Here I show some results for -19/20 <= log(h) <= -1/20
which means, that bases b are in the bounds 0.086<b<0.95
As in the previous post, h = h(b) or h^(1/h)=b, and -1<log(h)<0,
so we have complex eigenvalues for the half-iterate.
I used 32 terms; for log(h)<-4/5 the convergence was bad, so I used
Euler-summation of hand-selected orders to sum up the terms
(see third table)
Here are results for the half-iterates:
Here are the results in terms of polar-coordinates:
In the next table are the first 6 terms [update:reduced from7] for the evaluation at each log(h).
Let the coefficients in the first row be a_k, the coefficients in the second row be b_k (only the first six are documented),
then the result (the value for the half-iterate) is sum (a_k*b_k)
We see, that convergence is good for some log(h)>-4/5, but the first examples
(smaller log(h)) are "un-nice", and the convergence must be accelerated to get
usable results with 32 terms only.
Hmm, it may be worth noting, that these terms are exact (in the sense as we assume logarithms,
powers of logarithms and exp(x) are exact and their finite linear combinations) due to my
analytic solution.
They don't change when the dimension of matrices is increased, only their number increases
with the higher dimensions.
Gottfried
[update: I reduced the number of terms because the wide format also affected the readability of the other postings on my screen]
which means, that bases b are in the bounds 0.086<b<0.95
As in the previous post, h = h(b) or h^(1/h)=b, and -1<log(h)<0,
so we have complex eigenvalues for the half-iterate.
I used 32 terms; for log(h)<-4/5 the convergence was bad, so I used
Euler-summation of hand-selected orders to sum up the terms
(see third table)
Here are results for the half-iterates:
Code:
Table1:
b h log(h) f°0.5(b,1)
0.085740998 0.38674102 -19/20 0.36418555+0.18927488*I // only about 4 or 5 digits correct
0.10930176 0.40656966 -9/10 0.36934282+0.25207493*I
0.13687327 0.42741493 -17/20 0.38255541+0.28789034*I
0.16856519 0.44932896 -4/5 0.40234273+0.30722539*I // down from here the digits seem to be all correct
0.20438496 0.47236655 -3/4 0.42722347+0.31456470*I
0.24423439 0.49658530 -7/10 0.45586756+0.31265441*I
0.28791167 0.52204578 -13/20 0.48716903+0.30346945*I
0.33511843 0.54881164 -3/5 0.52028454+0.28854799*I
0.38547106 0.57694981 -11/20 0.55462781+0.26913667*I
0.43851527 0.60653066 -1/2 0.58983992+0.24626917*I
0.49374283 0.63762815 -9/20 0.62574943+0.22081962*I
0.55060925 0.67032005 -2/5 0.66233139+0.19354617*I
0.60855156 0.70468809 -7/20 0.69967007+0.16513107*I
0.66700506 0.74081822 -3/10 0.73792748+0.13622047*I
0.72541864 0.77880078 -1/4 0.77731772+0.10746647*I
0.78326786 0.81873075 -1/5 0.81808620+0.079576360*I
0.84006573 0.86070798 -3/20 0.86049207+0.053380211*I
0.89537103 0.90483742 -1/10 0.90479236+0.029949503*I
0.94879402 0.95122942 -1/20 0.95122645+0.010892129*IHere are the results in terms of polar-coordinates:
Code:
Table 2:
log(h) abs(f°0.5) arg(f°0.5)
--------------------------------------
-19/20 0.41043403 0.47929966
-9/10 0.44716428 0.59888134
-17/20 0.47877917 0.64512822
-4/5 0.50622832 0.65214291
-3/4 0.53053825 0.63467538
-7/10 0.55278207 0.60116254
-13/20 0.57395764 0.55710531
-3/5 0.59494197 0.50636535
-11/20 0.61647916 0.45178327
-1/2 0.63918670 0.39551685
-9/20 0.66356888 0.33924555
-2/5 0.69003115 0.28430356
-7/20 0.71889254 0.23177151
-3/10 0.75039522 0.18254374
-1/4 0.78471133 0.13738207
-1/5 0.82194734 0.096966309
-3/20 0.86214618 0.061955126
-1/10 0.90528790 0.033088886
-1/20 0.95128881 0.011450115In the next table are the first 6 terms [update:reduced from7] for the evaluation at each log(h).
Let the coefficients in the first row be a_k, the coefficients in the second row be b_k (only the first six are documented),
then the result (the value for the half-iterate) is sum (a_k*b_k)
We see, that convergence is good for some log(h)>-4/5, but the first examples
(smaller log(h)) are "un-nice", and the convergence must be accelerated to get
usable results with 32 terms only.
Code:
Table 3:
log(h)=-19/20
1.0000000 1.5857097 2.5144751 3.9872275 6.3225851
0.38674102 0.37694852*I -0.089495839-0.091820794*I 0.043600537-0.56587507*I 0.26178981+0.42070266*I
log(h)=-18/20
1.0000000 1.4596031 2.1304412 3.1095987 4.5387799
0.40656966 0.38570585*I -0.086663533-0.091351385*I 0.041051147-0.25818786*I 0.10983200+0.19013148*I
log(h)=-17/20
1.0000000 1.3396469 1.7946537 2.4042022 3.2207819
0.42741493 0.39405710*I -0.083461429-0.090526630*I 0.038347143-0.15505064*I 0.060177000+0.11303772*I
log(h)=-16/20 = -4/5
1.0000000 1.2255409 1.5019506 1.8407019 2.2558555
0.44932896 0.40189204*I -0.079880705-0.089309343*I 0.035502535-0.10320191*I 0.036178774+0.074471706*I
log(h)=-15/20
1.0000000 1.1170000 1.2476890 1.3936687 1.5567279
0.47236655 0.40908143*I -0.075916053-0.087660307*I 0.032535451-0.072006681*I 0.022502197+0.051450545*I
log(h)=-14/20
1.0000000 1.0137527 1.0276946 1.0418281 1.0561561
0.49658530 0.41547307*I -0.071566706-0.085538574*I 0.029468643-0.051267237*I 0.014028081+0.036315835*I
log(h)=-13/20
1.0000000 0.91554083 0.83821501 0.76742006 0.70260440
0.52204578 0.42088676*I -0.066837679-0.082901938*I 0.026329995-0.036623995*I 0.0085533520+0.025784195*I
...
log(h)=-1/20
1.0000000 0.051271096 0.0026287253 0.00013477763 0.0000069101968
0.95122942 0.21270137*I -0.0011324160-0.0050643182*I 0.000053924571+0.000067905019*I -0.0000013853114-0.00000016343552*IHmm, it may be worth noting, that these terms are exact (in the sense as we assume logarithms,
powers of logarithms and exp(x) are exact and their finite linear combinations) due to my
analytic solution.
They don't change when the dimension of matrices is increased, only their number increases
with the higher dimensions.
Gottfried
[update: I reduced the number of terms because the wide format also affected the readability of the other postings on my screen]
Gottfried Helms, Kassel