Re: Confluent hypergeometric function of matrix argument

*To*: mathgroup at smc.vnet.net*Subject*: [mg24611] Re: Confluent hypergeometric function of matrix argument*From*: Roland Franzius <Roland.Franzius at uos.de>*Date*: Fri, 28 Jul 2000 17:23:47 -0400 (EDT)*References*: <200007242128.e6OLSAO943732@graphics.lcs.mit.edu>*Sender*: owner-wri-mathgroup at wolfram.com

Hi Matt, For a diagonal matrix just the normal 1F1-Function will do (in Version 4), if you map it to the diagonal elements only. Probably the following is not what you expect In[1]= Hypergeometric1F1[1/2, 3/2, {{0, 0, 0}, {0, 7, 0}, {0, 0, -2}}] Out[1]= {{1, 1, 1}, {1, 1/2*Sqrt[Pi/7]*Erfi[Sqrt[7]], 1}, {1, 1, 1/2*Sqrt[Pi/2]*Erf[Sqrt[2]]}} but this will be what you need In[2]= hypergeometric1F1[a_, b_, x_List] := DiagonalMatrix@ Hypergeometric1F1[a, b, Table[x[[i, i]], {i, 1, Length[x]}]] /; MatrixQ[x] In[3]= hypergeometric1F1[1/2, 3/2, {{0, 0, 0}, {0, 7, 0}, {0, 0, -2}}] Out[3]= {{1, 0, 0}, {0, 1/2*Sqrt[Pi/7]*Erfi[Sqrt[7]], 0}, {0, 0, 1/2*Sqrt[Pi/2]*Erf[Sqrt[2]]}} Perhaps anybody has a more elegant way to map to the diagonal only. good luck roland Matt Antone schrieb: > > Hi, thanks for your reply. I think I should have been a bit more clear > in my message; actually I want 1F1(a; b; M) where a and b are real > scalar values (for example, a = 1/2 and b = 3/2) and M is the matrix > argument, for example a diagonal matrix with one zero diagonal entry. > > So it's > > 1F1(1/2; 3/2; diag(0, d1, d2)) where d1 and d2 are real scalars. > > I've tried to use the standard scalar-argument 1F1 but can't get it to > work the way I want. -- Roland Franzius +++ exactly <<n>> lines of this message have value <<FALSE>> +++