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Matrix-method: compare use of different fixpoints - Printable Version +- Tetration Forum (https://math.eretrandre.org/tetrationforum) +-- Forum: Tetration and Related Topics (https://math.eretrandre.org/tetrationforum/forumdisplay.php?fid=1) +--- Forum: Computation (https://math.eretrandre.org/tetrationforum/forumdisplay.php?fid=8) +--- Thread: Matrix-method: compare use of different fixpoints (/showthread.php?tid=83) |
Matrix-method: compare use of different fixpoints - Gottfried - 11/04/2007 I thought, I'll check my matrix-method and compare, what happens, if I used different fixpoints for initialization. I used s=sqrt(2) as base-parameter, because of simpliness of the two fixpoints t1 =2 and t2=4 This is the reference matrix CHK = Bs , based on s=sqrt(2). It is constructed the simple method by Bs = dV(log(s)) * B CHK = Bs and the terms of the second column, taken as coefficients for powers of x give the series for s^x = sum {r=0,inf} log(s)^r / r! * x^r or, iterated s^s^x = sum {r=0,inf} log(s)^r / r! * (s^x)^r ------------------------------------------------------------------------ Code: chk = This matrix should also result by my analytic matrix-composition whatever initial fixpoint I use. My parameters for composition are s : the base, here s = sqrt(2) t=h(s) : one of its fixpoints, such that s = t^(1/t), here I use 2 and 4 as examples tl = log(t), principal branch I compose Bs analytically by its eigensystem Bs = W * D * W^-1 where D is diagonal diag(tl^0,tl^1,tl^2,...) and W may have complex entries, if the fixpoint is complex. Furthermore, I construct W and W^-1 by the composition W = dV(1/t) *(P^-1)~ * X W^-1 = X^-1 * P~ * dV(t) where P is the pascal-matrix and X is triangular. and compute X^-1 by my analytical description using the parameters t and tl, and then X by (X^-1) numerical inversion. Note, that in W = dV(1/t) *(P^-1)~ * X we may have divergent summation, which is numerically problematic, although I use Euler-summation, but only the same order for all implicite vectorial products (I'll have to improve that...). But it seems, that my procedure overcame this problem, anyway, the errors seem to cancel out nicely. Only few rows and columns are displayed in the following. I used matrix-dimension 24 and 32, float-precision 200 digits A) Construction using Fixpoint 4 ============================================= (t=4, tl=log(4), s=t^(1/t) = sqrt(2) APT_Init1(tl,t,s,24) Code: X= The eigenmatrices: Code: W = (questionable due to divergent summation in computation of entries) Code: BS = W*D*W^-1 Code: chk = B) Construction using Fixpoint 2 ============================================= (t=2, tl=log(t), s=t^(1/t) = sqrt(2) APT_Init1(tl,t,s,24) Code: X= The Eigenmatrices Code: W= (the implicite summation due to vectorproducts is convergent here) Code: Bs= W * D * W^-1 Code: chk = Different eigensystems compose the same tetration-matrix Bs; hmm. The different fixpoints are reflected in the first row of W^-1 . This is necessarily so, since by the eigen-composition Bs = W * D * W^-1 we have W^-1 * Bs = D * W^-1 and for the first row of W^-1, associated with the first eigenvalue e0=1 , this means that it remains unchanged by right-multiplication with Bs, which is just the property of the (powerseries of the) fixpoint. Gottfried RE: Matrix-method: compare use of different fixpoints - bo198214 - 11/04/2007 I dont get it. Isnt the matrix decomposition independent of the fixed point? RE: Matrix-method: compare use of different fixpoints - Gottfried - 11/04/2007 bo198214 Wrote:I dont get it. Isnt the matrix decomposition independent of the fixed point? It means, we can have different eigenmatrices / sets of eigenvectors resulting in the same composition for the infinite matrix Bs. This agrees with the fact, that we have multiple fixpoints for a base s. For e^-e < s < e^(1/e) we have even two real fixpoints; for other s the fixpoints are complex. So we may write ( I assume the continuation to more than two solutions) Bs = W0 * D0 * W0^-1 = W1 * D1 * W1^-1 = W2 * D2 * W2^-1 where W0 and D0 are different to W1 and D1 (and assumed to W2 and D2,...) according to the different initialization with different fixpoints RE: Matrix-method: compare use of different fixpoints - bo198214 - 11/04/2007 Gottfried Wrote:It means, we can have different eigenmatrices / sets of eigenvectors resulting in the same composition for the infinite matrix Bs. Oh you mean for the infinite matrices! But the truncated matrices have a unique decomposition. And you use truncated matrices to approximate the solution. So how does this fit into your considerations? RE: Matrix-method: compare use of different fixpoints - Gottfried - 11/04/2007 bo198214 Wrote:Gottfried Wrote:It means, we can have different eigenmatrices / sets of eigenvectors resulting in the same composition for the infinite matrix Bs. ... approximately... ;-) You may have different series, which asymptotically compute gamma(something), truncated give good approximations as well. The terms of X*D*X^-1 are finite polynomials in t and tl, and - a guess for answering - they may be expressed differently by the different roots of these polynomials... Then they are also summed by weights of binomial-coefficients and finally weighted by powers of 1/t to produce the terms for Bs. [update] A situation, where two different infinite series occur, is the case of 1/3 = lim 1/2 - 1/4 + 1/8 - 1/16 ... = 1/2 / (1 + 1/2) 1/3 = lim 1 -2 +4 - 8 +16 ... // analytic continuation for 1 / (1 + 2) Possibly this analogy is better, since in the example we have an alternating convergent series for t=2 and an alternating divergent series for t=4 for the final construction of terms. [/update] RE: Matrix-method: compare use of different fixpoints - bo198214 - 11/07/2007 I simply dont see how you get fixed points involved. You have Carleman matrix Where are the fixed points used? RE: Matrix-method: compare use of different fixpoints - Gottfried - 11/07/2007 bo198214 Wrote:I simply dont see how you get fixed points involved. Assume the eigensystem-decomposition Bs = W0 * D0 * W0^-1 where the small s indicates the baseparameter s=b=base Now this means also W0^-1 * Bs = D0 * W0^-1 Now look at the first row of W0^-1, call this vector Y0. Then, since I assume the first eigenvalue d0_0 =1 we have Y0 * Bs = 1 * Y0 = Y0 Now I observed, that Y0 has the form of a powerseries of the scalar parameter y0, so I may note, V(y0)~ * Bs = V(y0)~ But on the other hand I know by construction of Bs that V(x)~ * Bs = V(s^x)~ so the previous means V(y0)~ * Bs = V(y0)~ = V(s^y0) ~ thus y0 = s^y0 and y0 is obviously a fixpoint. The observation actually was: the first row in W^-1 is the powerseries in the first fixpoint y0. Now consider this backways. Using another fixpoint y1 this relation holds again, y1 = s^y1 V(y1)~ * Bs = V(s^y1)~ = V(y1)~ First row of W1^-1 is V(y1)~ and W0^-1 <> W1^-1 in their first rows and supposedly the same for the whole matrices W0 <> W1 and W1^-1 * Bs = 1 * W1^-1 is another solution, provided that the form of the first row in W1^-1 is still that of a powerseries. And in fact; if I introduce the coefficients u1 = alpha + beta*I and t1=exp(u1) in my analytical eigensystem-constructor as parameters, I get valid eigen-matrices W1 and W1^-1 based on these other fixpoints for the cases I've checked where it holds that Bs = W1*D1*W1^-1 (but see note 1) These had in fact the powerseries of the second fixpoint in the first row in W1^-1 - and I suppose, this will be the same with all other fixpoints (however the numerical computations become too difficult to do a general conjecture based on and backed by reliable approximations) Gottfried --------------- (1) This is directly related to your introductory posting in the "Bummer"-thread, and we must solve the discrepancy between these two statements! RE: Matrix-method: compare use of different fixpoints - bo198214 - 11/07/2007 Gottfried Wrote:Now consider this backways. Using another fixpoint y1 this relation holds again, Sorry, the Eigensystem decomposition for Bs truncated to n is unqiue (up to permutations of eigenvalues), how can there be two solutions??? RE: Matrix-method: compare use of different fixpoints - Gottfried - 11/07/2007 bo198214 Wrote:Sorry, the Eigensystem decomposition for Bs truncated to n is unqiue (up to permutations of eigenvalues), how can there be two solutions??? Hmm, we are talking about truncated matrices only as approximations, no? When I compute the eigensystem not by numerical eigensystem-solver for finite matrices (as implemented in a software) but based on an analytical description of each entry, then I actually work with finite truncations of an assumed infinite matrix, which may provide multiple solutions for the same composed theoretical result matrix. 1/2 - 1/4 + 1/8 - ... = 1 - 2 + 4 -8 ... independently of the fact, that all finite truncations of lhs and rhs are different (and even do not converge to the same value) Don't we agree in this? RE: Matrix-method: compare use of different fixpoints - bo198214 - 11/07/2007 Gottfried Wrote:Hmm, we are talking about truncated matrices only as approximations, no?But this is than another method. The matrix operator method is to use truncated Matrices with unique Eigensystem decomposition and hence with a unique limit. That was the absolute good thing about this method as we dont have to take fixed points (with the attached question of which to choose) into consideration. For the infinite matrices there is not even only one solution for each fixed point but also all the other possible solutions come into play, i.e. the non-regular Schroeder functions. So how should this help? If we know there are some solutions out there which we also want to have and than modify our method (that was designed to be unique) also to include the other solutions??? |