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