Re: Taylor Series in R^n
- To: mathgroup at smc.vnet.net
- Subject: [mg8105] Re: Taylor Series in R^n
- From: "Steven T. Hatton" <hattons at cpkwebser5.ncr.disa.mil>
- Date: Tue, 12 Aug 1997 00:54:45 -0400
- Organization: Logicon supporting DISA
- Sender: owner-wri-mathgroup at wolfram.com
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.