Re: general nth term of series

*To*: mathgroup at smc.vnet.net*Subject*: [mg63018] Re: general nth term of series*From*: Peter Pein <petsie at dordos.net>*Date*: Sun, 11 Dec 2005 04:56:40 -0500 (EST)*References*: <dnbmun$5qm$1@smc.vnet.net> <dnedel$cn$1@smc.vnet.net>*Sender*: owner-wri-mathgroup at wolfram.com

Peter Pein schrieb: > Hi Ash, > > SeriesTerm[(1 - 2*x*u + u^2)^(-1/2), {u, 0, n}] > gives LegendreP[n,x] without any Gammas in Version 5.1 > > Use SeriesTerm with care. It is quite buggy for general n: > > Series[Sin[x]/(1 + x), {x, 0, 5}]//Normal > --> x - x^2 + (5*x^3)/6 - (5*x^4)/6 + (101*x^5)/120 > > SeriesTerm[Sin[x]/(1 + x), {x, 0, 5}] > --> 101/120 > > is OK, but: > > SeriesTerm[Sin[x]/(1 + x), {x, 0, n}] /. n -> 5 > --> Sqrt[Pi/2]*BesselJ[1/2, 1] > > N[120 %] > --> 100.977 > > _Incidentally_ almost good... > > SeriesTerm gives for this example (-(-1)^n)*Sqrt[Pi/2]*BesselJ[1/2, 1] > as coefficient of x^n. :-\ > > Peter > I should have pointed out that Sqrt[Pi/2]BesselJ[1/2,1]==Sin[1] which is not too far from the coefficient of x^(2k+1) in f[x]=(1 - x)Sum[Round[Sin[1](2k + 1)!]/(2k + 1)!*x^(2k + 1),{k, 0, â??}]. I have no idea how Mathematica should derive the representation f[x]=Sum[(-1)^(k)*Sum[(-1)^j/(2*j + 1)!, {j, 0, Floor[k/2]}]x^(k + 1), {k, 0, â??}], because the result of RSolve[{a[0]==0,a[1]==1,a[2]==0,(k+1)(k+2)(a[k+1]+a[k+2])+a[k]+a[k-1]==0},a,k] is spectacular ;-) but doesn't really help :-(. Peter

**Follow-Ups**:**Re: Re: general nth term of series***From:*Devendra Kapadia <dkapadia@wolfram.com>