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RE: complex iteration (complex "height") - Gottfried - 03/22/2008
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 RE: complex iteration (complex "height") - Ivars - 03/22/2008
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. 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 RE: complex iteration (complex "height") - Gottfried - 03/22/2008
Ivars Wrote:simplest are Xnumbers for Excel. Where can I fetch Your GUI to make life even simpler?Hi 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 RE: complex iteration (complex "height") - bo198214 - 03/23/2008
Gottfried Wrote:with higher precision I recomputed the value and got in the 92..96 partial-sums the following approximations: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` RE: complex iteration (complex "height") - Gottfried - 03/23/2008
bo198214 Wrote:Gottfried Wrote:with higher precision I recomputed the value and got in the 92..96 partial-sums the following approximations:Haha but I just used low precision so here is my 20 digits result: 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: `´` [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: `´` RE: complex iteration (complex "height") - bo198214 - 03/23/2008
Gottfried Wrote:I just computed this with the function But I still dont get how you calculated the iterates. Let me try to reconstruct: You consider the function , where is the fixed point of , so . (Its good to stay a bit with useful conventions: is the letter for the base, is mostly used for the fixed point, is mostly used as the iteration variable.) If you now use the transformation , then . And then you compute the regular iteration powerseries of at its fixed point . And so get the regular iteration of as . 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 , 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 , you know is used for the inversion of , but what you meant was ! If you dont use tex you need another presentation of iteration. The non-superscripted has the completely different meaning of composition! RE: complex iteration (complex "height") - Gottfried - 03/23/2008
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 , you know is used for the inversion of , but what you meant was ! If you dont use tex you need another presentation of iteration. The non-superscripted 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 (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. RE: complex iteration (complex "height") - Ivars - 03/23/2008
bo198214 Wrote:Specialized to our case using the lower fixed point we get the base by and , so: Please just give still an explanation of what is this iteration -I am still not 100% sure I understand iteration symbols correctly. Thank you in advance, Ivars RE: complex iteration (complex "height") - Ivars - 03/23/2008
I tried to do analytically one case: lower fixed point we get the base by if then but : 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: And for all n, so which equals 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: Then I have to go through the whole procedure again. In this case, instead of we will have 0, and + before tower as: this seems to equal 1. So with negative height: Ivars RE: complex iteration (complex "height") - Ivars - 03/23/2008
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 or and if we have chosen complex or real then we can find such that : So every time E.g. lower fixed point we get the base by if then but : this seems to equal 1. So with Imaginary height: Few more interesting outcome : Which seems wrong.So where all others, then... Or actually a=e and base 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 , we may propose that: Or, if But it was good tex training... Ivars |