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?