2 problems found
In this question, the following theorem may be used without proof. Let \(u_1, u_2, \ldots\) be a sequence of real numbers. If the sequence is
Let \(a_{1}=\cos x\) with \(0 < x < \pi/2\) and let \(b_{1}=1\). Given that \begin{eqnarray*} a_{n+1}&=&{\textstyle \frac{1}{2}}(a_{n}+b_{n}),\\[2mm] b_{n+1}&=&(a_{n+1}b_{n})^{1/2}, \end{eqnarray*} find \(a_{2}\) and \(b_{2}\) and show that \[a_{3}=\cos\frac{x}{2}\cos^{2}\frac{x}{4} \ \quad\mbox{and}\quad \ b_{3}=\cos\frac{x}{2}\cos\frac{x}{4}.\] Guess general expressions for \(a_{n}\) and \(b_{n}\) (for \(n\ge2\)) as products of cosines and verify that they satisfy the given equations.
Solution: \begin{array}{c|c|c} n & a_n & b_n \\ \hline 1 & \cos x & 1 \\ \hline 2 & \frac12(1 + \cos x) & \sqrt{a_2} \\ &=\frac12(1+2\cos^2 \frac{x}{2}-1)& \sqrt{a_2} \\ &= \cos^2 \frac{x}{2} & \cos \frac{x}{2} \\ \hline 3 & \frac12(\cos^2 \frac{x}{2}+\cos \frac{x}{2}) & \sqrt{a_3\cos \frac{x}{2}} \\ &= \cos \frac{x}{2} \cdot \frac12 (\cos \frac{x}{2}+1) & \sqrt{a_3\cos \frac{x}{2}} \\ &= \cos \frac{x}{2} \cos^2 \frac{x}{4} & \cos \frac{x}{2} \cos \frac{x}{4} \end{array} Claim: \(\displaystyle a_n = \cos \frac{x}{2^{n-1}}\prod_{k=1}^{n-1} \cos \frac{x}{2^k}\), \(\displaystyle b_n = \prod_{k=1}^{n-1} \cos \frac{x}{2^k}\) Claim: \(a_{n+1} = \frac12(a_n + b_n)\) Proof: \begin{align*} && \frac12(a_n + b_n) &= \frac12 \left ( \cos \frac{x}{2^{n-1}}\prod_{k=1}^{n-1} \cos \frac{x}{2^k} + \prod_{k=1}^{n-1} \cos \frac{x}{2^k} \right) \\ &&&= \prod_{k=1}^{n-1} \cos \frac{x}{2^k} \frac12\left (\cos \frac{x}{2^{n-1}} + 1 \right) \\ &&&= \left ( \prod_{k=1}^{n-1} \cos \frac{x}{2^k} \right) \cos^{2} \frac{x}{2^n} \\ &&&= \cos \frac{x}{2^n} \prod_{k=1}^{n} \cos \frac{x}{2^k} \\ &= a_{n+1} \end{align*} Claim: \(b_{n+1} = \sqrt{a_{n+1}b_n}\) Proof: \begin{align*} && \sqrt{a_{n+1}b_n} &= \sqrt{ \cos \frac{x}{2^n} \prod_{k=1}^{n} \cos \frac{x}{2^k} \cdot \prod_{k=1}^{n-1} \cos \frac{x}{2^k} }\\ &&&= \prod_{k=1}^{n-1} \cos \frac{x}{2^k} \sqrt{\cos ^2\frac{x}{2^{n}}} \\ &&&= \prod_{k=1}^{n} \cos \frac{x}{2^k} \\ &&&= b_{n+1} \end{align*}