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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:

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

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    @WillJagy It does. Thanks for you help. Did you receive my response to your mail?2012-02-25

1 Answers 1

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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.

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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_8859-1.2012-02-25