# Solve Difference Equation for c_n

• foxjwill
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
foxjwill

## Homework Statement

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

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

## The Attempt at a Solution

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

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 $x_{n+1}=x_n+b_n$

Rainbow Child said:
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.

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.

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?

Last edited:
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?

## 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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