I'm surprised.
The matrix-operator-method promises for this version of tetration (x --> exp(x)-1 ) an even better behave than that of general tetration itself.
Assume the constant matrix-operator (for parameter s=exp(1))
which is just a factorial-scaling of the matrix of Stirling-numbers 2'nd kind:
Now parametrize C with s, by premultiplying it with the diagonalmatrix
of powers of log(s)
Denote the entries of the second column of Cs as c with index r (beginning at zero)
Then
sum{r=0..inf} c_r = s - 1
Example:
s = sqrt(2)
The partial sums of V(1)~ * Cs converge very good, only the first few partial sums are given:
so
s - 1 ~ 0.41421356
---------------------------------------------------------------------------
Take Cs to any integral power to perform iteration, say Cs^y, denote the
entries of the second column as cy_r (they are meant to be different for
different y!)
y=2
sum{r=0..inf} cy_r = s^(s - 1)-1
The partial sums of V(1)~ * Cs^2 converge still very good, the first few partial sums:
and the result is in the second column:
s^(s-1)-1 = 0.15437086
-----------------------------------------------------------------------------------------
y=3
sum{r=0..inf} cy_r = s^(s^(s - 1)-1)-1
and so on.
==========================================================================
Continuous version.
Since Cs has triangular structure, its eigensystem is very simple.
The eigenvalues are just the powers of log(s), and the matrices of
eigenvectors are triangular. Also it appears, that their entries
do not change with increasing dimensions.
Let
Cs = Q * D * Q^-1
then for the example s=sqrt(2)
where D and Q are stabile for higher dimensions.
Now compute the half-iterate of CS by simply using the square-root
of D
Cs^0.5 = Q * D^0.5 * Q^-1
And the half-iterate of the function is
and the interesting result is in 2'nd column
half-iterate = 0.65549414
The next half-iterate is then, as expected:
next(half-iterate) = 0.41421356
---------------------------------------------------------------
I don't know, for what reason the continuous version is assumed as impossible?
Gottfried
(edit of the introductional remark)
(update 2: edit of the introductoy tables, 2'nd (now third) table corrected (I forgot the column-scaling))
The matrix-operator-method promises for this version of tetration (x --> exp(x)-1 ) an even better behave than that of general tetration itself.
Assume the constant matrix-operator (for parameter s=exp(1))
Code:
C=
1.0000000 . . . . . . .
0 1.0000000 . . . . . .
0 0.50000000 1.0000000 . . . . .
0 0.16666667 1.0000000 1.0000000 . . . .
0 0.041666667 0.58333333 1.5000000 1.0000000 . . .
0 0.0083333333 0.25000000 1.2500000 2.0000000 1.0000000 . .
0 0.0013888889 0.086111111 0.75000000 2.1666667 2.5000000 1.0000000 .
0 0.00019841270 0.025000000 0.35833333 1.6666667 3.3333333 3.0000000 1.0000000
.. ...
or
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 1
which is just a factorial-scaling of the matrix of Stirling-numbers 2'nd kind:
Code:
St2=
1 . . . . . . .
0 1 . . . . . .
0 1 1 . . . . .
0 1 3 1 . . . .
0 1 7 6 1 . . .
0 1 15 25 10 1 . .
0 1 31 90 65 15 1 .
0 1 63 301 350 140 21 1
Now parametrize C with s, by premultiplying it with the diagonalmatrix
of powers of log(s)
Code:
Cs = diag(1,log(s),log(s)^2,...) * C
Denote the entries of the second column of Cs as c with index r (beginning at zero)
Then
sum{r=0..inf} c_r = s - 1
Example:
s = sqrt(2)
Code:
Cs=
1.0000000 . . . . . . .
0 0.34657359 . . . . . .
0 0.060056627 0.12011325 . . . . .
0 0.0069380136 0.041628081 0.041628081 . . . .
0 0.00060113307 0.0084158630 0.021640790 0.014427194 . . .
0 0.000041667369 0.0012500211 0.0062501054 0.010000169 0.0050000843 . .
0 0.0000024068016 0.00014922170 0.0012996729 0.0037546105 0.0043322429 0.0017328972 .
0 0.00000011916198 0.000015014410 0.00021520654 0.0010009607 0.0020019213 0.0018017292 0.00060057639
The partial sums of V(1)~ * Cs converge very good, only the first few partial sums are given:
Code:
1.0000000 . . . . .
1.0000000 0.34657359 . . . .
1.0000000 0.40663022 0.12011325 . . .
1.0000000 0.41356823 0.16174133 0.041628081 . .
1.0000000 0.41416936 0.17015720 0.063268872 0.014427194 .
1.0000000 0.41421103 0.17140722 0.069518977 0.024427362 0.0050000843
1.0000000 0.41421344 0.17155644 0.070818650 0.028181973 0.0093323272
1.0000000 0.41421356 0.17157146 0.071033857 0.029182933 0.011334249
1.0000000 0.41421356 0.17157277 0.071063777 0.029393679 0.011984698
1.0000000 0.41421356 0.17157287 0.071067386 0.029430750 0.012150514
1.0000000 0.41421356 0.17157287 0.071067771 0.029436389 0.012185671
... ... ... ... ... ...
= (s-1)^0 (s-1)^1 (s-1)^2 (s-1)^3 (s-1)^4 (s-1)^5
s - 1 ~ 0.41421356
---------------------------------------------------------------------------
Take Cs to any integral power to perform iteration, say Cs^y, denote the
entries of the second column as cy_r (they are meant to be different for
different y!)
y=2
sum{r=0..inf} cy_r = s^(s - 1)-1
Code:
Cs^2=
1.0000000 . . . . . . .
0 0.12011325 . . . . . .
0 0.028027638 0.014427194 . . . . .
0 0.0051933906 0.0067329815 0.0017328972 . . . .
0 0.00087258195 0.0020331386 0.0012130805 0.00020814392 . . .
0 0.00013909595 0.00050073425 0.00050784250 0.00019427606 0.000025000843 . .
0 0.000021254738 0.00010929866 0.00016468480 0.00010399801 0.000029168911 0.0000030029326 .
0 0.0000031256524 0.000021966331 0.000045603341 0.000041826549 0.000019017611 0.0000042042874 0.00000036069200
Code:
1.0000000 . . . . .
1.0000000 0.12011325 . . . .
1.0000000 0.14814089 0.014427194 . . .
1.0000000 0.15333428 0.021160175 0.0017328972 . .
1.0000000 0.15420686 0.023193314 0.0029459776 0.00020814392 .
1.0000000 0.15434596 0.023694048 0.0034538201 0.00040241997 0.000025000843
1.0000000 0.15436721 0.023803347 0.0036185049 0.00050641798 0.000054169754
1.0000000 0.15437034 0.023825313 0.0036641083 0.00054824453 0.000073187365
1.0000000 0.15437078 0.023829461 0.0036754279 0.00056227479 0.000082316680
1.0000000 0.15437085 0.023830207 0.0036780174 0.00056641656 0.000085912777
1.0000000 0.15437085 0.023830335 0.0036785730 0.00056752723 0.000087142908
1.0000000 0.15437086 0.023830357 0.0036786861 0.00056780338 0.000087521005
1.0000000 0.15437086 0.023830360 0.0036787081 0.00056786794 0.000087627785
1.0000000 0.15437086 0.023830361 0.0036787123 0.00056788228 0.000087655927
... ... ... ... ... ...
= s^(s-1)-1 (s^(s-1)-1)^2 ...
and the result is in the second column:
s^(s-1)-1 = 0.15437086
-----------------------------------------------------------------------------------------
y=3
sum{r=0..inf} cy_r = s^(s^(s - 1)-1)-1
and so on.
==========================================================================
Continuous version.
Since Cs has triangular structure, its eigensystem is very simple.
The eigenvalues are just the powers of log(s), and the matrices of
eigenvectors are triangular. Also it appears, that their entries
do not change with increasing dimensions.
Let
Cs = Q * D * Q^-1
then for the example s=sqrt(2)
Code:
D = [ 1.0000000 0.34657359 0.12011325 0.041628081 ... ]
Q =
1.0000000 . . .
0 1.0000000 . .
0 0.26519711 1.0000000 .
0 0.058953682 0.53039422 1.0000000
0 0.012370448 0.18823687 0.79559133
0 0.0025334009 0.056009589 0.38784957
0 0.00051044437 0.015103553 0.14956860
0 0.00010163175 0.0038231569 0.050149006
Now compute the half-iterate of CS by simply using the square-root
of D
Cs^0.5 = Q * D^0.5 * Q^-1
Code:
Cs^0.5=
1.0000000 0 0 0 0
0 0.58870501 0 0 0
0 0.064212553 0.34657359 0 0
0 0.0026279354 0.075604503 0.20402961 0
0 -0.000053277975 0.0072174095 0.066763125 0.12011325
0 0.00000075647862 0.00027476287 0.010014456 0.052405048
And the half-iterate of the function is
Code:
1.0000000 . . . . .
1.0000000 0.58870501 . . . .
1.0000000 0.65291756 0.34657359 . . .
1.0000000 0.65554550 0.42217809 0.20402961 . .
1.0000000 0.65549222 0.42939550 0.27079273 0.12011325 .
1.0000000 0.65549298 0.42967027 0.28080719 0.17251830 0.070711274
1.0000000 0.65549420 0.42967122 0.28161261 0.18323707 0.10927517
1.0000000 0.65549417 0.42967248 0.28164602 0.18451886 0.11926607
1.0000000 0.65549414 0.42967260 0.28164764 0.18461316 0.12084026
1.0000000 0.65549414 0.42967258 0.28164786 0.18461814 0.12100355
1.0000000 0.65549414 0.42967257 0.28164786 0.18461850 0.12101552
1.0000000 0.65549414 0.42967257 0.28164786 0.18461852 0.12101631
... ... ... ... ... ...
half-iterate = 0.65549414
The next half-iterate is then, as expected:
Code:
1.0000000 0.41421356 0.17157288 0.071067812 0.029437252 0.012193309
next(half-iterate) = 0.41421356
---------------------------------------------------------------
I don't know, for what reason the continuous version is assumed as impossible?
Gottfried
(edit of the introductional remark)
(update 2: edit of the introductoy tables, 2'nd (now third) table corrected (I forgot the column-scaling))
Gottfried Helms, Kassel