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Why is this series of square root of twos equal $\pi$?

Find the limit of the expression $2^{n+1}\sqrt{2-t_n}$ as $n\rightarrow\infty$, where $t_1=\sqrt{2}$, $t_2=\sqrt{2+\sqrt{2}}$, $t_3=\sqrt{2+\sqrt{2+\sqrt{2}}}$ and so on.

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marked as duplicate by David Mitra, Henry T. Horton, Thomas, Rahul, rschwieb Jan 2 '13 at 20:55

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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What is the context? What have you tried? Where did the question come from? –  Old John Jan 2 '13 at 19:59
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See here. –  David Mitra Jan 2 '13 at 20:06
    
The answer for $n = 10$ is, 3.1415923455701177423403759941574 –  Rajesh K Singh Jan 2 '13 at 20:10
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Quote: What is the context? What have you tried? Where did the question come from? Unquote. –  Did Jan 2 '13 at 20:30

2 Answers 2

The result is a "clever" disguise of the double-angle formulas.

Note that, if $t_n=2\cos(2u)$ for some $n$ and $u$ then $t_{n+1}=\sqrt{2+2\cos(2u)}=2\cos(u)$ (recall that, for every $v$, $\cos(2v)=2\cos^2(v)-1$). Hence, if $t_0=2\cos(u)$ for some $u$, then $t_n=2\cos(u/2^n)$ for every $n$.

Call $x_n=2^{n+1}\sqrt{2-t_n}$, then $x_n=2^{n+1}\sqrt{2-2\cos(u/2^n)}=2^{n+2}\sin(u/2^{n+1})$ (recall that, for every $v$, $\cos(2v)=1-2\sin^2(v)$), and in particular, $\lim\limits_{n\to\infty}x_n=2u$ (recall that $\sin(v)\sim v$ when $v\to0$).

If $t_1=\sqrt2$, then $u/2=\pi/4$ hence $\lim\limits_{n\to\infty}x_n=\pi$.

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$L_n = 2^{n+1} \sqrt{2-t_n}$. Then $L_n = 2^{n+1} \sqrt{2 - \sqrt{2+t_{n-1}}}$. Now $$2 - \sqrt{2+t_{n-1}} = \dfrac{4 - (2+t_{n-1})}{2 + \sqrt{2+t_{n-1}}} = \dfrac{2-t_{n-1}}{2 + t_n} = \dfrac{2-t_{n-1}}{t_{n+1}^2}$$ Hence, $$L_n = 2^{n+1} \sqrt{\dfrac{2-t_{n-1}}{t_{n+1}^2}} = \dfrac{2L_{n-1}}{t_{n+1}}$$ We have $L_0 = 2 \sqrt{2}$. Hence, $$L_n = \dfrac{2}{t_{n+1}} \cdot \dfrac{2}{t_{n}} \cdot \dfrac{2}{t_{n-1}} \cdots \dfrac{2}{t_{2}} L_0 = \dfrac{2^n}{\displaystyle \prod_{k=2}^{n+1} t_k} 2 \sqrt{2} = 2 \dfrac{2^{n+1}}{\displaystyle \prod_{k=1}^{n+1} t_k}$$ Hence, we need to evaluate $$S_n = \dfrac{\displaystyle \prod_{k=1}^n t_k}{2^n}$$ Recall that $t_n \uparrow 2$. Let $t_ n = 2 \cos(\theta_n)$. We then have $$\cos(\theta_{n+1}) = \dfrac{t_{n+1}}2 = \dfrac{\sqrt{2+t_n}}2 = \dfrac{\sqrt{2+2 \cos(\theta_n)}}2 = \cos(\theta_n/2)$$ Hence, we get that $$\cos(\theta_{n+1}) = \cos(\theta_1/2^n)$$ Hence, $$S_n = \dfrac{\displaystyle \prod_{k=1}^n t_k}{2^n} = \prod_{k=0}^{n-1} \cos(\theta_1/2^k) = \dfrac{\displaystyle \sin(\theta_1/2^{n-1})\prod_{k=0}^{n-1} \cos(\theta_1/2^k)}{\sin(\theta_1/2^{n-1})} = \dfrac1{2^n} \dfrac{\sin(2\theta_1)}{\sin(\theta_1/2^{n-1})}$$ Note that $\cos(\theta_1) = \dfrac1{\sqrt{2}}$, which implies $\theta_1 = \dfrac{\pi}4$. Hence, $$S_n = \dfrac1{2^n \sin(\pi/2^{n+1})}$$ Hence, $S_n \to \dfrac2{\pi}$. Hence, $$L_n \to \pi$$

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