Solve Difference Equation for c_n

[foxjwill]

Homework Statement

Find a closed-form expression for $$c_n$$.

$$c_{n+1} = \frac{c_0 (3c_n + c_0)}{2c_n + c_0}$$

Homework Equations

The Attempt at a Solution

Besides finding $$c_1, c_2, c_3, \ldots$$ and looking for a pattern, I have absolutely no idea.

 

[Rainbow Child]

Every recurrence equation of the form
$$c_{n+1}=\frac{\kappa\,c_n+\lambda}{\mu\,c_n+\nu},\quad \kappa\,\nu-\mu\,\lambda\neq0$$
can be brought into the linear form $x_{n+1}=x_n+b_n$ with $b_n$ known.
The transformation that does that, is
$$c_n=\frac{\alpha^{-n}}{x_n}+\beta$$
Plug this to your equation and choose $\alpha,\beta$ in order to arrive to $x_{n+1}=x_n+b_n$

 

[foxjwill]

Every recurrence equation of the form
$$c_{n+1}=\frac{\kappa\,c_n+\lambda}{\mu\,c_n+\nu},\quad \kappa\,\nu-\mu\,\lambda\neq0$$
can be brought into the linear form $x_{n+1}=x_n+b_n$ with $b_n$ known.
The transformation that does that, is
$$c_n=\frac{\alpha^{-n}}{x_n}+\beta$$
Plug this to your equation and choose $\alpha,\beta$ in order to arrive to $x_{n+1}=x_n+b_n$

How do I get that to simplify into the required form? I tried plugging it in like this:

$$\frac{ \alpha ^{-n-1} }{x_{n+1} }+ \beta =\left( \frac{1}{x_{0} }+ \beta \right)\left[ \frac{3\left( \frac{ \alpha ^{-n} }{x_{n} }+ \beta \right)+\frac{1}{x_{0} }+ \beta }{2\left( \frac{ \alpha ^{-n} }{x_{n} }+ \beta \right)+\frac{1}{x_{0} }+ \beta } \right]$$

but I couldn't figure out how to simplify it without being left with a $$x_n x_{n+1}$$ term.

 

[Rainbow Child]

You don't change $c_0$ into $\frac{1}{x_0}+\beta$ because it is a constant. Just write

$$\frac{ \alpha ^{-n-1} }{x_{n+1} }+ \beta =c_0 \frac{3\left( \frac{ \alpha ^{-n} }{x_{n} }+ \beta \right)+c_0}{2\left( \frac{ \alpha ^{-n} }{x_{n} }+ \beta \right)+c_0}$$

and calculate $\alpha,\beta$. No term $x_n\,x_{n+1}$ must survive.

 

[foxjwill]

Here's what I have so far:

After making the substitution you suggested, I solved for $$\beta$$ by noticing that in order for the $$x_n x_{n+1}$$ term to disappear, either $$\alpha=0$$ or $$c_0^2 + 2c_0\beta -2\beta^2 = 0$$. Solving, I got $$\beta=\pm \frac{c_0}{2} (1+\sqrt{3}).$$ For convenience, I only used $$\beta = \frac{c_0}{2} (1+\sqrt{3}).$$ Am I allowed to do that?

So, substituting in, I got

$$x_{n+1} = \left ( \frac{2+\sqrt{3}}{2-\sqrt{3}} \right ) \frac{x_n}{\alpha} + \frac{2}{c_0\alpha^{n+1} (2-\sqrt{3})}.$$​

This is where I get "stuck". I figured out a way to solve for $$x_n$$ that's analagous to finding an integrating factor for a linear ODE, but it involves really nasty algebra. Is there a better way?

 

[Rainbow Child]

You are almost there!
Choose for
$$\alpha=\frac{2+\sqrt{3}}{2-\sqrt{3}}$$
in order to eliminate the coefficient of $x_n$. Can you solve the resulting equation?