Prove that $ x_{k}=2^{k} \cdot \sin \frac{\pi}{2^{k}}$ equals $ x_{1}'=2, x'_{2}=2 \sqrt{2}, x_{k+1}=x_{k} \sqrt{\frac{2x_{k}}{x_{k}+x_{k-1}}}$.

This is what I've managed:

$x_{k+1}=x_{k} \sqrt{\frac{2x_{k}}{x_{k}+x_{k-1}}}= 2^{k} \cdot \sin \frac{\pi}{2^{k}} \sqrt{\frac{2^{k+1} \cdot \sin \frac{\pi}{2^{k}}}{2^{k} \cdot \sin \frac{\pi}{2^{k}}+2^{k-1} \cdot \sin \frac{\pi}{2^{k}}}}=2^{k} \cdot \sin \frac{\pi}{2^{k}} \sqrt{\frac{2}{1+\cos \frac{\pi}{2^{k}}}}$

And I don't see how to proceed....

  • 1
    $\begingroup$ I see a mistake: by writing out $x_{k-1}$ the exponent of $2$ must change into $k-1$ instead of $k$ $\endgroup$
    – drhab
    Oct 16, 2013 at 13:24

1 Answer 1


You can use that

$$\cos (2x) = \cos^2 x - \sin^2 x,$$

and hence

$$1 - \cos \frac{\pi}{2^k} = 1 - \cos^2 \frac{\pi}{2^{k+1}} + \sin^2 \frac{\pi}{2^{k+1}} = 2\sin^2 \frac{\pi}{2^{k+1}}.$$

Then write

$$\sqrt{\frac{2}{1+\cos \frac{\pi}{2^k}}} = \sqrt{\frac{2(1 - \cos \frac{\pi}{2^k})}{1 - \cos^2 \frac{\pi}{2^{k}}}} = \sqrt{\frac{4\sin^2 \frac{\pi}{2^{k+1}}}{\sin^2 \frac{\pi}{2^k}}} = \frac{2\sin \frac{\pi}{2^{k+1}}}{\sin \frac{\pi}{2^k}},$$

since all the sines involved are non-negative.


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