Bifurcations and periods in chaos with HP50G

[Gil]

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

[Gil]

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...

[Thomas Klemm]

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.

[Gil]

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

[Thomas Klemm]

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:

[Gil]

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

[Thomas Klemm]

(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