What does 0^0 equal?  ALG48, and 49g on approx mode says one.  Ti-89
gives one, but with the message, " warning 0^0 replaced by 1". HP49 on
exact gives ?.

You've stumbled upon one of Math's many mysteries ;-)

http://mathforum.org/dr.math/faq/faq.0.to.0.power.html




"chris heaton" <cmhthethird@yahoo.com> wrote in message
news:89d0a867.0310011839.18b9f546@posting.google.com...
> What does 0^0 equal?  ALG48, and 49g on approx mode says one.  Ti-89
> gives one, but with the message, " warning 0^0 replaced by 1". HP49 on
> exact gives ?.

cmhthethird@yahoo.com (chris heaton) wrote in message news:<89d0a867.0310011839.18b9f546@posting.google.com>...
> What does 0^0 equal?  ALG48, and 49g on approx mode says one.  Ti-89
> gives one, but with the message, " warning 0^0 replaced by 1". HP49 on
> exact gives ?.

Hello,

0^0 has been discussed endless times, search for it in the newsgroups.
It is often regarded as an indeterminate expression because there are
2 different limits (0^x, x->0 = 0 whereas x^0, x->0 = 1). There are
some people who define it as 1, probably because the function x^x has
that limit.
It might make sense in some aspects.

All in all it's a question of definition.

Regards
Axel

Axel Bodemann wrote:
> > What does 0^0 equal?  ALG48, and 49g on approx mode says one.  Ti-89
> > gives one, but with the message, " warning 0^0 replaced by 1". HP49 
> > exact gives ?.
> There are
> some people who define it as 1, probably because the function x^x has
> that limit.

Not for that. Moreover, because lim a^x = 1 as x -> 0 for an arbitrary
real parameter a including 0. On the 49, 0^0 depends on Flag -3.

If flag -3 is clear it yields ?, also a good idea. What is "?" ? Its an
algebraic evaluating to an equally named rompointer (XLIB 788 137) which
either errors or evals FPTR 6 2E5 which again generates a symbolic. If
flag -3 is set (Function -> num) 0^0 yields 1 (a warning as on TI 89 is
not a bad idea).

A full understanding of a possible definition or non-definition of 0^0
can only be reached by investigation the analytic function x^y which has
a bad singularity at (0,0), no matter whether regarded as a two-variable
real or complex function. Draw the real function (x,y) -> x^y in 3D in
the unit circle and look at it from all sides. This strange function is
growing exponential along the Y-axis, but polynomial along the X-axis.

- Wolfgang

Axel Bodemann wrote:
> > What does 0^0 equal?  ALG48, and 49g on approx mode says one.  Ti-89
> > gives one, but with the message, " warning 0^0 replaced by 1". HP49 
> > exact gives ?.
> There are
> some people who define it as 1, probably because the function x^x has
> that limit.

Not for that. Moreover, because lim a^x = 1 as x -> 0 for an arbitrary
real parameter a including 0. That lim x^x = 1 is a special case of the
more general result lim f(x)^g(x) =1 if x approaches 0 from the right
(in the complex plane), provided z |-> f(z) and z |-> g(z) are global
analytic functions with limit 0 as z approaches 0 (Rotando/Korn The
Indeterminate Form 0^0 . Mathematics Magazine, Vol. 50 (1977), pp.
41-42. This result supports in some sense that in most cases it is
reasonable to set 0^0 = 1, but not more.

On the 49, 0^0 depends on Flag -3. If flag -3 is clear it yields ?, also
a good idea. What is "?" ? Its an algebraic evaluating to an equally
named rompointer (XLIB 788 137) which either errors or evals FPTR 6 2E5
which again generates a symbolic. If flag -3 is set (Function -> num)
0^0 yields 1 - a warning as on TI 89 is not a bad idea!

A full understanding of a possible definition or non-definition of 0^0
can only be reached by investigation the analytic function x^y which has
a bad singularity at (0,0), no matter whether regarded as a two-variable
real or complex function. Draw the real function (x,y) -> x^y in 3D in
the unit circle (best in cylindric coordinates) and look at it from all
sides. This strange function is growing exponential along the Y-axis,
but polynomial along the X-axis.

- Wolfgang

I wrote:
> ... Draw the real function (x,y) -> x^y in 3D in
> the unit circle (best in cylindric coordinates) and look at it from all
> sides. This strange function is growing exponential along the Y-axis,
> but polynomial along the X-axis.

Sorry, (x,y) |-> x^y is a real function as long as y is not negative.
Hence, better draw it in a square with (0,0) at the lower left corner or
in the middle of the left-hand side of the square, in 3D XYZ
coordinates. 

- Wolfgang

HP32sII returns: Invalid y^x

I think that the answer depends on what kind of poit of view you choose. The
limit is 1.
I think it is a good to return a error or a warning. This has been discussed
in many
NGs many times.

Niclas

"chris heaton" <cmhthethird@yahoo.com> wrote in message
news:89d0a867.0310011839.18b9f546@posting.google.com...
> What does 0^0 equal?  ALG48, and 49g on approx mode says one.  Ti-89
> gives one, but with the message, " warning 0^0 replaced by 1". HP49 on
> exact gives ?.

"Wolfgang Rautenberg" <raut@math.fu-berlin.de> wrote in message
news:3F7C0697.8644BDF9@math.fu-berlin.de...
> Sorry, (x,y) |-> x^y is a real function as long as y is not negative.

I don't think so.  (-1.2)^3.4 = a real value?

Rick



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Axel Bodemann <bodemann@web.de> wrote:

> It is often regarded as an indeterminate expression because there are
> 2 different limits (0^x, x->0 = 0 whereas x^0, x->0 = 1). 

If you define, as usual, exp(x)=sum(n=0 to infinite)(x^n/n!) then you
have to define 0^0=1 if you want to have exp(0)=1....
Last year my teacher told me that there was no problem with 0^0 because
of this definition of exp.

-- 
Thomas Deniau
"Unix is user friendly. It's just selective when choosing friends."

Yes, well noted. BUT, if you take y negative, then the result is complex.

ttfn

JasonG

Rick Nungester wrote:
> "Wolfgang Rautenberg" <raut@math.fu-berlin.de> wrote in message
> news:3F7C0697.8644BDF9@math.fu-berlin.de...
> 
>>Sorry, (x,y) |-> x^y is a real function as long as y is not negative.
> 
> 
> I don't think so.  (-1.2)^3.4 = a real value?
> 
> Rick
> 
> 
> 
> -----= Posted via Newsfeeds.Com, Uncensored Usenet News =-----
> http://www.newsfeeds.com - The #1 Newsgroup Service in the World!
> -----==  Over 100,000 Newsgroups - 19 Different Servers! =-----

Wolfgang Rautenberg <raut@math.fu-berlin.de> wrote in message news:<3F7BF530.5C37D105@math.fu-berlin.de>...

> Not for that. Moreover, because lim a^x = 1 as x -> 0 for an arbitrary
> real parameter a including 0. On the 49, 0^0 depends on Flag -3.

Not for a=0. It's easy to see, that lim 0^x = 0 as x -> 0.

Fabian

Fabian <fahasch@yahoo.de> wrote:
>Wolfgang Rautenberg <raut@math.fu-berlin.de> wrote:
>
>> Not for that. Moreover, because lim a^x = 1 as x -> 0 for an arbitrary
>> real parameter a including 0. On the 49, 0^0 depends on Flag -3.
>
>Not for a=0. It's easy to see, that lim 0^x = 0 as x -> 0.
>
>Fabian

Or, to be absolutely pedantic about it, lim 0^x = 0 as x -> 0+; 
unfortunately, lim 0^x as x->0- is undefined.  Since it's clearly not 
continuous, one has to question the validity of using limits on this 
function at all.
--
                             -- With Best Regards,
                                Matthew Funke (mff@hopper.unh.edu)

First of all I want to thank for various corrections. I wish students in
my course on foundations of analysis were so attentive...

Thomas Deniau wrote:
> If you define, as usual, exp(x)=sum(n=0 to infinite)(x^n/n!) then you
> have to define 0^0=1 if you want to have exp(0)=1....
> Last year my teacher told me that there was no problem with 0^0 because
> of this definition of exp.

You see, teachers may be mistaken :-) There are many definitions of the
exponential function. The one which you said to be the usual one, is not
particularly convenient because familiarity with the theory of infinite
series is needed to understand it.

The most elementary although abstract definition was given by Alfred
Tarski in his textbook from 1937. He first proves the basic theorem that
- up to isomorphism - there is one and only one continuously ordered
group. Since both the additive group of reals and the multiplicative
group of positive reals are continuously ordered, there must be an
isomorphism (which clearly maps 0 onto 1). It is uniquely determined if
its value for 1 is chosen, a (positive) real b, say. This is precisely
the function x |-> exp_b(x). Tarski's definition made it clear (without
the ordinary stepwise definition of b^x) why the logarithmic reduction
of real multiplication to addition so phantastically works. 

Since an isomorphism is always a one-to-one mapping, exp_b has trivially
an inverse, and this is just the function log_b.      
 
- Wolfgang

Wolfgang Rautenberg <raut@math.fu-berlin.de> wrote:

> Not for that. Moreover, because lim a^x = 1 as x -> 0 for an arbitrary
> real parameter a including 0. That lim x^x = 1 is a special case of the
> more general result lim f(x)^g(x) =1 if x approaches 0 from the right
> (in the complex plane), provided z |-> f(z) and z |-> g(z) are global
> analytic functions with limit 0 as z approaches 0 (Rotando/Korn The
> Indeterminate Form 0^0 .

Clearly  h(x) := exp(1/x)^x  is a counterexample where at least f is not
analytic, but  lim_(x->0) h(x) = e != 1.  

> This result supports in some sense that in most cases it is
> reasonable to set 0^0 = 1, but not more.

Simmply *define* for arbitrary reell a recursively:

    a^0 := 1 and a^(n + 1) := a*a^n.

Then -- with a = 0 -- 0^0 = 1 and 0^n = 0 if n > 0.
 
> A full understanding of a possible definition or non-definition of 0^0
> can only be reached by investigation the analytic function x^y
      ^^^^

No.  Avoid all limites and live long and prosper with the definition
given above.

Michael

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