Re: extracting fractional powers of series expansion?

• To: mathgroup at smc.vnet.net
• Subject: [mg78294] Re: [mg78225] extracting fractional powers of series expansion?
• From: DrMajorBob <drmajorbob at bigfoot.com>
• Date: Wed, 27 Jun 2007 05:41:31 -0400 (EDT)
• References: <26733444.1182852184761.JavaMail.root@m35>
• Reply-to: drmajorbob at bigfoot.com

```For example:

Series[Sqrt[Sin[x]], {x, 0, 10}]
% // FullForm

Sqrt[x]-x^(5/2)/12+x^(9/2)/1440-x^(13/2)/24192-(67
x^(17/2))/29030400+O[x]^(21/2)

SeriesData[x,0,List[1,0,0,0,Rational[-1,12],0,0,0,Rational[1,1440],0,0,0,Rational[-1,24192],0,0,0,Rational[-67,29030400]],1,21,2]

The first argument to SeriesData is the variable (x in this case) that's
raised to powers in the sum. The last argument, 2, is the denominator of
all the powers... so we're really talking about powers of Sqrt[x]. Those
powers range from 1 to 21 (the fourth and fifth arguments). The third
argument is the list of coefficients.

You can puzzle things out from there, if necessary, but this also works:

SeriesCoefficient[Sqrt[Sin[x]], {x, 0, 21/2}]

-1/5677056

Here's a more complicated example:

Series[x!, {x, Infinity, 5}]
% // FullForm

E^SeriesData[\$CellContext`x, Infinity, {-1 - Log[\$CellContext`x^(-1)]},
-1, 6, 1]*SeriesData[\$CellContext`x, Infinity, {Sqrt[2*Pi], 0, Rational[1,
6]*Sqrt[Rational[1, 2]*Pi], 0, Rational[1, 144]*Sqrt[Rational[1, 2]*Pi],
0, Rational[-139, 25920]*Sqrt[Rational[1, 2]*Pi], 0, Rational[-571,
1244160]*Sqrt[Rational[1, 2]*Pi], 0, Rational[163879,
104509440]*Sqrt[Rational[1, 2]*Pi]}, -1, 11, 2]

Times[Power[E,SeriesData[x,DirectedInfinity[1],List[Plus[-1,Times[-1,Log[Power[x,-1]]]]],-1,6,1]],SeriesData[x,DirectedInfinity[1],List[Power[Times[2,Pi],Rational[1,2]],0,Times[Rational[1,6],Power[Times[Rational[1,2],Pi],Rational[1,2]]],0,Times[Rational[1,144],Power[Times[Rational[1,2],Pi],Rational[1,2]]],0,Times[Rational[-139,25920],Power[Times[Rational[1,2],Pi],Rational[1,2]]],0,Times[Rational[-571,1244160],Power[Times[Rational[1,2],Pi],Rational[1,2]]],0,Times[Rational[163879,104509440],Power[Times[Rational[1,2],Pi],Rational[1,2]]]],-1,11,2]]

The first argument of Times is, of course, a function to be multiplied by
the second argument:

E^SeriesData[x, Infinity, {-1 - Log[x^(-1)]}, -1, 6, 1]

\[ExponentialE]^(-1-Log[1/x])/(1/x)+O[1/x]^6

The second argument of Times is a series with fractional powers, much like
the first example, but now the simple extraction method doesn't work, since

SeriesCoefficient[x!, {x, Infinity, 9/2}]

returns unevaluated.

I'm not sure there's any method that works in every case, but you can see,
I hope, how you'd have to proceed in an individual case. Good luck!

Bobby

On Tue, 26 Jun 2007 03:32:16 -0500, Tanim Islam <tanim.islam at gmail.com>
wrote:

> Hi:
>
> How do I extract a given fractional power term in a series expansion? For
> instance, I do a series expansion of a function about infinity. How do I
> extract the -1/2 power term with, say, SeriesCoefficient. I am using
> Mathematica 6.
>
> Tanim Islam
>
>

--
DrMajorBob at bigfoot.com

```

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