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Math Help - Prove that Cos(2Pi/n) is algebraic for n a nonzero integer.

  1. #1
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    Prove that Cos(2Pi/n) is algebraic for n a nonzero integer.

    I can't figure out what to do. I am supposed to prove that \cos{\frac{2\pi}{n} is algebraic for non-zero integers n

    This is what I have done so far. I don't know if I'm even on the right track.




    The question marks are there because the upper limit will depend on whether n is even or odd.
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  2. #2
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    Quote Originally Posted by CropDuster View Post
    I can't figure out what to do. I am supposed to prove that \cos{\frac{2\pi}{n} is algebraic for non-zero integers n

    This is what I have done so far. I don't know if I'm even on the right track.



    The question marks are there because the upper limit will depend on whether n is even or odd.


    Read this: Re: algebraic order of cyclotomic numbers

    Tonio
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  3. #3
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    Unfortunately that link didn't really help me. I don't know what a cyclomatic field is, or how I would use it for my proof. Thanks though.
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  4. #4
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    let's take a detour. let's prove we can write cos(nx) as a polynomial in u = cos(x). for n = 1, this is trivial. now we assume we can do so for all 0 < k < n.

    cos(2x) = 2cos^2(x) - 1 = 2u^2 - 1

    cos(3x) = 4cos^3(x) - 3cos(x) = 4u^3 - 3u.

    now for n ≥ 3, cos(nx) = cos(x)cos((n-1)x) - cos((n-2)x) = cos(x)[p(cos(x))] - q(cos(x)) = up(u) - q(u), by our induction assumption

    (expand the RHS of the trig identity to prove the equality).

    since up(u) - q(u) is clearly a polynomial in u, the assertion is shown.

    in particular, this means we can write 1 = cos(2π) = f(cos(2π/n)), for some polynomial f.

    hence if we set g(x) = f(x) - 1, then g(cos(2π/n)) = f(cos(2π/n)) - 1 = 1 - 1 = 0.

    evidently cos(2π/n) is a root of g(x), so cos(2π/n) is algebraic.

    (note: although this seems to only show a proof for natural numbers n, it actually shows it for any non-zero integer

    since cos(2π/n) = cos(2π/-n), if n is nonzero).
    Last edited by Deveno; March 31st 2011 at 11:40 PM.
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  5. #5
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    I figured out a fairly straight forward solution. Seeing as the text in images doesn't show up as results when searching the forum; I'm going to restate the problem and give my proof:

    Prove that for all non-zero integers n, \cos(2\pi/n) is an algebraic number.

    Proof: From Euler's identity we know that e^{2\pi i}=1 and therefore \left(e^{\frac{2\pi i}{n}}\right)^{n}=1. And from Euler's Formula we know that

    \left(e^{\frac{2\pi i}{n}}\right)^n=\left(\cos\left(2\pi/n\right)+i\sin\left(2\pi/n\right)\right)^{n}.<br />

    So if we apply the Binomial theorem to the right, and Euler's identity to the left we have

    1=\sum _{k=0}^n\binom{n}{k}(i)^k\left(\sin\left(2\pi/n\right)\right)^k\left(\cos\left(2\pi/n\right)\right)^{n-k}

    Seperating this sum into two sums: one with terms having only even exponents, and another with terms having only odd exponents:

    \begin{array}{cl}<br />
1&=\sum_{k=0}^p\binom{n}{2 k}(i)^{2 k}\sin ^{2 k}\left(2\pi/n\right)\cos\left(2\pi/n\right)^{n-2 k}\\&\qquad+\sum_{k=0}^q\binom{n}{2 k+1}(i)^{2k+1}\sin^{2 k}\left(2\pi/n\right) \cos\left(2\pi/n\right)^{n-2k-1}\\<br />
\end{array}

    where p+q=n. This sum consists of n+1 terms. If n is odd, p=q=\frac{n+1}{2}. Otherwise p=\frac{n}{2}+1 while q=\frac{n}{2}. So after some algebra,

    \begin{array}{cl}<br />
1&=\sum_{k=0}^p\binom{n}{2 k}(-1)^{k}\sin ^{2k}\left(2\pi/n\right)\cos\left(2\pi/n\right)^{n-2k}\\<br />
&\qquad+(i)\sum_{k=0}^q\binom{n}{2k+1}(-1)^{k}\sin^{2k}\left(2\pi/n\right) \cos\left(2\pi/n\right)^{n-2k-1}.<br />
\end{array}

    But the right side of this equation is a complex number of the form a+ib where a,b\in\mathbb{R} and the lefthand side is of the form 1+0i. Therefore

    0=\sum_{k=0}^q\binom{n}{2k+1}(-1)^{k}\sin^{2k}\left(2\pi/n\right) \cos\left(2\pi/n\right)^{n-2k-1}.

    Hence,
    \begin{array}{cl}<br />
1&=\sum_{k=0}^p\binom{n}{2 k}(-1)^{k}\sin ^{2k}\left(2\pi/n\right)\cos\left(2\pi/n\right)^{n-2k}\\<br />
&=\sum_{k=0}^p\binom{n}{2 k}(-1)^{k}(\sin^{2}\left(2\pi/n\right))^k\cos\left(2\pi/n\right)^{n-2k}\\<br />
\end{array}

    and applying the trigonometric identity \sin^2(x)=1-\cos^2(x) yields:

    1=\sum_{k=0}^p\binom{n}{2 k}(-1)^{k}(1-\cos^{2}\left(2\pi/n\right))^k\cos\left(2\pi/n\right)^{n-2k}.

    Replacing \cos\left(2\pi/n\right), with a variable y gives

    1=\sum_{k=0}^p\binom{n}{2 k}(-1)^{k}(1-y^{2})^k y^{n-2k}

    And after simplifying:

    0&=y^{n}\sum_{k=0}^p\binom{n}{2k}(1-y^{-2})^k-1

    This is a polynomial with rational coefficients that evaluates to zero at y=\cos(2\pi/n) for all non-zero integers n.
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