Given the sequence $x_1 = 0$, $x_{n+1} = \sqrt{2+x_n}$, proove: $$\lim_{n\to\infty} 2^n \sqrt{2-x_n} = \pi$$

I have used the relations $$2\cos({\frac x 2}) = \sqrt{2 + 2\cos(x)}$$ and $$2\sin({\frac x 2}) = \sqrt{2 - 2\cos(x)}$$

Observe: $x_1 = 0 = 2\cos(\alpha) \iff \alpha = \frac \pi 2$

I then hoped we can set $x_n = 2\cos(\frac{\alpha}{2^{n-1}})$ but if we can do this, why is that?

I assummed we can, so:

$\lim_{n\to\infty} 2^n \sqrt{2-x_n} = \lim_{n\to\infty} 2^n \sqrt{2-2\cos(\frac{\alpha}{2^{n-1}})} = \lim_{n\to\infty} 2^n \sqrt{4 \sin^2(\frac{\alpha}{2^n})}$

We know $$\alpha = \frac \pi 2$$

By substituting:

$ = \lim_{n\to\infty} 2^{n+1} \sin(\frac{\pi}{2^{n+1}})$

Again, by substituting $$\frac{\pi}{2^{n+1}} = \frac 1 m$$

$= \lim_{m\to\infty} \pi m \sin(\frac 1 m)= \pi$

Is this proof correct, and if not, how to prove it?

The even more important question is the why is that? part mentioned above.

  • $\begingroup$ It's absolutely correct. +1 $\endgroup$ – Paramanand Singh Nov 24 '17 at 5:21
  • $\begingroup$ @ParamanandSingh I have two questions in my post, the even more important one is the why is that? part $\endgroup$ – Math for fun Nov 24 '17 at 5:25
  • $\begingroup$ I would rather say "why do you doubt your solution"? If you want to be extra formal then you can prove that $x_{n} =2\cos(\pi/2^{n})$ using mathematical induction. Apart from that I don't know what else could be done to improve your smart solution. $\endgroup$ – Paramanand Singh Nov 24 '17 at 5:28
  • $\begingroup$ Ookay so my intuition was right about that "inductive thinking". But this is unusual because I took something quite arbitrarily which matches the recursive formula of the sequence and plugged it in. It works only because $x_1=0$. Are my intuition and my argumentation correct? $\endgroup$ – Math for fun Nov 24 '17 at 5:31
  • $\begingroup$ Mathematical induction is one of the simplest ways to turn your guess into a correct proof. $\endgroup$ – Paramanand Singh Nov 24 '17 at 5:32

$\cos(x) = \cos(2\cdot \frac{x}{2})= \cos^2(\frac{x}{2}) - \sin^2(\frac{x}{2}) = -1+ 2\cos^2(\frac{x}{2})$ and hence $\cos(\frac{x}{2})= \sqrt{\frac{1}{2}+\frac{1}{2}\cos(x)} = \frac{1}{2}\sqrt{2+2\cos(x)}$. That is the inductive step for the calculation.

Geometrically, you can check that $2^n \sqrt{2-x_n}$ is half the perimeter of a regular $2^{n+1}$-gon which is inscribed in the unit circle. Which for $n \to \infty$ approaches half the circumference of a unit circle, which is $\pi$.

Added: About your question what happens when you choose a different starting value $x_1 \in [-2,2]$ (note that outside this interval, at least one of the square roots in the calculation becomes unreal, and accordingly, the geometric interpretation breaks down).

  • In the calculation: You now get $\alpha = \displaystyle \arccos\left(\frac{x_1}{2}\right)$. The rest of the computation stays literally the same, just at the end you get the limit $\lim_{n \to \infty} 2^{n+1}\cdot \sin\displaystyle\left(\frac{\alpha}{2^n}\right) = 2\alpha$.
  • Geometrically: You get a polygonal chain of $2^n$ pieces, each of which is a chord within a circular segment of arc length $\displaystyle\frac{\alpha}{2^{n-1}}$. So the total arc in which this polygonal chain is inscribed has arc length $2\alpha$. As $n$ grows, the polygonal chain approximates the arc which gives the geometric reason for the limit being just $2\alpha$.

(Archimedes would be so happy with this.)

  • $\begingroup$ This exercise got me thinking: if $x_0$ had another value, we would only have to rotate the polygon, right? $\endgroup$ – Math for fun Nov 24 '17 at 6:26

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