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 Solving tetration for base 0 < b < e^-e Gottfried Ultimate Fellow Posts: 764 Threads: 118 Joined: Aug 2007 09/13/2009, 06:14 AM (09/12/2009, 09:04 PM)mike3 Wrote: So then since neither of those worked, it seems all we're left with is the Ansus formula and the Cauchy integral (but determining the correct contours and asymptotic behavior, now that's the rub...). I'm also curious: What about $b = e^{-e}$ exactly? You said it converges slowly, but how do you iterate it at all? What is the asymptotic as the tower $x \rightarrow \infty$? I used a complex fixpoint and could generate the matrices for regular iteration (in context of diagonalization). The fixpoint I used is t0 = -0.1957457524880764 - 1.691199920910569*I one of its logarithms is u0 = 0.5320921219863799 + 4.597158013302573*I where u0 = log(t0) + 2*Pi*I // log giving the principal branch With this I create the triangular Bell-matrix and diagonalize. The series has complex terms and is very difficult to evaluate - I accelerate slow converging series usually with Euler-summation, but the series has complex terms and it seems I need also complex order for Euler-summation. With 128 terms I could at least get results which reproduced the integer iteration to such an approximate that I'm confident that the series can be used in principle. However, the fractional iterates behave even worse, and two half-iterates reproduce the integer iterate just to two decimals... The schröder-term s for schr(x') and x'=x/t0 - 1 at x=1 is, according to the last three partial sums of the series (128 terms): Code:[126]  -0.4119542792176348+1.439754774257274*I [127]  -0.4119542792176264+1.439754774257268*I [128]  -0.4119542792176181+1.439754774257268*I ...where I assume s~ -0.411954279217... +1.439754774257...*I as correct decimals. From here we can compute y' = schr°-1(s * u0^h ) and with h=1 I reproduce exp_b°1(1) = b =exp(-exp(1)) to 15 digits exact. The last three partial sums of the series: Code:´ [126]   -1.004456441437337+0.03850266639539727*I [127]   -1.004456441437337+0.03850266639539727*I [128]   -1.004456441437337+0.03850266639539727*I ...From here y' = -1.004456441437337+0.03850266639539727*I, y=(y'+1)*t0 , y-b = -7.405812 E-17 + 2.57894 E-15*I is an acceptable approximation. However - it is only the integer-iteration. I've not yet found en Euler-order or another convergence acceleration which stabilizes the partial sums for the half-iterate to an acceptable degree. Gottfried Gottfried Helms, Kassel mike3 Long Time Fellow Posts: 368 Threads: 44 Joined: Sep 2009 09/13/2009, 07:24 AM Why do you need to use a complex fixed point for $b = e^{-e}$? It converges to a real fixed point as the tower approaches infinity (namely, $\frac{1}{e}$). bo198214 Administrator Posts: 1,389 Threads: 90 Joined: Aug 2007 09/13/2009, 08:08 AM (This post was last modified: 09/13/2009, 08:09 AM by bo198214.) (09/12/2009, 09:04 PM)mike3 Wrote: So then since neither of those worked, it seems all we're left with is the Ansus formula and the Cauchy integral (but determining the correct contours and asymptotic behavior, now that's the rub...). I dont think, one can say that the regular iteration didnt work. Its rather that it doesnt match your requirement to have a singularity at -2. Though this requirement is quite apparent for real-valued functions, because we necessarily use the real branch of the logarithm, the necessity is not so clear for complex valued functions, where there is free choice of the branch of the logarithm. It sounds anyway strange to prefer a singular tetrational over an entire tetrational if there is not the demand of real values. Btw.: Kouznetsov's Contour integral method is rather applicable between two conjugate non-real fixed points. I doubt that you can use the idea for a real 2-cycle. mike3 Long Time Fellow Posts: 368 Threads: 44 Joined: Sep 2009 09/13/2009, 09:49 AM (This post was last modified: 09/13/2009, 09:51 AM by mike3.) Why not use the entire? Because the behavior is not consistent. It seems stranger to imagine a function that has singularities for only real bases $b \geq e^{-e}$, then suddenly it becomes free of them and entire for all $b < e^{-e}$. It's two wildly different analytical behaviors and that doesn't make much sense. Ideally it would be nice to be able to interpret the tetration at $0 < b < e^{-e}$ to be what you'd get if you did an analytical continuation in the base from $b > e^{-e}$ through the complex plane. Finally, at the integer towers of this base, we are still dealing with real numbers: when we take the log of $^0 b = 1$ for $0 < b < e^{-e}$ to get $^{-1} b$ as 0, we are still using a real logarithm of a real number to a real base. So why not continue using this principal real logarithm for the rest of the tet function at this base? Gottfried Ultimate Fellow Posts: 764 Threads: 118 Joined: Aug 2007 09/13/2009, 09:57 AM (This post was last modified: 09/13/2009, 09:59 AM by Gottfried.) (09/13/2009, 07:24 AM)mike3 Wrote: Why do you need to use a complex fixed point for $b = e^{-e}$? It converges to a real fixed point as the tower approaches infinity (namely, $\frac{1}{e}$). The problem with the real fixpoint is, that the triangular Bell-matrices have (alternating signed) units on its diagonal. This prevents the computation of a matrix-logarithm as well of the diagonalization - at least in my implementations. If I have no option for one of those, I can approximate tetration only via diagonalization of the square Bell-matrices (means: omitting the fixpoint-shifting). But here all coefficients depend on the size of the used matrix, they are in my view unpredictable and may only serve as rough approximations for a "first impression". But well, let's see. It's surely not the highest summit of wisdom... and we also have the Newton-binomial-formula and others... Gottfried Gottfried Helms, Kassel bo198214 Administrator Posts: 1,389 Threads: 90 Joined: Aug 2007 09/13/2009, 11:23 AM (09/13/2009, 09:57 AM)Gottfried Wrote: The problem with the real fixpoint is, that the triangular Bell-matrices have (alternating signed) units on its diagonal. This prevents the computation of a matrix-logarithm as well of the diagonalization - at least in my implementations. That should not pose a problem. You need the first row of $B^t$, where $B$ is the Bell matrix. You make a Jordan decomposition $B = S J S^{-1}$ and then $B^t = S J^t S^{-1}$ where for each Jordanblock $J_m$ for eigenvalue $\lambd_m$ with multiplicity $M_m$ one sets $J_m^t = \sum_{n=0}^{M_m} \left(t\\n\right) (J_m-\lambda_m I)^n$. The sum is finite because $J_m-\lambda_m I$ is nilpotent: $J_m^{M_m}=0$. Unfortunately the Jordan decompostion is flawed in Sage so I could not try it myself. Quote:But well, let's see. It's surely not the highest summit of wisdom... and we also have the Newton-binomial-formula and others... I dont think that the Newton formula helps. It is real-valued and hence can not return a suitable solution for a decreasing base function. Gottfried Ultimate Fellow Posts: 764 Threads: 118 Joined: Aug 2007 09/13/2009, 11:44 AM (09/13/2009, 11:23 AM)bo198214 Wrote: Unfortunately the Jordan decompostion is flawed in Sage so I could not try it myself. arrgh... I knew, one day I'd have to look at the jordan decomposition... why not now... Gottfried Helms, Kassel Gottfried Ultimate Fellow Posts: 764 Threads: 118 Joined: Aug 2007 09/13/2009, 01:34 PM (This post was last modified: 09/13/2009, 03:46 PM by Gottfried.) (09/13/2009, 06:14 AM)Gottfried Wrote: The series has complex terms and is very difficult to evaluate - I accelerate slow converging series usually with Euler-summation, but the series has complex terms and it seems I need also complex order for Euler-summation. With 128 terms I could at least get results which reproduced the integer iteration to such an approximate that I'm confident that the series can be used in principle. However, the fractional iterates behave even worse, and two half-iterates reproduce the integer iterate just to two decimals... The schröder-term s for schr(x') and x'=x/t0 - 1 at x=1 is, according to the last three partial sums of the series (128 terms): Code:[126]  -0.4119542792176348+1.439754774257274*I [127]  -0.4119542792176264+1.439754774257268*I [128]  -0.4119542792176181+1.439754774257268*I ...where I assume s~ -0.411954279217... +1.439754774257...*I as correct decimals. The general precision can drastically be improved if we insert a "stirling-transform" of the schröder- and the inverse schröder-function. Code:[126]  -0.4119542792176179+1.439754774257279*I [127]  -0.4119542792176179+1.439754774257279*I [128]  -0.4119542792176179+1.439754774257279*I ... ps[128]-ps[127]=-1.432629629141992 E-55 - 1.024270322737871 E-55*ISo the partial sums in that region differ only by values of order 1e-55 and we get a reliable value for the schröder-function up to at least 50 digits. We compose the coefficients of the Schröder-function using the factorially scaled Stirling-numbers 2'nd kind (just pre-multiply the Bell-matrix of the schröder-function by the Bell-matrix of exp(x)-1 and postmultiply the Bell-matrix of the inverse schröder-function by the Bell-matrix of log(1+x), in my notation fS2F*W and WI*fS1F ) If we call the new schröder-function eschr(log(1+x)) = schr(x) then I got much better precision: denote the function f(h) the so constructed sexp-function f(h) = exp_b°h(1), u0 the log of the complex fixpoint, second branch (=log(t0) + 2*Pi*I) Code:bl = log(b) // = -2.718... x' = log(x/t0) y'=eschr°-1 (u0^h*eschr(x')) y = exp(y')*t0 then f(1)     - b    = -2.244868624099733 E-60 + 3.153343574801035 E-60*I  // order for Eulersum: 1.4 - 0.1*I f(2)     - b°2  =  1.922763821393618 E-13 + 1.897277601645949 E-13*I  // order for Eulersum: 2.5-1.2*I f(-1)*bl - 0    =  4.089597929065041 E-56 + 6.283185307179586*I       // order for Eulersum: 1 (direct sum, no acceleration needed)However, as we see at the f(-1)-entry we needed a correction factor to get the correct real value - but with this we have the imaginary part differing with 2*Pi*I. I don't know yet how to include this smoothly into the formula, so it does not yet make sense to try to improve the fractional iterates with this... Gottfried Gottfried Helms, Kassel Gottfried Ultimate Fellow Posts: 764 Threads: 118 Joined: Aug 2007 09/13/2009, 07:34 PM (This post was last modified: 09/13/2009, 09:37 PM by Gottfried.) (09/13/2009, 01:34 PM)Gottfried Wrote: then I got much better precision: denote the function f(h) the so constructed sexp-function f(h) = exp_b°h(1), u0 the log of the complex fixpoint, second branch (=log(t0) + 2*Pi*I) Code:bl = log(b) // = -2.718... x' = log(x/t0) y'=eschr°-1 (u0^h*eschr(x')) y = exp(y')*t0 then f(1)     - b    = -2.244868624099733 E-60 + 3.153343574801035 E-60*I  // order for Eulersum: 1.4 - 0.1*I f(2)     - b°2  =  1.922763821393618 E-13 + 1.897277601645949 E-13*I  // order for Eulersum: 2.5-1.2*I f(-1)*bl - 0    =  4.089597929065041 E-56 + 6.283185307179586*I       // order for Eulersum: 1 (direct sum, no acceleration needed)However, as we see at the f(-1)-entry we needed a correction factor to get the correct real value - but with this we have the imaginary part differing with 2*Pi*I. I don't know yet how to include this smoothly into the formula, so it does not yet make sense to try to improve the fractional iterates with this... However, I just tried the (positive h=+0.5) half-iterate (hoping that it may not be required for positive heights to include this additional correction-factor). I got an half-iterate, which -when iterated- reproduced the integer-iterate well, I got about b°0.5(b°0.5(1)) - b ~ 5.0e-18 Code:\\ half-iterate, u=log(t0)+2*Pi*I U_05=W*dV(u^0.5)*WI;  \\ Bell-matrix for halfiterate; stirling-transformation is already included in W and WI (=W^-1) x=1 ESum(1.7-0.7*I)  *dV(log(x/t0))  *  U_05   \\ Use complex Euler-order for summation of 1.7-0.7*I, manually optimized   \\ y' = -3.903870288432298   + 1.032427916941400  *I   \\ untransformed value as result of matrix-sum y = exp(y')*t0   \\ y  =  0.02725344094115782 - 0.02087339685842267*I  \\ transformed, final value x = y    \\ use found value and iterate again ESum(2.4-0.6*I)  *dV(log(x/t0))  * U_05  \\ another Euler-order required    \\ y' = -3.250373950445425 - 4.597158013302573*I  \\ untransformed matrix-sum y = exp(y') *t0    \\ y  =  0.06598803584531253 + 5.687696437325240 E-18*I  \\ transformed, final value y - b =  -4.706607418746630 E-18 + 5.687696437325240 E-18*I    \\error <1e-17So the half-iterate from 1 by regular iteration using fixpoint t0 is y ~= 0.02725344094115782 - 0.02087339685842267*I . Seems, one can find at least any solution using regular tetration for this base, however difficult. What makes me still headscratch is the required correction-term for the one negative height example [update] I could improve this to err~1e-25 when simply computed the schröder-values separately as described in the mail before. Then I even did not need complex-euler-orders; just for one summation (of four) I needed a small Euler-acceleration of order 1.3 (which is nearly untransformed partial sums) The above value for the half-iterate is confirmed and has even more stable digits in the partial sums using around 120 to 128 terms. [/update] [update 2] The graph below is only a proposal. Only h=0.5 was used to reproduce the integer iteration so far. Other rational fractions should be verified, too.     [/update 2]     Gottfried Gottfried Helms, Kassel Gottfried Ultimate Fellow Posts: 764 Threads: 118 Joined: Aug 2007 09/14/2009, 02:02 PM (This post was last modified: 09/14/2009, 02:06 PM by Gottfried.) Hmm, I could verify the fractional heights for h=1/2 , h=1/3, ... h=1/6 by reinserting the values in the according powerseries (using the stirling-transformation). Iterations with h=1/8 reproduce up to h=4/8 = 1/2 correctly, and fail at h=6/8 or h=7/8 . (This does not neccessarily mean, that the results for the fractional iterates of the previous posting are wrong, but the series may be useless for multiple repeated applications) With finer fractional heights there seem to be generally trouble which I didn't try to investigate yet. I'm nearly sure this is due to the modulus 2*Pi*I . In another investigation I took especially care for the effect of the exp-function, which reduces the windings to (mod 2*Pi*I) and got an correct answer where the use of the exp-function led to a wrong result, so I'm beginning to consider whether we should build a library of functions/operators, where the operations keep track of the integer multiples of 2*Pi*I as well. Maybe this will give another improvement for the difficult bases. Don't know, whether I can proceed here... Gottfried Gottfried Helms, Kassel « Next Oldest | Next Newest »

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