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 Parabolic Iteration, again andydude Long Time Fellow Posts: 509 Threads: 44 Joined: Aug 2007 04/30/2008, 10:03 AM (This post was last modified: 05/01/2008, 03:55 AM by andydude.) 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 « Next Oldest | Next Newest »

 Messages In This Thread Parabolic Iteration, again - by andydude - 04/30/2008, 10:03 AM RE: Parabolic Iteration, again - by Ivars - 04/30/2008, 10:19 AM RE: Parabolic Iteration, again - by andydude - 04/30/2008, 04:30 PM RE: Parabolic Iteration, again - by Ivars - 04/30/2008, 07:02 PM RE: Parabolic Iteration, again - by andydude - 04/30/2008, 07:34 PM RE: Parabolic Iteration, again - by andydude - 05/03/2008, 08:10 PM RE: Parabolic Iteration, again - by bo198214 - 05/04/2008, 07:14 AM RE: Parabolic Iteration, again - by andydude - 05/05/2008, 05:26 AM RE: Parabolic Iteration, again - by andydude - 05/07/2008, 12:57 AM RE: Parabolic Iteration, again - by andydude - 05/05/2008, 08:30 AM RE: Parabolic Iteration, again - by bo198214 - 05/05/2008, 05:33 PM RE: Parabolic Iteration, again - by andydude - 05/14/2008, 02:56 AM

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