1
$\begingroup$

First some background info. Feel free to skip to the bold text below if you already know this.

I) The derivative of the infinite tetration of $x$ (I guess that's how you say it) or ${}^\infty x$ can be found using the following method: $$y=x\color{red}{{}^{x^{x^{.^{.^.}}}}},\quad y=\color{red}{x^{x^{x^{.^{.^.}}}}} \therefore \quad y=x^y$$ Thanks blackpenredpen! Now take the natural log of both sides and implicity derive the equation: $$\frac{\mathrm d}{\mathrm dx}\left(\ln(y)=y\cdot \ln(x)\right) \Rightarrow \frac{1}{y}\frac{\mathrm dy}{\mathrm dx} = \frac{y}{x} + \ln x \frac{\mathrm dy}{\mathrm dx}$$ Then solve for $\frac{\mathrm dy}{\mathrm dx}$ to get the following equation: $$\frac{\mathrm dy}{\mathrm dx}=\frac{y^2}{x(1-y\cdot \ln x)} = \frac{\left(x^{x^{x^{.^{.^.}}}}\right)^2}{x(1-x^{x^{x^{.^{.^.}}}}\cdot \ln x)}$$

II) Now lets determine the derivative of the nth tetration of x or ${}^nx$ by looking at the trivial case of ${}^3x$ or $x^{x^x}$: $$\frac{\mathrm d}{\mathrm dx}({}^3x)=\frac{\mathrm d}{\mathrm dx}(e^{\ln({}^3x)})=\frac{\mathrm d}{\mathrm dx}(e^{{}^2x\cdot \ln x})$$ Now we get to the actual derivation with chain rule and product rule: $$=e^{{}^2x\cdot \ln x}\left(\frac{{}^2x}{x} + \ln x \cdot \frac{\mathrm d}{\mathrm dx}\left({}^2x\right)\right)$$ After deriving $\frac{\mathrm d}{\mathrm dx}({}^2x)$ and $\frac{\mathrm d}{\mathrm dx}({}^1x)$ (which is just 1) and using the substitution $e^{{}^2x\cdot \ln x} = {}^3x$ we get: $$\frac{\mathrm d}{\mathrm dx}({}^3x)={}^3x\left(\frac{{}^2x}{x} + \ln x\Bigl({}^2x(\frac{^1x}{x}+\ln x)\Bigr)\right)$$ which upon further review can be summarized with summation and product notation: $$=\frac{1}{x}\left({}^3x{}^2x\right)+\frac{\ln x}{x}\left({}^3x{}^2x{}^1x\right)+\frac{\ln^2x}{x}\left({}^3x{}^2x{}^1x{}^0x\right)=\sum_{i=1}^{3}\frac{(\ln x)^{i-1}}{x}\prod_{j=0}^{i}{}^{3-j}x$$ where 3 = n for the non-trivial derivative of $\; {}^nx$ (shoutout to Prateek Bhurkay for the proof).

Here is where my question begins for the people that didn't need the background info. I want to find a way to show that, for n = $\infty$, the above summation is equivalent to the derivation in part 1 which uses implicit differentiation. Symbolically, $$\sum_{i=1}^{\infty}\frac{(\ln x)^{i-1}}{x}\prod_{j=0}^{i}{}^{\infty-j}x = ... = \frac{\left(x^{x^{x^{.^{.^.}}}}\right)^2}{x(1-x^{x^{x^{.^{.^.}}}}\cdot \ln x)} \qquad \checkmark$$

I really have now clue where to even start with this. When I was researching this I kept running into the Lambert W function so maybe that has something to do with it (probably not). Hopefully I explained my problem well enough so that one of you smart people might be able to help me. This is my first post here so if you have any suggestions of how I should approach posts in the future please let me know. Thanks!

$\endgroup$
  • $\begingroup$ Well, $\infty-j=\infty$. So your last formula is just a simple geometric series. $\endgroup$ – metamorphy Aug 2 at 16:01
  • $\begingroup$ So you're saying $\prod_{j=0}^{i}{}^{\infty-j}x$ simplifies to ${}^\infty x$ . That makes sense I guess but that also brings up another question. How should $({}^\infty x)^2$ be represented? Usually you just multiply the top exponent by 2 but there is no top term so I'm not sure what to make of this or if $({}^\infty x)^2$ simplifies to ${}^\infty x$. That geometric progression idea is certainly simplifies things, thanks metamorphy! $\endgroup$ – John Preston Aug 3 at 2:50
  • $\begingroup$ $\prod_{j=0}^{i}{}^{\infty-j}x$ simplifies to $({}^\infty x)^{i+1}$. So, with $({}^\infty x)^2/x$ taken out, you're left with $\sum_{i\geq 1}({}^\infty x\ln x)^{i-1}$. $\endgroup$ – metamorphy Aug 3 at 3:21
  • $\begingroup$ Now we have $\sum_{i=1}^{\infty}(x^{x^{x^{.^{.^.}}}}\cdot \ln x)^{i-1}=\frac{1}{1-x^{x^{x^{.^{.^.}}}}\cdot \ln x}$. So I guess a simple proof would to test that this equation is equivalent on bounds of where x converges for ${}^\infty x$ (although I would like something more concrete). We know ${}^\infty x$ converges when $x\in[1/e^e,e^{1/e}]$ and that an infinite geometric series only converges when the ratio r$=(x^{x^{x^{.^{.^.}}}}\cdot \ln x)^{i-1}$ is within the bounds $(-1,1)$ with a sum $S_n=\frac{a_1}{1-r}$ ($a_1$ is 1 since we have no principle term). No room to type cont. in next post $\endgroup$ – John Preston Aug 3 at 13:57
  • $\begingroup$ Cont. post) Therefore the summation only converges to a finite value upon the interval $[1/e^e,1)$. However the right side of the equation produces a finite value when $x\in[1/e^e,e^{1/e}]$. This disparity appears to make the two formulas not equal as (in the interval $[1,e^{1/e}]$) the summation produces an infinite value when the other side produces a finite value. Am I flawed in this reasoning somehow? $\endgroup$ – John Preston Aug 3 at 14:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.