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For a long time I've been messing around with the factorial function (I've discovered a way to define it for all real numbers!! but that's beside the point.), and in a math "project" (I guess you could call it a project), the following function came up:

$$f(x) = x!^{1/x}$$

If you put that into Desmos, you can see that as $x$ becomes greater and greater, $f(x)$ seems to become more and more like a line, like there is a slanted asymptote. My goal was then to find the slope of that asymptote. I can approximate it by taking the derivative of $f(x)$, and plugging in an arbitrarily large $x$.

In math terms, the slope of the asymptote is defined as follows:

$$m = \lim_{x\to\infty} \frac d{dx} x!^{1/x}$$

Through Wolfram Alpha's magic, I have learned that the slope of that asymptote is $1/e$. However, I have not been able to prove that the derivative of $f(x)$ converges at that value, or even that it converges at all.

Could someone please show me a proof that it converges?

Thanks!

Polygon
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    Have you heard of Sterling's Approximation? – JimmyK4542 Aug 11 '16 at 02:49
  • When you say you extended it to all of $\Bbb R$, do you mean differently from the Gamma function? – Gregory Grant Aug 11 '16 at 03:08
  • Playing around with Mathematica, your definition seems to coincide with the Gamma function definition for $x>0$. (For $x<0$ it seems to give the absolute value of the Gamma function, but that's outside the scope of your question.) – Semiclassical Aug 11 '16 at 03:14
  • Why don't you want the gamma function in an answer? – Caleb Stanford Aug 11 '16 at 03:35
  • My mistake. It turns out that my definition is the gamma function. I thought that the gamma function could only be defined one way. I will change it when I have time. – Polygon Aug 11 '16 at 03:38
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    Your definition (as well as the gamma function), is not defined "for all real numbers." It has singularities at the negative integers. – shalop Aug 11 '16 at 03:51
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    @JimmyK4542 I've heard of Stirling's approximation, and of Sterling Archer, but not of Sterling's approximation. – Pedro Aug 11 '16 at 04:08
  • ^Wow, I can't spell. Thanks for catching that. – JimmyK4542 Aug 11 '16 at 05:15
  • Everything you need to solve this rigorously is on the WP pages about the Gamma function and the polygamma function. Note first that $$\log f(x)=\frac1x\log\Gamma(x+1)$$ hence $$f'(x)=f(x)\cdot\frac1x\cdot\left(-\frac1x\log\Gamma(x+1)+\frac{\Gamma'(x+1)}{{}{}{}{}\Gamma(x+1)}\right)$$ Furthermore, a simple version of the asymptotics of the polygamma function $\psi^{(0)}$ is that $$\frac{\Gamma'(x+1)}{\Gamma(x+1)}=\log x+o(1)$$ and a simple consequence ... – Did Aug 14 '16 at 12:02
  • ... of Stirling formula is that $$\log\Gamma(x+1)=x\log x-x+o(x)$$ hence $$\frac1x\cdot f(x)=\frac1e+o(x)$$ and finally, as you guessed, $$\lim_{x\to\infty}f'(x)=\frac1e$$ – Did Aug 14 '16 at 12:02
  • @Polygon. Please, add an answer with the solution. – Claude Leibovici Aug 17 '16 at 03:02
  • @ClaudeLeibovici I added it. It's really long, though, but I tried to make it as understandable as possible. – Polygon Aug 18 '16 at 03:00
  • Usually one would define $x!$ to be $\displaystyle \int_0^\infty t^x e^{-t}, dt. \qquad$ – Michael Hardy Aug 18 '16 at 03:00
  • @MichaelHardy Yeah, that's how I see it most often. The reason I don't define it like that is because no one has ever been able to show my why it works. That's why I came up with my own definition. – Polygon Aug 18 '16 at 03:02
  • @Polygon: Why it works is integration by parts. $\qquad$ – Michael Hardy Aug 18 '16 at 03:03
  • @Polygon This is odd: 1. You "do not define x! as" everybody else does because... what, already? You say: "no one has ever been able to show m(e) why it works" Really? Proof is a one-liner. 2. You keep "accepted" an answer providing at best heuristics. 3. You simply overlook my comments, although they provide a complete, mathematically rigorous, answer. 4. You end up posting some highly dubious "solution" (the step "I can replace n with x as x becomes unbounded" being a hair-raiser to any decent mathematician...). // Any reasons for these? Sure this is how you want to use the site? – Did Aug 18 '16 at 07:36
  • @Did I'm sorry. I'll try to explain myself a bit. 1. I decided to keep my different definition of x! because it seemed to have no impact on anything, even if it was unnecessary. I've deleted it now, because, well, it's unnecessary. The highest math class I've ever had was Calc I, so I had never before heard of integration by parts. I'm not saying that I don't trust the usual definition, or that I think mine is better, but believe me wen I say that I've searched for proofs that the usual one works, and haven't gotten a good one until now. 2. tbh I don't even remember accepting that answer... – Polygon Aug 18 '16 at 17:34
  • I un-accepted it now. 3. I agree that I should have payed more attention to your comments. I only have a phone most of the time, and the LaTeX was all messed up on it, so I ignored it, but it seems to work fine on a computer. 4. After looking at the statement about x and n, I can see why it is incorrect. I'm sorry about that. I've deleted my answer until I can think of a better one (Which is unlikely). In the future, I will be more careful about what I write. – Polygon Aug 18 '16 at 17:37

1 Answers1

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As JimmyK4542 commented, the simplest could be Stirling approximation.

Let $$y=(x!)^{\frac 1x}\implies \log(y)=\frac 1x\log(x!)$$ $$x!\approx \sqrt{2\pi x}\left(\frac xe \right)^x\implies \log(x!)\approx (x+\frac 12)\log(x)-x+\log(\sqrt{2\pi})$$ So, $$\log(y)=\log (x)+\frac{\log (2 \pi x)}{2 x}-1$$ Differentiating both sides $$\frac {y'}y=\frac{2 x-\log (2 \pi x)+1}{2 x^2}$$ and since $$y\approx (2\pi x)^{\frac 1{2x}}\frac x e$$ then $$y'\approx (2\pi x)^{\frac 1{2x}}\frac x e \frac{2 x-\log (2 \pi x)+1}{2 x^2}=(2\pi x)^{\frac 1{2x}}\frac{2 x-\log (2 \pi x)+1}{2 e x}$$ For large values of $x$, the first term tends to $1$ and the second behaves as $\frac{2x}{2ex}=\frac 1 e$.

For illustration purposes, at $x=10$, the exact derivative would be $\approx 0.381007$ while the above formula gives $\approx 0.381443$ while $\frac 1e\approx 0.367879$.

Claude Leibovici
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  • Thanks! I don't quite understand why Stirling's Approximation works, though? I've looked at Wikipedia and Wolfram Mathworld, and I can't make sense of any of it. Do you think you could explain it for me? – Polygon Aug 11 '16 at 15:21
  • You are very welcome. Let me give you a (good) trick : whenever you see a factorial in any problem related to limits, just think Stirling + Stirling + Stirling + ... – Claude Leibovici Aug 11 '16 at 15:31
  • ...Except that approximations of a function, such as Stirling formula, cannot in any way provide a control of their derivative -- hence I am afraid that nothing in this post after "Differentiating both sides" is proven. – Did Aug 14 '16 at 11:41
  • @Did. I fully agree with your statement about the lack of control of the derivative. – Claude Leibovici Aug 15 '16 at 03:50