x↑↑x = -1 KingDevyn Junior Fellow Posts: 2 Threads: 2 Joined: May 2014 05/28/2014, 04:07 AM What are some possible answers to the equation x↑↑x = -1? Must a new type of number be conceptualized similar to the answer to the equation x*x = -1? Or can it be proved that this answer lies within the real and complex planes? sheldonison Long Time Fellow Posts: 638 Threads: 22 Joined: Oct 2008 05/28/2014, 03:46 PM (This post was last modified: 05/28/2014, 08:42 PM by sheldonison.) (05/28/2014, 04:07 AM)KingDevyn Wrote: What are some possible answers to the equation x↑↑x = -1? Must a new type of number be conceptualized similar to the answer to the equation x*x = -1? Or can it be proved that this answer lies within the real and complex planes? edit update. It turns out that at least one of the solutions to the Op's question, is in the range that I can deal with numerically. For x~=2.6918099719192 + 0.62660048483655i, sexp_x(x)~=-1. The simplest algorithm I could come up with is to treat x^^x as an analytic function, and then iteratively use Newton's method to solve for x^^x=-1. For Newton's method to converge, it helps to get somewhere nearby first, which I did by hunt and peck, looking for bases where x^^x was near -1. I used my complex base tetration program. I wrote a program that explored complex base tetration, link here, but it is limited in terms of what bases it works for, due to problems at both exp(1/e), and at the Shell Thron boundary, and problems due to figuring out a consistent logarithmic branch to use... The Taylor series below can accurately calculate z=tet(b-2). Then iterate z=b^z twice, to get -1. Code:base = 2.6918099719192010269 + 0.62660048483655230254*I; {tet=    1.0000000000000000000 +x^ 1* ( 1.1102737979673991396 + 0.13444952486116713626*I) +x^ 2* ( 0.28425372561063406460 + 0.20168515612498045279*I) +x^ 3* ( 0.20142378046918946453 + 0.12287970799981736460*I) +x^ 4* ( 0.056631358539842508401 + 0.082262149871280378960*I) +x^ 5* ( 0.031626588237768922149 + 0.042664820582471964536*I) +x^ 6* ( 0.0060514513770401941396 + 0.023085772637730528515*I) +x^ 7* ( 0.0029937024980130884380 + 0.010708203875456837542*I) +x^ 8* (-0.00043235417834169701508 + 0.0051416902992490886284*I) +x^ 9* (-0.00017877732679188096499 + 0.0021743297255104601779*I) +x^10* (-0.00042253710857487390916 + 0.00095593830941724313184*I) +x^11* (-0.00016984009539643333991 + 0.00036797431151789026196*I) +x^12* (-0.00013598497262565877056 + 0.00014934164052065933115*I) +x^13* (-0.000050768369172865507021 + 0.000050790667372515377858*I) +x^14* (-0.000032617521009029439342 + 0.000018777349770437327073*I) +x^15* (-0.000011146610242226519358 + 0.0000050006190101552048081*I) +x^16* (-0.0000066270855458264950034 + 0.0000015446107355450406250*I) +x^17* (-0.0000020122113800518364907 + 0.000000068654818145175922999*I) +x^18* (-0.0000011963396468454392777 - 0.000000050931639762404521931*I) +x^19* (-0.00000030238855924614592898 - 0.00000012539726639748157461*I) +x^20* (-0.00000019689839845097420283 - 0.000000057637500744298040916*I) +x^21* (-0.000000035153434458995604130 - 0.000000041641278224258482194*I) +x^22* (-0.000000030197104715047471141 - 0.000000015975908532546115455*I) +x^23* (-0.0000000019166266427945385066 - 0.0000000095872390599983285641*I) +x^24* (-0.0000000044548959632689572594 - 0.0000000032492643102921963649*I) +x^25* ( 0.00000000051723734539716253138 - 0.0000000018638631462340237533*I) +x^26* (-0.00000000067021649772138737120 - 0.00000000054652627375747029680*I) +x^27* ( 0.00000000024760346059064288247 - 0.00000000032605780533942069098*I) +x^28* (-1.1228394170747083309 E-10 - 7.6537918519681082716 E-11*I) +x^29* ( 7.1268952289713966228 E-11 - 5.3189341538283539220 E-11*I) +x^30* (-2.2369708781921881068 E-11 - 8.0625867943287913385 E-12*I) +x^31* ( 1.7517436693666143597 E-11 - 8.3668572196268659348 E-12*I) +x^32* (-5.1777236965884120904 E-12 - 2.5187031197558940487 E-13*I) +x^33* ( 4.0336539933777435031 E-12 - 1.3265926389475846122 E-12*I) +x^34* (-1.2900297482939488422 E-12 + 1.8193947317104394501 E-13*I) +x^35* ( 9.0792695347242281651 E-13 - 2.2446482170229203341 E-13*I) +x^36* (-3.2656612422347855209 E-13 + 7.4097030312727709098 E-14*I) +x^37* ( 2.0437092576263104990 E-13 - 4.2548590115227614080 E-14*I) +x^38* (-8.1938431592790255774 E-14 + 2.0924876810646870153 E-14*I) +x^39* ( 4.6518922470943806673 E-14 - 9.0908340973673717718 E-15*I) +x^40* (-2.0262143014199117716 E-14 + 5.2297859307372668790 E-15*I) +x^41* ( 1.0743467122240305126 E-14 - 2.1121380807455872169 E-15*I) +x^42* (-4.9476280628213598972 E-15 + 1.2446927622317468161 E-15*I) +x^43* ( 2.5145731619654922803 E-15 - 5.1105870278086611369 E-16*I) +x^44* (-1.1976391479208810646 E-15 + 2.9141474535196681368 E-16*I) +x^45* ( 5.9468603024404147363 E-16 - 1.2519578605776942842 E-16*I) +x^46* (-2.8845734316656989916 E-16 + 6.8168421844852586540 E-17*I) +x^47* ( 1.4166399462777367792 E-16 - 3.0654575399052642127 E-17*I) +x^48* (-6.9322821884199600055 E-17 + 1.6036264175328628801 E-17*I) +x^49* ( 3.3908470426705261004 E-17 - 7.4731310274704053355 E-18*I) +x^50* (-1.6653357776046968983 E-17 + 3.7997687551447424638 E-18*I) +x^51* ( 8.1416335448137121457 E-18 - 1.8140427842389372934 E-18*I) +x^52* (-4.0032414617874811484 E-18 + 9.0620279617655147611 E-19*I) +x^53* ( 1.9590024443115400557 E-18 - 4.3905474199233700091 E-19*I) +x^54* (-9.6344349916460340144 E-19 + 2.1720467909141856696 E-19*I) +x^55* ( 4.7210606401205367545 E-19 - 1.0610621969521044634 E-19*I) +x^56* (-2.3217937825693896625 E-19 + 5.2248674247269975946 E-20*I) +x^57* ( 1.1392068860099503909 E-19 - 2.5632761528515730279 E-20*I) +x^58* (-5.6028249121768971146 E-20 + 1.2600230734782307737 E-20*I) +x^59* ( 2.7520884846767807057 E-20 - 6.1943235049722404547 E-21*I) +x^60* (-1.3537732547522302047 E-20 + 3.0441749210418233788 E-21*I) } - Sheldon tommy1729 Ultimate Fellow Posts: 1,370 Threads: 335 Joined: Feb 2009 05/28/2014, 10:34 PM So why dont we have Re(x) << 0 ? Well probably because Re(x) << 0 implies taking many logs , semilogs etc hence our value has " growth difficulties " in the sense that it gets too small. Similarly for Re(x) >> 0 then we have the supergrowth problem , it gets too big. I therefore guess all solutions have 0.5 < Re(x) < 4. However for x with very (or sufficiently ) large Imaginary parts this might no longer hold as a good argument ... Notice that log interations have stronger growth issues than exp iterations , because of the chaotic nature of exp iterates ... hence the previous sentense. Since For x~=2.6918099719192 + 0.62660048483655i, sexp_x(x)~=-1 apparantly , I wonder about 2 things -- apart from the fact that I need to learn more about complex base iterations ! -- 1) Is this x~=2.6918099719192 + 0.62660048483655i the only solution ?? 2) What happens with x~=2.6918099719192 - 0.62660048483655i ?? Also there should be a proof that x^^x = y Always has a complex solution. Since every nonconstant analytic function takes every value apart from possibly one value the REAL ( ahum ) question is there such an y such that x^^x has no complex solution ? Is there something intresting about oo ^ ^ oo for complex infinity ?? But the main question is : is x^^x analytically continuable to the entire complex plane or does it have a natural boundary ? And as suggested by question nr 2 : x^^x = conj( conj(x)^^conj(x) ) or similar ?? But thats not all , maybe I need to learn more about complex base iterations , but I know there are multiple solutions right ? So what answers do the other methods give ? Many questions ... I understand the kneser method for real bases > eta , but not for complex. for real bases we map the reals to the real line , but what do we do for the complex bases ?? Or is that an irrelevant question ? Sorry complex bases still confuse me. regards tommy1729 sheldonison Long Time Fellow Posts: 638 Threads: 22 Joined: Oct 2008 05/28/2014, 11:18 PM (This post was last modified: 05/30/2014, 03:59 AM by sheldonison.) (05/28/2014, 10:34 PM)tommy1729 Wrote: .... Since For x~=2.6918099719192 + 0.62660048483655i, sexp_x(x)~=-1 apparantly , I wonder about 2 things -- apart from the fact that I need to learn more about complex base iterations ! -- 1) Is this x~=2.6918099719192 + 0.62660048483655i the only solution ?? 2) What happens with x~=2.6918099719192 - 0.62660048483655i ?? .... And as suggested by question nr 2 : x^^x = conj( conj(x)^^conj(x) ) or similar ?? The conjugate of x would also be a solution, with x^^x = -1. The f=conj(f(conj(x))) equation is correct since the function is real valued at the real axis. The solution at hand for sexp_x(conj(x))~=3000+5500i Though that probably wasn't your question... I would expect an infinite number of solutions. Seeing as individually each tetration base solution takes on all values, then probably x^^x=y always has a solution, but finding them isn't easy. Quote:.... But the main question is : is x^^x analytically continuable to the entire complex plane or does it have a natural boundary ? We might want to move this to the tetcomplex.gp thread. This complex base tetration stuff was originally inspired by Mike3 solving one particular complex base. I took it a little farther, to try and explore the branch point at exp(1/e), and understand how tetration would work for real baseseta, you can go almost one full revolution in either direction. Adrien Douady conjectured a similar boundary for iterations of: $x^2+\frac{1}{4}+\epsilon \;\;$ See post#17, http://math.eretrandre.org/tetrationforu...416&page=2 Beyond that, there are other known singularities at b=1 and b=0. I got as far as conjecturing a solution for e^^-e, but I have never solved a a base with real(b) negative. Quote:But thats not all , maybe I need to learn more about complex base iterations , but I know there are multiple solutions right ? So what answers do the other methods give ? Many questions ... I understand the kneser method for real bases > eta , but not for complex. for real bases we map the reals to the real line , but what do we do for the complex bases ?? Or is that an irrelevant question ? Sorry complex bases still confuse me. As for as we can tell, Kouznetsov's method agrees with Kneser. The Kneser numeric method is a lot easier to extend to arbitrarily high precision levels. For complex tetration, we are interesting in analytically continuing Kneser so there is theoretically only one solution. The complex base solution is pretty weak, theoretically, though Henryk says you can rigorously show it should exist using the measurable Riemann mapping Theorem. Theoretically, research in complex dynamics might be applicable here too. Basically, you still have two fixed points. I use a pair of theta mapping for the two primary fixed points. The solution you're looking for is 1) analytically continuable from real base tetration, as you slowly modify the base, avoiding the singularity at eta. 2) sexp(-1)=0, sexp(0)=1, sexp(1)=b 3) converges to the superfunction for one fixed point in the upper half of the complex plane (theta mapping number 1) 4) converges to the superfunction for the other fixed point in the lower half of the complex plane (theta mapping number 2) The programming for tetcomplex.gp is not nearly as rock stable as the real base kneser.gp program. There are lots of numerical problems at or near the ShellThron boundary, for example, and with using the wrong logarithmic branch, and at the singularity at eta, and with the algorithm sometimes not converging as precision is increased. - Sheldon sheldonison Long Time Fellow Posts: 638 Threads: 22 Joined: Oct 2008 05/29/2014, 01:31 PM (This post was last modified: 05/29/2014, 02:47 PM by sheldonison.) The op skipped an operation. What about $x\uparrow x=-1$? $x^x=-1$ $\ln(x)x=(2n-1)\pi i\;\;\;$ taking ln of both sides, for all ln(-1) $ye^y = W^{-1}(y) = (2n-1)\pi i\;\;\;$ where $y=\ln(x)\;\;\;$ and W is the LambertW function $y=W((2n-1)\pi i)\;\;\; x=\exp(W((2n-1)\pi i))\;\;\;$ Solution with the LambertW function for n=1, I get x=1.6903867571636 + 1.8699079640268i, which I solved numerically with Newton's method. Oh, and then x=-1 is the obvious solution as well! I missed the obvious solution the first time; where $W(\pi i)=-\pi i$. Maybe that trivial solution of x=-1 is why the Op skipped asking about $x\uparrow x=-1$, but the complex solutions are analogous to the infinite number of complex solutions of $x\uparrow \uparrow x=-1$ - Sheldon tommy1729 Ultimate Fellow Posts: 1,370 Threads: 335 Joined: Feb 2009 05/29/2014, 04:37 PM (This post was last modified: 05/29/2014, 04:41 PM by tommy1729.) How does tetration work for base 3 + i ? I looked at some posts and there was talk about parabolic fixpoints and perturbated (parabolic) fixpoints. And stuff about merging in some way. The number of people involved in those threads and/or the number of posts was low, so I suspect im not the only one with questions ... Apart from the Shell-Tron region I fail to see the connection of base 3 + i with parabolic fixpoints or perturbations... and maybe there is none ( intended ). There was some stuff mentioned for complex bases involving riemann mappings and fourier series. It seems related to Kneser , but its unclear what is mapped where , and why. pseudocode is nice , but without explainations its " mystic " to the big audience I think. Also a dead link does not help ofcourse. W're are talking about analytic tetration , not just continu right ? I know having more than one fixpoint prevents an analytic solution that agrees on both fixpoints. I dont know how that makes sense then, for kneser we mapped to the real line ... because we wanted real-valued tetration for real bases. But what do we want and do with complex bases ? Sorry if I have forgotten stuff posted many years ago, I feel a bit like a noob now. Then again I cannot imagine a young new member/reader without a degree in dynamics too fully understand this from the first time (s)he reads it ... regards tommy1729 sheldonison Long Time Fellow Posts: 638 Threads: 22 Joined: Oct 2008 05/29/2014, 08:05 PM (This post was last modified: 06/04/2014, 01:19 PM by sheldonison.) (05/29/2014, 04:37 PM)tommy1729 Wrote: How does tetration work for base 3 + i ? ... Start with a base>eta, and rotate counterclockwise till you get to 3+i, or any other base. Lets say we have a complex tetration for b=3+i. $\text{tet}_b(0)=1$ As we rotated, we follow the two fixed points, L_1, L_2, which started out as L and L*. 1 refers to the upper half of the complex plane 2 refers to the lower half. $L_1=b^{L_1}\;\;\;$ fixed point, upper $L_2=b^{L_2}\;\;\;$ fixed point, lower $\lambda_1 = \ln(L_1)\;\;\;$ derivative at the fixed point, upper $\lambda_2 = \ln(L_2)\;\;\;$ derivative at the fixed point, lower Let as assume the most straightforward case, that both $|\lambda_1|>1\;$ and $|\lambda_2|>1\;$, so both fixed points are repelling. This is true at the real axis for bases>exp(1/e). On the ShellThron boundary, $|\lambda_1|=1$, and the Schroeder function does not converge if $\lambda_1^n=1$ for some integer n. For some mysterious reason, complex tet(z) is still defined and analytic for these bases; see this question on math overflow. But it is easier to start by assuming that the magnitude of both lambda's is greater than 1. $S_1(b^z) = \lambda_1(S_1(z))\;\;S_1(L_1)=0\;\;\;$ formal Schroeder equation, upper $S_2(b^z) = \lambda_2(S_2(z))\;\;S_2(L_2)=0\;\;\;$ formal Schroeder equation, lower $\alpha_1(z) = \ln(S_1(z))/\ln(\lambda_1)\;\;\;$ formal Abel function, upper $\;\alpha_1(b^z)=\alpha_1(z)+1$ $\alpha_2(z) = \ln(S_2(z))/\ln(\lambda_2)\;\;\;$ formal Abel function, lower $\;\alpha_2(b^z)=\alpha_2(z)+1$ So the conjecture is if Tet(z) is the bipolar complex analytic tetration function, then in the upper half of the complex plane, $\alpha_1(\text{tet}_b(z)) = \alpha_1(\text{tet}_b(z+1))+1$ $\alpha_1(\text{tet}_b(z)) = z + \theta_1(z)$ $\text{tet}_b(z)=\alpha_1^{-1}(z+\theta_1(z))\;\;\;\theta_1(z) = \sum_{n=0}^{\infty} a1_n \exp(2\pi i z)^n$ And in the lower half of the complex plane $\alpha_2(\text{tet}_b(z)) = \alpha_2(\text{tet}_b(z+1))+1$ $\alpha_2(\text{tet}_b(z)) = z + \theta_2(z)$ $\text{tet}_b(z)=\alpha_2^{-1}(z+\theta_2(z))\;\;\;\theta_2(z) = \sum_{n=0}^{\infty} a2_n \exp(-2\pi i z)^n$ If both $|\lambda_1|>1\;$ and $|\lambda_2|>1\;$, then the conjectured tetration solution will have $\theta_1(z)$ analytic in the upper half of the complex plane, with a singularity at integers at the real axis, and decaying to a constant as imag(z) goes to +infinity. Also, $\theta_2(z)$ is analytic in the lower half of the complex plane, with a singularity at integers at the real axis, and decaying to a constant as imag(z) goes to -infinity. Then tet(z) decays to L1 as imag(z) goes to +infinity, and decays to L2 as imag(z) goes to -infinity. The pseudo periodicity in the upper half of the complex plane will be $\frac{2\pi i}{\ln(\ln(L_1))}\;$ as tet(z) decays to $\;\alpha_1^{-1}(z)$ And in the lower half of the complex plane the pseudo periodicity will be $\frac{2\pi i}{\ln(\ln(L_2))}\;$ as tet(z) decays to $\;\alpha_2^{-1}(z)$ The tetcomplex.gp program calculates the two theta(z) functions which can it turns out can be represented with an analytic Taylor series, $\theta_1(z), \; \theta_2(z),$ as well as a Taylor series for tet(z) centered at z=0, where the tet(z) Taylor series has a radius of convergence of 2, since there is a singularity at z=-2. These three analytic functions, along with the inverse Abel functions, allows calculating tet(z) anywhere in the complex plane. The tet(z) taylor series at z=0 is important, since it allows getting accurate numeric results at the real axis, where the two theta(z) functions have a singularity, and it is difficult to get accurate results for tet(z) at the real axis using just the theta(z) functions, even though the three analytic representations are equal to each other at the real axis. Unlike real tetration, there is no Riemann mapping equivalent for determining the theta(z) functions, and all I can do is conjecture based on numeric evidence, so tetcomplex is fairly weak from a theoretical point of view. Moreover, as you rotate further you get to the ShellThron boundary, and there the Schroder function doesn't even converge, and then you continue on to where the upper |lambda_1|<1, and you have an attracting fixed point, which further complicates things... Things get even more complicated at the real axis for baseseta, and rotate counterclockwise till you get to 3+i, or any other base. Lets say we have a complex tetration for b=3+i. $\text{tet}_b(0)=1$ As we rotated, we follow the two fixed points, L_1, L_2, which started out as L and L*. 1 refers to the upper half of the complex plane 2 refers to the lower half. $L_1=b^{L_1}\;\;\;$ fixed point, upper $L_2=b^{L_2}\;\;\;$ fixed point, lower $\lambda_1 = \ln(L_1)\;\;\;$ derivative at the fixed point, upper $\lambda_2 = \ln(L_2)\;\;\;$ derivative at the fixed point, lower Let as assume the most straightforward case, that both $|\lambda_1|>1\;$ and $|\lambda_2|>1\;$, so both fixed points are repelling. This is true at the real axis for bases>exp(1/e). On the ShellThron boundary, $|\lambda_1|=1$, and the Schroeder function does not converge if $\lambda_1^n=1$ for some integer n. For some mysterious reason, complex tet(z) is still defined and analytic for these bases; see this question on math overflow. But it is easier to start by assuming that the magnitude of both lambda's is greater than 1. $S_1(b^z) = \lambda_1(S_1(z))\;\;S_1(L_1)=0\;\;\;$ formal Schroeder equation, upper $S_2(b^z) = \lambda_2(S_2(z))\;\;S_2(L_2)=0\;\;\;$ formal Schroeder equation, lower $\alpha_1(z) = \ln(S_1(z))/\ln(\lambda_1)\;\;\;$ formal Abel function, upper $\;\alpha_1(b^z)=\alpha_1(z)+1$ $\alpha_2(z) = \ln(S_2(z))/\ln(\lambda_2)\;\;\;$ formal Abel function, lower $\;\alpha_2(b^z)=\alpha_2(z)+1$Ok. So i use the koenigs function to solve the schroeder equation and from that I compute the abel function. But that gets me 2 abel functions ... What to do with them ? I cannot simply use one for the upper plane and one for the lower plane , can i ?? Quote:So the conjecture is if Tet(z) is the bipolar complex analytic tetration function, then in the upper half of the complex plane, $\alpha_1(\text{tet}_b(z)) = \alpha_1(\text{tet}_b(z+1))+1$ $\alpha_1(\text{tet}_b(z)) = z + \theta_1(z)$ Should that not be $\alpha_1(\text{tet}_b(z)) = \alpha_1(\text{tet}_b(z+1))-1$ ? Quote:And in the lower half of the complex plane $\alpha_2(\text{tet}_b(z)) = \alpha_2(\text{tet}_b(z+1))+1$ $\alpha_2(\text{tet}_b(z)) = z + \theta_2(z)$ Now I still have 2 abel functions. Not 1. And I have neither tet_b(z) or theta(z). This equation just shows how tet_b(z) and theta(z) are related. Since I have the abel function I can just plug in some theta(z) and get some tet_b(z). But still 2 of those and I see no reason to prefer some theta resp tet_b over another. Why dont we just take the inverse of the abel as superfunction(s) ? hmm. I see no motivation or uniqueness condition. And picking one function for the upper and one for the lower seems partially arbitrary ? Yeah ok that matches the fixpoints at imaginary infinity , but there are others too right. How do we merge the two theta functions ?? We do merge , to avoid ill defined parts of the complex plane right ? Quote:If both $|\lambda_1|>1\;$ and $|\lambda_2|>1\;$, then the conjectured tetration solution will have $\theta_1(z)$ analytic in the upper half of the complex plane, with a singularity at integers at the real axis, and decaying to a constant as imag(z) goes to +infinity. Also, $\theta_2(z)$ is analytic in the lower half of the complex plane, with a singularity at integers at the real axis, and decaying to a constant as imag(z) goes to -infinity. Then tet(z) decays to L1 as imag(z) goes to +infinity, and decays to L2 as imag(z) goes to -infinity. The pseudo periodicity in the upper half of the complex plane will be $\frac{2\pi i}{\ln(\ln(L_1))}\;$ as tet(z) decays to $\;\alpha_1^{-1}(z)$ And in the lower half of the complex plane the pseudo periodicity will be $\frac{2\pi i}{\ln(\ln(L_2))}\;$ as tet(z) decays to $\;\alpha_2^{-1}(z)$ The tetcomplex.gp program calculates the two theta(z) functions which can it turns out can be represented with an analytic Taylor series, $\theta_1(z), \; \theta_2(z),$ as well as a Taylor series for tet(z) centered at z=0, where the tet(z) Taylor series has a radius of convergence of 2, since there is a singularity at z=-2. These three analytic functions, along with the inverse Abel functions, allows calculating tet(z) anywhere in the complex plane. The tet(z) taylor series at z=0 is important, since it allows getting accurate numeric results at the real axis, where the two theta(z) functions have a singularity, and it is difficult to get accurate results for tet(z) at the real axis using just the theta(z) functions, even though the three analytic representations are equal to each other at the real axis. Unlike real tetration, there is no Riemann mapping equivalent for determining the theta(z) functions, and all I can do is conjecture based on numeric evidence, so tetcomplex is fairly weak from a theoretical point of view. Moreover, as you rotate further you get to the ShellThron boundary, and there the Schroder function doesn't even converge, and then you continue on to where the upper |lambda_1|<1, and you have an attracting fixed point, which further complicates things... Things get even more complicated at the real axis for bases