What is the derivative of $$x!^{x!^{x!^{x!^{x!^{x!^{x!^{.{^{.^{.}}}}}}}}}}$$

My effort:

Let $$g(x)=x!^{x!^{x!^{x!^{x!^{x!^{x!^{.{^{.^{.}}}}}}}}}}\implies g(x)=x!^{g(x)}$$ Taking natrual log on both sides, $$\ln(g(x))=g(x)\cdot\ln(x!)$$ Differentiating, $$\frac{1}{g(x)}\cdot g'(x)=g'(x)\cdot\ln(x!)+g(x)\cdot\frac{1}{x!}\cdot x!\cdot\psi^{(0)}(x+1)$$ $$\implies g'(x)\left[\frac{1}{g(x)}-ln(x!)\right]=g(x)\cdot\psi^{(0)}(x+1)$$ So does isolating $g'(x)$ give me the correct solution? If not, how can I solve for the differential?

Edit: The gamma function is indeed implicitly assumed when the factorial function is used.

  • $\begingroup$ @Brenton no, he isnt. He multiplies by the derivative to the right, which is correct. He says the derivative is $\frac{1}{x!}\cdot x!\cdot\psi^{(0)}(x+1)$ $\endgroup$ Dec 27 '15 at 5:37
  • $\begingroup$ @BrevanEllefsen Sorry, I missed the $\psi$ term. $\endgroup$
    – Brenton
    Dec 27 '15 at 5:38
  • 1
    $\begingroup$ What is $x!$ for $x$, say, $\sqrt2$? $\endgroup$ Dec 27 '15 at 5:40
  • $\begingroup$ @PrzemysławScherwentke $x!=\Gamma (x+1)$ (Gamma function). $\endgroup$
    – user236182
    Dec 27 '15 at 5:40
  • $\begingroup$ @PrzemysławScherwentke he clearly defines it as $\Gamma(\sqrt{2}+1)$... he uses the derivative of the gamma function in his proof $\endgroup$ Dec 27 '15 at 5:41

Your definition of $g(x)$ does not make sense for $x! \gt e^{(1/e)} \approx 1.44467$ because it does not converge as seen in the answer to this question and this question. It is less than this in about the ranges $-4.970 \lt x \lt -4.103$ and $-0.380 \lt x \lt 1.614$. Within those ranges, you are doing fine.

  • $\begingroup$ Clear. Thanks for the heads up! $\endgroup$
    – Kugelblitz
    Dec 27 '15 at 11:21
  • $\begingroup$ What if he wanted to allow complex numbers for $x$? $\endgroup$
    – GEdgar
    Dec 27 '15 at 15:04
  • $\begingroup$ @GEdgar: I don't know where the tower converges in the complex plane, but presumably somebody does. Once it converges you have to worry about branches of the log function as well. $\endgroup$ Dec 27 '15 at 15:07
  • $\begingroup$ @GEdgar if$$|W(-\ln(x!))|<1$$it converges, where $W$ is the Lambert $W$ function. math.stackexchange.com/questions/1674027/… $\endgroup$ Jun 30 '16 at 16:29

I would say that everything looks fine, as long as you define that $x! = \Gamma(x+1)$.

  • $\begingroup$ That indeed is my implicit assumptions. Thanks! $\endgroup$
    – Kugelblitz
    Dec 27 '15 at 11:24

If you define $x!$ to be equal to $\int_{0}^{\infty}e^{-t}t^x \mathrm{d}t$, then you can say the following:

$y = (x!)^{y} = (\int_{0}^{\infty}e^{-t}t^x \mathrm{d}t)^y$

Raising both sides to the power of $\frac{1}{y}$:

$y^\frac{1}{y} = \int_{0}^{\infty}e^{-t}t^x \mathrm{d}t$.

Since the derivative of $y^\frac{1}{y}$ with respect to $y$ is $-y^\frac{1}{y}(\ln(y)-1)$, and by the Leibniz rule, the derivative of $\int_{0}^{\infty}e^{-t}t^x \mathrm{d}t$ with respect to $x$ is $\int_{0}^{\infty}\ln(t)e^{-t}t^x \mathrm{d}t$, implicit differentiation tells us that:

$-y^\frac{1}{y}(\ln(y)-1)\mathrm{d}y = (\int_{0}^{\infty}\ln(t)e^{-t}t^x \mathrm{d}t)\mathrm{d}x$

We can solve for $\frac{\mathrm{dy}}{\mathrm{dx}}$:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\int_{0}^{\infty}\ln(t)e^{-t}t^x \mathrm{d}t}{-y^\frac{1}{y}(\ln(y)-1)}$

We can simplify the denominator. Since $y=(x!)^y$, we can conclude that $y^\frac{1}{y}=x!$ and also that $\ln(y)=y\ln(x!)=x!^{x!^{x!^{.^{.^.}}}}\ln(x!)$. Substituting this into the denominator gives:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\int_{0}^{\infty}\ln(t)e^{-t}t^x \mathrm{d}t}{-x!(x!^{x!^{x!^{.^{.^.}}}}\ln(x!)-1)}$

And there's the derivative.


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