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

Question

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.

@Brenton no, he isnt. He multiplies by the derivative to the right, which is correct. He says the derivative is $\frac{1}{x!}\cdot x!\cdot\psi^{(0)}(x+1)$ — Brevan Ellefsen, Dec 27 '15 at 05:37
@PrzemysławScherwentke he clearly defines it as $\Gamma(\sqrt{2}+1)$... he uses the derivative of the gamma function in his proof — Brevan Ellefsen, Dec 27 '15 at 05:41
^He probably is playing loose with the definition $\Gamma(x+1) = x!$. — JimmyK4542, Dec 27 '15 at 05:41
@JimmyK4542 True, but his steps seem fine as long as one defines the factorial that way. Integer values should match at the very minimum — Brevan Ellefsen, Dec 27 '15 at 05:41
@user236182 I know this relation. I was simply afraid that it may be a formal differentiation on naturals. — Przemysław Scherwentke, Dec 27 '15 at 05:44
@PrzemysławScherwentke perhaps, but it should hold anyway given that definition. I can't find anything wrong with it simply due to the use of the Gamma function — Brevan Ellefsen, Dec 27 '15 at 05:45
Right. I primarily wrote my working based on the fact that Euler's Gamma function and factorial function evaluate to the same value for positive integers (With a difference of 1 in the argument of course... since $\Gamma(n+1)=n!$) — Panglossian Oporopolist, Dec 27 '15 at 11:24

score 7 · Accepted Answer · edited Apr 13 '17 at 12:20

7

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.

edited Apr 13 '17 at 12:20

Community

1

answered Dec 27 '15 at 06:01

Ross Millikan

374,822

Clear. Thanks for the heads up! – Panglossian Oporopolist Dec 27 '15 at 11:21
What if he wanted to allow complex numbers for $x$? – GEdgar Dec 27 '15 at 15:04
@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. – Ross Millikan Dec 27 '15 at 15:07
@GEdgar if$$|W(-\ln(x!))|<1$$it converges, where $W$ is the Lambert $W$ function. http://math.stackexchange.com/questions/1674027/proving-that-if-w-ln-z-1-then-zzzz-is-convergent/1696352#1696352 – Simply Beautiful Art Jun 30 '16 at 16:29

score 4 · Answer 2 · answered Dec 27 '15 at 05:44

4

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

answered Dec 27 '15 at 05:44

Brevan Ellefsen

14,793

That indeed is my implicit assumptions. Thanks! – Panglossian Oporopolist Dec 27 '15 at 11:24

score 0 · Answer 3 · answered Jul 06 '18 at 11:50

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.

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

3 Answers3