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Re: A Series test


In[8]:=
$Version
rho = (x + a*I)/(x - a*I)
R = Abs[rho]
s = Series[R, {x, 0, 4}]
FullSimplify[s, a >= 0 && x >= 0]

Out[8]=
"5.2 for Microsoft Windows (June 20, 2005)"
Out[9]=
(I*a + x)/((-I)*a + x)
Out[10]=
Abs[(I*a + x)/((-I)*a + x)]
Out[11]=
Abs[(I*a + x)/((-I)*a + x)]
Out[12]=
(a - I*x)/(a + I*x)

Using ComplexExpand

In[18]:=
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[19]=
(I*a + x)/((-I)*a + x)
Out[20]=
1
Out[21]=
1
Out[22]=
1

But I am not sure if this has any relevance.

Dimitris

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.


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