Hmm, I solved another problem and killed two birds with one stone.
My coefficient for the z terms is the same as the coefficient for the z^2 term in the Sloane series, save for a discrepancy in the factorial denominator. And I have as the base of the logarithm at each singularity, the value of the point itself.
Well, if I divide the series through by z, then I get the coefficients to correspond correctly. This can be done safely, because the function equals 0 at z=0, so we can use l'Hopital's rule to find the value at z=0, and everywhere else it's well-defined. Anyway, that's the first bird.
Second bird: by dividing by z, I effectively make each logarithmic singularity have as its base the value at that point. To find the logarithm for a given base, first find the natural logarithm, then divide by the base. In this case, dividing a logarithmic singularity by z when -z is a fixed point, creates the effect of having used the logarithm with a base of exp(-z), which is still just -z in this case.
To put it all together, evaluating at the additive inverse of an arbitrary fixed point of natural exponentiation:
First, let \( z = w-c_k \)
\(
\begin{eqnarray}
F(z) & = & \frac{1}{z}\ln\left(1+z e^{z}\right) {\HUGE |}_{z = w-c_k}\\
& = & \frac{1}{z}\ln\left(1-\left(-z e^{z}\right)\right) {\HUGE |}_{z = w-c_k} \\
& = & \frac{1}{z}\ln\left(1-\frac{-z}{e^{-z}}\right) {\HUGE |}_{z = w-c_k} \\
& = & \frac{1}{w-c_k}\ln\left(1-\frac{c_k-w}{e^{c_k-w}}\right) {\HUGE |}_{w \to 0} \\
\end{eqnarray}
\)
At this point, we can use our knowledge of the behavior at the fixed point:
\(
\begin{eqnarray}
F(z) & = & \frac{1}{w-c_k}\ln\left(1-\frac{c_k-w}{e^{c_k-w}}\right) {\HUGE |}_{w \to 0} \\
& = & \frac{-1}{c_k}\ln\left(\frac{e^{c_k-w}-\left(c_k-w\right)}{e^{c_k-w}}\right) {\HUGE |}_{w \to 0} \\
& = & \frac{-1}{c_k}\ln\left(\frac{e^{c_k-w}-c_k+w}{e^{c_k}}\right) {\HUGE |}_{w \to 0} \\
& = & \frac{-1}{c_k}\ln\left(\frac{c_k-(c_k-1)w+\mathcal{O}(w^2)-c_k+w}{c_k}\right) {\HUGE |}_{w \to 0} \\
& = & -\frac{\ln\left(\frac{{1-(c_k-1)}w}{c_k}\right)}{\ln\left(c_k\right)} {\HUGE |}_{w \to 0} \\
& = & -\log_{c_k}\left(\frac{c_k}{c_k}w\right) {\HUGE |}_{w \to 0} \\
& = & -\log_{c_k}(w) {\HUGE |}_{w \to 0} \\
\end{eqnarray}
\)
And there you have it! There is an infinite set of fixed points, and hence an infinite number of such logarithms, which when summed get us back to this simple function. I'm wondering if some similar technique waits to be leveraged for the slog!
And, er, I haven't thoroughly checked my work (it's quite tedious), so if you find a mistake, please let me know.
My coefficient for the z terms is the same as the coefficient for the z^2 term in the Sloane series, save for a discrepancy in the factorial denominator. And I have as the base of the logarithm at each singularity, the value of the point itself.
Well, if I divide the series through by z, then I get the coefficients to correspond correctly. This can be done safely, because the function equals 0 at z=0, so we can use l'Hopital's rule to find the value at z=0, and everywhere else it's well-defined. Anyway, that's the first bird.
Second bird: by dividing by z, I effectively make each logarithmic singularity have as its base the value at that point. To find the logarithm for a given base, first find the natural logarithm, then divide by the base. In this case, dividing a logarithmic singularity by z when -z is a fixed point, creates the effect of having used the logarithm with a base of exp(-z), which is still just -z in this case.
To put it all together, evaluating at the additive inverse of an arbitrary fixed point of natural exponentiation:
First, let \( z = w-c_k \)
\(
\begin{eqnarray}
F(z) & = & \frac{1}{z}\ln\left(1+z e^{z}\right) {\HUGE |}_{z = w-c_k}\\
& = & \frac{1}{z}\ln\left(1-\left(-z e^{z}\right)\right) {\HUGE |}_{z = w-c_k} \\
& = & \frac{1}{z}\ln\left(1-\frac{-z}{e^{-z}}\right) {\HUGE |}_{z = w-c_k} \\
& = & \frac{1}{w-c_k}\ln\left(1-\frac{c_k-w}{e^{c_k-w}}\right) {\HUGE |}_{w \to 0} \\
\end{eqnarray}
\)
At this point, we can use our knowledge of the behavior at the fixed point:
\(
\begin{eqnarray}
F(z) & = & \frac{1}{w-c_k}\ln\left(1-\frac{c_k-w}{e^{c_k-w}}\right) {\HUGE |}_{w \to 0} \\
& = & \frac{-1}{c_k}\ln\left(\frac{e^{c_k-w}-\left(c_k-w\right)}{e^{c_k-w}}\right) {\HUGE |}_{w \to 0} \\
& = & \frac{-1}{c_k}\ln\left(\frac{e^{c_k-w}-c_k+w}{e^{c_k}}\right) {\HUGE |}_{w \to 0} \\
& = & \frac{-1}{c_k}\ln\left(\frac{c_k-(c_k-1)w+\mathcal{O}(w^2)-c_k+w}{c_k}\right) {\HUGE |}_{w \to 0} \\
& = & -\frac{\ln\left(\frac{{1-(c_k-1)}w}{c_k}\right)}{\ln\left(c_k\right)} {\HUGE |}_{w \to 0} \\
& = & -\log_{c_k}\left(\frac{c_k}{c_k}w\right) {\HUGE |}_{w \to 0} \\
& = & -\log_{c_k}(w) {\HUGE |}_{w \to 0} \\
\end{eqnarray}
\)
And there you have it! There is an infinite set of fixed points, and hence an infinite number of such logarithms, which when summed get us back to this simple function. I'm wondering if some similar technique waits to be leveraged for the slog!
And, er, I haven't thoroughly checked my work (it's quite tedious), so if you find a mistake, please let me know.
~ Jay Daniel Fox
