Well, given my model of tetration.
Remember, that I doubt that it is reasonable to assume the results of my definition of tetration and yours should be equal, although by the sheer notation with equal parameters the formulae look identical. But it may be instructive anyway.
I just hacked a dimension-32 approximation accroding to the general formula {a,x}^^h given your parameters according to your problem a=I,x=1,h=I.
The first three columns of Bs = dV(log(I)) * B are
\( \hspace{24}
\begin{matrix} {rrr}
1.00000000000 & 1.00000000000 & 1.00000000000 \\
0 & 1.57079632679*I & 3.14159265359*I \\
0 & -1.23370055014 & -4.93480220054 \\
0 & -0.645964097506*I & -5.16771278005*I \\
0 & 0.253669507901 & 4.05871212642 \\
0 & 0.0796926262462*I & 2.55016403988*I \\
0 & -0.0208634807634 & -1.33526276885 \\
0 & -0.00468175413532*I & -0.599264529321*I \\
0 & 0.000919260274839 & 0.235330630359 \\
0 & 0.000160441184787*I & 0.0821458866111*I \\
0 & -0.0000252020423731 & -0.0258068913900 \\
0 & -0.00000359884323521*I & -0.00737043094571*I \\
0 & 0.000000471087477882 & 0.00192957430940 \\
0 & 0.0000000569217292197*I & 0.000466302805768*I \\
0 & -0.00000000638660308379 & -0.000104638104925 \\
0 & -0.000000000668803510981*I & -0.0000219153534478*I \\
0 & 6.56596311498E-11 & 0.00000430306958703 \\
0 & 6.06693573110E-12*I & 0.000000795205400148*I \\
0 & -1.00000000000E-12 & -0.000000138789524622 \\
0 & 0 & -0.0000000229484289973*I \\
0 & 0 & 0.00000000360473079746 \\
0 & 0 & 0.000000000539266466261*I \\
0 & 0 & -7.70070713060E-11 \\
0 & 0 & -1.05184717169E-11*I
\end{matrix} \)
To get the (trivial) value of I = {I,1}^^1 the terms of the second column must be summed as powerseries in x with x=1. Here I show the partial sums, which nicely converge to the expected result y=I:
\( \hspace{24}
\begin{matrix} {rrr}
0.909090909091 \\
0.991735537190+1.29817878248*I \\
0.0723512020014+1.53421128838*I \\
-0.179756007234+1.12519536891*I \\
-0.0681469673065+0.968659579810*I \\
-0.00351032851781+0.977624337253*I \\
0.00436018003194+0.995580017724*I \\
0.00156814228411+1.00015357146*I \\
0.000218934491618+1.00030493211*I \\
-0.0000162526513523+1.00008325134*I \\
-0.0000151845153453+1.00001025056*I \\
-0.00000375533374982+0.999999385186*I \\
-0.000000479000915892+0.999999405083*I \\
0.00000000440878968775+0.999999851342*I \\
0.0000000185010573405+0.999999978737*I \\
0.00000000512110205692+0.999999999027*I \\
0.000000000844337715795+1.00000000042*I \\
7.67095502124E-11+1.00000000015*I \\
-4.07641970796E-12+1.00000000003*I \\
-3.36962348090E-12+1.00000000000*I \\
-1.00000000000E-12+1.00000000000*I \\
1.00000000000*I \\
1.00000000000*I \\
1.00000000000*I
\end{matrix} \)
Ok, no obvious error up to here
Now to construct the I'th power of Bs I do numerically an eigendecomposition. The computed eigenvalues are
\( \hspace{24}
\begin{matrix} {rrr}
-5.47974865102E17-2.37151639248E18*I \\
-1.40947398634E16+1.11407483052E16*I \\
2.46479084446E14+1.28514676152E14*I \\
558111745700.-6.74160002940E12*I \\
-220611525960.+65636453372.0*I \\
6058523291.41+8279445712.41*I \\
337747205.316-464618483.215*I \\
-36794031.6180-13535297.2580*I \\
-358363.072707+3168208.25176*I \\
302209.998163-27260.0213372*I \\
-8395.97431462-32222.7753783*I \\
-3862.15905348+1553.24166668*I \\
267.408153438+522.463483061*I \\
79.9487788351-46.6197593935*I \\
-8.44497709779-13.8323341991*I \\
-2.69468450558+1.59521959157*I \\
-0.153139121702-0.779903738266*I \\
-0.434207067913-0.0300570687829*I \\
-0.566417336767+0.688453222928*I \\
-0.584898170382+0.238979921688*I \\
0.166722386790-0.538414193535*I \\
0.253868227388-0.289262673366*I \\
0.0621911678425+0.0678908098416*I \\
0.327850649567+0.623039459463*I \\
0.271875740531+0.406308595179*I \\
0.623667274970+0.336312783860*I \\
1.00000000000 \\
0.000214033167476+0.00585473360451*I \\
-0.000171842783002+0.000232593700346*I \\
-0.00000000174878768422-0.00000000220727624373*I \\
-0.000000233830386869-0.0000000768729721930*I \\
-0.0000100176583851+0.00000302704107609*I
\end{matrix} \)
Note, that these eigenvalues are *not* constant with higher dimesions. They are just constructed from the empirical (truncated wwith dim=32) matrix-operator for base I
Anyway, this reproduces the basic matrixoperator with h=1 perfectly.
Taken to the I'th power (by exponentiating the eigenvalues) Pari/GP gives this matrix as Bs^I (only the finally relevant second column is given here)
\( \hspace{24}
\begin{matrix} {rrr}
2.06145217150-1.02926728869*I \\
5.81300594285+19.5645999807*I \\
-126.124737459-30.4110746405*I \\
470.620324532-351.161359046*I \\
-224.452590018+1917.76776945*I \\
-3037.77318290-3791.12832155*I \\
9716.74257623+1145.91813696*I \\
-13067.3202330+9858.26388713*I \\
4092.40268289-23076.9591360*I \\
15598.5659208+24912.6967922*I \\
-31488.4281518-9341.69517397*I \\
30138.7445858-13658.7995071*I \\
-12822.8086718+27435.3139921*I \\
-6715.89914642-24385.0097082*I \\
16020.8287524+10805.4109019*I \\
-13423.5109290+1685.41825964*I \\
5663.37890805-6617.80298683*I \\
328.580050705+5202.17407548*I \\
-2283.97893527-1905.01648008*I \\
1663.72812201-239.296020348*I \\
-554.785800953+780.887552013*I \\
-58.9789199840-530.120309375*I \\
185.617911945+203.010458380*I \\
-120.661742137-35.9711040163*I \\
48.9129882830-8.69936241104*I \\
-13.8034038806+9.04489016501*I \\
2.68261460172-3.61800295858*I \\
-0.316133256749+0.928004747822*I \\
0.00895779796132-0.163077449477*I \\
0.00364944523306+0.0191872578491*I \\
-0.000587487763491-0.00137114979404*I \\
0.0000308282641305+0.0000451105887704*I
\end{matrix} \)
The terms used as coefficients for a powerseries in x with x=1 has to be evaluated to get y={I,1}^^I. I approximate this by the partial-sums using Euler-summatiion (order 1.3 suffices)
\( \hspace{24}
\begin{matrix} {rrr}
1.58573243962-0.791744068222*I \\
5.39131918181+10.6022321413*I \\
-50.3444128154+2.06106295233*I \\
75.2580631596-128.633476610*I \\
148.738829992+270.527508926*I \\
-469.621461225+14.0551655027*I \\
197.268710310-509.011209276*I \\
382.544924169+325.914251348*I \\
-319.220819985+210.623903212*I \\
-86.6444368232-236.149488090*I \\
147.094103735-24.8153704764*I \\
1.87967578010+83.3028710164*I \\
-45.5253002777-3.29677555442*I \\
2.07384793001-24.3565866239*I \\
12.1754155038+0.365227776672*I \\
-0.0797957239470+5.77003813168*I \\
-3.21904307851-0.0361632958617*I \\
-0.617106650046-1.91201801927*I \\
0.489205540885-0.837213872318*I \\
0.102947554342-0.210707955957*I \\
-0.231195060161-0.311521742399*I \\
-0.232100049827-0.476259731040*I \\
-0.159094876028-0.502092239406*I \\
-0.135412161873-0.474546607718*I \\
-0.142918083910-0.459322945360*I \\
-0.150850150122-0.459521341773*I \\
-0.152427969793-0.462866323691*I \\
-0.151401051174-0.464246402460*I \\
-0.150603568878-0.464150373228*I \\
-0.150448471688-0.463811394143*I \\
-0.150542606888-0.463666152744*I \\
-0.150623503560-0.463664450950*I
\end{matrix} \)
So the result given here is the last row with
y={I,1}^^I ~ 0.1506... -0.4636...*I
But I don't trust this solution, since in the third column (not documented here) we should have the square, which doesn't appear:
0.873379283287-2.89383467237*I
but y^2 = -0.192297283250 + 0.139677528157*I
So, before one could say, y is a reasonable approximation one had to show
a) that the solutions y for values h->I are consistent and continuously approximating the given value
b) the result makes sense if consistent with further algebraic operations.
The remaining problem for my matrix-method is simply the missing hypothese, how to deal with the multivalued logarithms of complex eigenvalues and their complex powers. (Note that it may be interesting to read Jay's recent post, which adresses the analoguous problem when applying his method (I think, it is essentially the same))
Gottfried
Remember, that I doubt that it is reasonable to assume the results of my definition of tetration and yours should be equal, although by the sheer notation with equal parameters the formulae look identical. But it may be instructive anyway.
I just hacked a dimension-32 approximation accroding to the general formula {a,x}^^h given your parameters according to your problem a=I,x=1,h=I.
The first three columns of Bs = dV(log(I)) * B are
\( \hspace{24}
\begin{matrix} {rrr}
1.00000000000 & 1.00000000000 & 1.00000000000 \\
0 & 1.57079632679*I & 3.14159265359*I \\
0 & -1.23370055014 & -4.93480220054 \\
0 & -0.645964097506*I & -5.16771278005*I \\
0 & 0.253669507901 & 4.05871212642 \\
0 & 0.0796926262462*I & 2.55016403988*I \\
0 & -0.0208634807634 & -1.33526276885 \\
0 & -0.00468175413532*I & -0.599264529321*I \\
0 & 0.000919260274839 & 0.235330630359 \\
0 & 0.000160441184787*I & 0.0821458866111*I \\
0 & -0.0000252020423731 & -0.0258068913900 \\
0 & -0.00000359884323521*I & -0.00737043094571*I \\
0 & 0.000000471087477882 & 0.00192957430940 \\
0 & 0.0000000569217292197*I & 0.000466302805768*I \\
0 & -0.00000000638660308379 & -0.000104638104925 \\
0 & -0.000000000668803510981*I & -0.0000219153534478*I \\
0 & 6.56596311498E-11 & 0.00000430306958703 \\
0 & 6.06693573110E-12*I & 0.000000795205400148*I \\
0 & -1.00000000000E-12 & -0.000000138789524622 \\
0 & 0 & -0.0000000229484289973*I \\
0 & 0 & 0.00000000360473079746 \\
0 & 0 & 0.000000000539266466261*I \\
0 & 0 & -7.70070713060E-11 \\
0 & 0 & -1.05184717169E-11*I
\end{matrix} \)
To get the (trivial) value of I = {I,1}^^1 the terms of the second column must be summed as powerseries in x with x=1. Here I show the partial sums, which nicely converge to the expected result y=I:
\( \hspace{24}
\begin{matrix} {rrr}
0.909090909091 \\
0.991735537190+1.29817878248*I \\
0.0723512020014+1.53421128838*I \\
-0.179756007234+1.12519536891*I \\
-0.0681469673065+0.968659579810*I \\
-0.00351032851781+0.977624337253*I \\
0.00436018003194+0.995580017724*I \\
0.00156814228411+1.00015357146*I \\
0.000218934491618+1.00030493211*I \\
-0.0000162526513523+1.00008325134*I \\
-0.0000151845153453+1.00001025056*I \\
-0.00000375533374982+0.999999385186*I \\
-0.000000479000915892+0.999999405083*I \\
0.00000000440878968775+0.999999851342*I \\
0.0000000185010573405+0.999999978737*I \\
0.00000000512110205692+0.999999999027*I \\
0.000000000844337715795+1.00000000042*I \\
7.67095502124E-11+1.00000000015*I \\
-4.07641970796E-12+1.00000000003*I \\
-3.36962348090E-12+1.00000000000*I \\
-1.00000000000E-12+1.00000000000*I \\
1.00000000000*I \\
1.00000000000*I \\
1.00000000000*I
\end{matrix} \)
Ok, no obvious error up to here
Now to construct the I'th power of Bs I do numerically an eigendecomposition. The computed eigenvalues are
\( \hspace{24}
\begin{matrix} {rrr}
-5.47974865102E17-2.37151639248E18*I \\
-1.40947398634E16+1.11407483052E16*I \\
2.46479084446E14+1.28514676152E14*I \\
558111745700.-6.74160002940E12*I \\
-220611525960.+65636453372.0*I \\
6058523291.41+8279445712.41*I \\
337747205.316-464618483.215*I \\
-36794031.6180-13535297.2580*I \\
-358363.072707+3168208.25176*I \\
302209.998163-27260.0213372*I \\
-8395.97431462-32222.7753783*I \\
-3862.15905348+1553.24166668*I \\
267.408153438+522.463483061*I \\
79.9487788351-46.6197593935*I \\
-8.44497709779-13.8323341991*I \\
-2.69468450558+1.59521959157*I \\
-0.153139121702-0.779903738266*I \\
-0.434207067913-0.0300570687829*I \\
-0.566417336767+0.688453222928*I \\
-0.584898170382+0.238979921688*I \\
0.166722386790-0.538414193535*I \\
0.253868227388-0.289262673366*I \\
0.0621911678425+0.0678908098416*I \\
0.327850649567+0.623039459463*I \\
0.271875740531+0.406308595179*I \\
0.623667274970+0.336312783860*I \\
1.00000000000 \\
0.000214033167476+0.00585473360451*I \\
-0.000171842783002+0.000232593700346*I \\
-0.00000000174878768422-0.00000000220727624373*I \\
-0.000000233830386869-0.0000000768729721930*I \\
-0.0000100176583851+0.00000302704107609*I
\end{matrix} \)
Note, that these eigenvalues are *not* constant with higher dimesions. They are just constructed from the empirical (truncated wwith dim=32) matrix-operator for base I
Anyway, this reproduces the basic matrixoperator with h=1 perfectly.
Taken to the I'th power (by exponentiating the eigenvalues) Pari/GP gives this matrix as Bs^I (only the finally relevant second column is given here)
\( \hspace{24}
\begin{matrix} {rrr}
2.06145217150-1.02926728869*I \\
5.81300594285+19.5645999807*I \\
-126.124737459-30.4110746405*I \\
470.620324532-351.161359046*I \\
-224.452590018+1917.76776945*I \\
-3037.77318290-3791.12832155*I \\
9716.74257623+1145.91813696*I \\
-13067.3202330+9858.26388713*I \\
4092.40268289-23076.9591360*I \\
15598.5659208+24912.6967922*I \\
-31488.4281518-9341.69517397*I \\
30138.7445858-13658.7995071*I \\
-12822.8086718+27435.3139921*I \\
-6715.89914642-24385.0097082*I \\
16020.8287524+10805.4109019*I \\
-13423.5109290+1685.41825964*I \\
5663.37890805-6617.80298683*I \\
328.580050705+5202.17407548*I \\
-2283.97893527-1905.01648008*I \\
1663.72812201-239.296020348*I \\
-554.785800953+780.887552013*I \\
-58.9789199840-530.120309375*I \\
185.617911945+203.010458380*I \\
-120.661742137-35.9711040163*I \\
48.9129882830-8.69936241104*I \\
-13.8034038806+9.04489016501*I \\
2.68261460172-3.61800295858*I \\
-0.316133256749+0.928004747822*I \\
0.00895779796132-0.163077449477*I \\
0.00364944523306+0.0191872578491*I \\
-0.000587487763491-0.00137114979404*I \\
0.0000308282641305+0.0000451105887704*I
\end{matrix} \)
The terms used as coefficients for a powerseries in x with x=1 has to be evaluated to get y={I,1}^^I. I approximate this by the partial-sums using Euler-summatiion (order 1.3 suffices)
\( \hspace{24}
\begin{matrix} {rrr}
1.58573243962-0.791744068222*I \\
5.39131918181+10.6022321413*I \\
-50.3444128154+2.06106295233*I \\
75.2580631596-128.633476610*I \\
148.738829992+270.527508926*I \\
-469.621461225+14.0551655027*I \\
197.268710310-509.011209276*I \\
382.544924169+325.914251348*I \\
-319.220819985+210.623903212*I \\
-86.6444368232-236.149488090*I \\
147.094103735-24.8153704764*I \\
1.87967578010+83.3028710164*I \\
-45.5253002777-3.29677555442*I \\
2.07384793001-24.3565866239*I \\
12.1754155038+0.365227776672*I \\
-0.0797957239470+5.77003813168*I \\
-3.21904307851-0.0361632958617*I \\
-0.617106650046-1.91201801927*I \\
0.489205540885-0.837213872318*I \\
0.102947554342-0.210707955957*I \\
-0.231195060161-0.311521742399*I \\
-0.232100049827-0.476259731040*I \\
-0.159094876028-0.502092239406*I \\
-0.135412161873-0.474546607718*I \\
-0.142918083910-0.459322945360*I \\
-0.150850150122-0.459521341773*I \\
-0.152427969793-0.462866323691*I \\
-0.151401051174-0.464246402460*I \\
-0.150603568878-0.464150373228*I \\
-0.150448471688-0.463811394143*I \\
-0.150542606888-0.463666152744*I \\
-0.150623503560-0.463664450950*I
\end{matrix} \)
So the result given here is the last row with
y={I,1}^^I ~ 0.1506... -0.4636...*I
But I don't trust this solution, since in the third column (not documented here) we should have the square, which doesn't appear:
0.873379283287-2.89383467237*I
but y^2 = -0.192297283250 + 0.139677528157*I
So, before one could say, y is a reasonable approximation one had to show
a) that the solutions y for values h->I are consistent and continuously approximating the given value
b) the result makes sense if consistent with further algebraic operations.
The remaining problem for my matrix-method is simply the missing hypothese, how to deal with the multivalued logarithms of complex eigenvalues and their complex powers. (Note that it may be interesting to read Jay's recent post, which adresses the analoguous problem when applying his method (I think, it is essentially the same))
Gottfried
Gottfried Helms, Kassel