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Pratik Parikh wrote:

> If 0 to the power of 0 = indeterminate quantity then why do some
> calculators
> like the one we get with MS Windows and Linux show 0^0=1?

See the sci.math FAQ.  0^0 is often considered to be 1 for convenience:


http://db.uwaterloo.ca/~alopez-o/math-faq/mathtext/node14.html#SECTION005300000000000000000

Sometimes it's considered an indeterminate form, particularly in cases
thta involve lim [f(x)/g(x)] where lim f(x) = lim g(x) = 0.  In other
cases, x^0 = 1 holds for all x, even x = 0.  It's a case where it
depends on what you prefer to emphasize.

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Erik Max Francis wrote:
> 
> x^0 = 1 holds for all x

OTOH 0^x = 0, and you can probably argue for anything
between 0 and 1 by similar arguments.

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Kasper Dupont wrote:

> Erik Max Francis wrote:
>
> > x^0 = 1 holds for all x
> 
> OTOH 0^x = 0, and you can probably argue for anything
> between 0 and 1 by similar arguments.

This is a very weak argument, because for one thing, 0^x = 0 is simply
not true; consider x = -1.  (And y^x isn't defined at _all_ for negative
irrational x, regardless of y!)  The properties of exponentiation are
defined in terms of the exponent, not the base; nowhere is there a rule
that 0^x = 0 for all x.  You are taking a very small pattern (certainly
it's true for positive numbers) and extending it into domains where it
does not belong.  That 0^x = 0 for some x doesn't tell us anything about
the situation.

There are very strong algebraic reasons for considering 0^0 = 1 (for
instance, making the binomial theorem pretty), just as there are strong
reasons for, say, not considering 1 to be prime (namely, unique prime
factorization).  We're making the definitions, we get to choose how they
work in the edge cases.

As I've already pointed out, the FAQ is the place to go for the complete
rundown on this information for mathematicians and by mathematicians.

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In comp.os.linux.development.apps Erik Max Francis <max@alcyone.com> wrote:
> Kasper Dupont wrote:
> 
>> Erik Max Francis wrote:
>>
>> > x^0 = 1 holds for all x
>> 
>> OTOH 0^x = 0, and you can probably argue for anything
>> between 0 and 1 by similar arguments.
> 
> This is a very weak argument, because for one thing, 0^x = 0 is simply
> not true; consider x = -1.  (And y^x isn't defined at _all_ for negative
> irrational x, regardless of y!)

Wouldn't that make e^x a noncontinous function? I was sure that e^(-Pi)
is well-defined - certainly e^(-j.Pi) is.

	- Kevin.