Re: Series problem

*To*: mathgroup at smc.vnet.net*Subject*: [mg2119] Re: Series problem*From*: "Brian J. Albright" <albright at physics.ucla.edu>*Date*: Wed, 5 Jun 1996 01:38:32 -0400*Organization*: UCLA Department of Physics and Astronomy*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. > Computing the series in terms of x don't work, because the exponetial and > the divisor will be expanded as well, when the series of the Cosine is > multiplied by the other factors. > Making the series expansion in terms of b don't work, because Mathematica > can't integrate the resulting series expansion in terms of x. > Are there any way to handle this except by introducing a new representation > for function series. > (The problem I'm currently interested in is finding a series for the function > f[b_,k_] = Integrate[Cos[b x] Exp[-x^2]/(k^2+x^2),{x,-Infinity,Infinity}], > which is valid for small b) > -- > ------------------------------------------------------------------------- > Tommy Nordgren "Home is not where you are born, > Royal Institute of Technology but where your heart finds peace." > Stockholm Tommy Nordgren - The dying old crone > f85-tno at nada.kth.se Hi. Here's a quick way to get the nth term in a Taylor expansion in b for your function, provided that the contour you want is the one along the real axis, and provided that k is chosen in such a manner as to make the integrand regular along the contour (like, e.g., k real and nonzero). g[n_] := Integrate[ Normal[ Series[ Cos[b x] Exp[-x^2]/(k^2+x^2),{b,0,n}]], {x,-Infinity,Infinity} ] Be careful of the poles; for example, k=0 leads to a divergent integral. Hope this helps. -Brian -- Brian J. Albright Department of Physics and Astronomy, UCLA albright at physics.ucla.edu http://bohm.physics.ucla.edu/~albright ==== [MESSAGE SEPARATOR] ====