Solve Difference Equation for c_n

In summary, the recurrence equation c_{n+1} = \frac{c_0 (3c_n + c_0)}{2c_n + c_0} can be transformed into the linear form x_{n+1}=x_n+b_n with the transformation c_n=\frac{\alpha^{-n}}{x_n}+\beta. By solving for \alpha and \beta, we can obtain the expression x_{n+1} = \frac{2+\sqrt{3}}{2-\sqrt{3}} \frac{x_n}{\alpha} + \frac{2}{c_0\alpha^{n+1} (2-\sqrt{3})}, which can then be simplified to find a
  • #1
foxjwill
354
0

Homework Statement


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]


Homework Equations





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.
 
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  • #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
[tex]c_n=\frac{\alpha^{-n}}{x_n}+\beta[/tex]
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]
 
  • #3
Rainbow Child said:
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
[tex]c_n=\frac{\alpha^{-n}}{x_n}+\beta[/tex]
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]

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.
 
  • #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.
 
  • #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:
  • #6
You are almost there! :smile:
Choose for
[tex]\alpha=\frac{2+\sqrt{3}}{2-\sqrt{3}}[/tex]
in order to eliminate the coefficient of [itex]x_n[/itex]. Can you solve the resulting equation?
 

Related to Solve Difference Equation for c_n

1. What is a difference equation?

A difference equation is a mathematical equation that describes the relationship between a sequence of values. It shows how the values of a sequence change from one term to the next, usually based on a given mathematical rule or formula.

2. How do you solve a difference equation for c_n?

To solve a difference equation for c_n, you must first determine the type of difference equation it is (e.g., linear, nonlinear, homogeneous, non-homogeneous). Then, you can use various techniques such as substitution, iteration, or generating functions to find a general solution for c_n.

3. What is the difference between a difference equation and a differential equation?

The main difference between a difference equation and a differential equation is that a difference equation deals with discrete values (e.g., whole numbers), while a differential equation deals with continuous values (e.g., real numbers). Another difference is that a difference equation describes the relationship between a sequence of values, while a differential equation describes the relationship between a function and its derivatives.

4. Can you provide an example of solving a difference equation for c_n?

Sure, let's say we have the difference equation c_n = 2c_n-1 + 3 with initial condition c_0 = 1. We can solve this using iteration by plugging in the initial condition to find c_1, then using that value to find c_2, and so on. In this case, we would get c_1 = 5, c_2 = 13, c_3 = 29, and so on. The general solution for this difference equation is c_n = 2^n + 3/2.

5. What are some real-world applications of solving a difference equation for c_n?

Difference equations are commonly used in fields such as physics, engineering, and economics to model and predict the behavior of systems over time. For example, a difference equation can be used to model population growth, the spread of diseases, or the fluctuation of stock prices. Solving these equations for c_n can provide valuable insights and help make informed decisions in these industries.

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