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 Has anyone solved iterations of z+Γ(z)? Leo.W Junior Fellow  Posts: 38 Threads: 4 Joined: Apr 2021 09/29/2021, 03:15 PM Just out of curiosity. Has anyone solved the complex continuous iterations of the function ? This will be very challenging because has no zeros, and the zero in the directed complex infinity is non-Botcher-constructable. My thought is using a function to map the fixed point at to 0 with a specific function, may be the inverse of . Ember Edison Fellow   Posts: 72 Threads: 7 Joined: May 2019 09/29/2021, 07:52 PM (This post was last modified: 03/21/2022, 04:47 PM by Ember Edison.) A wild guess without basis, maybe this function is just a special case of Fox H  tommy1729 Ultimate Fellow     Posts: 1,605 Threads: 363 Joined: Feb 2009 10/03/2021, 09:13 PM I see no reasons why iterations of this type should not be possible with the same techniques we used for tetration and related ones. I once started a thread here about the superfunction of exp(z)+z. Notice exp(z)+z also has no FINITE fixpoint just like your z + Γ(z). But exp(z) + z has a fixpoint at negative infinity. On the other hand , exp(z) + z has no poles. And exp(z) + z has a nice asymp to z. So the situation is different. One could also wonder about the generalization z +  Γ(z,v) for various v. In particular v = 1 for obvious reasons. On the other hand Γ(z,1) does have zero's. ---- the infinite composition/functional equation : f(s+1) = t(s) f(s) + Γ( t(s) f(s) ) - where t(s) is like with the gaussian method - should work , not ? Interesting question, thank you. regards tommy1729 tommy1729 Ultimate Fellow     Posts: 1,605 Threads: 363 Joined: Feb 2009 10/03/2021, 09:17 PM Again , nice question ! Leo.W Junior Fellow  Posts: 38 Threads: 4 Joined: Apr 2021 01/05/2022, 06:23 PM (This post was last modified: 01/05/2022, 06:28 PM by Leo.W.) (10/03/2021, 09:17 PM)tommy1729 Wrote: Again , nice question !OMG this forum has upgraded so hugelyyyyyyyy my final exams just ended i'm back and find me unable to catch up on anyway Happy new year, Tommy! This question is exactly not the same as original ones that do map the fixed point at infinity to local one and then solvable, you see, given that Gamma function has this asymptotic expansion at infinity with a_n coefficients: $$\Gamma(z)=\sqrt{\frac{2\pi}{z}}\bigg(\frac{z}{e}\bigg)^z\sum_{n\ge0}{\frac{a_n}{z^n}}$$ And recall back on the technique we used in the case \begin{align}f(z)=z+e^z\\g(z)=ln(z)\\g^{-1}(f(g(z)))=ze^z\\\text{The fixed point at }-\infty\text{ is mapped to }g^{-1}(-\infty)\end{align} And we notice that since Gamma function has zeroes at directed infinity $$e^{i\theta}\infty$$ where $$\frac{\pi}{2}\le|\theta|\le\pi$$ And then the function f(z)=z+Gamma(z) must have fixed point at these infinities* We may pick some function g to map that fixed point to local ones, 0 as preferred. First attempt, let $$g(z)=-log(z)e^{i \theta}, g^{-1}(z)=e^{-e^{-i \theta}z}$$ and hence the infinity in the direction at e^(i \theta) will be mapped to 0. However this leads to such asymptotic expansion at z=0: \begin{align}\text{Let sgnL(z)=}(-1)^{2+\left\lfloor \frac{\arg \left(-e^{i \theta } \log (z)\right)+\pi }{2 \pi }\right\rfloor } 2^{\frac{1}{2}-\left\lfloor \frac{\arg \left(-e^{i \theta } \log (z)\right)+\pi }{2 \pi }\right\rfloor }\sqrt{\pi }\\\times\left(e^{i \theta } \log (z) (-1)^{1+\left\lfloor \frac{\arg \left(-e^{i \theta } \log (z)\right)+\pi }{2 \pi }\right\rfloor }\right)^{\frac{1}{2}-e^{i \theta } \log (z)} \left(-\csc \left(\pi e^{i \theta } \log (z)\right)\right)^{\left\lfloor \frac{\arg \left(-e^{i \theta } \log (z)\right)+\pi }{2 \pi }\right\rfloor}\\g^{-1}(f(g(z)))=ze^{\frac{\text{sgnL(z)}e^{i\theta}}{log(z)}+P_\theta\small(\frac{1}{log(z)}\small)}\end{align} where P is another Taylor series Even if its series can be calculated at a specific direction, we won't be able to apply multiplier method: Assuming z is positive and theta is not \pi, then \begin{align}g^{-1}(f(g(z)))=ze^{-\sqrt{2\pi}e^{-log(z)P\small(\frac{1}{log(z)}\small)}}\\=z e^{-\sqrt{2 \pi } z^{-P\left(\frac{1}{\log (z)}\right)}}\end{align} It doesn't have a formal series at 0 at all Second attempt, let g(z)=1/z, this method will map all infinities* to 0: $$g^{-1}(f(g(z)))=z\frac{1}{1+e^{-\frac{1+log(z)}{z}}\sqrt{z^3}(1+O(z))}$$ It has no computable expansion at z=0 either. So we have to find a map g(z) that has these very properties: g has to map the fixed point at infinity to a local point, say, 0 $$g^{-1}(e^{i\theta}\infty)=0$$ After the mapping, we can find a series at z=0 only containing z (or abracadabra we have a much more complicated issue than ever!) That is, we can find the second derivative, third derivative at 0 and so on. (And they can't be all zeroes) $$F(z)=g^{-1}(f(g(z))),\forall{k\in\mathbb{N}_+},\exists{\mathcal{D}_z^k{F}(z)}|_{z=0}$$ Let's calculate the second derivative where p is some function of z $$g(0)=e^{i\theta}\infty,g^{-1}(e^{i\theta}\infty)=0,f'(e^{i\theta}\infty)=1,\mathcal{D}_z^2{F}(z)|_{z=0}=\lim_{z\to 0,p\to e^{i\theta}\infty}{f''(p)g'(z)}$$ And the third derivative $$\mathcal{D}_z^3{F}(z)|_{z=0}=\lim_{z\to 0,p\to e^{i\theta}\infty}{f'''(p)g'(z)^2}$$ Forth derivative $$\mathcal{D}_z^4{F}(z)|_{z=0}=\lim_{z\to 0,p\to e^{i\theta}\infty}{f^{(4)}(p)g'(z)^3-2f''(p)g'''(z)+2g'(z)g''(z)f'''(p)+3f''(p)g''(z)^2g'(z)^{-1}-3g'(z)f''(p)^2g''(z)}$$ You can see that all derivatives of F would contain some information from f, but not about its first derivative=1, and since all derivatives of f at very infinities* are all 0 (Easily proved), it is almost impossible to find such function g. Also it tests why the two maps above are no-use, the log map leads that all derivatives oscillates at z=0, thus has no specific derivatives, even if assumed z>0, whether the derivative diverges, or all derivatives are 0. The 1/z map makes all derivatives are 0. The reason why this is so hard is that the fundamental term, or leading term in the Gamma(z) expansion is $$\Gamma(z)\sim\Gamma_0(z)=\sqrt{\frac{2\pi}{z}}\bigg(\frac{z}{e}\bigg)^z$$ *maybe it's easier to deal with this function? How will you solve iterations of $$f^*(z)=z+\Gamma_0(z)=z+\sqrt{\frac{2\pi}{z}}\bigg(\frac{z}{e}\bigg)^z$$ without using fixed point 0 (since it is excluded in the z+gamma case)? *and maybe equally hard to use z=0 looooooooooooool? Again, happy new year!!! Leo JmsNxn Long Time Fellow    Posts: 729 Threads: 104 Joined: Dec 2010 01/07/2022, 08:15 AM (This post was last modified: 01/07/2022, 10:09 AM by JmsNxn.) Very cool, Leo! Very cool! I think for the most part you are correct; and I believe you answered your own question. We just have to consider $$z+\Gamma(z)$$ with the fixed point at $$-\infty$$; and additionally ignore the singularities; by treating the upper and lower half planes separately. Then it becomes a game of Schroder fixed point theory; just a special case analysis. For $$\Im(z) > 0$$ we are assured that $$z=\infty$$ is a fixed point; as $$\Im(z) \to \infty$$ this function diverges, and as $$\Re(z) \to \pm \infty$$ this function diverges. Then since $$\Im(z) > 0$$ is a simply connected domain, we can apply Schroder. Especially because $$\infty$$ is in the interior of $$\Im(z) > 0$$. Choose a linear fractional transformation such that $$h:\Im(z) > 0 \to |z|<1$$ such that $$h(\infty) = 0$$; and now we are doing schroder iterations about $$f(z) = h(z+\Gamma(z))$$; which should be doable about $$f\approx 0$$; or for large z arguments for $$z+\Gamma(z)$$. Great job! James « Next Oldest | Next Newest »

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