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Re: A Series test


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.

I'll leave the 5.2 run to others. Here is what happens in the 
development kernel of Mathematica.

rho = (x+a*I)/(x-a*I);
absrho = Abs[rho];

In[20]:= InputForm[s0 = Series[absrho, {x,0,4}]]
Out[20]//InputForm= Abs[(I*a + x)/((-I)*a + x)]

In[21]:= InputForm[s0b = FullSimplify[s0,Assumptions->a>=0&&x>=0]]
Out[21]//InputForm= 1

Other possibile routes include using

absrho2 = Sqrt[rho*Conjugate[rho]];

or calling ComplexExpand explicitly to get your absolute value. This 
last will give 1 immediately. Also will do that in version 5.2, if you 
use Abs rather than Sqrt[rho*Conjugate[rho]].


Daniel Lichtblau
Wolfram Research



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