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MathGroup Archive 2005

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Re: general nth term of series

  • To: mathgroup at smc.vnet.net
  • Subject: [mg62933] Re: general nth term of series
  • From: dh <dh at metrohm.ch>
  • Date: Thu, 8 Dec 2005 06:25:37 -0500 (EST)
  • Organization: Cablecom Newsserver
  • References: <dn8qqq$hiq$1@smc.vnet.net>
  • Sender: owner-wri-mathgroup at wolfram.com

Hi Ash,
simply remember your math course:
n-th term = (1/n!) n-thederivative(f) (x-x0)^n

therefore, if f is an expression in x and we want the n-th term of the 
taylor series around x0 we may say:

TaylorTerm[f_, n_, x_, x0_] := (1/n!)(x - x0)^n (D[f, {x, n}] /. x -> x0)

Daniel

n00dle0 at yahoo.com wrote:
> Hi,
> 
> Is there a way in mathematica to obtain the general term of a taylor
> series expansion?
> 
> \!\(G[u_, x_] = 1\/v\((1 - 2*x*u + u\^2)\)\)
> 
> Series[G[u, 0], {u, 0, 8}]
> 
> 
> Thanks,
> Ash
> 


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