Demonstrating an equivalent formula for $\sin(x)/x$ using power series

Question

Would appreciate some ideas for the following:

"Prove that $\frac{\sin{x}}{x}=\prod_{n=1}^{\infty}\cos{\frac{x}{2^n}}$ using power series."

I'm aware this identity can be shown using trig identities and a telescoping product. Also, you can get another proof using the infinite product expressions $\sin{x} = x\prod_{k=1}^{\infty} (1-\frac{x^2}{k^2 \pi^2})$ and $\cos{x}=\prod_{k=1, \ k \ \text{odd}}^\infty (1-\frac{4x^2}{k^2 \pi^2})$.

However, since the question explicitly mentions power series, I was wondering if there is a proof that directly uses power series? I've tried calculating some derivatives and coefficients but they seem to get pretty nasty.

use Taylor series for sin at zero? – Dole Dec 23 '18 at 06:26 — Dole, Dec 23 '18 at 06:26
yeah, but the RHS is giving me trouble – Merk Zockerborg Dec 23 '18 at 06:28 — Merk Zockerborg, Dec 23 '18 at 06:28
Is there a nice power series for $\log(\cos x)$? – mathworker21 Dec 23 '18 at 06:44 — mathworker21, Dec 23 '18 at 06:44

score 1 · Answer 1 · answered Dec 23 '18 at 07:00

Using the fact that $\sin(2x)=2\sin(x)\cos(x)$ we have $\cos(x)=\frac{\sin(2x)}{2\sin(x)}$.

Define $p_n(x):=\prod\limits_{j=1}^n \cos(\frac{x}{2^j})$ and note that $\cos(\frac{x}{2^j})=\dfrac{\sin(\frac{x}{2^{j-1}})}{2\sin(\frac{x}{2^j})}$ for each $j$.

Now, $p_n(x)=\frac{1}{2^n}\frac{\sin(x)}{\sin(\frac{x}{2^n})}=\frac{\sin(x)}{x}\dfrac{\frac{x}{2^n}}{\sin(\frac{x}{2^n})}$ and $p_n(x)\to_n \frac{\sin(x)}{x}$

score 0 · Answer 2 · answered Dec 23 '18 at 10:08

0

HINT

$\cos z= e^{iz}(1+e^{-2iz})/2$ so $ln \cos z=\cdots$

answered Dec 23 '18 at 10:08

G Cab

35,272

Demonstrating an equivalent formula for $\sin(x)/x$ using power series

2 Answers2

Linked