Proper way of notating nested expression whose value is 1

Question

Trying to express $x=1$ in the following way, replacing its value recursively, but it's not clear that it "converges" properly

$$x=-e^{\pi\sqrt{e^{\pi\sqrt{e^{\pi\sqrt{e^{\pi\sqrt{...}}}}}}}}$$

any thoughts would be appreciated

Answer 1

You use $~i^{1/i}=e^{\pi/2}$ . $~$ Let $~T(x)~$ the $~$infinite power tower$~$

(e.g. see Tetration and PowerTower) then for real $~x~$ it's $~T(x^{1/x})=x~$ if $~\frac{1}{e}<x<e$ .

Now we can $~$formally$~$ write $~T(i^{1/i})=i~$ and therefore:

$$-e^{\pi\sqrt{e^{\pi\sqrt{e^{\pi\sqrt{e^{\pi\sqrt{...}}}}}}}}=-T(e^{\pi/2})^2=-T(i^{1/i})^2=-i^2=1$$

Note:

$\displaystyle T(x)=\frac{W(-\ln x)}{-\ln x}~$ where $W$ represents Lambert's W function .

Here you can read, that $~\displaystyle W\left(-\frac{\pi}{2}\right)=i\frac{\pi}{2}~$ as a unique assignment so that $~\displaystyle T(e^{\pi/2})=-i~$ .

Answer 2

Let $x_0=e^\pi$ and define the sequence $x_n$ recursively by

$$x_{n+1}=e^{\pi\sqrt{x_n}}$$

If you assume the limit $L=\lim_{x\to\infty}x_n$ exists, then it must satisfy the equation

$$L=e^{\pi\sqrt{L}}$$

Now $L=-1$ does satisfy that equation, since $e^{\pi\sqrt{-1}}=e^{\pi i}=-1$, so it's tempting to conclude, as the OP wants, that

$$1=-e^{\pi\sqrt{e^{\pi\sqrt{e^{\pi\sqrt{e^{\pi\sqrt{...}}}}}}}}$$

However, this conclusion contradicts the very assumption it's predicated upon. That's because the numbers $x_n$ are, if you use the usual convention that the square root of a positive real is a positive real, all positive reals, and such a sequence cannot only have a nonnegative real number as a limit.

We could ask if the equation $L=e^{\pi\sqrt{L}}$ has some other solution with $L\ge0$. With a bit of calculus, it's possible to show it doesn't. So the sequence of $x_n$'s does not have a limit; it diverges.