1. N sin^2(x) versus sin^2(Nx)

Hi guys, I'm working through a problem on violations of Bell's inequality using polarised photons passing through filters. I've managed to get right through to the end but for my work to be correct the following statement must be true:

$\displaystyle \forall N \geqslant 2$ $\displaystyle \exists \theta$ such that $\displaystyle N \sin^2{\theta} - \sin^2{(N \theta)} < 0$

I know this is true, and I'd probably get away with just stating this as fact, but I'm not satisfied with that. It should also be really easy to prove and I'm probably missing something obvious, but any help would be appreciated.

Here's what I've tried so far.

Let $\displaystyle f(\theta) = N \sin^2{\theta} - \sin^2{(N \theta)}$

Then extrema of f occur at $\displaystyle \frac{df}{d \theta} = 0$ $\displaystyle \Rightarrow$ $\displaystyle 2 N \sin{\theta} \cos{\theta} - 2 N \sin{(N \theta)} \cos{(N \theta)} = 0$

$\displaystyle N \sin{2 \theta} - N \sin{(2 N \theta)} = 0$

So extrema occur at $\displaystyle \theta = \frac{k \pi}{N+1}$ and $\displaystyle \frac{(\frac{1}{2} + k) \pi}{N-1}$

My idea was to show that a minimum occurs under the theta axis, but I can't seem to be able to show this

pomp.

2. I think I have it now, though I'm sure a simpler solution exists.

Let $\displaystyle \phi_k =\frac{(\frac{1}{2} + k) \pi }{n+1}$

Then we have the extrema:

$\displaystyle f(\phi_0) = N \sin^2{\left( \frac{\pi}{2N+2} \right)} - \sin^2{\left( \frac{N \pi}{2N+2} \right)}$

so, $\displaystyle f(\phi_0) = N \sin^2{(\phi_0)} - \cos^2{(\phi_0)} = (N+1) \sin^2{(\phi_0)} - 1$

$\displaystyle f(\phi_0) = \frac{\pi}{2} \frac{\sin^2{(\phi_0)}}{\phi_0} - 1$

So as N increases, phi-zero decreases and we can use small angle approximations to get

$\displaystyle f(\phi_0) = \frac{\pi}{2} \frac{(\phi_0 + O(n^3))^2}{\phi_0} - 1$

So as N increases, the minimum value of f decreases to -1, so if we can show that when N=2, $\displaystyle f(\phi_0)<0$ we are done.

So.. when N=2,

f$\displaystyle (\phi_0) = 3 \sin^2{\left( \frac{\pi}{6} \right)} - 1 = - \frac{1}{4}$

Therefore $\displaystyle \forall$ $\displaystyle N \geq 2$ $\displaystyle \exists$ $\displaystyle \theta$ such that $\displaystyle f(\theta) < 0$

If anyone can see an easier solution, please let me know.

Thanks,
pomp.

3. Originally Posted by pomp
Hi guys, I'm working through a problem on violations of Bell's inequality using polarised photons passing through filters. I've managed to get right through to the end but for my work to be correct the following statement must be true:

$\displaystyle \forall N \geqslant 2$ $\displaystyle \exists \theta$ such that $\displaystyle N \sin^2{\theta} - \sin^2{(N \theta)} < 0$

I know this is true, and I'd probably get away with just stating this as fact, but I'm not satisfied with that. It should also be really easy to prove and I'm probably missing something obvious, but any help would be appreciated.

Here's what I've tried so far.

Let $\displaystyle f(\theta) = N \sin^2{\theta} - \sin^2{(N \theta)}$

Then extrema of f occur at $\displaystyle \frac{df}{d \theta} = 0$ $\displaystyle \Rightarrow$ $\displaystyle 2 N \sin{\theta} \cos{\theta} - 2 N \sin{(N \theta)} \cos{(N \theta)} = 0$

$\displaystyle N \sin{2 \theta} - N \sin{(2 N \theta)} = 0$

So extrema occur at $\displaystyle \theta = \frac{k \pi}{N+1}$ and $\displaystyle \frac{(\frac{1}{2} + k) \pi}{N-1}$

My idea was to show that a minimum occurs under the theta axis, but I can't seem to be able to show this

pomp.
This is equivalent to proving that $\displaystyle \frac{N\sin^2(\theta)}{\sin^2(N\theta)}<1$, but $\displaystyle \sin^2\left(N\theta\right)\underset{\theta\to0}{\s im}N^2\theta^2$ and so $\displaystyle \lim_{\theta\to0}\frac{N\sin^2(\theta)}{\sin^2\lef t(N\theta\right)}=\frac{1}{N}\lim_{\theta\to0}\fra c{\sin^2(\theta)}{\theta^2}=\frac{1}{N}<1$. Thus, there exists some $\displaystyle \theta$ such that $\displaystyle \left|\frac{N\sin^2(\theta)}{\sin^2(N\theta)}-\frac{1}{N}\right|<\frac{N-1}{N}\implies \frac{N\sin^2(\theta)}{\sin^2(N\theta)}<1$. The conclusion follows.

4. Originally Posted by Drexel28
This is equivalent to proving that $\displaystyle \frac{N\sin^2(\theta)}{\sin^2(N\theta)}<1$, but $\displaystyle \sin^2\left(N\theta\right)\underset{\theta\to0}{\s im}N^2\theta^2$ and so $\displaystyle \lim_{\theta\to0}\frac{N\sin^2(\theta)}{\sin^2\lef t(N\theta\right)}=\frac{1}{N}\lim_{\theta\to0}\fra c{\sin^2(\theta)}{\theta^2}=\frac{1}{N}<1$. Thus, there exists some $\displaystyle \theta$ such that $\displaystyle \left|\frac{N\sin^2(\theta)}{\sin^2(N\theta)}-\frac{1}{N}\right|<\frac{N-1}{N}\implies \frac{N\sin^2(\theta)}{\sin^2(N\theta)}<1$. The conclusion follows.
Thanks, I prefer your solution, it's very tidy.