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from formulas to pari implementations - Xorter - 05/27/2020
Hey! Almost everybody knows Schröder's equation: Psi o y(x) = s × Psi(x) where s = (d/dx y) o y^o plus/minus infinity where y^o plus/minus infinity is the fixpoint of the function y. Almost everybody knows some form of the Abel equation: Y o (1+x) o Y^o-1 = y(x) where Y is the superfunction of y. The connection of the two equation: Y^o-1 = log Psi(x) / log s Y = Psi^o-1 o s^x y(x) = Psi^o-1 o (s × Psi) y and s are usually given, we look for its iterated function or superfunction Y. Some of you knows the Carleman matrix representations of analytic functions, I will notate it with [ f(x) ] or just [ f ]. It can be usefull, when you look for a concrate functional power/root of a function, beacuse: [ y^o n ] = [ y ]^n for every n, it is derived from: [f o g] = [f]×[g] The bigger is the matrix, the more precise is the represantation. My pari/gp implementation of this conversions: Code: `D(f,n)=for(k=1,n,f=deriv(f));f;` How to iterate? I also show you my tetration implementation: Code: `h=10;` It is much less precise as Sheldon's and others' fatou.gp programme; it works without using Schröder and Abel equations but using Carleman matrices and diagonalization. It is not enough for me, lets search a better method. What is about determining the Psi function? We said that: Y = Psi^o-1 o s^x So to determine inverse of Psi(x) seems promising. Let me notate P = [ Psi(x) ]. The formula says: P^-1 × (s × P) = [ y(x) ], here are the first product which is matrix-matrix and the second product which is scalar-matrix. P^-1 × s^x' × P = [ Y(x') ] It looks like a diagonalized form, so what if I identify P^-1 as mateigen([ y(x) ])? I have already tested it for y = 2x-1 for which it succeded on a small interval but no success with determining the tetration for base 2. How would you approximate function Psi? And sorry not for using latex. I forget how to. :( RE: from formulas to pari implementations - Gottfried - 05/29/2020
(05/27/2020, 11:22 PM)Xorter Wrote: Hey! Hmm, I don't really get it what you are after. For one: perhaps my very introductiory essay ContinuousfunctionalIteration is enough to explain the method which leads to the Carleman-method and Schroeder-function (where I did not know in my first writing that already known/used terms and concepts and thus had rediscovered them on my own), and for second, this is all done for examples with a not too large set of procedures (mostly matrix-functions) in Pari/GP. RE: from formulas to pari implementations - Xorter - 06/01/2020
(05/29/2020, 02:35 PM)Gottfried Wrote: Hmm, I don't really get it what you are after. I read your paper, it is great, understandable. Thank you for sharing. In fact, this year in january I have (re)invented an iteration method which is actually the functional form of 2.1.5. b) but I discarded the idea because I misunderstood it. It was the following: (y^ox')(x) = oexp(x' × olog(y(x))) Substitute the nullelement to x and x to x', then you give the superfunction of y(x), where oexp(y) = lim (x + y/h)^oh = sum from k=0 to infinity y^ok / k! olog(y) = lim h×(y^o(1/h) - x) = sum from k=1 to infinity (-1)^(k+1) y^ok / (k+1) And I now know that [ oexp(y) ] = exp [y] and [ olog(y) ] = log [y], so I was right. But I thought that: oexp^ot(x' × olog^ot (y)) is the t-th superfunction of y and t could be real or even complex. It was a mistake, and this is why I had beleived I was wrong until now, despite I could calculate some superfunctions by hand. Lets try your methods with trigonometric function in no pari/gp but Mathematica because I could not write the code of matrix logarithm in pari/gp. Here are some codes and plots: definitions, sucos (eigen decomposition), susin (matrix log method) and some wrong results. It seems almost correct but it is not. Some idea to fix it? RE: from formulas to pari implementations - Xorter - 06/04/2020
oexp(x' × olog(y(x))) = y^ox' =: super(y, 1) oexp(oexp(x'' × x' × olog(olog(y)))) = super(y, 2) oexp^op ((prod k=1 to p x[k]) × olog^op(y)) = super(y, p) What if there is a osexp and a oslog operators for which: osexp(f(x',oslog(y))) = super(y, x') Let f(a,b) be a+b, so: osexp(x' + oslog(y)) = super(y, x') E. g.: osexp(0 + oslog(y)) = super(y, 0) = y So osexp is inverse of oslog and vica versa. osexp(1 + oslog(y)) = oexp(x' × olog(y)) = super(y, 1) osexp(-1 + oslog(y)) = y o (1+x) o y^o-1 = super(y, -1) We know that [ oexp(y) ] = exp [y], [ olog(y) ] = log [y], but what is about [ osexp(y) ] and [ oslog(y) ]? If I knew what these functions are, then I could determine Hyper(a, x, b) = a[x]b = osexp(x + oslog(a+x') o b) and it would be a really new thing in mathematics. Ofc, I checked and [ osexp(y) ] cannot be sexp [y]... |