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Math Help - A nice identity of Viette and Euler

  1. #1
    MHF Contributor Bruno J.'s Avatar
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    A nice identity of Viette and Euler

    Show that for 0<\theta<\pi,

    \theta = \frac{\sin\theta}{\cos\left(\frac{\theta}{2}\right  )\cos\left(\frac{\theta}{4}\right)\cos\left(\frac{  \theta}{8}\right)\cos\left(\frac{\theta}{16}\right  )...}
    Last edited by Bruno J.; August 10th 2009 at 07:22 PM.
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    Let [tex] T_n = \cos(\theta/2)\cos(\theta/4)\cos(\theta/8)...\cos(\theta/2^n)[/maTH]

     = \frac{1}{2\sin(\theta/2^n)}\cos(\theta/2)\cos(\theta/4)\cos(\theta/8)...\cos(\theta/2^n) (2\sin(\theta/2^n))

     = \frac{1}{2\sin(\theta/2^n)}\cos(\theta/2)\cos(\theta/4)\cos(\theta/8)...\cos(\theta/2^{n-1}) \sin(\theta/2^{n-1}))

     = \frac{ T_{n-1} \sin(\theta/2^{n-1})}{2 \sin(\theta/2^n)}

    when n tend to infinity

      T_n = \frac{T_1 \sin(\theta/2)}{2^{n-1} \sin(\theta/2^n)}

    For n is very large  \sin(\theta/2^n) = \theta/2^n
     T_n = 2 \sin(\theta/2)\cos(\theta/2) /\theta =\sin(\theta)/\theta
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    MHF Contributor Bruno J.'s Avatar
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    Here is my solution : consider the function f(\theta)=\theta\cos\left(\frac{\theta}{2}\right)\  cos\left(\frac{\theta}{4}\right)\cos\left(\frac{\t  heta}{8}\right)\cos\left(\frac{\theta}{16}\right).  ... Its zeroes are \theta = 0 and \theta=2^n(k\pi+\pi/2) for k\in \mathbb{Z}, n\in \mathbb{N^+} Thus its zeroes are all k\pi, k \in \mathbb{Z}, which are the zeroes of \sin\theta.

    Moreover we know f(\theta) must be an entire function because it is a product of entire functions. By the identity theorem for entire functions, we must have f(\theta)=m\sin\theta for some m \in \mathbb{C}. If we consider the coefficient of \theta in the series expansion of f(\theta) about \theta=0, we see that it is equal to \sum_{j=1}^\infty\frac{1}{2^j}=1, and on the left side as well. Hence we must have k=1 and f(\theta)=\sin\theta.
    Last edited by Bruno J.; August 10th 2009 at 09:52 PM.
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    Quote Originally Posted by Bruno J. View Post
    Here is my solution : consider the function f(\theta)=\theta\cos\left(\frac{\theta}{2}\right)\  cos\left(\frac{\theta}{4}\right)\cos\left(\frac{\t  heta}{8}\right)\cos\left(\frac{\theta}{16}\right).  ... Its zeroes are \theta = 0 and \theta=2^n(k\pi+\pi/2) for k\in \mathbb{Z}, n\in \mathbb{N^+} Thus its zeroes are all k\pi, k \in \mathbb{Z}, which are the zeroes of \sin\theta.

    Moreover we know f(\theta) must be an entire function because it is a product of entire functions. By the identity theorem for entire functions, we must have f(\theta)=m\sin\theta for some m \in \mathbb{C}. If we consider the coefficient of \theta in the series expansion of f(\theta) about \theta=0, we see that it is equal to \sum_{j=1}^\infty\frac{1}{2^j}=1, and on the left side as well. Hence we must have k=1 and f(\theta)=\sin\theta.

    Perfect ! The important thing is the order of its zeros (  f(\theta) = 0 ) or the degree of all factors is zero , then  f(\theta) = m \sin(\theta)
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