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Let $\phi$ be the golden ratio. We know it has a beautiful infinite nested radical,

$$\sqrt{1+\sqrt{1+\sqrt{1+\sqrt{1+\dots}}}}=\phi$$

However, it is also the case that,

$$3+\sqrt{11+\sqrt{11+\sqrt{11+\sqrt{11+\dots}}}}=\phi^4$$

$$5+\sqrt{31+\sqrt{31+\sqrt{31+\sqrt{31+\dots}}}}=\phi^5$$

$$\tfrac{17}2+\sqrt{\tfrac{319}4+\sqrt{\tfrac{319}4+\sqrt{\tfrac{319}4+\sqrt{\tfrac{319}4+\dots}}}}=\phi^6$$


Q: How do we show that, in general

$$a_n+\sqrt{b_n+\sqrt{b_n+\sqrt{b_n+\sqrt{b_n+\dots}}}}=\phi^n$$

where,

$$a_n = \frac{L_n-1}2,\quad b_n = \frac{5F_n^2-1}4$$

with Lucas numbers $L_n$ and Fibonacci numbers $F_n$?

Tito Piezas III
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  • I knew the expression $5F_n^2$ looked familiar. Turns out $L_n^2-5F_n^2 = 4(-1)^n$, so $a_n,b_n$ can be expressed purely in terms of the Lucas numbers. – Tito Piezas III Mar 22 '18 at 14:52
  • This article on the [Mandelbrot Set](https://en.wikipedia.org/wiki/Mandelbrot_set) might be if use. – Mark Viola Mar 22 '18 at 15:06

1 Answers1

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If we plug in the given formulas we get the famous formula. We note: $$\sqrt{b_n+x}=x \Rightarrow x=\frac{1+\sqrt{1+4b_n}}{2}=\frac{1+F_n\sqrt{5}}{2};$$ Hence: $$\frac{L_n-1}{2}+\frac{1+F_n\sqrt{5}}{2}=\frac{L_n+F_n\sqrt{5}}{2}=\phi^n.$$ This is a famous formula that relates the two sequences. See: https://en.m.wikipedia.org/wiki/Lucas_number

farruhota
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