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 Rational sums of inverse powers of fixed points of e jaydfox Long Time Fellow Posts: 440 Threads: 31 Joined: Aug 2007 11/20/2007, 07:55 PM (This post was last modified: 11/21/2007, 01:24 AM by jaydfox.) This post doesn't have much if anything to do with tetration, but I discovered it while analyzing logarithms at the fixed points. I had mentioned this previously: http://math.eretrandre.org/tetrationforu...php?tid=59 The fixed points of e are irrational. However, the sums of the inverse n-th powers are rational. For example, the sum of the inverse third powers of all the fixed points of e is -1/2: $\sum_{k=0}^{\infty} \left(\frac{1}{\hspace{5}(c_k)^3\hspace{5}}+\frac{1}{\hspace{5} (\overline{c_k})^3 \hspace{5}}\right)=\frac{-1}{2}$ Note: Here, $c_k$ is the fixed point in the kth branch of the natural logarithm, i.e., $\ln(z)+2\pi k i$. If there were a single fixed point in the 0th branch of the natural logarithm, we could use $c_{-k}=\overline{c_k}$ and run our sum from negative infinity to positive. Alas, with two fixed points in the 0th branch, I thought it clearer to use complex conjugates, but perhaps we could still run the sum from negative to positive infinity if we had a way to account for both points in the 0th branch. Continuing with a couple more examples, the sum of the inverse fifth powers is 3/8. The sum of the inverse eighth powers is -29/630. Well, as it turns out, the denominators roughly increase as factorials. For example, the denominator of the 7th sum (i.e., the sum of the inverse 8th powers) is 630. This is 7!/8. The denominator of the 10th sum is 10!/3. So, here's a list of the numerators for the first 26 sums (inverse 2nd to 27th powers), when using the factorials as the denominators (starting with 1!, not 0!): Code:-1 -1 2 9 -6 -155 -232 3969 20870 -118779 -1655028 1610257 1436977220 522358005 -13332842416 -138189937791 1128293525646 29219838555781 -17274118159180 -5993074252801839 -38541972209299966 1179892974640047669 19460524014823618872 -187430579164971912575 -7757827302592891768426 -3443729183055238386555 I've currently got a program running to calculate the first 100,000 fixed points to 5120 bits of precision, to allow me to calculate the first few hundred sums. I'll then use continued fraction expansion, and look for the first "very large" number, which indicates that the previous fraction would most likely be exact. Using only the first 5,000 fixed points, I had typically seen situations where the subsequent convergent fraction would have a denominator with three to four times as many digits, e.g., 14 digits versus 50. I can't provide a specific example at the moment because SAGE is still crunching numbers (takes a good fraction of a second to calculate a fixed point to the desired precision, so 100,000 of them will take a good fraction of a day [Edit: It took 47200 seconds!]). With 100,000 digits, I'm hoping to make the difference something like 14 digits versus 60 or 70. 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. ~ Jay Daniel Fox jaydfox Long Time Fellow Posts: 440 Threads: 31 Joined: Aug 2007 11/20/2007, 08:11 PM (This post was last modified: 11/20/2007, 08:29 PM by jaydfox.) Well, surprise, surprise, there is already a sequence that gives the terms, though the signs don't always line up: http://www.research.att.com/~njas/sequences/A009306 0, 1, 1, -1, -2, 9, 6, -155, 232, 3969, -20870, -118779, 1655028, 1610257, -143697722, ... Okay, so now it's time to sit back and try to figure out why the sequences are related. According to the description, the Sloane series is the "Expansion of ln(1+exp(x).x)", or in Mathematica: Log[ 1+Exp[ x ]*x ] ~ Jay Daniel Fox Gottfried Ultimate Fellow Posts: 764 Threads: 118 Joined: Aug 2007 11/20/2007, 08:14 PM Hi Jay - interesting! Unfortunately, I cannot guess the sequence of results, since if you document the numerator only without a fixed rule for the complete denominator it is impossible to look for further rules. Could you provide the full rationals, maybe reduced by the factorial part of the denominator? I'd like to try to find some more rules. Gottfried Gottfried Helms, Kassel jaydfox Long Time Fellow Posts: 440 Threads: 31 Joined: Aug 2007 11/20/2007, 08:31 PM Sorry, those numerators were with respect to factorial denominators, e.g.: Code:-1/1 -1/2 2/6 9/24 -6/120 -155/720 -232/5040 ~ Jay Daniel Fox jaydfox Long Time Fellow Posts: 440 Threads: 31 Joined: Aug 2007 11/20/2007, 08:43 PM Just at a casual glance, I'm having trouble understanding the relationship. I mean, besides the obvious use of the natural logarithm. For example, my series is related to the power series of the sum of the logarithms at the fixed points, using as the base for each logarithm the associated fixed point itself. I would have been less surprised by something of the form ln(1-exp(x)/x), as but an example. In that case, we'd have singularities everywhere that exp(x)=x, which is of course at the fixed points. Changing things around, ln(1-exp(x)/x) is the same thing as ln(1+exp(-1/w)*w), where w=-1/x. But this still gets me no closer to the form used in the Sloane series. ~ Jay Daniel Fox jaydfox Long Time Fellow Posts: 440 Threads: 31 Joined: Aug 2007 11/20/2007, 09:31 PM Aha! $\ln\left(1-\frac{-z}{\hspace{5} e^{-z} \hspace{5}}\right)$ $\ln\left(1-\left(-ze^{z}\right)\right)$ $\ln\left(1+ze^{z}\right)$ So this is related to my previous findings! When z is the additive inverse of a fixed point, we get: $ \begin{eqnarray} 1+ze^{z}|_{\small z=-c_k} & = & -c_k e^{-c_k} \\ & = & 1+\frac{-c_k}{c_k} \\ & = & 1-1 \\ & = & 0 \end{eqnarray}$ Therefore, at the additive inverses of the fixed points, we'll get logarithmic singularities. I'm not sure about the bases though, and there are a few other oddities I need to figure out, but it's cool to see even this level of relationship. ~ Jay Daniel Fox jaydfox Long Time Fellow Posts: 440 Threads: 31 Joined: Aug 2007 11/21/2007, 01:15 AM (This post was last modified: 11/21/2007, 01:31 AM by jaydfox.) 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. ~ Jay Daniel Fox jaydfox Long Time Fellow Posts: 440 Threads: 31 Joined: Aug 2007 11/21/2007, 08:14 AM jaydfox Wrote:$ \begin{eqnarray} F(z) & = & \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{\left(1-(c_k-1)\right)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}$ Argh, I made a math error. Two, in fact. The first is that I subtracted -1 from 1 and got 0, when I should have gotten 2: $\left(1-(c_k-1)\right)w \ne c_k w$ Second, I incorrectly expanded an exponentiation. I didn't have paper and pencil handy to calculate, so I used Excel to get the answer by approximation. But I was subtracting the wrong set of numbers and got the wrong coefficient: $e^{c_k-w} \ne c_k-(c_k-1)w+\mathcal{O}(w^2)$ Okay, so, just to be clear, if w is a small number going to 0, then: $ \begin{eqnarray} e^{c_k-w} & = & e^{c_k}e^{-w} \\ & = & c_k\left(1-w+\frac{w^2}{2}+\dots\right) \\ & = & c_k-c_k w+\mathcal{O}(w^2) \end{eqnarray}$ Fortunately, this doesn't affect my initial derivation too badly: $ \begin{eqnarray} F(z) & = & \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 w+\mathcal{O}(w^2)-c_k+w}{c_k}\right) {\HUGE |}_{w \to 0} \\ & = & -\frac{\ln\left(\frac{\left(1-c_k\right)w}{c_k}\right)}{\ln\left(c_k\right)} {\HUGE |}_{w \to 0} \\ & = & -\log_{c_k}\left(\frac{1-c_k}{c_k}w\right) {\HUGE |}_{w \to 0} \\ & = & -\log_{c_k}(w)-\log_{c_k}\left(\frac{1-c_k}{c_k}\right) {\HUGE |}_{w \to 0} \\ \end{eqnarray}$ So I'm off by a constant factor, and I wasn't calculating the constant term anyway. So my initial result still holds. ~ Jay Daniel Fox Gottfried Ultimate Fellow Posts: 764 Threads: 118 Joined: Aug 2007 11/21/2007, 08:50 AM (This post was last modified: 11/21/2007, 08:58 AM by Gottfried.) jaydfox Wrote:This post doesn't have much if anything to do with tetration, but I discovered it while analyzing logarithms at the fixed points. I had mentioned this previously: http://math.eretrandre.org/tetrationforu...php?tid=59 The fixed points of e are irrational. However, the sums of the inverse n-th powers are rational.Hmm, how did you compute this actually? I tried it with Henryks fixed-point-formula, getting the reciprocals of the first few fixpoints of e Code:.         0.168376379087-0.707754188785*I // branch 0        0.0333484363677-0.122713337947*I // branch 1       0.0131593793714-0.0691856926303*I      0.00718943124116-0.0482569114994*I      0.00458333618210-0.0370546869791*I      0.00319972064661-0.0300725070538*IDoes this agree with your data? Gottfried Gottfried Helms, Kassel jaydfox Long Time Fellow Posts: 440 Threads: 31 Joined: Aug 2007 11/21/2007, 09:35 AM The first one looks about right, meaning it looks familiar at a glance. Hmm, going through Excel, the next few look correct, so at least we're on the same page there. Now, to get the first sum, we'll sum the inverse 2nd powers. To start, square each of the values you listed: Code:-0.472565386708 - 0.238338175183*I -0.013946445102 - 0.008184595884*I -0.004613490799 - 0.001820881553*I -0.002277041586 - 0.000693879494*I -0.001352042857 - 0.000339668175*I -0.000894117468 - 0.000192447243*I And of course, we'll add the conjugates as well. Essentially then, just double the real parts and sum them. The second sum is the sum of the inverse 3rd powers, so cube the values you listed and sum (including conjugates). ~ Jay Daniel Fox « Next Oldest | Next Newest »

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