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 Real-analytic tetration uniqueness criterion? mike3 Long Time Fellow Posts: 368 Threads: 44 Joined: Sep 2009 06/15/2014, 09:59 AM (This post was last modified: 06/15/2014, 10:04 AM by mike3.) (06/15/2014, 04:51 AM)sheldonison Wrote: (06/14/2014, 10:24 PM)tommy1729 Wrote: (06/14/2014, 05:15 AM)sheldonison Wrote: For tetration at the real axis, the nearest singularity is at x=-2, and there are no other singularities to the right of that anywhere in the complex plane.Well that depends on what type of tetration we use. Kneser seems to have this property. But so do many theta variations of Kneser. ( the analytic theta's ) Conjecture; Kneser is the only solution with no singularities in the upper/lower halves of the complex plane, and this is a uniqueness criterion. For all the of the entire theta functions, we know there will be an infinite number of singularities in the upper half of the complex plane, where z+theta(z)=-2,-3,-4 ..... I would also conjecture that these singularities will be in the right half of the complex plane as well. But either way, Kneser has this special property, so Kneser's so at any value of z, there are negative even derivatives for large enough (2n). This doesn't prove that all of the odd derivatives are always positive, but that is also a conjectured uniqueness criterion. However, didn't you disprove this conjecture with the construction of the tetration function from the alternate fixed point here: http://math.eretrandre.org/tetrationforu...hp?tid=452 http://math.eretrandre.org/tetrationforu...452&page=2 or does this also qualify as a "Kneser"? But it's not a unique function if that's the case. However, just from looking at the graphs on that second page, it's quite obvious this function fails the criterion given in my OP. I wonder what the $\theta(z)$ mapping carrying the "good" Kneser solution to that thing looks like. I suspect it'll be multivalued, with branch singularities instead of just poles or whatever, which significantly complicates the composition $\mathrm{tet}(z + \theta(z))$ in the complex plane -- although on the real line it will, of course, be single-valued. On the other hand, your "max at the real axis" criterion would seem to rule out this function. sheldonison Long Time Fellow Posts: 644 Threads: 22 Joined: Oct 2008 06/15/2014, 01:46 PM (This post was last modified: 06/17/2014, 06:54 PM by sheldonison.) (06/15/2014, 09:59 AM)mike3 Wrote: However, didn't you disprove this conjecture with the construction of the tetration function from the alternate fixed point here: http://math.eretrandre.org/tetrationforu...hp?tid=452 http://math.eretrandre.org/tetrationforu...452&page=2 or does this also qualify as a "Kneser"? But it's not a unique function if that's the case. However, just from looking at the graphs on that second page, it's quite obvious this function fails the criterion given in my OP.Yeah, we need to add a criteria that not only is the tet(x) function increasing from -2->infinity, but also that the first derivative is positive. The alternative fixed point has a zero derivative at integer values, -1,1,2,3 etc. This is equivalent to requiring that tet(z) have an inverse at the real axis; that the slog be analytic at the real axis. Quote:I wonder what the $\theta(z)$ mapping carrying the "good" Kneser solution to that thing looks like. I suspect it'll be multivalued, with branch singularities instead of just poles or whatever, which significantly complicates the composition $\mathrm{tet}(z + \theta(z))$ in the complex plane -- although on the real line it will, of course, be single-valued. On the other hand, your "max at the real axis" criterion would seem to rule out this function. Yeah, $z+\theta(z)=\text{slog}(\text{tet}_{\text{alt}}(z))$ would have a cube root branch at integers, so yeah, theta is not analytic at the real axis. Ooops, not correct; edit $\theta(z)=\text{slog}(\text{tet}_{\text{alt}}(z))-z$ is 1-cyclic analytic function at the real axis. - Sheldon tommy1729 Ultimate Fellow Posts: 1,374 Threads: 337 Joined: Feb 2009 06/15/2014, 02:45 PM (06/15/2014, 04:51 AM)sheldonison Wrote: (06/14/2014, 10:24 PM)tommy1729 Wrote: (06/14/2014, 05:15 AM)sheldonison Wrote: For tetration at the real axis, the nearest singularity is at x=-2, and there are no other singularities to the right of that anywhere in the complex plane.Well that depends on what type of tetration we use. Kneser seems to have this property. But so do many theta variations of Kneser. ( the analytic theta's ) Conjecture; Kneser is the only solution with no singularities in the upper/lower halves of the complex plane, and this is a uniqueness criterion. For all the of the entire theta functions, we know there will be an infinite number of singularities in the upper half of the complex plane, where z+theta(z)=-2,-3,-4 ..... I would also conjecture that these singularities will be in the right half of the complex plane as well. But either way, Kneser has this special property, so Kneser's so at any value of z, there are negative even derivatives for large enough (2n). This doesn't prove that all of the odd derivatives are always positive, but that is also a conjectured uniqueness criterion. (06/14/2014, 10:24 PM)tommy1729 Wrote: Also why would the maximum value be at z0 ? This is not in general true for polynomials UNLESS all derivatives are nonnegative. But here we have some negatives. Well even that depends on parameters and conjectures.Yes, I guess that's another conjecture, that for all real(z)>~=0.5, if you make a line from -imag infinity to +imag(infinity) at real(z), the maximum absolute value occurs at the real axis. This is also supported by empirical evidence, but I can't think of any obvious way to prove it. This would also mean that the maximum magnitude on any circle on the real axis occurs at the real axis, so long as its bigger than about 0.5. This is related to the conjecture that all of the odd derivatives are positive; you might be able to prove one from the other. Assuming all of the odd derivatives are positive, it would be interesting to try to prove that any sexp(z+sin(z)/k) mapping has negative odd derivatives. Notice your singularity uniqueness criterion is ACTUALLY equivalent to the boundedness criterion. Both are proven by the little picard theorem. So uniqueness of both has already been shown ! See TPID 4 to which you agreed as being proven. Thats solves that. Notice that I believe existance has also been proven and that this is kneser solution. Also notice that this is WHY i believe cauchy's method ( kouznetsov ?) is trying to construct Kneser's solution rather than being a distinct solution !! What I meant by " other solution " is intended not on the complex plane , but on the related Taylor radiuses. In other words the desired property of the OP is not trivially garanteed by the singularity/boundedness uniqueness criterion ! Even the domination of the log(x+2) is not proven. What is proven - assuming kneser has the boundedness criterion - is that kneser has the domination of the log(x+2) , but without uniqueness or other stuff proven. http://math.eretrandre.org/tetrationforu...hp?tid=747 Your conjecture about the "attained max at the real line" is a partial strenghtening of the boundedness uniqueness condition. It is very nearly true that the " attained max at the real line " implies the boundedness uniqueness but not exactly ... However notice that if we can prove it on a strip Re(z)=R -> Re(z)=R+1 , then by induction it holds for all real Q > R. ... Im am tempted to say this might be done by investigating the Riemann mapping. regards tommy1729 tommy1729 Ultimate Fellow Posts: 1,374 Threads: 337 Joined: Feb 2009 06/15/2014, 03:34 PM An inspired thread for periodic functions : http://math.eretrandre.org/tetrationforu...hp?tid=886 regards tommy1729 sheldonison Long Time Fellow Posts: 644 Threads: 22 Joined: Oct 2008 06/15/2014, 10:00 PM (06/15/2014, 04:51 AM)sheldonison Wrote: Yes, I guess that's another conjecture, that for all real(z)>~=0.5, if you make a line from -imag infinity to +imag(infinity) at real(z), the maximum absolute value occurs at the real axis. This is also supported by empirical evidence, but I can't think of any obvious way to prove it. This would also mean that the maximum magnitude on any circle on the real axis occurs at the real axis, so long as its bigger than about 0.5. I did some experiments and this holds if z>=z0 where z0=0.47823520737667784466. What is special about this particular number, z0? Well, sexp''(z0-1)=0, which means that $\Re(\text{sexp}(z0-1+k i))\approx 0.4777430947666662352 -0.021068926393682 k^4$ near the real axis, where these are the a0 and a4 Taylor series coefficients. - Sheldon tommy1729 Ultimate Fellow Posts: 1,374 Threads: 337 Joined: Feb 2009 06/15/2014, 10:17 PM (06/15/2014, 10:00 PM)sheldonison Wrote: (06/15/2014, 04:51 AM)sheldonison Wrote: Yes, I guess that's another conjecture, that for all real(z)>~=0.5, if you make a line from -imag infinity to +imag(infinity) at real(z), the maximum absolute value occurs at the real axis. This is also supported by empirical evidence, but I can't think of any obvious way to prove it. This would also mean that the maximum magnitude on any circle on the real axis occurs at the real axis, so long as its bigger than about 0.5. I did some experiments and this holds if z>=z0 where z0=0.47823520737667784466. What is special about this particular number, z0? Well, sexp''(z0-1)=0, which means that $\Re(\text{sexp}(z0-1+k i))\approx 0.4777430947666662352 -0.021068926393682 k^4$ near the real axis, where these are the a0 and a4 Taylor series coefficients. z0 is an intresting number imho. I cant help to ask : z0 = a0 ? you say a0 and a4 are Taylor series coefficients. But expanded where ? The way I understand it is that you are saying expansion at z0 of sexp(z0 - 1 + k i) in the direction k. and then apparantly z0 = a0 although your values differ ... and then sexp ' (z0 - 1 + ki) = sexp " (z0 - 1 + ki) = sexp "' (z0 - 1 + ki) = 0 ? for some mysterious reason ? (sexp " (z0 - 1 + ki) = 0 was trivial) regards tommy1729 « Next Oldest | Next Newest »

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