Results 1 to 4 of 4

Thread: A nice identity of Viette and Euler

  1. #1
    MHF Contributor Bruno J.'s Avatar
    Joined
    Jun 2009
    From
    Canada
    Posts
    1,266
    Thanks
    1
    Awards
    1

    A nice identity of Viette and Euler

    Show that for $\displaystyle 0<\theta<\pi$,

    $\displaystyle \theta = \frac{\sin\theta}{\cos\left(\frac{\theta}{2}\right )\cos\left(\frac{\theta}{4}\right)\cos\left(\frac{ \theta}{8}\right)\cos\left(\frac{\theta}{16}\right )...}$
    Last edited by Bruno J.; Aug 10th 2009 at 07:22 PM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    Joined
    Jan 2009
    Posts
    715
    Let [tex] T_n = \cos(\theta/2)\cos(\theta/4)\cos(\theta/8)...\cos(\theta/2^n)[/maTH]

    $\displaystyle = \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)) $

    $\displaystyle = \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})) $

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

    when n tend to infinity

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

    For n is very large $\displaystyle \sin(\theta/2^n) = \theta/2^n$
    $\displaystyle T_n = 2 \sin(\theta/2)\cos(\theta/2) /\theta =\sin(\theta)/\theta$
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor Bruno J.'s Avatar
    Joined
    Jun 2009
    From
    Canada
    Posts
    1,266
    Thanks
    1
    Awards
    1
    Here is my solution : consider the function $\displaystyle f(\theta)=\theta\cos\left(\frac{\theta}{2}\right)\ cos\left(\frac{\theta}{4}\right)\cos\left(\frac{\t heta}{8}\right)\cos\left(\frac{\theta}{16}\right). ..$. Its zeroes are $\displaystyle \theta = 0 $ and $\displaystyle \theta=2^n(k\pi+\pi/2)$ for $\displaystyle k\in \mathbb{Z}$, $\displaystyle n\in \mathbb{N^+}$ Thus its zeroes are all $\displaystyle k\pi$, $\displaystyle k \in \mathbb{Z}$, which are the zeroes of $\displaystyle \sin\theta$.

    Moreover we know $\displaystyle f(\theta)$ must be an entire function because it is a product of entire functions. By the identity theorem for entire functions, we must have $\displaystyle f(\theta)=m\sin\theta$ for some $\displaystyle m \in \mathbb{C}$. If we consider the coefficient of $\displaystyle \theta$ in the series expansion of $\displaystyle f(\theta)$ about $\displaystyle \theta=0$, we see that it is equal to $\displaystyle \sum_{j=1}^\infty\frac{1}{2^j}=1$, and on the left side as well. Hence we must have $\displaystyle k=1$ and $\displaystyle f(\theta)=\sin\theta$.
    Last edited by Bruno J.; Aug 10th 2009 at 09:52 PM.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member
    Joined
    Jan 2009
    Posts
    715
    Quote Originally Posted by Bruno J. View Post
    Here is my solution : consider the function $\displaystyle f(\theta)=\theta\cos\left(\frac{\theta}{2}\right)\ cos\left(\frac{\theta}{4}\right)\cos\left(\frac{\t heta}{8}\right)\cos\left(\frac{\theta}{16}\right). ..$. Its zeroes are $\displaystyle \theta = 0 $ and $\displaystyle \theta=2^n(k\pi+\pi/2)$ for $\displaystyle k\in \mathbb{Z}$, $\displaystyle n\in \mathbb{N^+}$ Thus its zeroes are all $\displaystyle k\pi$, $\displaystyle k \in \mathbb{Z}$, which are the zeroes of $\displaystyle \sin\theta$.

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

    Perfect ! The important thing is the order of its zeros ( $\displaystyle f(\theta) = 0 $ ) or the degree of all factors is zero , then $\displaystyle f(\theta) = m \sin(\theta) $
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. [SOLVED] Euler Identity Algebra Problem
    Posted in the Differential Geometry Forum
    Replies: 4
    Last Post: Aug 30th 2011, 11:53 AM
  2. Replies: 0
    Last Post: Feb 20th 2010, 08:26 AM
  3. verify Euler's sine and cosine Identity
    Posted in the Calculus Forum
    Replies: 1
    Last Post: Dec 5th 2009, 12:39 PM
  4. Replies: 0
    Last Post: Sep 17th 2009, 06:44 PM
  5. Replies: 4
    Last Post: Feb 21st 2009, 06:10 PM

Search Tags


/mathhelpforum @mathhelpforum