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
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