Another Calculus Challenge!

Prove the following identity -

r=1nΓ(rn+1)=(2π)nn+1\displaystyle \prod_{r=1}^{n}{\Gamma\left(\frac{r}{n+1}\right)} = \sqrt{\frac{{(2\pi)}^{n}}{n+1}}

Please be original. I hope you will not copy from sites like MSE(where it might be discussed).

Note by Kartik Sharma
5 years, 2 months ago

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Lemma :

r=1n1sin(rπn)=n2n1\prod_{r=1}^{n-1}\sin\left(\dfrac{r\pi}{n}\right)=\dfrac{n}{2^{n-1}}

Proof:

sin(x)=12i(eixeix)\because\sin(x) = \frac{1}{2i}(e^{ix}-e^{-ix})

    k=1n1sin(kπn)=(12i)n1k=1n1(ekπinekπin)\displaystyle\implies\prod_{k=1}^{n-1} \sin\left(\dfrac{k\pi}{n}\right) = \left(\dfrac{1}{2i}\right)^{n-1}\prod_{k=1}^{n-1} \left(e^{\frac{k\pi i}{n}} - e^{\frac{-k\pi i}{n}}\right)

=(12i)n1 (k=1n1ekπin)k=1n1(1e2kπin)\displaystyle= \left(\dfrac{1}{2i}\right)^{n-1} \ \left(\prod_{k=1}^{n-1} e^{\frac{k\pi i}{n}} \right) \prod_{k=1}^{n-1} \left(1-e^{\frac{-2k\pi i}{n}} \right)

=(12i)n1×in1k=1n1(1e2kπin)\displaystyle= \left(\dfrac{1}{2i}\right)^{n-1} \times i^{n-1} \prod_{k=1}^{n-1} \left(1-e^{\frac{-2k\pi i}{n}} \right)

=12n1k=1n1(1e2kπin)\displaystyle= \dfrac{1}{2^{n-1}} \prod_{k=1}^{n-1} \left(1-e^{\frac{-2k\pi i}{n}} \right)

Consider the factorisation,

xn1+xn2++x+1=k=1n1(xe2πkn)x^{n-1}+x^{n-2}+ \ldots + x + 1 = \prod_{k=1}^{n-1}\left(x-e^{\frac{-2\pi k}{n}}\right)

Putting x=1x=1 in the above identity gives,

k=1n1(1e2kπin)=n \displaystyle \prod_{k=1}^{n-1} \left(1-e^{\frac{-2k\pi i}{n}} \right)=n

r=1n1sin(rπn)=n2n1\displaystyle \therefore \prod_{r=1}^{n-1}\sin\left(\dfrac{r\pi}{n}\right)=\dfrac{n}{2^{n-1}}

Now, consider

S=r=1nlog(Γ(rn+1))\text{S}=\sum_{r=1}^{n}\log \left(\Gamma{\left(\dfrac{r}{n+1}\right)}\right)

=r=1nlog(Γ(n+1rn+1)) (r=abf(r)=r=abf(a+br))\displaystyle = \sum_{r=1}^{n}\log \left(\Gamma{\left(\dfrac{n+1-r}{n+1}\right)}\right) \ \left(\because \sum_{r=a}^{b}f(r)=\sum_{r=a}^{b}f(a+b-r)\right)

=r=1nlog(Γ(1rn+1))\displaystyle = \sum_{r=1}^{n}\log \left(\Gamma{\left(1-\dfrac{r}{n+1}\right)}\right)

By Euler's Reflection Formula,

S=r=1nlog(πΓ(rn+1)sin(rπn+1))\displaystyle \text{S} = \sum_{r=1}^{n}\log \left(\dfrac{\pi}{\Gamma{\left(\dfrac{r}{n+1}\right)}\sin \left(\dfrac{r\pi}{n+1}\right)}\right)

=log(πn)r=1nlog(Γ(rn+1))r=1nlog(sin(rπn+1))\displaystyle =\log ({\pi}^n) - \sum_{r=1}^{n}\log\left(\Gamma{\left(\dfrac{r}{n+1}\right)}\right) - \sum_{r=1}^{n} \log \left(\sin \left(\dfrac{r\pi}{n+1}\right)\right)

    2S=log(πn)r=1nlog(sin(rπn+1))\displaystyle \implies 2\text{S} = \log({\pi}^n) - \sum_{r=1}^{n} \log \left(\sin \left(\dfrac{r\pi}{n+1}\right)\right)

Using the Lemma, we have,

2S=log(πn)log(n+12n)2\text{S}=\log({\pi}^n) - \log\left(\dfrac{n+1}{2^n}\right)

    S=log((2π)nn+1)\displaystyle \implies \text{S} = \log \left(\sqrt{\dfrac{(2\pi)^n}{n+1}}\right)

    r=1nlog(Γ(rn+1))=log((2π)nn+1)\displaystyle \implies \sum_{r=1}^{n}\log \left(\Gamma{\left(\dfrac{r}{n+1}\right)}\right) = \log \left(\sqrt{\dfrac{(2\pi)^n}{n+1}}\right)

    log(r=1nΓ(rn+1))=log((2π)nn+1)\displaystyle \implies \log\left(\prod_{r=1}^{n}\Gamma{\left(\dfrac{r}{n+1}\right)}\right) = \log \left(\sqrt{\dfrac{(2\pi)^n}{n+1}}\right)

r=1nΓ(rn+1)=(2π)nn+1\boxed {\therefore \displaystyle \prod_{r=1}^{n}\Gamma{\left(\dfrac{r}{n+1}\right)}=\sqrt{\dfrac{(2\pi)^n}{n+1}}}

Ishan Singh - 5 years, 1 month ago

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Yep. That's correct! That is also the way I did it. I am curious to know if there is any other approach. Maybe using convolution theorem?

Kartik Sharma - 5 years, 1 month ago

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Another way is to use the Multiplication Theorem. To prove the multiplication theorem, the limit definition of gamma function can be used.

Ishan Singh - 5 years, 1 month ago

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@Ishan Singh Oh! That's nice! I didn't know about it.

To prove the multiplication theorem, the limit definition of gamma function can be used.

I don't understand what's the limit definition?

Kartik Sharma - 5 years, 1 month ago

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@Kartik Sharma Γ(z)=limnnzn!z(z+1)(z+2)(z+n)\displaystyle\Gamma(z)=\lim_{n\to \infty}\dfrac{n^z n!}{z(z+1)(z+2)\ldots(z+n)}

Ishan Singh - 5 years, 1 month ago

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@Ishan Singh Oh! Okay fine, I get it.

Kartik Sharma - 5 years, 1 month ago

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