the inconsistency depending on fixpoint-selection

[Gottfried]

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:

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

and consequently

Code:

´
G1 = W1 * D * W1^-1 = J * W0 * D * W0^-1 * J
    = J * G0 * J

and this is also valid for powers (by powers of D).

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')

Then also

Code:

´
  g1(x´)´´ = (-g0(x'))´´
           = (-g0(x')-1)/-4
           = (g0(x')+1)/4
           = g0(x')"

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