03/07/2008, 03:16 AM
Hi Henryk -
I gave it some more time to document the two fixpoint-solutions. Here is the eigendecomposition in a synopsis , left for using fixpoint t0=1/4, right for using fixpoint t1=-1/4.
Definitions according to previous post:
Here the half power-matrices:
What we see is, that, using J = diag(1,-1,1,-1,...)
and consequently
and this is also valid for powers (by powers of D).
Now the transformations of x->x' for G0 and x->x´ are
But since the two substitutions for g0(x) and g1(x) mean, that x´ = - x'
we have
Then also
and we have identity for the height 1 for the two fixpoint-solutions.
Since the two diagonals D0 and D1 are equal, also their powers are equal and thus the whole matrix-expression must lead to equal results.
The half-iterate-matrices show the irrational numbers based on b=sqrt(3/2); but since all coefficients of b are rational I don't see space for relevant approximation-errors or result-differences.
So since this -well: basic deal- seems to give equality. It would be good, if I could go further and get into the computations, with which you found the discrepancies.
Gottfried
I gave it some more time to document the two fixpoint-solutions. Here is the eigendecomposition in a synopsis , left for using fixpoint t0=1/4, right for using fixpoint t1=-1/4.
Definitions according to previous post:
Code:
´
G0^h = W0 * D0^h * W0^-1
G1^h = W1 * D1^h * W1^-1
W0= |W1=
1 . . . . . | 1 . . . . .
0 1 . . . . | 0 1 . . . .
0 -1/3 1 . . . | 0 1/3 1 . . .
0 2/15 -2/3 1 . . | 0 2/15 2/3 1 . .
0 -49/855 17/45 -1 1 . | 0 49/855 17/45 1 1 .
0 158/6175 -58/285 11/15 -4/3 1 | 0 158/6175 58/285 11/15 4/3 1
D0= |D1=
1 3/2 9/4 27/8 81/16 243/32 | 1 3/2 9/4 27/8 81/16 243/32
W0^-1= |W1^-1
1 . . . . . | 1 . . . . .
0 1 . . . . | 0 1 . . . .
0 1/3 1 . . . | 0 -1/3 1 . . .
0 4/45 2/3 1 . . | 0 4/45 -2/3 1 . .
0 52/2565 13/45 1 1 . | 0 -52/2565 13/45 -1 1 .
0 2048/500175 256/2565 3/5 4/3 1 | 0 2048/500175 -256/2565 3/5 -4/3 1
G0= |G1=
1 . . . . . | 1 . . . . .
0 3/2 . . . . | 0 3/2 . . . .
0 1/4 9/4 . . . | 0 -1/4 9/4 . . .
0 0 3/4 27/8 . . | 0 0 -3/4 27/8 . .
0 0 1/16 27/16 81/16 . | 0 0 1/16 -27/16 81/16 .
0 0 0 9/32 27/8 243/32 | 0 0 0 9/32 -27/8 243/32
Here the half power-matrices:
Code:
´
G0^0.5=
1.0000000 . . . . .
0 1.2247449 . . . .
0 0.091751710 1.5000000 . . .
0 -0.0067347010 0.22474487 1.8371173 . .
0 0.0010135508 -0.0080782047 0.41288269 2.2500000 .
0 -0.00019936640 0.0012468417 0.00062493491 0.67423461 2.7556760
G1^0.5
1.0000000 . . . . .
0 1.2247449 . . . .
0 -0.091751710 1.5000000 . . .
0 -0.0067347010 -0.22474487 1.8371173 . .
0 -0.0010135508 -0.0080782047 -0.41288269 2.2500000 .
0 -0.00019936640 -0.0012468417 0.00062493491 -0.67423461 2.7556760
What we see is, that, using J = diag(1,-1,1,-1,...)
Code:
´
W1 = J * W0 * J
W1^-1 = J * W0^-1 * J
D1 = D0 = D
Code:
´
G1 = W1 * D * W1^-1 = J * W0 * D * W0^-1 * J
= J * G0 * J
Now the transformations of x->x' for G0 and x->x´ are
Code:
´
using e1=1/4 e2 = -1/4
d1=1/4 / e1 =1 d2 = 1/4 / e2 = -1
x' = 4x - 1 x´ = -4x + 1
x" = (x+1)/4 x´´= (x-1)/-4
But since the two substitutions for g0(x) and g1(x) mean, that x´ = - x'
we have
Code:
´
V(x´)~ * G1 = V(-x')~ * J G0 J
= V(x')~ * G0 * J
and
g1(x´) = - g0(x')
Code:
´
g1(x´)´´ = (-g0(x'))´´
= (-g0(x')-1)/-4
= (g0(x')+1)/4
= g0(x')"
Since the two diagonals D0 and D1 are equal, also their powers are equal and thus the whole matrix-expression must lead to equal results.
The half-iterate-matrices show the irrational numbers based on b=sqrt(3/2); but since all coefficients of b are rational I don't see space for relevant approximation-errors or result-differences.
So since this -well: basic deal- seems to give equality. It would be good, if I could go further and get into the computations, with which you found the discrepancies.
Gottfried
Gottfried Helms, Kassel