Re: making 0^0=1 and not indet
- To: mathgroup at smc.vnet.net
- Subject: [mg94868] Re: [mg94821] making 0^0=1 and not indet
- From: Bob Hanlon <hanlonr at cox.net>
- Date: Mon, 29 Dec 2008 06:38:53 -0500 (EST)
- Reply-to: hanlonr at cox.net
f1[n_ /; n == 0, x_ /; x == 0] = 1;
f1[n_, x_] := x^n;
f2[n_, x_] := If[n == x == 0, 1, x^n];
f3[n_, x_] := (x + DiscreteDelta[n, x])^n;
f4[n_, x_] := (x + KroneckerDelta[0, n, x])^n;
And @@
Flatten[
Table[
f1[n, x] == f2[n, x] == f3[n, x] == f4[n, x],
{n, 0, 3}, {x, -3, 3}]]
True
Bob Hanlon
On Sat, Dec 27, 2008 at 9:45 PM , Affan wrote:
> I have another question. One of the terms in my formulation involves
> computing 0^0. This indeterminate form is widely accepted to be equal
> to 1 and if this form was used all of my equation can be solved
> separately (without having to worry about some corner case). Is there
> some way to force mathematica to use 0^0=1?
>
>
> Thanks
>
> Affan