Observations on power series involving logarithmic singularities

[Gottfried]

[updated]

Gottfried, I must admit that I haven't put enough study into your matrix methods, having focussed all my attention on Andrew's solution.

Can you point me to a discussion that mentions how to compute the coefficients of S2, P, etc., so that I can be sure I'm looking at the right matrices? If there is a relationship between your matrices and the ones I'm using, I'd like to understand it.

Jay,

I give a short list of the used matrices. I'm preparing a code-pad for Pari/gp with which you may then experiment with these matrices. Perhaps tomorrow or friday.

In general
V(x)~ = [1,x,x^2,x^3,...]
A prefix d declares this as diagonalmatrix
So, for instance
V(2)~ = [1,2,4,8,16,...]

F = [0!,1!,2!,...], dF arranges this as diagonal dF^-1 contains the reciprocals (to construct the exponential-series, for instance)

P (binomialmatrix)
\( \hspace{24}
\begin{matrix} {rrrrr}
1 & . & . & . & . & . \\
1 & 1 & . & . & . & . \\
1 & 2 & 1 & . & . & . \\
1 & 3 & 3 & 1 & . & . \\
1 & 4 & 6 & 4 & 1 & . \\
1 & 5 & 10 & 10 & 5 & 1
\end{matrix} \)

Two properties of this matrix are important here.

1) Application of the binomial-rules, when postmutiplied by a formal powerseries
\( \hspace{24}
P * V(x) = V(x+1) \\
P^m * V(x) = V(x + m) \)

2) Derivatives
\( \hspace{24}
V(x)\sim * P = Y\sim \)
Y contains then scaled derivatives for \( \hspace{24} f(x)= sum(k=0,inf,x^k) \)



St2 tirling kind 2, version 1 (Abramowitsch&Stegun)
\( \hspace{24}
\begin{matrix} {rrrrr}
1 & . & . & . & . & . \\
1 & 1 & . & . & . & . \\
1 & 3 & 1 & . & . & . \\
1 & 7 & 6 & 1 & . & . \\
1 & 15 & 25 & 10 & 1 & . \\
1 & 31 & 90 & 65 & 15 & 1
\end{matrix} \)

St2 tirling kind 2, version 2, Wikipedia
\( \hspace{24}
\begin{matrix} {rrrrr}
1 & . & . & . & . & . \\
0 & 1 & . & . & . & . \\
0 & 1 & 1 & . & . & . \\
0 & 1 & 3 & 1 & . & . \\
0 & 1 & 7 & 6 & 1 & . \\
0 & 1 & 15 & 25 & 10 & 1
\end{matrix} \)

I use this version here




S2 : factorial scaled version of St2: dF^-1 * St2 * dF
\( \hspace{24}
\begin{matrix} {rrrrr}
1 & . & . & . & . & . \\
0 & 1 & . & . & . & . \\
0 & 1/2 & 1 & . & . & . \\
0 & 1/6 & 1 & 1 & . & . \\
0 & 1/24 & 7/12 & 3/2 & 1 & . \\
0 & 1/120 & 1/4 & 5/4 & 2 & 1
\end{matrix} \)

This version performs U-exponentiation for a powerseries
V(x)~ * S2 = V(exp(x)-1)~
(see Abramowitsch & Stegun)

Since Input and output are of the form of a powerseries, one can iterate to implement the "x-> (exp(x)-1)"" - iteration







St1 : Stirlingnumbers 1'st kind = inverse of St2
\( \hspace{24}
\begin{matrix} {rrrrr}
1 & . & . & . & . & . \\
0 & 1 & . & . & . & . \\
0 & -1 & 1 & . & . & . \\
0 & 2 & -3 & 1 & . & . \\
0 & -6 & 11 & -6 & 1 & . \\
0 & 24 & -50 & 35 & -10 & 1
\end{matrix} \)


S1 : factorial scaled St1, inverse of S2
\( \hspace{24}
\begin{matrix} {rrrrr}
1 & . & . & . & . & . \\
0 & 1 & . & . & . & . \\
0 & -1/2 & 1 & . & . & . \\
0 & 1/3 & -1 & 1 & . & . \\
0 & -1/4 & 11/12 & -3/2 & 1 & . \\
0 & 1/5 & -5/6 & 7/4 & -2 & 1
\end{matrix} \)

V(x)~ * S1 = V(log(1+x))~

Since this is the inverse of S2, it performs x->log(1+x), and since Input and Output are of the form of a powerseries, this can be iterated
(See also A&S)






B : my base-matrix for T-iteration
\( \hspace{24}
\begin{matrix} {rrrrr}
1 & 1 & 1 & 1 & 1 & 1 \\
0 & 1 & 2 & 3 & 4 & 5 \\
0 & 1/2 & 2 & 9/2 & 8 & 25/2 \\
0 & 1/6 & 4/3 & 9/2 & 32/3 & 125/6 \\
0 & 1/24 & 2/3 & 27/8 & 32/3 & 625/24 \\
0 & 1/120 & 4/15 & 81/40 & 128/15 & 625/24
\end{matrix} \)

This matrix can be understood in two ways:

As
\( \hspace{24}
B= matrix(r,c,c^r/r!) = dF^{-1} * matrix(c^r) = dF^{-1} * VZ \)
just the same way as in your code-snippet

Or
\( \hspace{24}
B = S2 * P\sim
\)



Application is
\( \hspace{24}
V(x)\sim * B = V(e^x)\sim
\)
and since input and output are of the form of powerseries, this can be iterated.

Note that since B = S2 * P~ we have, that
V(x)~ * S2 = V(e^x-1)~
and (see binomial-rules using P -transposed as above))
V(e^x-1)~ * P~ = V((e^x-1)+1)~ = V(e^x)~
we have
V(x)~ * S2 * P~ = V(e^x-1)~ *P~ = V(e^x)~
which is the same as
V(x)~ * B = V(e^x)~



To apply another base s, different from e, such that s<>e, this matrix must be premultiplied by powers of logarithms of s. I use lambda here for brevity. The decomposed description

Bs
\( \hspace{24}
\begin{matrix} {rrrrr}
0^0/0!*\lambda^0 & 1^0/0!*\lambda^0 & 2^0/0!*\lambda^0 & 3^0/0!*\lambda^0 & 4^0/0!*\lambda^0 & 5^0/0!*\lambda^0 \\
0^1/1!*\lambda^1 & 1^1/1!*\lambda^1 & 2^1/1!*\lambda^1 & 3^1/1!*\lambda^1 & 4^1/1!*\lambda^1 & 5^1/1!*\lambda^1 \\
0^2/2!*\lambda^2 & 1^2/2!*\lambda^2 & 2^2/2!*\lambda^2 & 3^2/2!*\lambda^2 & 4^2/2!*\lambda^2 & 5^2/2!*\lambda^2 \\
0^3/3!*\lambda^3 & 1^3/3!*\lambda^3 & 2^3/3!*\lambda^3 & 3^3/3!*\lambda^3 & 4^3/3!*\lambda^3 & 5^3/3!*\lambda^3 \\
0^4/4!*\lambda^4 & 1^4/4!*\lambda^4 & 2^4/4!*\lambda^4 & 3^4/4!*\lambda^4 & 4^4/4!*\lambda^4 & 5^4/4!*\lambda^4 \\
0^5/5!*\lambda^5 & 1^5/5!*\lambda^5 & 2^5/5!*\lambda^5 & 3^5/5!*\lambda^5 & 4^5/5!*\lambda^5 & 5^5/5!*\lambda^5
\end{matrix} \)

Application is
\( \hspace{24}
V(x)\sim * dV(\log(s))*B = V(s^x)\sim
\)
and since input and output are of the form of powerseries, this can be iterated.




Bs numerically
\( \hspace{24}
\begin{matrix} {rrrrr}
1 & 1 & 1 & 1 & 1 & 1 \\
0 & \lambda & 2*\lambda & 3*\lambda & 4*\lambda & 5*\lambda \\
0 & 1/2*\lambda^2 & 2*\lambda^2 & 9/2*\lambda^2 & 8*\lambda^2 & 25/2*\lambda^2 \\
0 & 1/6*\lambda^3 & 4/3*\lambda^3 & 9/2*\lambda^3 & 32/3*\lambda^3 & 125/6*\lambda^3 \\
0 & 1/24*\lambda^4 & 2/3*\lambda^4 & 27/8*\lambda^4 & 32/3*\lambda^4 & 625/24*\lambda^4 \\
0 & 1/120*\lambda^5 & 4/15*\lambda^5 & 81/40*\lambda^5 & 128/15*\lambda^5 & 625/24*\lambda^5
\end{matrix} \)


The terms, that you use can -if at all- mostly be found in the second-column of the matrices or result, since you need only the final scalar result, and not the additional powers, which occur in the next coumns.

So much for short

Gottfried