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
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