proof of this:

x^x^x^.^.^. (infinite exponentiation) only converges for e^(-e)<x<e^(1/e)

Please halp, for it is beyond me :(.

This isn't true. x = 1.2, for instance, will certainly diverge under infinite exponentiation, but e^(-e) < 1.2 < e^(1/e). Further, x = (0.01)^(100) will converge to 0.01 under infinite exponentiation, but 0.01 < e^(-e).

>>2
Disregard that, I don't know how to exponentiate.

>>1
Man, that is fucking PhD math. GFTO.
Along with doing any root/ratio test, you'd also have to prove that when doing so, dividing an infinite series by a term is valid. And /that/ is the hard part.

>>4
Shouldn't that depend on if the series is uniformly convergent, or something gay like that?

>>4
This isn't "PhD Math".

Suppose the sequence {x, x^x, x^x^x,...} converges to L. Then x^L is the limit of the sequence {x^x, x^x^x, x^x^x^x,...}. But clearly the limit of that sequence is also L, since it only chops off the first element of the original sequence. So x^L = L, so x = L^(1/L). This obtains a maximum value of e^(1/e) over positive real numbers L. I'm not sure, however, where the e^(-e) comes from; there isn't anything apparent in that function, except that if x = e^(-e), L = 1/e. (And if x = e^(1/e), L = e).

Although it is "useless math"

>>7
There's no such thing.

Like that's going to help you at your job at McDonald's...

>>9

OP here.

HOW THE FUCK did you know?

>>7
Just because a purpose haven't been found, doesn't mean it is useless.

>>9

People who take that attitude in general toward learning (math, literature, history) often tend to be the people ending up at McDonalds, etc.

Is x^(infinity) even defined?

>>13

$\lim_{A\rightarrow\infty} x^A$ ?

>>13
For x != 0, it is. For positive x, if x < 1 it's 0, otherwise it's infinity. For negative x, the other way around.

6[super]9[/super]

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