What Are the Key Characteristics of Henon Attractors?

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Henon attractors are analyzed through the Henon map, defined recursively with specific equations for x and y. To find the fixed point, all variables must equal a common value, x_b, leading to a quadratic equation. The discussion highlights confusion regarding an extra term in the quadratic equation, specifically the -1 in the b-1 term, which arises from rearranging the equation. Additionally, participants share tips on formatting LaTeX equations, emphasizing the use of "\\" for new lines. Understanding these aspects is crucial for studying the chaotic behavior of Henon attractors.
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1. Homework Statement

I am studying Henon Attractors. The Henon map is recursively defined as follows:

x_{t+1} = a - x^2_{t} + by_{t}<br /> y_{t+1} = x_{t}

I am supposed to find the fixed point (may be unstable) that is contained with the chaotic behavior

The Attempt at a Solution



It is clear that to find the fixed point would be when x_{t+1}, x_{t}, y_{t}
are all equal (Lets call them all x_{b}). It seemed obvious to just plug in
x_{b} and solve the quadratic, but the book had an extra term in the answer and I do not know where it came from:
-x^2_{b} + (b-1) x_{b} +a = 0. Where did the -1 in the b-1 term come from?
 
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If you just plug in x_b for x_t, y_t, and x_{t+1}, then the equation you get is exactly what the book says. Do your math again.
 
Oh, yes, the x_{b} on the left side is subtracted from the right to make it equal to 0. Duh!

Could you now how I insert a new line in the Latex equation? I can't seem to keep those two first equations from running together.
 
MathAmateur said:
Oh, yes, the x_{b} on the left side is subtracted from the right to make it equal to 0. Duh!

Could you now how I insert a new line in the Latex equation? I can't seem to keep those two first equations from running together.

The latex symbol for new line is \\
 
I did the following and got the following result:

x_{t+1} = a - x^2_{t} + by_{t} \\
y_{t+1} = x_{t}

x_{t+1} = a - x^2_{t} + by_{t}\\<br /> y_{t+1} = x_{t}
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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