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the inconsistency depending on fixpoint-selection
#13
Hi Henryk -

I'm rereading the "bummer"-posts and I just tested your example-function



with my matrix-analysis. However, it is not clear to me,

* how the different fixpoints are involved here (update:but maybe I've got it now)
* what the plot in your first msg describes exactly.

What I've got so far is a good result for the half-iterate, which seems to be exact to more than 20 digits. I've also used a shifting of x->x' to get a triangular matrix with exact terms, where I denote



and with the above then



The matrix-operator G for g(x) is then triangular.

Results:


[update]
If I understood the fix-point-problem correctly, then the other transformations x', x", and the second function g2(x) are


Then I get identical results for the back-shifted solutions (checked up to 20 digits):
g2(x) =


----------------------------------------------------------------------
Documents:

Matrix-operator F for f(x)
Code:
´
  1  -1/16  1/256  -1/4096  1/65536  -1/1048576
  0      1   -1/8    3/256  -1/1024     5/65536
  0      1    7/8  -45/256  23/1024  -155/65536
  0      0      2      5/8   -13/64     35/1024
  0      0      1    45/16   35/128   -405/2048
  0      0      0        3     13/4     -17/128
Because this operator is not rowfinite, I use the substitution x->x' and f(x)=g(x')" , replacing the original matrix-multiplication



by



and finally apply re-substitution.


Matrix-operator G for g(x)
Code:
´
  1    .     .      .      .       .
  0  3/2     .      .      .       .
  0  1/4   9/4      .      .       .
  0    0   3/4   27/8      .       .
  0    0  1/16  27/16  81/16       .
  0    0     0   9/32   27/8  243/32

The fractional powers for g(x) are expressed by fractional powers of G, so we need the eigensystem-decomposition G = W * D * W^-1

W =
Code:
´
  1         .        .      .     .  .
  0         1        .      .     .  .
  0      -1/3        1      .     .  .
  0      2/15     -2/3      1     .  .
  0   -49/855    17/45     -1     1  .
  0  158/6175  -58/285  11/15  -4/3  1

D = diag() (which is a divergent sequence of eigenvalues!)
Code:
´
  1  3/2  9/4  27/8  81/16  243/32

W^-1 =
Code:
´
  1            .         .    .    .  .
  0            1         .    .    .  .
  0          1/3         1    .    .  .
  0         4/45       2/3    1    .  .
  0      52/2565     13/45    1    1  .
  0  2048/500175  256/2565  3/5  4/3  1

The half-power G^0.5 =
Code:
´
  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

And using the second column of G^0.5 with x'=3 according to

we get the diverging sequence of terms of the powerseries (first 64 terms)
Code:
´
               0
       3.6742346
      0.82576539
     -0.18183693
     0.082097618
    -0.048446035
     0.032977002
    -0.024500419
     0.019276236
...
      -1.1067517
       2.6939736
      -5.1977227
       9.0095267
      -14.654213
       22.823197
      -34.410169
       50.545198
      -72.619728
       102.28920
      -141.43120
       192.02334
      -255.88447
       334.19212
      ...

which seems to be Euler-summable to


so

The next iteration is even more divergent, but again seems to be Euler-summable to get


The backtransformation of these values gives


Appendix: [update]
The half-power G2^0.5 (developed using the second fixpoint) is
Code:
´
  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

where apparently only the signs have changed, but the numerical digits are the same up to the 20'th digit.


========================================================================


Now, at some place I need the idea, how to introduce the other fixpoint correctly (if I didn't get it this way), and then, what computation your plot shows.

Gottfried
Gottfried Helms, Kassel
Reply


Messages In This Thread
RE: the inconsistency depending on fixpoint-selection - by Gottfried - 03/06/2008, 04:06 PM

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