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This is going to be a long question; please bear with me.

We are familiar with the notations $\Sigma^{n}_{k=0} a_k$ and $\Pi^{n}_{k=0}a_k$ for the sum and product of the finite sequence $\{a_n\}$. I've recently been looking into whether we could extend this to exponentiation.

[In this context, we will assume $a^{b^c}$ = $a^{(b^c)}$.] We define $$\Delta^n_{k=1}a_k = \large{a_1^{a_2^{a_3^{^\cdots}}}}$$ Hence we see that $$\Delta^n_{k=1}a_k = a_1^{\LARGE{\Delta^n_{k=2}a_k}}$$ I've worked out derivatives etc. for this. It (this notation, or this way of looking at it) also seems to be helpful in finding the derivative of $a$ tetrated to $b$ (analogous to $\frac{d}{dx}f(x)^{g(x)}$).

My question is, what does this limit evaluate to? (Does it even exist?) $$\lim_{x\rightarrow\infty}\Large{\Delta^x_{k=1}}\normalsize{\sin \frac{k\pi}{2x}}$$

In the spirit of Math.SE, hints are all I ask for. I feel (probably mistakenly) that this delta thing is something important I've stumbled upon :)

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  • $\begingroup$ Are you familiar with the limits on the convergent values of $x^{x^{x^{^\cdots}}}$? $\endgroup$
    – abiessu
    Sep 26, 2013 at 15:14
  • $\begingroup$ I seem to recall having read a PDF that said something about them. IIRC they were something like $e^{-1}$ or $e^{-e}$ or something like that. $\endgroup$
    – Soham Chowdhury
    Sep 26, 2013 at 15:16
  • $\begingroup$ Your notation and idea seem reasonable to me. I expect that due to the aforementioned limits the set of convergent series of this form will be severely limited... $\endgroup$
    – abiessu
    Sep 26, 2013 at 15:21
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    $\begingroup$ In tetration the "exponent" has to be integer. How would you got about differentiating a function that is only defined on integers? $\endgroup$
    – Marc van Leeuwen
    Sep 26, 2013 at 15:28
  • $\begingroup$ Maybe we could say the derivative is only defined at integer values of x. $\endgroup$
    – Soham Chowdhury
    Sep 26, 2013 at 15:34

1 Answer 1

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after writing a quick script i can tell you that it evaluates to 1. i don't know how to proof it though.

edit: think i made a mistake, will post after i checked something

edit2: http://pastebin.com/fsSbj83k

also, this actually kind of makes sense. every individual term is between 0 and 1. something between 0 and 1 to the power something between zero and 1 will get closer to 1.

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  • $\begingroup$ Could you post the script? $\endgroup$
    – Soham Chowdhury
    Sep 26, 2013 at 15:38
  • $\begingroup$ Actually, this analysis is false. In particular, note how $0.5^{0.5^{0.5}}$ is smaller in value than $0.5^{0.5}$. $\endgroup$
    – abiessu
    Sep 26, 2013 at 18:05

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