Approximation to Stirling's Formula for Gamma Function

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Theorem

Let:

$D_\epsilon = \set {z \in \C : \cmod {\Arg z} < \pi - \epsilon,\ \cmod z > 1}$

where:

$\cmod {\Arg z}$ denotes the absolute value of the principal argument of $z$

$\cmod z$ denotes the modulus of $z$

$\epsilon \in \R_{>0}$.

Then for all $z \in D_\epsilon$, the gamma function of $z$ satisfies:

$\map \Gamma z = \sqrt {\dfrac {2 \pi} z} \paren {\dfrac z e}^z \paren {1 + \map \OO {z^{-1} } }$

where $\map \OO {z^{-1} }$ denotes big-O of $z^{-1}$.

Proof

From Gamma Function is Unique Extension of Factorial:

\(\text {(1)}: \quad\)

\(\ds \paren {y + n}^y n!\)

\(=\)

\(\ds \paren {y + n}^y \map \Gamma {n + 1}\)

Gamma Function Extends Factorial

\(\ds \)

\(\le\)

\(\ds \map \Gamma {y + n + 1}\)

\(\ds \)

\(\le\)

\(\ds \paren {n + 1}^y \map \Gamma {n + 1}\)

\(\ds \)

\(=\)

\(\ds \paren {n + 1}^y n!\)

for $0 < y \le 1$ and $n \in \N$.

Let $x$ be given.

Let $n + 1$ be the largest natural number such that $n + 1 \le x$.

This article, or a section of it, needs explaining. In particular: So it was implicitly assume that $x \in \R_{\ge 1}$. Write this explicitly. You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by explaining it. To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{Explain}} from the code.

Let $x = y + n + 1$, and thus $0 < y \le 1$.

Then:

\(\ds \frac {\map \Gamma x} {\sqrt {2 \pi} x^x x^{-1/2} e^{-x} }\)

\(\le\)

\(\ds \frac {\paren {n + 1}^y n!} {\sqrt {2 \pi} x^x x^{-1/2} e^{-x} }\)

from $(1)$

\(\ds \)

\(\sim\)

\(\ds \frac {\paren {n + 1}^y n! \sqrt {2 \pi} n^n n^{1/2} e^{-n} } {\sqrt {2 \pi} x^x x^{-1/2} e^{-x} }\)

Stirling's Formula

\(\ds \)

\(=\)

\(\ds \frac {\paren {n + 1}^y n! n^n n^{1/2} e^{-n} } {\paren {y + n + 1}^{y + n} \paren {y + n + 1}^{1/2} e^{-y - n - 1} }\)

\(\ds \)

\(=\)

\(\ds \paren {\frac {n + 1} {y + n + 1} }^y \paren {\frac n {y + n + 1} }^{1/2} \paren {1 + \frac {y + 1} n}^{-n} e^{y + 1}\)

\(\ds \)

\(\to\)

\(\ds 1 \cdot 1 \cdot \frac 1 {e^{y + 1} } e^{y + 1} \text { as } n \to \infty\)

\(\ds \)

\(=\)

\(\ds 1\)

Similarly for the right hand side.

The result follows from Gamma Function Extends Factorial.

The validity of the material on this page is questionable. In particular: Proved only for $x \in \R_{\ge 1}$. What about for other $z \in D_\epsilon$? You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by resolving the issues. To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{Questionable}} from the code. If you would welcome a second opinion as to whether your work is correct, add a call to {{Proofread}} the page.

$\blacksquare$

Also see

• Logarithmic Approximation of Error Term of Stirling's Formula for Gamma Function

Sources

• 1977: K.G. Binmore: Mathematical Analysis: A Straightforward Approach ... (previous) ... (next): $\S 17.7 \ (2)$