Results 1 to 4 of 4

Math Help - Problem 42

  1. #1
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    10

    Problem 42

    1)Let f(x) be any polynomial in \mathbb{C} (complex numbers). Let g(x) be the conjugate polynomial of f(x). For example, say f(x) = (1+i)x^2+2x+i then g(x) = (1-i)x^2+2x-i (just replace a+bi by a-bi). Prove that f(x)g(x) is a polynomial in \mathbb{R} (real coefficients).

    2)Find the infinite product: \prod_{n=0}^{\infty} \cos (2^n x).
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Junior Member
    Joined
    Nov 2007
    Posts
    33
    1) Proceed by induction on the degree of f(x). The statement is obviously true for degree 0. Now assume it is true for degree \leq n. Let f(x) be a polynomial of degree n+1. Then it can be written as
    f(x)=f'(x)+zx^{n+1}
    for some polynomial f'(x) with degree n(or less). Similarly, let
    g(x)=g'(x)+\overline{z}x^{n+1}
    where g'(x) is the conjugate polynomial of f'(x). Now,
    f(x)g(x)=(f'(x)+zx^{n+1})(g'(x)+\overline{z}x^{n+1  })=f'(x)g'(x)+x^{n+1}(\overline{z}f'(x)+zg'(x))+z\  overline{z}x^{2n+2}
    By induction, f'(x)g'(x) has real coefficients. Next, by the multiplicative property of conjugation, zg'(x) is still the conjugate of \overline{z}f(x). Thus, when added, the imaginary parts cancel out. Lastly, z\overline{z} is obviously real. Thus, f(x)g(x) must also have real coefficients.

    2) If x\neq\frac{k\pi}{2^n} for some integer n and odd integer k, then the infinite product certainly does not converge as |\cos(2^nx)|<1 for all n. Now if x=\frac{k\pi}{2^n} for some n>0, then \cos(2^{n-1}x) is 0, so the product diverges. Finally, if x=\frac{k\pi}{2^n} for n\leq0, then the product converges to 1 if k/2^n is even and -1 if k/2^n is odd.
    Last edited by math sucks; November 27th 2007 at 04:38 PM.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    10
    1)Lemma: If z_0 is a zero of f(x) then \bar z_0 is a zero of g(x). The proof is really simple say f(x) = a_nx^n+...+a_1x+a_0 then a_nz_0^n+...+a_1z_0+a_0=0. Now take the complex conjugate of both sides, but the complex conjugate preserves sums and products thus \bar a_n (\bar z_0)^n+...+\bar a_1 (\bar z_0)+\bar a_0 = 0. Thus, \bar z_0 is a zero of g(x). Now consider the product h(x)=f(x)g(x). Now if z_0 is a root of this polynomial h(x) then z_0 is a zero of f(x) WLOG, thus \bar z_0 is a zero of g(x) and so \bar z_0 is a zero of h(x). We have show that any zero of h(x) also has its complex cunjugate as a zero. Thus, h(x) is a polynomial with real coeffcients. (This problem was an excercise problem in an abstract algebra book).

    2)This is an infinite series that I have seen several times before. Suppuse we want to simplifiy \cos x \cos (2x) \cos (4x) = y. Multiply both sides by \sin x thus \sin x \cos x \cos (2x)\cos (4x) = y\sin x. Thus, \frac{1}{2}\sin 2x \cos 2x \cos 4x = y\sin x. Thus, \frac{1}{2^2}\sin 4x \cos 4x = y\sin x. Thus, \frac{1}{2^3}\sin 8x = y\sin x. Then supposing that \sin x \not = 0 (that is a special case) we have y = \frac{\sin 8x}{2^3 \sin x}. So in general if P_n is the n-th partial product then P_n = \frac{\sin 2^{n+1} x}{2^n \sin x} if \sin x \not = 0. Now it is easy to take the limit.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor Bruno J.'s Avatar
    Joined
    Jun 2009
    From
    Canada
    Posts
    1,266
    Thanks
    1
    Awards
    1
    Here is another simple proof of the nice theorem (1) :

    Denote \overline{f(x)} the complex conjugate of f(x). Then for any two functions f(x), g(x) \in \mathbb{C}[x] we have \overline{f(x)g(x)}=\overline{f(x)}\: \: \overline{g(x)}.
    This follows directly from the similar property of conjugation for elements of \mathbb{C} and from the definition of a polynomial product.

    Then \overline{f(x)\overline{f(x)}} = \overline{f(x)}\: \: \overline{\overline{f(x)}} = \overline{f(x)}f(x) = f(x)\overline{f(x)}.

    Hence f(x)\overline{f(x)} is its own conjugate, i.e. f(x)\overline{f(x)} \in \mathbb{R}[x].
    Follow Math Help Forum on Facebook and Google+

Search Tags


/mathhelpforum @mathhelpforum