In response to the thread Solving tetration for base 0 < b < e^-e. I will feature here what one could call regular superlogarithm.
The idea is the following: One can obtain the regular Abel function
of
by the Julia function
:
(*)=\int \frac{1}{j(z)})
The Julia function (or also called iterative logarithm) can be obtained by
(**)=\left.\frac{\partial f^{\circ t}}{\partial t}\right|_{t=0})
where we mean
to be the *regular* iteration of
.
It is similar to the logarithm:
)
And we have a way to obtain the regular iteration
by matrix powers.
Before we begin we reduce the problem of iterating
to iterating
. This is a linear conjugation with
and
, i.e.
. It moves the fixed point from
to
.
has the powerseries coefficients:
Then we take the Carleman matrix of this series (I truncate it to 10 here):
)
And then we consider the first line of the matrix power
. These are the coefficients of
:
^{t}\\<br />
-\frac{1}{4} \, \left(-1\right)^{t} + \frac{1}{4}\\<br />
\frac{1}{12} \, t \left(-1\right)^{{(t - 1)}} + \frac{1}{8} \, \left(-1\right)^{t} - \frac{1}{8}\\<br />
-\frac{1}{16} \, t \left(-1\right)^{{(t - 1)}} - \frac{7}{96} \, \left(-1\right)^{t} - \frac{1}{24} \, t + \frac{7}{96}\\<br />
\frac{1}{96} \, {(t - 1)} t \left(-1\right)^{{(t - 2)}} + \frac{7}{160} \, t \left(-1\right)^{{(t - 1)}} + \frac{3}{64} \, \left(-1\right)^{t} + \frac{1}{24} \, t - \frac{3}{64}\\<br />
-\frac{5}{384} \, {(t - 1)} t \left(-1\right)^{{(t - 2)}} + \frac{1}{144} \, {(t - 1)} t - \frac{1}{32} \, t \left(-1\right)^{{(t - 1)}} - \frac{41}{1280} \, \left(-1\right)^{t} - \frac{181}{5760} \, t + \frac{41}{1280}\\<br />
\frac{5}{3456} \, {(t - 2)} {(t - 1)} t \left(-1\right)^{{(t - 3)}} + \frac{1}{90} \, {(t - 1)} t \left(-1\right)^{{(t - 2)}} - \frac{1}{96} \, {(t - 1)} t + \frac{365}{16128} \, t \left(-1\right)^{{(t - 1)}} + \frac{263}{11520} \, \left(-1\right)^{t} + \frac{263}{11520} \, t - \frac{263}{11520}\\<br />
-\frac{35}{13824} \, {(t - 2)} {(t - 1)} t \left(-1\right)^{{(t - 3)}} - \frac{1}{864} \, {(t - 2)} {(t - 1)} t - \frac{11}{1280} \, {(t - 1)} t \left(-1\right)^{{(t - 2)}} + \frac{43}{4608} \, {(t - 1)} t - \frac{769}{46080} \, t \left(-1\right)^{{(t - 1)}} - \frac{901}{53760} \, \left(-1\right)^{t} - \frac{1807}{107520} \, t + \frac{901}{53760}\\<br />
\frac{35}{165888} \, {(t - 3)} {(t - 2)} {(t - 1)} t \left(-1\right)^{{(t - 4)}} + \frac{259}{103680} \, {(t - 2)} {(t - 1)} t \left(-1\right)^{{(t - 3)}} + \frac{1}{432} \, {(t - 2)} {(t - 1)} t + \frac{37691}{5806080} \, {(t - 1)} t \left(-1\right)^{{(t - 2)}} - \frac{1}{144} \, {(t - 1)} t + \frac{5197}{414720} \, t \left(-1\right)^{{(t - 1)}} + \dots<br />
)
Each coefficient is a mixture of polynomials in
containing
. It has 0 convergence radius.
We apply (**) and get the coefficients of the Julia function
:
Now we apply (*):
When we take the reciprocal of the julia function we get a formal Laurent series starting at -1. The coefficient at -1 is
, the other coefficients are:
 \, \frac{1}{\pi}\\<br />
\left(\frac{1}{16} I\right) \, \frac{1}{\pi} - \frac{1}{12} \, \frac{1}{\pi^{2}}\\<br />
-\left(\frac{5}{192} I\right) \, \frac{1}{\pi} - \frac{1}{48} \, \frac{1}{\pi^{2}}\\<br />
\left(\frac{11}{768} I\right) \, \frac{1}{\pi} + \frac{1}{320} \, \frac{1}{\pi^{2}} + \left(\frac{1}{144} I\right) \, \frac{1}{\pi^{3}}\\<br />
-\left(\frac{137}{15360} I\right) \, \frac{1}{\pi} - \frac{13}{3840} \, \frac{1}{\pi^{2}} + \left(\frac{1}{576} I\right) \, \frac{1}{\pi^{3}}\\<br />
\left(\frac{367}{61440} I\right) \, \frac{1}{\pi} + \frac{1697}{967680} \, \frac{1}{\pi^{2}} - \left(\frac{1}{11520} I\right) \, \frac{1}{\pi^{3}} + \frac{1}{1728} \, \frac{1}{\pi^{4}}\\<br />
-\left(\frac{21557}{5160960} I\right) \, \frac{1}{\pi} - \frac{1669}{1290240} \, \frac{1}{\pi^{2}} + \left(\frac{53}{138240} I\right) \, \frac{1}{\pi^{3}} + \frac{1}{6912} \, \frac{1}{\pi^{4}}\\<br />
\left(\frac{62171}{20643840} I\right) \, \frac{1}{\pi} + \frac{42083}{46448640} \, \frac{1}{\pi^{2}} - \left(\frac{10943}{58060800} I\right) \, \frac{1}{\pi^{3}} + \frac{1}{138240} \, \frac{1}{\pi^{4}} - \left(\frac{1}{20736} I\right) \, \frac{1}{\pi^{5}}\\<br />
\left(\frac{52187}{7077888} I\right) \, \frac{1}{\pi} - \frac{401003}{309657600} \, \frac{1}{\pi^{2}} + \left(\frac{12203}{77414400} I\right) \, \frac{1}{\pi^{3}} + \frac{67}{1658880} \, \frac{1}{\pi^{4}} - \left(\frac{1}{82944} I\right) \, \frac{1}{\pi^{5}}\\<br />
-\left(\frac{4720619}{4954521600} I\right) \, \frac{1}{\pi} + \frac{95027}{176947200} \, \frac{1}{\pi^{2}} - \left(\frac{309629}{2786918400} I\right) \, \frac{1}{\pi^{3}} - \frac{487}{25804800} \, \frac{1}{\pi^{4}} - \left(\frac{1}{552960} I\right) \, \frac{1}{\pi^{5}} - \frac{1}{248832} \, \frac{1}{\pi^{6}})
When integrating the coefficient of
becomes the coefficient of the logarithm. The other coefficients of the Abel function are obtained by formal integration. The coefficients of the Abel function
of
are:
:\quad -I\,\frac{1}{\pi}\\<br />
z:\quad-\left(\frac{1}{4} I\right) \, \frac{1}{\pi}\\<br />
z^{2}:\quad\left(\frac{1}{32} I\right) \, \frac{1}{\pi} - \frac{1}{24} \, \frac{1}{\pi^{2}}\\<br />
z^{3}:\quad-\left(\frac{5}{576} I\right) \, \frac{1}{\pi} - \frac{1}{144} \, \frac{1}{\pi^{2}}\\<br />
z^{4}:\quad\left(\frac{11}{3072} I\right) \, \frac{1}{\pi} + \frac{1}{1280} \, \frac{1}{\pi^{2}} + \left(\frac{1}{576} I\right) \, \frac{1}{\pi^{3}}\\<br />
z^{5}:\quad-\left(\frac{137}{76800} I\right) \, \frac{1}{\pi} - \frac{13}{19200} \, \frac{1}{\pi^{2}} + \left(\frac{1}{2880} I\right) \, \frac{1}{\pi^{3}}\\<br />
z^{6}:\quad\left(\frac{367}{368640} I\right) \, \frac{1}{\pi} + \frac{1697}{5806080} \, \frac{1}{\pi^{2}} - \left(\frac{1}{69120} I\right) \, \frac{1}{\pi^{3}} + \frac{1}{10368} \, \frac{1}{\pi^{4}}\\<br />
z^{7}:\quad-\left(\frac{21557}{36126720} I\right) \, \frac{1}{\pi} - \frac{1669}{9031680} \, \frac{1}{\pi^{2}} + \left(\frac{53}{967680} I\right) \, \frac{1}{\pi^{3}} + \frac{1}{48384} \, \frac{1}{\pi^{4}}\\<br />
z^{8}:\quad\left(\frac{62171}{165150720} I\right) \, \frac{1}{\pi} + \frac{42083}{371589120} \, \frac{1}{\pi^{2}} - \left(\frac{10943}{464486400} I\right) \, \frac{1}{\pi^{3}} + \frac{1}{1105920} \, \frac{1}{\pi^{4}} - \left(\frac{1}{165888} I\right) \, \frac{1}{\pi^{5}}\\<br />
z^{9}:\quad\left(\frac{52187}{63700992} I\right) \, \frac{1}{\pi} - \frac{401003}{2786918400} \, \frac{1}{\pi^{2}} + \left(\frac{12203}{696729600} I\right) \, \frac{1}{\pi^{3}} + \frac{67}{14929920} \, \frac{1}{\pi^{4}} - \left(\frac{1}{746496} I\right) \, \frac{1}{\pi^{5}})
The superlogarithm is then just
.
Notes:
1. The provided values are just for comparison if you want to reproduce this approach. I worked with Sage and my own powerseries libary. I can provide more coefficients as numerics if you specify some format. I didnt test my computations against obvious errors nor plotted the corresponding graph, nor do I know whether the series converges (thoug it looks quite so).
2. The powers
in the first row of the matrix power
are of course ambigous. I choose the standard branch
of the logarithm in
.
3. In the formula for
you can of course choose arbitrary cuts and branches for the logarithm. The corresponding cut is then the cut for the Abel function and the cut after application of
is the cut of the superlogarithm in the point
.
4. You can add arbitrary constants to the superlogarithm and obtain again a superlogarithm.
The idea is the following: One can obtain the regular Abel function
(*)
The Julia function (or also called iterative logarithm) can be obtained by
(**)
where we mean
It is similar to the logarithm:
And we have a way to obtain the regular iteration
Before we begin we reduce the problem of iterating
Code:
0, -1, 1/2, -1/6, 1/24, -1/120, 1/720, -1/5040, 1/40320, -1/362880, ...
Then we take the Carleman matrix of this series (I truncate it to 10 here):
And then we consider the first line of the matrix power
Each coefficient is a mixture of polynomials in
We apply (**) and get the coefficients of the Julia function
Code:
0, I*pi, -1/4*I*pi, 1/8*I*pi - 1/12, -7/96*I*pi + 1/48, 3/64*I*pi - 1/80, -41/1280*I*pi + 17/2880, 263/11520*I*pi - 41/12096, -901/53760*I*pi + 229/120960, 3245/258048*I*pi - 227/207360
Now we apply (*):
When we take the reciprocal of the julia function we get a formal Laurent series starting at -1. The coefficient at -1 is
When integrating the coefficient of
The superlogarithm is then just
Notes:
1. The provided values are just for comparison if you want to reproduce this approach. I worked with Sage and my own powerseries libary. I can provide more coefficients as numerics if you specify some format. I didnt test my computations against obvious errors nor plotted the corresponding graph, nor do I know whether the series converges (thoug it looks quite so).
2. The powers
3. In the formula for
4. You can add arbitrary constants to the superlogarithm and obtain again a superlogarithm.