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 Fractals from calculations of 2^I, 2^(2^I), 2^(2^(2^I).. a^(a^(...a^I) Gottfried Ultimate Fellow Posts: 764 Threads: 118 Joined: Aug 2007 05/27/2008, 09:23 AM (This post was last modified: 05/27/2008, 10:12 AM by Gottfried.) Ivars Wrote:Gottfried Wrote:AS(2,I) = I - 2^I + 2^2^I - 2^2^2^I + ... -... = ?? My proposal is AS(2,I) = -0.440033096027+0.928380628227*I What do you think? Gottfried I can only say it diverges, and I understand You want me to find a method to sum it and a result. (...) Ivars Well, diagonalization seems to give an answer here. Remember, in my matrix-notation, we have $\hspace{24} V(x)\sim * B_b = V(b^x)\sim$ or more general $\hspace{24} V(T_b^{{o}h}(x)) )\sim * B_b = V(T_b^{{o}h+1}(x)) )\sim$ and using digonalization $\hspace{24} B_b^h = W *^dV(u^h) * W^{-1}$ Now the alternating sum is (using the small letter "i" for the imaginary unit to prevent confusion with "I" as the identity matrix) $\hspace{24} AS(2,i) = V(i)\sim * B_b^0 - V(I)\sim *B_b^1 + V(I)\sim*B_b^2 - ... + ...$ $\hspace{24} = V(i)\sim*(B_b^0 - B_b^1 + B_b^2 - ... +...)$ $\hspace{24} = V(i)\sim*W *(^dV(u^0)-^dV(u^1)+^dV(u^2)-...+...)*W^{-1}$ $\hspace{24} = V(i)\sim*W *(X)*W^{-1}$ where we have to determine the entries of the diagonal-matrix X. But they can all be determined by the rule of geometric series: $\hspace{24} X[0] = (u^0)^0 - (u^1)^0+(u^2)^0 ... = 1/(1+u^0) = 1/(1+1)=1/2$ $\hspace{24} X[1] = (u^0)^1 - (u^1)^1+(u^2)^1 ... = 1/(1+u^1)$ $\hspace{24} X[2] = (u^0)^2 - (u^1)^2+(u^2)^2 ... = 1/(1+u^2)$ ... (Note, that we need no fractional powers, so the multiplications in the exponents are commutative and can be reordered) Then $\hspace{24} X = diag([1/(1+u^0),1/(1+u^1),1/(1+u^2),...])$ which can be described by $\hspace{24} X = (I +^dV(u))^{-1}$ // "I" means here the identity-matrix so the eigenvalues of $\hspace{24} AS_b = W * X * W^{-1}$ are not a vandermonde-vector, btw. In most short form we may simply write, again invoking the identities of diagonalization $\hspace{24} AS(2,i) = V(i)\sim *\(I + B_b)^{-1} [,1]$ // [,1] means: use column 1 (the second) only for coefficients The nice aspect now is, that even for b>e^(1/e), or abs(u)>1, which leads to divergent trajectories, the eigenvalues of the matrix (I+Bb)^-1 form a convergent sequence (well, I've to check for the precise range) and AS() gives a reasonable result. I've given a plot for some bases, where x=1 (instead of I as in the case here), which compares the (cesaro-and) Euler-summable bases and the Shanks-summable bases with values of the diagonalization-summable bases, where the diagonalization extends the summablity to arbitrary high bases. (Note: this is an old picture, for instance I used "s" for base and "matrix-method" for "diagonalization") [attachment=77] The result looks pretty smooth... Because of the nice eigenvalue-configuration, you may arrive at these results even without fixpoint-shifts and expression by the Ut-matrix; simply invert -using an empirical Bb-matrix- I+Bb of reasonable size, say 32x32 or 64x64, to get good approximations. Gottfried Gottfried Helms, Kassel « Next Oldest | Next Newest »

 Messages In This Thread Fractals from calculations of 2^I, 2^(2^I), 2^(2^(2^I).. a^(a^(...a^I) - by Ivars - 05/19/2008, 09:02 PM RE: Calculations of 2^I, 2^(2^I), 2^(2^(2^I).. a^(a^(...a^I) - by bo198214 - 05/20/2008, 06:12 AM RE: Calculations of 2^I, 2^(2^I), 2^(2^(2^I).. a^(a^(...a^I) - by Ivars - 05/20/2008, 06:43 AM RE: Calculations of 2^I, 2^(2^I), 2^(2^(2^I).. a^(a^(...a^I) - by Ivars - 05/20/2008, 04:46 PM RE: Calculations of 2^I, 2^(2^I), 2^(2^(2^I).. a^(a^(...a^I) - by Ivars - 05/20/2008, 04:59 PM RE: Fractals from calculations of 2^I, 2^(2^I), 2^(2^(2^I).. a^(a^(...a^I) - by Ivars - 05/20/2008, 08:36 PM RE: Fractals from calculations of 2^I, 2^(2^I), 2^(2^(2^I).. a^(a^(...a^I) - by Ivars - 05/20/2008, 09:08 PM RE: Fractals from calculations of 2^I, 2^(2^I), 2^(2^(2^I).. a^(a^(...a^I) - by bo198214 - 05/21/2008, 06:33 AM RE: Fractals from calculations of 2^I, 2^(2^I), 2^(2^(2^I).. a^(a^(...a^I) - by Ivars - 05/21/2008, 10:29 AM RE: Fractals from calculations of 2^I, 2^(2^I), 2^(2^(2^I).. a^(a^(...a^I) - by Ivars - 05/22/2008, 08:13 PM RE: Fractals from calculations of 2^I, 2^(2^I), 2^(2^(2^I).. a^(a^(...a^I) - by Ivars - 05/23/2008, 06:51 AM RE: Fractals from calculations of 2^I, 2^(2^I), 2^(2^(2^I).. a^(a^(...a^I) - by bo198214 - 05/23/2008, 11:44 AM RE: Fractals from calculations of 2^I, 2^(2^I), 2^(2^(2^I).. a^(a^(...a^I) - by Ivars - 05/24/2008, 08:13 AM RE: Fractals from calculations of 2^I, 2^(2^I), 2^(2^(2^I).. a^(a^(...a^I) - by bo198214 - 05/24/2008, 08:55 AM RE: Fractals from calculations of 2^I, 2^(2^I), 2^(2^(2^I).. a^(a^(...a^I) - by Ivars - 05/24/2008, 04:09 PM RE: Fractals from calculations of 2^I, 2^(2^I), 2^(2^(2^I).. a^(a^(...a^I) - by Ivars - 05/25/2008, 08:28 AM RE: Fractals from calculations of 2^I, 2^(2^I), 2^(2^(2^I).. a^(a^(...a^I) - by Gottfried - 05/25/2008, 10:48 AM RE: Fractals from calculations of 2^I, 2^(2^I), 2^(2^(2^I).. a^(a^(...a^I) - by Ivars - 05/25/2008, 06:08 PM RE: Fractals from calculations of 2^I, 2^(2^I), 2^(2^(2^I).. a^(a^(...a^I) - by Gottfried - 05/25/2008, 10:12 PM RE: Fractals from calculations of 2^I, 2^(2^I), 2^(2^(2^I).. a^(a^(...a^I) - by Ivars - 05/26/2008, 07:43 AM RE: Fractals from calculations of 2^I, 2^(2^I), 2^(2^(2^I).. a^(a^(...a^I) - by Gottfried - 05/26/2008, 09:01 AM RE: Fractals from calculations of 2^I, 2^(2^I), 2^(2^(2^I).. a^(a^(...a^I) - by Ivars - 05/26/2008, 10:30 AM RE: Fractals from calculations of 2^I, 2^(2^I), 2^(2^(2^I).. a^(a^(...a^I) - by Ivars - 05/26/2008, 01:11 PM RE: Fractals from calculations of 2^I, 2^(2^I), 2^(2^(2^I).. a^(a^(...a^I) - by Gottfried - 05/25/2008, 10:51 PM RE: Fractals from calculations of 2^I, 2^(2^I), 2^(2^(2^I).. a^(a^(...a^I) - by Ivars - 05/27/2008, 08:10 AM RE: Fractals from calculations of 2^I, 2^(2^I), 2^(2^(2^I).. a^(a^(...a^I) - by Gottfried - 05/27/2008, 09:23 AM RE: Fractals from calculations of 2^I, 2^(2^I), 2^(2^(2^I).. a^(a^(...a^I) - by Ivars - 05/27/2008, 11:45 AM RE: Fractals from calculations of 2^I, 2^(2^I), 2^(2^(2^I).. a^(a^(...a^I) - by tommy1729 - 04/11/2014, 10:54 PM RE: Fractals from calculations of 2^I, 2^(2^I), 2^(2^(2^I).. a^(a^(...a^I) - by Gottfried - 04/12/2014, 01:10 AM

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