Descartes Sangaku : The Golem

Descartes' circle theorem gives a nice recursion for the radius of circle $R_{n+1}$ squeezed between two larger circles, $R_n$ and $R_{n-1}$ as

$\dfrac{1}{R_n} = \dfrac{1}{R_{n-1}} + \dfrac{1}{R_{n-2}} + \dfrac{2}{\sqrt{R_{n-1} R_{n-2}}}, \quad n \ge 3$ with $R_1 = 4, R_2 = 1$ .

We observe the first six values of $R_n$ using the recursion above

Since the numbers in the denominator seem to be squares of $1,2,3,5,8,13$ respectively, we make the following claim:

$R_n = \dfrac{4}{F_{n+1}^2}$ where $F_n$ denotes the $n^{th}$ Fibonacci number.

We prove this by strong induction. The base case is already taken care of as shown above. We now assume the above holds true $\forall i \in \mathbb{N} : i \le k$ . We now prove that $k+1$ case holds true.

To prove this, we use the recursion obtained at the beginning to get

$\begin{aligned} \dfrac{1}{R_{k+1}} &= \dfrac{1}{R_{k}} + \dfrac{1}{R_{k-1}} + \dfrac{2}{\sqrt{R_{k} R_{k-1}}} \\ &= \dfrac{F^2_{k+1}}{4} + \dfrac{F^2_{k}}{4} + \dfrac{F_k F_{k+1}}{2} \\ &= \left( \dfrac{F_k}{2} + \dfrac{F_{k+1}}{2} \right)^2 \\ &= \dfrac{F_{k+2}^2}{4} \end{aligned}$

thus the induction holds $\forall n \in \mathbb{N}$ and the formula holds true.

Hence,

$\lim_{n \to \infty} \sqrt{\dfrac{R_n}{R_{n+1}}} = \lim_{n \to \infty} \dfrac{F_{n+2}}{F_{n+1}} = \varphi$

Thus,

$\begin{aligned} \left( \lim_{n \to \infty} \sqrt{\dfrac{R_n}{R_{n+1}}} - \lim_{n \to \infty} \sqrt{\dfrac{R_{n+1}}{R_{n}}} \right)^{{2}/{3}} &= \left( \varphi - \frac{1}{\varphi} \right)^{{2}/{3}} \\ &= 1^{{2}/{3}} \\ &= 1 \end{aligned}$

• Using Descartes' Circle Theorem we find easily the following radii :
• $\boxed{R_1=1}$
• $\boxed{R_2=4}$
• $\boxed{R_3=\frac{4}{9}}$
• $\boxed{R_4=\frac{4}{25}}$
• $\boxed{R_5=\frac{1}{16}}$
• $\boxed{R_6=\frac{4}{169}}$
• $\boxed{R_7=\frac{4}{441}}$

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• They can be rewritten as :
• $\boxed{R_1=\frac{4}{1^{2}}}$
• $\boxed{R_2=\frac{4}{2^{2}}}$
• $\boxed{R_3=\frac{4}{3^{2}}}$
• $\boxed{R_4=\frac{4}{5^{2}}}$
• $\boxed{R_5=\frac{4}{8^{2}}}$
• $\boxed{R_6=\frac{4}{13^{2}}}$
• $\boxed{R_7=\frac{4}{21^{2}}}$

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• What notice that the consecutive denominators are the fibonacci numbers squared !
• We can conclude that $\boxed{R_n=\frac{2^{2}}{F_n^{2}}}$
• Thus $\boxed{\lim\limits_{n\to\infty}\sqrt{\frac{R_n}{R_{n+1}}}=\lim\limits_{n\to\infty}\frac{F_{n+1}}{F_{n}}}$ and $\boxed{\lim\limits_{n\to\infty}\sqrt{\frac{R_{n+1}}{R_{n}}}=\lim\limits_{n\to\infty}\frac{F_{n}}{F_{n+1}}}$
• Moreover, we know this famous identity $\boxed{\lim\limits_{n\to\infty}\frac{F_{n+1}}{F_n}=\phi=\frac{1+\sqrt{5}}{2}}$
• Then : $\boxed{(\lim\limits_{n\to\infty}\sqrt{\frac{R_n}{R_{n+1}}}-\lim\limits_{n\to\infty}\sqrt{\frac{R_{n+1}}{R_n}})^{\frac{2}{3}}=\left(\left(\frac{1+\sqrt{5}}{2}\right)^{\:}-\left(\frac{\sqrt{5}-1}{2}\right)^{\:}\right)^{\frac{2}{3}}=1}$