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Oops, messed up Stirling's formula itself
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Douglas Zare
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$\pi$ shows up in at least two different ways related to factorials.


The $\pi$ in Stirling's Formula $n! \approx (\frac{n}{e})^n /\sqrt{2 \pi n}$$n! \approx (\frac{n}{e})^n \times\sqrt{2 \pi n}$ comes down to Wallis's Formula

$$\frac {\pi}{2} = \prod_{n=1}^\infty \frac{2n\times 2n}{2n-1 \times 2n+1}$$

which follows from the infinite product for sine

$$\sin x = x \prod_{n=1}^\infty \bigg( 1 - \frac {x^2}{\pi^2 n^2}\bigg)$$

evaluated at $x=\frac\pi2$. I guess that comes down to the circle after all.


Also, $(-1/2)! = \Gamma(1/2) = \sqrt \pi.$ That can be seen from Euler's reflection formula,

$$\Gamma(x)\Gamma(1-x) = \frac {\pi}{\sin \pi x}$$

$\pi$ shows up in at least two different ways related to factorials.


The $\pi$ in Stirling's Formula $n! \approx (\frac{n}{e})^n /\sqrt{2 \pi n}$ comes down to Wallis's Formula

$$\frac {\pi}{2} = \prod_{n=1}^\infty \frac{2n\times 2n}{2n-1 \times 2n+1}$$

which follows from the infinite product for sine

$$\sin x = x \prod_{n=1}^\infty \bigg( 1 - \frac {x^2}{\pi^2 n^2}\bigg)$$

evaluated at $x=\frac\pi2$. I guess that comes down to the circle after all.


Also, $(-1/2)! = \Gamma(1/2) = \sqrt \pi.$ That can be seen from Euler's reflection formula,

$$\Gamma(x)\Gamma(1-x) = \frac {\pi}{\sin \pi x}$$

$\pi$ shows up in at least two different ways related to factorials.


The $\pi$ in Stirling's Formula $n! \approx (\frac{n}{e})^n \times\sqrt{2 \pi n}$ comes down to Wallis's Formula

$$\frac {\pi}{2} = \prod_{n=1}^\infty \frac{2n\times 2n}{2n-1 \times 2n+1}$$

which follows from the infinite product for sine

$$\sin x = x \prod_{n=1}^\infty \bigg( 1 - \frac {x^2}{\pi^2 n^2}\bigg)$$

evaluated at $x=\frac\pi2$. I guess that comes down to the circle after all.


Also, $(-1/2)! = \Gamma(1/2) = \sqrt \pi.$ That can be seen from Euler's reflection formula,

$$\Gamma(x)\Gamma(1-x) = \frac {\pi}{\sin \pi x}$$

Added Gamma(1/2)
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Douglas Zare
  • 28k
  • 6
  • 90
  • 130

$\pi$ shows up in at least two different ways related to factorials.


The $\pi$ in Stirling's Formula $n! \approx (\frac{n}{e})^n /\sqrt{2 \pi n}$ comes down to Wallis's Formula

$$\frac {\pi}{2} = \prod_{n=1}^\infty \frac{2n\times 2n}{2n-1 \times 2n+1}$$

which follows from the infinite product for sine

$$\sin x = x \prod_{n=1}^\infty \bigg( 1 - \frac {x^2}{\pi^2 n^2}\bigg)$$

evaluated at $x=\frac\pi2$. I guess that comes down to the circle after all.


Also, $(-1/2)! = \Gamma(1/2) = \sqrt \pi.$ That can be seen from Euler's reflection formula,

$$\Gamma(x)\Gamma(1-x) = \frac {\pi}{\sin \pi x}$$

The $\pi$ in Stirling's Formula $n! \approx (\frac{n}{e})^n /\sqrt{2 \pi n}$ comes down to Wallis's Formula

$$\frac {\pi}{2} = \prod_{n=1}^\infty \frac{2n\times 2n}{2n-1 \times 2n+1}$$

which follows from the infinite product for sine

$$\sin x = x \prod_{n=1}^\infty \bigg( 1 - \frac {x^2}{\pi^2 n^2}\bigg)$$

evaluated at $x=\frac\pi2$. I guess that comes down to the circle after all.

$\pi$ shows up in at least two different ways related to factorials.


The $\pi$ in Stirling's Formula $n! \approx (\frac{n}{e})^n /\sqrt{2 \pi n}$ comes down to Wallis's Formula

$$\frac {\pi}{2} = \prod_{n=1}^\infty \frac{2n\times 2n}{2n-1 \times 2n+1}$$

which follows from the infinite product for sine

$$\sin x = x \prod_{n=1}^\infty \bigg( 1 - \frac {x^2}{\pi^2 n^2}\bigg)$$

evaluated at $x=\frac\pi2$. I guess that comes down to the circle after all.


Also, $(-1/2)! = \Gamma(1/2) = \sqrt \pi.$ That can be seen from Euler's reflection formula,

$$\Gamma(x)\Gamma(1-x) = \frac {\pi}{\sin \pi x}$$

Post Made Community Wiki
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Douglas Zare
  • 28k
  • 6
  • 90
  • 130

The $\pi$ in Stirling's Formula $n! \approx (\frac{n}{e})^n /\sqrt{2 \pi n}$ comes down to Wallis's Formula

$$\frac {\pi}{2} = \prod_{n=1}^\infty \frac{2n\times 2n}{2n-1 \times 2n+1}$$

which follows from the infinite product for sine

$$\sin x = x \prod_{n=1}^\infty \bigg( 1 - \frac {x^2}{\pi^2 n^2}\bigg)$$

evaluated at $x=\frac\pi2$. I guess that comes down to the circle after all.