       Re: A Series test

• To: mathgroup at smc.vnet.net
• Subject: [mg72766] Re: A Series test
• From: "dimitris" <dimmechan at yahoo.com>
• Date: Fri, 19 Jan 2007 00:53:16 -0500 (EST)
• References: <eonhr6\$i8\$1@smc.vnet.net>

```In:=
\$Version
rho = (x + a*I)/(x - a*I)
R = Abs[rho]
s = Series[R, {x, 0, 4}]
FullSimplify[s, a >= 0 && x >= 0]

Out=
"5.2 for Microsoft Windows (June 20, 2005)"
Out=
(I*a + x)/((-I)*a + x)
Out=
Abs[(I*a + x)/((-I)*a + x)]
Out=
Abs[(I*a + x)/((-I)*a + x)]
Out=
(a - I*x)/(a + I*x)

Using ComplexExpand

In:=
Clear["Global`*"]
rho = (x + a*I)/(x - a*I)
R = ComplexExpand[Abs[rho]]
s = Series[R, {x, 0, 4}]
FullSimplify[s, a >= 0 && x >= 0]

Out=
(I*a + x)/((-I)*a + x)
Out=
1
Out=
1
Out=
1

But I am not sure if this has any relevance.

Dimitris

> 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[Abs][-1]))/a^2,
> (((2*I)/3)*(3 + 6*Derivative[Abs][-1] - 2*Derivative[Abs][-1]))/
>     a^3, (2*(3 + 9*Derivative[Abs][-1] - 6*Derivative[Abs][-1] +
> Derivative[Abs][-1]))/(3*a^4)}, 0, 5, 1]
>
> The correct answer is 1. (The result with Simplify is more
> complicated.)  Thanks.

```