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 complex iteration (complex "height") Gottfried Ultimate Fellow Posts: 758 Threads: 117 Joined: Aug 2007 03/22/2008, 08:23 PM (This post was last modified: 03/22/2008, 08:24 PM by Gottfried.) Ivars Wrote:My brain is already overstretched... I think I need some software that can calculate infinitely many logarithmsIvars - besides the answer of Henryk, which is quite right - why don't you use Pari/GP? It's free (and I 've made a free GUI (Windows) for it) search for "Pari" or "Paritty". Using Paritty you could also get my matrix-routines for a first start. Gottfried Gottfried Helms, Kassel Ivars Long Time Fellow Posts: 366 Threads: 26 Joined: Oct 2007 03/22/2008, 08:56 PM (This post was last modified: 03/22/2008, 09:29 PM by Ivars.) Gottfried Wrote:Ivars Wrote:My brain is already overstretched... I think I need some software that can calculate infinitely many logarithmsIvars - besides the answer of Henryk, which is quite right - why don't you use Pari/GP? It's free (and I 've made a free GUI (Windows) for it) search for "Pari" or "Paritty". Using Paritty you could also get my matrix-routines for a first start. Gottfried Thank You , that is practical. Downloaded version 2.3.3, Windows. Seems simpler to start with than Octave,Scilab, and more comlicated then gnuplot (those I have tried, I actually liked gnuplot best as it gives pictures of functions of exponents (like Lambert( x^(1/x) easily). The simplest are Xnumbers for Excel. Where can I fetch Your GUI to make life even simpler? I already Love PARI; intuitively easy, great tutorial, gives power series if You type log(1+x) or exp(x). Just this Lambert function-but if its not there, we have to maneuvre around it-maybe its worth it? Ivars Gottfried Ultimate Fellow Posts: 758 Threads: 117 Joined: Aug 2007 03/22/2008, 10:36 PM Ivars Wrote:simplest are Xnumbers for Excel. Where can I fetch Your GUI to make life even simpler? IvarsHi Ivars - just go to Pari-TTY there are also some introductory screenshots. If you need some basic matrix-functions email me, although I've my libraries not developed for distribution and so its not really much documented and little structured- just as it grows with a bad-organizing-mind as I am ... Ah - I remember, I already prepared a short basic introduction in a communication with Jay Fox, but don't remember, which thread... Gottfried Gottfried Helms, Kassel bo198214 Administrator Posts: 1,389 Threads: 90 Joined: Aug 2007 03/23/2008, 01:28 AM (This post was last modified: 03/23/2008, 01:34 AM by bo198214.) Gottfried Wrote:with higher precision I recomputed the value and got in the 92..96 partial-sums the following approximations: Code:´   ...   1.210309025559961+0.5058275713618201*I ---------------------------------------------   1.210309011      + .5058275611          *I  <--- yoursSo this differs from the 7'th digit;Haha but I just used low precision so here is my 20 digits result: Code:1.2103090255599614766+.50582757136182013605*I update: once for later time, 50 Digits! You see the last some digits are always unreliable. Code:1.2103090255599614779588104735397176784037341102467+.50582757136182013700589565226517951794360523253345*I Gottfried Ultimate Fellow Posts: 758 Threads: 117 Joined: Aug 2007 03/23/2008, 06:43 AM (This post was last modified: 03/23/2008, 07:56 AM by Gottfried.) bo198214 Wrote:Gottfried Wrote:with higher precision I recomputed the value and got in the 92..96 partial-sums the following approximations: Code:´   ...   1.210309025559961+0.5058275713618201*I ---------------------------------------------   1.210309011      + .5058275611          *I  <--- yoursSo this differs from the 7'th digit;Haha but I just used low precision so here is my 20 digits result: Code:1.2103090255599614766+.50582757136182013605*I update: once for later time, 50 Digits! You see the last some digits are always unreliable. Code:1.2103090255599614779588104735397176784037341102467+.50582757136182013700589565226517951794360523253345*I The last few partial sums computed with internal precision of 400, displayed precision 50, Euler-sum of order 2.538, where order=1 means direct summation without transformation (the last few of partial sums of 128 terms now) Code:´    1.21030902555996147 05286038734660188979105600443644+0.50582757136182012 502296304591137652670054111308587*I    1.21030902555996147 44685314155510971433670516046106+0.50582757136182012 835671840643903386779427797897960*I    1.21030902555996147 66583812490070529178600987551324+0.50582757136182013 098572869953873370073216724647955*I    1.21030902555996147 77865734208122443540107642898914+0.50582757136182013 294937569862301423395913769512951*I    1.21030902555996147 82987507553564562088347731401270+0.50582757136182013 435542605993548055302116773994320*I    1.21030902555996147 84733868997136185883982264653880+0.50582757136182013 532695350500506166087181069651300*I    1.21030902555996147 84778386937626628200500840930051+0.50582757136182013 597695153820676613393345071380896*I    ------------------------------------------------------------------------------------------------------------------------    1.21030902555996147 79588104735397176784037341102467+ .50582757136182013 700589565226517951794360523253345*I <--- your valueThe non-vanishing increase means, I'd use more precise Euler-order of a bit higher order. I'll see, whether I can find a better transformation. [update] A better transformation uses a Stirling kind 2 transformation first, which is also regular. Then I don't need high Euler-orders, and get for example the last few partial-sums (128 terms): Code:´   1.210309025559961477958810473539717649 4269005980356+0.5058275713618201370058956522651794 9881594263547436*I   1.210309025559961477958810473539717660 1745711120037+0.5058275713618201370058956522651795 1511709385013158*I   1.210309025559961477958810473539717668 9726354895139+0.5058275713618201370058956522651795 2061256599110112*I   1.210309025559961477958810473539717674 4384709020618+0.5058275713618201370058956522651795 2132025558347280*I   1.210309025559961477958810473539717677 2354562276144+0.5058275713618201370058956522651795 2045256215306036*I   1.210309025559961477958810473539717678 3946530537033+0.5058275713618201370058956522651795 1941438348397801*I   1.210309025559961477958810473539717678 7249247611291+0.5058275713618201370058956522651795 1865625158284700*I -------------------------------------------------------------   1.210309025559961477958810473539717678 4037341102467+ .5058275713618201370058956522651795 1794360523253345*I <--- your valueAgain, the change of sign of change in differences in partial sums shows, that the summation is still not optimal. Gottfried Helms, Kassel bo198214 Administrator Posts: 1,389 Threads: 90 Joined: Aug 2007 03/23/2008, 08:47 AM (This post was last modified: 03/23/2008, 08:56 AM by bo198214.) Gottfried Wrote:I just computed this with the function U_t (x) = t^x - 1 T_b(x) = b^x U_t°h(x) and T_b°h(x) as their iterates of general height h and the shift T_b°h(x) = (U_t°h(x/t-1) +1)*t and the height-iteration by applying powers to the diagonal of the diagonalized operator for U_t(x) with 96 terms, using b=sqrt(2),t=2,x=1 But I still dont get how you calculated the iterates. Let me try to reconstruct: You consider the function $U(x)=a^x-1$, where $a$ is the fixed point of $b^x$, so $b=a^{1/a}$. (Its good to stay a bit with useful conventions: $b$ is the letter for the base, $a$ is mostly used for the fixed point, $t$ is mostly used as the iteration variable.) If you now use the transformation $\tau(x)=x/a-1$, $\tau^{-1}(x)=a(x+1)$ then $\tau^{-1}(U(\tau(x)) = (a^{x/a-1}-1+1)t=a^{x/a}=b^x$. And then you compute the regular iteration powerseries of $U$ at its fixed point $0$. And so get the regular iteration of $b^x$ as $\exp_b^{\circ t}=\tau^{-1}\circ U^{\circ t}\circ \tau$. It is regular as we have seen that additional (to translating the fixed point to 0) multiplicative constants in the transformation do not change the result. My formula also computes the regular iteration at the lower fixed point, but in an iterative way, not by power series. So at a theoretical level both results must be equal! The differences are of purely numeric nature. Euler summation should only be needed for $a=e$, $U(x)=e^x-1$ otherwise it should converge/is analytic at the fixed point. (Which does not mean that you shouldnt use it for acceleration of convergence ) edit: Ah now I see, what puzzled me was U_t°h(x) and T_b°h(x). Gottfried, this is very prone to misinterpretation, my first reading was $U_t\circ h (x)$, you know $h(x)$ is used for the inversion of $x^{1/x}$, but what you meant was $U_t^{\circ h}(x)$! If you dont use tex you need another presentation of iteration. The non-superscripted $\circ$ has the completely different meaning of composition! Gottfried Ultimate Fellow Posts: 758 Threads: 117 Joined: Aug 2007 03/23/2008, 10:16 AM (This post was last modified: 03/23/2008, 10:22 AM by Gottfried.) bo198214 Wrote:edit: Ah now I see, what puzzled me was U_t°h(x) and T_b°h(x). Gottfried, this is very prone to misinterpretation, my first reading was $U_t\circ h (x)$, you know $h(x)$ is used for the inversion of $x^{1/x}$, but what you meant was $U_t^{\circ h}(x)$! If you dont use tex you need another presentation of iteration. The non-superscripted $\circ$ has the completely different meaning of composition! Yes, I'm getting a bit sloppy with this, sorry. In Tex it seems impossible to get this tiny circle, well I got it using $U^{oh}$ (alas - I had it already - shame... ) I'd prefer a more serial notation for text, like {b,x}^^h and even better x {[4],b} h , because this would then be concatenable: x {[4],b} h1 {[4],b} h2 = x {[4],b} (h1+h2) and had adapted our current notation a [4] b, if... if the start-parameter (x in my notation) would be existent and the base parameter would not occupy the place, where only a concatenation is possible. So all easy ascii-available notations that I can think of are somehow exotic ... If we would have x [4,b] h instead of b[4]h, or - well again I used h for height, so x [4,b]t instead of b[4]t, which seems to evolve as a standard currently, I would immediately adapt this notation --- Numerics: What surpries me a bit is, that acceleration by Euler-summation in this case does not yield much, even if the pre-transformation by Stirling matrix makes things much better. Many times I sat down to get a better insight in the characteristics of these powerseries and a good adaption by summation-methods, but this was not yet a fundamental enhancement. There is still something waiting to be discovered/characterized here. Gottfried Helms, Kassel Ivars Long Time Fellow Posts: 366 Threads: 26 Joined: Oct 2007 03/23/2008, 04:06 PM bo198214 Wrote:Specialized to our case using the lower fixed point $a$ we get the base by $b=a^{1/a}$ and $f'(a)=(b^x)'(a)=\ln(b)b^a=\ln(b)a=\ln(b^a)=\ln(a)$, so: $b[4]z=\exp_b^{\circ z}(1)=\lim_{n\to\infty} \log_b^{\circ n}(a(1-\ln(a)^z) + \ln(a)^z \exp_b^{\circ n}(1))$ Please just give still an explanation of what is this iteration -I am still not 100% sure I understand iteration symbols correctly. $\exp_b^{\circ n}(1))$ Thank you in advance, Ivars Ivars Long Time Fellow Posts: 366 Threads: 26 Joined: Oct 2007 03/23/2008, 07:27 PM (This post was last modified: 03/23/2008, 09:25 PM by Ivars.) I tried to do analytically one case: lower fixed point $a=\Omega=0.567143..$ we get the base by $b={1/e}=\Omega^{1/\Omega}$ if $z={1/\Omega}$ then ${1/e}[4]{1/\Omega}=\exp_{1/e}^{\circ 1/\Omega}(1)=\lim_{n\to\infty} \log_{1/e}^{\circ n}(\Omega*(1-\ln(\Omega)^{1/\Omega}) + \ln(\Omega)^{1/\Omega }\exp_{1/e}^{\circ n}(1))$ but : $\ln(\Omega)=-\Omega$ $\ln(\Omega)^{1/\Omega }=({1/\Omega })*\ln(\Omega)={-\Omega/\Omega }=-1$ ${1/e}[4]{1/\Omega}=\lim_{n\to\infty} \log_{1/e}^{\circ n}(\Omega*(1-(-1)) -1*\exp_{1/e}^{\circ n}(1))$ ${1/e}[4]{1/\Omega}=\lim_{n\to\infty} \log_{1/e}^{\circ n}(\Omega*2 -1*\exp_{1/e}^{\circ n}(1))$ At this point I do not know what to do with limits -how do You apply n to iterating logarithms, and at the same time to iteration of exponentiation of 1/e inside it? If we could move limit inside, than: $\lim_{n\to\infty} \exp_{1/e}^{\circ n}(1))=h({1/e)}=\Omega$ And $2*\Omega-\Omega=\Omega$ $\log_{1/e}(\Omega)= \Omega$ for all n, so ${1/e}[4]{1/\Omega}=\Omega$ which equals ${\Omega^{1/\Omega}[4]{1/\Omega}=\Omega$ I know it is not complex but real... Must be a mistake with the limits, but looks nice still... If someone could explain me on this example how I should have proceeded after the place where I got to limit taking inside iterated logarithm, I promise never to make the same mistake. If we take the same but: $z={-1/\Omega}$ Then I have to go through the whole procedure again. In this case, instead of $2*\Omega$ we will have 0, and + before tower as: $\ln(\Omega)^{-1/\Omega }=(-{1/\Omega })*\ln(\Omega)={-\Omega/-\Omega }=1$ ${1/e}[4]{-1/\Omega}=\lim_{n\to\infty} \log_{1/e}^{\circ n}(1*\exp_{1/e}^{\circ n}(1))$ this seems to equal 1. So with negative height: ${1/e}[4]{-1/\Omega}= 1$ ${\Omega^{1/\Omega}[4]{-1/\Omega}=1$ Ivars Ivars Long Time Fellow Posts: 366 Threads: 26 Joined: Oct 2007 03/23/2008, 10:08 PM (This post was last modified: 03/24/2008, 04:34 PM by Ivars.) Some complex value tetration: If I have understood right (if not, tell me please) , basic idea to be able to have some analytics relations is to make $ln(a)^z = 1$ or $z*ln(a) =1$ and if we have chosen complex or real $z$ then we can find $a$ such that : $ln(a) =1/z , b= a^{(1/a)}$ So every time $a^{(1/a)} [4] {1/ln(a)} = 1$ E.g. lower fixed point $a=e^-I=cos(1) -I*sin(1)=0.5403023-I*0.8414098$ we get the base by $b=(e^-I)^{(e^I)}$ if $z=I$ then $(e^{-I})^{(e^I)}[4]I =\exp_b^{\circ I}(1)=\lim_{n\to\infty} \log_b^{\circ n}((e^{-I})*(1-\ln(e^{-I})^I) + \ln(e^{-I})^I \exp_b^{\circ n}(1))$ but : $\ln(e^{-I})=-I$ $\ln(e^{-I})^I=I*\ln(e^{-I})=I*-I=1$ $(e^{-I})^{(e^I)}[4]I =\exp_b^{\circ I}(1)=\lim_{n\to\infty} \log_b^{\circ n}((e^{-I})*(1-1) + 1* \exp_b^{\circ n}(1))$ $(e^{-I})^{(e^I)}[4]I =\exp_b^{\circ I}(1)=\lim_{n\to\infty} \log_b^{\circ n}(\exp_b^{\circ n}(1))$ this seems to equal 1. So with Imaginary height: $(e^{-I})^{(e^I)}[4] I= 1$ Few more interesting outcome : $I^{1/I} = e^{\pi/2}[4] {-I*2/\pi}=1$ $e^{1/e}[4] 1 =1$ Which seems wrong.So where all others, then... Or actually a=e and base $b=e^{1/e}$ seems to be rather unique point in selfroots as it is the only one for a>1 which has only 1 value, so h(e^(1/e))=e and that is the only value ( all other points for a>1 has 2 values for h(a^(1/a))), because 2 different a on different sides of a=e have the same selfroot values). So if we speculate a little to continue with 2 values as a demand for $h(a^{1/a}) a>=1$ , we may propose that: $e^{1/e}[4] 1 =1$ $e^{1/e}[4] 1 =e^{1/e}$ $h( e^{1/e}) =e$ $h( e^{1/e}) =1$ $h(1)=1$ $h(1)=\infty$ Or, if $ln(e) = 1+I*2\pi*k$ $e^{1/e}[4] 1 =e^{1/e}$ $e^{1/e}[4] {1/(1+2\pi*I*k)} =1$ $1^1[4] \infty =1$ $1^1[4] {1/(2\pi*I*k)} =1$ But it was good tex training... Ivars « Next Oldest | Next Newest »

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