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I'm trying to find a solution to the following recurrence:

$$ R_n=A+\frac{2AR_{n-1}}{2A+R_{n-1}} $$

I don't see any simple way to find the solution to this, but WolframAlpha gives the following elegant solution (unfortunately without steps):

$$ R_n=A\left(2-\frac{3}{c_14^{n+1}+1}\right) $$

For some constant $c_1$ I can find setting $R_0$. I tried to linearize it so far without success. Is there some other method I miss here? Thanks in advance.

Francesco Sollazzi
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    Do you just want a proof of the result or do you want to know how to come up with such a solution (if you get a slightly different problem in the future)? Because the proof is trivial- you just plug in the resulting expression in both sides – Albert Paradek Oct 17 '23 at 09:57
  • @AlbertParadek No, I can prove it easily by induction, but I want the reasoning on how to get from the recurrence to the solution. – Francesco Sollazzi Oct 17 '23 at 09:59
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    https://math.stackexchange.com/questions/3885418/finding-the-closed-form-of-a-recursive-sequence/3885693#3885693 – Claude Leibovici Oct 17 '23 at 10:42

2 Answers2

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$$R_n - 2A = \frac{2AR_{n-1}}{2A + R_{n-1}} -A = \frac{AR_{n-1} - 2A^2}{2A + R_{n-1}} = \frac{A(R_{n-1}-2A)}{(R_{n-1}-2A) +4A}$$ $$\implies \frac{1}{R_n - 2A} = \frac{(R_{n-1}-2A) +4A}{A(R_{n-1}-2A)} = \frac{1}{A} + 4\cdot \frac{1}{R_{n-1} - 2A}$$

By denoting $x_n = \frac{1}{R_n - 2A}$, we have:

$$\begin{align} & x_n = 4 x_{n-1} + \frac{1}{A}\\ &\implies \left(x_n + \frac{1}{3A} \right) = 4\left(x_{n-1} +\frac{1}{3A} \right)\\ & \implies \left(x_n + \frac{1}{3A} \right) = 4\left(x_{n-1} +\frac{1}{3A} \right) =...= 4^n \left(x_0 + \frac{1}{3A} \right)\\ & \implies x_n = 4^n \left(x_0 + \frac{1}{3A} \right) - \frac{1}{3A} \tag{1} \end{align}$$

From $(1)$, we deduce easily $R_n$ via this expression

$$\frac{1}{R_n -4A} = 4^n \left(\frac{1}{R_0 -4A} + \frac{1}{3A} \right) - \frac{1}{3A}$$ and $c_1$ must be equal to

$$c_1 = \frac{1}{R_0 -4A} + \frac{1}{3A}$$

NN2
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I have tried the solution for $c_1$ proffered by @NN2 in the equation provided by the OP. I find that they do not conform to a direct calculation of the recurrence relation.

I have two problems here. In the first place, I do not agree with the WA solution. I have found consistent error with WA where they have exponents of $n+1$ where it should be $n$. Thus, I suggest that the solution should be

$$ R_n=A\left(2-\frac{3}{c_14^n+1}\right) $$

This is consistent with my own derivation of the closed form. In that case, we can find $c_1$ directly from this equation as follows

$$ \begin{align} &\frac{R_0}{A}-2=-\frac{3}{c_1+1}\\ &\frac{c_1+1}{3}=-\frac{A}{R_0-2A}\\ &c_1=-\frac{3A}{R_0-2A}-1=-\frac{R_0+A}{R_0-2A}\\ \end{align} $$

This result has been tested against (a) the recurrence relation, (b) my own closed-form solution, and (c) two formulations of the Rational/Ricotti Difference Equation, as per Wikipedia, which I have programmed previously. I have tested these solutions with random complex numbers for both $R_0$ and $A$.

Cye Waldman
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