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ThePerfectHacker · #3 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.