Re: Why Indeterminate?
- To: mathgroup at smc.vnet.net
- Subject: [mg118522] Re: Why Indeterminate?
- From: Bill Rowe <readnews at sbcglobal.net>
- Date: Sun, 1 May 2011 06:20:58 -0400 (EDT)
On 4/30/11 at 5:53 AM, tmatsoukas at me.com (Themis Matsoukas) 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.
But there is an indeterminate term. When x = 0.5, 1 - 2 x is 0.
This term appears both in the numerator and denominator of your
expression as a multiplicative term giving 0/0 an indeterminate
expression as reported by Mathematica.
But when you do the sum symbolically it reduces to:
In[25]:= A[a, x] // Simplify
Out[25]= ((x - 1)*x*(4*x - 3))/(6*x - 2)
which has no zero term when x = .5
A couple of asides. First, your post would be much easier to
read had you first converted your expression to input form
before pasting in to the email. That is:
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 = Range[3];
is much easier to read and understand than what you posted.
Second, the way I quickly found the problem was by making use of
Trace. When you run into problems like this, going through the
output of Trace will often make the problem more apparent.