The problem:

To make things easier on us so we don't have to use an underbrace, I'm going to declare:

$$\Psi(x,n) := \underbrace{x^{x^{x^{x^{\dots}}}}}_n$$

And we evaluate $f$ from the topright to the bottom left (like a calculator does).

Question: What is $$\Psi'(x,n) = \frac{\partial}{\partial x}\Psi(x,n) = \text{?}$$

My attempt:

Since we know that

$$\begin{align}\frac{\partial}{\partial x}f(x)^x &= \frac{\partial}{\partial x} e^{x \ln f(x)} = \\ &= e^{x \ln f(x)} \left(\frac{x f'(x)}{f(x)} + \ln f(x)\right) = \\ &= f(x)^x \left(\frac{x f'(x)}{f(x)} + \ln f(x)\right),\end{align}$$

We can substitute

$$f(x) := \Psi(x,n-1).$$

So we get

$$\begin{align} \frac{\partial}{\partial x}\Psi(x,n) &= \frac{\partial}{\partial x}\Psi(x,n-1)^x = \\ &= \Psi(x,n-1)^x \left(\frac{x \cdot \frac{\partial}{\partial x} \Psi(x,n-1)}{\Psi(x,n)} + \ln \Psi(x,n-1)\right) = \\ &= \Psi(x,n) \left(\frac{x \cdot \frac{\partial}{\partial x} \Psi(x,n-1)}{\Psi(x,n)} + \ln\underbrace{x^{x^{x^{x^{\dots}}}}}_{n-1}\right) = \\ &= x \cdot \frac{\partial}{\partial x} \Psi(x,n-1) + \Psi(x,n)x^{n-2}\ln x .\end{align}$$

We were able to reduce $n$ by one and get the derivative for $\Psi(x,n)$ in terms of $\Psi(x,n-1)$. If we keep repeating this for $n-1,n-2,\dots,2,1$, we should get the full derivative. However, this is what I'm not quite sure of:

$$\begin{align} \frac{\partial}{\partial x}\Psi(x,n) &= x \cdot \frac{\partial}{\partial x} \Psi(x,n-1) + \Psi(x,n)x^{n-2}\ln x = \\ &= x \cdot \frac{\partial}{\partial x} \Psi(x,n-2) + x \cdot \Psi(x,n-1)x^{n-3}\ln x + \Psi(x,n)x^{n-2}\ln x = \dots \\ \dots &= x \cdot \frac{\partial}{\partial x} \Psi(x,1) + \Psi(x,2)x^{n-2}\ln x + \dots + \Psi(x,n)x^{n-2}\ln x = \\ &= x + \ln x \cdot x^{n-2} \cdot \sum_{k=2}^n \Psi(x,k) = x + \ln \Psi(x,n-1) \sum_{k=2}^n \Psi(x,k) \end{align}$$

So rewriting it with the underbrace notation we get: $$\frac{\partial}{\partial x} \underbrace{x^{x^{x^{x^{\dots}}}}}_n = x + \ln \underbrace{x^{x^{x^{x^{\dots}}}}}_{n-1} \sum_{k=2}^n \underbrace{x^{x^{x^{x^{\dots}}}}}_k$$


$1$. Is this correct? Are my calculations right?

$2$. Is there a better way to write the final answer?

$3$. Is there a name for this $(\Psi)$ function?

$4$. Can we generalize $\Psi(x,n)$ for rational (and perhaps irrational) $n$'s?

$5$. What could be the derivative of $\Psi(x,n)$ for such rational / irrational $n$'s?

  • 1
    $\begingroup$ I would suggest looking into tetration. $\endgroup$ – Peter Foreman Apr 12 at 18:23
  • $\begingroup$ @PeterForeman Thank you! I was searching all over the internet but Google didn't like my search term of "x^x^x..." .I guess you just really need to know the name of things to be able to find them :) $\endgroup$ – Daniel P Apr 12 at 18:25
  • $\begingroup$ I don't think it is correct. $\endgroup$ – Thomas Andrews Apr 12 at 18:58
  • $\begingroup$ The problem is that if $\Psi(x,n)$ is evaluated from the top right to left, your recursion is $$\Psi(x,n)=x^{\Psi(x,n-1)}$$ and not $$\Psi(x,n)=(\Psi(x,n-1))^x.$$ With your recursion, $\Psi(x,n)=x^{x^{n-1}},$ which is much easier to take the derivative of. But this is the left-to-right interpretation. $\endgroup$ – Thomas Andrews Apr 12 at 19:01

I think your answer is incorrect. Your recursion $$\Psi(x,n)=\left(\Psi(x,n-1)\right)^x$$ defines the left-to-right interpretation of your expression. The value of the left-to-right interpretation yields a simple $\Psi(x,n)=x^{(x^{n-1})}.$ The derivative in this left-to-right case is $$\frac{\partial}{\partial x}\Psi(x,n)=\Psi(x,n)\left(\frac{1}{x}x^{n-1}+(n-1)x^{n-2}\log x \right)=\Psi(x,n)x^{n-2}\left(1+(n-1)\log x\right)$$

But your question asked for a right-to-left reading, which yields the recursion: $$\Psi(x,n)=x^{\Psi(x,n-1)}$$

The derivative here is going to be messier.

Write it as $\Psi_n(x)$ for slight ease. Then, for $n=0,$ $\Psi_0(x)=1$ and for $n>0,$ $\Psi_n(x)=x^{\Psi_{n-1}(x)}=e^{\log(x)\Psi_{n-1}(x)}$. Then we apply the chain rule: $$\Psi_n'(x)=\left(\frac{1}{x}\Psi_{n-1}(x)+\log(x)\Psi_{n-1}'(x))\right)\Psi_n(x)$$

If we then write this in terms of $$\Phi_n(x)=\frac{\Psi_n'(x)}{\Psi_n(x)}$$ (which we can see as the derivative of $\log\Psi_n(x).$) Then $\Phi_0(x)=0$ and when $n>0$:


This is going to get ugly, but:

$$\begin{align}\Phi_n(x)&=\Psi_{n-1}(x)\left(\frac{1}{x}+\log(x)\Psi_{n-2}(x)\left(\frac{1}{x}+\Psi_{n-3}(x)\log(x)\left(\frac{1}{x}+\cdots\right)\right)\right)\\ &=\frac{1}{x}\sum_{k=1}^{n}\left(\log^{k-1}(x)\Psi_{n-1}(x)\Psi_{n-2}(x)\cdots \Psi_{n-k}(x)\right) \end{align}$$

Then $$\Psi_n'(x)=\Psi_n(x)\Phi_n(x)=\frac{1}{x}\sum_{k=1}^{n}\left(\log^{k-1}(x)\Psi_n(x)\Psi_{n-1}(x)\Psi_{n-2}(x)\cdots \Psi_{n-k}(x)\right)$$

For example, $$\begin{align}\Psi_2'(x)&=\frac{1}{x}\Psi_2(x)\Psi_1(x)\left(1+ \log(x)\Psi_0(x)\right)\\ &=\frac1x \cdot x^x\cdot x(1+\log(x)) \\&=x^x \left(1+\log(x)\right) \end{align}$$

and $$\begin{align}\Psi_3'(x) &= \frac{1}{x}\Psi_3(x)\Psi_2(x)\left(1+\log(x)\Psi_1(x)+\log^2(x)\Psi_1(x)\Psi_0(x)\right)\\ &=\frac{1}{x}\cdot x^{x^x}\cdot x^x\left(1+x\log(x)+x\log^2(x)\right) \end{align}$$


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