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 [Regular tetration] [Iteration series] norming fixpoint-dependencies Gottfried Ultimate Fellow     Posts: 786 Threads: 121 Joined: Aug 2007 07/28/2010, 03:04 PM (This post was last modified: 12/10/2012, 10:47 PM by Gottfried.) If we do regular tetration having a "nice" base 1y1         x2->y2       x3->y3``` For some x1 of seg1 we may iterate with infinite height to some y1, but y1 will remain in seg1. However, negative heights may result in imaginary results or even in the log(0)-singularity, so I didn't include that range in the sketch above. For some x2 in seg2 we can iterate with arbitrary heights to some y2, but again are confined to y2 in seg2. For some x3 in seg3 we can iterate with arbitrary negative height to some y3 and -in principle- also to arbitrary positive heigt, but practically encounter numerical overflow very soon. The powerseries for regular tetration can be developed around fp0 or fp1. Let's call that tet0 and tet1 for shortness. Then if we look at the schroeder-function of tet0 for all x1 we get negative values and for all x2 we get positive values. Thus we can map the set of seg1 to that of seg2 by negating the schroeder-value. This means for instance, that x1 = 1 gets mapped to x2 = 2.467914... and because the height-limits of seg2 are -infinity and +infinity we could use that value x2 to define a norm for that segment, so in seg2 the value x2=2.467914 could be said has (real) height 0 by definition. We have a similar problem with seg3: here too we have infinity at both height-limits. But we can repeat the norming-process, now using the schroeder-values of tet1. We compute the schroeder-value of x2 using tet1 and compute x3 by negating that value. However, without further measures we get infinity here. If we reduce the height of x2 by 2, then we get x3 = 417.234406762 So we have the segments with the normed heights Code:```.            seg1                   seg2                 seg3 .     --------------------|-----------------------|------------- .             .. 1 ..   (fp0)   ..              (fp1)   ..            (+oo) .             .. 1 ..    (2)    .. 2.467 ...     (4)    .. 417 ...    (+oo)     // b=sqrt(2) .      -------------------|-----------------------|------------- .              h(x1)=0  oo|oo     h(x2)=0      -oo|***************    // set norms for tet0 .        *****************|oo     h(x2)=0      -oo|-oo  h(x3)=?=-2    // set norms for tet1 .``` Unfortunately this has two asymmetries: the tet0 and tet1 have somehow opposite sign; but more inconvenient is, that we cannot have the same height-norm. What we can do is to shift left and use x1=0 as reference. We get then Code:```.         x1 = 0     x2 = 2.606584    (x3=417.2344) . .            seg1                   seg2                 seg3 .     --------------------|-----------------------|------------- .        (0)  .. 0 ..   (fp0)   ..              (fp1)   ..            (+oo) .        (0)  .. 0 ..    (2)    .. 2.606 ...     (4)    .. 417 ...    (+oo)     // b=sqrt(2) .      -------------------|-----------------------|------------- .             h0(x1)=-1 oo|oo    h0(x2)=-1     -oo|***************    // set norms for tet0 .        *****************|oo    h1(x2)=-1     -oo|-oo h1(x3)=?=-2    // set norms for tet1 .``` and still x3 computed by x2 seems to become infinite. We may reduce again x2 by height 1 to get the usable x3-value of 417.2344... We cannot reduce x1 by one more height, but my proposal here is to use x1 = b^^-1.5 as reference value. Then we have, for base b=sqrt(2) the reference-values for height -1.5 Code:```.       x1 = -1.33729937324   x2 =  2.68345013524     x3 =  3465302.30778           . .            seg1                   seg2                 seg3 .     -----------------------|-----------------------|------------- .             ..      ..   (fp0)   ..              (fp1)   ..             (+oo) .             ..-1.33 ..    (2)    .. 2.683 ...     (4)    .. 3465302. ...(+oo)     // b=sqrt(2) .      ----------------------|-----------------------|------------- .           h0(x1)=-1.5    oo|oo    h0(x2)=-1.5   -oo|***************    // set norms for tet0 .          ******************|oo    h1(x2)=-1.5   -oo|-oo h1(x3)=-1.5    // set norms for tet1 .``` The inversion of sign of the schroeder-function-value is essentially the iteration with an imaginary height. For notation I introduce now u0 = ln(fp0) and u1 = ln(fp1) If we have, for some x, the schroeder-value s, then the schroeder-value of the h'th iterate of x is s*u^h and the negation of sign can be achieved by supplying the according complex value in h. Using the different fixpoints and different u0 and u1 we can state this norming more explicitely Code:```.        x1 = tet0(1,   -1.5)     .        x2 = tet0(x1, Pi*I/ln(u0)) .        x3 = tet1(x2, Pi*I/ln(u1))```which define the heights -1.5 for the two tetrations in the three segments. What is now interesting is, whether the observed wobbling of the tetrates in seg2 using the different fixpoints changes in some interesting way. I remember that the shifting of the height by a half-unit made some significant change in the wobbling when I considered the infinite alternating iteration series (tetra-series) in one of my older msgs, I'll have a look at it soon. Gottfried Gottfried Helms, Kassel Gottfried Ultimate Fellow     Posts: 786 Threads: 121 Joined: Aug 2007 07/28/2010, 09:49 PM (This post was last modified: 07/29/2010, 01:48 AM by Gottfried.) I have added a plot of the principle of norming. For the plot I took my old norming where I identify height h=0 in the middle segment by Code:```.    x0 = 1                                     // height(x0) is zero y initial definition    x1 = tet0 (x0,  Pi*i / ln(fp0) )      // real part of height(x2) is zero       x1_2 = log_b(log_b(x1))            // shift height of x1 to avoid overflow in the next    x2 = tet1 (x1_2, Pi*I/ln(fp1))     // height(x2) = -2  using tet1-functions (using fp1)``` Gottfried Gottfried Helms, Kassel bo198214 Administrator Posts: 1,391 Threads: 90 Joined: Aug 2007 07/29/2010, 04:53 AM (07/28/2010, 09:49 PM)Gottfried Wrote: I have added a plot of the principle of norming. Ah, that makes your idea more accessible. Gottfried Ultimate Fellow     Posts: 786 Threads: 121 Joined: Aug 2007 07/29/2010, 11:36 AM (This post was last modified: 07/29/2010, 11:43 AM by Gottfried.) Another picture which shows the wobble of different values for the regular height-function when fp0 or fp1 is used. Example: base b=sqrt(2). Then the value "normzero" is the map of 1 into the segment seg1 between 2 and 4 using fp0-powerseries and has reference-height 0 in that segment. It is that value of 2.46... = tet0(1,Pi*I/ln(fp0)) To have computations numerically nearer at the fixpointvalue 2, I increase its height (tetrate it using tet0) by 13.5. Then I generate a set of x-coordinates in small steps in the height-interval hgh0(x)= 13... 17. Now I determine the heights of these x-coordinates using the hgh1()-function which employs the second fixpoint fp1. Then the height-values using hgh0(x) and hgh1(x) differ periodically by small differences of about 1e-25. This is the basic idea of the curves in the plot. But we find, that the norming process has more implications. If we connect the tet0 and tet1-functions using a common x at a fractional iterate from "normzero", then the difference-curve becomes asymmetric. Examples: if we use the connection-value at tet0(normzero,+0.25) all differences are positive, if at tet0(normzero,+0.5) we have nearly the same curve as with tet0(normzero,0) itself, and if we connect tet0 and tet1 at tet0(normzero,+0.75) all differences become negative. So the selection of the connection-point for the norming is an important aspect. However, the matter is not yet satisfactorily solved: still we have a small (but seemingly constant ~ 2e-26) difference of the curves for connection-point tet0(normzero,+0) and tet0(normzero,+0.5). So the wobbling is not exact the same even at half-integer steps of the connection-point. Gottfried Gottfried Helms, Kassel Gottfried Ultimate Fellow     Posts: 786 Threads: 121 Joined: Aug 2007 08/29/2010, 10:31 PM (This post was last modified: 08/31/2010, 10:05 AM by Gottfried.) Just came across an older subject and thought it would fit into this "norming"-thread. As older fellows here may remember, nearly my first contact with tetration was the question of alternating iteration-series for which I worked out some interesting heuristics. (see  and ) [update] I should explain, that for convenient ascii-notation of the tetration I "misuse" here the common notation. With z^^h I mean in the context of a given fixed base b, the value of z^^h := [/update] Using base b=sqrt(2) we have the real-valued interval 2..4 for which we may find iteration heights from -inf to +inf if we start at some value z, say z=3, in this interval. Because in both direction the values of z^^h are finite we can compute a value for the alternating series of that values. So using Pari/Gp we can compute f(z) = sumalt(h=0,(-1)^h*iter(z,h)) + sumalt(h=0,(-1)^h*iter(z,-h) ) - z to evaluate the alternating iteration-series with center at the chosen z. It is clear that this series is periodic for z in the interval z..z^^2 . But what's interesting is, that in general the f(z) is "small" and even we find f(z)=0 Because this is a remarkable result (and matches, for instance, the analogue problem when applied to a doubly-infinite geometric series by analytic continuation) this value z (where f(z)=0) introduces itself gently as candidate for a norm-value, at which the height is defined to be zero or at least an integer. Here is a picture of the sinusoidal curve f(z) when z is moved from z to z^^2 beginning at some arbitrary value z0: We see that astonishing approximation to a sine-curve, where the amplitude should be normed. Actually the deviance from the sine-curve is of the order of 1e-3 : I mean, if the height-parameter of this curve is compared with the abscissa of the sine-curve after the two curves are matched (for instance by binary search of the same y-values). I'm not experienced with Fourier-analysis, but I think, it would be profitable to try to describe the f(z)-function by a fourier-decomposition. Analoguously this could be done for the other bases 1 eta ?? Gottfried Ultimate Fellow     Posts: 786 Threads: 121 Joined: Aug 2007 08/30/2010, 10:25 AM (This post was last modified: 08/30/2010, 10:28 AM by Gottfried.) (08/30/2010, 09:32 AM)tommy1729 Wrote: let me recap A is carleman of z. B is carleman of base^z. carleman f(z) = A/(1+B) but why is 1+B invertible for bases > eta ?? Let M=(I+B), W = M^-1; ß some eigenvalue of B |ß|>1 , µ =ß+1 eigenvalue of M w = 1/(1+ß) the according eigenvalue of W Heuristically it converges when size is increased (no proof yet). I suppose: because all |ß|>=1, ==> µ =/= zero, or equivalently 0 < |1/(1+ß)| < 1 for all ß No proof yet Gottfried Helms, Kassel tommy1729 Ultimate Fellow     Posts: 1,438 Threads: 349 Joined: Feb 2009 08/30/2010, 11:08 AM ok that makes some sense. now explain me what you are doing in this thread and WHY ? and WHY is f(z) periodic ??? how to compute its period ?? this reminds me of my thread " 452 pi " where i conjectured a period. maybe related ?? regards tommy1729 Gottfried Ultimate Fellow     Posts: 786 Threads: 121 Joined: Aug 2007 08/30/2010, 12:34 PM (This post was last modified: 09/01/2010, 12:14 PM by Gottfried.) (08/30/2010, 11:08 AM)tommy1729 Wrote: ok that makes some sense. now explain me what you are doing in this thread and WHY ?Using base b= sqrt(2) what is the height-representation of 3 ? By Kousnetzov? By Robbins slog? By regular-tetration using fixpoint 2? using fixpoint 4? Besides the difficulty to determine "the" height of 3 by the above methods - there is still room to find an optimal solution for tetration. My idea gives one which is only depending on the integer heights, thus for instance independent of the selection of the fixpoint. Quote:and WHY is f(z) periodic ??? Code:```´ f(z)         = ... z^^-3 + z^^-2 - z^^-1 + z - z^^1 + z^^2 - z^^3 +... let y=b^b^z = z^^2 f(y)        = ... y^^-3 + y^^-2 - y^^-1 + y    - y^^1 + y^^2 - y^^3 +...             = ... z^^-1 + z     - z^^ 1 + z^^2 - z^^3 + ...             = f(z)``` f(z) and f(y)=f(z^^2) are equal => f(z) is periodic in terms of the height-parameter Quote:how to compute its period ?? f(z) = f(z^^2) where z^^2 = b^b^z ==> period-length is delta_h = 2 It is more difficult to find the amplitude Gottfried Helms, Kassel tommy1729 Ultimate Fellow     Posts: 1,438 Threads: 349 Joined: Feb 2009 08/30/2010, 11:14 PM ahh i think .... f(z) is NOT periodic , ONLY in the interval [z,z^^2] IF f(z) = 0. meaning that not only f(z) = f(z^^2) = 0 but also f(z+(z^^2-z)/2) = = f(z/2 + z^^2 /2) = f(z) = 0. hmm. f(z) = f(z^^2). for all z. f(z) = - f(z^z) right ? but if f(z) = f(z^^2) = 0 then - f(z^z) must be 0 but z^z is not in the middle of z and z^^2. i.e. z<> z^z => z^z =/= (z + z^^2)/2. or ... is that the equation to solve for f(z) = 0 ?? man your f(z) is weird ! « Next Oldest | Next Newest »

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