# 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?

## Answer

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

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 – 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
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.

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