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Given $$a_n=\sqrt{2-a_{n-1}},a_1=\sqrt{2}$$

I calculated $a_1$ to $a_5$ $$\sqrt{2}, \sqrt{2-\sqrt{2}}, \sqrt{2-\sqrt{2-\sqrt{2}}}, \\ \sqrt{2-\sqrt{2-\sqrt{2-\sqrt{2}}}}, \\ \sqrt{2-\sqrt{2-\sqrt{2-\sqrt{2-\sqrt{2}}}}}$$

which made me think of $\sin/\cos$. So I divided each by $2$, calculated $\arcsin(x)$ and got $$\left\{\frac{\pi }{4},\frac{\pi }{8},\frac{3 \pi }{16},\frac{5 \pi }{32},\frac{11 \pi }{64}\right\}$$

I found that this could be a formula $$\frac{\pi}{3} \cdot (-1)^n \cdot\frac{(-2)^n-1}{2^{n+1}}$$

So $$\sin \left(\frac{\pi}{3} \cdot (-1)^n \cdot\frac{(-2)^n-1}{2^{n+1}}\right)$$ is $a_n/2$, but this way seems not natural. Is there a more natural way?

rtybase
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AsukaMinato
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2 Answers2

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If the question is about finding the limit, let's consider $a_{n+1}=f(a_n)$, where $f(x)=\sqrt{2-x}$. Then we have

If $0\leq x \leq \sqrt{2}$ then $0\leq f(x)\leq\sqrt{2}$

Indeed $0\leq x\leq\sqrt{2} \Rightarrow 0\geq-x \geq -\sqrt{2} \Rightarrow 2\geq 2-x \geq 2-\sqrt{2}>0 \Rightarrow \sqrt{2}\geq \sqrt{2-x}=f(x)\geq 0$.

Now, let's use Banach fixed-point theorem, and MVT, given $f'(x)=-\frac{1}{2\sqrt{2-x}}$, for $\forall x,y \in[0,\sqrt{2}], x<y$, there $\exists\varepsilon\in (x,y)$ s.t. $$|f(x)-f(y)|=|f'(\varepsilon)|\cdot |x-y|= \frac{1}{2\sqrt{2- \varepsilon}}\cdot|x-y|< \frac{1}{2\sqrt{2-\sqrt{2}}}\cdot|x-y|$$ since $\varepsilon\in[0,\sqrt{2}]$ as well. It's not to difficult to check that $0<\frac{1}{2\sqrt{2-\sqrt{2}}}<1$.

So, the limit exists and you can legitimately use $L=\sqrt{2-L}$ to find it, considering that $L\in[0,\sqrt{2}]$ of course, since all $(a_n)_{n>0} \subset[0,\sqrt{2}]$.

Remark: More interesting results here.

The trick that is typically applied for the $+$ with $\cos$ may not easily apply for $\arcsin$ and $\sin$ since: $$\sin{\frac{\pi}{8}}=\frac{1}{2}\sqrt{2-\sqrt{2}}$$ $$\sin{\frac{\pi}{16}}=\frac{1}{2}\sqrt{2-\sqrt{2\color{red}{+}\sqrt{2}}}$$ I'd rather try induction, given that $$\sin{\frac{\color{red}{1}\cdot\pi}{4}}=\frac{\sqrt{2}}{2}$$ $$\sin{\frac{\color{red}{1}\cdot\pi}{8}}=\frac{1}{2}\sqrt{2-\sqrt{2}}$$ $$\sin{\frac{\color{red}{3}\cdot\pi}{16}}=\frac{1}{2}\sqrt{2-\sqrt{2-\sqrt{2}}}$$ $$\sin{\frac{\color{red}{5}\cdot\pi}{32}}=\frac{1}{2}\sqrt{2-\sqrt{2-\sqrt{2-\sqrt{2}}}}$$ $$\sin{\frac{\color{red}{11}\cdot\pi}{64}}=\frac{1}{2}\sqrt{2-\sqrt{2-\sqrt{2-\sqrt{2-\sqrt{2}}}}}$$ $$\sin{\frac{\color{red}{21}\cdot\pi}{128}}=\frac{1}{2}\sqrt{2-\sqrt{2-\sqrt{2-\sqrt{2-\sqrt{2-\sqrt{2}}}}}}$$

where $\{1,1,3,5,11,21\}$ is the begining of the Jacobsthal sequence, according to OEIS.

And of course: $$\sin^2\left(\frac{\pi}{4}-\frac{x}{2}\right)=\frac{1}{2}\left(1-\sin{x}\right)\tag{1}$$

Jacobsthal sequence is $J_{n+1}=2^n-J_n$ and, assuming induction hypothesis, we have $$\sqrt{\frac{1}{2}\left(1-\sin\left(\color{red}{J_n}\frac{\pi}{2^{n+1}}\right)\right)}=\\ \sqrt{\frac{1}{2}\left(1-\frac{1}{2}\sqrt{2-\sqrt{2-...-\sqrt{2-\sqrt{2}}}}\right)}=\\ \sqrt{\frac{1}{4}\left(2-\sqrt{2-\sqrt{2-...-\sqrt{2-\sqrt{2}}}}\right)}=\\ \frac{1}{2}\sqrt{2-\sqrt{2-\sqrt{2-...-\sqrt{2-\sqrt{2}}}}}\overset{(1)}{=}\\ \sin\left(\frac{\pi}{4}-J_n\frac{\pi}{2^{n+2}}\right)= \sin\left(\left(2^n-J_n\right)\frac{\pi}{2^{n+2}}\right)= \sin\left(\color{red}{J_{n+1}}\frac{\pi}{2^{n+2}}\right)$$ The positive sign of the square root above, while appling $(1)$, is justified by $\frac{\pi}{4}>\frac{\pi}{4}-J_n\frac{\pi}{2^{n+2}}>0$, where the $\sin$ function is positive. As a result: $$2\sin\left(J_{n}\frac{\pi}{2^{n+1}}\right)=\underbrace{\sqrt{2-\sqrt{2-\sqrt{2-...-\sqrt{2-\sqrt{2}}}}}}_{n\text{ times}} \tag{2}$$

Jacobsthal sequence also has a closed form of $$ J_n = \frac{2^n - (-1)^n}{3}$$ which can be solved using, for example, characteristic polynomials (more than half of the work is done here), leading to

$$2\sin\left(\frac{2^n-(-1)^n}{2^{n+1}}\cdot\frac{\pi}{3}\right)=\underbrace{\sqrt{2-\sqrt{2-\sqrt{2-...-\sqrt{2-\sqrt{2}}}}}}_{n\text{ times}} \tag{3}$$

Remark: It is worth noting this question is not duplicating this family of questions.

rtybase
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1

Let $x_n$ for $n\in\Bbb N$ be the sequence defined by \begin{align} x_0&=\frac\pi 2& x_n&=\frac{\pi-x_{n-1}}2 \end{align} and $a_n=2\cos(x_n)$. Then $0\leq x_n\leq\frac\pi 2$, hence $0\leq a_n\leq 2$. Moreover: \begin{align} a_n^2 &=4\cos^2(x_n)\\ &=2\cos(2x_n)+2\\ &=2\cos(\pi-x_{n-1})+2\\ &=-2\cos(x_{n-1})+2\\ &=-a_{n-1}+2 \end{align} so that $a_n=\sqrt{2-a_{n-1}}$. On the other hand $x_n=\frac\pi 3+(-2)^{-n}(x_0-\frac\pi 3)$, hence $$a_n=2\cos\Bigl(\frac\pi 3+(-2)^{-n}\frac\pi 6\Bigr)$$ gives an explicit formula for $a_n$.

Fabio Lucchini
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