In a recent post I gave a forumla for the second diagonal of iterated-dxp (e^x-1) and this line of research has led to some interesting discoveries. I have since generalized the approach to "interpolating" the diagonals of these series, and I found generating functions for the first three diagonals of parabolic iteration. I would like to present my findings and see if there is anything like this already out there...
It all started with noticing that the first diagonal is:
\( f^{\circ t}(x) = \sum_{k=0}^{\infty} t^k x^{k+1} f_2^k + \cdots \)
where the function being iterated is of the form \( f(x) = x + \sum_{k=2}^{\infty} f_k x^k \). This naturally lead to investigating the second diagonal, which I found is not too different than that of iterated-dxp:
\( - \sum_{k=0}^{\infty} t^k x^{k+2} f_2^k H_k^{(2)}
\left(f_2 - \frac{f_3}{f_2}\right) \)
but it involves a little more than just harmonic numbers. So I began looking at the third diagonal, and found some patterns, but I was only able to interpolate the coefficients up to a point, then I was stuck with a sequence of rational numbers I had no idea what to do with, then I eventually found A130894/A130895 which solved the problem I was having. Before I went to OEIS, I had found the coefficients of the third diagonal to be:
\( \sum_{k=0}^{\infty} t^k x^{k+3} f_2^k \left({k+1 \atop 2}\right)
\left(A^2 C_k + D\right) \)
where A and D are constants (described later), and \( C_k \) was the rational sequence [0, 1, 3/2, 71/36, 29/12, 638/225, 349/108, ...], which according to OEIS is equivalent to
\( C_n = \frac{1}{n} \sum_{k=1}^{n} \frac{H^{(2)}_k}{n+1-k} \)
where H is Conway and Guy's harmonic numbers (not the usual generalized harmonic numbers). Once I had the generating functions from OEIS, then I could being playing the game of generatingfunctionology. So I took this huge expression (D is "large" when written out) for the third diagonal, and played with derivatives and integrals until it was a recognizable function that generated the right coefficients. Maybe I'll post a more in-depth discussion of the techniques I used later on, but for now, I just want to show the results.
Going back to the first diagonal:
\( \sum_{k=0}^{\infty} t^k x^{k+1} f_2^k = \frac{x}{1- f_2 t x} \)
and according to OEIS the generating function of the 2nd degree harmonic numbers is \( -\frac{\log(1-x)}{(1-x)^2} \), which means the second diagonal is:
\( - \sum_{k=0}^{\infty} t^k x^{k+2} f_2^k H_k^{(2)}
A = x^2 \frac{\log(1 - f_2 t x)}{(1 - f_2 t x)^2} A \)
where \( A = \left(f_2 - \frac{f_3}{f_2}\right) \) and using the new generating functions for \( C_k \), we find the generating function for the third diagonal is:
\( \sum_{k=0}^{\infty} t^k x^{k+3} f_2^k \left({k+1 \atop 2}\right)
\left(A^2 C_k + D\right) = x^3 \frac{A^2(\log(1 - f_2 t x)^2 - \log(1 - f_2 t x)) + (f_2 t x) D}{(1 - f_2 t x)^3} \)
and finally, written out in full:
\(
\begin{tabular}{rl}
f^{\circ t}(x) &
= \frac{x}{z}
\\ & + \left(\frac{x}{z}\right)^2 \left(f_2 - \frac{f_3}{f_2}\right) \log(z)
\\ & + \left(\frac{x}{z}\right)^3 \left[
\left(f_2 - \frac{f_3}{f_2}\right)^2 (\log(z)^2 - \log(z)) +
(1-z)\left(\frac{f_2}{2} \left(f_2 - \frac{f_3}{f_2}\right) - \left(\frac{f_3}{f_2}\right)^2 + \frac{f_4}{f_2} \right)
\right]
\\ & + \cdots
\end{tabular}
\)
where \( z = 1 - f_2 t x \).
The most fascinating part, though, is that t only appears in z.
Andrew Robbins
It all started with noticing that the first diagonal is:
\( f^{\circ t}(x) = \sum_{k=0}^{\infty} t^k x^{k+1} f_2^k + \cdots \)
where the function being iterated is of the form \( f(x) = x + \sum_{k=2}^{\infty} f_k x^k \). This naturally lead to investigating the second diagonal, which I found is not too different than that of iterated-dxp:
\( - \sum_{k=0}^{\infty} t^k x^{k+2} f_2^k H_k^{(2)}
\left(f_2 - \frac{f_3}{f_2}\right) \)
but it involves a little more than just harmonic numbers. So I began looking at the third diagonal, and found some patterns, but I was only able to interpolate the coefficients up to a point, then I was stuck with a sequence of rational numbers I had no idea what to do with, then I eventually found A130894/A130895 which solved the problem I was having. Before I went to OEIS, I had found the coefficients of the third diagonal to be:
\( \sum_{k=0}^{\infty} t^k x^{k+3} f_2^k \left({k+1 \atop 2}\right)
\left(A^2 C_k + D\right) \)
where A and D are constants (described later), and \( C_k \) was the rational sequence [0, 1, 3/2, 71/36, 29/12, 638/225, 349/108, ...], which according to OEIS is equivalent to
\( C_n = \frac{1}{n} \sum_{k=1}^{n} \frac{H^{(2)}_k}{n+1-k} \)
where H is Conway and Guy's harmonic numbers (not the usual generalized harmonic numbers). Once I had the generating functions from OEIS, then I could being playing the game of generatingfunctionology. So I took this huge expression (D is "large" when written out) for the third diagonal, and played with derivatives and integrals until it was a recognizable function that generated the right coefficients. Maybe I'll post a more in-depth discussion of the techniques I used later on, but for now, I just want to show the results.
Going back to the first diagonal:
\( \sum_{k=0}^{\infty} t^k x^{k+1} f_2^k = \frac{x}{1- f_2 t x} \)
and according to OEIS the generating function of the 2nd degree harmonic numbers is \( -\frac{\log(1-x)}{(1-x)^2} \), which means the second diagonal is:
\( - \sum_{k=0}^{\infty} t^k x^{k+2} f_2^k H_k^{(2)}
A = x^2 \frac{\log(1 - f_2 t x)}{(1 - f_2 t x)^2} A \)
where \( A = \left(f_2 - \frac{f_3}{f_2}\right) \) and using the new generating functions for \( C_k \), we find the generating function for the third diagonal is:
\( \sum_{k=0}^{\infty} t^k x^{k+3} f_2^k \left({k+1 \atop 2}\right)
\left(A^2 C_k + D\right) = x^3 \frac{A^2(\log(1 - f_2 t x)^2 - \log(1 - f_2 t x)) + (f_2 t x) D}{(1 - f_2 t x)^3} \)
and finally, written out in full:
\(
\begin{tabular}{rl}
f^{\circ t}(x) &
= \frac{x}{z}
\\ & + \left(\frac{x}{z}\right)^2 \left(f_2 - \frac{f_3}{f_2}\right) \log(z)
\\ & + \left(\frac{x}{z}\right)^3 \left[
\left(f_2 - \frac{f_3}{f_2}\right)^2 (\log(z)^2 - \log(z)) +
(1-z)\left(\frac{f_2}{2} \left(f_2 - \frac{f_3}{f_2}\right) - \left(\frac{f_3}{f_2}\right)^2 + \frac{f_4}{f_2} \right)
\right]
\\ & + \cdots
\end{tabular}
\)
where \( z = 1 - f_2 t x \).
The most fascinating part, though, is that t only appears in z.
Andrew Robbins