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Math Help - N sin^2(x) versus sin^2(Nx)

  1. #1
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    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:

    \forall N \geqslant 2    \exists \theta  such that 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 f(\theta) = N \sin^2{\theta} - \sin^2{(N \theta)}

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

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

    So extrema occur at \theta = \frac{k \pi}{N+1} and \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.
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  2. #2
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    I think I have it now, though I'm sure a simpler solution exists.

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

    Then we have the extrema:

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

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

    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

    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, f(\phi_0)<0 we are done.

    So.. when N=2,

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

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


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

    Thanks,
    pomp.
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  3. #3
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by pomp View Post
    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:

    \forall N \geqslant 2    \exists \theta  such that 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 f(\theta) = N \sin^2{\theta} - \sin^2{(N \theta)}

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

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

    So extrema occur at \theta = \frac{k \pi}{N+1} and \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 \frac{N\sin^2(\theta)}{\sin^2(N\theta)}<1, but \sin^2\left(N\theta\right)\underset{\theta\to0}{\s  im}N^2\theta^2 and so \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 \theta such that \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.
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  4. #4
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    Quote Originally Posted by Drexel28 View Post
    This is equivalent to proving that \frac{N\sin^2(\theta)}{\sin^2(N\theta)}<1, but \sin^2\left(N\theta\right)\underset{\theta\to0}{\s  im}N^2\theta^2 and so \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 \theta such that \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.
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