2012-06-18

One of the fundamental relations satisfied by the gamma function is Euler’s so-called reflection formula,

\displaystyle \Gamma(z)\Gamma(1-z) = \frac{\pi}{\sin(\pi z)}.

We also know the Euler product formula from complex analysis,

\displaystyle \sin(\pi z) = \pi z \prod_{n=1}^\infty \left( 1 - \frac{z^2}{n^2} \right).

The gamma function may be defined by

\displaystyle \Gamma(z) := \frac{1}{z} \prod_{n=1}^\infty \frac{(1+\frac{1}{n})^z}{1+\frac{z}{n}}.

Therefore the relation we mentioned first reads

\displaystyle \frac{\pi}{\sin(\pi z)} = \left( \frac{1}{z} \prod_{n=1}^\infty \frac{(1+\frac{1}{n})^z}{1+\frac{z}{n}} \right) \left( \frac{1}{1-z} \prod_{n=1}^\infty \frac{(1+\frac{1}{n})^{1-z}}{1+\frac{1-z}{n}} \right).

That is,

\displaystyle \sin(\pi z) = \pi z(1-z) \left( \prod_{n=1}^\infty \frac{1+\frac{z}{n}}{(1+\frac{1}{n})^z} \right) \left( \prod_{n=1}^\infty \frac{1+\frac{1-z}{n}}{(1+\frac{1}{n})^{1-z}} \right).

In view of the Euler infinite product, this seems to suggest that

\displaystyle (1-z) \left( \prod_{n=1}^\infty \frac{1+\frac{z}{n}}{(1+\frac{1}{n})^z} \right) \left( \prod_{n=1}^\infty \frac{1+\frac{1-z}{n}}{(1+\frac{1}{n})^{1-z}} \right) = \prod_{k=1}^\infty \left( 1 - \frac{z^2}{k^2} \right).

We have

\displaystyle (1-z) \left( \prod_{n=1}^\infty \frac{1+\frac{z}{n}}{(1+\frac{1}{n})^z} \right) \left( \prod_{n=1}^\infty \frac{1+\frac{1-z}{n}}{(1+\frac{1}{n})^{1-z}} \right) = (1-z) \prod_{n=1}^\infty \frac{1 + \frac{1}{n} + \frac{z(1-z)}{n^2}}{1 + \frac{1}{n}}.

This in turn is equal to… ah, you know what, I’m too tired. I’m almost certain that Euler’s infinite product formula can be derived from this alternative definition of the gamma function, but I’m too lazy to prove it right now.

EDIT: Here you go.

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About mlbaker

just another guy trying to make the diagrams commute.
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