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 Taylor series of upx function Zagreus Newbie Posts: 1 Threads: 1 Joined: Sep 2008 09/07/2008, 10:56 AM Hello, I'm a hobby mathematician, taking fun on working on tetration. I would like to know if a Taylor series development exists for the ultra exponential function. In my mind (correct me if I'm wrong), upx function as defined as the following: - upx(1) = e - upx(x+1) = exp ( upx(x) ) Thank you for your help ! bo198214 Administrator Posts: 1,389 Threads: 90 Joined: Aug 2007 09/10/2008, 01:56 PM Zagreus Wrote:I'm a hobby mathematician, taking fun on working on tetration. I would like to know if a Taylor series development exists for the ultra exponential function. In my mind (correct me if I'm wrong), upx function as defined as the following: - upx(1) = e - upx(x+1) = exp ( upx(x) ) Hi Zagreus, unfortunately there is no *the* tetration. There are several approaches how to define a Taylor-Series for tetration and it is not even clear whether some of them are equal as the coefficients are defined by limits and and even different solutions can come very close. For the different approaches have a look into the FAQ and/or search the forum. andydude Long Time Fellow Posts: 509 Threads: 44 Joined: Aug 2007 10/23/2008, 10:16 PM Since there is more than one tetration, however, we do know a little bit about each (regular tetration, and natural tetration are 2 of them). If you wanted to know more about natural tetration, here is an approximate power series: ${}^{x}e = \text{sexp}(x) = 1.0917(x+1) - 0.3244(x+1)^2 + 0.3498(x+1)^3 - 0.2308(x+1)^4 + \cdots$ One of the problems with this method, though, is that the coefficients of the power series are not exact. You can find out more about natural tetration and its power series from this thread and this thread for the inverse function. If you wanted to know more about regular tetration, then I suggest you read this thread, since it seems to be the first definition of it on this forum. Regular tetration is nice, because the coefficients are exact, but the problem is that it the power series is not an expansion of sexp(x), but an expansion of $\exp^x(z)$ about z, which is a completely different power series. If we simply set z=1, then it doesn't always converge, so we are stuck with a different problem. I hope that helps clarify a bit. Andrew Robbins Kouznetsov Fellow Posts: 151 Threads: 9 Joined: Apr 2008 11/16/2008, 01:40 PM (This post was last modified: 12/01/2008, 01:43 AM by Kouznetsov.) andydude Wrote:Since there is more than one tetration, however, we do know a little bit about each (regular tetration, and natural tetration are 2 of them). If you wanted to know more about natural tetration, here is an approximate power series: ${}^{x}e = \text{sexp}(x) = 1.0917(x+1) - 0.3244(x+1)^2 + 0.3498(x+1)^3 - 0.2308(x+1)^4 + \cdots$ One of the problems with this method, though, is that the coefficients of the power series are not exact. You can find out more about natural tetration and its power series from this thread and this thread for the inverse function. If you wanted to know more about regular tetration, then I suggest you read this thread, since it seems to be the first definition of it on this forum. Regular tetration is nice, because the coefficients are exact, but the problem is that it the power series is not an expansion of sexp(x), but an expansion of $\exp^x(z)$ about z, which is a completely different power series. If we simply set z=1, then it doesn't always converge, so we are stuck with a different problem. I hope that helps clarify a bit. Andrew Robbins There is EXACT Tailor series, if we insist that upx(z^*) = upx(z)^* and upx(z) is holomorphic outside the part of the negative part of axis, id est, holomorphic everywhere except $z\le 2$. Then the solution is unique. You can calculate so many derivatives as you like in any regular point, and, in particular, at z=0. The algorithm of evaluation is described in http://www.ils.uec.ac.jp/~dima/PAPERS/2008analuxp99.pdf P.S. Dear Andrew. 11-16-2008 11:58 PM , I did not have the table of derivatives in hands. Now I copypast the evaluation of first 10 terms: $\text{sexp}[z]= 1.09176735(z+1) -0.32449476(z+1)^{2} +0.34983627(z+1)^{3} -0.23085443(z+1)^{4} +0.20133021(z+1)^{5} -0.16435217(z+1)^{6} +0.14283634(z+1)^{7} -0.12469499(z+1)^{8} +0.11107354(z+1)^{9} -0.09995457(z+1)^{10}$ I compare it to your ${}^{x}e = \text{sexp}(x) = 1.0917(x+1) - 0.3244(x+1)^2 + 0.3498(x+1)^3 - 0.2308(x+1)^4 + \cdots$ and I see, the precision is not sufficient to declare the difference; Bo already had discussed this below. 1. Could you calculate more terms and more digits for the comarison? 2. All the "extensions" look similar along the real axis. The difference, if any, should grow exponentially in the direction of imaginary axis. Even it at the real axis the difference is of order of $10^{-24}$, we'll see it outside the real axis. Does your solution approach the fixed point of logarithm as the imaginary part of the argument grows up? 3. We may compare also the $\exp^x(z)$ ; the figures are posted at http://math.eretrandre.org/tetrationforu...hp?tid=206 and http://en.citizendium.org/wiki/Tetration, fig.9, 10. Do your figures look similar? 4. If you expand at minus unity, the radius of convergence is only unity. It is better to expand at zero; then the rarius of convergence is 2; the terms decay faster; the $n$th coefficient becomes smaller than $2^{-n}$ bo198214 Administrator Posts: 1,389 Threads: 90 Joined: Aug 2007 11/16/2008, 07:17 PM (This post was last modified: 11/16/2008, 07:21 PM by bo198214.) Kouznetsov Wrote:There is EXACT Tailor series, if we insist that upx(z^*) = upx(z)^* and upx(z) is holomorphic outside the part of the negative part of axis, id est, holomorphic everywhere except $z\le 2$. Then the solution is unique. Dmitrii, that is not yet proven. Even with what you call lemma about almost identical functions, we still need conditions for the slog, to make something unique here. I do not yet see a uniqueness criterion that does not depend on the shape of $\text{upx}(S)$. We have to make a waterproof proof before announcing something wrong! See my thread Universal Uniqueness Criterion II about what we really have in the moment. sheldonison Long Time Fellow Posts: 641 Threads: 22 Joined: Oct 2008 11/17/2008, 12:21 AM (This post was last modified: 11/17/2008, 12:26 AM by sheldonison.) bo198214 Wrote:Kouznetsov Wrote:There is EXACT Tailor series, if we insist that upx(z^*) = upx(z)^* and upx(z) is holomorphic outside the part of the negative part of axis, id est, holomorphic everywhere except $z\le 2$. Then the solution is unique. Dmitrii, that is not yet proven. Even with what you call lemma about almost identical functions, we still need conditions for the slog, to make something unique here. I do not yet see a uniqueness criterion that does not depend on the shape of $\text{upx}(S)$. We have to make a waterproof proof before announcing something wrong! See my thread Universal Uniqueness Criterion II about what we really have in the moment.Can this solution be verified to be different from the approximation that Andrew came up with using linear equations for slog? The values for Dmitri's this solution appear very close to Andrew's approximation. To me, it would seem as if the first couple of derivatives at x=-1,0 were the same for both solutions, than it is likely that they're the same solution, though the two could still diverge in the higher derivatives. Kouznetsov Fellow Posts: 151 Threads: 9 Joined: Apr 2008 11/17/2008, 04:11 AM sheldonison Wrote:... To me, it would seem as if the first couple of derivatives at x=-1,0 were the same for both solutions, than it is likely that they're the same solution, though the two could still diverge in the higher derivatives. I had submitted the evaluation of the first derivative with 14 (I hope) decimal digits. $\mathrm{tet}'(-1)=\mathrm{tet}'(0)\approx 1.09176735125832$ In the similar way, the second derivative can be evaluated, and so on. But I like the approach by Bo: it is better to prove, than to compare the numerical evaluations. We already have proof that no other tetration is allowed to be holomorphic in the complex plane with cutted out set $\{x~\in~ \mathbb{R} ~:~x\le -2 \}$. By the way, how about to add some small amount of finction $d(z)= \left\{ \begin{array}{ccc} \exp\!\Big(-0.1/x^2 - 0.1/(z+1)^2\big) &,& (x+1)x \ne 0\\ 0 &,& (x+1)x = 0 \end{array}\right.$?THis function has all the derivatives along the real axis, and all of them are zero at points 0 and -1. Consider the modification of tetration $\mathrm{tet}(z)$ to $\mathrm{tem}(z)=\mathrm{tet}(z)+d(z)$ with corresponding extension from the range [-1,0] usind the recurrent equation. At the real axis, the modified tetration tem has the same "expansion" as the tetration in these points; however, such a modification destroys the holomorphism. bo198214 Administrator Posts: 1,389 Threads: 90 Joined: Aug 2007 11/18/2008, 11:49 AM (This post was last modified: 11/18/2008, 11:50 AM by bo198214.) sheldonison Wrote:Can this solution be verified to be different from the approximation that Andrew came up with using linear equations for slog? The values for Dmitri's this solution appear very close to Andrew's approximation. To me, it would seem as if the first couple of derivatives at x=-1,0 were the same for both solutions, than it is likely that they're the same solution, though the two could still diverge in the higher derivatives. Thats always the difficulty with those numerical comparison. If the difference is small does that mean that they are equal? Here is a really good counterexample. There are two methods two compute an iterational square root of $\sqrt{2}^x$: regular iteration at the lower fixed point and regular iteration at the upper fixed point. The difference is in the order of $10^{-24}$. So you wouldnt have noticed it by usual numeric approximation. Kouznetsov Fellow Posts: 151 Threads: 9 Joined: Apr 2008 12/02/2008, 12:03 PM (This post was last modified: 12/02/2008, 12:49 PM by Kouznetsov.) Zagreus Wrote:.. I would like to know if a Taylor series development exists for the ultra exponential function. In my mind (correct me if I'm wrong), upx function as defined as the following: - upx(1) = e - upx(x+1) = exp ( upx(x) ) ..Dear Zagreus. If you are still interested... I have expressed the holomorphic function which satisfies the equation you wrote through the Cauchi integral. It allows the straigihtforward differentiation. I claim that such a function is unique, if we insist that $\mathrm{upx}(z^*)= \mathrm{upx}(z)^*$ $\mathrm{upx'}(x)>0 \forall x>-2$ upx is holonorphic in the complex plane except real values smaller or equal 2. I have evaluated several coefficients in the Tailor expansion: $\mathrm{upx}(z)=\sum_{n=0}^\infty c_n z^n ~\forall z\in \mathbb{C}: |z|<2$ Here is the table of the coefficients $c_0, c_1,c_1,..$: 1. , 1.091767351258322138 , 0.271483212901696469 , 0.212453248176258214 , 0.069540376139988952 , 0.044291952090474256 , 0.014736742096390039 , 0.008668781817225539 , 0.002796479398385586 , 0.001610631290584341 , 0.000489927231484419 , 0.000288181071154065 , 0.000080094612538551 , 0.000050291141793809 , 0.000012183790344901 , 0.000008665533667382 , 0.000001687782319318 , 0.000001493253248573 , 0.000000198760764204 , 0.000000260867356004 , 0.000000014709954143 , 0.000000046834497327 ,-0.000000001549241666 , 0.000000008741510781 ,-0.000000001125787310 , 0.000000001707959267 ,-0.000000000377858315 , 0.000000000349577877 ,-0.000000000105377012 , 0.000000000074590971 ,-0.000000000027175982 , 0.000000000016460766 ,-0.000000000006741873 , 0.000000000003725329 ,-0.000000000001639087 , 0.000000000000858364 ,-0.000000000000394374 , 0.000000000000200252 ,-0.000000000000094420 , 0.000000000000047121 The partial sum gives the polunomial approximation; while modulus of argument is of order of unity or smaller, it gives of order of 14 correct decimal digits. I plot the partial sum $F(z)=\sum_{n=0}^{25} c_n z^n$ below:     In the upper right corner, in the complex $z$ plane, the function $f= F(z)$ is shown with lines of constant real and lines of constant imaginary part. Levels $\Re(f)=-2,-1,0,1,2,3,4$ are shown with thick black curves. Levels $\Re(f)=-1.8,-1.6,-1.4,-1.2,-0.8,-0.6,-0.4,-0.2$ are shown with thin red curves. Levels $\Re(f)=1.8,1.6,1.4,1.2,0.8,0.6,0.4,0.2$ are shown with thin thin blue curves. Levels $\Im(f)=-2,-1$ are shown with thick red curves. Levels $\Im(f)=1,2$ are shown with thick blue curves. Levels $\Im(f)=\pm \pi,\pm 3\pi$ are shown with thick pink curves. Levels $\Im(f)=\pm 1.8, \pm 1.6, \pm 1.4, \pm 1.2, \pm 0.8, \pm 0.6, \pm 0.4,\pm 0.2$ are shown with thin green curves. In the upper left corner, I plot $f=\log(F(z))$ in the same notations. Singularities of log, id est, zeros of functions $F$, appear as small dence loops. You may interpret the left graphic as contour plot of constant $|F(z)|$ and constant arg$(F(z))$. Due to the second of your equations, in the central part, the left upper plot is just displacement od the upper right plot. If you print the pics at the transparencies you can see that the central parts overlap well. For those who have no overhead projector, I show the overlap of the two pics above in the botton left picture. For comparison, at the bottom right I show the central part of plot of tetration from my paper, which is expected to appear in Mathematics of Computation. Tetration is shown with lines of constant modulus and lines of constant phase. (If you displace the x-axis for unity, they become lines of constant real and constant imaginaty part) You can copypast the coefficients imto your program and plot similar pics by yourself. Enjoy! P.S. It is amaising how the polynomial tries to reproduce the behavior of lines $\Im(f)=\pm 1$ in vicinity of $z=2$, a little bit outside of the range of convergence of the series. If we take more terms, this detail disappears, although the lines incide the circle become more perfect. sheldonison Long Time Fellow Posts: 641 Threads: 22 Joined: Oct 2008 12/03/2008, 09:11 PM Kouznetsov Wrote:Dear Zagreus. If you are still interested... I have expressed the holomorphic function which satisfies the equation you wrote through the Cauchi integral. It allows the straigihtforward differentiation. I claim that such a function is unique, if we insist that $\mathrm{upx}(z^*)= \mathrm{upx}(z)^*$ $\mathrm{upx'}(x)>0 \forall x>-2$ upx is holonorphic in the complex plane except real values smaller or equal 2. I have evaluated several coefficients in the Tailor expansion: $\mathrm{upx}(z)=\sum_{n=0}^\infty c_n z^n ~\forall z\in \mathbb{C}: |z|<2$ Here is the table of the coefficients $c_0, c_1,c_1,..$: .....Thank you for your table! About as far as I could go was to plug the taylor series coefficients into excel and show that the function and its first derivative match for several values for upx(x) and upx(x+1). What does this equation mean? $\mathrm{upx}(z^*)= \mathrm{upx}(z)^*$ I was expecting more along the lines of upx(x+1) = exp ( upx(x) ) The next line is also intriguing.... $\mathrm{upx'}(x)>0 \forall x>-2$ Does this mean that other holomorphic solutions exist, but they all must have a negative derivative at some point x>-2? I started working on the algebra to generate the general equations for all 5th and 6th order upx approximations that have smooth 1st, 2nd, and 3rd derivatives, but not necessarily smooth 4th and 5th derivatives. You would claim that all members of that set, if extended with higher order coefficients so that all derivatives would be smooth, would have a negative 1st derivative somewhere? Thanks again for your post! - Sheldon « Next Oldest | Next Newest »

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