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Re: 0^0=1
Posted:
May 2, 2012 9:22 AM
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Dan Christensen <Dan_Christensen@sympatico.ca> wrote:
> On May 1, 6:56=A0pm, David W. Cantrell <DWCantr...@sigmaxi.net> wrote:
> > "Jesse F. Hughes" <je...@phiwumbda.org> wrote:
> >
> > > I wonder if you [Dan] can find a calculus text which explicitly
> > > leaves 0^0 undefined.
> >
> > I have several introductory calculus texts on the shelves.
> >
> > Some authors feel the need to review pre-calculus topics. Thomas &
> > Finney, 7th ed., do this in an appendix. They say that, if a is
> > nonzero, a^0 = 1.
> > But of course, they give no reason for excluding a = 0.
> >
> > But unless the author feels the need to review pre-calculus topics,
> > there may be no mention of 0^0 until the section on power series.
> > Perhaps Leithold, 5th ed., is typical. [Thomas & Finney do something
> > similar regarding power series.] Soon after defining power series,
> > Leithold states
> >
> > "(Note that we take (x - a)^0 =3D 1, even when x =3D a, for convenience
> > in writing the general term.)"
> >
> > I had more-or-less expected to find such a statement, but was surprised
> > by a similarly phrased statement on the next page:
> >
> > "When n! is used in representing the nth term of a power series, we
> > take 0! = 1 so that the expression for the nth term will hold when n = 0."
>
> Again, I have no trouble with 0! = 1
Of course, 0! = 1. But to be consistent, since both 0^0 and 0! are empty
products, you should either accept or reject that both are 1. (Of course,
you and Leithold _should_ accept that both are 1.)
> We define the ! function as:
> 1! = 1
> (x+1)! = x! * (x+1) for any natural number
From what I see below, you intend "natural number" to include 0. That's
fine; I normally do that myself.
But is that a _definition_ of the factorial? We put the definiendum (the
thing _being_ defined) on the left and the definiens on the right. If you
have done that, then your first line defines 1! to be 1. But then your
second line, when x = 0, attempts to REdefine 1! in terms of something
which has yet to be defined, namely, 0!. Surely you see that that is a
problem!
To solve that problem, I suggest using the classic recursive definition:
0! = 1
(n + 1)! = n! * (n + 1) for natural n
> No need for any empty products.
>
> We could then legitimately define the ! function as:
>
> x! = {1 for x = 0
> {(x-1)! * x for x > 0
"then"? It wouldn't have been illegitimate even before then.
But thinking of 0! as the empty product provides motivation for its
definition.
For natural n (including 0), I like to think of n! as being simply
| The product of the first n positive integers.
And then naturally this gives us 0! = 1.
David W. Cantrell
> > Wow! Horrible! But at least Leithold is consistent: Both 0^0 and 0! are
> > empty products and he treats them in essentially same way, making it
> > seem that they aren't _really_ 1, but will just be taken to be 1 in the
> > context of power series, for convenience. Aaargh.
>
> Interesting...
>
> Dan
> Download my DC Proof 2.0 software at http://www.dcproof.com
> Also see video demo
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