrow\uparrow (n+c)$, where $c$ is the least integer satisfying $A \le (x\uparrow)^c 1 = x\uparrow\uparrow c$.

Here are some particular cases:

x   y   c   A
--  --  --  ----------------
2   3   2   2.44402146148920
2   4   2   3.17037617633756
2   5   2   3.68091002494335
2   6   3   4.07723742182623
e   3   1   1.22172930187025
3   4   1   1.51107202382304   <-- your case
3   5   1   1.85474212525557
4   5   1   1.28188454071981
$\endgroup$
1
$\begingroup$

Try to prove

$3 \uparrow \uparrow n+1 > \log_3(4) \cdot 4 \uparrow \uparrow n$ for all $n \in \mathbb{N}$

and use the following facts about $\log$

  • $\log_a (x) = \frac{\log_b(x)}{\log_b(a)}$
  • $\ln : \mathbb{R}_+ \rightarrow \mathbb R$ is strictly increasing
$\endgroup$
  • 2
    $\begingroup$ How would proving this help? $\endgroup$ – Tim Vermeulen May 29 '13 at 19:42
  • $\begingroup$ Oh, I didn't mean to take the fraction there. I corrected it. $\endgroup$ – user79202 May 29 '13 at 19:47
  • $\begingroup$ If I assume this is true for some $k \in \mathbb{N}$ and I raise 3 to the power of what is left of the inequality sign and what is right of it (and simplify), I get that $3 \uparrow \uparrow (k + 2) > 4 \uparrow \uparrow (k + 1)$, which is not a proof, because I lose the log term. What am I doing wrong? $\endgroup$ – Tim Vermeulen May 30 '13 at 16:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.