How is $\,(e^{\pi})^{{{\left(e^{\pi/2}\right)}^{\left(e^{\pi/2}\right)}}^{\ldots\infty}}=-1\,?$

Question

So, here's how I've arrived to that conclusion.

My first question was: can $\,e^{i\pi}\,$ be written as $\,(e^\pi)^i\,?$

Well, by taking $\,\log_e$ to both sides of the equation $\,{e^{i\pi}}=-1\,,\,$ we get $\,\log_e{i}=i\pi/2\,$ which is true. And similarly doing it for the other equation $({e^\pi})^i = -1\,,\,$ it holds true.
So they are equivalent forms.

And again $\,(e^\pi)^i=(e^\pi)^{e^{{i\pi}/2}}=-1\,.$

Now if we repeatedly replace $i$ with $e^{i\pi/2}$, we obtain

$\,(e^{\pi})^{{{\left(e^{\pi/2}\right)}^{\left(e^{\pi/2}\right)}}^{\ldots\infty}}=-1\,.$

Is this just true or have I done something wrong here?

Answer 1

This kind of depends on exactly what you mean. When you ask if $(e^\pi)^{(e^{\pi/2})^\cdots}=-1$, I interpret it as you asking if \begin{align*} \lim_{n\rightarrow \infty} (e^\pi)^{(e^{\pi/2})^{\text{...($n$ iterations)...}^{(e^{\pi/2})}}}=-1. \end{align*} Every term in this sequence is a positive real number, so clearly this cannot hold. However, if we toss that $i$ up at the top, then sure, every term equals $-1$, so we are ultimately dealing with a constant sequence and of course it converges to $-1$. My point is that if one sees the expression $(e^\pi)^{(e^{\pi/2})^\cdots}$ in a vacuum, your notation would not suggest there's an $i$ hidden in the question.

Answer 2

In general, if $f(x,y)$ is a function of two variables, then $$f(x,f(x,f(x,\dots)))\tag1$$ is not really well-defined.

The best we can do is define a sequence, by choosing $y_0,$ and then defining $y_{n+1}=f(x,y_n)$ and get a limit. But the limit depends on $y_0.$ So it is an abuse of notation to write the limit as $(1),$ which doesn't not appear to depend on another value.

If every $y_0$ gives the same limit, then $(1)$ can be justified.

What you've shown is that if there is one limit for all complex $y_0,$ then the limit of this expression is $-1.$ But that just means there isn't one limit.

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It might even be clearer if we define $g(y)=f(x,y)$ for fixed $x.$ Then you'd write:

$$g(g(g(g(\cdots))))$$ which clearly means nothing. The limit of the sequence will sometimes exist and be independent of the the starting value - say, when $g$ is a contraction - but those cases are very rare.

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If $g(y)=-y,$ the only starting value which converges is $y_0=0.$ But would we really say:

$$-(-(-(\dots)))=0?$$ It seems strange to write that, even though any $y_0$ gives a sequences whose Cesàro mean converges to $0.$