2
$\begingroup$

$\lim _{r\to \infty}\frac{\left(\prod_{n=1}^{r}\sin\left(nx\right)\right)}{\left(\frac{1}{r}\right)}$

I tried using the product sin formula but got nowhere.& even after multipling and dividing by $2cos (x)$, answer couldnt be obtained as only multiples of two cut out themselves.
it is a 0/0 indeterminate form.
Also I didn't get the answer by using the l's-hopital rule.
I even tried graphing it on desmos, but the graph was strange---like this( I think even desmos couldn't compute it further)
please help.
thanks in advanced..............

$\endgroup$
3
  • $\begingroup$ You seem to have the wrong function, the term in the product $\prod_{n-1}^r\sin(rx)$ doesn't depend on $n$... $\endgroup$
    – Isaac Ren
    May 8, 2020 at 10:36
  • $\begingroup$ thanks. corrected that $\endgroup$
    – Aatmaj
    May 8, 2020 at 11:17
  • $\begingroup$ The product is zero if $x$ is a rational multiple of $\pi$. This would lead me to believe the product tends to zero for any real $x$ by a density argument, though I can't quite prove it. I don't believe the product has a known closed-form either, which doesn't help. $\endgroup$
    – Integrand
    May 10, 2020 at 1:58

2 Answers 2

3
$\begingroup$

If $x$ is a rational multiple of $\pi$, then for some integer $N > 0$, $\sin(Nx) = 0$. This forces $\prod_{n=1}^r \sin(nx) = 0$ whenever $r \ge N$. In this case, the limit is $0$.

Otherwise, $x$ is not a rational multiple of $\pi$ and $|\cos x| < 1$. Notice

$$|\sin(nx)\sin(n+1)x| = \frac{|\cos x - \cos((2n+1)x)|}{2} \le \mu \stackrel{def}{=}\frac{1 + |\cos x|}{2}$$

By grouping the factors in the numerator in pairs, we have following bound for the weighted product at finite $r$.

$$r\left|\prod_{n=1}^r \sin(nx)\right| \le r\prod_{k=1}^{\lfloor r/2\rfloor} |\sin((2k-1)x)\sin(2kx)| \le r\mu^{\lfloor r/2\rfloor} $$ Since $\mu < 1$, the limit of the weighted product is again $0$.

$\endgroup$
1
$\begingroup$

If $x=\frac pq \pi$ for $p,q\in \mathbb Z, (q\ne 0)$, then the limit would evaluate to zero as for any $q$ we have the term $\sin(qx)$ in our product. If $x$ is not a rational multiple of $\pi$, then no term in the numerator equals zero and we have to take the limit.

$\because -1 \lt \sin(mx) \lt 1$ doesn’t vary much, the limit can be reasonably approximated as:

$$\lim_{r\to\infty} r\prod_{n=1}^r \sin(nx) \sim \lim_{r\to\infty} r (\sin x)^r \\ =\lim_{r\to\infty} \frac{r}{(\csc x)^r} =0$$

$ \because|\csc x| \gt 1$, the limit equals zero due to the fact exponentials grow way faster.

$\endgroup$
4
  • $\begingroup$ Having spent some time on this, could you please elaborate on passing the product to $r$ powers of $-\sin(x)$? $\endgroup$
    – Integrand
    May 10, 2020 at 22:06
  • $\begingroup$ @Integrand Sorry, the $(-1)^r$ wasn’t meant to be there. $\endgroup$
    – Vishu
    May 10, 2020 at 22:38
  • $\begingroup$ Even the passing to $r$ powers is not obvious. For instance, the locations of extrema and zeroes change. $\endgroup$
    – Integrand
    May 10, 2020 at 23:37
  • $\begingroup$ @Integrand That’s true, but it’s a good approximation , especially as $r$ gets large. The idea is that a bunch of numbers ranging from $-1$ to $1$ when multiplied together infinitely many times produce the same result as multiplying one particular number among them, infinitely many times. $\endgroup$
    – Vishu
    May 11, 2020 at 9:28

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .