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 Parabolic Iteration, again andydude Long Time Fellow Posts: 509 Threads: 44 Joined: Aug 2007 05/07/2008, 12:57 AM Also, just to be clear, if you were wondering where the 6 came from in: $ (e^x-1)^{\circ t}(x) = \text{dxp}^{\circ t}(x) = \sum_{k=0}^{\infty} \frac{t^k x^{k+1}}{2^k} + \sum_{k=0}^{\infty} \frac{t^k x^{k+2}H^{(2)}_k}{-6\cdot2^k} + \cdots$ in this post, I thought I'd explain. According to the generating function above, this should all be the same, only with $f_2=1/2!$ and $f_3=1/3!$. So if we plug these into the formula above we find that: $\left(f_2 - \frac{f_3}{f_3}\right) = \left(\frac{1}{2} - \frac{1/6}{1/2}\right) = \left(\frac{3}{6} - \frac{2}{6}\right) = \frac{1}{6}$ so thats where the 6 comes from. The negative actually comes from the harmonic numbers, since: $\frac{\log(1-x)}{(1-x)} = -\sum_{n=0}^{\infty} H_n x^n$ and $\frac{\log(1-x)}{(1-x)^2} = -\sum_{n=0}^{\infty} H_n^{(2)} x^n$. $\frac{\log(1-x)}{(1-x)^k} = -\sum_{n=0}^{\infty} H_n^{(k)} x^n$. I think this might actually be the generating function for all $H_n^{(k)}$ but I can't find a reference for this. The only place I have found that goes into depth into this kind of generalized harmonic number is MathWorld, although apparently Conway and Guy go into detail, I'll have to read their book again to find out... Andrew Robbins andydude Long Time Fellow Posts: 509 Threads: 44 Joined: Aug 2007 05/14/2008, 02:56 AM (This post was last modified: 05/14/2008, 02:58 AM by andydude.) bo198214 Wrote:There are no denominators depending on the value of any $f_k$. You're right about $f_2=0$, and I was wrong. I was thinking of A. A. Bennett "The Iteration of Functions of one Variable" where he says (page 30) Bennett Wrote:The only other cases which can arise are the singular ones, and in these, $a_{11} = 0$.Which is refering to the Carleman matrix of a function with a fixed point at zero. So if $f(0) = 0$ and $f'(0) = 0$ then the Carleman matrix is not invertible (i.e. singular). This is completely different than what I remember reading, and it doesn't apply to $f''(0)$ at all. This means that the case $f''(0) = 0$ is a singularity of my generating functions, and not a singularity of parabolic iteration. Andrew Robbins « Next Oldest | Next Newest »

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