# Solving recurrences of form $x_{n+1} = ax_n+\frac{b}{x_n}$?

Given the recurrence: $$x_{n+1} = ax_n+\frac{b}{x_n}$$ How do I determine the general solution? Is it possible to solve by generating functions? If I could have a small hint that would be great. Thanks.

• There is not going to be a general solution. Just the asymptotics, maybe. – Ivan Neretin Oct 20 '17 at 10:16
• Nah, there is going to be a solution. Or I will make one. – mtheorylord Oct 20 '17 at 10:39
• I like your attitude. As to the solution, let's wait for some third opinion. – Ivan Neretin Oct 20 '17 at 10:55
• For $\,a\neq 1\,$ it's enough to discuss the special case $\,\displaystyle z_{n+1} = az_n+\frac{1-a}{z_n}\,$ because with $\,\displaystyle z_n:=x_n/\sqrt{\frac{b}{1-a}}\,$ for all $\,n\,$ one gets the initial recurrence. – user90369 Oct 20 '17 at 12:24

## 1 Answer

Hint:

Let $$x_n=ku_n$$ ,

Then $$ku_{n+1}=aku_n+\dfrac{b}{ku_n}$$

$$u_{n+1}=au_n+\dfrac{b}{k^2u_n}$$

$$u_{n+1}=a\left(u_n+\dfrac{b}{ak^2u_n}\right)$$

Case $$1$$: $$ab>0$$

Take $$k=\dfrac{\sqrt b}{\sqrt a}$$ , the recurrence becomes

$$u_{n+1}=a\left(u_n+\dfrac{1}{u_n}\right)$$

• I'm sorry, I don't see how this is any easier than the original – mtheorylord Oct 23 '17 at 19:39