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Evaluate $$\lim_{n\rightarrow \infty}\frac{\sqrt{1-x_0^2}}{x_1x_2...x_n}$$ where $x_{r+1}=\sqrt{\frac{1+x_r}{2}}; 0\leq r<n;\space r,n\in \mathbb{Z}$

My teacher says to substitute $x_0=\cos(\theta)$ but I don't understand why that is the case.

Is the $\cos(\theta)$ substitution arbitrary or is there any underlying logic to it?

Boy
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    It would be helpful in the recurrence relation. $(1+\cos{2\theta})/2 = \cos^2\theta$ – Nikunj Jun 30 '20 at 17:11
  • I see it makes the problem much more simpler. But how do we know that $x_0$ IS $\cos(\theta)$. – Boy Jun 30 '20 at 17:48
  • You're defining it to be $\cos\theta$, you could've defined it to be $\sin\theta$ too. – Nikunj Jun 30 '20 at 17:51

1 Answers1

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Take $x_0 = \cos \theta_0$

Note that $\implies x_r = \cos(\theta_0/2^r)$

(Using $(1+\cos2\theta) = 2\cos^2\theta$)

Your problem then becomes, $$\lim_{n\rightarrow\infty}\frac{\sin\theta_0}{\cos(\theta_0/2).\cos(\theta_0/4)...\cos(\theta_0/2^n)}$$ Note that $\sin\theta_0 = 2\sin(\theta_0/2)\cos(\theta_0/2)$

Plugging this in and cancelling the $\cos\theta_0/2$ in the denominator gives:

$$\lim_{n\rightarrow\infty}\frac{2\sin\theta_0/2}{\cos(\theta_0/4)...\cos(\theta_0/2^n)}$$

If you keep on repeating this process, you'll be left with $$\lim_{n\rightarrow\infty}2^n\sin(\theta_0/2^n)$$

Which is simple enough as this is just $$\lim_{n\rightarrow\infty}\frac{\theta_0\sin(\theta_0/2^n)}{\theta_0/2^n}$$

As $n \rightarrow \infty, (\theta_0/2^n) \rightarrow 0$

So this limit evaluates to $\theta_0$

Which is just $\cos^{-1}x_0$

Nikunj
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