# Problem 42

• Nov 27th 2007, 08:56 AM
ThePerfectHacker
Problem 42
1)Let $\displaystyle f(x)$ be any polynomial in $\displaystyle \mathbb{C}$ (complex numbers). Let $\displaystyle g(x)$ be the conjugate polynomial of $\displaystyle f(x)$. For example, say $\displaystyle f(x) = (1+i)x^2+2x+i$ then $\displaystyle g(x) = (1-i)x^2+2x-i$ (just replace $\displaystyle a+bi$ by $\displaystyle a-bi$). Prove that $\displaystyle f(x)g(x)$ is a polynomial in $\displaystyle \mathbb{R}$ (real coefficients).

2)Find the infinite product: $\displaystyle \prod_{n=0}^{\infty} \cos (2^n x)$.
• Nov 27th 2007, 02:49 PM
math sucks
1) Proceed by induction on the degree of $\displaystyle f(x)$. The statement is obviously true for degree 0. Now assume it is true for degree $\displaystyle \leq n$. Let $\displaystyle f(x)$ be a polynomial of degree $\displaystyle n+1$. Then it can be written as
$\displaystyle f(x)=f'(x)+zx^{n+1}$
for some polynomial $\displaystyle f'(x)$ with degree $\displaystyle n$(or less). Similarly, let
$\displaystyle g(x)=g'(x)+\overline{z}x^{n+1}$
where $\displaystyle g'(x)$ is the conjugate polynomial of $\displaystyle f'(x)$. Now,
$\displaystyle f(x)g(x)=(f'(x)+zx^{n+1})(g'(x)+\overline{z}x^{n+1 })=f'(x)g'(x)+x^{n+1}(\overline{z}f'(x)+zg'(x))+z\ overline{z}x^{2n+2}$
By induction, $\displaystyle f'(x)g'(x)$ has real coefficients. Next, by the multiplicative property of conjugation, $\displaystyle zg'(x)$ is still the conjugate of $\displaystyle \overline{z}f(x)$. Thus, when added, the imaginary parts cancel out. Lastly, $\displaystyle z\overline{z}$ is obviously real. Thus, $\displaystyle f(x)g(x)$ must also have real coefficients.

2) If $\displaystyle x\neq\frac{k\pi}{2^n}$ for some integer $\displaystyle n$ and odd integer $\displaystyle k$, then the infinite product certainly does not converge as $\displaystyle |\cos(2^nx)|<1$ for all $\displaystyle n$. Now if $\displaystyle x=\frac{k\pi}{2^n}$ for some $\displaystyle n>0$, then $\displaystyle \cos(2^{n-1}x)$ is 0, so the product diverges. Finally, if $\displaystyle x=\frac{k\pi}{2^n}$ for $\displaystyle n\leq0$, then the product converges to 1 if $\displaystyle k/2^n$ is even and -1 if $\displaystyle k/2^n$ is odd.
• Dec 3rd 2007, 07:37 PM
ThePerfectHacker
1)Lemma: If $\displaystyle z_0$ is a zero of $\displaystyle f(x)$ then $\displaystyle \bar z_0$ is a zero of $\displaystyle g(x)$. The proof is really simple say $\displaystyle f(x) = a_nx^n+...+a_1x+a_0$ then $\displaystyle a_nz_0^n+...+a_1z_0+a_0=0$. Now take the complex conjugate of both sides, but the complex conjugate preserves sums and products thus $\displaystyle \bar a_n (\bar z_0)^n+...+\bar a_1 (\bar z_0)+\bar a_0 = 0$. Thus, $\displaystyle \bar z_0$ is a zero of $\displaystyle g(x)$. Now consider the product $\displaystyle h(x)=f(x)g(x)$. Now if $\displaystyle z_0$ is a root of this polynomial $\displaystyle h(x)$ then $\displaystyle z_0$ is a zero of $\displaystyle f(x)$ WLOG, thus $\displaystyle \bar z_0$ is a zero of $\displaystyle g(x)$ and so $\displaystyle \bar z_0$ is a zero of $\displaystyle h(x)$. We have show that any zero of $\displaystyle h(x)$ also has its complex cunjugate as a zero. Thus, $\displaystyle h(x)$ 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 $\displaystyle \cos x \cos (2x) \cos (4x) = y$. Multiply both sides by $\displaystyle \sin x$ thus $\displaystyle \sin x \cos x \cos (2x)\cos (4x) = y\sin x$. Thus, $\displaystyle \frac{1}{2}\sin 2x \cos 2x \cos 4x = y\sin x$. Thus, $\displaystyle \frac{1}{2^2}\sin 4x \cos 4x = y\sin x$. Thus, $\displaystyle \frac{1}{2^3}\sin 8x = y\sin x$. Then supposing that $\displaystyle \sin x \not = 0$ (that is a special case) we have $\displaystyle y = \frac{\sin 8x}{2^3 \sin x}$. So in general if $\displaystyle P_n$ is the $\displaystyle n$-th partial product then $\displaystyle P_n = \frac{\sin 2^{n+1} x}{2^n \sin x}$ if $\displaystyle \sin x \not = 0$. Now it is easy to take the limit.
• Jun 17th 2009, 03:01 PM
Bruno J.
Here is another simple proof of the nice theorem (1) :

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

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

Hence $\displaystyle f(x)\overline{f(x)}$ is its own conjugate, i.e. $\displaystyle f(x)\overline{f(x)} \in \mathbb{R}[x]$.