Question

Does 0^0=1?

Answer 1

nope

Answer 2

its undefined :D

Answer 3

I don't believe you exactly.

Answer 4

lolz why :O

Answer 5

for any n \(\neq 0\)

n^0 =1

Answer 6

the limit of x^0 is 1

the limit of 0^x is 0

0^0 has no determinable value ....

Answer 7

let y = 0^0

ln(y) = 0 ln(0)


ln(y)
---- = ln(0)
0

you simply get a bunch of indeterminable forms that have no specific value

Answer 8

Well, suppose you have a geometric series.

\[S=x^0+x^1+x^2+...\]\[xS=x^1+x^2+x^3...\]\[xS+x^0=+x^0+x^1+x^2+x^3...\]\[xS+x^0=S\]\[S(x-1)=-x^0\]\[S=\frac{x^0}{1-x}\]

Actually following through with it as x^0 instead of writing a 1 there has made it obvious that it is indeterminate. Plug in x=0 and you get: \[S=\frac{0^0}{1-0}=0^0\] It's sort of its own value or something.

Is it related to the kronecker delta, or what kind of interesting things does 0^0 have going for it?

Answer 9

x^0 is not the same as 0^0

x^0 limits to 1 as x approaches 0.

Answer 10

\[\large (a+b)^n = \sum \limits_{k=0}^n \binom{n}{k} a^{n-k}b^k\]

\[\large (0+1)^1 = \sum \limits_{k=0}^1 \binom{1}{0} (0)^{1-1}1^1\]

\[\large 1 = 0^0\]

Answer 11

0^x doesnt have a 2 sided limit :) i was considering the right side limit

Answer 12

@amistre64 I'm not even evaluating a limit at all in my result. I've simply just shown that algebraically there's no conflict with the geometric series. Usually I'd have just assumed x^0=1 but by leaving it as x^0 when I plug in 0, the geometric series just gives 0^0 which is just whatever it is... Undefined though.

Answer 13

the sum of an infinite geometric series IS a limiting function ....

Answer 14

otherwise we would have stuff like:\[1+2+4+8+...=\frac{1}{1-2}\]

Answer 15

Yeah @ganeshie8 I still feel like there are other reasons why 0^0 should be 1 too. I know people do it for certain reasons but maybe it's just a common simplification.

I guess I want to understand that whole 0^0 thing more to understand what's going on there.

@amistre64 sure, but in this case I wasn't taking the limit of either part of 0^0... It was untouched and undefined.

Answer 16

It's even consistent both ways you define it.

If 0^0=0 then
\[0^0+0^1+0^2+0^3+... = 0\]
if 0^0=1 then
\[0^0+0^1+0^2+0^3+... = 1\]
But I haven't even gone so far as to plug in either value, the sum is 0^0 in either case you see?

Answer 17

lets try a power rule on:
\[\frac d{dx}x^0=0/x\]
now if x=0, we have a value that simply cannot be determined. 0/0

just because a value is not able to be determined in general does not mean that it has no useful purposes to us.

Answer 18

confused xD

Answer 19

Wait, so are you saying that it might be possible for the derivative of a constant to be nonzero at the origin? Or wait, I'm interested at least lol.

Answer 20

your summation shows that you get different results depending on however you define 0^0

therefore youre results are proof that 0^0 has no specific value attributable to it.

Answer 21

Interesting, you want to define 0^0 based on this :

\(0^0+0^1+0^2+0^3+... = 0^0\)

Answer 22

the derivative of a constant cannot be reliably determined by the power rule

Answer 23

Yeah, but @ganeshie8 that can't define it since it'll give you the same result either way. It's consistent with its inconsistency so it sort of just sneaks through. Kinda cool. But w/e I just want to understand it, because I think it's a hint at something really cool if we can just figure it out.

Answer 24

the application of 1/0, even tho it has no specific value; we know that in certain cases, it helps define the slope of a line .... as a vertical slope.

Answer 25

but as for 0/0

assume there is some value: n = k/0

in order for this to have a chance of working, then: 0n=k, therefore k=0, so what is n?

well, 0n = 0 is true for all real values of n, so there is no way to determine a unique value for it. this is why uniqueness is an important concept in mathematics

Answer 26

True, that's also pretty weird and has always bugged me. It sort of makes y=mx+b feel kind of made up when it fails for one particular type of slope when you'd think all slopes should be more or less considered equal. I mean we can have parameterized curves but it still never really sat well with me. Maybe I'm getting off topic though.

Answer 27

Well I'm not disagreeing with you. I guess when I think of taking derivatives, I just feel like it's just how we define dividing by zero. I think in general division by zero is defined, but it relies on a context to give it meaning whether it by the binomial theorem or a derivative/limit kind of thing.

I guess it's not that important really, feels more like semantics more than anything. But still feels weird to me.

Answer 28

0^0 has no unique value to attribute to it; therefore its true value does not lie within its quantity, but the qualitative nature in which it resides.

using it as a quantity is the same as asking what time the movie start, and getting ' i dunno' in reply :)

Answer 29

Yeah, I am in agreement with you. But I feel like this is sort of dodging the point. Why is it so special? 2^2 can't be sneaky and be undefined for instance. It's just not fair! lol

Answer 30

mathematics is a science that has to be founded on sound logical principles. the other sciences can be wishy washy and make assumptions, but that is not the case in mathematics. thrms have to be precise and applicable in order for the work that relies upon them to be true .... to be true.

the theories and hypotheses of the other sciences are fabrications of the mind and come and go with the direction of the wind, but mathematics has to stand firm and resolute.

Answer 31

2^2 has a specific/unique value, sqrt(4) does not ;) unless we restrict its meaning

Answer 32

So what you're saying is 0^0 is equal to 0 and 1 just like a quadratic equation has two answers basically.

Answer 33

0^0 has no equality, its value is determined if anything by the nature in which it exists within a solution attempt.

Answer 34

Yeah you're right. \[\lim_{x \rightarrow 0}\frac{a \ \sin x}{x}=a\] So that pretty much means it's anything in my mind. Hmm. Well I'm still going to be thinking about 0 and infinty I guess haha, thanks.

Answer 35

:) good luck

Answer 36

Thoughts @ganeshie8 ?