non-neglectable Infinitesimals in the Bell-matrix? Gottfried Ultimate Fellow Posts: 898 Threads: 130 Joined: Aug 2007 11/26/2007, 06:31 PM (This post was last modified: 11/28/2007, 09:43 AM by Gottfried.) Hi folks - the counterexample of G.A.Edgar to my "tetra-series"-conjecture made me rethink and investigate the Bell-matrix to some extreme. I think, I have an idea how to improve this object for the use of its inverse, and in this special case, for its use in infinite series (as done in my Tetra-conjecture). See the attached article as discussion, somehow brainstorming; although my feeling is, that this way of consideration might be fruitful, when settled with more care and depth. The article evolved out of the manuscript for my more general "tetration by matrix-method" in work, so the introductory remarks may be felt a bit inappropriate here, but... See: [update3] an updated version is available[/update3] Update I'm also removing the attachment due to many updates. I'll reattach, if my textversion becomes stable[/update] Gottfried Gottfried Helms, Kassel Gottfried Ultimate Fellow Posts: 898 Threads: 130 Joined: Aug 2007 11/28/2007, 09:16 PM (This post was last modified: 11/28/2007, 09:34 PM by Gottfried.) The discrepancy between the serial summation to the infinite alternating series of powertowers and the matrix-based summation is going to make me crazy... In my other discussion I considered the tetration x->b^x and there are some caveats, like impossibility of inversion of the infinite Bell-matrix. But the discrepancy occurs also with the x->exp(x)-1 tetration, resp. its inverse x-> log(1+x). This is even more curious, since we have a row-finite operator, for which an exact reciprocal (and its unchanged truncation) can be computed and used. But I ran out of ideas, where the source of the discrepancy may be. Here I give a list of results of the serial summation and matrix-based summation of b=1.3 and b=eta=e^(1/e) for ub(x) = b^x - 1 lb(x) = log(1+x)/log(b) which are the inverse operations and are computed by V(x)~ * S2b = V(ub(x))~ V(x)~ * S1b = V(lb(x))~ where S2b = dV(log(b))*S2 and S1b = S1*dV(1/log(b)) , parametrized versions of the Stirlingmatrices (see appendix) where the results come out correctly and satisfy the definitions of the Stirling-numbers of the two kinds. -------- The alternating series Su = x - ub(x) + ub(ub(x)) - ... +... Sl = x - lb(x) + lb(lb(x)) - ... + ... are implemented by the matrixoperators Mu = I - S2b + S2b^2 - S2b^3 + ... = (I + S2b)^-1 Ml = I - S1b + S1b^2 - S1b^3 + ... = (I + S1b)^-1 and the matrix-computation for Su(x) and Sl(x) is V(x)~ * Mu = V(Su(x)) ~ V(x)~ * Ml = V(Sl(x)) ~ This comes out to be correct for Su(x), when crosschecked by serial summation, but differ for Sl(x) Moreover, by matrixtheory it is Mu + Ml = I so the sum of the two results are V(x)~ * Mu + V(x)~* Ml = V(x)~ *( Mu + Ml) = V(x)~ * I = V(x)~ which again comes out correctly to given real-precision, when numerically checked. But the serial computation of Su(x) + Sl(x) =/= x, and the reason is in the occuring discrepancy for Sl(x) only. Also the differences for the Sl(x)-versions are relatively small and also roughly periodic with varying freqency, so I assume a systematic small error in the definition of S1 , and then also in S2. Or - for any other reason the geometric-series-based computation of Ml is somehow incompatible with the serial summation of the lb()-towers I have currently no idea, where to look into... Gottfried ---- Appendix ---------- Here are two plots:         Code:base=1.3, Sl(x)        x               matrix-based       serial            difference     1.00000000000    0.235301957128     0.245204215222  -0.00990225809450    0.975000000000    0.228701914313     0.239151131819   -0.0104492175061    0.950000000000    0.222141129686     0.233094325597   -0.0109531959119    0.925000000000    0.215619404337     0.227027895542   -0.0114084912053    0.900000000000    0.209136539762     0.220945608398   -0.0118090686363    0.875000000000    0.202692337870     0.214840897906   -0.0121485600364    0.850000000000    0.196286600996     0.208706869038   -0.0124202680427    0.825000000000    0.189919131913     0.202536308811   -0.0126171768976    0.800000000000    0.183589733845     0.196321705675   -0.0127319718303    0.775000000000    0.177298210472     0.190055280039   -0.0127570695670    0.750000000000    0.171044365948     0.183729029165   -0.0126846632164    0.725000000000    0.164828004906     0.177334790567   -0.0125067856610    0.700000000000    0.158648932469     0.170864329198   -0.0122153967297    0.675000000000    0.152506954259     0.164309455153   -0.0118025008943    0.650000000000    0.146401876408     0.157662180530   -0.0112603041216    0.625000000000    0.140333505569     0.150914926535   -0.0105814209661    0.600000000000    0.134301648917     0.144060795071  -0.00975914615380    0.575000000000    0.128306114167     0.137093923193  -0.00878780902604    0.550000000000    0.122346709574     0.130009944135  -0.00766323456122    0.525000000000    0.116423243947     0.122806585589  -0.00638334164115    0.500000000000    0.110535526655     0.115484444901  -0.00494891824556    0.475000000000    0.104683367633     0.108047992534  -0.00336462490093    0.450000000000   0.0988665773892     0.100506870001  -0.00164029261148    0.425000000000   0.0930849670147    0.0928775672741  0.000207399740560    0.400000000000   0.0873383481879    0.0851855875555   0.00215276063243    0.375000000000   0.0816265331820    0.0774682336608   0.00415829952120    0.350000000000   0.0759493348711    0.0697781768756   0.00617115799554    0.325000000000   0.0703065667364    0.0621879866737   0.00811858006277    0.300000000000   0.0646980428720    0.0547957847775   0.00990225809450    0.275000000000   0.0591235779909    0.0477320835127    0.0113914944782    0.250000000000   0.0535829874301    0.0411675435700    0.0124154438601    0.225000000000   0.0480760871563    0.0353205246358    0.0127555625205    0.200000000000   0.0426026937709    0.0304611391266    0.0121415546443    0.175000000000   0.0371626245147    0.0269031449478    0.0102594795668    0.150000000000   0.0317556972729    0.0249614784615   0.00679421881140    0.125000000000   0.0263817305798    0.0248181823565   0.00156354822330    0.100000000000   0.0210405436227    0.0261459357564  -0.00510539213364   0.0750000000000   0.0157319562467    0.0270897957341   -0.0113578394875   0.0500000000000   0.0104557889581    0.0217178308705   -0.0112620419124   0.0250000000000  0.00521186292883  -0.00152823441650   0.00674009734534 Code:S2   1      .     .    .  .  .   0      1     .    .  .  .   0    1/2     1    .  .  .   0    1/6     1    1  .  .   0   1/24  7/12  3/2  1  .   0  1/120   1/4  5/4  2  1 [\code] [code] S1   1     .      .     .   .  .   0     1      .     .   .  .   0  -1/2      1     .   .  .   0   1/3     -1     1   .  .   0  -1/4  11/12  -3/2   1  .   0   1/5   -5/6   7/4  -2  1 Gottfried Helms, Kassel Gottfried Ultimate Fellow Posts: 898 Threads: 130 Joined: Aug 2007 11/30/2007, 06:42 PM I tried with various modifications of the series for Sl to eliminate possible effects of divergence or only conditional convergence. Using b as base , lb(x) = log(1+x)/log(b) Sla(x) = x - lb(x)/d + lb(lb(x))/d^2 - lb(lb(lb(x)))/d^3 + ... - ... Slb(x) = x/0! - lb(x)/1! + lb(lb(x))/2! - lb(lb(lb(x)))/3! + ... - ... where I used 2 <= d <=32 for Sla(x) The observed differences between serial and matrix-based summation are still occuring with no apparent diminuation, so I think, the effect is based already in very early terms of the matrix-based coefficients for the powerseries. Note, that the matrix-based results are all consistent between each other - using the matrix for iterated logarithm, for iterated exponentiation, the sums, the reciprocity of the sum-matrices, the summing of Sl(x)+Su(x) = x , even the sum Su(x) via matrix- and serial summation. Only the Sl-summation between serial and matrix-summations differs, and the difference is small, sometimes zero and of roughly sinusoidal dilated periodicity. So, one needs really a new idea Gottfried Helms, Kassel andydude Long Time Fellow Posts: 510 Threads: 44 Joined: Aug 2007 11/30/2007, 07:53 PM (This post was last modified: 11/30/2007, 07:56 PM by andydude.) Well, here's one. You know how Taylor series are like an infinite vector space that represents a function with the basis $x^k$ and Fourier series are like an infinite vector space that represents a function with the basis $e^{ikx}$? Well you could think of it not in terms of your function specifically, but instead think of it in terms of these basis representations. This is also a more general approach, rather than a specific-function-motivated approach. In fact, one could even extend Bell and Carleman matrices to such basis systems. For Taylor Series: $ \begin{tabular}{rl} f(x) & = \sum_{k=0}^{\infty} f_k x^k \\ f(x)^j & = \sum_{k=0}^{\infty} M[f]_{jk} x^k \end{tabular}$ For Fourier Series: $ \begin{tabular}{rl} f(x) & = \sum_{k=0}^{\infty} g_k \exp(ikx) \\ \exp(ijf(x)) & = \sum_{k=0}^{\infty} (FourierM[f]_{jk}) \exp(ikx) \end{tabular}$ For Tetration Series: $ \begin{tabular}{rl} f(x) & = \sum_{k=0}^{\infty} (TetraM[f]_{1k}) ({}^{k}x) \\ {}^{j}(f(x)) & = \sum_{k=0}^{\infty} (TetraM[f]_{jk}) ({}^{k}x) \end{tabular}$ For Iterated Exponential Series: $ \begin{tabular}{rl} f(x) & = c_0 + \sum_{k=0}^{\infty} (IterExpM[f]_{0k}) \exp_b^{k}(x) \\ \exp_b^{j}(f(x)) & = c_1 + \sum_{k=0}^{\infty} (IterExpM[f]_{jk}) \exp_b^{k}(x) \end{tabular}$ All of these matrices convert function composition into matrix multiplication, or equivalently, function iteration into matrix powers. Although I haven't investigated what these alternate Carleman matrices look like, I'm sure they would be much more complicated than their Taylor counterparts, since there is not nice way of inverting them. For Teylor series, inverting, or "finding coefficients" is just a matter of taking derivatives, for Fourier series "finding coefficients" is just a matter of taking a Cauchy integral. For the last two series, though, I can't think of any way of finding coefficients aside from doing a change-of-basis between these 4 basis systems. Andrew Robbins Gottfried Ultimate Fellow Posts: 898 Threads: 130 Joined: Aug 2007 12/14/2007, 08:26 PM (This post was last modified: 12/14/2007, 08:27 PM by Gottfried.) I just added another small observation trying to narrow down the reason for the problem of inconsistency in infinite alternating tetra-series... See http://go.helms-net.de/math/tetdocs/Tetr..._short.pdf @Andrew - please excuse, that I did not yet comment on your previous post. It's not ignorance, but bad health in the last three weeks. I'll come to it soon. Gottfried Gottfried Helms, Kassel « Next Oldest | Next Newest »

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