Re: Why Indeterminate?
- To: mathgroup at smc.vnet.net
- Subject: [mg118526] Re: Why Indeterminate?
- From: Bob Hanlon <hanlonr at cox.net>
- Date: Sun, 1 May 2011 06:21:41 -0400 (EDT)
A[a_List, x_] := x*(1 - x)*Sum[(a[[j]]*(1 - 2*x)^(j - 1))/(1 - a[[3]]*(1 - 2*x)),
{j, 1, 2}]
A[Array[a,3],1/2]
Power::indet: Indeterminate expression 0^0 encountered. >>
Indeterminate
A[Array[a,3],x]/.x->1/2
a[1]/4
Looks like a bug. The workaround is to define the function with the substitution that you used.
A[a_List, x_] := Module[{t},
t*(1 - t)*Sum[(a[[j]]*(1 - 2*t)^(j - 1))/(1 - a[[3]]*(1 - 2*t)), {j, 1, 2}] /.
t -> x]
A[Array[a,3],1/2]
a[1]/4
Bob Hanlon
---- Themis Matsoukas <tmatsoukas at me.com> wrote:
=============
Consider this expression:
A[a_List, x_] := x (1 - x) \!\(
\*UnderoverscriptBox[\(\[Sum]\), \(j = 1\), \(2\)]
\*FractionBox[\(a[[j]]
\*SuperscriptBox[\((1 - 2\ x)\), \(j - 1\)]\), \(1 -
a[[3]] \((1 - 2 x)\)\)]\)
a = Range[3];
Evaluation at x=0.5 gives
A[a, 0.5]
Indeterminate
..but I can get the right answer if I use
A[a, x] /. x -> 0.5
0.25
What puzzles me is that there is no obvious indeterminacy in the original expression at x=0.5.
Thanks
Themis