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MathGroup Archive 1996

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Re: Series problem

  • To: mathgroup at smc.vnet.net
  • Subject: [mg4123] Re: Series problem
  • From: Paul Abbott <paul at earwax.pd.uwa.edu.au>
  • Date: Wed, 5 Jun 1996 01:39:14 -0400
  • Organization: University of Western Australia
  • Sender: owner-wri-mathgroup at wolfram.com

Tommy Nordgren wrote:
 
>         I want to expand the Cosine only in the expression:
>Cos[b x] Exp[-x^2]/(k^2+x^2) into a taylor series around 0.

How about

Cos[b x] Exp[-x^2]/(k^2+x^2) /. Cos[a_] :> Normal[Cos[a] + O[x]^3]

       2  2
      b  x
  1 - -----
        2
-------------
  2
 x    2    2
E   (k  + x )

Note that delayed replacement (:>) is required for otherwise the series 
on the right-hand side of the rule is evaluated befor the replacement 
operation is carried out.

> Making the series expansion in terms of b don't work, because 
> Mathematica
> can't integrate the resulting series expansion in terms of x.

Yes it can:

Cos[b x] Exp[-x^2]/(k^2+x^2) + O[b]^3

                      2  2
      1              x  b             3
------------- - --------------- + O[b]
  2                 2
 x    2    2       x    2    2
E   (k  + x )   2 E   (k  + x )

It is best to Map the integration operation over each term in this 
expression:

(Integrate[#, {x,-Infinity,Infinity}]& /@
	Normal[%]) // PowerExpand

  2
 k
E   Pi (1 - Erf[k])
------------------- + 
         k
 
        2
    2  k                                   1       2
   b  E   k Sqrt[Pi] (2 Sqrt[Pi] + Gamma[-(-), 0, k ])
                                           2
   ---------------------------------------------------
                            4

> Are there any way to handle this except by introducing a new 
> representation for function series.

Another way is to use parametric differentiation.  Noting that

D[Exp[-a x^2]/(k^2+x^2),a]

         2
        x
-(---------------)
      2
   a x    2    2
  E     (k  + x )

then the integral:

gen[a_] = Integrate[Exp[-a x^2]/(k^2+x^2),
	{x,-Infinity,Infinity}] // PowerExpand

    2
 a k
E     Pi (1 - Erf[Sqrt[a] k])
-----------------------------
              k

generates all the terms in the series expansion of Cos[b x]:

gen[1]

  2
 k
E   Pi (1 - Erf[k])
-------------------
         k

gen'[1]

              2
             k
-Sqrt[Pi] + E   k Pi (1 - Erf[k])

Hence another (simpler) representation for the integral is

Sum[b^(2n)/(2n)! Derivative[n][gen][1], {n,0,1}]

                  2                        2
 2               k                        k
b  (-Sqrt[Pi] + E   k Pi (1 - Erf[k]))   E   Pi (1 - Erf[k])
-------------------------------------- + -------------------
                  2                               k

Cheers,
	Paul 
_________________________________________________________________ 
Paul Abbott
Department of Physics                       Phone: +61-9-380-2734 
The University of Western Australia           Fax: +61-9-380-1014
Nedlands WA  6907                         paul at physics.uwa.edu.au 
AUSTRALIA                           http://www.pd.uwa.edu.au/Paul
_________________________________________________________________

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