Right now I am trying to figure it out as well.
What you have stated is
What is interesting is that also
EDIT:
Aparently, [LaTeX ERROR: Convert failed] from formula 23 on this page Sine identities
Right now I am trying to figure it out as well.
What you have stated is
What is interesting is that also
EDIT:
Aparently, [LaTeX ERROR: Convert failed] from formula 23 on this page Sine identities
We need to consider a general term where is a positive integer, is a positive integer between and inclusive and is a positive real number. We will derive an identity that is similar to yours but features where we will take the limit as approaches to get your answer.
After plugging in in identity 41 Here and rearranging the sine and the gamma functions, we get
Now lets perform the following calculation
We want to eventually use the Gauss Multiplication formula. Now we break up the product in two pieces and we'll deal with the latter first.
[LaTeX ERROR: Convert failed]
Where the last two follow from flipping the sum around. Now we use the identity from here to get
[tex]\displaystyle \pi^{n-1} \prod_{k=1}^{n-1}\frac{1}{\Gamma\left(z+\frac{k}{n}\right)\Gamma\ left(1-z-\frac{k}{n}\right)} = \pi^{n-1}\prod_{k=1}^{n-1}\frac{1}{\Gamma\left(z+\frac{k}{n}\right)}\prod_ {k=1}^{n-1}\frac{1}{\Gamma\left(-z+\frac{k}{n}\right)} = [/tex]
[LaTeX ERROR: Convert failed]
Thus we have an identity:
[tex]\displaystyle \prod_{k=1}^{n-1} \sin(\pi (z + k/n)) = \frac{\sin(\pi n z)}{2^{n-1}\sin(\pi z)}[/Math]
The limit on the right hand side is [LaTeX ERROR: Convert failed] with L'Hopital's rule. Thus we get
We are almost there. Consider n = 2m +1, that is it's odd. Then
Hence by taking the square root of both sides we get
Now when we plug in m = 13, we get
And this completes the proof!
Thanks mattpi I just found another proof inspired from the cyclotomic equation.
Consider and let be one of its roots, then
.
Evaluate by taking the limit on the LHS, we get
.
Now express the RHS using trigonometric functions and simplify by double-angle formula
Combine all these together we get
Rearrange and notice the symmetric property of sine function, we get
So the proof is archieved by taking square root on both sides:
That's a lot better than going through all that Gamma function pain! I was thinking of doing this problem with complex exponentials because of their nice multiplicative properties but I ended up getting stuck on the fact that I started with the sine functions and not with the cyclotomic polynomial.