3

I am trying to solve the rational difference equation $\displaystyle x_{n+1}=\frac{a^2}{2a-x_{n}}$ using the roots of the characteristic equation.

The characteristic equation is $\displaystyle \lambda=\frac{a^2}{2a-\lambda}$, which can be rewritten as $\lambda^2-2a\lambda+a^2=0$ and $a$ is the repeated root.

Since the roots of the characteristic equation is a repeated root, I do not know how to proceed.

If the rational difference equation is $\displaystyle x_{n+1}=\frac{ab}{(a+b)-x_{n}}$, then the roots of the characteristic equation would be $a$ and $b$ and the substitution $\displaystyle y_n=\frac{x_n-a}{x_n-b}$ should work.

Is there any trick in the substitution? Or this approach does not work at all?

For reference, the solution of $\displaystyle x_{n+1}=\frac{a^2}{2a-x_{n}}$ is $\displaystyle x_n=a-\frac{a(a-x_1)}{na-(n-1)x_1}$.

Yuta
  • 1,493
  • For reference the solution for $a=1$ is $$x(n)=\frac{c_{1}(1+\sqrt{2})^{n}+(1-\sqrt{2})^{n}}{c_{1}(1+\sqrt{2})^{n+1}+(1-\sqrt{2})^{n+1}}$$ as computed by W|A here https://www.wolframalpha.com/input/?i=solve+x%5Bn%2B1%5D+%3D+1%2F%282%2Bx%5Bn%5D%29 – Calvin Khor May 09 '20 at 13:15
  • The concept of characteristic equation applies only to constant coefficient recurrences, of which yours is not an example. – Gerry Myerson May 09 '20 at 13:32

1 Answers1

3

Set $y_n=\dfrac{1}{x_n-a}$. After few computations, you find $y_{n+1}=y_n -a^{-1}$, which is easily solved.

GreginGre
  • 15,028
  • Thank you! And I am surprised to see that $y_n=\frac{1}{x_n-a}$ for $x_{n+1}=\frac{a^2}{2a-x_n}$ gives an arithmetic sequence while $y_n=\frac{x_n-a}{x_n-b}$ for $x_{n+1}=\frac{ab}{(a+b)-x_n}$ gives a geometric sequence in contrast. – Yuta May 09 '20 at 14:36
  • This is not so surprising if you know the theory hidden behind (conjugacy classes of homographies). – GreginGre May 09 '20 at 15:46