Waste less time on Facebook — follow Brilliant.
×

Another Calculus Challenge!

Prove the following identity -

\[\displaystyle \prod_{r=1}^{n}{\Gamma\left(\frac{r}{n+1}\right)} = \sqrt{\frac{{(2\pi)}^{n}}{n+1}}\]

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

Note by Kartik Sharma
1 year, 5 months ago

No vote yet
1 vote

Comments

Sort by:

Top Newest

Lemma :

\[\prod_{r=1}^{n-1}\sin\left(\dfrac{r\pi}{n}\right)=\dfrac{n}{2^{n-1}}\]

Proof:

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

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

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

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

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

Consider the factorisation,

\[x^{n-1}+x^{n-2}+ \ldots + x + 1 = \prod_{k=1}^{n-1}\left(x-e^{\frac{-2\pi k}{n}}\right)\]

Putting \(x=1\) in the above identity gives,

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

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

Now, consider

\[\text{S}=\sum_{r=1}^{n}\log \left(\Gamma{\left(\dfrac{r}{n+1}\right)}\right)\]

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

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

By Euler's Reflection Formula,

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

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

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

Using the Lemma, we have,

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

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

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

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

\[\boxed {\therefore \displaystyle \prod_{r=1}^{n}\Gamma{\left(\dfrac{r}{n+1}\right)}=\sqrt{\dfrac{(2\pi)^n}{n+1}}}\] Ishan Singh · 1 year, 5 months ago

Log in to reply

@Ishan Singh 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 · 1 year, 5 months ago

Log in to reply

@Kartik Sharma Another way is to use the Multiplication Theorem. To prove the multiplication theorem, the limit definition of gamma function can be used. Ishan Singh · 1 year, 5 months ago

Log in to reply

@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 · 1 year, 5 months ago

Log in to reply

@Kartik Sharma \(\displaystyle\Gamma(z)=\lim_{n\to \infty}\dfrac{n^z n!}{z(z+1)(z+2)\ldots(z+n)}\) Ishan Singh · 1 year, 5 months ago

Log in to reply

@Ishan Singh Oh! Okay fine, I get it. Kartik Sharma · 1 year, 5 months ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...