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 the inconsistency depending on fixpoint-selection Gottfried Ultimate Fellow Posts: 787 Threads: 121 Joined: Aug 2007 03/04/2008, 11:32 AM bo198214 Wrote:Hey Gottfried, it is a quite well known result, that in most cases the regular iteration of an analytic function at different fixed points gives different (usually only slightly differing) functions. Ecalle referred me even to an article about this phenomenon "Etude theorique et numerique de la fonction de Karlin-McGregor", Serge Dubuc, Journal d'Analyse Math.. Vol. 42 (1982 / 83)Is this also available in english? (I don't speak french... ) Well, it should be not too difficult to construct examples... Quote:and I checked it numerically in the bummer thread. ...as you did it. But I didn't recognize a final conclusion, usable for our case. Damn, I've much too little experience... So this requires more reading and testing first. Thanks for the hints! Gottfried Gottfried Helms, Kassel bo198214 Administrator Posts: 1,391 Threads: 90 Joined: Aug 2007 03/04/2008, 11:43 AM Gottfried Wrote:Quote:and I checked it numerically in the bummer thread. ...as you did it. But I didn't recognize a final conclusion, usable for our case. Damn, I've much too little experience... Hm, the conclusion is a) dont trust numerical comparisons b) We still have a uniqueness problem if we base the tetration extension on regular iteration of $b^x$ at a fixed point (i.e. which fixed point to choose). This is besides the problem that there are no real fixed points for $b>e^{1/e}$. Gottfried Ultimate Fellow Posts: 787 Threads: 121 Joined: Aug 2007 03/06/2008, 04:06 PM (This post was last modified: 03/06/2008, 05:15 PM by Gottfried.) Hi Henryk - I'm rereading the "bummer"-posts and I just tested your example-function $\hspace{24} f(x) = x^2 + x - 1/16$ with my matrix-analysis. However, it is not clear to me, * how the different fixpoints are involved here (update:but maybe I've got it now) * what the plot in your first msg describes exactly. What I've got so far is a good result for the half-iterate, which seems to be exact to more than 20 digits. I've also used a shifting of x->x' to get a triangular matrix with exact terms, where I denote $\hspace{24} \begin{matrix} x' &=& x/(1/4) - 1 \\ x'' &=& (x+1)*1/4 \\ g(z)&=& 1/4 z^2 + 3/2 z \end{matrix} $ and with the above then $\hspace{24} f(x) = g(x')''$ The matrix-operator G for g(x) is then triangular. Results: $\hspace{24} \begin{matrix} {llll} & f^{(h)}(x) & g^{(h)}(x) \\ h=0 & 1.0000000000000000000 & 3 \\ h=0.5 & 1.3427402879772577347 & 4.3709611519090309387 \\ h=1 & 1.9375000000000000000 & 6.7500000000000000000 \end{matrix}$ [update] If I understood the fix-point-problem correctly, then the other transformations x', x", and the second function g2(x) are $\hspace{24} \begin{matrix} x' &=& x/(-1/4) + 1 \\ x'' &=& (x-1)*(-1/4) \\ g2(z)&=& -1/4 z^2 + 3/2 z \end{matrix} $ Then I get identical results for the back-shifted solutions (checked up to 20 digits): g2(x) = $\hspace{24} \begin{matrix} {r} 1.0000000000000000000 & -3 \\ 1.3427402879772577347 & -4.3709611519090309387 \\ 1.9375000000000000000 & -6.7500000000000000000 \end{matrix}$ ---------------------------------------------------------------------- Documents: Matrix-operator F for f(x) Code:´   1  -1/16  1/256  -1/4096  1/65536  -1/1048576   0      1   -1/8    3/256  -1/1024     5/65536   0      1    7/8  -45/256  23/1024  -155/65536   0      0      2      5/8   -13/64     35/1024   0      0      1    45/16   35/128   -405/2048   0      0      0        3     13/4     -17/128Because this operator is not rowfinite, I use the substitution x->x' and f(x)=g(x')" , replacing the original matrix-multiplication $\hspace{24} V(x)\sim * F = V(f(x))\sim$ by $\hspace{24} V(x')\sim * G = V(g(x'))\sim$ and finally apply re-substitution. Matrix-operator G for g(x) Code:´   1    .     .      .      .       .   0  3/2     .      .      .       .   0  1/4   9/4      .      .       .   0    0   3/4   27/8      .       .   0    0  1/16  27/16  81/16       .   0    0     0   9/32   27/8  243/32 The fractional powers for g(x) are expressed by fractional powers of G, so we need the eigensystem-decomposition G = W * D * W^-1 W = Code:´   1         .        .      .     .  .   0         1        .      .     .  .   0      -1/3        1      .     .  .   0      2/15     -2/3      1     .  .   0   -49/855    17/45     -1     1  .   0  158/6175  -58/285  11/15  -4/3  1 D = diag() (which is a divergent sequence of eigenvalues!) Code:´   1  3/2  9/4  27/8  81/16  243/32 W^-1 = Code:´   1            .         .    .    .  .   0            1         .    .    .  .   0          1/3         1    .    .  .   0         4/45       2/3    1    .  .   0      52/2565     13/45    1    1  .   0  2048/500175  256/2565  3/5  4/3  1 The half-power G^0.5 = Code:´   1.0000000               .              .              .           .          .           0       1.2247449              .              .           .          .           0     0.091751710      1.5000000              .           .          .           0   -0.0067347010     0.22474487      1.8371173           .          .           0    0.0010135508  -0.0080782047     0.41288269   2.2500000          .           0  -0.00019936640   0.0012468417  0.00062493491  0.67423461  2.7556760 And using the second column of G^0.5 with x'=3 according to $\hspace{24} V(x')\sim * G^{0.5}$ we get the diverging sequence of terms of the powerseries (first 64 terms) Code:´                0        3.6742346       0.82576539      -0.18183693      0.082097618     -0.048446035      0.032977002     -0.024500419      0.019276236 ...       -1.1067517        2.6939736       -5.1977227        9.0095267       -14.654213        22.823197       -34.410169        50.545198       -72.619728        102.28920       -141.43120        192.02334       -255.88447        334.19212       ... which seems to be Euler-summable to $\hspace{24} V(3)\sim * G^{0.5} = V(4.3709611519090309387...)\sim$ so $\hspace{24} g^{(0.5)}(3)=4.3709611519090309387...$ The next iteration is even more divergent, but again seems to be Euler-summable to get $\hspace{24} V(4.3709611519090309387...)\sim * G^{0.5} = V(6.750)\sim$ The backtransformation of these values gives $\hspace{24} \begin{matrix} 3'' &=& 1 \\ 4.37096115...'' &=& 1.3427... \\ 6.750'' &=& 1.9375 \end{matrix}$ Appendix: [update] The half-power G2^0.5 (developed using the second fixpoint) is Code:´   1.0000000               .              .              .            .          .           0       1.2247449              .              .            .          .           0    -0.091751710      1.5000000              .            .          .           0   -0.0067347010    -0.22474487      1.8371173            .          .           0   -0.0010135508  -0.0080782047    -0.41288269    2.2500000          .           0  -0.00019936640  -0.0012468417  0.00062493491  -0.67423461  2.7556760 where apparently only the signs have changed, but the numerical digits are the same up to the 20'th digit. ======================================================================== Now, at some place I need the idea, how to introduce the other fixpoint correctly (if I didn't get it this way), and then, what computation your plot shows. Gottfried Gottfried Helms, Kassel bo198214 Administrator Posts: 1,391 Threads: 90 Joined: Aug 2007 03/06/2008, 04:58 PM Gottfried Wrote:What I've got so far is a good result for the half-iterate, which seems to be exact to more than 20 digits. I've also used a shifting of x->x' to get a triangular matrix with exact terms, where I denote $\hspace{24} \begin{matrix} x' &=& x/(1/4) - 1 \\ x'' &=& (x+1)*1/4 \\ g(z)&=& 1/4 z^2 + 3/2 z \end{matrix}$ and with the above then $\hspace{24} f(x) = g(x')''$ Wah thats strong tobacco, I just had used a *translation*, for example $x' = x-\frac{1}{4}$ $x'' = x+\frac{1}{4}$ this $g_1(x)=f(x')''$ is $f$ with the fixed point $\frac{1}{4}$ moved to 0. And $g_2(x)=f(x'')'$ corresponds to moving the fixed point $-\frac{1}{4}$ to 0. Computing the half iterate with your triangular matrices corresponds to computing the regular fractional iterate at the corresponding fixed point (to get $f^{\circ 1/2}$ you of course have to move back the fixed point 0 of $g_1$ or $g_2$ to its original place. The half iterate at both fixed points is different. And from here you already see variants: 1. You dont need to use a translation. You can use any bijective (analytic) function $\tau$ which moves the fixed point to 0 (you used a *linear* transformation). You then take the fractional iterate of $g=\tau^{-1}\circ f\circ \tau$ and get the fractional iteration of $f$ as $f^{\circ t}=\tau\circ g^{\circ t}\circ \tau^{-1}$. 2. The matrix operator method is more general. It does not only work at fixed points but also in between where you use the triangulated finite matrices as approximating sequence. So you can use *any* translation and then apply the matrix operator method. And it is quite sure that all those results differ (by a super small amount). 3. And then you can go even a step further and use any (analytic) bijective $\tau$ before applying the matrix operator method. Gottfried Ultimate Fellow Posts: 787 Threads: 121 Joined: Aug 2007 03/06/2008, 05:25 PM (This post was last modified: 03/06/2008, 05:26 PM by Gottfried.) Upps - my updating and your answering crossed. I've included the fixpoint-2 computation. Well, I developed the substitution-formula using x'=x/e - d with the only restriction, that e*d = 1/4. I chose d=1. Your selection were e=1 and d=1/4 , I think I wouldn't get different results by this. Since you got differing results for the different fixpoints with your analysis: in which digits/which precision did the difference occur? I'd like to increase the numerical precision in my analysis to that degree to check this... Gottfried Gottfried Helms, Kassel Gottfried Ultimate Fellow Posts: 787 Threads: 121 Joined: Aug 2007 03/07/2008, 03:16 AM Hi Henryk - I gave it some more time to document the two fixpoint-solutions. Here is the eigendecomposition in a synopsis , left for using fixpoint t0=1/4, right for using fixpoint t1=-1/4. Definitions according to previous post: Code:´ G0^h = W0 * D0^h * W0^-1 G1^h = W1 * D1^h * W1^-1 W0=                                          |W1=   1         .        .      .     .  .       |    1         .       .      .    .  .           0         1        .      .     .  .       |    0         1       .      .    .  .           0      -1/3        1      .     .  .       |    0       1/3       1      .    .  .           0      2/15     -2/3      1     .  .       |    0      2/15     2/3      1    .  .           0   -49/855    17/45     -1     1  .       |    0    49/855   17/45      1    1  .           0  158/6175  -58/285  11/15  -4/3  1       |    0  158/6175  58/285  11/15  4/3  1         D0=                                          |D1=                                                   1  3/2  9/4  27/8  81/16  243/32           |    1  3/2  9/4  27/8  81/16  243/32           W0^-1=                                       |W1^-1                                                                                             1            .         .    .    .  .      |    1            .          .    .     .  .     0            1         .    .    .  .      |    0            1          .    .     .  .     0          1/3         1    .    .  .      |    0         -1/3          1    .     .  .     0         4/45       2/3    1    .  .      |    0         4/45       -2/3    1     .  .     0      52/2565     13/45    1    1  .      |    0     -52/2565      13/45   -1     1  .     0  2048/500175  256/2565  3/5  4/3  1      |    0  2048/500175  -256/2565  3/5  -4/3  1   G0=                                          |G1=                                                   1    .     .      .      .       .         |    1     .     .       .      .       .         0  3/2     .      .      .       .         |    0   3/2     .       .      .       .         0  1/4   9/4      .      .       .         |    0  -1/4   9/4       .      .       .         0    0   3/4   27/8      .       .         |    0     0  -3/4    27/8      .       .         0    0  1/16  27/16  81/16       .         |    0     0  1/16  -27/16  81/16       .         0    0     0   9/32   27/8  243/32         |    0     0     0    9/32  -27/8  243/32 Here the half power-matrices: Code:´ G0^0.5=           1.0000000               .              .              .           .          .           0       1.2247449              .              .           .          .           0     0.091751710      1.5000000              .           .          .           0   -0.0067347010     0.22474487      1.8371173           .          .           0    0.0010135508  -0.0080782047     0.41288269   2.2500000          .           0  -0.00019936640   0.0012468417  0.00062493491  0.67423461  2.7556760 G1^0.5   1.0000000               .              .              .            .          .           0       1.2247449              .              .            .          .           0    -0.091751710      1.5000000              .            .          .           0   -0.0067347010    -0.22474487      1.8371173            .          .           0   -0.0010135508  -0.0080782047    -0.41288269    2.2500000          .           0  -0.00019936640  -0.0012468417  0.00062493491  -0.67423461  2.7556760 What we see is, that, using J = diag(1,-1,1,-1,...) Code:´   W1    = J * W0    * J   W1^-1 = J * W0^-1 * J   D1    = D0     = Dand consequently Code:´ G1 = W1 * D * W1^-1 = J * W0 * D * W0^-1 * J     = J * G0 * Jand this is also valid for powers (by powers of D). Now the transformations of x->x' for G0 and x->x´ are Code:´ using e1=1/4                                e2 = -1/4         d1=1/4 / e1 =1                        d2 = 1/4 / e2 = -1 x' =  4x - 1                               x´ = -4x + 1 x" =  (x+1)/4                              x´´= (x-1)/-4 But since the two substitutions for g0(x) and g1(x) mean, that x´ = - x' we have Code:´   V(x´)~ * G1 = V(-x')~ * J G0 J               = V(x')~ * G0 * J and   g1(x´) = - g0(x')Then also Code:´   g1(x´)´´ = (-g0(x'))´´            = (-g0(x')-1)/-4            = (g0(x')+1)/4            = g0(x')"and we have identity for the height 1 for the two fixpoint-solutions. Since the two diagonals D0 and D1 are equal, also their powers are equal and thus the whole matrix-expression must lead to equal results. The half-iterate-matrices show the irrational numbers based on b=sqrt(3/2); but since all coefficients of b are rational I don't see space for relevant approximation-errors or result-differences. So since this -well: basic deal- seems to give equality. It would be good, if I could go further and get into the computations, with which you found the discrepancies. Gottfried Gottfried Helms, Kassel bo198214 Administrator Posts: 1,391 Threads: 90 Joined: Aug 2007 03/07/2008, 10:56 AM Gottfried Wrote:Code:G0=                                          |G1=                                                   1    .     .      .      .       .         |    1     .     .       .      .       .         0  3/2     .      .      .       .         |    0   3/2     .       .      .       .         0  1/4   9/4      .      .       .         |    0  -1/4   9/4       .      .       .         0    0   3/4   27/8      .       .         |    0     0  -3/4    27/8      .       .         0    0  1/16  27/16  81/16       .         |    0     0  1/16  -27/16  81/16       .         0    0     0   9/32   27/8  243/32         |    0     0     0    9/32  -27/8  243/32 But Gottfried, you didnt use the translation! For applying the regular iteration you have to *move* the fixed point to zero. The function graph is just translated so that the fixed point is situated at 0. No stretching or whatever is allowed. Here that would be $g_0(z)=(z-\frac{1}{4})^2+ (z-\frac{1}{4})-\frac{1}{16} + \frac{1}{4}=z^2-2\frac{1}{4}z+\frac{1}{16}+z-\frac{1}{4}-\frac{1}{16}+\frac{1}{4}=z^2+\frac{1}{2}z$ $g_1(z)=(z+\frac{1}{4})^2+(z+\frac{1}{4})-\frac{1}{16} - \frac{1}{4} = z^2+2\frac{1}{4}z+\frac{1}{16}+z+\frac{1}{4}-\frac{1}{16}-\frac{1}{4}=z^2+\frac{3}{2}z$ This is because regular half iteration is characterized as *the* half iteration that has at the fixed point $a$ the slope $f'(a)^{1/2}$. If you change the slope at the fixed point by the transformation, and you do by your transformation $g_1'(0)=(-\frac{1}{4}z^2+\frac{3}{2}z)'|_{z=0}=\frac{3}{2}\neq\frac{1}{2}=(2z+1)|_{z=-\frac{1}{4}}=f'(-\frac{1}{4})$, it is no more *regular* half iteration. Gottfried Ultimate Fellow Posts: 787 Threads: 121 Joined: Aug 2007 03/07/2008, 07:17 PM (This post was last modified: 03/07/2008, 07:30 PM by Gottfried.) Well, Henryk, that opended eyes, thanks. With the following parametrization I got the difference. Starting with the general x' = x/e-d, x"=(x+d)*e, e*d=fixpoint and fixing e=1 e= 1 , d0=t0=fixpoint_0, d1=t1=fixpoint_1 and development of the matrices accordingly I got differences, too. So thanks for the hints! Now I'd like to understand, what is the source of the difference of selection e=1 vs d=1. The most obvious difference was in the given case f(x) = x^2 + x - 1/16 = x + (x - 1/4) (x + 1/4) = x + (x-t0)*(x-t1) that I kept the eigenvalues in both developments equal. Maybe my transformation introduces somehow a triviality - but note, that in my tetration-examples I also used d=1, e0=t0 versus e1=t1 for the definition of the triangular matrices, just this selection, that you say is "illegal" I tried with a general definition of a quadratic function (fixpoints p,q) f(x) = x + (x - p) (x - q) = x^2 + (1 - (p+q))x + pq g(z) = a*z^2 + b*z z = x/e-d f(x) = (g(z) + d)*e By comparing coefficients at powers of x in f(x) and g(z) I get conditions for a and b in g(z): a = e b = 1 - (p+q) + 2de , b-1 = 2de - (p+q) and then for d and e the condition occurs (de - p) (de - q) = 0 where either e or d is free. It seems then, if b equals for each selection of fixpoints, such that either de=p or de=q and the resulting b's are equal, then the resulting powerseries seem to be compatible and give equal results (I've to check that with more examples) But again: maybe this introduces some triviality... On the other hand: it is intriguing, that using d=1 , e=fixpoint is the appropriate selection to convert the tetration-matrices into triangular matrices for decremented iterated exponentiation. Hmmm. Gottfried Gottfried Helms, Kassel bo198214 Administrator Posts: 1,391 Threads: 90 Joined: Aug 2007 03/07/2008, 08:30 PM (This post was last modified: 03/09/2008, 08:51 PM by bo198214.) Gottfried Wrote:Now I'd like to understand, what is the source of the difference of selection e=1 vs d=1. The most obvious difference was in the given case Hm, I had another look at your transformations (which indeed puzzled me). I came up with this: Your first transformation: $x' = 4 x -1$, $x'' = \frac{1}{4} x + \frac{1}{4}$, $g_0(x)=f(x'')'$ gives the same result as the regular iteration at the upper fixed point, because: $x'' = \tau\circ \mu(x)$ where $\tau(x)=x+\frac{1}{4}$ and $\mu(x)=\frac{1}{4}x$ so $g_0(x)=\mu^{-1}\circ \tau^{-1}\circ f\circ\tau\circ\mu=\mu^{-1}\circ h_0 \circ \mu$ and $h_0$ is the function with upper fixed point moved to 0. But now it is easily derivable that the regular iteration $(\mu^{-1}\circ h\circ \mu)^{\circ t}=\mu^{-1}\circ h^{\circ t}\circ \mu$, where $h^{\circ t}$ is the regular iteration of $h$. This has to do with that the regular iteration is the only one which posses the limit $\lim_{x\downarrow 0} (f^{\circ t})'(x)=\left(\lim_{x\downarrow 0} f'(x)\right)^t$. So we have $f_0^{\circ t}= \tau\circ \mu\circ g_0^{\circ t}\circ \mu^{-1}\circ \tau^{-1}=\tau\circ h_0^{\circ t}\circ \tau^{-1}$. However your second transformation $x' = -4 x +1$, $x'' = -\frac{1}{4} x + \frac{1}{4}$, $g_1(x)=f(x'')'$ has as translation $\tau(x)=x+\frac{1}{4}$ the same translation as in your previous transformation! Only $\mu$ changes to $\mu(x)=-\frac{1}{4}x$. So this second transformation moves also the *upper* fixed point to 0. And we have seen that we can dismiss any multiplicative conjugation $\mu^{-1}\circ f\circ \mu$ with respect to regular iteration. So it has of course the same result (fractional iteration of $f$) as your first transformation. Gottfried Ultimate Fellow Posts: 787 Threads: 121 Joined: Aug 2007 03/07/2008, 09:38 PM (This post was last modified: 03/07/2008, 09:46 PM by Gottfried.) bo198214 Wrote:However your second transformation $x' = -4 x +1$, $x'' = -\frac{1}{4} x + \frac{1}{4}$, $g_1(x)=f(x'')'$ has as translation $\tau(x)=x+\frac{1}{4}$ the same translation as in your previous transformation! Only $\mu$ changes to $\mu(x)=-\frac{1}{4}x$. So this second transformation moves also the *greater* fixed point to 0. And we have seen that we can dismiss any multiplicative conjugation $\mu^{-1}\circ f\circ \mu$ with respect to regular iteration. So it has of course the same result (fractional iteration of $f$) as your first transformation. Yepp - I just found the position of the error, grmmpff. It is in Code:´ using e1=1/4                                e2 = -1/4         d1=1/4 / e1 =1                        d2 = 1/4 / e2 = -1 x' =  4x - 1                               x´ = -4x + 1 x" =  (x+1)/4                              x´´= (x-1)/-4 Although I had already that not e or d, but e*d = fixpoint, in the computation of d2, I used the same (1/4) as for d1. For clarity, I should have written a symbol for the selected fixpoint: Code:´ involving fixpoint p=1/4                    fixpoint q=-1/4 using e1=1/4                                e2 = -1/4         d1=p / e1 =1                          d2 = q / e2 = 1 x' =  4x - 1                               x´ = -4x - 1 x" =  (x+1)/4                              x´´= (x+1)/-4 Hope this corrects things, didn't check it again yet, but it seems that this reflects just the simple sign-changing in the matrices. The previous G2-matrix and the subsequent considerations in that posting are on a false assumption... Gottfried Gottfried Helms, Kassel « Next Oldest | Next Newest »

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