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

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In summary, the conversation revolves around finding the solution for an iteration where x' becomes x again each time, starting with x=1 and variables a and b. The solution involves a continued fraction and quadratic equations, and the final result is x = [1 - a - b + sqrt((1 - a - b)^2 + 4 b)] / (2 b), with the condition that the sequence is increasing.
  • #1
grav-universe
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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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  • #2


grav-universe said:
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?
 
  • #3


chiro said:
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.
 
  • #4


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


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


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


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


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


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 +.
 

1. How do you determine what an iteration works out to?

In order to determine what an iteration works out to, you must first understand the concept of an iteration. An iteration is a repetitive process that uses a set of instructions to generate a sequence of values. To determine what an iteration works out to, you would need to follow the instructions and track the sequence of values until the iteration reaches its stopping point.

2. What is the purpose of finding out what an iteration works out to?

The purpose of finding out what an iteration works out to is to understand the behavior and outcome of a repetitive process. This information can be used to analyze and improve the process, or to make predictions about future values in the sequence.

3. Can someone determine what an iteration works out to without knowing the initial value?

No, it is not possible to determine what an iteration works out to without knowing the initial value. The initial value is a crucial factor in the process and without it, the sequence of values will not be accurate. It is important to have all necessary information in order to accurately determine what an iteration works out to.

4. Is there a specific formula or method to determine what an iteration works out to?

There is no one specific formula or method to determine what an iteration works out to. It depends on the specific instructions and stopping point of the iteration. However, there are common mathematical and logical principles that can be applied to analyze and determine the outcome of an iteration.

5. Can the outcome of an iteration be predicted before it reaches its stopping point?

In some cases, it may be possible to predict the outcome of an iteration before it reaches its stopping point. This depends on the complexity and randomness of the process. With simple and predictable instructions, it may be possible to make accurate predictions. However, with more complex and unpredictable instructions, it may not be possible to accurately predict the outcome.

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