# Gamma reflection

Here's another cool identity from Euler. You can start by defining the factorial of a number, denoted $n!$, as that number multiplied by all the numbers below it. Mathematically, this means $n! = n \cdot (n-1) \cdot (n-2) \cdots 1$. We can also write this recursively, as

$n! = n \cdot (n-1)! \qquad 0! = 1$

Taking $n \in \mathbb{N}$, the first few values indexed by $n$ are $\{1,2,6,24,125,750, 5250, \dots\}$. Taking the factorial as a function, it seems that it can grow far more quickly than any exponential function. This can be easily checked by using the taylor series of $f(x)=e^x$:

$e^n = \displaystyle\sum_{k=0}^\infty \frac{n^k}{k!}$ $e^n > \frac{n^n}{n!} \quad \to \quad n! > \frac{n^n}{e^n} = e^{n \ln(n) - n}$

This sets a lower bound on the growth of $n!$. An obvious upper bound is $n! < n^n$, so we can roughly say that the factorial grows like $n^n$ as $n$ approaches $\infty$. However, $n^n$ doesn't fully satisfy the functional equation $n! = n(n-1)!$, so there must be lower-order terms. These can be found with Stirling's approximation.

Here's a challenge: can we define the factorial for all real numbers? How about complex ones as well? It turns out we can, and this is where the gamma function can be introduced:

$\Gamma(x) = \displaystyle\int_0^\infty t^{x-1}e^{-t}dt$

This seemingly comes out of nowhere, but it relies on a following neat observation:

$\displaystyle\int_0^\infty t^{n}e^{-t}dt = \displaystyle\int_0^\infty n t^{n-1}e^{-t}dt$

Thus it satisfies the same functional formula, $\Gamma(x) = x \Gamma(x-1)$. Evaluating the full integral gives $\Gamma(x) = (n-1)!$ whenever $x$ is an integer. We now have a factorial function, the Gamma function, which is continuously defined. Due to the functional equation, this function blows up at all negative values. However, it stays finite for non-integer negative values. Thanks to the Weierstrass factorization theorem, we can factorize this function (or it's reciprocal) as a product of its poles (or zeroes) as:

$\frac{1}{\Gamma(x)} = xe^{\gamma x} \displaystyle\prod_{n=1}^\infty (1 + x/n)e^{-x/n}$

Where $\gamma= \lim_{k \to \infty} \sum_{n=1}^k 1/n - \ln k \approx 0.57722$. This will take a bit of work to justify. We can start with the peculiar identity

$\Gamma(x) = \lim_{p \to \infty} \frac{p! p^x (x-1)!}{(x+p)!}$

Proof:

To prove this, we can expand $(p+x)!$ as $(x-1)! (x)(x+1)\cdots (x+p)$ and introduce a strange factor $p^x p!$ to get

$(x-1)! = \frac{(p+x)!}{ x(x+1)\cdots (x+p)} = \frac{(p+x)!}{p^x p!} \frac{p^x p!}{ x(x+1)\cdots (x+p)}$

The leftmost fraction expands to

$\frac{(p+x)!}{p^x p!} = \frac{p!}{p!} \frac{(p+1)(p+2)\cdots(p+x)}{p^x} = (1 + 1/p)(1 + 2/p) \cdots (1 + x/p)$

Which approaches $1$ as $p \to \infty$. This must mean that

$(x-1)! = \frac{(p+x)!}{p^x p!} \frac{p^x p!}{ x(x+1)\cdots (x+n)} = \lim_{p \to \infty} \frac{p^x p!}{ x(x+1)\cdots (x+p)}$

Or $\Gamma(x) = \lim_{p \to \infty} \frac{p! p^x (x-1)!}{(x+p)!}$

It takes some more work to prove that this works with any argument (it's definitely a challenging exercise). I'll skip the details here though. From here we can find the Weierstrass factorization

$\Gamma(x) = \lim_{p \to \infty} \frac{e^{x \ln p}}{x(1+x)(1+x/2) \cdots (1+x/p)}$ $= \lim_{p \to \infty} \frac{e^{x (\ln p - 1 - 1/2 - 1/3 - \dots - 1/p)}}{x} \displaystyle\prod_{n=1}^p \frac{1}{1 + x/n} e^{x/n}$ $\Gamma(x) = \frac{e^{- \gamma x}}{x} \displaystyle\prod_{n=1}^\infty \frac{1}{1 + x/n} e^{x/n}$

This leads to a really interesting functional equation for the gamma function, known as Euler's reflection formula. The Weierstrass factorization of $\sin (\pi x)$ is given by:

$\sin (\pi x) = \pi x \displaystyle\prod_{n \in \mathbb{Z}} (1-x/n) = \pi x \left( \displaystyle\prod_{n =1}^\infty (1 - x/n) \right) \left( \displaystyle\prod_{n =1}^\infty (1 + x/n)\right)$

The big terms suggest that we can relate this with $\Gamma(x)$ in some way. We can derive:

$\displaystyle\prod_{n =1}^\infty (1 + x/n) = \frac{\epsilon(x)}{xe^{\gamma x} \Gamma (x)}$ $\displaystyle\prod_{n =1}^\infty (1 - x/n) = \frac{-e^{\gamma x}}{x \epsilon (x) \Gamma (-x)}$

Where $\epsilon(x) = \displaystyle\prod_{n =1}^\infty e^{-x/n}$. Replacing the separated terms in the Weierstrass factorization of $\sin (\pi x)$ yields:

$\sin (\pi x) = \pi x \left( \frac{\epsilon(x)}{xe^{\gamma x} \Gamma (x)} \right) \left( \frac{-e^{\gamma x}}{x \epsilon (x) \Gamma (-x)} \right) = \frac{- \pi}{x \Gamma(x) \Gamma(-x)}$

Appealing to the functional equation $\Gamma(x+1) = x\Gamma(x)$ leads us to simplify $x \Gamma(-x)$ as $-\Gamma(1-x)$, giving us the full reflection identity:

$\large \sin (\pi x) = \frac{\pi}{\Gamma(x) \Gamma(1-x)}$ Note by Levi Walker
1 year, 2 months ago

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