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Re: 0^0=1
Posted:
Apr 28, 2012 2:12 AM
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Dan Christensen wrote:
>
> On Apr 27, 4:45 pm, Frederick Williams <freddywilli...@btinternet.com>
> wrote:
> > Dan Christensen wrote:
> >
> > > "They" have defined the product of finite lists of numbers. You need
> > > at least 2 numbers to multiply,
> >
> > So do you now think that x^1 is also undefined?
> >
>
> Not undefined. But I'm beginning to have doubts about 0^1. I might be
> able to justify using it as a starting point since it doesn't have the
> same problem in the limiting case that 0^0 does. As Virgil pointed
> out:
>
> lim(x->0):x^0 = 1 and lim(y->0):0^y = 0
>
> On the other hand:
>
> lim(x->0):x^1 = lim(y->1):0^y = 0
>
> I will have to give it more thought.
>
> > > but lists can have 1 or even 0 numbers
> > > in them, so "they" arbitrarily define the product of any empty list to
> > > be 1. Why not 0? Why not -1? Why not just say it is undefined? It is
> > > an arbitrary convention to make some recursive definition work out
> > > nicely (with no nasty exceptions). What kind of basis is this for
> > > deciding to assign a value to 0^0?
> >
> > "to make some recursive definition work out nicely (with no nasty
> > exceptions)" - so not arbitrary at all, unless arbitrary means
> > exceedingly sensible.
> >
>
> It just seems so ad hoc,
"to make some recursive definition work out nicely (with no nasty
exceptions)" - so not ad hoc at all, unless ad hoc means exceedingly
sensible.
> especially given the problems in the limiting
> case (see above).
Note that the question of the value of 0^0 has got nothing to do with
the limits of x^y. I write _limits_ because the answer will depend on
the relative rates at which x and y tend to 0.
--
When a true genius appears in the world, you may know him by
this sign, that the dunces are all in confederacy against him.
Jonathan Swift: Thoughts on Various Subjects, Moral and Diverting
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