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#Math #Calculus
Intro
The Gamma function \Gamma(x) is a way to extend the factorial function, where \Gamma(n + 1) = n!. This gives us two conditions defining \Gamma (x):
\Gamma(1) = 1 \\
\Gamma(x + 1) = x \Gamma (x)
However, by adding a third condition stating \Gamma (x) is logarithimically convex (\log \circ \space \Gamma is convex), we can prove that \Gamma (x) is unique!
Proof
Let G be a function with the properties above. Since G(x + 1) = xG(x), we can define any G(x + n), where n \in \mathbb{N} as:
G(x + n) = G(x)\prod _{i = 0}^{n - 1}(x + i)
This means that it is sufficient to define G(x) on x \in (0, 1] for a unique G(x).
Let S(x_1, x_2) be defined as \frac {\log (\Gamma(x_2)) - \log (\Gamma(x_1))} {x_2 - x_1}. Observe that by log-concavity, for all 0 \lt x \leq 1 and n \in \mathbb{N}:
S(n - 1, n) \leq S(n, n +x) \leq S(n, n + 1) \\
\log (G(n))) - \log (G(n-1)) \leq \frac {\log (G(n + x)) - \log (G(n))} {x} \leq \log (G(n + 1)) - \log (G(n)) \\
\log ((n - 1)!) - \log ((n-2)!) \leq \frac {\log (G(x + n)) - \log ((n - 1)!)} {x} \leq \log (n!) - \log ((n - 1)!) \\
\log(n - 1) \leq \frac {\log (\frac{G(x + n)}{(n - 1)!})} {x} \leq \log(n) \\
\log((n - 1)^x) \leq \log (\frac {G(x + n)}{(n - 1)!})\leq \log(n^x) \\
Raising to the n:
(n - 1)^x \leq \frac {G(x + n)}{(n - 1)!}\leq n^x \\
(n - 1)^x(n - 1)! \leq G(x + n)\leq n^x(n - 1)! \\
Using the above work to expand G(x + n):
\frac{(n - 1)^x(n - 1)!} {\prod _{i = 0}^{n - 1}(x + i)} \leq G(x) \leq \frac{n^x(n - 1)!} {\prod _{i = 0}^{n - 1}(x + i)} \\
\frac{(n - 1)^x(n - 1)!} {\prod _{i = 0}^{n - 1}(x + i)} \leq G(x) \leq \frac{n^xn!} {\prod _{i = 0}^n(x + i)}(\frac {n + x} n) \\
Of course, taking the limit as n goes to infinity on both sides by brute force will produce the value of G(x), however I will present a more elegant solution. Notice we can take the inequalities separately, resulting in:
\frac{(n_1 - 1)^x(n_1 - 1)!} {\prod _{i = 0}^{n_1 - 1}(x + i)} \leq G(x)\\
G(x) \leq \frac{n_2^xn_2!} {\prod _{i = 0}^{n_2}(x + i)}(\frac {n_2 + x} {n_2}) \\
This shows that no matter n_1 and n_2, the equality still holds!
Now we can sub in n_1 = n + 1, n_2 = n, to get:
\frac{n^xn!} {\prod _{i = 0}^{n}(x + i)} \leq G(x) \leq \frac{n^xn!} {\prod _{i = 0}^{n}(x + i)} (\frac {n + x} n)\\
Taking a limit to infinity on both sides:
\lim _{n \to \infty} \frac{n^xn!} {\prod _{i = 0}^{n}(x + i)} \leq G(x) \leq \lim _{n \to \infty} \frac{n^xn!} {\prod _{i = 0}^{n}(x + i)} (\frac {n + x} n)\\
\lim _{n \to \infty} \frac{n^xn!} {\prod _{i = 0}^{n}(x + i)} \leq G(x) \leq \lim _{n \to \infty} \frac{n^xn!} {\prod _{i = 0}^{n}(x + i)} \\
Exercise to the Reader
Prove that the definition:
\Gamma(n) = \int _{0}^{\infty} x^{n - 1} e^{-x} dx
is valid.