How do you find this product?

Question

Is there a way to find the exact value of the product $$P=\displaystyle\prod_{n=1}^{1007} \sin {\left(\dfrac{n\pi}{2015}\right)}$$

Also, http://math.stackexchange.com/questions/70231/how-to-prove-those-curious-identities?rq=1 — lab bhattacharjee, Jul 12 '14 at 09:09
@DavidRicherby Stuck on a test because you can't find the product? Well, with Amazon's (brand new) Yesterday Shipping™, you'll already have the product! ;) — Cole Tobin, Jul 12 '14 at 18:27

score 9 · Accepted Answer · answered Jul 12 '14 at 08:10

9

By symmetry, your product is just: $$ P = \sqrt{\prod_{n=1}^{2014}\sin\frac{n \pi}{2015}}=\sqrt{\frac{1}{2^{2014}}\prod_{n=1}^{2014}\left(\exp\left(\frac{2\pi i n}{2015}\right)-1\right)},$$ where the innermost product is the product of the roots of the polynomial: $$ \frac{(x+1)^{2015}-1}{x}.$$ Hence, by Vieta's theorem: $$ P = \sqrt{\frac{2015}{2^{2014}}} = \frac{\sqrt{2015}}{2^{1007}}.$$

answered Jul 12 '14 at 08:10

Jack D'Aurizio

353,855

Or more roughly put, $P \approx 0$. – Cole Tobin Jul 12 '14 at 18:23

score 5 · Answer 2 · answered Jul 12 '14 at 08:05

5

For all positive integers $n\in\mathbb{N}$, the following finite product identity holds:

$$\prod_{k=1}^{\lfloor\frac{n-1}{2}\rfloor}\sin{\left(\frac{\pi\,k}{n}\right)}=2^{\frac{1-n}{2}}\,\sqrt{n}~.$$

Since you have the good fortune that the pair of integers $1007$ and $2015$ satisfy the necessary arithmetic relationship, $\lfloor\frac{2015-1}{2}\rfloor=1007$, the above product identity applies.

answered Jul 12 '14 at 08:05

David H

29,921

Any idea about how to prove this identity? – user1001001 Jul 12 '14 at 08:07
@user157130 Good old fashioned induction is one option. – David H Jul 12 '14 at 08:30
@user157130 Also, see http://math.stackexchange.com/questions/8385/prove-that-prod-k-1n-1-sin-frack-pin-fracn2n-1 – David H Jul 12 '14 at 08:52

How do you find this product?

2 Answers2