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 Imaginary iterates of exponentiation jaydfox Long Time Fellow Posts: 440 Threads: 31 Joined: Aug 2007 09/12/2007, 07:13 AM (This post was last modified: 09/12/2007, 07:36 AM by jaydfox.) Okay, first of all, we need a name for performing the ith iteration of exponentiation. The 1st iteration is exponentiation, the -1st iteration is the logarithm. What's the ith iteration? Notice that the -ith iteration is the inverse of the ith iteration, but unlike the logarithm/exponentiation duality, imaginary iterations and their inverses are conjugates of each other for real numbers and real bases. For complex numbers and/or bases, we've picked one side or the other of the real line, so they might not necessarily be conjugates anymore (indeed, they probably aren't). While I'm on the subject of naming, I'm searching for a better verb for performing logarithms. To perform an exponentiation is to exponentiate. To perform a logarithm is to...??? To perform an exorcism (on an evil spirit) is to exorcize (the spirit). To give criticism (to a person or about a movie) is to criticize (the person or movie). I'm tempted at times to say that to perform a logarithm is to logarithe, but it doesn't quite roll off the tongue... Okay, back on subject. We need an ith iteration of exponentiation. We could start with a tetration solution and derive one, but I'd very much prefer to derive one from basic principles instead, and use it to derive and/or validate a tetration solution. First, some observations. The ith iteration of natural exponentiation of base e has two primary fixed points at $0.318131505... \pm 1.337235701...i$. If we momentarily call the ith iteration "imaginary exponentiation" and the -ith iteration "imaginary logarithm", the the upper fixed point is an attracting fixed point for imaginary exponentiation, and the lower fixed point is a repelling fixed point. For imaginary logarithms, the upper fixed point is repelling and the lower is attracting. I'm wondering if we could possibly experiment with exponentiation of complex bases to study this effect. I've experimented and found that complex bases (the few I played with) have both attracting and repelling fixed points of exponentiation, so that logarithms will tend to one and exponentiations will tend to the other, unless they happen to escape first (we could say that infinity is an attracting "fixed point" of exponentiation or logarithms in these cases...). I had been hoping to find a complex base with a fixed point at 1.337-0.318i, which is -i times the fixed point of natural exponentiation. However, I haven't found a way to find a base that has a specific fixed point. Even if I found such a base, I'm not sure what I'd do with it, other than play around with it and see if I get any inspiration... ~ Jay Daniel Fox andydude Long Time Fellow Posts: 509 Threads: 44 Joined: Aug 2007 01/16/2008, 05:47 AM $b=-0.670847 + 1.116757i$ is such a base. If you know a fixed point $a = b^a$ then $b = a^{1/a}$ quickfur Junior Fellow Posts: 22 Threads: 1 Joined: Feb 2008 02/21/2008, 08:51 PM jaydfox Wrote:[...] While I'm on the subject of naming, I'm searching for a better verb for performing logarithms. To perform an exponentiation is to exponentiate. To perform a logarithm is to...??? To perform an exorcism (on an evil spirit) is to exorcize (the spirit). To give criticism (to a person or about a movie) is to criticize (the person or movie). I'm tempted at times to say that to perform a logarithm is to logarithe, but it doesn't quite roll off the tongue... [...]What about ... logarithmatize?? Or, just "log (something)" for short. But "log" is already too overloaded. Hmm. Logarithate, maybe? or logarithmate. Bah, you're right, there is no good term for it. Ivars Long Time Fellow Posts: 366 Threads: 26 Joined: Oct 2007 02/26/2008, 08:12 AM (This post was last modified: 03/07/2008, 04:31 PM by Ivars.) jaydfox Wrote:First, some observations. The ith iteration of natural exponentiation of base e has two primary fixed points at $0.318131505... \pm 1.337235701...i$. If we momentarily call the ith iteration "imaginary exponentiation" and the -ith iteration "imaginary logarithm", the the upper fixed point is an attracting fixed point for imaginary exponentiation, and the lower fixed point is a repelling fixed point. For imaginary logarithms, the upper fixed point is repelling and the lower is attracting. Hi Jaydfox, This may be relevant: Omega constant =0.5671432904097838729999686622 is defined by 1=Omega*(e^Omega): So selfroot of Omega: Omega^(1/Omega) = e^ln(Omega^(1/Omega) = e^((1/Omega)*(ln(Omega))=e^((1/Omega)*(-Omega)) =e^(-1)=1/e=0,367879441 Infinite tetration of selfroot of Omega: h(Omega^(1/Omega))=h(1/e) = -W(-ln(1/e))/(ln(1/e))= -W(1)/-1=Omega=0,56714329=-ln(Omega), Square superroot of (Omega^1/Omega) : ssrt(Omega^(1/Omega) = ln(1/e)/W(ln(1/e))= -1/W(-1)= -1/(-0.318131505204764 +- 1.337235701430689*I) = 0.16837688705553+-0.707755195958823*I. W(-1) = 0.318131505204764 +- 1.337235701430689*I So why not call it Omegation? Ivars Ivars Long Time Fellow Posts: 366 Threads: 26 Joined: Oct 2007 03/20/2008, 07:47 AM (This post was last modified: 03/20/2008, 10:32 AM by Ivars.) Just to clarify for myself: i -th exponentiation of anything (e.g.) 2 would be: 2^i = 2*2*2*2*2............ i times = 0.76923890..+I*0.63896127...=e^(i*ln2) i-th root of this number would be 2. (1/i=-i ) Learning, thanks. ith exponentiation of a function is not i-th iteration of a function. See next post by administrator which puts it straight. Ivars bo198214 Administrator Posts: 1,389 Threads: 90 Joined: Aug 2007 03/20/2008, 10:24 AM (This post was last modified: 03/20/2008, 10:27 AM by bo198214.) Ivars, please dont post completely unrelated things! Before you reply you should at least have a base understanding what you are replying to! You can not just throw in your ideas because of similar sounding words like "imaginary exponentiation" that even are explained in the same paragraph to have a different meaning. Taking powers with complex exponents is a well explored area, there is nothing mysterious about it, its even tought at highschool. However taking the I-th iteration of a function $f^{\circ I}$ is a rather unknown area and barely connected to the I-th power. And *this* was discussed by Jay here in this thread for $f(z)=b^z$. « Next Oldest | Next Newest »

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