Re: A Series test

• To: mathgroup at smc.vnet.net
• Subject: [mg72778] Re: A Series test
• From: Jean-Marc Gulliet <jeanmarc.gulliet at gmail.com>
• Date: Fri, 19 Jan 2007 01:41:30 -0500 (EST)
• Organization: The Open University, Milton Keynes, UK
• References: <eonhr6\$i8\$1@smc.vnet.net>

```carlos at colorado.edu wrote:
> Just curious. Could somebody pls run this script on the
> latest Mathematica user version (I think it's 5.2) under
> Windows or Unix  and report the results:
>
> rho=(x+a*I)/(x-a*I); R=Abs[rho];
> s=Series[R,{x,0,4}];
> Print[FullSimplify[s,a>=0&&x>=0]//InputForm];
>
> My 5.0 answer (Mac G5 under OS 10.4.8) is
>
> SeriesData[x, 0, {1, (-2*I)/a, (-2*(1 + Derivative[2][Abs][-1]))/a^2,
> (((2*I)/3)*(3 + 6*Derivative[2][Abs][-1] - 2*Derivative[3][Abs][-1]))/
>     a^3, (2*(3 + 9*Derivative[2][Abs][-1] - 6*Derivative[3][Abs][-1] +
> Derivative[4][Abs][-1]))/(3*a^4)}, 0, 5, 1]
>
> The correct answer is 1. (The result with Simplify is more
> complicated.)  Thanks.
>

In[1]:=
\$Version

Out[1]=
"5.2 for Microsoft Windows (June 20, 2005)"

In[2]:=
rho = (x + a*I)/(x - a*I); R = Abs[rho];
s = Series[R, {x, 0, 4}];
Print[InputForm[FullSimplify[s, a >= 0 && x >= 0]]];

From In[2]:=
(a - I*x)/(a + I*x)

(* Why not using ComplexExpand? *)

In[5]:=
ComplexExpand[s]

Out[5]=
1

Regards,
Jean-Marc

```

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