Math Help - Getting around interacting sequences

1. Getting around interacting sequences

The problem I'm currently facing is how to produce and expression to describe the path of a particle in order that it's position can be calculated with a given time. So similar to your suvat equations of motion, this is all fine and good until you start trying to derrive equations where the next x coordinate, for example, is dependent somehow on the last y an . The same is true for y. One instance of this is gravity between two bodies, the displacement on x and y of a body is dependent on the distance between the two, which sure enough can only be calculated from the previous x,y displacement and if i try to rearrange this I end up with an infinite cycle of substitution. Mandelbrot and attractors springs to mind.

This seems to shows up everywhere in physics and I've not yet got my head round it. Effectively you have two interacting sequences, I've diluted this idea into a simple sequence where p and q are constants and a and b begin at 0:

$
a_{n+1} = a_{n} * b_{n} +p
$

$
b_{n+1} = 2*a_{n} * (b_{n} - 3) +q
$

I've graphed this in quickly flash, excuse the scaleless graph! Fork it and add scales if you into as3 flash on 2010-5-19 | wonderfl build flash online
so if you follow that link you'll see a spiral slowly changing before tightening up and apparently exploding. What happening is I'm iterating over that sequence 500 times and plotting each point (a = x, b=y), then once it's been plotted it's displayed, q is increased by 0.001 and it is calculated and replotted. This is happening 30 times a second so you can see a gradual change. Is this an attractor? I'm certainly new to all this stuff

So my question is how can work around this? Say I wanted to derive and expression for the x & y of an paritcle if its is being attracted by a static body? Is this possible without iterating over each value?

Thanks, apologies for the long post!

2. Originally Posted by Geo877
The problem I'm currently facing is how to produce and expression to describe the path of a particle in order that it's position can be calculated with a given time. So similar to your suvat equations of motion, this is all fine and good until you start trying to derrive equations where the next x coordinate, for example, is dependent somehow on the last y an . The same is true for y. One instance of this is gravity between two bodies, the displacement on x and y of a body is dependent on the distance between the two, which sure enough can only be calculated from the previous x,y displacement and if i try to rearrange this I end up with an infinite cycle of substitution. Mandelbrot and attractors springs to mind.

This seems to shows up everywhere in physics and I've not yet got my head round it. Effectively you have two interacting sequences, I've diluted this idea into a simple sequence where p and q are constants and a and b begin at 0:

$
a_{n+1} = a_{n} * b_{n} +p
$

$
b_{n+1} = 2*a_{n} * (b_{n} - 3) +q
$

I've graphed this in quickly flash, excuse the scaleless graph! Fork it and add scales if you into as3 flash on 2010-5-19 | wonderfl build flash online
so if you follow that link you'll see a spiral slowly changing before tightening up and apparently exploding. What happening is I'm iterating over that sequence 500 times and plotting each point (a = x, b=y), then once it's been plotted it's displayed, q is increased by 0.001 and it is calculated and replotted. This is happening 30 times a second so you can see a gradual change. Is this an attractor? I'm certainly new to all this stuff

So my question is how can work around this? Say I wanted to derive and expression for the x & y of an paritcle if its is being attracted by a static body? Is this possible without iterating over each value?

Thanks, apologies for the long post!
With a bit of manipulation you can find a second order non-linear recurrence for $a_n$, if my algebra is error free it is:

$a_{n+1}=2a_n^2-6a_na_{n-1}+a_a(q-2p)+p$

I am not familiar with solving such recurrences, but there may be a method of doing so.

CB