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 Tetra-series Gottfried Ultimate Fellow Posts: 770 Threads: 120 Joined: Aug 2007 11/20/2007, 12:47 PM (This post was last modified: 11/20/2007, 03:21 PM by Gottfried.) I have another strange, but again very basic result for the alternating series of powertowers of increasing height (I call it Tetra-series, see also my first conjecture at alternating tetra-series ) Assume a base "b", and then the alternating series Code:.       Sb(x) = x - b^x + b^b^x - b^b^b^x +... - ... and for a single term, with h for the integer height (which may also be negative) Code:.   Tb(x,h) = b^b^b^...^x     \\ b occurs h-times which -if h is negative- actually means (where lb(x) = log(x)/log(b) ) Code:.   Tb(x,-h) = lb(lb(...(lb(x))...)   \\ lb occurs h-times ------------------------------------------------------- My first result was, that these series have "small" values and can be summed even if b>e^(1/e) (which is not possible with conventional summation methods). For the usual convergent case e^(-e)1        = sum r=0..inf  x^r * mb[r,1]   serial notation        = sum h=0..inf  (-1)^h* Tb(x,h)  \\ only possible for e^(-e) < b < e^(1/e)                                         \\ Euler-summation required ------------------------------------------------------- Now if I extend the series Sb(x) to the left, using lb(x) = log(x)/log(b) for log(x) to base b, then define Code:.    Rb(x) = x - lb(x) + lb(lb(x) - lb(lb(lb(x))) +... - ... This may be computed by the analoguous formula above to that for Mb from the inverse of Bb: Code:.    Lb = (I + Bb^-1)^-1 I get for the sum of both by my matrix-method Code:.   Sb(x) + Rb(x) = V(x)~ *Mb[,1] + V(x)~ * Lb[,1]                 = V(x)~ * (Mb + Lb)[,1]                 = V(x)~ *    I [,1]                 = V(x)~ *   [0,1,0,0,...]~                   = x   Sb(x) + Rb(x) = xor, and this looks even more strange (but even more basic) Code:.   0 = ... lb(lb(x)) - lb(x) + x - b^x + b^b^x - ... + ... x cannot assume the value 1, 0 or any integral height of the powertower b^b^b... since at a certain position we have then a term lb(0), which introduces a singularity. Using the Tb()-notation for shortness, then the result is $\hspace{24} 0 = \sum_{h=-\infty}^{+\infty} T_b(x,h)$ and is a very interesting one for any tetration-dedicated... Gottfried ------------------------------------------------------- An older plot; I used AS(s) with x=1,s=b for Sb(x) there. (a bigger one AS     Gottfried Helms, Kassel andydude Long Time Fellow Posts: 509 Threads: 44 Joined: Aug 2007 11/21/2007, 07:14 AM Have you tried computing or plotting $AS(x^{1/x})$ or $AS(x)^{1/AS(x)}$ yet? I would do this but I don't have any code yet for AS(x), and I'm lazy. Andrew Robbins andydude Long Time Fellow Posts: 509 Threads: 44 Joined: Aug 2007 11/21/2007, 07:51 AM Gottfried Wrote:I get for the sum of both by my matrix-method Code:.   Sb(x) + Rb(x) = V(x)~ *Mb[,1] + V(x)~ * Lb[,1]                 = V(x)~ * (Mb + Lb)[,1]                 = V(x)~ *    I [,1]                 = V(x)~ *   [0,1,0,0,...]~                   = x   Sb(x) + Rb(x) = x This is what I have the most trouble understanding. First what is your [,1] notation mean? I understand "~" is transpose, and that Bb is the Bell matrix $Bb = B_x[s^x]$. Second, what I can't see, or is not obvious to me at least, is why: $(I + Bb^{-1})^{-1} + (I + Bb)^{-1} = I$ Is there any reason why this should be so? Can this be proven? Wait, I just implemented it in Mathematica, and you're right! (as right as can be without a complete proof). Cool! This may just be the single most bizarre theorem in the theory of tetration and/or divergent series. Andrew Robbins Gottfried Ultimate Fellow Posts: 770 Threads: 120 Joined: Aug 2007 11/21/2007, 09:41 AM (This post was last modified: 11/21/2007, 10:45 AM by Gottfried.) andydude Wrote:Gottfried Wrote:I get for the sum of both by my matrix-method Code:.   Sb(x) + Rb(x) = V(x)~ *Mb[,1] + V(x)~ * Lb[,1]                 = V(x)~ * (Mb + Lb)[,1]                 = V(x)~ *    I [,1]                 = V(x)~ *   [0,1,0,0,...]~                   = x   Sb(x) + Rb(x) = x This is what I have the most trouble understanding. First what is your [,1] notation mean? I understand "~" is transpose, and that Bb is the Bell matrix $Bb = B_x[s^x]$. Second, what I can't see, or is not obvious to me at least, is why: $(I + Bb^{-1})^{-1} + (I + Bb)^{-1} = I$ Is there any reason why this should be so? Can this be proven? Wait, I just implemented it in Mathematica, and you're right! (as right as can be without a complete proof). Cool! This may just be the single most bizarre theorem in the theory of tetration and/or divergent series. Andrew Robbins Hi Andrew - first: I appreciate your excitement! Yepp! :-) second: (The notation B[,1] refers to the second column of a matrix B) Yes, I just posed the question, whether (I+B)^-1 + (I+B^-1)^-1 = I in the sci.math- newsgroup. But the proof for finite dimension is simple. You need only factor out B or B^-1 in one of the expressions. Say C = B^-1 for brevity Code:.    (I + B)^-1 + (I + C)^-1   = (I + B)^-1 + (CB + C)^-1 = (I + B)^-1 + (C(B + I))^-1 = (I + B)^-1 + (B + I)^-1*C^-1 = (I + B)^-1 + (B + I)^-1*B = (I + B)^-1 *(I + B) = I As long as we deal with truncations of the infinite B and these are well conditioned we can see this identity in Pari or Mathematica with good approximation. However, B^-1 in the infinite case is usually not defined, since it implies the inversion of the vandermonde matrix, which is not possible. On the other hand, for infinite lower *triangular* matrices a reciprocal is defined. The good news is now, that B can be factored into two triangular matrices, like B = S2 * P~ where P is the pascal-matrix, S2 contains the stirling-numbers of 2'nd kind, similarity-scaled by factorials S2 = dF^-1 * Stirling2 * dF (dF is the diagonal of factorials diag(0!,1!,2!,...) ) Then, formally, B^-1 can be written B^-1 = P~^-1 *S2^-1 = P~^-1 * S1 (where S1 contains the stirling-numbers of 1'st kind, analoguously factorial rescaled, and S1 = S2^-1 even in the infinite case) B^-1 cannot be computed explicitely due to divergent sums for all entries (rows of P~^-1 by columns of S1), and thus is not defined. However, in the above formulae for finite matrices we may rewrite C in terms of its factors P and S1, and deal with that decomposition-factors only and arrive at the desired result (I've not done this yet, pure lazyness...) third: This suggests immediately new proofs for some subjects I've already dealt with, namely all functions, which are expressed by matrix-operators and infinite series of these matrix-operators. For instance, I derived the ETA-matrix (containing the values for the alternating zeta-function at negative exponents) from the matrix-expression Code:. ETA = (P^0 - P^1 + P^2 ....)      = (I + P)^-1If I add the similar expression for the inverses of P I arrive at a new proof for the fact, that each eta(2k) must equal 0 for k>0. Yes- this is a very beautiful and far-reaching fact, I think ... Gottfried Gottfried Helms, Kassel Ivars Long Time Fellow Posts: 366 Threads: 26 Joined: Oct 2007 11/21/2007, 03:58 PM hej gottfried, iI would sincerely like to understand more about these matrixes, i have a feeling its important, but I could not find in You texts what is I-i suppose it is identityu matrix, but how does it look like? Best regards, Ivars Gottfried Ultimate Fellow Posts: 770 Threads: 120 Joined: Aug 2007 11/21/2007, 04:37 PM Ivars Wrote:hej gottfried, iI would sincerely like to understand more about these matrixes, i have a feeling its important, but I could not find in You texts what is I-i suppose it is identityu matrix, but how does it look like? Best regards, Ivars Hi Ivars - just the matrix containing 1 on its diagonal. So multiplying by it doesn't change a matrix, like multiplication by 1 does not change the multiplicand. Gottfried Gottfried Helms, Kassel Gottfried Ultimate Fellow Posts: 770 Threads: 120 Joined: Aug 2007 11/21/2007, 06:59 PM (This post was last modified: 11/22/2007, 03:58 PM by Gottfried.) Gottfried Wrote:Yes- this is a very beautiful and far-reaching fact, I think ...I've just received an answer in the newsgroup sci.math by Prof G.A.Edgar who states a numerical discrepancy between my matrix-based conjecture and termwise evaluation of the series. I cannot resolve the problem completely - the problem doesn't affect the Mb-matrix related conjectures (also of earlier date) but the problem of representation of the alternating series of powers of the reciprocal of Bb by the analoguous expression. I don't have an idea currently, how to cure this and how to correctly adapt my conjecture. So - sigh - I have to retract it for the moment. [update] I should mention, that this concerns only the Bb-matrix, which is not simply invertible. The application of the idea of the formula to other matrix-operators may be still valid; especially for triangular matrices like P the observation is still valid; I assume, it is also valid for the U-iteration x->exp(x)-1 , since the matrix-operator is the triangular Stirling-matrix. I'll check that today [/update] [update2] The problem occurs also with the U-iteration and its series of negative heights. Looks like the reciprocal matrix needs some more consideration [/update2] Gottfried P.s. I'll add the conversation here later as an attachment. [update3] A graph which shows perfect match between serial and matrix-method-summation for the Tb-series; and periodic differences between the Rb-series [/update3]     Gottfried Helms, Kassel andydude Long Time Fellow Posts: 509 Threads: 44 Joined: Aug 2007 11/21/2007, 07:24 PM Ok, so the identity: $(I + Bb^{-1})^{-1} + (I + Bb)^{-1} = I$ holds true for all matrices, not just the Bell matrix of exponentials. Good to know. Andrew Robbins Gottfried Ultimate Fellow Posts: 770 Threads: 120 Joined: Aug 2007 11/21/2007, 07:49 PM andydude Wrote:Ok, so the identity: $(I + Bb^{-1})^{-1} + (I + Bb)^{-1} = I$ holds true for all matrices, not just the Bell matrix of exponentials. Good to know. Andrew Robbins Regularly invertible matrices (for instance of finite size); and I think, that even infinite matrices can be included, if they are triangular (row- or column-finite) or, if not triangular, at least if some other condition holds (on their eigenvalues or the like). I'll have to perform some more tests... Gottfried Helms, Kassel andydude Long Time Fellow Posts: 509 Threads: 44 Joined: Aug 2007 11/21/2007, 08:39 PM Gottfried Wrote:You need only factor out B or B^-1 in one of the expressions. Say C = B^-1 for brevity Code:.    (I + B)^-1 + (I + C)^-1   = (I + B)^-1 + (CB + C)^-1 = (I + B)^-1 + (C(B + I))^-1 = (I + B)^-1 + (B + I)^-1*C^-1 = (I + B)^-1 + (B + I)^-1*B = (I + B)^-1 *(I + B) = I` This completes the proof in my view. Good job. Andrew Robbins « Next Oldest | Next Newest »

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