Looking for General solution for a difference equation

[ivyhawk]

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

 

[ivyhawk]

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.

 

[epenguin]

I don't think it is as easy as it looks.

It doesn't even look like anything.

Is your second one meant to be

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

 

[ivyhawk]

It doesn't even look like anything.

Is your second one meant to be

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

Yes That is what I mean. And it is just algebra manipulation of the original equation

 

[Mute]

Your starting form is

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

$$A_{t+1} = 1 - \frac{1}{A_t+r+1}$$

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)

 

[ivyhawk]

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.

 

[elibj123]

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

$$A_{k}=A+p_{k}$$

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:

$$A+p_{k+1}=1-\frac{1}{A+r+1}\frac{1}{1+\frac{p_{k}}{A+r+1}}=1-\frac{1}{A+r+1}(1-\frac{p_{k}}{A+r+1})$$

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

$$A+p_{k+1}=A+(1-A)^{2}p_{k}$$

Or

$$p_{k+1}-(1-A)^{2}p_{k}=0$$

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

$$A_{k}=A+(1-A)^{2k}$$

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.

 

[ivyhawk]

How did u expand it?

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

$$A_{k}=A+p_{k}$$

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:

$$A+p_{k+1}=1-\frac{1}{A+r+1}\frac{1}{1+\frac{p_{k}}{A+r+1}}=1-\frac{1}{A+r+1}(1-\frac{p_{k}}{A+r+1})$$

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

$$A+p_{k+1}=A+(1-A)^{2}p_{k}$$

Or

$$p_{k+1}-(1-A)^{2}p_{k}=0$$

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

$$A_{k}=A+(1-A)^{2k}$$

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.

 

[elibj123]

This is a geometrical series:

$$\frac{1}{1-q}=1+q+q^{2}+q^{3}+...$$

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

$$\frac{1}{1+q}=1-q$$