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+--- Thread: Superlog determinant factored! (/showthread.php?tid=381)
Superlog determinant factored! - andydude - 11/07/2009
I will write the proof as a story, so as to make it more interesting, but first, some definitions.
Let \( \mathbf{A}[f]^{(z_0)}_n \) be the nth truncated Abel matrix of f expanded at \( z_0 \), and let \( d_{nb}(z_0) = \det(\mathbf{A}[\exp_b]^{(z_0)}_n) \). As I wrote about in my 2005 paper, each nth solution is unique iff \( d_{nb}(z_0) \) is nonzero. These solutions can be used as the coefficients of the superlog as \( \text{slog}_b(z) = y_0 + \sum_{k=1} a_k (z - z_0)^k \).
Let \( f(b) \overset{n}{=} g(b) \) indicate that \( f^{(k)}(1) = g^{(k)}(1) \) for all \( 0 \le k \le n \). In some textbooks, this is referred to as "f and g have contact of order n at 1". In order to keep the notation small, every occurance of this notation will refer to f and g as functions of b, not any other variable, and the expansion point will always be 1.
Theorem.
\( \begin{tabular}{|c|}
\hline{}
\ \\
d_{nb}(z_0) \overset{n}{=} \frac{(F(b) + 1 - z_0)^n}{G(b)} \\
\ \\
\hline\end{tabular} \)
Proof.
So one day, I was playing around with the determinants \( d_{nb}(z_0) \), and began differentiating them. I was bored, and stumbled upon the quotient:
\( \frac{d_{nb}'(z_0)}{d_{nb}(z_0)} \)
and I noticed that the series expansions of this expression at b=1 increased with every approximation number n, so I then looked at the quotient:
\(
\frac{d_{nb}'(z_0)}{n d_{nb}(z_0)}
\overset{n}{=}
\sum_{k=0}^{\infty} \frac{(b-1)^k}{(z_0-1)^{k+1}}
\sum_{j=0}^{k-1} c_{jk} z_0^j
\)
and to my surprise, the series expansions did not change at all with n, except for the coefficients of \( (b - 1)^{n+1} \), which are beyond the claim of \( f\overset{n}{=}g \). So after looking at the quotient a bit more, I saw that it was the logarithmic derivative of \( d_{nb}(z_0) \), so I integrated both sides.
\(
\frac{1}{n}\ln(d_{nb}(z_0))
\overset{n}{=}
\ln(z_0 - 1) +
\sum_{k=0}^{\infty} \frac{(b-1)^k}{(z_0-1)^{k}}
\sum_{j=0}^{k-1} c_{jk} z_0^j
+ \ln(-1)
- g(b)/n
\)
The extra terms are to make up for the constant of integration, and for the integral of \( \frac{1}{z_0 - 1} \) in the big summation on the right. So after some simplification, I realized that
\(
\sum_{k=0}^{\infty} \frac{(b-1)^k}{(z_0-1)^{k}}
\sum_{j=0}^{k-1} c_{jk} z_0^j = f(b, z_0) = \ln\left(\frac{F(b)}{1 - z_0} + 1\right)
\)
or in other words, the \( f(b, z_0) \) can be factored into a function of b and a function of \( z_0 \). Putting it all together:
\(
\begin{tabular}{rl}
ln(d_{nb}(z_0))
& \overset{n}{=} n\ln(1 - z_0) + n f(b, z_0) - g(b) \\
d_{nb}(z_0)
& \overset{n}{=} \exp(n\ln(1 - z_0) + n f(b, z_0) - g(b)) \\
& \overset{n}{=} (1 - z_0)^n \frac{\exp(n f(b, z_0))}{\exp(g(b))} \\
& \overset{n}{=} (1 - z_0)^n \frac{(\frac{F(b)}{1 - z_0} + 1)^n}{\exp(g(b))} \\
& \overset{n}{=} \frac{\left[(1 - z_0)(\frac{F(b)}{1 - z_0} + 1)\right]^n}{\exp(g(b))} \\
& \overset{n}{=} \frac{\left(F(b) + (1 - z_0))\right)^n}{\exp(g(b))} \\
& \overset{n}{=} \frac{\left(F(b) + (1 - z_0))\right)^n}{G(b)} \\
\end{tabular}
\)
where
\(
\begin{tabular}{rl}
F(b+1)
&= 0 + b + b^2 + \frac{3}{2} b^3 + \frac{7}{3} b^4 + 4 b^5 + \frac{283}{40} b^6 + \frac{4681}{360} b^7 + \cdots \\
G(b+1)
&= 1 + b + \frac{5}{2} b^2 + \frac{35}{6} b^3 + \frac{173}{12} b^4 + \frac{2077}{60} b^5 + \frac{6079}{72} b^6 + \cdots
\end{tabular} \)
Although it is not as exciting as I hoped at first, it is still interesting. The formula can be verified because it asserts something that is true for all n, and so if it is true for \( n < 20 \) it will probably be true for more n. I have tried to use this formula to find information about base-e, but these functions don't converge out there. I'm thinking it has something to do with the fact that base-1 is divergent. I'm hoping that this kind of formula could help with the higher bases, but so far no such luck. I have also tried to find a similar formula for the same matrix with the first column replaced with (1, 0, 0, 0, ...) which represents the numerator of Cramer's rule for the first coefficient/derivative of superlog.
One final remark. Notice that n has been isolated. This makes it obvious that
\( \lim_{n\to\infty} d_{nb}(z_0) \)
diverges to infinity if \( F(b) > z_0 \) and converges to 0 if \( F(b) < z_0 \), but is this only for bases near 1?
Andrew Robbins
RE: Superlog determinant factored! - andydude - 11/09/2009
OMG, I just realized:
\( F(b) = {}^{\infty}b - 1 \).
RE: Superlog determinant factored! - bo198214 - 11/10/2009
(11/09/2009, 09:39 PM)andydude Wrote: OMG, I just realized:
\( F(b) = {}^{\infty}b - 1 \).
ui!
RE: Superlog determinant factored! - andydude - 11/10/2009
So I guess I should rephrase the statement as:
\( d_{nb}(z_0) \overset{n}{=} \frac{({}^{\infty}b - z_0)^n}{G(b)} \)
which may provide a useful shortcut to showing whether or not intuitive/natural iteration even works. Suppose we make a Cramer's rule matrix \( \mathbf{E} \) with all the same entries of \( \mathbf{A}[\exp_b] \), except the first column is replaced with (1, 0, 0, 0, ...) so we can solve for \( \text{slog}_b'(z_0) \). Let \( c_{nb}(z_0) = \det(E) \), then if the limit exists, \( \text{slog}_b'(z_0) = \lim_{n\to\infty} \frac{c_{nb}(z_0)}{d_{nb}(z_0)} \) which is a rational polynomial mess.
If we take into account the above formula, then
\( \lim_{n\to\infty} \frac{c_{nb}(z_0)}{
({}^{\infty}b - z_0)^n} = \frac{\text{slog}_b'(z_0)}{G(b)} \)
which may provide a clear path to answering the question of convergence.