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Thread: building complex polynomials

  1. #1
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    building complex polynomials

    Hi folks,

    By considering the roots of cos 5t = 0 prove that

    $\cos 5t = 16 \cos t (\cos t - \cos \frac{\pi}{10}) (\cos t - \cos \frac{3\pi}{10}) (\cos t - \cos \frac{7\pi}{10}) (\cos t - \cos \frac{9\pi}{10})$

    in the first part of the problem we had to prove using de Moivre's theorem, that $\cos 5t = 16 \cos^5 t - 20 \cos^3 t + 5 \cos t$ I don't know if this is relevant for the second part.

    The roots of $\cos 5t = 0$ are $5t = \cos^{-1} (0)$ i.e. $5t = \dfrac{\pi}{2} + n\pi$ which gives

    t = $\dfrac{\pi}{10}$ , $\dfrac{3\pi}{10}$, $\dfrac{\pi}{2}$, $\dfrac{7\pi}{10}$, $\dfrac{9\pi}{10}$

    The form of the answer suggests that I can build a 5th order polynomial using these roots in the same way that for a quadratic equation in x, I could build it from the 2 roots $\alpha$ and $\beta$ as
    $(x - \alpha)(x - \beta)$

    but this would give

    $ (\cos t - \dfrac{\pi}{10}) $ ... etc not

    $(\cos t - \cos \dfrac{\pi}{10}) $ ....

    I am not sure whether I am on track or not. Can anyone supply a hint?

    thanks
    Last edited by s_ingram; Apr 3rd 2018 at 08:25 AM.
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  2. #2
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    Re: building complex polynomials

    Quote Originally Posted by s_ingram View Post
    Hi folks,

    By considering the roots of cos 5t = 0 prove that

    $\cos 5t = 16 \cos t (\cos t - \cos \frac{\pi}{10}) (\cos t - \cos \frac{3\pi}{10}) (\cos t - \cos \frac{7\pi}{10}) (\cos t - \cos \frac{9\pi}{10})$

    in the first part of the problem we had to prove using de Moivre's theorem, that $\cos 5t = 16 \cos^5 t - 20 \cos^3 t + 5 \cos t$ I don't know if this is relevant for the second part.

    The roots of $\cos 5t = 0$ are $5t = \cos^{-1} (0)$ i.e. $5t = \dfrac{\pi}{2} + n\pi$ which gives

    t = $\dfrac{\pi}{10}$ , $\dfrac{3\pi}{10}$, $\dfrac{\pi}{2}$, $\dfrac{7\pi}{10}$, $\dfrac{9\pi}{10}$

    The form of the answer suggests that I can build a 5th order polynomial using these roots in the same way that for a quadratic equation in x, I could build it from the 2 roots $\alpha$ and $\beta$ as
    $(x - \alpha)(x - \beta)$

    but this would give

    $ (\cos t - \dfrac{\pi}{10}) $ ... etc not

    $(\cos t - \cos \dfrac{\pi}{10}) $ ....

    I am not sure whether I am on track or not. Can anyone supply a hint?

    thanks
    If $t=\dfrac{\pi}{10}$, then $\cos t = \cos \dfrac{\pi}{10}$.

    Yes, you are on the right track. You will wind up with:

    $$\cos (5t) = k\cos t \left( \cos t - \cos \dfrac{\pi}{10} \right) \left( \cos t - \cos \dfrac{3\pi}{10} \right) \left( \cos t - \cos \dfrac{7\pi}{10} \right) \left( \cos t - \cos \dfrac{9\pi}{10} \right)$$

    Then, just solve for $k$ when you plug in some known value for $t$. For example, $t = 0$.

    $$1 = k\left(1-\cos \dfrac{\pi}{10} \right)\left(1- \cos \dfrac{3\pi}{10} \right) \left( 1 - \cos \dfrac{7\pi}{10} \right) \left(1 - \cos \dfrac{9\pi}{10} \right)$$

    Solving for $k$ gives:

    $$k = \dfrac{1}{\left(1-\cos \dfrac{\pi}{10} \right)\left(1- \cos \dfrac{3\pi}{10} \right) \left( 1 - \cos \dfrac{7\pi}{10} \right) \left(1 - \cos \dfrac{9\pi}{10} \right) }$$
    Last edited by SlipEternal; Apr 3rd 2018 at 09:00 AM.
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  3. #3
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    Re: building complex polynomials

    But I don't see how to justify putting $(cos - cos (root)) instead of (cos - root). I am confused by trigonometric equations. They don't seem to work like algebraic equations.
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  4. #4
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    Re: building complex polynomials

    It works exactly like algebraic expressions. You can use $$\cos(5t)=K\prod_{n\in \mathbb{Z}} \left( t-\dfrac{(2n+1)\pi}{10} \right) $$

    But this is an infinite product. $\cos t $ has the advantage of being $2\pi $ cyclic.
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  5. #5
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    Re: building complex polynomials

    Not sure why it's an infinite product. There are only 5 terms in this case. But building up (cos - root) makes sense. It is analogous to algebraic equations. I don't see how you can just change the root to the cos of the root even if cos is 2 pi cyclic.
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  6. #6
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    Re: building complex polynomials

    Quote Originally Posted by s_ingram View Post
    Not sure why it's an infinite product. There are only 5 terms in this case. But building up (cos - root) makes sense. It is analogous to algebraic equations. I don't see how you can just change the root to the cos of the root even if cos is 2 pi cyclic.
    The term must be zero when $t$ equals the root. At $t = \dfrac{\pi}{10}$, you do not have $\cos t = \dfrac{\pi}{10}$. You have $\cos t = \cos \dfrac{\pi}{10}$. Term by term:

    $$\cos t - \cos \dfrac{\pi}{10} = 0\text{ when }t = \dfrac{\pi}{10} + n\pi$$
    $$\cos t - \cos \dfrac{3\pi}{10} = 0\text{ when }t = \dfrac{3\pi}{10} + n\pi$$
    $$\cos t = 0\text{ when }t = \dfrac{\pi}{2} + n\pi$$
    $$\cos t - \cos \dfrac{7\pi}{10} = 0\text{ when }t = \dfrac{7\pi}{10} + n\pi$$
    $$\cos t - \cos \dfrac{9\pi}{10} = 0\text{ when }t = \dfrac{9\pi}{10} + n\pi$$

    Assume you want to use $\sin t$ instead of $\cos t$.

    $$\sin \left( \dfrac{\pi}{10} + n\pi \right) = \sin \dfrac{\pi}{10} \cos (n\pi) + \sin (n\pi) \cos \dfrac{\pi}{10} = (-1)^n \sin \dfrac{\pi}{10}$$

    So, this will not generate all of the roots over $\cos 5t$.
    Last edited by SlipEternal; Apr 3rd 2018 at 10:05 AM.
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  7. #7
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    Re: building complex polynomials

    Let

    $\displaystyle p(x)=16x^5-20x^3+5x$

    where $\displaystyle p(\cos t)=\cos 5t$

    now

    $\displaystyle p\left(\cos \frac{(2k-1)\pi }{10}\right)=\cos \text{ }\frac{(2k-1)\pi }{2}=0$

    so $\displaystyle \cos \frac{(2k-1)\pi }{10}$ , $\displaystyle k=1,2,3,4,5$ are all roots of $\displaystyle p(x)$

    therefore we can factor

    $\displaystyle p(x)=16\prod _{k=1}^5 \left(x-\cos \frac{(2k-1)\pi }{10}\right)$

    now replace $\displaystyle x=cos t$

    $\displaystyle \cos 5t=p(\cos t)=16\prod _{k=1}^5 \left(\cos t-\cos \frac{(2k-1)\pi }{10}\right)$
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  8. #8
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    Re: building complex polynomials

    I have been plodding throughout this, trying to figure out what you are trying to drill into me, but it isn't working!

    I am perfectly happy (now!) that
    $\cos 5t = \cos t (\cos t - \cos \frac{\pi}{10}) (\cos t - \cos \frac{3\pi}{10}) (\cos t - \cos \frac{7\pi}{10}) (\cos t - \cos \frac{9\pi}{10})$

    but where is the k springing from?
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