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