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Re: Why Indeterminate?

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On 4/30/11 at 5:53 AM, tmatsoukas@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.

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Re: Why Indeterminate? Bill Rowe <readnews@sbcglobal.net> - 2011-05-01 10:21 +0000

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