Rational sums of inverse powers of fixed points of e Gottfried Ultimate Fellow Posts: 887 Threads: 129 Joined: Aug 2007 11/21/2007, 12:52 PM (This post was last modified: 11/21/2007, 12:53 PM by Gottfried.) jaydfox Wrote:The first one looks about right, meaning it looks familiar at a glance. Ahh, just I got it myself. I got for the first few approximations (upscaled by factorials) Code:-1           -1            2            9           -6         -155         -232         3969        20870      -118779     -1655028so I have it now. But for the first n=96 entries I needed ~120 secs, using also only 96 fixpoints (Eulersumming them). Gottfried Gottfried Helms, Kassel jaydfox Long Time Fellow Posts: 440 Threads: 31 Joined: Aug 2007 11/21/2007, 06:20 PM Yes, they converge very rapidly. The fixed points increase by approximately 2*pi*I as we move away from the real line, so they form an arithmetic sequence to a first approximation. Therefore, they'll converge about as fast as the partial sums of the zeta function. So yes, we shouldn't need very many fixed points to calculate the coefficients of the rational numbers, though I haven't bothered to figure out if some minimum number of fixed points would suffice for all sums. My main interest in calculating 100,000 fixed points was to generate continued fractions for each sum, and try to use those to provide strong numerical evidence that the infinite sums indeed converge on rational numbers. ~ Jay Daniel Fox Gottfried Ultimate Fellow Posts: 887 Threads: 129 Joined: Aug 2007 11/22/2007, 06:25 PM (This post was last modified: 11/22/2007, 06:28 PM by Gottfried.) jaydfox Wrote:Anyway, in the meantime, I wanted to try to figure out how to calculate the numerators. It seems to me that they deserve their own Sloane sequence, considering how surprising it was (to me) to get rational sums. Hmm, this may not be of much help, at least I didn't find out a useful simplification. Anyway. If I use my formula for complex fixpoint for real base s, with the single independent parameter beta: u = alpha + beta*i where alpha=beta/sin(beta)*cos(beta) t = exp(u) = a + b*i s = exp(u/t) and in your example s=e, then these formulae can possibly be reversed to determine the allowed values for beta. The only thing I know already is, that the different allowed beta_k are roughly periodic at k*2*pi+eps_k with eps_k decreasing towards zero. So for any t of all t_k, omitting the index, t = exp( alpha + beta*i) = exp(alpha)*cos(beta) + exp(alpha)*sin(beta)*I and 1/t = exp(-alpha - beta*i) = exp(-alpha)/(cos(beta)+sin(beta)*i) = exp(-alpha)*(cos(beta)-sin(beta)*i) Now you consider the sum of 1/t and 1/conj(t) as one term v: v=1/t + conj(1/t) = 2*exp(-alpha)*cos(beta) v=2 * exp(-beta/sin(beta)*cos(beta))*cos(beta) and ask, whether the sum of all v_k add up to a rational... I don't know, how to proceed from here; the most difficult thing is surely the reverse determination of the possible beta's from the given base-parameter s=e. Hmmm ... an apple without vitamins... Gottfried Gottfried Helms, Kassel jaydfox Long Time Fellow Posts: 440 Threads: 31 Joined: Aug 2007 11/22/2007, 08:32 PM Hmm, I didn't follow your post, but it's early and I've been thinking about the singularity at 0 that I only just noticed. Actually, I haven't quite been following the maths in the Imaginary zeros of f(z)= z^(1/z) discussion. (Not because you didn't explain the maths well, but because I haven't devoted the time yet.) As for the numerators, I already found a Sloane series (A009306), which by changing every other sign and removing the first couple terms gives me the series I found. A separate Sloane series probably isn't required, merely the simple rules for how to get the numerators from the existing series. The calculation of the terms in the Sloane series is done with basic functional composition of power series with rational coefficients, so determining an arbitrary term should be straightforward, and from the matrices involved in the composition we could derive a (not necessarily simple) formula. ~ Jay Daniel Fox bo198214 Administrator Posts: 1,616 Threads: 102 Joined: Aug 2007 11/23/2007, 08:22 AM This is indeed an interesting connection. Now my quick 2 cents about it. First, we want the explicit function $f$ of which the sums of the powers of the inverted fixed points are the coefficients. We compare Jay's beginnin with index 1 in the first row with Sloane's beginning at index 1 in the second row: $\begin{pmatrix} -1 & -1 & 2 & 9 & -6 & -155 & -232 & 3969 & 20870 & -118779 & -1655028\\ 1 & 1 & -1 & -2 & 9 & 6 & -155 & 232 & 3969 & -20870 & -118779 & 1655028\end{pmatrix}$ Obviously we have to move the lower row to the left which is the same as dividing Sloane's function $S(z)=\ln\left(1+ze^z\right)$ by $z$. We get then $\begin{pmatrix} -1 & -1 & 2 & 9 & -6 & -155 & -232 & 3969 & 20870 & -118779 & -1655028\\ 1 & -1 & -2 & 9 & 6 & -155 & 232 & 3969 & -20870 & -118779 & 1655028\end{pmatrix}$ and see that the sign is swapped for each uneven power, which can be achieved by using $-z$ instead of $z$. So we get $f(z)=-\frac{1}{z} \ln\left(1-\frac{z}{e^z}\right)$ with $f_n = \sum_{k=0} \left(\frac{1}{\overline{c_k}^n}+\frac{1}{c_k^n}\right)$. $f(z)=\sum_{n=1}^\infty z^n\sum_{k=0}^\infty \left(\frac{1}{\overline{c_k}^n}+\frac{1}{c_k^n}\right) =\sum_{k=0}^\infty \sum_{n=1}^\infty \left(\frac{z}{\overline{c_k}}\right)^n+\left(\frac{z}{c_k}\right)^n =\sum_{k=0}^\infty \frac{\frac{z}{\overline{c_k}}}{1-\frac{z}{\overline{c_k}}}+ \frac{ \frac{z}{c_k} }{ 1-\frac{z}{c_k}}=\sum_{k=0}^\infty \frac{z}{\overline{c_k}-z}+\frac{z}{c_k-z}$ for $|z|<|c_k|$. If we transform this further via $e^{-zf(z)}=1-z/e^z$ we get $\prod_{k=0}^\infty e^{\frac{z^2}{z-\overline{c_k}}} e^{ \frac{z^2}{z-c_k}}=1-z/e^z$ to prove. Looks strange, perhaps I made an error somewhere. « Next Oldest | Next Newest »

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