Is i = \sqrt{e^{\pi\sqrt{e^{\pi\sqrt\ldots}}}}i = \sqrt{e^{\pi\sqrt{e^{\pi\sqrt\ldots}}}}?

I was messing with the identity e^{i\pi}=-1 and I got that i = \sqrt{e^{\pi\sqrt{e^{\pi\sqrt\ldots}}}} and on. I plugged it in to a calculator and it was infinite. It grew very fast. Does that make i solvable?


I’m assuming to reach that conclusion, you took the square root of both sides and got:
which is correct. Then, on the left hand side, you substituted for i using this equality to get
which is still okay – and then you tried doing this infinitely often, and lo and behold, it broke!

The main issue is that, suppose we let
f(x)=\sqrt{e^{\pi x}}.
What’s you’ve shown amount to the fact that f has what is called a fixed-point at i – that is
And this implies things like
or, where we let f^n(i) be f applied to i repeatedly n times, we get
So far so good. However, if we wanted to make sense of an expression which is basically an infinite nest of the function f with no starting value, it would be that, no matter what x we started with, the sequence (x,f(x),f^2(x),f^3(x),\ldots) – called the orbit of x under f – would always converge to i – and as you see, this is not the case.

A simple example of the difference between “has a fixed point” and “converges to something” would be if we took
which has a fixed point at x=0. If we wished to think of the expression 2\cdot (2\cdot (2\cdots)))
we would hope that f would have that (x,f(x),f^2(x),\ldots) converged for any x – but it only converges for x=0 – otherwise, the sequence has f^n(x)=2^nx which diverges. Your function acts similarly, but it is harder to visualize, as it takes place on the complex plane. So, in fact, you cannot “compute” i using this method – the series will simply diverge. However, for any finite number of applications, your identity is correct.

Source : Link , Question Author : user2888499 , Answer Author : user26486

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