Re: Why Indeterminate?
- To: mathgroup at smc.vnet.net
- Subject: [mg118524] Re: Why Indeterminate?
- From: Stefan <wutchamacallit27 at gmail.com>
- Date: Sun, 1 May 2011 06:21:20 -0400 (EDT)
- References: <ipgm80$8tb$1@smc.vnet.net>
On Apr 30, 5:54 am, Themis Matsoukas <tmatsou... 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 Themis, The message generated is Power::indet: Indeterminate expression 0.^0 encountered. >> A closer look shows that in your sum, when the index j is = 1, you have the expression (1-2*0.5)^(1-1) = 0^0 this is indeterminate, since we dont know anything about how x and j are related to eachother to do any sort of L'Hopital calculation to resolve the problem. When you run it for arbitrary x, it evaluates (1-2x)^(1-1) as (1-2x)^0, and evaluates this to 1, a perfectly good assumption for all (1-2x) =/= 0. This is the source of the error, but its hard to say if its actually a problem or not without context. -Stefan Salanski