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Bifurcations and periods in chaos with HP50G - Gil - 09-09-2022 11:14 PM 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 RE: Bifurcations and periods in chaos with HP50G - Gil - 09-10-2022 12:21 AM 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... RE: Bifurcations and periods in chaos with HP50G - Thomas Klemm - 09-10-2022 05:32 AM 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 } 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. RE: Bifurcations and periods in chaos with HP50G - Gil - 09-10-2022 09:29 AM 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 RE: Bifurcations and periods in chaos with HP50G - Thomas Klemm - 09-11-2022 09:43 PM 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: RE: Bifurcations and periods in chaos with HP50G - Gil - 09-11-2022 10:55 PM 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 RE: Bifurcations and periods in chaos with HP50G - Thomas Klemm - 09-12-2022 05:56 AM (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 |