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

Solution 1:

I'm assuming to reach that conclusion, you took the square root of both sides and got: $$\sqrt{e^{i\pi}}=i$$ which is correct. Then, on the left hand side, you substituted for $i$ using this equality to get $$\sqrt{e^{\pi\sqrt{e^{i\pi}}}}=i$$ 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 $$f(i)=i.$$ And this implies things like $$f(f(i))=i$$ $$f(f(f(i)))=i$$ or, where we let $f^n(i)$ be $f$ applied to $i$ repeatedly $n$ times, we get $$f^n(i)=i.$$ 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 $$f(x)=2x$$ 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.

Solution 2:

If you take $$\exp(i\pi)=-1=i^2\implies i=\sqrt{\exp(i\pi)}=\sqrt{\exp(\pi\sqrt{\exp(i\pi)})}$$ So the correct identity should be: $$i=\sqrt{\exp(\pi\sqrt{\exp(\pi\sqrt{\exp(\pi\ldots)})})}$$ You can't evaluate from left to right, you need to evaluate it from right to left. Note that last term in an case would be $\exp(i\pi)$ which normals calculators can't evaluate.