grav

2012-Apr-08, 03:30 AM

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)

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

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grav

2012-Apr-08, 03:30 AM

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)

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

frankuitaalst

2012-Apr-08, 08:45 AM

Hard to tell Grav , I think . Depends strongly on the coefficients a,b . Fi . If b=1 ; a=0 you'll get x'=1 all over .

There is a good chance that you will end up at the roots of the polynomial , which is a kwadratic : x*(b(1+x)+a)-(1+x)=0

There is a good chance that you will end up at the roots of the polynomial , which is a kwadratic : x*(b(1+x)+a)-(1+x)=0

antoniseb

2012-Apr-08, 12:46 PM

Just trying some values:

- If b is more than 1, and a is positive, then for positive initial x it approaches zero.

- If b is less than 1 and a is small than it approaches infinity

- if b is 1 and a is positive, then for positive initial x is approaches zero.

- For initial x = -(b+a)/b you can't start.

Perhaps there is some convergence if b is near one, and a has some special value WRT b. Also, I obviously didn't look at negative numbers for any of a, b, or x-prime.

- If b is more than 1, and a is positive, then for positive initial x it approaches zero.

- If b is less than 1 and a is small than it approaches infinity

- if b is 1 and a is positive, then for positive initial x is approaches zero.

- For initial x = -(b+a)/b you can't start.

Perhaps there is some convergence if b is near one, and a has some special value WRT b. Also, I obviously didn't look at negative numbers for any of a, b, or x-prime.

grav

2012-Apr-08, 04:19 PM

Thanks, frankuitaalst and antisoneb. I found that for a=1 and b=1, antisoneb, we get x_infinity = (sqrt(5) - 1) / 2. For a=1, we get

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

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

and for b=1, we get

x_inf^2 + a x_inf - 1 = 0

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

but combining the two with

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

is still only correct for a=1 and b=1, so there had to be something like (1 - a) (1 - b) in there somewhere, which worked 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)

But the second to last equation turns out to be just the same as the polynomial quadratic root frankuitaalst gave, though, which looks like it could have been found by simply rearranging the original equation if I had thought about it, since x' = x at the limit, so it is too bad I spent half the day finding it by trial and error lol. Finding it for finite x would also be very handy, but I might be able to use x_inf for my purposes, I'll see. Thank you both very much for your help. :)

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

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

and for b=1, we get

x_inf^2 + a x_inf - 1 = 0

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

but combining the two with

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

is still only correct for a=1 and b=1, so there had to be something like (1 - a) (1 - b) in there somewhere, which worked 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)

But the second to last equation turns out to be just the same as the polynomial quadratic root frankuitaalst gave, though, which looks like it could have been found by simply rearranging the original equation if I had thought about it, since x' = x at the limit, so it is too bad I spent half the day finding it by trial and error lol. Finding it for finite x would also be very handy, but I might be able to use x_inf for my purposes, I'll see. Thank you both very much for your help. :)

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