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 >