What is 0^0?

You heard me, is it 0 because 0*anything=0 or is it 1 because anything^0=1?

It is none.

0^0 is an undefined expression.

http://mathworld.wolfram.com/DivisionbyZero.html

But you can graph n/0

I say 0.5, just to piss my math teachers off. xD

read: http://www.faqs.org/faqs/sci-math-faq/specialnumbers/0to0/

mIRC:
//say $calc(0^0)
<CCFreak2K> 1

Hmm...

N^0 = N/N.  Therefore 0^0 = 0/0, which is undefined.

Perl says it's 1.

>>9

If Perl told you to jump off a cliff, would you?

>>10
I would

it's one okay? because there's exactly one way to order zero things in zero positions.

http://www.faqs.org/faqs/sci-math-faq/specialnumbers/0to0/

>>13
Yesssss.

>>2
>>8
Nooooo.  0/0, 0^0, 1^infinity, etc. are all /indeterminate./  Indeterminate quanitities do not exist, that's true, but they exist in an entirely different ways from undefined quantities.  There is no value that satisfies an undefined quantity, whereas there are infinitely many values that satisfy an indeterminate form.

0^0 = 0/0 = Any real number at all.

On the other hand,

n/0 (n != 0) = Nothing.

They both don't exist, but they are very, very different, and not just in theory.

lim(x^x,x,0) = 1.  that's good enough for me.

>>14

what about in the complex number set?

Don't even go there.  sqrt(-1)=i.  i^2=-1.  (2+i)^2=2^2+2(2i)+i^2=4+4i-1=3+4i.  sqrt(-8)=2isqurt(2).  Just forget complex numbers.

>>15

That's incorrect.  The lim(x^x,x,0) = 1 only from the RHS.  If the absolute limit is to be defined for when x->0 then the limit from the LHS must equal RHS in which case it does not.  Trust me on this, I have a degree in math.

lim(x^x,x,-0) is undefined, hence no limit at all.

0^0 is often defined to be 1, because zero over zero is one (i.e. there's only one way to order nothing)

but 0^0^0 = 0

OH SHIT YOU GUYS BETTER STOP DIVIDING BY ZERO!!!!

I personally think of the exponential in that case as a kind of special function which is just defined to be 1 at zero.

Not very formal, but I never was.

>>19
0/0 is indeterminate; read up on limits for more detail.

Consensus says 0^0 = 1, but there seems to be no proof as yet.

>>23

oops, i thought the binomial coefficient read n over k, not n choose k. but i'm excused because english isn't my native language.

>>23

btw in mathematics you can just define things to be something if it's convenient. you don't need to prove it

>>25
But through expansion of the meaning of a^n in such a way that it still holds true for all values a != 0, you can achieve a solution that will also work for a=0.

Well, multiplication is iterated addition, right?

Therefore...

2+0 = 2x1 = 2

2+2 = 2x2 = 4

2+2+2 = 2x3 = 6

So exponents are iterated multiplication, which is iterated addition.

2+2 = 2^2 = 4

2+2+2+2 = 2^3 = 8

2+2+2+2+2+2+2+2 = 2

Therefore, should 0^0 not be 0x0 zero times?

Idunno, I'm tired and have the flu... hehe

>>27

No, becuase x^0 = 1 for all x.  They way you're looking at it, you're thinking 2^0 = 0, but you're mistaken.

( )O( ) = fudge for everyone

bump

Windows calculator says 1, although it could be horribly wrong.

>>9
>>10

This made me laugh. Thank you anon.

>>10

Python says its 1 as well, so, yeah. Probably.

Oh Knuth, you devil, you!

Some textbooks leave the quantity 0^0 undefined, because the functions x^0 and 0^x have different limiting values when x decreases to 0. But this is a mistake. We must define x^0=1 for all x, if the binomial theorem is to be valid when x = 0, y = 0, and/or x = -y. The theorem is too important to be arbitrarily restricted! By contrast, the function 0^x is quite unimportant.

0 multiplyed by 0, 0 times.

>>35
facepalm

C says it's 0.


In[1]:= 0^0

                                        0
Power::indet: Indeterminate expression 0  encountered.

Out[1]= Indeterminate

OMG .. it is 1 !!!

Search for 'empty set' on Wiki. For Wiki, the article explains it well.

Prelude> 0^0
1
Prelude> 0**0
1.0

>>39
See >>38

This question pops up on /sci/ pretty regularly, and this is the last time I'm going to answer it.

To set the record straight the correct answer is "It depends". The Mathematics of numbers is basically a set of logical "models" (analogous to the mathematical models of physical situations) to things we intuitively know.

To clarify, the basic operations of arithmetic are addition, subtraction, exponentiation, multiplication, etc. are all intuitive for the nonzero finite numbers we know and love, but not always otherwise (for example, it is not clear what, if anything 0^0 should be), so the mathematical models can disagree.

I'll give the two main models and what happens. It is clear that many people are arguing based (possibly unknowingly) on one or the other model, but I must emphasize that these are models so there is no universal truth other than "it depends".

Modelling numbers in the cardinal sense (eg. seeing numbers as meaning the "size" of a set of distinct things) gives us the answer 0^0. Because in this model 0 is (usually) defined as {}, x^y is (usually) defined as the cardinality of the set of functions from y to x. Hence 0^0={}^{}=1 because there is only one map from {} to {} (namely {}, ie. the map that has an empty domain and codomain).

So 0^0=1. This result actually makes sense in a huge number of cases where counting is involved or numbers are treated in the discrete sense.

But let's model numbers in a different way now. The model above does not encompass negative, rational, or irrational numbers. Now, let's view numbers in the model often given in Analysis courses (one that does everything we expect and gives us all real numbers); that is, as the completion of the field of rationals (the set of rationals being defined as the quotient set of all ordered pairs of integers over the appropriate equivalence relation, and integers being defined in the usual way -- often on the definition given above). Usually, in this model we have the definition x^y=log (y exp x) (or ln, if you prefer that notation), where exp is defined in terms of its Taylor Series and log is defined as the inverse of exp. Now, in this case 0^0 is completely undefined and frankly meaningless; which is good, because if 0^0 is defined in this model then it would ruin much of Analysis!

So, it depends on your model -- and that's all there is to it.

>>43
>x^y=log (y exp x)
lol wut?

x^y = e^{y ln(x)}


In[1]:= e^(0 ln[0])

Out[1]= 1


In[2]:= Limit[x^0, x -> 0]

Out[2]= 1

In[4]:= Limit [0^y, y -> 0]

Out[4]= 0

\lim_{x \to 0}x^{\frac{\ln(k)}{\ln(x)} = k

For k = 0, the old 0^x trick works.

Thus, indeterminate.

X = 2+2

lim_{x \to 0}x^{\frac{\ln(k)}{\ln(x)}} = k

0^0=1

You are all wrong. 0^0 is a tasty sandwich. Not because it is logically arguable, but because it can be argued.

1, niggers

so whats 0/0

x/0 = 0
0/0 = 0

JESUS CHRIST GUYS

FINAL ANSWER:

DISCRETE MATHEMATICS—0^0=1
CONTINUOUS MATHEMATICS—0^0=¿

---END OF LINE---

0^0 is the face I made when I read this.

>>56
Even in anime, one's nose is not at the same altitude as one's eyes.

>>57
That's a mouth.

>>58
That's even moar anatomically incorrect.

>>55
That's a very discrete answer

hey fags

>>61
hey

>>59
He's raging probably.

1 >C