x↑↑x = -1 sheldonison Long Time Fellow Posts: 614 Threads: 22 Joined: Oct 2008 05/29/2014, 11:44 PM (This post was last modified: 05/30/2014, 10:38 AM by sheldonison.) (05/29/2014, 11:41 PM)tommy1729 Wrote: (05/29/2014, 11:34 PM)sheldonison Wrote: For a complex base, tet is not a real valued function at the real axis. But if the base is real, then yes, the two theta functions agree naturally due to Schwarz reflection involving L and L*. So the complex tet function naturally becomes the Kneser solution for a real valued base. Im aware of that. I mean mapping a continu line on the complex plane to the positive reals. For instance mapping f(4+3i) to g(1) as a point example. ... it seems weird that 2 analytic functions can agree on the real line , yet be different off the real line... so f(1) = g(1) = a , f(2) = g(2) = b , ... for all positive reals ... yet f and g are different ?? regards tommy1729They don't disagree, they just become less accurate. In particular, theta1 has a singularity at the real axis, so you can't use it in the bottom half of the complex plane, and it isn't very accurate right at the real axis, but you can use theta2 in the bottom half of the complex plane, but not too close to the real axis, where it is not that accurate, and not the upper half of the complex plane. But the two theta mappings agree with the third function, which is a unit radius Taylor series, in their respective halves of the complex plane. The tet 3+I Taylor series in post#7, is accurate to a little better than 13 decimal digits, inside a circle radius=0.85, and agrees with the theta mappings to better than 13 decimal digits if abs(imag(z))>0.2i. So all three functions are in excellent mutual agreement wherever they converge well, and one can easily double the pari-gp precision, or quadruple it, so results are accurate and consistent to 32 decimal digits, or 66 decimal digits. The motivation for complex tetration is that it is conjectured that all of the following become analytic functions, with complex tetration, where z is the tetration base. For example, the half iterate of 0 becomes an analytic function with respect to the tetration base. $f(z)=\text{tet}_z(-0.5)\;\;f(z)=\text{tet}_z(i)\;\;f(z)=\text{tet}_z(-i+1)\;\;$ or any other value, as you vary the tetration base z Numeric evidence also supports this conjecture; see post#9. And in fact, there are bases on the ShellThron boundary that don't have the required analytic upper Abel function and theta mapping, but we can accurately calculate tetration for those bases indirectly by assuming tetration is analytic as you vary the base. Sometimes a couple of pictures can help The first picture is tet base=3+i at the real axis from -1.99 to +2. The second picture is a complex plane graph, from -4 to +16 in the reals, and -6 to +4 in the imaginary, with grids every 2 units. In the lower graph, you can see how the upper and lower halves of the complex plane have different pseudo periods. Yet the conjecture is the function is analytic in the upper and lower halves of the complex plane, with singularities at the -integers, less than or equal to -2. You can see the zero at -1, and the logarithmic singularity at -2, -3. The cutpoints for z<-2 are kind of random ... I find the complex plane graphs of some of these tetcomplex functions to be truly beautiful.         - Sheldon tommy1729 Ultimate Fellow Posts: 1,354 Threads: 328 Joined: Feb 2009 05/30/2014, 09:29 PM (This post was last modified: 05/30/2014, 09:30 PM by tommy1729.) $\alpha_1(\text{tet}_b(z)) = z + \theta_1(z)$ $\alpha_2(\text{tet}_b(z)) = z + \theta_2(z)$ for b = 3 + i. This is the main part and I agree on that. No arguing there. Forget about the Riemann mapping comments. After some consideration this is not so bad afterall. How do we call this ? system of 2 functional equations ? coupled functional equations ? But the core question is : how do we solve this : $\alpha_1(\text{tet}_b(z)) = z + \theta_1(z)$ $\alpha_2(\text{tet}_b(z)) = z + \theta_2(z)$ for b = 3 + i. ? The equation seemed familiar to me, so I looked in my very old notebook and ... I found the related following 2 equivalent equations : 1) [A_1(f(z)) - A_1(f(z-1))] / [A_2(f(z)) - A_2(f(z-1))] = 1 and 2) [A_1(f(z)) - A_1(f(z-1))] - [A_2(f(z)) - A_2(f(z-1))] = 0 where the A_i where given functions. (the [] were written not as brackets but by a difference symbol equivalent to newton's though that does not matter) Now it turns out I tried to solve 2) by using a (truncated) taylor series for A_1,A_2 and f(z). In more modern context I guess that means using truncated carleman matrices ! Afterall since equation 2) only has -,+ and composition , carleman matrices seem perfect ! A bit later I wrote the comment : " fibonacci like " ? Which still makes me wonder today. Is there a better method than carleman/taylor in the style of fibonacci ? And of course, the reason I write all this : Do sheldon and mike use the same method to solve this " coupled functional equation " ? It appears the equations 1) and 2) are more general than the original equations ... ... for instance say A_1(f(z)) - A_1(f(z-1)) = z^2 = A_2(f(z)) - A_2(f(z-1)) then we also get : [A_1(f(z)) - A_1(f(z-1))] / [A_2(f(z)) - A_2(f(z-1))] = 1 and [A_1(f(z)) - A_1(f(z-1))] - [A_2(f(z)) - A_2(f(z-1))] = 0 But NOT a solution we want. All a bit confusing ... But very intresting ! Reminds me a little bit of " the fermat superfunction " too. 2 fixpoints are intresting it seems. http://math.eretrandre.org/tetrationforu...hp?tid=809 regards tommy1729 tommy1729 Ultimate Fellow Posts: 1,354 Threads: 328 Joined: Feb 2009 05/31/2014, 08:31 PM (This post was last modified: 05/31/2014, 09:59 PM by tommy1729.) However there might be a problem ... tommy1729 Ultimate Fellow Posts: 1,354 Threads: 328 Joined: Feb 2009 05/31/2014, 09:23 PM (This post was last modified: 05/31/2014, 09:58 PM by tommy1729.) $\alpha_1(\text{tet}_b(z)) = z + \theta_1(z)$ $\alpha_2(\text{tet}_b(z)) = z + \theta_2(z)$ SO $\alpha_1(\text{tet}_b(z)) - \alpha_1(\text{tet}_b(z-1)) = 1$ And $\alpha_2(\text{tet}_b(z)) - \alpha_2(\text{tet}_b(z-1)) = 1$ Thus $\alpha_1(\text{tet}_b(z)) - \alpha_1(\text{tet}_b(z-1)) - \alpha_2(\text{tet}_b(z)) + \alpha_2(\text{tet}_b(z-1)) = 0$ Let $G(z) = \G(z) = \alpha_1(z) - \alpha_2(z)$ Then we get $\G(\text{tet}_b(z)) - \G(\text{tet}_b(z-1))= 0$ or $\G(\text{tet}_b(z)) = \G(\text{tet}_b(z-1))$ This implies : $\text{tet}_b(z) = G^{[-1]}( \G(\text{tet}_b(z-1)) )$ Hence because G^[-1](G(z)) = id(z) is absurd we get ( by branches ) one of the following potential conclusions. 1) G^[-1](G(z)) =/= b^z And this implies that $\text{tet}_b(z)$ is a superfunction of 2 functions !!?? 2) G^[-1](G(z)) = b^z or G^[-1](G(z)) = b^z + 2pi i / ln(b) Very unlikely. Since both potential conclusions are very very likely wrong , this implies $\alpha_1(\text{tet}_b(z)) = z + \theta_1(z)$ $\alpha_2(\text{tet}_b(z)) = z + \theta_2(z)$ has no solution. And this idea can easily be generalized to other functions then b^z , such as entire functions with 2 repelling fixpoints. Hence my pessimistic attitude. ... ------------------------------------------------------------------ But then what did mike and sheldon compute !?? Another fake function ??? Hmmm. ------------------------------------------------------------------ Lets take another look : $\G(\text{tet}_b(z)) - \G(\text{tet}_b(z-1))= 0$ The only hope seems G(z) = p(inv tet_b(z)). where inv tet_b is the inverse function of tet_b and p is a 1-periodic function. Or said differently : $\alpha_1(\text{tet}_b(z)) - \alpha_2(\text{tet}_b(z)) = p(z)$ But this seems to bring us back at the beginning : $p(z) = \theta_1(z) - \theta_2(z)$ Lets analyze further : in the pessimistic case we wrote G(tet(z)) = G(tet(z-1)) the key is that if G(z) = p(tet^[-1](z)) We get p(z) = p(z-1). THIS IS IMPORTANT because we got the hidden paradox : p(z) = p(z-1) => z = p^[-1](p(z-1)) => z = z - 1. So the step $\G(\text{tet}_b(z)) = \G(\text{tet}_b(z-1))$ This implies : $\text{tet}_b(z) = G^{[-1]}( \G(\text{tet}_b(z-1)) )$ Might not be as valid as it might seem. Hmm. regards tommy1729 sheldonison Long Time Fellow Posts: 614 Threads: 22 Joined: Oct 2008 05/31/2014, 09:48 PM (This post was last modified: 05/31/2014, 09:50 PM by sheldonison.) (05/31/2014, 09:23 PM)tommy1729 Wrote: $\alpha_1(\text{tet}_b(z)) = z + \theta_1(z)$ $\alpha_2(\text{tet}_b(z)) = z + \theta_2(z)$ .... $\text{tet}_b(z) = G^{[-1]}( \G(\text{tet}_b(z-1)) )$ Hence because G^[-1](G(z)) = id(z) is absurd we get ( by branches ) one of the following potential conclusions. 1) G^[-1](G(z)) =/= b^z And this implies that $\text{tet}_b(z)$ is a superfunction of 2 functions !!?? 2) G^[-1](G(z)) = b^z or G^[-1](G(z)) = b^z + 2pi i / ln(b) Very unlikely. Since both potential conclusions are very very likely wrong , this implies $\alpha_1(\text{tet}_b(z)) = z + \theta_1(z)$ $\alpha_2(\text{tet}_b(z)) = z + \theta_2(z)$ has no solution. .... But then what did mike and sheldon compute !??Why would your logic apply only to complex bases? Your logic should apply equally to Kneser's real valued tetration which also has two Abel functions, with different periodicities, etc, for L and L*. In fact, if I plug in a real valued base to tetcomplex, it calculates an Abel function and theta mapping for each fixed point, ignorant of the fact that they are complex conjugates of each other. By symmetry, the solution is a real valued Schwarz reflection, and is the same as Kneser's solution. Later, I will find the flaw in the logic of this contradiction, but Tommy will probably figure it out first. - Sheldon tommy1729 Ultimate Fellow Posts: 1,354 Threads: 328 Joined: Feb 2009 05/31/2014, 10:11 PM (05/31/2014, 09:48 PM)sheldonison Wrote: [quote='tommy1729' pid='7091' dateline='1401567839'] Later, I will find the flaw in the logic of this contradiction, but Tommy will probably figure it out first. It is done. I figured it out. In fact I edited before you replied. You are correct about your Kneser remark. However Kneser is a special case with the riemann mapping and conjugate fixpoint pair ... For existance I bet on Truncated Carleman matrices ... Still lots to do as usual. My question was , how do you and mike do it ? regards tommy1729 sheldonison Long Time Fellow Posts: 614 Threads: 22 Joined: Oct 2008 06/01/2014, 01:04 AM (This post was last modified: 06/01/2014, 05:00 PM by sheldonison.) (05/31/2014, 10:11 PM)tommy1729 Wrote: .... My question was , how do you and mike do it ? regards tommy1729 The algorithm is very much like the real valued kneser.gp algorithm. But first, I start with the observation that if you have an exact Tetration Taylor series with a radius of convergence=2, centered at z=0, then you also have an exact theta, where the the theta(z) Taylor series coefficients can by calculated by Fourier series, or by substitution, as a Cauchy integral around a circle. $\theta_1 = \alpha_1(\text{tet}(z))-z$ $\theta_1(z) = \sum_{n=0}^{\infty} a1_n \exp(2\pi i z)^n$ $\theta_2 = \alpha_2(\text{tet}(z))-z$ $\theta_2(z) = \sum_{n=0}^{\infty} a2_n \exp(-2\pi i z)^n$ Viewed in this way, the theta series has a Taylor series representation with a radius of convergence=1. To calculate the theta(z) an series coeffieients as a Cauchy integral, the only rule is that the radius must be less or equal to 1. Now, back to the exact Taylor series for tet, centered at 0, radius of convergence=2. Then one can exactly calculate the a1_n and a2_n theta(z) Taylor series. The tetcomplex code does just that, at $z=-0.5+i0.175 .. 0.5+i0.175$, which corresponds to a radius=0.333, so the theta series is very well behaved at that point. Also, there is a major computational hassle of figuring out which logarithm to use as you take the Abel function of tet(z), and as |Im(z)| get bigger, the problem is observed to go away. Now what if the initial tet series wasn't exact? Well, we calculate the theta series anyway. I use some tricks to improve convergence of both the Tet series, and the theta(z) series. For example, the tet(z) series a1 & a2 coeffients are modified to make tet(0.5+0.175i)=b^(tet(-0.5+0.175i)), and also tet(0.5-0.175i)=b^(tet(-0.5-0.175i)). For the theta(z) series, I throw away the Laurent 1/z terms. Then I use a combination of the initial tet series, along with the two theta series, to create a new tetration series by generating a unit_circle approximation. See post #13, equations for Kneser for a picture showing how the three functions are combined to generate a new unit_circle approximation, using $\alpha_1^{-1}(z+\theta_1(z))$ and $\alpha_2^{-1}(z+\theta_2(z))$ and exp(tet), and log(tet), splicing together what is expected to be the most accurate of these approximations at each part of the unit_circle. We treat this unit_circle as if it were an analytic function and use Cauchy to generate all the Taylor series coefficients of the new tet(z) approximation; we ignore the fact that unit_circle is discontinuous at the splice points, and actually has a Laurent series, defined only on the boundary, and we throw away the x^-n terms, and we force set a_0=1, to prevent the function from drifting off tet(0). $a_n = \frac{1}{2\pi} \int_{-\pi i}^{\pi i} \exp(-nx)\text{unitcircle}(\exp(x)) dx \;\;$ radius=1 Cauchy integral formula on the unit_circle function to generate next tet approximation $\text{ tet}_{i+1} = 1 + \sum_{n=1}^{\infty} a_n x^n\;\;$ from the tet function, we generate the next pari of theta(z) Taylor series, with similar equations And it turns out, the algorithm tetcomplex.gp uses will give you about 8 binary bits more precision, for the new Tet(z) series, than the initial one, if you use all of the tricks that I use. Also, you need an algorithm to get some sort of initial Tet(z) seed approximation as well, that is good enough to start convergence. After that, keep iterating, until you have your desired precision of your Tetration series. To estimate the error as you go, compare the tetration taylor series with the theta(z) approximations, and see how closely they match. - Sheldon tommy1729 Ultimate Fellow Posts: 1,354 Threads: 328 Joined: Feb 2009 06/02/2014, 11:17 PM (This post was last modified: 06/02/2014, 11:19 PM by tommy1729.) $\alpha_1(\text{tet}_b(z)) = z + \theta_1(z)$ $\alpha_2(\text{tet}_b(z)) = z + \theta_2(z)$ I feel like this is overdetermined. For K = 1,2,3 : let $\theta_K(z)$ satisfy 1) $\theta_K(z) = \theta_K(z+1)$ 2) $\theta_K(n) = 0$ for every integer $n$. 3) $\theta'_K(n) = 0$ for every integer $n$. 4) $z + \theta_K(z)$ is univalent for z near the real line. 5) $1 + \theta'_K(z) =/= 0$ for z near the real line. Then $\alpha_1(\text{tet}_b(z + \theta_3(z) )) = T(z)$ $\alpha_2(\text{tet}_b(z + \theta_3(z) )) = T_2(z)$ Where T(z) and T_2(z) are Taylor series that satisfy T_i(n) = n for all integer n and i. ( Since the LHS is a Taylor series , so is the RHS ) Let $\alpha_2( \alpha^{[-1]}_1(z) )) = D(z)$ Then $D(T(z)) = T_2(z)$ Now it seems we have more freedom. However we require - by the above - : $D(n) = n$ Fortunately that seems no problems since the the two abel functions agree on integer iterates (alpha_i(b^z) = alpha_i(z)+1 ). So we end up with : $\alpha_1(\text{tet}_b(z + \theta_3(z) )) = T(z)$ such that $T(n)=n$ This seems to be relaxable to $\alpha_1(\text{tet2}_b(z)) = T(z)$ such that $T(n)=n$ I suggest z is a positive real to solve the equation. -------- now if z is a positive real : $\alpha_1(\text{tet2}_b(z)) - \alpha_1(\text{tet2}_b(z-1)) = T(z)-T(z-1)$ And maybe T(z)-T(z-1) is a periodic function ? -------- *** I notice the resemblance of $\alpha_1(\text{tet2}_b(z)) = T(z)$ towards $\alpha_1(\text{tet}_b(z)) = z + \theta_1(z)$ *** So when are the 5 conditions valid ?? It seems things are simplified. But I would like to add that the radius of D(z) matters ... or that of D(T(z)). Equations are only meaningfull within the domain of convergeance. Maybe that is a criterion to the more general case ( apart from tetration , wheither we can agree upon 2 fixpoints ). So it seems our attention should go to D and T perhaps rather than tet(z) in order to solve the equation. Maybe this can be converted into carleman matrices equations. Still thinking... regards tommy1729 sheldonison Long Time Fellow Posts: 614 Threads: 22 Joined: Oct 2008 06/02/2014, 11:44 PM (This post was last modified: 06/03/2014, 02:22 PM by sheldonison.) (06/02/2014, 11:17 PM)tommy1729 Wrote: I feel like this is overdetermined. For K = 1,2,3 : let $\theta_K(z)$ satisfy 1) $\theta_K(z) = \theta_K(z+1)$ 2) $\theta_K(n) = 0$ for every integer $n$. 3) $\theta'_K(n) = 0$ for every integer $n$. 4) $z + \theta_K(z)$ is univalent for z near the real line. 5) $1 + \theta'_K(z) =/= 0$ for z near the real line. ....Just remember that for both theta functions, $\theta(n)$ has a really nasty singularity at integer values of n, and is definitely not a constant. So the only place the two $\alpha^{-1}(z+\theta(z))$ functions agree is right at their analytic boundaries. I suppose you could probably extend the boundary at non-integers, but besides the singularities, its going to behave pretty poorly, and $\alpha1(\text{tet}(z))-z$ is no longer 1-cyclic if $\Im(z)<0$ so analytic continuation isn't helpful. Also, at the real axis, even with thousands of series terms, theta(z) would still limit precision to a handful of decimal digits. So, we also avoid the real axis, and use a third function, tet(z), to glue together the two theta(z) representations, to have any hope of getting any accuracy at all for results. - Sheldon tommy1729 Ultimate Fellow Posts: 1,354 Threads: 328 Joined: Feb 2009 06/03/2014, 12:16 PM Somewhat of a followup : We can say D(z + theta_A(z)) = z + theta_B(z). Then by substitution u = z + theta_A(z) , z + theta_A(z) = u + theta_C(u) we get : D(u) - u = theta_C(u) rewrite u as z we get : D(z) - z = theta(z) Now this is a yes/no question. IS D(z) - z a 1periodic holomorphic function that satisfies the 5 conditions or not ? If the answer is yes , we have existance and a way to compute. Not sure about uniqueness. It seems we do not have uniqueness but we can make it unique by adding another theta shift and require local boundedness. If the answer is no then .... euh ... I dont know what to conclude. ... If we weaken the 5 conditions then the steps are no longer necc valid. Assuming yes , for analysis it seems intesting to investigate the extrema of D(z) - z , if we find a single max and no min we can use carleman matrices to solve the equation. Hence D ' (z) - 1 = 0 is an intresting equation. Cauchy can then be used too btw. Still thinking ... regards tommy1729 « Next Oldest | Next Newest »