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Re: Taylor Series in R^n

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  • Subject: [mg8105] Re: Taylor Series in R^n
  • From: "Steven T. Hatton" <hattons at>
  • Date: Tue, 12 Aug 1997 00:54:45 -0400
  • Organization: Logicon supporting DISA
  • Sender: owner-wri-mathgroup at

SHAZAM! You're one smart feller {;-{)>.  Let me see if I get this:

Plus@@:  This applies Plus to the subsequent list.
Thread[{x,y,z}->eps {x,y,z}: This just makes {x -> eps x, y -> eps y, z
-> eps z}
Series[f[x,y,z] /. Thread[{x,y,z}->eps {x,y,z}],{eps,0,2}]: This
generates a "Taylor" series in eps about the origin in R.
[[3]]:  This little feller here just picks out the coefficients in that
series for Plus@@ to chew on.
In other words the series in eps that is generated about 0 has as its
coefficients the terms of the Taylor series in {x, y, z}.  That's pretty
slick.  I call that a Maclaurin series.  (See Calculus, by Howard
Anton.)  A Maclaurin series is just the special case of a Taylor series
expanded about 0.

The other part of what you said, the part about the Taylor series not
being generated by the recursive method you have described, I take to
mean that the method that MMA uses to evaluate Series[{x1, x2, ...
xn}...] does not really generate a true Taylor series.

Thank you for your contribution to my understanding of both MMA and
math.  Deciphering your one-liner was a worthwhile exercise.  One day I
hope to beguile newbies with these tricky one-liners.

Andre Deprit wrote:

> 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.
> 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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