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Can someone determine what this iteration works out to, where x' 
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#1
Apr712, 10:24 PM

P: 429

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) 


#2
Apr712, 11:07 PM

P: 4,579




#3
Apr812, 12:42 AM

P: 429




#4
Apr812, 12:44 AM

P: 429

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.



#5
Apr812, 12:56 AM

P: 429

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 


#6
Apr812, 01:24 AM

P: 429

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) 


#7
Apr812, 02:40 AM

P: 429

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. 


#8
Apr812, 11:43 AM

P: 429

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. 


#9
Apr812, 01:35 PM

Math
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Thanks
PF Gold
P: 39,682

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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