Re: Apart question
- To: mathgroup at smc.vnet.net
- Subject: [mg73070] Re: Apart question
- From: "dimitris" <dimmechan at yahoo.com>
- Date: Thu, 1 Feb 2007 03:10:54 -0500 (EST)
- References: <epn9uv$cn3$1@smc.vnet.net><epp6g5$d5o$1@smc.vnet.net>
Dear Paul,
I really appreciate your response. Thanks a lot!
Best Regards
Dimitris Anagnostou
Ï/Ç Paul Abbott Ýãñáøå:
> In article <epn9uv$cn3$1 at smc.vnet.net>,
> "dimitris" <dimmechan at yahoo.com> wrote:
>
> > Dear All,
> >
> > In[317]:=
> > f[x_]:=(x^2+2*x+4)/(x^4-7*x^2+2*x+17)
> >
> > In[323]:=
> > Apart[f[x]]
> >
> > Out[323]=
> > (4 + 2*x + x^2)/(17 + 2*x - 7*x^2 + x^4)
> >
> > In[320]:=
> > Times@@Apply[#1[[1]] - #1[[2]] & , Solve[Denominator[f[x]] == 0, x],
> > 1]
> > Apart[(4 + 2*x + x^2)/%]
> > Map[FullSimplify, %, 1]
> >
> > Out[320]=
> > (I/2 - (1/2)*Sqrt[15 - 4*I] + x)*(-(I/2) - (1/2)*Sqrt[15 + 4*I] +
> > x)*((1/2)*(I + Sqrt[15 - 4*I]) + x)*
> > ((1/2)*(-I + Sqrt[15 + 4*I]) + x)
> >
> > Out[321]=
> > (4*I*((4 + 15*I) + (1 + 2*I)*Sqrt[15 - 4*I]))/(Sqrt[15 - 4*I]*(-2*I +
> > Sqrt[15 - 4*I] - Sqrt[15 + 4*I])*
> > (-2*I + Sqrt[15 - 4*I] + Sqrt[15 + 4*I])*(-I + Sqrt[15 - 4*I] -
> > 2*x)) - (4*I*((-4 - 15*I) + (1 + 2*I)*Sqrt[15 - 4*I]))/
> > (Sqrt[15 - 4*I]*(2*I + Sqrt[15 - 4*I] - Sqrt[15 + 4*I])*(2*I +
> > Sqrt[15 - 4*I] + Sqrt[15 + 4*I])*(I + Sqrt[15 - 4*I] + 2*x)) -
> > (4*((15 + 4*I) + (2 + I)*Sqrt[15 + 4*I]))/(Sqrt[15 + 4*I]*(-2*I +
> > Sqrt[15 - 4*I] - Sqrt[15 + 4*I])*
> > (2*I + Sqrt[15 - 4*I] + Sqrt[15 + 4*I])*(-I - Sqrt[15 + 4*I] +
> > 2*x)) + (4*((-15 - 4*I) + (2 + I)*Sqrt[15 + 4*I]))/
> > (Sqrt[15 + 4*I]*(-2*I - Sqrt[15 - 4*I] + Sqrt[15 + 4*I])*(-2*I +
> > Sqrt[15 - 4*I] + Sqrt[15 + 4*I])*(-I + Sqrt[15 + 4*I] + 2*x))
> >
> > Out[322]=
> > 1/(1 + Sqrt[-15 + 4*I] - 2*I*x) + 1/(1 - Sqrt[-15 - 4*I] + 2*I*x) + 1/
> > (1 + Sqrt[-15 - 4*I] + 2*I*x) -
> > 1/(-1 + Sqrt[-15 + 4*I] + 2*I*x)
> >
> > In[323]:=
> > Options[Apart]
> >
> > Out[323]=
> > {Modulus -> 0, Trig -> False}
> >
> > Why Apart cannot provide straightly the output Out[322]?
>
> Why does Factor not factor the denominator into linear factors? With the
> appropriate extension (Extension -> Automatic does not work), one can
> factor the denominator
>
> Factor[(x^2 + 2*x + 4)/(x^4 - 7*x^2 + 2*x + 17),
> Extension -> {Sqrt[-15 + 4*I], Sqrt[-15 - 4*I]}]
>
> and then obtain the desired result using Apart and FullSimplify.
>
> FullSimplify /@ Apart[%]
>
> Cheers,
> Paul
>
> _______________________________________________________________________
> Paul Abbott Phone: 61 8 6488 2734
> School of Physics, M013 Fax: +61 8 6488 1014
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