SOLUTION TO PROBLEM H-4

Observe that $x_0 = 2\cos\left(\frac{\pi}{6}\right)$. Then

$$2+x_0=2+2\cos\left(\frac{\pi}{6}\right)= 2(1+\cos \left(\frac{\pi}{6}\right) = 4\cos^2\left(\frac{\pi}{12}\right)$$

(formula used: $1 + \cos(x) = 2\cos^2(\frac{x}{2})$). Therefore $x_1 = 2\cos\left(\frac{\pi}{12}\right)=2\cos\left(\frac{\pi}{2^2\cdot 3} \right)$. By a similar computation $x_2 = 2\cos\left(\frac{\pi}{24}\right)= 2\cos\left(\frac{\pi}{2^3\cdot 3}\right)$.

Assuming by induction that

$$x_n=2\cos\left(\frac{\pi}{2^{n+1}\cdot 3}\right) \leqno(1)$$

then

$$x_{n+1}=\sqrt{2+x_n}=\sqrt{2(1+\cos\left(\frac{\pi}{2^{n+1}\cdot 3}\right))} = 2\cos\left(\frac{\pi}{2^{n+2}\cdot 3}\right)$$

So formula (1) is proved by induction. Now

$$2-x_n=2(1-\cos\left(\frac{\pi}{2^{n+1}\cdot 3}\right))=4\sin^2\left( \frac{\pi}{2^{n+2}\cdot 3}\right)$$

(formula used: $1-\cos(x) = 2\sin^2(\frac{x}{2})$). So

$$2^{n+1}\cdot\sqrt{2-x_n} = 2^{n+2}\sin\left(\frac{\pi}{2^{n+2}\cd... ...dot \frac{\pi }{3} = \frac{\pi }{3} \qquad \mbox{as} \quad n\rightarrow \infty$$

because $\frac{\pi}{2^{n+2}\cdot 3}\longrightarrow 0$ and $\lim_{x \rightarrow 0}\frac{\sin(x)}{x} = 1$.

So the answer is $\frac{\pi}{3}$.

Questions and/or comments should be directed to Judy Downey or Griff Elder

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Last modified:   Sun Feb 10 15:33:16 CST 2002