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.

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