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Old November 27th, 2007, 10:56 AM
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Default 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).
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Old November 27th, 2007, 04:49 PM
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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 05:38 PM.
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Old December 3rd, 2007, 09:37 PM
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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.
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Old June 17th, 2009, 05:01 PM
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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].
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