Can someone determine what this iteration works out to, where x'


by grav-universe
Tags: determine, iteration
grav-universe
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#1
Apr7-12, 10:24 PM
P: 434
Can someone determine what this iteration works out to, where x' becomes x again each time, starting with x=1 and a and b are variables?

x' = (1 + x) / (b (1 + x) + a)
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chiro
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#2
Apr7-12, 11:07 PM
P: 4,570
Quote Quote by grav-universe View Post
Can someone determine what this iteration works out to, where x' becomes x again each time, starting with x=1 and a and b are variables?

x' = (1 + x) / (b (1 + x) + a)
This looks like it is going to be some kind of continued fraction, but it might end up simplifying to be something less complicated. Are you aware of continued fractions?
grav-universe
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#3
Apr8-12, 12:42 AM
P: 434
Quote Quote by chiro View Post
This looks like it is going to be some kind of continued fraction, but it might end up simplifying to be something less complicated. Are you aware of continued fractions?
Yes, thanks, but it doesn't seem to work out that way.

grav-universe
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#4
Apr8-12, 12:44 AM
P: 434

Can someone determine what this iteration works out to, where x'


Here's something that might help. If I put in a=1 and b=1, it works out to x_infinity = (sqrt(5) - 1) / 2.
grav-universe
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#5
Apr8-12, 12:56 AM
P: 434
Ahah. With b=1, the solution for x_infinity seems to be

x_inf^2 + a x_inf - 1 = 0

x_inf = a (sqrt(4 / a^2 + 1) - 1) / 2
grav-universe
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#6
Apr8-12, 01:24 AM
P: 434
The solution for x_inf with a=1 is

(x_inf^2 + x_inf) b - 1 = 0

x_inf = [sqrt(b) sqrt(b + 4) - b] / (2 b)
grav-universe
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#7
Apr8-12, 02:40 AM
P: 434
Okay, combining those two equations with (x_inf^2 + a x_inf) b - 1 = 0 still only gives the correct result for a=1 or b=1 only, so I figured there had to be something like (1 - a) (1 - b) in there somewhere to make the rest zero. The full solution for x_inf works out to be

(x_inf^2 + a x_inf) b - 1 - (1 - a) (1 - b) x_inf = 0

x_inf^2 b - (1 - a - b) x_inf - 1 = 0

x_inf = [1 - a - b + sqrt((1 - a - b)^2 + 4 b)] / (2 b)

Now I just need the solution for x_n with finite n.
grav-universe
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#8
Apr8-12, 11:43 AM
P: 434
Um, yeah. It was brought to my attention elsewhere that where x_inf converges, we have x' = x at the limit, so we can just rearrange the original equation with

x' = (1 + x) / (b (1 + x) + a)

x (b (1 + x) + a) - (1 + x) = 0

That works out just the same as the middle equation in the last post. Too bad I spent much of the day getting that far. :) Oh well. I might be able to use x_inf for my purposes, although finite x would still be very handy.
HallsofIvy
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#9
Apr8-12, 01:35 PM
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That equation is just a quadratic:
[tex]bx^2+ (a+ b- 1)x- 1= 0[/tex]
and so, by the quadratic formula,
[tex]x= \frac{1- a- b\pm\sqrt{(a+ b- 1)^2- 4b}}{2b}[/tex]
If you can show that the sequence, starting with x= 1, is increasing, you know that that [itex]\pm[/itex] must be +.


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