Re: Difficulties with Complex-Modulus Series

• To: mathgroup at smc.vnet.net
• Subject: [mg72751] Re: [mg72712] Difficulties with Complex-Modulus Series
• From: Bob Hanlon <hanlonr at cox.net>
• Date: Wed, 17 Jan 2007 07:32:57 -0500 (EST)

```\$Version

5.2 for Mac OS X (June 20, 2005)

r = (2*I+x)/(2*I-x);

R=Abs[r];

Series[R,{x,0,4}]

Abs[(2*I + x)/(2*I - x)]

Series[R,{x,0,4},Assumptions->Element[x,Reals]]

SeriesData[x, 0, {-1, I, 1/2, -I/4, -1/8}, 0, 5, 1]

Series[FullSimplify[R,Element[x,Reals]],{x,0,4}]

1

Series[R,{x,0,4},Assumptions->x>=0]

SeriesData[x, 0, {-1, I, 1/2, -I/4, -1/8}, 0, 5, 1]

Series[FullSimplify[R,x>=0],{x,0,4}]

1

Series[ComplexExpand[R],{x,0,4}]

1

Series[R,{x,Infinity,4}]

Abs[(2*I + x)/(2*I - x)]

Series[R,{x,Infinity,4},Assumptions->Element[x,Reals]]

Abs[(2*I + x)/(2*I - x)]

Series[FullSimplify[R,Element[x,Reals]],{x,Infinity,4}]

1

Series[R,{x,Infinity,4},Assumptions->x>=0]

SeriesData[x, Infinity, {-1, -4*I, 8, 16*I, -32}, 0, 5, 1]

Series[FullSimplify[R,x>=0],{x,Infinity,4}]

1

Series[ComplexExpand[R],{x,Infinity,4}]

1

Hence, use assumptions to simplify an expression before using Series

Bob Hanlon

> Say I have  r = (2*I+x)/(2*I-x),  in which x is real and nonnegative.
>
> Series[r,{x,0,4}]  and  Series[r,{x,Infinity,4}]   work as expected.
>
> Introduce now  R=Abs[r] and try the same:
>
> Series[R,{x,0,4}]  and  Series[R,{x,Infinity,4}]
>
> Results are now "contaminated" with Abs'[-1], Abs''[-1], etc,
> I dont understand the presence of those derivatives.
> Anybody can explain the reason?  (I teach students that the
> derivative of a constant is zero, but perhaps that has changed
> with the new year)  BTW it would be nice to say
>
> Series[R,{x,0,4}, x>=0]  or Series[R,{x,0,4}, R>=0]  etc
>
> if that would get rid of the garbage,  but Mathematica 5
> does not allow Assumptions in Series. Note BTW that R=1 for
> any x, so the R series are in fact trivial to any order.
>

```

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