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Re: Why Indeterminate?

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On 30/04/2011 10:54, 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.
>
> Thanks
>
> Themis
>

I presume you realise that when j=1 and x=0.5, you get a term of the 
form 0^0 - which is indeterminate! The reason substituting x->0.5 
afterwords works, is that Mathematica performs the simplification

x^0 == 1

on the assumption that x doesn't equal 0. A lot of algebraic expressions 
have exceptional values at which the expression blows up, but the 
simplifications take place on the assumption that the variables don't 
have these critical values.

David Bailey
http://www.dbaileyconsultancy.co.uk

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Why Indeterminate? Themis Matsoukas <tmatsoukas@me.com> - 2011-04-30 09:54 +0000
  Re: Why Indeterminate? David Bailey <dave@removedbailey.co.uk> - 2011-05-01 10:20 +0000
  Re: Why Indeterminate? Stefan <wutchamacallit27@gmail.com> - 2011-05-01 10:21 +0000

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