Here I show some matrices, which allow to compute real iterates
for x -> exp(x)-1
The matrix, which performs the iterate is
\( \hspace{24} C = F^{-1} * St2 * F \)
This matrix allows to compute
\( \hspace{24} V(x)\sim * C = V(e^x - 1)\sim \)
Example 1: normal iterates
\( \hspace{24} V(1)\sim * C = V(e - 1)\sim \)
\( \hspace{24} V(1)\sim * C^2 = V(e^{e - 1}-1)\sim \)
-----------------------------------------------------------------------
To obtain real or complex iterates real or complex powers of
C are needed. The eigensystem of C is degenerate, so this cannot
be used. But there is the possibility of computing the matrix-
logarithm of C.
\( \hspace{24} LC = \log{\left(C\right)} \)
Of this we can compute arbitrary complex multiples, say for
a start, to realize the half-iterate means the square-root of
C, and this means 1/2*LC (not documented here, trivial)
Example 2: Square-root - 1/2-iterates
From this, the top left of C^0.5 is then
\( \hspace{24} C^{0.5} = \exp(0.5*LC) \)
(note, that the second column is just the sequence, Henryk provided us with in the first post of this thread)
and we get the first four half-iterates (the interesting result is in 2'nd column)
Example 3: Cube-root - 1/3-iterates
The top-left of the cube-root of C, C^(1/3) is
We get the first six 1/3-iterates
------------------------------------------------------------------------
The convergence of the series for fractional iterates was worse than that
of integral iterates.
It looks also suspicious, that with dim=32, in the matrix-logarithm the
entries first decrease to higher indexed rows, (which suggests good
approximability) but increase again from, say,
row 24, so I don't know about the general behaviour (see table below)
This may have the same consequences for the fractional powers of C. (see table below)
I arrived at the above values using 32 coefficients "by default" by Euler-
summation. May be this procedure is required in general, in case the
entries of the fractional powers of C diverge but alternate in sign.
(That is then the field of analytic continuation, I assume?)
Gottfried
-------------------------------------------------------------------------
Top left from the matrix-logarithm of C
Top left from C^(1/2) = Exp(1/2*log©)
for x -> exp(x)-1
The matrix, which performs the iterate is
\( \hspace{24} C = F^{-1} * St2 * F \)
Code:
C= (only 8 rows/columns shown)
1 . . . . . . .
0 1 . . . . . .
0 1/2 1 . . . . .
0 1/6 1 1 . . . .
0 1/24 7/12 3/2 1 . . .
0 1/120 1/4 5/4 2 1 . .
0 1/720 31/360 3/4 13/6 5/2 1 .
0 1/5040 1/40 43/120 5/3 10/3 3 1This matrix allows to compute
\( \hspace{24} V(x)\sim * C = V(e^x - 1)\sim \)
Example 1: normal iterates
\( \hspace{24} V(1)\sim * C = V(e - 1)\sim \)
\( \hspace{24} V(1)\sim * C^2 = V(e^{e - 1}-1)\sim \)
Code:
results
V(e - 1)= [1, e-1, (e-1)^2, (e-1)^3, ...] =
[ 1.0000000 1.7182818 2.9524924 5.0732098 8.7170908 14.976648 ...]
V(e^(e - 1)-1 )= [1, e^(e-1)-1, (e^(e-1)-1)^2, ...] =
[ 1.0000000 4.5749415 20.930084 95.751617 437.84444 1994.7177 ...]-----------------------------------------------------------------------
To obtain real or complex iterates real or complex powers of
C are needed. The eigensystem of C is degenerate, so this cannot
be used. But there is the possibility of computing the matrix-
logarithm of C.
\( \hspace{24} LC = \log{\left(C\right)} \)
Code:
LC =
0 . . . . . . .
0 0 . . . . . .
0 1/2 0 . . . . .
0 -1/12 1 0 . . . .
0 1/48 -1/6 3/2 0 . . .
0 -1/180 1/24 -1/4 2 0 . .
0 11/8640 -1/90 1/16 -1/3 5/2 0 .
0 -1/6720 11/4320 -1/60 1/12 -5/12 3 0Of this we can compute arbitrary complex multiples, say for
a start, to realize the half-iterate means the square-root of
C, and this means 1/2*LC (not documented here, trivial)
Example 2: Square-root - 1/2-iterates
From this, the top left of C^0.5 is then
\( \hspace{24} C^{0.5} = \exp(0.5*LC) \)
Code:
C^0.5=
1 . . . . . . .
0 1 . . . . . .
0 1/4 1 . . . . .
0 1/48 1/2 1 . . . .
0 0 5/48 3/4 1 . . .
0 1/3840 1/96 1/4 1 1 . .
0 -7/92160 11/11520 3/64 11/24 5/4 1 .
0 1/645120 -1/46080 23/3840 1/8 35/48 3/2 1(note, that the second column is just the sequence, Henryk provided us with in the first post of this thread)
and we get the first four half-iterates (the interesting result is in 2'nd column)
Code:
1/2-iterates
1.0000000 1.2710274 1.6155107 2.0533584 2.6098746 3.3172201
1.0000000 1.7182818 2.9524924 5.0732098 8.7170908 14.976648
1.0000000 2.5645129 6.5767163 16.865290 43.227699 110.47725
1.0000000 4.5749415 20.930084 95.751617 437.84444 1994.7177Example 3: Cube-root - 1/3-iterates
The top-left of the cube-root of C, C^(1/3) is
Code:
C^(1/3)=
1 . . . . . . .
0 1 . . . . . .
0 1/6 1 . . . . .
0 0 1/3 1 . . . .
0 0 1/36 1/2 1 . . .
0 1/4860 0 1/12 2/3 1 . .
0 -7/58320 1/2430 1/216 1/6 5/6 1 .
0 23/612360 -1/5832 1/1620 1/54 5/18 1 1We get the first six 1/3-iterates
Code:
1/3-iterates
1.0000000 1.1667861 1.3613899 1.5884509 1.8533825 2.1625009
1.0000000 1.3939228 1.9430208 2.7084211 3.7753299 5.2625167
1.0000000 1.7182818 2.9524924 5.0732137 8.7171961 14.978293
1.0000000 2.2116542 4.8914136 10.818048 23.923566 52.871434
1.0000000 3.0306304 9.1846792 27.831709 84.229412 253.25930
1.0000000 4.5749415 20.930084 95.751617 437.84444 1994.7177------------------------------------------------------------------------
The convergence of the series for fractional iterates was worse than that
of integral iterates.
It looks also suspicious, that with dim=32, in the matrix-logarithm the
entries first decrease to higher indexed rows, (which suggests good
approximability) but increase again from, say,
row 24, so I don't know about the general behaviour (see table below)
This may have the same consequences for the fractional powers of C. (see table below)
I arrived at the above values using 32 coefficients "by default" by Euler-
summation. May be this procedure is required in general, in case the
entries of the fractional powers of C diverge but alternate in sign.
(That is then the field of analytic continuation, I assume?)
Gottfried
-------------------------------------------------------------------------
Top left from the matrix-logarithm of C
Code:
1.0*LC=
0 . . .
0 0 . .
0 0.50000000 0 .
0 -0.083333333 1.0000000 0
0 0.020833333 -0.16666667 1.5000000
0 -0.0055555556 0.041666667 -0.25000000
0 0.0012731481 -0.011111111 0.062500000
0 -0.00014880952 0.0025462963 -0.016666667
0 -0.000045469577 -0.00029761905 0.0038194444
0 0.000019979056 -0.000090939153 -0.00044642857
0 0.000011321465 0.000039958113 -0.00013640873
0 -0.000011319378 0.000022642931 0.000059937169
0 -0.0000017266470 -0.000022638755 0.000033964396
0 0.0000070561217 -0.0000034532940 -0.000033958133
0 -0.0000010130178 0.000014112243 -0.0000051799409
0 -0.0000055954100 -0.0000020260355 0.000021168365
0 0.0000027850090 -0.000011190820 -0.0000030390533
0 0.0000055636225 0.0000055700180 -0.000016786230
0 -0.0000054890313 0.000011127245 0.0000083550269
0 -0.0000066929168 -0.000010978063 0.000016690867
0 0.000011612181 -0.000013385834 -0.000016467094
0 0.0000092209451 0.000023224363 -0.000020078750
0 -0.000028132640 0.000018441890 0.000034836544
0 -0.000013074482 -0.000056265279 0.000027662835
0 0.000079067854 -0.000026148964 -0.000084397919
0 0.000012559808 0.00015813571 -0.000039223446
0 -0.00025744463 0.000025119617 0.00023720356
0 0.000039946326 -0.00051488925 0.000037679425
0 0.00096515491 0.000079892652 -0.00077233388
0 -0.00047939822 0.0019303098 0.00011983898
0 -0.0041340004 -0.00095879644 0.0028954647
0 0.0036464303 -0.0082680007 -0.0014381947Top left from C^(1/2) = Exp(1/2*log©)
Code:
1.0*C^(1/2)=
1.0000000 . . .
0 1.0000000 . .
0 0.25000000 1.0000000 .
0 0.020833333 0.50000000 1.0000000
0 0 0.10416667 0.75000000
0 0.00026041667 0.010416667 0.25000000
0 -0.000075954861 0.00095486111 0.046875000
0 0.0000015500992 -0.000021701389 0.0059895833
0 0.000015404111 -0.000024026538 0.00048828125
0 -0.0000090745391 0.000028418485 -0.000018859540
0 -0.000000082819971 -0.000010314618 0.000032939608
0 0.0000036074073 -0.0000041006314 -0.0000054640997
0 -0.0000016951497 0.0000068018753 -0.0000091647597
0 -0.0000013308992 -0.0000015822591 0.0000083398934
0 0.0000017752144 -0.0000033661285 0.000000062080956
0 0.00000037035398 0.0000028157312 -0.0000054374338
0 -0.0000019147568 0.0000015749616 0.0000029115066
0 0.00000034467343 -0.0000035720802 0.0000033166038
0 0.0000024191341 -0.00000025294483 -0.0000046812506
0 -0.0000014770587 0.0000049320576 -0.0000016708263
0 -0.0000036046260 -0.0000017303876 0.0000071932744
0 0.0000042603060 -0.0000078480430 -0.00000075380949
0 0.0000061940178 0.0000066573185 -0.000012268458
0 -0.000012625293 0.000014369165 0.0000069893294
0 -0.000011736089 -0.000021977272 0.000023806922
0 0.000041395229 -0.000029528323 -0.000027370787
0 0.000022203030 0.000076399188 -0.000052087716
0 -0.00015310857 0.000064617129 0.00010284248
0 -0.000027832787 -0.00029339839 0.00012465733
0 0.00064101866 -0.00013129869 -0.00041348905
0 -0.00011130752 0.0012617648 -0.00030496184
0 -0.0030302667 0.000096739308 0.0018342749
Gottfried Helms, Kassel
