Gauss Multiplication Formula

      Today´s post is a long post, but worth it for sure. We will proof Gauss´s Multiplication Formula for the Gamma Function, namely:


\prod_{k=0}^{m-1}\Gamma\left(z+\frac{k}{m} \right)=(2 \pi)^{\frac{1}{2}(m-1)}m^{\frac{1}{2}-mz}\Gamma\left(mz \right)

       In order to make this post as full detailed and as self contained as possible, we will first provide the proofs of four Lemmas, and then finally proof the main theorem.

     I want to give a special thanks for Daniel Ángeles (@Javier_DanielAR on Twitter)for showing the proof of Lemma 4.

So lets get to it!

Lemma 1:

\sum_{k=0}^{m-1}k=\frac{m(m-1)}{2}

proof:

S_{m-1}=\sum_{k=0}^{m-1}k=0+1+2+ \cdots+(m-1)(L1.1)

S_{m-1}=\sum_{k=0}^{m-1}k=\underbrace{(m-1)+(m-2)+ \cdots + 2+1+0}_{\text{m terms}}(L1.2)

(L1.1) + (L1.2)

2S_{m-1}=(m-1)+\left(1+(m-2) \right)+ \cdots + \left( (m-1)+0 \right)

2S_{m-1}=\underbrace{(m-1)+(m-1)+ \cdots + (m-1)}_{\text{m terms}}

2S_{m-1}=m(m-1)

\boxed{\sum_{k=0}^{m-1}k=\frac{m(m-1)}{2}}

Lemma 2:

\Gamma(mz)= \lim_{n \rightarrow \infty}\frac {(mn-1)! \cdot (mn)^{zm} }{mz(mz+1)\cdots(mz+mn-1)}

Proof: Start by writting (zm-1)! as follows

(zm-1)!=\frac{zm}{zm} \cdot(zm-1) \cdot (zm-2)\cdots 3 \cdot 2 \cdot 1

(zm-1)!= \frac{1}{zm} \cdot 1 \cdot 2 \cdot 3 \cdots (zm-2) \cdot (zm-1) \cdot zm

(zm-1)!= \frac{1}{zm}\cdot \frac { 1 \cdot 2 \cdot 3 \cdots (zm-2) \cdot (zm-1) \cdot zm \cdot (zm+1)\cdot (zm+2) \cdots (zm+mn-1)}{(zm+1)\cdot (zm+1) \cdots (zm+mn-1)}

(zm-1)!= \frac { 1 \cdot 2 \cdot 3 \cdots (mn-1) \cdot (mn)\cdot (mn+1) \cdot (mn+2) \cdots (zm+mn-1)}{\prod_{k=0}^{mn-1}(mz+k)}

(zm-1)!= \frac {(mn-1)! \cdot (mn) \cdot (mn+1) \cdots (mn+zm-1)}{\prod_{k=0}^{mn-1}(mz+k)}

(zm-1)!= \frac {(mn-1)! \cdot (mn)^{zm} \cdot \frac{(mn)}{mn} \cdot \frac{(mn+1)}{mn} \cdots \frac{(mn+zm-1)}{mn}}{\prod_{k=0}^{mn-1}(mz+k)}

Note that

\lim_{n \rightarrow \infty} \frac{(mn)}{mn} \longrightarrow 1 , \,\,\lim_{n \rightarrow \infty} \frac{(mn+1)}{mn} \longrightarrow 1, \, \cdots, \, \lim_{n \rightarrow \infty} \frac{(mn+zm-1)}{mn} \longrightarrow 1

therefore

(zm-1)!= \lim_{n \rightarrow \infty}\frac {(mn-1)! \cdot (mn)^{zm} }{mz(mz+1)\cdots(mz+mn-1)}

\boxed{\Gamma(mz)= \lim_{n \rightarrow \infty}\frac {(mn-1)! \cdot (mn)^{zm} }{mz(mz+1)\cdots(mz+mn-1)}}

Lemma 3:

\prod_{k=1}^{m-1} \Gamma\left(\frac{k}{m}\right)=(2 \pi)^{\frac{1}{2}(m-1)} m^{-\frac{1}{2}}

Proof:

Lets first expand

\prod_{k=1}^{m-1} \Gamma\left(\frac{k}{m}\right)=\Gamma\left(\frac{1}{m}\right) \cdot \Gamma\left(\frac{2}{m}\right) \cdots \Gamma\left(\frac{m-1}{m}\right)

\prod_{k=1}^{m-1} \Gamma\left(\frac{k}{m}\right)=\Gamma\left(\frac{1}{m}\right) \cdot \Gamma\left(\frac{2}{m}\right) \cdots \Gamma\left(1-\frac{1}{m}\right)(L3.1)

\prod_{k=1}^{m-1} \Gamma\left(\frac{k}{m}\right)=\Gamma\left(1-\frac{1}{m}\right) \cdot \Gamma\left(1-\frac{2}{m}\right) \cdots \Gamma\left(\frac{1}{m}\right)(L3.2)

\left(\prod_{k=1}^{m-1} \Gamma\left(\frac{k}{m}\right)\right)^{2}=(L3.1) \times (L3.2)

=\Gamma\left(\frac{1}{m}\right) \cdot \Gamma\left(\frac{2}{m}\right) \cdots \Gamma\left(1-\frac{1}{m}\right) \times \Gamma\left(1-\frac{1}{m}\right) \cdot \Gamma\left(1-\frac{2}{m}\right) \cdots \Gamma\left(\frac{1}{m}\right)

Rearranging terms

\left(\prod_{k=1}^{m-1} \Gamma\left(\frac{k}{m}\right)\right)^{2} =\Gamma\left(\frac{1}{m}\right)\Gamma\left(1-\frac{1}{m}\right) \cdot \Gamma\left(\frac{2}{m}\right) \Gamma\left(1-\frac{2}{m}\right) \cdots

\begin{aligned} \left(\prod_{k=1}^{m-1} \Gamma\left(\frac{k}{m}\right)\right)^{2} &=\prod_{k=1}^{m-1} \Gamma\left(\frac{k}{m}\right) \Gamma\left(1-\frac{k}{m}\right) \\ \end{aligned}

And from the reflection formula for the Gamma function we get

\left(\prod_{k=1}^{m-1} \Gamma\left(\frac{k}{m}\right)\right)^{2}=\prod_{k=1}^{m-1} \frac{\pi}{\sin \left(\frac{\pi k}{m}\right)}(L3.3)

To evaluate (L3.3), we use the identity,

\frac{x^{m}-1}{x-1}=\sum_{k=0}^{m-1} x^{k}=\prod_{k=1}^{m-1}\left(x-e^{2 \pi i k / m}\right),

since the numbers \mathrm{e}^{2 \pi \mathrm{k} / \mathrm{n}}, 1 \leq \mathrm{k} \leq \mathrm{m}-1, are the roots of the polynomial on the left side. Letting x=1, we get

\begin{aligned} m &=\prod_{k=1}^{m-1}\left(1-e^{2 \pi i k / m}\right) \\ &=\prod_{k=1}^{m-1} e^{\pi i k / m} \prod_{k=1}^{m-1}\left(e^{-\pi i k / m}-e^{\pi i k / m}\right) \\ &=e^{(m-1) \pi i / 2} \prod_{k=1}^{m-1}\left(-2 i \sin \left(\frac{\pi k}{ m}\right)\right) \\ &=e^{(m-1) \pi i / 2} 2^{m-1} e^{-(m-1) \pi i / 2} \prod_{k=1}^{m-1} \sin \left(\frac{\pi k}{m}\right) \\ &=2^{m-1} \prod_{k=1}^{m-1}\sin \left(\frac{\pi k}{m}\right) \end{aligned}

\boxed{\prod_{k=1}^{m-1}\sin \left(\frac{\pi k}{m}\right)=\frac{m}{2^{m-1}}}(L3.4)

Plugging (L3.4)\,\tex{in}\, (L3.3) \, \text{we get:}

\left(\prod_{k=1}^{m-1} \Gamma\left(\frac{k}{m}\right)\right)^{2}=\frac{(2\pi)^{m-1}}{m}

\boxed{\prod_{k=1}^{m-1} \Gamma\left(\frac{k}{m}\right)=(2 \pi)^{\frac{1}{2}(m-1)}m^{-\frac{1}{2}} }(L3.5)

Lemma 4:

\frac{\Gamma \left(x+mn \right)}{\Gamma \left(x \right)}=m^{mn}\prod_{k=1}^{m}\frac{\Gamma \left(\frac{x+k-1}{m}+n \right)}{\Gamma \left(\frac{x+k-1}{m} \right)} \, \tag{1}(L4.1)

Proof:

Note that both sides of (L4.1), the ratio between Gamma functions, can be rewritten as Pocchammer symbols. By the recurrence equation of the Gamma function, namely

\Gamma(x+1)=x\Gamma(x)

we may write

\frac{\Gamma \left(x+mn \right)}{\Gamma \left(x \right)}=\frac{\left(x+mn-1 \right)\Gamma \left(x+mn-1 \right)}{\Gamma \left(x \right)}

\cdots

=\frac{\left(x+mn-1 \right)\left(x+mn-2 \right) \cdots x\Gamma \left(x \right)}{\Gamma \left(x \right)}

\frac{\Gamma \left(x+mn \right)}{\Gamma \left(x \right)}=x (x+1) \cdots\left(x+mn-2 \right)\left(x+mn-1 \right)

\frac{\Gamma \left(x+mn \right)}{\Gamma \left(x \right)}=\left( x\right)_{mn}\, \tag{2}(L4.2)

enforcing     x \longrightarrow \frac{x+k-1}{m} \, \, \text{and}\,\, mn \longrightarrow n      in (L4.2) above we get

\frac{\Gamma \left(\frac{x+k-1}{m}+n \right)}{\Gamma \left(\frac{x+k-1}{m} \right)}=\left(\frac{x+k-1}{m} \right)_{n}\, \tag{3}(L4.3)

plugging (L4.2) and (L4.3) in (L4.1) we get

\left( x\right)_{mn}=m^{mn}\prod_{k=1}^{m}\left(\frac{x+k-1}{m} \right)_{n}\tag{4}(L4.4)

Now, lets play a little with Pocchammer symbol to see whether we can draw something from it. By definition we have

\left( x\right)_{m}=\underbrace{x(x+1)(x+2) \cdots (x+m-2)(x+m-1)}_{\text{m terms}}

We have one group of m terms

\left( x\right)_{2m}=\color{red} {x(x+1)(x+2) \cdots (x+m-2)(x+m-1)}\color{blue}{(x+m-1+1)(x+m-1+2)\cdots(x+m-1+m)}

\left( x\right)_{2m}=\underbrace{\underbrace{\color{red} {x(x+1)(x+2) \cdots (x+m-2)(x+m-1)}}_{\text{m terms}}\underbrace{\color{blue}{(x+m)(x+m+1)\cdots(x+m+m-1)}}_{\text{m terms}}}_{\text{2m terms}}

we have 2 groups of m terms each

\left( x\right)_{3m}=\underbrace{\underbrace{\color{red} {x(x+1)(x+2) \cdots (x+m-2)(x+m-1)}}_{\text{m terms}}\underbrace{\color{blue}{(x+m)(x+m+1)\cdots(x+m+m-1)}}_{\text{m terms}}\color{green} {\underbrace{(x+2m)(x+2m+1) \cdots(x+2m+m-1)}_{\text{m terms}}}}_{\text{3m terms}}

We have 3 groups of m terms each

Generalizing we may write

\left( x\right)_{nm}=\color{red} {x(x+1)(x+2) \cdots (x+m-2)(x+m-1)} \color{black}{\cdots} \color{blue}{(x+(n-1)m)(x+(n-1)m+1)\cdots(x+(n-1)m+m-1)}

\left( x\right)_{nm}=\color{red} {x(x+1)(x+2) \cdots (x+m-2)(x+m-1)} \color{black}{\cdots} \color{blue}{(x+(n-1)m)(x+(n-1)m+1)\cdots(x+nm-1)}

We obtain n groups of m terms each.

Now, instead of writing each group horizontally, lets write each group in a different line getting sort of a matrix look. For (x)_{2m}

\begin{aligned} &\begin{array}{llcclll} (x)_{2m}=&x & (x+1) & (x+2) & \cdots & (x+m-1) \\ &\left(x+m\right) & \left(x+m+1\right) & \left(x+m+2\right) & \ldots & \left(x+m+m-1 \right) \\ \end{array}\\ \end{aligned}

\begin{aligned} &\begin{array}{llcclll} (x)_{2m}=&m^m& \frac{x}{m} & (\frac{x+1}{m}) & (\frac{x+2}{m}) & \cdots & (\frac{x+m-1)}{m} \\ &m^m&\left(\frac{x}{m}+1\right) & \left(\frac{x+1}{m}+1\right) & \left(\frac{x+2}{m}+1\right) & \ldots & \left(\frac{x+m-1}{m}+1 \right) \end{array}\\ \end{aligned}

Now multiply vertically along each column to obtain

(x)_{2m}=m^{2m} \cdot \left(\frac{x}{m}\right)_{2} \cdot\left(\frac{x+1}{m}\right)_{2} \cdot \left(\frac{x+2}{m}\right)_{2} \cdots \left(\frac{x+m-1}{m}\right)_{2}

And we may write

(x)_{2m}=m^{2m}\prod_{k=0}^{m-1}\left(\frac{x+k}{m} \right)_{2}

Similarly

\begin{aligned} &\begin{array}{llcclll} (x)_{3m}=&x & (x+1) & (x+2) & \cdots & (x+m-1) \\ &\left(x+m\right) & \left(x+m+1\right) & \left(x+m+2\right) & \ldots & \left(x+m+m-1 \right) \\ &\left(x+2m\right) & \left(x+2m+1\right) & \left(x+2m+2\right) & \ldots & \left(x+2m+m-1 \right) \end{array}\\ \end{aligned}

\begin{aligned} &\begin{array}{llcclll} (x)_{3m}=&m^m& \frac{x}{m} & (\frac{x+1}{m}) & (\frac{x+2}{m}) & \cdots & (\frac{x+m-1)}{m} \\ &m^m&\left(\frac{x}{m}+1\right) & \left(\frac{x+1}{m}+1\right) & \left(\frac{x+2}{m}+1\right) & \ldots & \left(\frac{x+m-1}{m}+1 \right) \\ &m^m&\left(\frac{x}{m}+2\right) & \left(\frac{x+1}{m}+2\right) & \left(\frac{x+2}{m}+2\right) & \ldots & \left(\frac{x+m-1}{m}+2 \right) \\ (x)_{3m}=&m^{3m}& \left(\frac{x}{m}\right)_{3}&\left(\frac{x+1}{m}\right)_{3}&\left(\frac{x+2}{m}\right)_{3}& \cdots &\left(\frac{x+m-1}{m}\right)_{3} \end{array}\\ \end{aligned}

And we may write

(x)_{3m}=m^{3m}\prod_{k=0}^{m-1}\left(\frac{x+k}{m} \right)_{3}

Generalizing, we may write

(x)_{nm}=A_{1} \cdot A_{2} \cdots A_{n}

where

\begin{array}{cccccc} A_{1}= & x & (x+1) & (x+2) & \ldots & (x+m-1) \\ A_{2}= & (x+m) & (x+m+1) & (x+m+2) & \ldots & (x+m+ m-1) \\ A_{3}= & (x+2 m) & (x+2 m+1) & (x+2 m+2) & \ldots & (x+2m+ m-1) \\ \vdots & \vdots & \vdots & \ldots & \ldots & \vdots \\ A_{n}= & (x+(n-1) m) & (x+(n-1) m+1) & (x+(n-1) m+2) & \ldots & (x+(n-1)m+ m-1) \end{array}

Factoring out m in each line

\begin{aligned} &\begin{array}{llcclll} A_{1}= & m^{m} & \frac{x}{m} & \left(\frac{x+1}{m}\right) & \left(\frac{x+2}{m}\right) & \ldots & \left(\frac{x+m-1}{m}\right) \\ A_{2}= & m^{m} & \left(\frac{x}{m}+1\right) & \left(\frac{x+1}{m}+1\right) & \left(\frac{x+2}{m}+1\right) & \ldots & \left(\frac{x+m-1}{m}+1\right) \\ A_{3}= & m^{m} & \left(\frac{x}{m}+2\right) & \left(\frac{x+1}{m}+2\right) & \left(\frac{x+2}{m}+2\right) & \ldots & \left(\frac{x+ m-1}{m}+2\right)\\ \vdots & \vdots & \vdots & \ldots & \ldots & \vdots\\ A_{n}=&m^{m} &\left(\frac{x}{m}+n-1\right)& \left(\frac{x+1}{m}+n-1\right) &\left(\frac{x+2}{m}+n-1\right)& \ldots &\left(\frac{x+m -1}{m}+n-1\right)\\ \end{array}\\ \end{aligned}

Which analogously as the above cases, we may then write

(x)_{nm}=m^{nm}\left(\frac{x}{m} \right)_{n}\left(\frac{x+1}{m} \right)_{n} \cdots \left(\frac{x+m-1}{m} \right)_{n}

and finally

(x)_{nm}=m^{nm}\prod_{k=0}^{m-1}\left(\frac{x+k}{m} \right)_{n}

or shifting the index we get exactly (L4.4)

(x)_{nm}=m^{nm}\prod_{k=1}^{m}\left(\frac{x+k-1}{m} \right)_{n}(L4.5)

Gauss Multiplication Formula

Theorem:

\prod_{k=0}^{m-1}\Gamma\left(z+\frac{k}{m} \right)=(2 \pi)^{\frac{1}{2}(m-1)}m^{-\frac{1}{2}-mx}\Gamma\left(mz \right)\, \qquad \,\,m=2,3,4,\cdots

Proof: Let

\varphi(z)=\frac{m^{mz}\prod_{k=0}^{m-1}\Gamma\left(z+\frac{k}{m} \right)}{\Gamma\left(mz \right)}(T.1)

From Gauss product for the Gamma function and the result of Lemma 2 we have

\varphi(z)=\lim_{n \rightarrow \infty}\frac{m^{mz}\prod_{k=0}^{m-1}\frac{(n-1)!n^{z+\frac{k}{m}}}{\left(z+\frac{k}{m} \right)\left(z+\frac{k}{m}+1 \right)\cdots\left(z+\frac{k}{m} +n-1\right)}}{\frac {(mn-1)! \cdot (mn)^{zm} }{mz(mz+1)\cdots(mz+mn-1)}}

\varphi(z)=\lim_{n \rightarrow \infty}\frac{m^{mz}\left((n-1)!\right)^{m}n^{mz}n^{\left(\sum_{r=0}^{m-1}r\right)/m}}{(mn-1)!(mn)^{mz}} \cdot\frac{mz(mz+1)\cdots(mz+mn-1)}{\prod_{k=0}^{m-1}\left(z+\frac{k}{m} \right)\left(z+\frac{k}{m}+1 \right)\cdots\left(z+\frac{k}{m} +n-1\right)}

By Lemma 1 we have

\varphi(z)=\lim_{n \rightarrow \infty}\frac{m^{mz}\left((n-1)!\right)^{m}n^{mz+\frac{1}{2}(m-1)}}{(mn-1)!(mn)^{mz}} \cdot\frac{mz(mz+1)\cdots(mz+mn-1)}{\prod_{k=0}^{m-1}\left(z+\frac{k}{m} \right)\left(z+\frac{k}{m}+1 \right)\cdots\left(z+\frac{k}{m} +n-1\right)}

Note that in the second ratio of the above equation we can rewrite the products in terms of Pocchamer symbols, i.e.

(mz)_{mn}=mz(mz+1)\cdots(mz+mn-1)

\left(\frac{zm+k}{m} \right)_{n}=\left(\frac{zm+k}{m} \right)\left(\frac{z+k}{m}+1 \right)\cdots\left(\frac{zm+k}{m} +n-1\right)

Therefore

\varphi(z)=\lim_{n \rightarrow \infty}\frac{m^{mz}((n-1)!)^{m}n^{mz+\frac{1}{2}(m-1)}}{(mn-1)!(mn)^{mz}}\cdot \frac{(mz)_{mn}}{\prod_{k=0}^{m-1}\left(z+\frac{k}{m} \right)_{n}}

\varphi(z)=\lim_{n \rightarrow \infty}\frac{m^{mz}((n-1)!)^{m}n^{mz+\frac{1}{2}(m-1)}}{(mn-1)!(mn)^{mz}}\cdot \frac{(mz)_{mn}}{\prod_{k=0}^{m-1}\left(\frac{mz+k}{m} \right)_{n}}

From Lemma 4 we have :

\frac{(x)_{nm}}{\prod_{k=0}^{m-1}\left(\frac{x+k}{m} \right)_{n}}=m^{nm}

Letting x=mz in the above equation we obtain

\varphi(z)=\lim_{n \rightarrow \infty}\frac{m^{mz}((n-1)!)^{m}n^{mz+\frac{1}{2}(m-1)}}{(mn-1)!(mn)^{mz}}\cdot m^{mn}

\varphi(z)=\lim_{n \rightarrow \infty}\frac{((n-1)!)^{m}n^{\frac{1}{2}(m-1)} m^{mn}}{(mn-1)!}(T.2)

We see that \varphi(z) is independent of z, and we can evaluate it by setting an appropriate value for z. If we let z=\frac{1}{m} in (T.1) we get

\varphi(z)=\frac{m\prod_{k=0}^{m-1}\Gamma\left(\frac{k+1}{m} \right)}{\Gamma\left(1 \right)}=m\prod_{k=1}^{m-1}\Gamma\left(\frac{k}{m}\right)

From Lemma 3 equation (L3.5) we get

\varphi(z)=(2 \pi)^{\frac{1}{2}(m-1)}m^{\frac{1}{2}}

and finally!

\boxed{\prod_{k=0}^{m-1}\Gamma\left(z+\frac{k}{m} \right)=(2 \pi)^{\frac{1}{2}(m-1)}m^{-\frac{1}{2}-mx}\Gamma\left(mz \right)}

Erdélyi, A.; Magnus, W.; Oberhettinger, F.; and Tricomi, F. G. Higher Transcendental Functions, Vol. 1. New York: Krieger, pp. 4-5, 1981.

Berndt BC., Rudiments of the theory of the gamma function, University of Chicago, 1976. Available from: http://www.math.uiuc.edu/˜berndt/theory gamma function.pdf.

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