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If $a$ and $b$ are positive real numbers such that $a<b$ and if
$$a_1=\frac{a+b}{2}, b_1=\sqrt{(a_1b)},..., a_n=\frac{a_{n-1}+b_{n-1}}{2},b_n=\sqrt{a_nb_{n-1}},$$ then show that

$$\lim_{n\to\infty}b_n=\frac{\sqrt{b^2-a^2}}{\arccos\frac{a}{b}}.$$
I tried to calculate explicitly the first few terms $a_2,b_2$ etc but the terms got too complicated really quickly and I couldn't spot any pattern.

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  • $\begingroup$ It seems that $\lim b_n = \lim a_n$. I don't know if that helps. $\endgroup$ – Guy May 29 '17 at 13:03
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If we set $a=b\cos\theta$, we can show by induction $$b_n=b\prod^n_{k=1}\cos\frac{\theta}{2^k},\quad a_n=b_n\cos\frac{\theta}{2^k},$$ using $$\frac{1+\cos\frac{\theta}{2^k}}{2}=\cos^2\frac{\theta}{2^{k+1}}.$$ But $$\prod^n_{k=1}\cos\frac{\theta}{2^k}=\frac{\sin\theta}{\theta}\,\frac{\frac{\theta}{2^n}}{\sin\frac{\theta}{2^n}},$$ as we can show by repeated application of the identity $\sin2\alpha=2\sin\alpha\cos\alpha$, so the limit of the product as $n\rightarrow\infty$ is $\sin\theta/\theta$. Since $$b\sin\theta=b\sqrt{1-\cos^2\theta}=\sqrt{b^2-a^2},$$ the result follows.

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hint

$$0 <a <b \implies a=b\cos (\theta) $$

for example $$a_1=\frac {a+b}{2}=b\cos^2 (\theta/2) $$ $$b_1=b\cos (\theta/2) $$

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  • $\begingroup$ How to prove that $\Pi_{i=1}^\infty \cos(x/2^i) = \sin(x)/x$? $\endgroup$ – Guy May 29 '17 at 13:16
  • $\begingroup$ Use $\sin (2A )=2\sin (A)\cos (A) $ $\endgroup$ – hamam_Abdallah May 29 '17 at 13:19
  • $\begingroup$ Wow, that is neat! +1. I don't think its obvious though, so add it to your answer maybe. $\endgroup$ – Guy May 29 '17 at 13:22

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