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.