Problem 42

ThePerfectHacker · #1 $Old$ November 27th, 2007, 10:56 AM

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: $\prod_{n=0}^{\infty} \cos (2^n x)$ .

math sucks · #2 $Old$ November 27th, 2007, 04:49 PM

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.

ThePerfectHacker · #3 $Old$ December 3rd, 2007, 09:37 PM

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.

Bruno J. · #4 $Old$ June 17th, 2009, 05:01 PM

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. $f(x)\overline{f(x)} \in \mathbb{R}[x]$ .