x=1.2872122

>>7734677
x=x

>>7734685
bingo

>>7734683
incorrect

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242 KB, 1600x900

>>7734683
>1.2872122

>>7734677
approximately 1.4335

>>7734677
1.0552728827024529267e0

>>7734697
>Failing this badly

I give up, how does one solve this problem algebrajifically?

>>7734938
whatever you do, you end up having a x and a ln(x) in the equation. I think it's imposible

>>7734677
If you're talking about the case with infinitely many x's, this is a pretty cool problem. In fact, I think there's a problem just like this in the infinite sequences chapter of Spivak.

If I remember correctly, such sequences only converge for x between 0 and e^(1/e).

I think the strategy is to show that it converges as the number of xs goes to infinity. Once you have convergence, you must have that it converges to some number a.
Then we must have
a = x^a
The solution of which is:
x = a^(1/a)

We have a = 2
Therefore, x = 2^(1/2), i.e. x = sqrt(2).

If you asked me to prove the convergence of x^x^x^x... these days, though, I don't know I could make it rigorous in any reasonable amount of time for a shitpost on sci.

>>7735220
Just do it

>>7735227
Maybe I will if I get bored today and this thread is still up.

After doing some research (googling) on the topic, I found that the convergence actually holds for e^-e<=x<=e^(1/e)

As a preliminary thing, if the series is monotonic, then the problem amounts to finding a bound on the series, but I'm not so confident that it is in fact monotonic.

7729321
Anonymous
Can equations such as x^x^x^x^x = 2 or x↑↑5 = 2 if you prefer be solved anylitically

->Archive

>>7735220
>>7735227
x^x^x^... converges because you can define the function f(x)=x^x and then observe that the graph of the function is dominated above and below by y=x in the appropriate places to force the iteration to those two rest points. No need to get fancy about it.