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Re: Why Indeterminate?

  • To: mathgroup at smc.vnet.net
  • Subject: [mg118533] Re: Why Indeterminate?
  • From: "Nasser M. Abbasi" <nma at 12000.org>
  • Date: Mon, 2 May 2011 06:50:25 -0400 (EDT)
  • References: <ipgm80$8tb$1@smc.vnet.net>
  • Reply-to: nma at 12000.org

On 4/30/2011 2:54 AM, 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];
>

To post something that one can read, I suggest the following:

1. select the cell, and convert it to inout form
2. copy the cell using COPY AS plain text
3. paste the result to your news reader. The above become

----------------------------------------
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];
---------------------------------------

Now one can see it.   

> Evaluation at x=0.5 gives
>
> A[a, 0.5]
>
> Indeterminate
>

But it clear why this is, Mathematica says so. zero raised to power zero
is indeterminate.

When x=0.5, and j=1, you get 0^0

> ..but I can get the right answer if I use
>
> A[a, x] /. x ->  0.5
>
> 0.25
>

Because when you substitute at the end, the expression which would
have resulted in 0^0 is no longer there due to the sum being
finalized, so the final expression is now the total sum.

> What puzzles me is that there is no obvious
>  indeterminacy in the original expression at x=0.5.
>
> Thanks
>
> Themis
>

Ofcourse there is. Mathematica even prints on the screen:

"Power::indet: Indeterminate expression {At Line = 127, the
input was:,A[a,0.5],0.^0} encountered. >>

--Nasser





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