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let sequence $\{a_{n}\},\{b_{n}\}$ such $$a_{1}=1,b_{1}=3$$ and $$\begin{cases} a_{n+1}=2+\dfrac{27a_{n}}{9a^2_{n}+4b^2_{n}}\\ b_{n+1}=\dfrac{27b_{n}}{9a^2_{n}+4b^2_{n}} \end{cases}$$

Find $a_{n},b_{n}$

My idea: since

$$\dfrac{a_{n+1}-2}{b_{n+1}}=\dfrac{a_{n}}{b_{n}}$$ then I can't.Thank you

1 Answers1

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As stated in my previous comments, by setting $z_n=3a_n+2i b_n$ we get: $$ z_{n+1} = 6+\frac{27}{\overline{z_n}}\tag{1} $$ with $z_1 = 3+6i$. By setting $z_n = 3\sqrt{3}\, w_n$ we get: $$w_{n+1} = \frac{2}{\sqrt{3}}+\frac{1}{\overline{w_n}}\tag{2} $$ with $w_1=\frac{1}{\sqrt{3}}(1+2i)$. $(2)$ gives that the sequence $\{w_n\}_{n\in\mathbb{N}}$ is the orbit of an element in $PSL(2,\mathbb{C})$. Iterating twice we can remove the complex conjugation and get: $$w_{n+2}=\frac{2}{\sqrt{3}}+\frac{1}{\frac{2}{\sqrt{3}}+\frac{1}{w_n}}.\tag{3}$$ In order to have $a_n$ and $b_n$ we only need to find a closed form for the sequence given by: $$t_{n+1} = \frac{2}{\sqrt{3}}+\frac{1}{t_n}.\tag{4}$$ Since the roots of the polynomial $x^2-\frac{2}{\sqrt{3}}x-1$ are $-\frac{1}{\sqrt{3}}$ and $\frac{3}{\sqrt{3}}$, with the choice $$ t_n = \frac{A\left(-\frac{1}{\sqrt{3}}\right)^{n+1}+B\left(\frac{3}{\sqrt{3}}\right)^{n+1}}{A\left(-\frac{1}{\sqrt{3}}\right)^{n}+B\left(\frac{3}{\sqrt{3}}\right)^{n}}$$ equation $(4)$ is fulfilled, and we only need to choice $A$ and $B$ such that $t_1$ and $t_2$ fit with our starting values.

Jack D'Aurizio
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    Great answer. How do you notice that relation in complex view? – Shuchang Feb 15 '14 at 03:15
  • I just put together the facts that the inverse of $x+iy$ is $\frac{x-iy}{x^2+y^2}$ and that the solution of a recurrence like $x_{n+1}=Cx_n+Dy_n, y_{n+1}=Ex_n+Fy_n$ is given by $w_n = M^n w_0$, where $w_n=(x_n,y_n)$ and $M$ is a suitable $2\times 2$ matrix. – Jack D'Aurizio Feb 15 '14 at 14:15