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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.

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2 Answers

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
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How would proving this help? –  timvermeulen May 29 '13 at 19:42
    
Oh, I didn't mean to take the fraction there. I corrected it. –  user79202 May 29 '13 at 19:47
    
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? –  timvermeulen May 30 '13 at 16:49
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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
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