Brainlet here, how do I solve this? It kinda looks like

>>8171839
>implying there is something to solve

>>8171839
It's divergent

>>8171848
We talked about euler's identuty in class and our professor gave us that infinite expression to solve.

>>8171852
>solve
>an expression

>>8171854
Find the value of that. Call it what you want, I'm a brainlet

>>8171839
It's actually i. Just square root Euler's identity, it should be pretty simple once you've done that.

>>8171839
note that if
x = sqrt(e^(pi*sqrt(e^pi...)))
then
x = sqrt(e^(pi*x))
x^2 = e^(pi*x)
2ln(x) = pi*x
*shrugs* dunno if there are solutions to that or not.

>>8171865
Sqrt(e^pi)=i

>>8171859
i or minus i, but yeah, /thread

>>8171839
It's divergent, but i, -i are fixed points of f(x) = e^(pi*x)/2

Related:

http://math.stackexchange.com/questions/45283/compute-lim-limits-n-to-infty-sin-sin-dots-sin-n/

https://en.wikipedia.org/wiki/Tetration

>>8171935
What are fixed points?

>>8171946
x is a fixed point of f if and only if f(x) = x

no real solution,
complex solutions are ±i
http://www.wolframalpha.com/input/?i=%28lnx%29%2Fx+%3D+pi%2F2&x=7&y=4

Put it in your calculator with the longest input it can handle. It'll be close enough.

>>8171848
Jeez you re a fucking cunt

>>8171865
I always had a problem with this way of thinking.
x + x + x + x ... = 2
Note that x + x + x ... = 2
x + 2 = 2
x = 0
same goes for multiplication
x*x*x*x*x ... = 2
x*2 = 2
x = 1
And exponentiation
x^(x^(x^(x^(x... = 2
x^2 = 2
x = sqrt(2)
I don't think your "solution" is right at all.

>>8173981
Hello retard. Welcome to epsilons. It's valid mathematics. Unless you work in axiomatic systems without them, then you have other problems.

>>8173989
Ok so explain it to me.

>>8173981
You're correct. It's total shit. Analysis was invented to avoid exactly that.

>>8171855
Find the value of e^2.

>>8173981
Any finite \sqrt{2}^\sqrt{2}^...^\sqrt{2}=2 so that one isn't that weird.

>>8173999
Okay, but why does it work for exponentiation towers but not infinite series? If exponentiation is glorified multiplication and multiplication is glorified addition, shouldn't it not work for any of them?

>>8171839
-2.02471401844 - 0.0455237761827*i

>>8173981
What's the problem with any of these?

>>8173981
there's a couple of things that keep you from saying x = 0 in the case of addition, for infinite series you cant rearrange terms in a divergent expression, (changing the order of evaluation counts as rearranging.) The theorem shows that if it is convergent, there is only 1 solution, if it is divergent, there are infinite solutions if rearrangement is allowed. slightly changing how its usually used for series you get in the case of x + x + x + ...= 2 that it can be x + 2 = 2 so x = 0, or x + 2 + 2 = 2, so x = -2, and so on. same in the case for x*x*x... = 2 you can get 1, or 1/2, or 1/4 and so on. The reason it works for x^x^x^... = 2 is because the equation is convergent.

tl;dr - if you know there is an answer, you can do what you did to find it, but you cant use it to determine if there is one.

If you want to determine if there is one you need to use something like the epsilon-delta definition of limits to determine if it converges, and for what values of x, then you can determine the range and find if 2 is in the range, if it is then you know a solution exists and you can do the replacement thing you did (or use a more riquress way if you want)

>>8174049
>cont.

there are instances that you may 2want to rearrange terms even if it diverges to get a finite answer that 'behaves like' the divergent case, that's called summation methods, and its how you get 1+2+3+4+... = -1/12. in the case of x + x + x + ... = 2 that gives 0, because the x has to 'behave like' 0, in this case that you have to be able to add it to x + x + ... without changing the value. THis is a pretty hand wavy explanation, but you can treat is more righteously, which allows you to use the results (like -1/12 in string theory)

>>8174049
>if it is divergent, there are infinite solutions if rearrangement is allowed
This is only true if the series converges conditionally but not absolutely, i.e., both positive and negative terms of the series should diverge.

>>8173994
https://en.wikipedia.org/wiki/Hyperreal_number