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Here I give some more numerical examples for the approximation
of the half-iterate for

f: x -> exp(x)-1

as discussed in my other article (url?)

The heuristic indicates, that indeed the approximation
seems to be impossible by conventional series-evaluation.
To get approximations anyway, Euler-summation is employed
here.

-----------------------------------------------------

Matrixlogarithm of C, first 64 rows

$\hspace{24} CL = \log{\left(C\right)}$

Code:
CL[0..63,0..2] (show the first 64 rows,first 3 columns):
0                     .                     .
0                     0                     .
0        0.500000000000                     0
0      -0.0833333333333         1.00000000000
0       0.0208333333333       -0.166666666667
0     -0.00555555555556       0.0416666666667
0      0.00127314814815      -0.0111111111111
0    -0.000148809523810      0.00254629629630
0   -0.0000454695767196    -0.000297619047619
0    0.0000199790564374   -0.0000909391534392
0    0.0000113214653145    0.0000399581128748
0   -0.0000113193776388    0.0000226429306290
0  -0.00000172664698128   -0.0000226387552776
0   0.00000705612171179  -0.00000345329396255
0  -0.00000101301777217    0.0000141122434236
0  -0.00000559541003662  -0.00000202603554435
0   0.00000278500897643   -0.0000111908200732
0   0.00000556362245182   0.00000557001795286
0  -0.00000548903126386    0.0000111272449036
0  -0.00000669291680506   -0.0000109780625277
0    0.0000116121812620   -0.0000133858336101
0   0.00000922094512360    0.0000232243625240
0   -0.0000281326395956    0.0000184418902472
0   -0.0000130744820957   -0.0000562652791912
0    0.0000790678539611   -0.0000261489641914
0    0.0000125598083643     0.000158135707922
0    -0.000257444626675    0.0000251196167287
0    0.0000399463258851    -0.000514889253349
0     0.000965154912485    0.0000798926517702
0    -0.000479398220790      0.00193030982497
0     -0.00413400035554    -0.000958796441580
0      0.00364643030317     -0.00826800071108
0       0.0200644656154      0.00729286060634
0      -0.0266575258227       0.0401289312307
0       -0.109433975865      -0.0533150516453
0        0.203327436701       -0.218867951729
0        0.665130619209        0.406654873401
0        -1.66290268242         1.33026123842
0        -4.46640908999        -3.32580536484
0         14.7351954384        -8.93281817999
0         32.8328822406         29.4703908769
0        -141.988015420         65.6657644812
0        -261.445537846        -283.976030840
0         1488.67144258        -522.891075691
0         2225.38586814         2977.34288517
0        -16965.9992298         4450.77173627
0        -19864.8060373        -33931.9984596
0         209816.100420        -39729.6120746
0         179952.210614         419632.200841
0        -2809660.13947         359904.421228
0        -1537617.51817        -5619320.27894
0         40646569.5053        -3075235.03634
0         9497336.70101         81293139.0106
0        -633756457.512         18994673.4020
0         55726510.4799        -1267512915.02
0         10624909421.6         111453020.960
0        -4479387389.03         21249818843.2
0        -191085306212.        -8958774778.05
0         147339758784.        -382170612425.
0      3.67833350467E12         294679517568.
0     -4.19319940120E12      7.35666700933E12
0     -7.56224969314E13     -8.38639880240E12
0      1.15852751376E14     -1.51244993863E14
0      1.65696934682E15      2.31705502752E14

Since the diagonal of the matrix-logarithm is zero, CL is nilpotent,
and the matrixexponential can be computed exactly for any finite
dimension (the number of terms involved to compute one entry of
the matrixexponential is finite for a finite dimension)

$\hspace{24} C_2 = \exp(0.5*CL)$

The first 64 rows and 2 columns of C^(1/2) are

Code:
C_2 = Exp(1/2*CL);
1.0*C_2[0..63,0..1]: // n=64 is the current dimension
1.00000000000                       .
.           1.00000000000
.          0.250000000000
.         0.0208333333333
.                       0
.       0.000260416666667
.     -0.0000759548611111
.     0.00000155009920635
.      0.0000154041108631
.    -0.00000907453910384
.  -0.0000000828199706170
.     0.00000360740727676
.    -0.00000169514972633
.    -0.00000133089916348
.     0.00000177521444910
.    0.000000370353976658
.    -0.00000191475684776
.    0.000000344673434042
.     0.00000241913411616
.    -0.00000147705874041
.    -0.00000360462602023
.     0.00000426030599723
.     0.00000619401781838
.     -0.0000126252925336
.     -0.0000117360887110
.      0.0000413952285774
.      0.0000222030302120
.      -0.000153108566769
.     -0.0000278327871472
.       0.000641018661899
.      -0.000111307516319
.       -0.00303026666274
.        0.00167666962999
.         0.0160951154545
.        -0.0157084159784
.        -0.0954804645039
.          0.139489606827
.          0.628520649401
.          -1.27694165810
.          -4.55956399021
.           12.4027724564
.           36.1854546816
.          -129.305594720
.          -311.608441223
.           1453.71643376
.           2883.75499733
.          -17648.6056027
.          -28323.2666121
.           231312.842070
.           289837.706925
.          -3269335.96562
.          -2992168.60724
.           49750634.1587
.           28980063.0330
.          -813616473.772
.          -201961594.949
.           14271686431.9
.          -1325490857.72
.          -267978508283.
.           119319788076.
.        5.37563669599E12
.       -4.37013046468E12
.       -1.14977800862E14
.        1.37951986894E14

We see the increasing absolute values of the entries.
To get the value for the half-iterate z, we have the formula

$\hspace{24} V(1)\sim * C_2 = V(z)\sim$

where in the second column of V(z) (z=V(z)[1]) is the numerical value
for the half iterate z. In conventional notation this means

$\hspace{24} z = \sum_{k=0}^{\infty} C_2 \left[k,1\right]$

If we extrapolate the tendency, this is a divergent series.
But since the signs are alternating, we may be able to assign
a value to this sum using Cesaro-, Euler- or Borel-summation.
The achievable precision of approximation is then depending
on the growth-rate of the absolute values; if the absolute
value of the quotient of two subsequent entries is bounded
by a constant q, we are able to assign an arbitrary precise
value to the sum via Euler-summation. If the absolute values
of the quotients increase (let's say, we have a factorial
growth), it may be possible to sum with Borel-summation
(but I don't know this exactly, I've never understand Borel-
summation so far to be able to apply it practically)
In this case, Euler-summation for finite dimensions may give
still an approximation, but is is principally limited to a
certain best precision.

The absolute values of the quotients are neither constant nor
seem to be bounded in this case, so the Euler-summation should
be limited to a best precision.

Code:
some quotients, sequentially listed (read:left-to-right, top-down)
beginning with C_2[5,1]/C_2[4,1]
-0.291666666667,   -0.0204081632653,      9.93750000000,    -0.589098532495,
0.00912663107948,  -43.5572151244,        -0.469908052038,    0.785121893839,
-1.33384594251,    0.208624922384,      -5.17007233197,    -0.180008983619,
7.01862655265,   -0.610573316519,       2.44040803633,    -1.18189958496,
1.45389035961,   -2.03830419992,        0.929569645991,   -3.52717413756,
0.536366894808,   -6.89584103195,        0.181784649511,  -23.0310625561,
-0.173641615970,   27.2242770564,        -0.553307618305,    9.59945547209,
-0.975974109836,    6.07829997851,       -1.46092300192,     4.50586006869,
-2.03166222036,    3.57069092490,       -2.72016633236,     2.91752951276,
-3.57341356790,    2.40986046967,       -4.66520235478,     1.98371218098,
-6.12000867585,    1.60484444209,       -8.16688432298,     1.25301174086,
-11.2798848718,    0.915222124219,     -16.6269487750,      0.582506404655,
-28.0750415499,    0.248227022756,     -70.6653482088,     -0.0928755591744,
202.173034028    -0.445258796463,      45.0523486731,    ...

With Euler-smmation of order 1.58 I get the following best approximation
for the first 64 coefficients. To see the progress of approximation
the partial sums are displayed, the last row includes all 64 coefficients.
The interesting result is again in the second column, the following
columns should contain powers of that value (note that the order 1.58
is not sufficient to aproximate the powers as good as the original value)

Code:
%print ESum1(1.58)*C_2[0..63 ,0..3]:
1.00000000000               .               .                .
1.00000000000  0.387596899225               .                .
1.00000000000  0.662520281233  0.150231356289                .
1.00000000000  0.855097858454  0.363350257070  0.0582292078638
1.00000000000  0.988604386938  0.588218066627   0.182135545528
1.00000000000   1.08035742605  0.797599657217   0.356816170837
1.00000000000   1.14294970244  0.979220300037   0.560892033232
1.00000000000   1.18537739995   1.12961228376   0.774392972307
1.00000000000   1.21397742575   1.25008128148   0.981917349700
1.00000000000   1.23316258822   1.34418757363    1.17323277772
1.00000000000   1.24597672252   1.41625824587    1.34266264470
1.00000000000   1.25450262124   1.47057084255    1.48803493204
1.00000000000   1.26015574574   1.51095525462    1.60960197897
1.00000000000   1.26389236729   1.54064314916    1.70911640931
1.00000000000   1.26635520485   1.56225442711    1.78912217850
1.00000000000   1.26797427412   1.57785205310    1.85245503567
1.00000000000   1.26903611888   1.58902457786    1.90191873922
1.00000000000   1.26973098932   1.59697368889    1.94009573710
1.00000000000   1.27018479030   1.60259528745    1.96925353488
1.00000000000   1.27048059776   1.60654920645    1.99131465764
1.00000000000   1.27067308026   1.60931638823    2.00786572117
1.00000000000   1.27079812407   1.61124424518    2.02018808901
1.00000000000   1.27087923274   1.61258176364    2.02929828999
1.00000000000   1.27093176755   1.61350614893    2.03599070570
1.00000000000   1.27096574876   1.61414273636    2.04087814968
1.00000000000   1.27098770085   1.61457967881    2.04442808173
1.00000000000   1.27100186495   1.61487866344    2.04699357079
1.00000000000   1.27101099359   1.61508265819    2.04883894862
1.00000000000   1.27101687054   1.61522146569    2.05016055479
1.00000000000   1.27102065019   1.61531567660    2.05110318836
1.00000000000   1.27102307863   1.61537946590    2.05177294664
1.00000000000   1.27102463745   1.61542255932    2.05224710921
1.00000000000   1.27102563717   1.61545160916    2.05258165906
1.00000000000   1.27102627777   1.61547115233    2.05281694778
1.00000000000   1.27102668793   1.61548427459    2.05298192472
1.00000000000   1.27102695034   1.61549306940    2.05309726902
1.00000000000   1.27102711810   1.61549895358    2.05317769251
1.00000000000   1.27102722526   1.61550288385    2.05323362215
1.00000000000   1.27102729368   1.61550550485    2.05327242174
1.00000000000   1.27102733732   1.61550725009    2.05329927444
1.00000000000   1.27102736515   1.61550841049    2.05331781705
1.00000000000   1.27102738288   1.61550918095    2.05333059375
1.00000000000   1.27102739417   1.61550969183    2.05333937934
1.00000000000   1.27102740136   1.61551003015    2.05334540863
1.00000000000   1.27102740593   1.61551025392    2.05334953852
1.00000000000   1.27102740884   1.61551040174    2.05335236222
1.00000000000   1.27102741068   1.61551049929    2.05335428948
1.00000000000   1.27102741186   1.61551056358    2.05335560269
1.00000000000   1.27102741260   1.61551060592    2.05335649603
1.00000000000   1.27102741307   1.61551063376    2.05335710280
1.00000000000   1.27102741337   1.61551065206    2.05335751431
1.00000000000   1.27102741356   1.61551066407    2.05335779299
1.00000000000   1.27102741368   1.61551067195    2.05335798145
1.00000000000   1.27102741376   1.61551067711    2.05335810872
1.00000000000   1.27102741381   1.61551068049    2.05335819456
1.00000000000   1.27102741384   1.61551068271    2.05335825238
1.00000000000   1.27102741386   1.61551068415    2.05335829128
1.00000000000   1.27102741387   1.61551068509    2.05335831742
1.00000000000   1.27102741388   1.61551068571    2.05335833497
1.00000000000   1.27102741388   1.61551068611    2.05335834673
1.00000000000   1.27102741388   1.61551068637    2.05335835460
1.00000000000   1.27102741389   1.61551068654    2.05335835987
1.00000000000   1.27102741389   1.61551068665    2.05335836340
1.00000000000   1.27102741389   1.61551068673    2.05335836575

Note also, that with each order of the Euler-summation we'll find
divergence of the sequence partial sums after a certain row, so
we have a maximal possible precision for each order.

I should explain also the rationale for the appropriateness
of this type of approximation, but I have it not at hand.
It may be found in "unendliche Reihen" of K.Knopp, or in
G.H.Hardy's monography on this subject. Perhaps I can add