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 Real-analytic tetration uniqueness criterion? mike3 Long Time Fellow Posts: 368 Threads: 44 Joined: Sep 2009 06/10/2014, 09:42 AM (This post was last modified: 06/10/2014, 09:44 AM by mike3.) Ah, so it gives n/2 correct digits (or thereabout), then. So at the 340 or so digits I was using, it should've had around 170 digits correct, which would probably have been enough to get the 32nd derivative provided the epsilon was not too small (I think I used 1e-4 or so, maybe 1e-5 but the last one may have been pushing it, although 1e-4 still produces a graphable approximation of the derivative). sheldonison Long Time Fellow Posts: 641 Threads: 22 Joined: Oct 2008 06/10/2014, 10:38 AM (This post was last modified: 06/10/2014, 11:02 AM by sheldonison.) (06/10/2014, 09:42 AM)mike3 Wrote: Ah, so it gives n/2 correct digits (or thereabout), then. So at the 340 or so digits I was using, it should've had around 170 digits correct, which would probably have been enough to get the 32nd derivative provided the epsilon was not too small (I think I used 1e-4 or so, maybe 1e-5 but the last one may have been pushing it, although 1e-4 still produces a graphable approximation of the derivative).There is an internal "xsexp" Taylor series polynomial at zero, that should work well for the derivatives. Earlier, I tried "\p 134", and that was starting to show significant errors (>0.1%) for the 100th derivative at x=0.5, since the higher order derivatives started to dominate the error term, and there wasn't enough precision in those higher order Taylor series terms. but that was at "\p 134". Also, you could use the internal function, sexptaylor(z0,r), where r is the sample radius for a 200 term polynomial generated via Cauchy approximation radius r centered at z0. The following algoritm will give accurate results for the 150th derivative, but starts to fall apart near the 175th derivative. Perhaps I could add the option for more sample points on sexptaylor when used with higher precision. Code:\p 134 init(exp(1)); dnz(n,z)={for (i=1,n,z=deriv(z));z} default(format,"g0.14") d150=dnz(150,sexptaylor(0.5,1)); print ("150th derivative at 0.5= "polcoeff(d150,0)); 150th derivative at 0.5= 2.4946223112445 E207 Alternatively, this algorithm works accurately to about the 370th derivative at x=0.5. Code:\p 539 init(exp(1)); /* wait about 2 hours */ dnz(n,z)={for (i=1,n,z=deriv(z));z} default(format,"g0.14") print ("360th derivative at 0.5: "subst(dnz(360,xsexp),x,0.5)); 360th derivative at 0.5: 2.1356744655704 E621 - Sheldon mike3 Long Time Fellow Posts: 368 Threads: 44 Joined: Sep 2009 06/10/2014, 01:46 PM (This post was last modified: 06/10/2014, 01:50 PM by mike3.) Well, this is certainly more efficient. I did manage to detect the oscillation produced by the $\theta$ mapping with an amplitude of $10^{-5}$. mike3 Long Time Fellow Posts: 368 Threads: 44 Joined: Sep 2009 06/13/2014, 09:29 AM (This post was last modified: 06/13/2014, 10:06 AM by mike3.) (06/09/2014, 12:54 PM)tommy1729 Wrote: Yes of course you are right. Sorry silly of me. But the argument still remains ... What is your opinion about the periodic functions that I tried ? Did you try those ? regards tommy1729 Exactly what $\theta(x)$ mapping did you try? What was its $\sup\ |\theta(x)|$? I tried $\sin(6 \pi x)^6$ (5 derivatives at $\frac{n}{6}$ 0, i.e. $|\theta^{(k)}(x)| = 0$ for $k < 6$ and $x = \frac{n}{6}$) divided by 100,000 and it was in fact much easier to pick that up than to pick up the usual sin mapping. EDIT: Just tried $\theta(x) = \frac{\sin(6 \pi x)^{25}}{100000}$. Darn easy to pick that up (got it on the 3rd derivative!) and it meets all your criteria (first 24 derivatives at every $\frac{n}{6}$ zero). I even tried it with $100000$ replaced by 1 billion (so $\sup\ |\theta(x)| = 10^{-9}$) and was still able to pick it up. I think I know why, too: any such function with lots of flat spots is going to have "steeper" (in a relative sense) peaks (the peaks are more "concentrated" in terms of the interval they cover regardless of their amplitude). That is, $|\theta'(x)|$ (and higher derivatives) is going to achieve bigger values for such functions at a given amplitude. Pure $\sin$ is probably the "softest" periodic function you can get, and so is actually the hardest one to detect. EDIT 2: Just tried a $\theta(x) = \frac{\sin(2\pi x) + \frac{1}{10} \sin(4 \pi x)}{K}$. I.e. like in my original post, only with an added harmonic. The added harmonic made it somewhat easier to pick up. So I conjecture that straight up $\sin(2\pi x)$ (divided by a $K$, of course) is the hardest periodic function to detect. And yes, you have to try $x < 1$. It looks it may still be detectable for larger $x$ but it would take a LOT more derivatives. tommy1729 Ultimate Fellow Posts: 1,370 Threads: 335 Joined: Feb 2009 06/13/2014, 10:47 PM Can you prove that if the property holds at $x = A$, it also holds at $x = A + B$ where $B$ is between 0 and 0.5 ? I believe that would be useful and could lead to induction which could prove the whole case. Also is there a value $C$ such that for $x > C$ we have $D^n tet(x) >= 0$ for all positive integer $n$ ? Maybe Im confused ... I once read about the same conditions as in the OP but for another function then $tet(x)$. So there must be some theory I think. I changed my mind, I think it might be true. regards tommy1729 sheldonison Long Time Fellow Posts: 641 Threads: 22 Joined: Oct 2008 06/14/2014, 05:15 AM (This post was last modified: 06/14/2014, 04:34 PM by sheldonison.) (06/13/2014, 10:47 PM)tommy1729 Wrote: Also is there a value $C$ such that for $x > C$ we have $D^n tet(x) >= 0$ for all positive integer $n$ ? Maybe Im confused ... I once read about the same conditions as in the OP but for another function then $tet(x)$. So there must be some theory I think. I changed my mind, I think it might be true. regards tommy1729 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. So, for all points on the real axis, the derivatives must eventually be governed by log(x+2), for large enough derivatives! And the even derivatives of log(x+2) are all negative. So I think that is a proof that for any point on the real axis, no matter how large, eventually the (2n)th derivative must be negative for large enough values of n. However, all of the odd derivatives are always positive. Now the other side of the conjecture is that for tet(z+theta(z)/k), eventually for large enough odd derivatives misbehave, and the odd derivatives are no longer all positive. I believe a possible avenue to proving this is to imagine a circle from (-1) to z0, where we are looking at the odd derivatives at a point (z0/2-1/2). Without the theta perturbation, the tetration function on the boundary of that circle has its maximum positive value at z0. Especially if we are dealing with a simple case like theta(z)=sin(2piz), then with the z+theta(z), permutation, we can prove that eventually as the circle gets bigger, the maximum absolute value occurs somewhere else on the circle, instead of at z0, since theta(z) grows exponentially and that exponential growth eventually overcomes the "k" in the denominator. Somewhere near or past that point, as the derivatives are now more or less randomly controlled by the phase of the maximum, and that means some of the odd derivatives must also misbehave. edit Making this is probably much more complicated (I don't know how yet). We may need to know how to approximate some of the derivatives at z0/2-1/2, which I'm also working on.... Another problem I'm working is how to actually approximate the very large derivatives for Tetration, that can't be easy calculated by Cauchy integral, so we can tell when the derivatives go from being dominated by super-exponential growth, to the derivatives being dominated by the singularity at x=-2, so that the even derivatives become negative. typos new prediction the 4904th derivative of sexp(z) at z=0.7 is the first negative derivative at z=0.7, with a Taylor series coefficient for a_4904 = -4.371521861406 E-2122. I need to post the algorithm later. Unfortunately, this is outside of the range that I can verify numericaly against a Cauchy series, so I tried again with z=0.5. This time the prediction was that the the 406th derivative would be the first negative derivative; with a Taylor series coefficient of a_406~=-9.3E-166. This is within the range that I could verify with a "\p 522" kneser Taylor series at 0.5, and the results mach with a_406 the first negative coefficient=-9.535 E-166. The algorithm is based on the algorithm in post#16, I used for the Taylor series of the entire half sexp approximation thread; it will take me a little while to write up the equations for this algorithm. But basically, we divide the sexp(z) into two parts and estimate the logarithmic singularity at -2 plus a second part, sexp(z)=exp(sexp(z-1)), where we look at the first and second derivative and values at z-1 to estimate the behavior at z, and use this to approximate the higher derivatives. Equations will be posted later, when I have time; plus I'm still working on simplifications to the algorithm, so I need more time. - Sheldon tommy1729 Ultimate Fellow Posts: 1,370 Threads: 335 Joined: Feb 2009 06/14/2014, 10:24 PM (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 ) Notice by adding the condition bounded in a strip we get an old uniqueness criterion. Quote: So, for all points on the real axis, the derivatives must eventually be governed by log(x+2), for large enough derivatives! And the even derivatives of log(x+2) are all negative. So I think that is a proof that for any point on the real axis, no matter how large, eventually the (2n)th derivative must be negative for large enough values of n. However, all of the odd derivatives are always positive. Hmm why not all of the odd derivatives are positive FOR LARGE ENOUGH values of n ?? Because if so , then it follows from sexp has a log singularity at -2 => f(x) = sexp(x) - log(x+2) is analytic for x > -3. now when expanded at y > -2 then f(x) has a larger radius than log(x+2) Thus the derivatives of log(x+2) expanded at y must eventually dominate the derivatives of f(x) expanded at y. SO expansion at y > -2 => sexp(x) = log(x+2) + f(x) DOMIN sexp(x) = DOMIN(log(x+2),f(x)) = DOMIN log(x+2). (QED) But again : Hmm why not all of the odd derivatives are positive FOR LARGE ENOUGH values of n ?? AND what happens with sexp(z+theta(z)) ? is f_2(z) = sexp(z+theta(z)) - log(z+theta(z)+2) still analytic for Re(z) > -3 ??? If sexp(z+theta(z)) is analytic for Re(z)>-2 then we seem to require log(z+theta(z)+2) ~ log(z+2) However theta(z) = 0 may have many solutions ... SO IN GENERAL it is NOT true ! We have a bad approximation and thus did not remove the singularity without adding another one. Notice : DOMIN sexp(z+theta(z)) = DOMIN(f_2(z),log(z+theta(z)+2)) Together with theta is analytic and 1 periodic with fixpoint 0 , this gives quite an intresting structure. Lets assume theta(z) = 0 only for one value z (= 0). then theta(z) is no longer periodic ! contradiction. But however no issue since these values are far from z = 0. SO we need to look again at : log(z+theta(z)+2) ~ log(z+2). log(z+theta(z)+2) - log(z+2) = g(z). Now Im convinced g'(0) = 1 is necc otherwise the approximation is not good. Hence this requires theta ' (0) = 0. IF NOT the singularity is not properly removed right ? And ofcourse if the singularities are not properly removed the Domination is not clear. This was also my motivation for having theta ' (0) = 0 in the past. ( such as sin^2(x)/K as theta where the derivative IS 0 , not just in the limit K -> oo ) Needs more investigation ... Quote:Now the other side of the conjecture is that for tet(z+theta(z)/k), eventually for large enough odd derivatives misbehave, and the odd derivatives are no longer all positive. I believe a possible avenue to proving this is to imagine a circle from (-1) to z0, where we are looking at the odd derivatives at a point (z0/2-1/2). Without the theta perturbation, the tetration function on the boundary of that circle has its maximum positive value at z0. See the comments above ! 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. I do not see then why it would be the case here. But maybe it requires just a bit more explication. This is not a zeta function afterall ! ( for the Riemann zeta we have that for Re(z) > 1 any circle that lies to the right of it , with real center A , reaches its max at the largest real on the real line touching the circle. This follows easily from the exp function. ) Also some fake function of ln(z+w(z)) seems to be a counterexample for some 1 periodic theta function w(z) that is choosen wisely ( to have signs for the nth derivative the way we want ). But fake function theory is overkill here. Quote: ... The algorithm is based on the algorithm in post#16, I used for the Taylor series of the entire half sexp approximation thread; it will take me a little while to write up the equations for this algorithm. Fake function theory again ! Lots of work to do again. regards tommy1729 " thruth is that what does not go away when you stop believing in it " tommy1729 Ultimate Fellow Posts: 1,370 Threads: 335 Joined: Feb 2009 06/14/2014, 10:41 PM Im looking for an old related thread but I cant find it ... tommy1729 Ultimate Fellow Posts: 1,370 Threads: 335 Joined: Feb 2009 06/14/2014, 11:22 PM (This post was last modified: 06/14/2014, 11:26 PM by tommy1729.) The idea of using Leibniz rule also comes to mind. sexp(x) = a0 + a1 x + a2 x^2/2 + ... theta(x) = t0 + t1 x + r2 x^2/2 + ... sexp2(x) = sexp(x+theta(x)) = b0 + b1 x + b2 x^2/2 + ... sexp2 ' (x) = sexp ' (x+theta(x)) * (1+theta '(x)) set x = 1 and use Leibniz rule to make a system of equations. also a0 = t0 = b0 = 0. Get a proof by contradiction that is theta(x) is 1-periodic from the system of equations. Needs more work ... regards tommy1729 sheldonison Long Time Fellow Posts: 641 Threads: 22 Joined: Oct 2008 06/15/2014, 04:51 AM (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. - Sheldon « Next Oldest | Next Newest »

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