Looking for General solution for a difference equation

In summary: So our case: q=\frac{p_{k}}{A+r+1}Alternatively you can use the explicit formula for the geometric series:1+q+q^{2}+q^{3}+...+q^{n}=\frac{1-q^{n+1}}{1-q}But the results is the same as above.
  • #1
ivyhawk
12
0
At+1=(At+r)/(At+r+1)

A1=constant

I know I can set At+1=At=A and solve for a special solution.

What would be a general solution?

I am not taking a course in Difference Equation, and this is not my homework but I encounter a similar question and I reduce it to this form.

Thanks
 
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  • #2
Or One step further, how to solve this one?

(1+r)A_t+1-A_t+A_t+1*A_t=r

A_t means A with t subsciptI don't think it is as easy as it looks.
 
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  • #3
ivyhawk said:
I don't think it is as easy as it looks.

It doesn't even look like anything. :tongue:

Is your second one meant to be

(1+r)At+1-At+At+1*At=r ?
 
  • #4
epenguin said:
It doesn't even look like anything. :tongue:

Is your second one meant to be

(1+r)At+1-At+At+1*At=r ?

Yes That is what I mean. And it is just algebra manipulation of the original equation
 
  • #5
Your starting form is

[tex]A_{t+1} = \frac{A_t + r}{A_t + r + 1},[/tex]
which is perhaps more convenient to write as

[tex]A_{t+1} = 1 - \frac{1}{A_t+r+1}[/tex]

What you could to do try to solve it is find A_2 in terms of A_1, plug that into A_3 to get it in terms of A_1, and so on, to see if you can guess what A_t for general t will look like. (I don't know if this one will turn out nice. I did the first couple of steps myself, and it looks like you may want to look into http://en.wikipedia.org/wiki/Continued_fraction (go to the page on periodic continued fractions too). There may be nice solutions for specific values of r or A_1. (Of course, A_1 will be irrelevant as t -> infinity)
 
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  • #6
Specific solution for A_t is easy to solve

Set A_t=A=A_t+1

We can solve that r=A/(1-A) or the other way, A=(1/2)(-r+(-)(r^2+4r)^1/2

But I still find it very hard to find a general solution.
 
  • #7
A general solution is hard to find.
But you can get some idea about the asymptotic behaviour when the sequence is approaching equilibrium A.

So assume your solution is

[tex]A_{k}=A+p_{k}[/tex]

Where the sequence p goes to zero as k increases. At the limit of large k, you can expand the RHS to a taylor series. Because we want a linear easy solution, we will only care about the first order. So:

[tex]A+p_{k+1}=1-\frac{1}{A+r+1}\frac{1}{1+\frac{p_{k}}{A+r+1}}=1-\frac{1}{A+r+1}(1-\frac{p_{k}}{A+r+1})[/tex]

Let A+r+1=b to make it look nicer. Also 1-1/b=A. So:

[tex]A+p_{k+1}=A+(1-A)^{2}p_{k}[/tex]

Or

[tex]p_{k+1}-(1-A)^{2}p_{k}=0[/tex]

So at large k's, the series will be:

[tex]A_{k}=A+(1-A)^{2k}[/tex]

This will be a real approximation only if abs(1-A)<1 (the solution is really stable), and that depends on what value of A you have chose, and what is r.
 
  • #8
How did u expand it?


elibj123 said:
A general solution is hard to find.
But you can get some idea about the asymptotic behaviour when the sequence is approaching equilibrium A.

So assume your solution is

[tex]A_{k}=A+p_{k}[/tex]

Where the sequence p goes to zero as k increases. At the limit of large k, you can expand the RHS to a taylor series. Because we want a linear easy solution, we will only care about the first order. So:

[tex]A+p_{k+1}=1-\frac{1}{A+r+1}\frac{1}{1+\frac{p_{k}}{A+r+1}}=1-\frac{1}{A+r+1}(1-\frac{p_{k}}{A+r+1})[/tex]

Let A+r+1=b to make it look nicer. Also 1-1/b=A. So:

[tex]A+p_{k+1}=A+(1-A)^{2}p_{k}[/tex]

Or

[tex]p_{k+1}-(1-A)^{2}p_{k}=0[/tex]

So at large k's, the series will be:

[tex]A_{k}=A+(1-A)^{2k}[/tex]

This will be a real approximation only if abs(1-A)<1 (the solution is really stable), and that depends on what value of A you have chose, and what is r.
 
  • #9
This is a geometrical series:

[tex]\frac{1}{1-q}=1+q+q^{2}+q^{3}+...[/tex]

setting q-> (-q) the terms will occur with alternating signs so the linear approximation will be

[tex]\frac{1}{1+q}=1-q[/tex]
 

1. What is a difference equation?

A difference equation is a mathematical equation that models the relationship between the values of a sequence or function at successive points. It is a discrete-time version of a differential equation and is often used in fields such as physics, engineering, and economics.

2. How is a difference equation different from a differential equation?

The main difference between a difference equation and a differential equation is that a difference equation deals with discrete values, while a differential equation deals with continuous values. This means that a difference equation is used to model systems that change in discrete steps, while a differential equation is used to model systems that change continuously.

3. What is the general solution for a difference equation?

The general solution for a difference equation is a formula that gives the values of the sequence or function at any point in time, based on the initial conditions and the equation itself. It is a way to express the sequence or function in a simplified form that can be used to find specific values or make predictions about the system.

4. How do you find the general solution for a difference equation?

To find the general solution for a difference equation, you need to solve the equation using methods such as substitution, iteration, or using a generating function. The specific method used will depend on the type of difference equation and its complexity.

5. What are some applications of difference equations in science?

Difference equations have many applications in science, including modeling population growth, chemical reactions, and electrical circuits. They are also used in fields such as economics, epidemiology, and computer science to study systems that change in discrete steps.

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