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 Taylor series of upx function 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 $2^{-n}$ « Next Oldest | Next Newest »

 Messages In This Thread Taylor series of upx function - by Zagreus - 09/07/2008, 10:56 AM RE: Taylor series of upx function - by bo198214 - 09/10/2008, 01:56 PM RE: Taylor series of upx function - by andydude - 10/23/2008, 10:16 PM RE: Taylor series of upx function - by Kouznetsov - 11/16/2008, 01:40 PM RE: Taylor series of upx function - by bo198214 - 11/16/2008, 07:17 PM RE: Taylor series of upx function - by sheldonison - 11/17/2008, 12:21 AM RE: Taylor series of upx function - by Kouznetsov - 11/17/2008, 04:11 AM RE: Taylor series of upx function - by bo198214 - 11/18/2008, 11:49 AM RE: Taylor series of upx function - by sheldonison - 03/05/2009, 06:48 PM RE: Taylor series of upx function - by Kouznetsov - 12/02/2008, 12:03 PM RE: Taylor series of upx function - by sheldonison - 12/03/2008, 09:11 PM RE: Taylor series of upx function - by Kouznetsov - 12/05/2008, 12:30 AM

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