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Bifurcations and periods in chaos with HP50G
09-09-2022, 11:14 PM (This post was last modified: 09-09-2022 11:15 PM by Gil.)
Post: #1
Bifurcations and periods in chaos with HP50G
I want to check the results of:

X[n+1] = a - X[n]²
for a=1.25 (bifurcation begin for supposed 4? fixed points) and n—> infinity,
starting with X[0]=0.4

X[1] = 1.25 - 0.4² = 1.25 - 0.16 = 1.09.

Then, with phone EMU48 HP50G
\<< .4 0. 'S' STO
WHILE S 1000000. \<=
REPEAT 1. 'S' STO+ 1.25 SWAP SQ - S \->STR \->TAG
END
\>>

we seem to get
our approximated fixed points :
X[1000001] = 1.20693554923
X[1000002] = -.20669342
X[1000003] = 1.20727783013
X[1000004] = -.20751975912.

Note : 4 points!

Carrying on, we get with the calculator:
X[1000005] =
1.20693554957
(almost, as expected, the same value
found for X[1000001]
X[1000006] = -.20669342
X[1000007] = 1.20727783013
X[1000008] = -.20751975912
Now we write
'X[5] = a - X[4]²'
'X[4] = a - X[3]²' SUBST

And we get
'X[5] = a - (a-X[3]²)²'

Then we write
'X[5] = a - (a-X[3]²)²'
'X[3] = a - X[2]²' SUBST
'X[2] = a - X[1]²' SUBST

Then we say that we want
X[n] = X[n+5]
Or 'X[5] = X[1]' SUBST

And we get
'X1=a-(a-(a-(a-X1^2)^2)^2)^2'

Or 'a-(a-(a-(a-X1^2)^
2)^2)^2-X1=0'.

Let's have a=1.25
1.25 'a' STO
EVAL

And we get

'-(X1^16.+-10.*X1^14.
+38.75*X1^12.+-71.875
*X1^10.+60.5859375*X1
^8.+-9.9609375*X1^6.+
-13.4033203125*X1^4.+
3.60107421875*X1^2.+
X1+7.78961181641E-2)'

The corresponding polynomial Matrix is:
:[ana0]:
[ -1 0 10 0 -38.75 0 71.875 0 -60.5859375 0 9.9609375 0 13.4033203125 0 -3.60107421875 -1 -7.78961181641E-2 ]

We use
SR+7 (Num Solver)
3 (Solve Polynom)

And get
:Roots:
[ (-.207100346975,1.02659356484E-5) (-.207100346975,-1.02659356484E-5) (-.207119649589,0.)
(.724744871331,0.)
(1.20727206925,0.)
(2.10497544283E-2,.651174406046) (2.10497544283E-2,-.651174406046) (-1.55013526807,9.17597298232E-2) (-1.55013526807,-9.17597298232E-2) (1.20702413682,1.43181059532E-4) (1.20702413682,-1.43181059532E-4) (-1.14449950134,.284479986425) (-1.14449950134,-.284479986425) (1.67358501524,2.74141263844E-2) (1.67358501524,-2.74141263844E-2) (-1.72474487121,0.) ]

We do have
f(.724744871331)=.724744871331
f(-1.72474487121)=-1.72474487121.

Great and normal.

But we do not get 4 alternating points, contrarily to what I thought by the partial results on my HP50G.
Is that really correct?

According to
https://en.m.wikipedia.org/wiki/Feigenbaum_constants
by a= 1.25 we should get 4 periods.

Do these 4 periods include the two values
724744871331 and -1.72474487121, with
f(.724744871331)=.724744871331
f(-1.72474487121)=-1.72474487121?

In other words, does the function
X[n+1] = 1.25 - X[n]²,
with X[0] = 0.4 and
with n—> infinity, "oscillates" between four different values or only between two different values?

Thanks in advance for your help.

Gil Campart
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09-10-2022, 12:21 AM
Post: #2
RE: Bifurcations and periods in chaos with HP50G
The solver solution of HP50G should be correct.

The bifurcation point "1.25“ gives 4 iterating points for a values = 1.25+epsilon, with epsilon tiny and ≠0, and not for a exactly =1.25.

Corresponding simplification with
X[n+3]=a-X[n+2]²
X[n+2]=a-X[n+1]²

And X[n+3]=X[n+1]

Then
'1.25-(1.25-X1^2)^2-X1' :
Roots: [ -.207106781187 .724744871392 1.20710678119 -1.72474487139 ]

And only 2 oscillation points:-0.207... and +1.207...
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09-10-2022, 05:32 AM
Post: #3
RE: Bifurcations and periods in chaos with HP50G
From WolframAlpha I have the following exact values with \(a=1.25\) for a solution of:

\(a-(a-(a-(a-x^2)^2)^2)^2=x\)

x = 1/2 (1 - sqrt(2)) ≈ -0.207107
x = 1/2 (1 + sqrt(2)) ≈ 1.20711
x = 1/2 (-1 - sqrt(6)) ≈ -1.72474
x = 1/2 (sqrt(6) - 1) ≈ 0.724745

If you iterate the following program you will notice that only two of them are attractive.
Code:
00 { 12-Byte Prgm }
01 XEQ 01
02▸LBL 01
03 XEQ 00
04▸LBL 00
05 RCL 00
06 X<>Y
07 X↑2
08 -
09 END

Examples

1.25 STO 00

1.20711
R/S R/S R/S …

It slowly converges to the exact value.

0.724745
R/S R/S R/S …

After only a few iteration it apparently starts converging to the other fixed-point as well.
Similarly for the other two values.

The derivative of the 4th iteration of the function \(a - x^2\) decides whether it is attractive.
A fixed-point \(x_0\) is attractive if \(|f\,'(x_{0})|<1\).
I'm leaving the verification up to you.
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09-10-2022, 09:29 AM
Post: #4
RE: Bifurcations and periods in chaos with HP50G
Thanks.

Suppose from my above example that I don't know the value a = 0.75 above which we start getting two fixed points.

Then how would you tackle the question to find anatically the solution a = 0.75?

I set
X3 = a - X2²
X2 = a- X1²
X3 = X1

And I get
'A-(A-X^2)^2-X'
A — (A² - 2AX² + X⁴) - X
—A² + A(2X²+1) -(X+X⁴)

But do I get further to find the a value = 0.75?

Thanks for your hint or help.

Regards,
Gil
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09-11-2022, 09:43 PM (This post was last modified: 09-12-2022 06:10 AM by Thomas Klemm.)
Post: #5
RE: Bifurcations and periods in chaos with HP50G
1st Iteration

Fixed-point equation:

\(
\begin{align}
x
&= f(x) \\
&= a - x^2 \\
\end{align}
\)

Solutions:

\(
\begin{align}
x = \frac{-1 \pm \sqrt{4a + 1}}{2}
\end{align}
\)

Derivative:

\(
\begin{align}
D[f(x)]
&= D[a - x^2] \\
&= -2x \\
\end{align}
\)

Attractive fixed-points:

\(
\begin{align}
|-2x| =
& 2\left|\frac{-1 + \sqrt{4a + 1}}{2}\right| < 1 \\
& |-1 + \sqrt{4a + 1}| < 1 \\
& |\sqrt{4a + 1}| < 2 \\
& 4a + 1 < 4 \\
\end{align}
\)

For the upper limit we get:

\(
\begin{align}
4a + 1 &= 4 \\
4a &= 3 \\
a &= \frac{3}{4} \\
\end{align}
\)

For the lower limit we get:

\(
\begin{align}
4a + 1 &= 0 \\
4a &= -1 \\
a &= -\frac{1}{4} \\
\end{align}
\)

Thus the range of \(a\) where this solution is attractive is:

\(
\begin{align}
-\frac{1}{4} < a < \frac{3}{4}
\end{align}
\)

For the other solution \(\frac{-1 - \sqrt{4a + 1}}{2}\) the value \(|2x| \geqslant 1\).
Therefore this solution is never attractive.

2nd Iteration

Fixed-point equation:

\(
\begin{align}
x
&= f^2(x) \\
&= f(f(x)) \\
&= a - (a - x^2)^2 \\
\end{align}
\)

Solutions:

Both solution of the 1st iteration are also solutions of the 2nd iteration.
Therefore we can split them off by division with the polynomial \(x^2 + x - a\):

\(
\begin{align}
\frac{a - (a - x^2)^2 - x}{x^2 + x - a} = -(-a + x^2 - x + 1)
\end{align}
\)

Therefore the additional solutions are:

\(
\begin{align}
x = \frac{1 \pm \sqrt{4a - 3}}{2}
\end{align}
\)

Derivative:

\(
\begin{align}
D[f(f(x))]
&= -2f(x) \cdot (-2x) \\
&= 4x(a - x^2) \\
&= 4x(1 - x) \\
&= 4(x - x^2) \\
&= 4(a - 1) \\
\end{align}
\)

Here we used that \(x\) is a solution of the following equation:

\(
\begin{align}
-a + x^2 - x + 1 = 0
\end{align}
\)

Attractive fixed-points:

For these solutions to be attractive we need:

\(
\begin{align}
|4(a - 1)| &< 1 \\
|a - 1| &< \frac{1}{4} \\
\end{align}
\)

This leads to:

\(
\begin{align}
\frac{3}{4} < a < \frac{5}{4}
\end{align}
\)

4th Iteration

We could continue with that but it gets tedious.
Thus I think this is a good point to stop.
From here on it's probably easier to use numerical methods.


The following diagram is from Chaos: An Introduction to Dynamical Systems:

[Image: attachment.php?aid=11094]


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09-11-2022, 10:55 PM (This post was last modified: 09-11-2022 10:58 PM by Gil.)
Post: #6
RE: Bifurcations and periods in chaos with HP50G
Great!

Many thanks for your remarkable and clear explanation, Thomas,
with full, comprehensive details for dummies like me.

By the way, did you use a special program like LATEX to get such a nice, clean presentation?

With my most thankful regards,
Gil
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09-12-2022, 05:56 AM
Post: #7
RE: Bifurcations and periods in chaos with HP50G
(09-11-2022 10:55 PM)Gil Wrote:  By the way, did you use a special program like LATEX to get such a nice, clean presentation?

For stuff I don't know or don't remember I often use this Online Equation Editor.
Other than that I just write it in the text field.
And often click [Preview Post] before finally hitting [Post Reply].

For a more sophisticated approach, I recommend: How I'm able to take notes in mathematics lectures using LaTeX and Vim
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