Question

Take x∈ℝ. Set x>0 and let the sequence (x_n) be defined by x_1=x, x_2=x^x_1, x_3=x^x_2, ... , x_n=x^(x_(n-1)) and so on.

For what values of x is the sequence convergent?

Answer 1

Intuitively, it seems that it would converge if \(x < 1 \), but I'm unsure of how to show that.

Answer 2

Well, I do have the answer. It'd be interesting to see an actual proof, though. :P Would you like the answer?

Answer 3

I would like the answer.

Answer 4

0<x≤e^(1/e)

Although, now that I consider it, there's a chance this answer is wrong... I don't think our class has solved this problem yet. XD

Answer 5

I'm sorry, I mistyped the "solution". It's 0<c^x≤e^(1/e). The other submitted solution is that there's no value for x such that the equation converges.

Interesting.

Answer 6

Well, now that I've looked at it longer, if x=1, it obviously converges to 1. If x<1, then it converges to a number less than 1.

Answer 7

That's what I intuitively think, but I don't know how to go about writing that stuff down.

Answer 8

Well, the definition I have written down is as follows:

"Let \(a_n \in \mathbb{R}^\mathbb{N}\) be a sequence of Real-valued numbers. Let \(S \in \mathbb{R}\) such that \(\forall \epsilon \in \mathbb{R},\quad |a_m - S| < \epsilon\qquad \forall m \in \mathbb{N}, \:\:\: m > n. \) If this is true, then \(a_n\) is said to converge.

Answer 9

And \(S\) is called the limit of \(a_n\).

Answer 10

For \(x = 1\) let \(S=1\). Then \(a_m-S = 0\) since \(\epsilon > 0\) this is smaller than \(\epsilon\). For \(x<1\) it's a little trickier, but \(S=0\) would probably work. As for the upper limit of \(x\), I really don't know how to show that right now.

Answer 11

I've got to sleep now, but good luck with this problem.

BTW, in my definition, I forgot to include that \(\epsilon > 0\). That's rather important to have in there.

Answer 12

Good night, Sir. I'm puzzling over this, but I like your approach.

Answer 13

Is this not equivalent to asking for what values positive \(a\) is it true that \(\lim_{n \to \infty}{}^na\ne \infty \)? (Note that \({}^na\) is the \(n\)th tetration of \(a\). See http://en.wikipedia.org/wiki/Tetration .)

I'm almost 100% sure this is true only for \(0 < a \le 1\) as a consequence of the fact that the tetration operator satisfies \({}^{n-1}a<{}^{n}a\) whenever \(a > 1.\)

Answer 14

Oh, I'm 4 months late again. Sorry about that.

Answer 15

Just using Wolfram, it would appear that it converges for \(\displaystyle0<x\leq e^{\frac{1}{e}}\). If I go any larger, it immediately rushes off to \(\infty\), but for \(x=e^{1/e}\), it does appear to converge. I'm not sure what it converges to however.

If I had to take a random guess, I might say that it converges to \(e^{1/e}+1\).

Answer 16

In fact, if you look at it, you get \[\huge {e^{\frac{1}{e} ^{e^{\frac{1}{e}^{e^{\frac{1}{e}^{e^{\frac{1}{e}^{...}}}}}}}}}
\]Notice that you get a lot of cancellation in the exponents, so it makes sense that this would converge.

Answer 17

I looked into this a bit more on Wolfram Mathworld, and found that the limits for convergence are \[\Large e^{-e}\leq x\leq e^{\frac{1}{e}}\] http://mathworld.wolfram.com/PowerTower.html