Hi Henryk -
I'm rereading the "bummer"-posts and I just tested your example-function
\( \hspace{24} f(x) = x^2 + x - 1/16 \)
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
\( \hspace{24}
\begin{matrix}
x' &=& x/(1/4) - 1 \\
x'' &=& (x+1)*1/4 \\
g(z)&=& 1/4 z^2 + 3/2 z
\end{matrix}
\)
and with the above then
\( \hspace{24} f(x) = g(x')'' \)
The matrix-operator G for g(x) is then triangular.
Results:
\( \hspace{24}
\begin{matrix} {llll}
& f^{(h)}(x) & g^{(h)}(x) \\
h=0 & 1.0000000000000000000 & 3 \\
h=0.5 & 1.3427402879772577347 & 4.3709611519090309387 \\
h=1 & 1.9375000000000000000 & 6.7500000000000000000
\end{matrix}
\)
[update]
If I understood the fix-point-problem correctly, then the other transformations x', x", and the second function g2(x) are
\( \hspace{24}
\begin{matrix}
x' &=& x/(-1/4) + 1 \\
x'' &=& (x-1)*(-1/4) \\
g2(z)&=& -1/4 z^2 + 3/2 z
\end{matrix}
\)
Then I get identical results for the back-shifted solutions (checked up to 20 digits):
g2(x) =
\( \hspace{24}
\begin{matrix} {r}
1.0000000000000000000 & -3 \\
1.3427402879772577347 & -4.3709611519090309387 \\
1.9375000000000000000 & -6.7500000000000000000
\end{matrix}
\)
----------------------------------------------------------------------
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
\( \hspace{24} V(x)\sim * F = V(f(x))\sim \)
by
\( \hspace{24} V(x')\sim * G = V(g(x'))\sim \)
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
\( \hspace{24} V(x')\sim * G^{0.5} \)
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
\( \hspace{24} V(3)\sim * G^{0.5} = V(4.3709611519090309387...)\sim \)
so \( \hspace{24} g^{(0.5)}(3)=4.3709611519090309387... \)
The next iteration is even more divergent, but again seems to be Euler-summable to get
\( \hspace{24} V(4.3709611519090309387...)\sim * G^{0.5} = V(6.750)\sim \)
The backtransformation of these values gives
\( \hspace{24}
\begin{matrix}
3'' &=& 1 \\
4.37096115...'' &=& 1.3427... \\
6.750'' &=& 1.9375
\end{matrix}
\)
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