Taylor Series in R^n

*To*: mathgroup at smc.vnet.net*Subject*: [mg8075] Taylor Series in R^n*From*: Andre Deprit <Deprit at his.com>*Date*: Tue, 5 Aug 1997 03:22:40 -0400*Organization*: Heller Information Services, Inc.*Sender*: owner-wri-mathgroup at wolfram.com

On August 2 Aug 1997, Steven T. Hatton left the following message: < I have been kicking around using MMA to generate the Taylor series < expansion of a single valued function of n variables. I think that < MMA can do this. I am not what one would call a power user at this < point. If anybody has done this, has seen it done or just knows off < the top of his or her head how to do this please share this insight. The problem is this: Let f[x,y,z] be a numerical function that is sufficiently differentiable at the origin. It is proposed to produce the Taylor formula for f at the origin to a given order, say 2. The one-line code below will do the job: Plus@@Series[f[x,y,z] /. Thread[{x,y,z}->eps {x,y,z}],{eps,0,2}][[3]] It amounts to multiplying the variables x, y and z by a scale factor eps, then initiating a Taylor series in eps at the origin. The coefficients of that Taylor series are stored as the third element of the structure SeriesData by which Mathematica represents a series expansion. Do not try to make the replacement eps-> 1 in the Series itself. Mathematica will protest, and rightly so. Here and now, I am not interested in converting this one-liner into a full-fledged code valid for any variables in any (finite!) dimension. I just wanted to convey the idea that the one-liner corresponds to what mathematicians define as the "Taylor Formula.", save for the remainder that the one-liner omits. Incidentally, a Taylor series for f[x,y,z] is not a Taylor series for f[x,y,z} at x=0 with coefficients that are themselves Taylor series at y = 0, and so on. A recursion of that sort does generate too many terms; besides, it breaks the homogeneity in order which is inherent to a Taylor formula in the mathematical sense.

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