**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 $.

Does this Gamma have anyting to do with the "Gamma" radiation in Chem?

ReplyDeletejust curious

Nope.

ReplyDeleteGamma is just a letter. =/

ReplyDeleteYou can also use the second definition of gamma function i.e. the one involving the euler mascheroni constant, to get to the step with the infinite product for sine

ReplyDeleteFYI, that is the Weierstrass form (the ultimate cheap way to prove the identity lol). It's not hard to go to that from from the Euler form though, which is the one I used.

ReplyDeleteoh i see what you're doing hahaha

ReplyDelete