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I am trying to find a general derivative for the function: $f(x)=x^{x^{x^{...^{x}}}}$however to do that I must find $f^{\prime }$ and $f^{\prime \prime}$...etc. I am now trying to write down a general expression for $f^{\prime \prime}$ and I have stumble opon the series in the picture below, I wonder, do anyone have an idea of what sort of expantion it migth be ? I have tried breaking it down into three different sums, but I wasn't very successful!

EDIT

So since this post has gotten some attention and the title has been change I will tell you what I am on about, although I already have a post just for the question in the title from before. I am as I said earlier trying to find a general derivative for the function $f(x)=x^{x^{x^{...^{x}}}}$ where $n$ is an arbitrary number, natural number and greater or equal to 2. I have already made success when I tried to find the first derivative, and I have also found some interesting pattern emerge when the derivatives as well as n incresses. But my question here was if anyone knew what this expation might be or look similar to be, since I have yet to be lernt series etc. I did not know that the question needed such background since it's straight a forward question.

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    $\begingroup$ what brings about this horror? $\endgroup$
    – Betty Mock
    Dec 24 '13 at 1:57
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    $\begingroup$ I am trying to find the second derivative of the function: $f(x)=x^{x^{x^{...^{x}}}}$ $\endgroup$ Dec 24 '13 at 2:00
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    $\begingroup$ That's the sort of thing you should have said in the first place. Hiding context from readers will only make things harder, not easier! $\endgroup$
    – anon
    Dec 24 '13 at 2:04
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    $\begingroup$ @Axcelneo : I would recommend Googling "tetration" and seeing what has been done with this already. You don't want to reinvent the wheel. tomasz makes a good point too. $\endgroup$ Dec 24 '13 at 2:38
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    $\begingroup$ It's Knuth's up-arrow notation. $\endgroup$ Dec 24 '13 at 12:28
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Note: This answer ended up much longer than I intended (supposed to be a "hint"). However, it was too fun to not investigate further, hence the large amount of edits :)

Hint: The derivative of $x\uparrow\uparrow n$ can be expressed in terms of the following recurrence relation $$\frac{d}{dx}\left(x\uparrow\uparrow n\right)=x\uparrow\uparrow n \left(\frac{x\uparrow\uparrow (n-1)}{x}+\log x\frac{d}{dx}\left(x\uparrow\uparrow (n-1)\right)\right)$$ You can use this relation to find a closed form solution

Edit 1: It turns out the closed form is quite horrid so I'll save you the bother and write it out here. You can verify it with mathematical induction if you ever get the time. $$\frac{d}{dx}\left(x\uparrow\uparrow n\right)=\left[\sum _{i=0}^{n-3} x^{-1+\sum _{k=1}^{i+2} x\uparrow\uparrow(n-k)}\log ^i(x) \right]+\left(\log ^{n-2}(x)+\log ^{n-1}(x)\right) x^{\sum _{k=1}^{n-1} x\uparrow\uparrow k}$$

Edit 2: I had some time on my hands so here is a proof of the above recurrence relation: \begin{align*}\frac{d}{dx}\left(x\uparrow\uparrow n\right) &=\frac{d}{dx}\left(x^{x\uparrow\uparrow (n-1)}\right)\\ &=\frac{d}{dx}\left(e^{x\uparrow\uparrow (n-1)\cdot\log x}\right)\\ &=e^{x\uparrow\uparrow (n-1)\cdot\log x}\left(\frac{x\uparrow\uparrow (n-1)}{x}+\log x\cdot\frac{d}{dx}\left(x\uparrow\uparrow (n-1)\right)\right)\\ &=x\uparrow\uparrow n\left(\frac{x\uparrow\uparrow (n-1)}{x}+\log x\cdot\frac{d}{dx}\left(x\uparrow\uparrow (n-1)\right)\right)\end{align*}

Edit 3: After playing around with the formula a little, it turned out that the closed form wasn't so bad after all. The odd looking $\log$'s don't have to awkwardly stand apart from the sum if we work with the convention that $x\uparrow\uparrow0=1$ and $x\uparrow\uparrow-1=0$. By modifying the equation accordingly we get the following much tidier formula: $$\frac{d}{dx}\left(x\uparrow\uparrow n\right)=f(n)=\frac{1}{x}\sum _{i=1}^n x^{\sum _{k=n-1-i}^{n-1} x\uparrow\uparrow k}\cdot\log ^{i-1}(x)$$

With this "new" formula we can now in fact more easily verify that $\frac{d}{dx}\left(x\uparrow\uparrow n\right)$ is indeed equal to $f(n)$. To do this we simply checking that $f(n)$ satisfies the recurrence relation for $\frac{d}{dx}\left(x\uparrow\uparrow n\right)$ and that it works for n=1.


Proof:

(1) If $n=1$ we have \begin{align*}f(1)&=\frac{1}{x}\sum _{i=1}^1 x^{\sum _{k=-i}^{0} x\uparrow\uparrow k}\cdot\log ^{i-1}x \\ &=\frac{1}{x}\left(x^{\sum _{k=-1}^{0} x\uparrow\uparrow k}\cdot\log ^{1-1}x\right) \\ &=\frac{1}{x}\cdot x^{(x\uparrow\uparrow -1)+(x\uparrow\uparrow 0)} \\ &=\frac{1}{x}\cdot x^{0+1}=1=\frac{d}{dx}x=\frac{d}{dx}\left(x\uparrow\uparrow 1\right)\end{align*} Therefore $\frac{d}{dx}\left(x\uparrow\uparrow n\right)=f(n)$ holds for $n=1$.

(2) Now we want to check that $f$ satisfies the recurrence relation:

\begin{align*}f(n+1)&=\frac1x \left(\sum _{j=1}^{n+1} x^{\sum _{k=n-j}^{n} x\uparrow\uparrow k}\cdot\log ^{j-1}(x)\right)\\ &=\frac1x \left(\sum _{i=0}^n x^{\sum _{k=n-1-i}^{n} x\uparrow\uparrow k}\cdot\log ^{i}(x)\right) \hspace{5mm} \{\text{reindexing such that }i=j-1\} \\ &=\frac1x \left(x^{\sum _{k=n-1}^{n} x\uparrow\uparrow k}+\left(\sum _{i=1}^n x^{\sum _{k=n-1-i}^{n} x\uparrow\uparrow k}\cdot\log ^{i}(x)\right)\right) \\ &=\frac1x \left(x^{x\uparrow\uparrow (n-1)+x\uparrow\uparrow n}+x^{x\uparrow\uparrow n}\left(\sum _{i=1}^n x^{\sum _{k=n-1-i}^{n-1} x\uparrow\uparrow k}\cdot\log ^{i}(x)\right)\right) \\ &=\frac1x \left((x\uparrow\uparrow n)\cdot (x\uparrow\uparrow (n+1))+x\uparrow\uparrow (n+1)\left(\sum _{i=1}^n x^{\sum _{k=n-1-i}^{n-1} x\uparrow\uparrow k}\cdot\log ^{i}(x)\right)\right) \\ &=\frac{x\uparrow\uparrow (n+1)}{x} \left(x\uparrow\uparrow n+\sum _{i=1}^n x^{\sum _{k=n-1-i}^{n-1} x\uparrow\uparrow k}\cdot\log ^{i}(x)\right)\\ &=x\uparrow\uparrow (n+1) \left(\frac{x\uparrow\uparrow n}{x}+\frac1x\sum _{i=1}^n x^{\sum _{k=n-1-i}^{n-1} x\uparrow\uparrow k}\cdot\log ^{i}(x)\right) \\ &=x\uparrow\uparrow (n+1) \left(\frac{x\uparrow\uparrow n}{x}+\log x\left(\frac1x\sum _{i=1}^n x^{\sum _{k=n-1-i}^{n-1} x\uparrow\uparrow k}\cdot\log ^{i-1}(x)\right)\right)\\ &=x\uparrow\uparrow (n+1) \left(\frac{x\uparrow\uparrow n}{x}+\log x\cdot f(n)\right) \blacksquare\end{align*} Final edit: upon request by the OP I'll demonstrate how to calculate the second derivative too. I will use the defintion of $f(n)$ that I used above. \begin{align*} \frac{d^2}{dx^2}\left(x\uparrow\uparrow n\right)&=\frac{d}{dx}\left(\sum _{i=1}^n x^{-1+\sum _{k=n-1-i}^{n-1} x\uparrow\uparrow k}\cdot\log ^{i-1}(x)\right)\\ &=\sum _{i=1}^n \frac{d}{dx}\left(x^{-1+\sum _{k=n-1-i}^{n-1} x\uparrow\uparrow k}\cdot\log ^{i-1}(x)\right)\\ &=\sum _{i=1}^n \left((i-1)\cdot x^{-2+\sum _{k=n-1-i}^{n-1} x\uparrow\uparrow k}\cdot\log ^{i-2}(x)+\color{red}{\frac{d}{dx}\left(x^{-1+\sum _{k=n-1-i}^{n-1} x\uparrow\uparrow k}\right)}\color{black}{ \cdot\log ^{i-1}(x)}\right) \end{align*} Now, \begin{align*} \color{red}{\frac{d}{dx}\left(x^{-1+\sum _{k=n-1-i}^{n-1} x\uparrow\uparrow k}\right)}&=\frac{d}{dx}\left(e^{\left(-1+\sum _{k=n-1-i}^{n-1} x\uparrow\uparrow k\right)\log x}\right)\\ &=\frac{x^{\sum _{k=n-1-i}^{n-1} x\uparrow\uparrow k}}{x}\left(\frac{-1+\sum _{k=n-1-i}^{n-1} x\uparrow\uparrow k}{x}+\log x\left(\sum _{k=n-1-i}^{n-1} f(k)\right)\right) \end{align*} Substitute for the red parts and you are done.

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  • $\begingroup$ Well, it's the purly closed form I am out after. I have already found the first derivative my problem now is with the second one. $\endgroup$ Dec 24 '13 at 11:28
  • $\begingroup$ @Axcelneo, I have found a nicer closer form. I'll post it in the morning with a proof. $\endgroup$
    – E.O.
    Dec 24 '13 at 15:08
  • $\begingroup$ I am still however having difficulities with finding the second derivative of the function, which was my problem to begin with. $\endgroup$ Dec 25 '13 at 1:14
  • $\begingroup$ @Axcelneo, now that you have the derivative of $x\uparrow\uparrow n$ you can easily find the second derivative by several applications of the product rule and chain rule. I am slightly reluctant to make many more edits (excessive bumping), but if I get time, I'll add one last equation for the second derivative as well. $\endgroup$
    – E.O.
    Dec 25 '13 at 1:25
  • $\begingroup$ Would you mind showing exactly how you would go about finding it? $\endgroup$ Dec 25 '13 at 1:33
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In case of the infinite tower powers, the dérivatives can be expressed on closed form thanks to the Lambert's W function : http://mathworld.wolfram.com/LambertW-Function.html enter image description here

The second derivative is obtained thanks to the chain rule : enter image description here

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