# Prove that $\vert\sin(x)\sin(2x)\sin(2^2x)\cdots\sin(2^nx)\vert < \left(\frac{\sqrt{3}}{2}\right)^n$

For $$x$$ in $$\mathbb R^*$$ and $$n$$ in $$\mathbb N$$, define

$$U_n = \sin(x)\sin(2x)\sin(2^2x)\cdots\sin(2^nx) = \prod_{k=0}^n\sin(2^k x)$$

Prove that $$\vert U_n\vert \leq \left(\frac{\sqrt{3}}{2}\right)^n$$

(Edited by @River Li) I found it was B6(c) in the 34th Putnam (1973):

B-6: On the domain $$0 \le \theta \le 2\pi$$:

(a) Prove that $$\sin^2\theta \sin 2\theta$$ takes its maximum at $$\pi/3$$ and $$4\pi/3$$ (and hence its maximum at $$2\pi/3$$ and $$5\pi/3$$).

(b) Show that $$\left|\sin^2\theta \, \Big[\sin^3(2\theta) \cdot \sin^3(4\theta) \cdots \sin^3(2^{n - 1}\theta)\Big]\, \sin (2^n\theta)\right|$$ takes its maximum at $$\theta = \pi/3$$. (The maximum may also be attained at other points.)

(c) Derive the inequality: $$\sin^2\theta \cdot \sin^2(2\theta) \cdot \sin^2(4\theta) \cdots \sin^2(2^n\theta)\le (3/4)^n.$$

See: The American Mathematical Monthly, Vol. 81, No. 10, Dec., 1974, (Page 1086-1095) or https://prase.cz/kalva/putnam/putn73.html

• The revised version of the question should be at least sensible now. What is still lacking is any sort of personal input from the OP.
– Did
Commented Dec 4, 2016 at 11:15
• i think 0 is allowed Commented Dec 4, 2016 at 11:16
• Yeah, but with $\leq$; not $<$.
– Did
Commented Dec 4, 2016 at 11:17
• multiply by $\cos(x)$ both sides and see what you get.
– user175968
Commented Dec 4, 2016 at 11:29
• Did you try to follow @frank000's suggestion? Please explain in details how things went when you did.
– Did
Commented Dec 4, 2016 at 14:15

Easy to see that $|U_1(x)|\leq\frac{\sqrt3}{2}$ and $|U_2(x)|\leq\frac{3}{4}$ for every $x$.

Assume that $|U_k(x)|\leq\left(\frac{\sqrt3}{2}\right)^k$ for every real $x$ and every $k\leq n$, for some $n\ge2$.

• If $|\sin{x}|\leq\frac{\sqrt3}{2}$, then $$|U_{n+1}(x)|=|\sin x|\cdot|U_n(2x)|\leq\frac{\sqrt3}{2}\cdot\left(\frac{\sqrt3}{2}\right)^n=\left(\frac{\sqrt3}{2}\right)^{n+1}$$

• If $|\sin{x}|\geq\frac{\sqrt3}{2}$, then $|\cos{x}|\leq\frac{1}{2}$ and $|\sin{x}\sin2x|=|2(1-\cos^2x)\cos{x}|\leq\frac{3}{4}$. Thus, $$|U_{n+1}(x)|=|\sin{x}\sin2x|\cdot|U_{n-1}(4x)|\leq\frac{3}{4}\cdot\left(\frac{\sqrt3}{2}\right)^{n-1}=\left(\frac{\sqrt3}{2}\right)^{n+1}$$

By induction on $n\ge1$, we are done.

• Nice split into cases. Commented Dec 23, 2016 at 19:51

Hint:

Let us assume it holds for $n=k$, that is: $|U_k| \leq \left(\frac{\sqrt{3}}{2}\right)^k$

Now what we want to show for the product for $n=k+1$ reduces to $\leq \left(\frac{\sqrt{3}}{2}\right)^{k-1} \sin(2^{k}x)\sin(2^{k+1}x)$, right? If we multiply it with $\cos(2^{k}x)$, what happens?

• (√3/2)^k-1 . 1/2 . sin(2^k+1 . x )^2 Commented Dec 4, 2016 at 21:47
• Yes, and how is that bounded? Commented Dec 4, 2016 at 21:49
• honnestly i don't know what to do next Commented Dec 4, 2016 at 21:58
• i've got no idea can u help please Commented Dec 4, 2016 at 22:03