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Re: Taylor Series in R^n

  • To: mathgroup at smc.vnet.net
  • Subject: [mg8106] Re: Taylor Series in R^n
  • From: Paul Abbott <paul at physics.uwa.edu.au>
  • Date: Tue, 12 Aug 1997 00:54:46 -0400
  • Organization: University of Western Australia
  • Sender: owner-wri-mathgroup at wolfram.com

Andre Deprit wrote:

> 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.

I think that using Normal is preferable to using Plus and [[3]] because,
among other things, the internal format of Series could change.  After
using Normal you can make the replacement eps-> 1

 In[1]:= series[v_List] :=  Expand[Normal[Series[f @@ v /. Thread[v ->
eps v], 
      {eps, 0, 2}]] /. eps -> 1]

> 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.

The above code works in any (finite!) dimension:

 In[2]:= series[{x, y, z}]
 Out[2]=

1  (2,0,0)           2    (1,0,0)
- f       [0, 0, 0] x  + f       [0, 0, 0] x + 
2
 
     (1,0,1)                 (1,1,0)
  z f       [0, 0, 0] x + y f       [0, 0, 0] x + 
 
                  (0,0,1)
  f[0, 0, 0] + z f       [0, 0, 0] + 
 
  1  2  (0,0,2)               (0,1,0)
  - z  f       [0, 0, 0] + y f       [0, 0, 0] + 
  2
 
       (0,1,1)            1  2  (0,2,0)
  y z f       [0, 0, 0] + - y  f       [0, 0, 0]
                          2

Cheers,
	Paul 

____________________________________________________________________ 
Paul Abbott                                   Phone: +61-8-9380-2734
Department of Physics                           Fax: +61-8-9380-1014
The University of Western Australia           
Nedlands WA  6907                     mailto:paul at physics.uwa.edu.au 
AUSTRALIA                              http://www.pd.uwa.edu.au/Paul

            God IS a weakly left-handed dice player
____________________________________________________________________


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