# Solving nonlinear recursive relation

I am trying to solve a nonlinear recursive relation that reads $$$$\frac{1}{f(n+1)} = \frac{1}{f(n)+A} + B, \quad \text{for} \quad n=1,2,\ldots$$$$ where $$A$$ and $$B$$ are two positive constants. Is it possible to solve these exactly and/or approximately (eg, for large $$n$$) and obtain a form for $$f(n)$$, assuming $$f(1)$$ is known?

Any input would be appreciated as I know very little about recursive equations.

$$\frac 1 {f_{n+1}}=\frac 1 {f_n+a}+b\implies f_{n+1}=\frac{f_n+a } {b\,f_n+(1+ab) }=\frac{\frac 1bf_n+\frac ab } {f_n+\frac{1+ab}b }$$

Let $$m=\frac 1b$$, $$x=\frac ab$$, $$y=\frac{1+ab}b$$ to make $$f_{n+1}=\frac {m\,f_n+x}{f_n+y }$$ and have a look to my answer to this question where all steps are detailed.

• This is great! I didn't fully read the references cited in the said Wiki page - could you possibly spell out what $K$ is in $A_n$ and $B_n$ in your other answer? Commented Mar 15, 2022 at 17:04
• @SaMaSo. It is the constant to be fixed by the value of $f_0$ Commented Mar 16, 2022 at 3:11

You can write it in the following way:

$$u_{n+1}=\frac{u_n+A}{Bu_n+C} \ \text{with} \ C:=AB+1$$

This kind of recurrence is called a homographic sequence, and there are standard ways to express its general term ; see for example here explaining the connection with geometric or arithmetic sequences. In your case, the "fixed point equation" is given by the quadratic:

$$Bx^2+ABx-A=0$$

has always 2 real roots because $$\Delta=(AB)^2+4AB >0$$.

A matrix treatment is also possible: see this old answer of mine here.

• Thank you. Minor issue, but is it not $B x^2 +AB x-A=0$ ? Commented Mar 15, 2022 at 17:02
• You are right. I correct it. Commented Mar 15, 2022 at 17:45