Math Help Forum - View Single Post

math sucks · #2 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.