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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>> +++
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