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Re: Why Indeterminate?

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

Hi.  Just 2 cents.  As others have mentioned, there are other combinations that give Indet as well.
Here's just one way to look at it:
Your equation does give Indeterminate...

A[Range[3],1/2]
Power::indet:

Do it symbolically for x:

A[Array[a,3],x]//Together
((-1+x) x (-a[1]-a[2]+2 x a[2])) / (1-a[3]+2 x a[3])

You don't want the denominator to be zero, else an error:

Reduce[Denominator[%]==0]
a[3]!=0 && x==(-1+a[3])/(2 a[3])

What this says is that a[3] can't be zero, or that x should not take on the value related to a[3].
Given your data, the last term being 3 in the array...

(-1+a[3])/(2 a[3]) /. a[3]->3
1/3

What this says is that given {1,2,3} for input (last value 3), then x should not be 1/3...
Indeed, 1/3 gives an error also:

A[{1,2,3},1/3]
Power::infy: Infinite expression 1/0 encountered.

A Quick plot:

Plot[A[{1,2,3},t],{t,0,1}]

= = = = = = = = = =
HTH  : >)
Dana DeLouis
= = = = =

On Apr 30, 5:54 am, Themis Matsoukas <tmatsou...@me.com> 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.

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Re: Why Indeterminate? Dana DeLouis <dana.del@gmail.com> - 2011-05-03 09:46 +0000

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