## Thursday, November 16, 2006

### Gammazzz. Topic: Calculus/S&S.

Definition: (Gamma function) $\displaystyle \Gamma(z) = \int_0^{\infty} t^{z-1}e^{-t} dt = \lim_{n \rightarrow \infty} \frac{n! \cdot n^z}{z(z+1)(z+2) \cdots (z+n)}$. Note that the Gamma function satisfies the property $\Gamma(z+1) = z\Gamma(z)$ and $\Gamma(k+1) = k!$ for nonnegative integers $k$.

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Problem: Prove that $\Gamma(1-z) \Gamma(z) = \frac{\pi}{\sin{\pi z}}$.

Solution: We use the second definition for $\Gamma$ given above to evaluate

$\displaystyle \Gamma(-z) \Gamma(z) = \lim_{n \rightarrow \infty} \frac{n! \cdot n^z}{z(z+1)(z+2) \cdots (z+n)} \cdot \lim_{n \rightarrow \infty} \frac{n! \cdot n^{-z}}{-z(-z+1)(-z+2) \cdots (-z+n)}$.

Divide the top and bottom by $n!$ in both limits to obtain

$\displaystyle \Gamma(-z) \Gamma(z) = \lim_{n \rightarrow \infty} \frac{ n^z}{z(1+\frac{z}{1})(1+\frac{z}{2}) \cdots (1+\frac{z}{n})} \cdot \lim_{n \rightarrow \infty} \frac{n^{-z}}{-z(1-\frac{z}{1})(1-\frac{z}{2}) \cdots (1-\frac{z}{n})}$.

Multiplying these limits together, we get

$\displaystyle \Gamma(-z) \Gamma(z) = \lim_{n \rightarrow \infty} \frac{ 1}{-z^2(1+\frac{z}{1})(1-\frac{z}{1})(1+\frac{z}{2})(1-\frac{z}{2}) \cdots (1+\frac{z}{n})(1-\frac{z}{n})}$.

Notice, however, that

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

because it has zeros at all of the integers and has first term $\pi z$ (by Taylor series). So

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

and

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

Lastly, by the property $\Gamma(1-z) = -z\Gamma(-z)$, we arrive at the identity

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

QED.

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Comment: The Gamma function is quite a useful function and actually shows up a lot in higher-level math. The proof above assumes the equivalence of the two forms of the function, which can be proven using the fact that the Gamma function is log-convex. In fact, in can be shown that only one such function exists that satisfies $f(x+1) = xf(x)$ and is log-convex.

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Practice Problem: Show that $\Gamma(x)$ is log-convex for $x > 0$, i.e. $\log{\Gamma(tx_1+(1-t)x_2)} \le t\log{\Gamma(x_1)}+(1-t)\log{\Gamma(x_2)}$ for all $0 \le t \le 1$.