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I'm watching this video, where D. Knuth explains the connection of $\pi$ and factorials, and other matters (it is very interesting). Almost at the end of the talk he says the area of the superellipse

$$x^{\frac{1}{\alpha}}+y^{\frac{1}{\alpha}}=1$$

is given by

$$A(\alpha) = \frac{2 \alpha \cdot\Gamma{(\alpha)}^2}{\Gamma{(2 \alpha)}}$$ whic would be

$$A(\alpha) = 2 \alpha B(\alpha,\alpha) = 2 \alpha\int_0^1(1-u)^{\alpha-1}u^{\alpha-1}du $$

I was trying to check this so I put

$$A\left( \alpha \right) = \int\limits_0^1 {{{\left( {1 - {x^{1/\alpha }}} \right)}^\alpha }dx} $$

Now let $x = {u^\alpha }$

$$A\left( \alpha \right) = \alpha \int\limits_0^1 {{{\left( {1 - u} \right)}^\alpha }{u^{\alpha - 1}}du} $$

What's going on?

The $2$ in Knuth's formula probably comes from the fact he considers the full figure and not only a fourth, as I am, but I don't know what I'm doing wrong here. If you want to check, it is at $1:21:00$ aproximately.

PS: Just as a curiosity, does Knuth have a stutter or is it he is just thinking about too many things in too little time?

So it was just OK:

$$A\left( \alpha \right) = \alpha \int\limits_0^1 {{{\left( {1 - u} \right)}^\alpha }{u^{\alpha - 1}}du} = \frac{{\alpha \Gamma \left( {\alpha + 1} \right)\Gamma \left( \alpha \right)}}{{\Gamma \left( {2\alpha + 1} \right)}} = \frac{{\Gamma {{\left( {\alpha + 1} \right)}^2}}}{{\Gamma \left( {2\alpha + 1} \right)}}$$

J. M. ain't a mathematician
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Pedro
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  • I did the case $\alpha = 2,$ a sort of nonconvex star shape. In the first quadrant it is just $y = 1 + x - 2 \sqrt x,$ with area $1/6,$ so the whole figure has area $2/3.$ Agrees with Dirichlet. – Will Jagy Feb 25 '12 at 20:37
  • @WillJagy It does. Thanks for you help. Did you receive my response to your mail? – Pedro Feb 25 '12 at 21:05

1 Answers1

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I wouldn't call it a stutter. It strikes me as closer to someone very famous trying to seem folksy; not quite the same as saying "um," more "I'm as uncertain as you."

Dirichlet wrote a generalization of this in Über eine neue Methode zur Bestimmung vielfacher Integrale. Collected Works, volume 1, page 389.

For integrating the constant 1 on $$ x \geq 0, \; y \geq 0, \; x^{1/\alpha} + y^{1/\alpha} \leq 1, $$ the result is $$ \frac{\alpha^2 \Gamma(\alpha)^2}{\Gamma(1 + 2 \alpha)} = \frac{ \Gamma(1 + \alpha)^2}{\Gamma(1 + 2 \alpha)}, $$ multiply by 4 to get the whole thing, $$ \frac{ 4 \; \Gamma(1 + \alpha)^2}{\Gamma(1 + 2 \alpha)}. $$ Sample points, $\alpha = 1/2$ is the circle, $\Gamma(3/2) = (1/2) \sqrt \pi $ and $\Gamma(2) = 1,$ so we get $\pi.$ With $\alpha = 1,$ we have a tilted square, $\Gamma(3) = 2,$ area is indeed $2.$ As $\alpha \rightarrow \infty,$ area goes to $0,$ it is not necessary to quote Stirling's to believe that $(\alpha!)^2 / (2 \alpha)! \rightarrow 0.$ Finally, as $\alpha \rightarrow 0,$ we approach the entire square, and the area approaches $4.$

I understand part of the Dirichlet technique is in Whittaker and Watson.

joriki
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Will Jagy
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  • I see. So I was on the right path. It just looked like a stutter, and although I watched the whole hour and 40 minutes, it made me a little nervous while watching... he just changed is words too many times. However, I was delighted by the exposition. – Pedro Feb 25 '12 at 20:20
  • @joriki, very nice, I have never known how to do umlauts. I do know better, but the article title is spelled incorrectly in my notes, including an illegible final e on integrale. I made my best estimate on which words are nouns. – Will Jagy Feb 25 '12 at 21:58
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    Yes, you got the noun part right. I couldn't tell you how to produce umlauts; I just press the key on my German keyboard :-) But you can always copy them from a German text on the net, e.g. the German Wikipedia, or from an ISO-Latin table, e.g. http://en.wikipedia.org/wiki/ISO/IEC_8859-1. – joriki Feb 25 '12 at 22:25
  • It's straightforward to compute asymptotics that might also be of interest: As $\alpha \searrow 0$, the area is $4 - \frac{2 \pi^2}{3} \alpha^2 + R_0(\alpha)$, where $R_0(\alpha) \in O(\alpha^3)$. As $\alpha \to +\infty$, the area is $\frac{1}{4^\alpha}\left(\sqrt{4\pi\alpha} + R_\infty(\alpha)\right)$, where $R_\infty(\alpha) \in O\left(\frac1{\sqrt\alpha}\right)$. – Travis Willse Jul 06 '23 at 06:05