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 Understanding Abel/Schroeder with matrix-expression Gottfried Ultimate Fellow Posts: 767 Threads: 119 Joined: Aug 2007 03/30/2008, 05:06 AM (This post was last modified: 05/24/2008, 10:01 AM by Gottfried.) Hmm, up to now I never could get a grip about the problems of the Abel-function. I merely took it as a notational formalism - perhaps the eigen-decomposition and your previous statements here helps me, to get a first step to such a grip now. If I write Code:´    V(x)~ * Bb^h = V(y)~where $y = \exp_b^{\circ h}(x)$ in Tex (or y = x {4,b} h in my partisan notation ), then writing Code:´    t = h(b) which makes t^(1/t) =b    u = log(t)and the diagonalization (I should better write W_u, since W_u is dependent on u but use W for shortness here) Code:´     Bb   = W * dV(u)   *W^-1     Bb^h = W * dV(u)^h *W^-1          = W * dV(u^h) *W^-1 then the second column of W provides the coefficients of a powerseries, which defines a function wu(x) Code:´     wu(x) = V(x)~*W [,1]     // [,1] means second column of W           = sum{k=0,inf} W[k,1]*x^kthen Code:´     V(x)~ * W * dV(u^h) * W^-1 = V(y)~     V(x)~ * W * dV(u^h)        = V(y)~ * W and from dV(u^h) being diagonal it follows (and it suffices to use column 1 only) Code:´ (V(x)~ * W * dV(u^h)) [,1]  = (V(y)~ * W) [,1] (V(x)~ * W [,1])* u^h        = V(y)~ * W  [,1]and then the functional relation Code:´ wu(x)* u^h = wu(y)from where , writing lg_b(x) = log(x)/log(b) Code:´ lg_u(wu(x)) + h = lg_u(wu(y)) or h = lg_u(wu(y)) - lg_u(wu(x))as I also noted recently in the older Abel-thread. Isn't then lg_u(wu(x)) such an Abel-function? [update] well, I've several times read the post of Andrew (or -another instance- of Henryk) which I never understood (and felt being left just to skim through after few lines, leaving it "at the expert's level") because I could not resolve the many unknowns into a system, which I'd been able to relate to my handwaved approach. But rereading it now, it seems to me, the above relation is identical to that considerations. At least the formal handling says this to me. Shame on me... [/update] [some more edit-updates in the original part above] Gottfried Helms, Kassel Gottfried Ultimate Fellow Posts: 767 Threads: 119 Joined: Aug 2007 05/18/2008, 02:36 PM (This post was last modified: 05/18/2008, 02:39 PM by Gottfried.) A sort of mind-breaking... I tried to step into the subject of Abel-/Schroeder-function from the functional approach, leaving the matrix-considerations aside, and stepped into a problem. If I assume $\hspace{24} f_b(x) = b^x$ $\hspace{24} f_b^{^oh}(x) = b^{...^{b^x}}$ (b occurs height h-times) replaced by a b-related function pb(x) and its inverse qb(x) such that $\hspace{24} f_b^{^oh}(x)= q_b( u^h*p_b(x))$ (as a sketch only - leaving for instance the fixpoint-shift question aside) and I assume $p_b(x)$ and $q_b(x)$ as b-parametrized powerseries in x (and "u" as the base of the assumed eigenvalues of the diagonalisation) then I observe the problem, that the scalar product $y = u^h*p_b(x)$, where $p_b(x)$ is a constant "c" for a chosen "x" and only "h" is variable, is independent of the order of its factors; so the function $q_b(y)$ "does not know" about the order of the "u"'s and "c". But anyway we expect the effect of "c" always on top of the powertower, which clearly implies some ordering. From this I question, whether we are correct in assuming the existence of a *scalar* function $p_b(x)$. Don't we need a notion of an object, which assures, that with different orders in $u*u*u*\ldots *u*p_b(x)$ the expression has different values? The matrix-notation gives this order implicitely, but a scalar expression like the above doesn't depend on any order... Gottfried Gottfried Helms, Kassel bo198214 Administrator Posts: 1,389 Threads: 90 Joined: Aug 2007 05/18/2008, 04:51 PM (This post was last modified: 05/18/2008, 04:52 PM by bo198214.) I am not 100% sure what you are meaning, but I think you convey the notion of $b^{\dots ^{b^c}}$ to $u*\dots *u * p_b( c)$. But thats probably not adequate. When you do regular iteration the ordering is already contained. It is about $h$ time applying a function to an argument. If the function is exponentiation (letting aside the fixed point shift) then this just means that the argument is on top of the tower and nowhere else. The method only works for this case. There is no freedom to rearrange the tower, say be rearring the $u$'s and $p_b( c)$. Was it that? Gottfried Ultimate Fellow Posts: 767 Threads: 119 Joined: Aug 2007 05/18/2008, 07:41 PM bo198214 Wrote:I am not 100% sure what you are meaning, but I think you convey the notion of $b^{\dots ^{b^c}}$ to $u*\dots *u * p_b( c)$. But thats probably not adequate. When you do regular iteration the ordering is already contained. It is about $h$ time applying a function to an argument. If the function is exponentiation (letting aside the fixed point shift) then this just means that the argument is on top of the tower and nowhere else. The method only works for this case. There is no freedom to rearrange the tower, say be rearring the $u$'s and $p_b( c)$. Was it that? Well, first I've to say, that possibly I'm completely off the road with my "brainstorm", since I've still far too little experience with the concept of the Schroeder-functional equation. I refer here to your article there bo198214 Wrote:For doing this we first compute the regular Schroeder function (note that the Schroeder function is determined up to a multiplicative constant and the Abel function is determined up to an additive constant). A Schroeder function $\sigma$ of a function $f$ is a function that satisfies the Schroeder equation $\sigma(f(x))=s\sigma(x)$ (...) This function particularly yields the regular iteration at 0, via $f^{\circ t}(x)=\sigma^{-1}(s^t\sigma(x))$. (...) and my notation of $p_b(x)$ is meant to be $\sigma(x)$, of $q_b(x)$ to be $\sigma^{-1}(x)$ and of $u^h$ to be $s^t$ . Here b, as usual, is the base, $t^{1/t}=b$, u=log(t) and I assume my formula-substitution to get the Schroeder-equation $f_b^{^oh}(x)= q_b( u^h*p_b(x)) =\sigma^{-1}(s^h\sigma(x))$ from the diagonalization-method. If I assume the transposed carleman-matrix Bb for x->b^x as diagonalizable, then I also assume formally for any iteration-height h $\hspace{24} B_b^h = W_b * dV(u)^h * W_b^{-1}$ where Wb contains the set of eigenvectors of Bb. Then the whole operation $\hspace{24} V(x)\sim * B_b^h = V(\exp_b^{^{o}h}(x))\sim$ can be splitted into $\hspace{24} V(x)\sim * W_b * dV(u)^h * W_b^{-1} = V(\exp_b^{^oh}(x))\sim$ Here it would be nice, if Wb behaves as an "operator on powerseries", such that $\hspace{24} V(x)\sim * W_b = V(y)\sim$ If this would be so, then also Wb^-1 would be such an operator and all argumentation simplifies. Empirically for some convergent cases this seems to be true, so assuming this we can split our computation into three parts: $\hspace{24} V(x)\sim * W_b = V(y)\sim$ $\hspace{24} V(y)\sim * dV(u)^h = V(z)\sim$ $\hspace{24} V(z)\sim * W_b^{-1} = V(\exp_b^{^oh}(x))\sim$ The value for y can be seen as evaluation of a powerseries in x, which uses the coefficients of the second column of Wb, and I denoted this as function $y=p_b(x)$, which I understood as the function $\sigma(x))$ in your text. Then $V(y)\sim *dV(u)^h$ is $V(p_b(x)*u^h)~$ because both V(y) and V(u)^h are vandermonde-vectors and dV(u)^h is diagonal. Then since Wb^-1 provides the coefficients for the inverse $q_b(x) = p_b^{-1}(x)$ we have finally $\hspace{24} V(p_b(x)*u^h)~ *W_b^{-1} = V(q_b(u^h*p_b(x)))~$ which looks like a perfect match with the Schroeder-formula $\hspace{24} = V(\sigma^{-1}(s^h * \sigma(x)))~$ --------------------------------- So, what is the relation to my previous post? I confess, I've difficulties myself to restate the problem as I've seen it in the morning. But let's try. I was considering a sequence of computations $\hspace{24} f_b^{{^o}1}(x)= \sigma^{-1}(s*\sigma(x))$ $\hspace{24} f_b^{{^o}2}(x)= \sigma^{-1}(s*\sigma(\sigma^{-1}(s*\sigma(x)))) = \sigma^{-1}(s*s*\sigma(x))$ ... and asked, why the heck does $\sigma^{-1}$ know, that $\sigma(x)$ has to appear at the top since all factors of this scalar equation have the same level and can arbitrarily be interchanged in the above sequence of computations... Hmm... Gottfried Helms, Kassel bo198214 Administrator Posts: 1,389 Threads: 90 Joined: Aug 2007 05/18/2008, 08:00 PM Ya, everything alright. However for me is the functional approach the more basic one because it works also for non-analytic functions, while the matrix approach is only applicable if you have the powerseries coefficients. And I really wonder whether there are also iterational formulas (i.e. formulas that dont rely on the powerseries expansion, but only on function values) to compute iterations at non-fixed points ... Well the thing about the Abel function is quite natural, it corresponds to the log and superlog functions, once I realized that I wondered how this ever could seem strange to me. Gottfried Ultimate Fellow Posts: 767 Threads: 119 Joined: Aug 2007 05/19/2008, 12:02 PM (This post was last modified: 05/20/2008, 04:33 PM by Gottfried.) I think I've resolved my brain-scattering concerning the ordering in my previous post now. It was just confusion, when I tried to get access to the understanding of the sequence of increasing heights "from the scratch" by composing appropriate and "near-appropriate" functions. Anyway, what is still surprising, is, that the $\sigma$-function seems to have an exponential characteristic. I'd expected something near a logarithmic characteristic instead. This is $\sigma(x)$ or "p_b(x)" (as denoted in the previous post) for b~1.345 (u=0.5, t=exp(u), b=exp(u/t)=t^(1/t))     Strange... Gottfried [update]: Plot updated ,code updated for better readability [/update] Code:n=96 \\ set matrix-size; 96 terms for powerseries fmt(200,12) \\ set internal float-precision to 200 dec digits \\ set base-parameter b [u = 0.5,t=exp(u),b=exp(u/t),bl = u/t] \\ b = t^(1/t) \\ the carleman-matrix Bb is not needed, instead we use the analytically \\ determined Bell-matrix UtA of the decomposition Bb = (P^(-t))~ * UtA * (P^t)~ \\ This decomposition is only correct for the untruncated case, so we have to determine \\ UtA analytically \\ create Bell-matrix for regular iteration  b^x using fixpoint t Ut  = dV(u)*fS2F;    \\ analytically; from factorially scaled Stirlingnumbers 2'nd kind          \\ Ut performs x -> t^x - 1          \\ uses fixpoint-shift " x'=x/t-1 " UtA = dV(1/t)* Ut * dV(t);    \\ modifies the fixpoint-shift to " x'=x-t "          \\ so UtA performs x -> b^(x+t) - t   \\ diagonalization of Bell-matrix, returns parameters and X,dV(u),X^-1 UtAKenn = APT_Init2EW(UtA); X = UtAKenn[2];     \\ the eigenvector-matrix of UtA \\ note, that the eigenvector-matrix W of the Carleman-matrix Bb is W= P~^t * X \\ Schroeder-function-plot evec=ESum(2.5)[n,];   \\ create coefficients-vector for Eulersummation order 2.5                   \\ this was checked against smaller orders, differences occur 3.2plot list \\ from box "plot" transfer coordinates to Excel "Schroederfunction.xls"  in csv-format Gottfried Helms, Kassel bo198214 Administrator Posts: 1,389 Threads: 90 Joined: Aug 2007 05/19/2008, 06:14 PM Gottfried Wrote:Anyway, what is still surprising, is, that the $\sigma$-function seems to have an exponential characteristic. I'd expected something near a logarithmic characteristic instead. Hm, the logarithm of a Schroeder function is an Abel function: $\sigma(f(x))=u\sigma(x)$ $(\log_u\circ \sigma)(f(x))=1+(\log_u\circ\sigma)(x)$ So the logarithm should look like an slog. Gottfried Ultimate Fellow Posts: 767 Threads: 119 Joined: Aug 2007 05/19/2008, 07:31 PM (This post was last modified: 05/19/2008, 08:41 PM by Gottfried.) Looking a bit deeper at the sigma-function the above plot looked a bit strange to me, when I thought in terms of "height of powertowers": because at x>=t we should have infinite heights, but where is the infinity here? But this resolves (and possibly there is nothing new here for the sigma-experienced reader...) Let me restate the use of standard variable-names: Code:base = b, t = h(b) such that b=t^(1/t) , u = log(t) h = height of the powertower Also I use the fixpoint-shift x/t-1 instead of x-t, but that doesn't matter besides the same scaling in the results. For the example I use here as in the previous msg Code:u=0.5, t=exp(0.5)~1.64872127070, b~1.35427374603 From the functional description it should be, that the function sigma decomposes its parameter x into a power of u; additionally it provides some constant c for sigma(1). If x is seen as a powertower of height h to base b, such that $x = 1 \lbrace b \rbrace h$ then formally $\hspace{24} \sigma(x ) = \sigma(1 \lbrace b \rbrace h) = u^h * \sigma(1)$ where sigma(1) is a constant (call it "c") where c=-0.327639416789 (If the fixpoint-shift "x-t" is done then c1=-0.540186075580 = t*c) So the sigma-function decomposes x, replaces b by u and replaces height by exponent. Thus we can get the height-parameter h simply by $\log(\sigma(x)/c)/ \log(u)$ Obviously, for x increasing to $(1 \lbrace b \rbrace \infty) ((= t))$ we get $\hspace{24} \lim_{x->t} \sigma(x) = \sigma(1 \lbrace b \rbrace \infty)$ $\hspace{24} =\lim_{h->\infty} u^h * c = 0$ and for all u with -1 < u < 1 this converges to zero, so in the plot we have the crossing with y-axis at x = t ((~ 1.6487)) The nice feature is, that the height-parameter does not occur explicitely in the sigma-function; otherwise we were lost in the case where x>t; this would require then a more-than-infinite height. If we test one x>t we see, that sigma(x) gets positive, so sigma(x)/c is negative and the height h becomes complex due to log of a negative value. But the plot seems to be smooth in the vicinity of x=t; so if |u|<1 there seems to be a compensation of small u against the change of infinite real height to complex heights - I'll look at it later... For bases b, where u>1 we need no infinite real heights nor complex heights to arrive at an arbitrary high real x; here all heights should be real. So far for this msg.     (the data for the plot were computed in steps of h where h=0,1,2,...,32 and x=1 {b} 0,1 {b} 1, 1 {b} 2,...) Gottfried Helms, Kassel Gottfried Ultimate Fellow Posts: 767 Threads: 119 Joined: Aug 2007 05/19/2008, 07:45 PM (This post was last modified: 05/19/2008, 08:23 PM by Gottfried.) bo198214 Wrote:Gottfried Wrote:Anyway, what is still surprising, is, that the $\sigma$-function seems to have an exponential characteristic. I'd expected something near a logarithmic characteristic instead. Hm, the logarithm of a Schroeder function is an Abel function: $\sigma(f(x))=u\sigma(x)$ $(\log_u\circ \sigma)(f(x))=1+(\log_u\circ\sigma)(x)$ So the logarithm should look like an slog.Yes; may be I didn't see the equality since I had always a different powerseries due to a different fixpoint-shift. In the light of my last msg it seems to me, that possibly the application of the Schroeder-function has a benefit over the Abel-function because we do not explicitely deal with logarithms; this is of significance for all infinite heights and complex heights (log(0),log(-x)), which occurs if x is "out-of-bounds" for b in the range e^(-e) < b < e^(1/e) and x>t . I didn't work with the slog up to now, so I can't say anything about this. [update] Concerning the expectation of a "logarithmic characteristic": I meant this, because x is an iterated exponential which is converted into a product. This conversion to a lower operation is usually connected with some logarithm-function. So I thought, the Schröder-function would be roughly similar to a logarithmic function... [/update] Gottfried Helms, Kassel Gottfried Ultimate Fellow Posts: 767 Threads: 119 Joined: Aug 2007 05/20/2008, 10:59 AM (This post was last modified: 05/20/2008, 11:15 AM by Gottfried.) A corrected updated image of the Schroeder-function for b~1.354 (u=0.5) there is an asymptote at the second fixpoint (x~5.79173098837) Overview     Detail     Gottfried Helms, Kassel « Next Oldest | Next Newest »

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