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Difference equation

  1. Feb 14, 2008 #1
    1. The problem statement, all variables and given/known data
    Find a closed-form expression for [tex]c_n[/tex].

    [tex]c_{n+1} = \frac{c_0 (3c_n + c_0)}{2c_n + c_0}[/tex]

    2. Relevant equations

    3. The attempt at a solution
    Besides finding [tex]c_1, c_2, c_3, \ldots[/tex] and looking for a pattern, I have absolutely no idea.
  2. jcsd
  3. Feb 14, 2008 #2
    Every recurrence equation of the form
    [tex]c_{n+1}=\frac{\kappa\,c_n+\lambda}{\mu\,c_n+\nu},\quad \kappa\,\nu-\mu\,\lambda\neq0[/tex]
    can be brought into the linear form [itex]x_{n+1}=x_n+b_n[/itex] with [itex]b_n[/itex] known.
    The transformation that does that, is
    Plug this to your equation and choose [itex]\alpha,\beta[/itex] in order to arrive to [itex]x_{n+1}=x_n+b_n[/itex]
  4. Feb 14, 2008 #3
    How do I get that to simplify into the required form? I tried plugging it in like this:

    [tex]\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][/tex]

    but I couldn't figure out how to simplify it without being left with a [tex]x_n x_{n+1} [/tex] term.
  5. Feb 15, 2008 #4
    You don't change [itex]c_0[/itex] into [itex]\frac{1}{x_0}+\beta[/itex] because it is a constant. Just write

    [tex]\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}[/tex]

    and calculate [itex]\alpha,\beta[/itex]. No term [itex]x_n\,x_{n+1}[/itex] must survive.
  6. Feb 15, 2008 #5
    Here's what I have so far:

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

    So, substituting in, I got
    [tex]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})}.[/tex]​

    This is where I get "stuck". I figured out a way to solve for [tex]x_n[/tex] that's analagous to finding an integrating factor for a linear ODE, but it involves really nasty algebra. Is there a better way?
    Last edited: Feb 15, 2008
  7. Feb 15, 2008 #6
    You are almost there! :smile:
    Choose for
    in order to eliminate the coefficient of [itex]x_n[/itex]. Can you solve the resulting equation?
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