Defining Success: What Does it Mean?

[pivoxa15]

Is it defined? if not why not?

 

[TD]

I'm sure you'll find one or more topics about that if you use the search function.

When defined, it is usually defined as 1 (and not as 0) for practical reasons (making certain summations and other formula/notations easier because you don't have the consider the case with 0 separately). Looking at the graph y = x^x for x approaching 0 will make it clear why this is the more logical choice (above 0). Try taking the limit of x^x for x->0 for example.

It is often left undefined though.

 

[Hurkyl]

The most common use of 0^0 = 1 is a result of de-abstraction.

In a polynomial ring, or power series ring, one might have a symbol x. x does not stand for a number; it is a number in its own right! And since x is not zero, or a zero divisor, we would properly say that x^0 = 1.

One of the typical uses of a polynomial ring, or a power series ring, is that we can substitute a number for x. I.E. if we have the polynomial f = x^2 + x + 1, we can make a substitution f(2) = 2^2 + 2 + 1.

Now, if f = x^0 = 1, then f(0) = 1 as well, but formally it looks like we said f(0) = 0^0 = 1!


When exponents can only refer to repeated multiplication, then 0^0 = 1 is usually right.


But when we are working with real numbers, none of the above applies. A very important thing about exponentiation is that it's continuous, and exponentiation cannot be continuous at 0^0, therefore it is undefined there.

 

[NEWO]

it seems also similar to the 0/0 paradox. theory states that anything divided by zero is infinte, theory also states that anything to the power of 0 is 1. Therefore, 0^0 =1.

 

[arildno]

it seems also similar to the 0/0 paradox. theory states that anything divided by zero is infinte,

No, it doesn't. Neither is there a paradox here.

theory also states that anything to the power of 0 is 1. Therefore, 0^0 =1.

No, it doesn't and isn't.

 

[NEWO]

no offense, I am sure your credentials are much better than mine but that is how I've been taught and it works for me.

 

[NEWO]

i suppose because zero is strange that 0^0 is not defined.

 

[AlphaNumeric]

no offense, I am sure your credentials are much better than mine but that is how I've been taught and it works for me.

As with so many subjects in education, there are lies everywhere. In physics you're told "electrons spin around the nucleus like planets around the sun" only to then find out they form 'distributions'. You are often given the impression at high school level that all functions can be differentiated or integrated nicely, then you try to integrate $$e^{-x^{2}}$$ indefinitely.

Particularly at high school level the notion of proof is often missing so 'just because it works for you' doesn't mean there aren't more powerful results and concepts which should be applied :)

 

[matt grime]

no offense, I am sure your credentials are much better than mine but that is how I've been taught and it works for me.

Who taught you that "0/0 is a paradox"? What does thet even mean? Theory does not state that anything divided by zero is infinite. Theory does state that the operation x --->1/x is not a well defined function on R, but merely on R\{0}. Theory states that there are extensions of R (compactifications) where 1/x does extend to a function whose domain includes 0, and the symbol 1/0 is commonly labelled $\infty$. It still does not provide a way to define 0/0 though, as can be readily demonstrated, but it is not a paradox.

 

[arildno]

i suppose because zero is strange that 0^0 is not defined.

this is a good thought on your part!
It is precisely because of the properties of zero (which you perceive as "strange"), and because of what we want exponentiation to mean ,that 0^0 is left undefined (usually).

 

[JCCol]

Who taught you that "0/0 is a paradox"? What does thet even mean? Theory does not state that anything divided by zero is infinite. Theory does state that the operation x --->1/x is not a well defined function on R, but merely on R\{0}. Theory states that there are extensions of R (compactifications) where 1/x does extend to a function whose domain includes 0, and the symbol 1/0 is commonly labelled $\infty$. It still does not provide a way to define 0/0 though, as can be readily demonstrated, but it is not a paradox.

From this am i supposed to gather that 1/0 is infinity or is that wrong?

 

[matt grime]

You can gather what you will. 1/0=$\infty$ is true in the extended real numbers or the extended complex plane. The phrase 1/x tends to infinity as x tends to zero is true in the real numbers, where 'tends to infinity' has a very specific meaning which is often abbreviated in a convenient shorthand to 1/0=[itex]\infty[/tex].

 

[pivoxa15]

The most common use of 0^0 = 1 is a result of de-abstraction.

In a polynomial ring, or power series ring, one might have a symbol x. x does not stand for a number; it is a number in its own right! And since x is not zero, or a zero divisor, we would properly say that x^0 = 1.

One of the typical uses of a polynomial ring, or a power series ring, is that we can substitute a number for x. I.E. if we have the polynomial f = x^2 + x + 1, we can make a substitution f(2) = 2^2 + 2 + 1.

Now, if f = x^0 = 1, then f(0) = 1 as well, but formally it looks like we said f(0) = 0^0 = 1!


When exponents can only refer to repeated multiplication, then 0^0 = 1 is usually right.


But when we are working with real numbers, none of the above applies. A very important thing about exponentiation is that it's continuous, and exponentiation cannot be continuous at 0^0, therefore it is undefined there.

Why can't exponentiation be continuous at 0^0?

 

[Hurkyl]

Why can't exponentiation be continuous at 0^0?

Because 0^x = 0 and x^0 = 1 for positive real numbers x. If exponentiation was continuous at 0^0, then it would be equal to the limit of both of those expressions as x goes to zero.

 

[rbj]

i suppose because zero is strange that 0^0 is not defined.

often the question is what is

$$\lim_{x \rightarrow \ 0} x^x \$$

and the answer to that is

$$\lim_{x \rightarrow \ 0} x^x = 1 \$$

 

[pivoxa15]

Because 0^x = 0 and x^0 = 1 for positive real numbers x. If exponentiation was continuous at 0^0, then it would be equal to the limit of both of those expressions as x goes to zero.

Also I realized that the function x^x is discontinous at x=0 because there is a limit going from positive x to 0 but no limit going from negative x to 0. So limit as x->0 doesn't exist.

So if a function is continuous at a point, we define the value of the function to be that value at that point. If f is not continuous there, we defne f to be the limit at that point. But if f dosen't have a limit there than we are in trouble like 0^0. Although the next best thing would be to define f by one side of the limit if it exists which is 1 or 0 in this case. Hence we define 0^0 to be 0 or 1 at different times depending on which is more suitable but its safer to leave it undefined?

 

[Hurkyl]

You've caught a trick there -- depending on what you're doing, you might not even want to define 0^x and x^0 for that very reason!

But at least if we restrict ourselves to the closed first quadrant, but with the origin removed, x^y would be everywhere continuous, so it is still reasonable to leave x^0 and 0^x defined for positive x.

So if a function is continuous at a point, we define the value of the function to be that value at that point.

Minor nitpick -- if f is continuous at a point, then its value at that point is the limit, and if is discontinuous there, then its value is something other than the limit.

But I think your idea is right -- you're talking about making a new function whose value at that point is equal to the limit. (And, by abuse of notation, calling that function f too)


But for 0^0, in practice I've only ever really seen it defined when there's something sneaky lurking in the background.

 

[pivoxa15]

But at least if we restrict ourselves to the closed first quadrant, but with the origin removed, x^y would be everywhere continuous, so it is still reasonable to leave x^0 and 0^x defined for positive x.

This usually occurs in power series? For example, the Taylor expansion of cos(x) about x=0 http://mathworld.wolfram.com/Cosine.html where the first term would be 0^0 when expanding cos(0). But in that case they have assumed it to equal 1 without explicitly stating it. Is it usually the case that if 0^0 occurs but they don't mention its value, just assume it equals 1?

 

[Gokul43201]

Ooh, cos(0) is a veritable playground for the "zeroists" - the Taylor expansion even has the famous 0! in it. And seeing as how it's the lim(x->0) sinx/x, you can throw in the more famous 0/0 into the ring.

Now back to your serious discussion...

 

[shmoe]

But in that case they have assumed it to equal 1 without explicitly stating it. Is it usually the case that if 0^0 occurs but they don't mention its value, just assume it equals 1?

You shouldn't have to assume it's anything- any text should explain this notation if they use it. I don't know if Mathworld explicitly mentions them sticking to this usage, but they do mention it http://mathworld.wolfram.com/ExponentLaws.html and the context it comes up in makes it clear what they mean. Some texts will avoid this completely, writing power series instead as

$$a_0+\sum_{n=1}^{\infty}a_{n}x^n$$

 

[arildno]

The mathworld power series definition of cos(x) is incomplete; it ought to have had, say, the following definition appended to it:
$$0^{0}\equiv{1}$$
It should be obvious why such a local definition of the symbol is needed.

 

[Universe_Man]

I asked a simular question a week or two ago. Here's my interpretation (I apologize if it is incorrect, I am but a mere HS student):

As I understand it:

$$n^-1 = 1/n$$ and $$n^1= n$$

and

$$(n^-1)(n^1) = n^0 = (1/n)(n/1) = 1$$

correct?

so if n=0:

$$(0^-1)(0^1) = (1/0)(0/1) = 1 = 0^0$$

So given that the following is true, I would come to the conclusion that $$0^0$$ is defined. But there is something weird about that $$1/0$$.

 

[Hurkyl]

$$n^-1 = 1/n$$
...
correct?

Nope. That identity is false. You get one guess as to why. The actual identity has another clase you left out.

 

[Universe_Man]

Really? I didn't know that. I was always taught that n^-1= 1/n. I asked my teachers why of course (they just said, "that's the way it is, we can't discuss it now, I have material to cover"). But my calculator came out with an agreeable answer when I plugged a number in for n. So, please enlighten me if you will (Your mathematic credentials are very likey far better than mine), as I have no earthly idea, and probably could not give you a rational guess.

 

[Universe_Man]

Oh, well it would be undefined of course raising 0 to a negative exponent, I just assumed that for the sake of argument, you could notate it as 1/0, but obviously my assumption is incorrect. thank you for your information.

 

[Hurkyl]

I was always taught that n^-1= 1/n.

You were (or should have been) taught that n^(-1) = 1/n... when n is nonzero. You should never have been taught that the above equation is true, let alone is well-defined, for every n.

Most of these silly paradoxes (and some other mistakes) arise when people ignore these little details.

 

[pivoxa15]

$$(0^-1)(0^1) = (1/0)(0/1) = 1 = 0^0$$

So given that the following is true, I would come to the conclusion that $$0^0$$ is defined. But there is something weird about that $$1/0$$.

You are doing mathematical operations on entities that they are undefined such as 1/0. That is not allowed. But this argument even though false is sort of intuitively reasonable so makes it more plausible to define 0^0 as 1 rather than any other number.

 

[**bouncey!!**]

no, I am only at secondary school but it makes more sense for the answer to be 0.
the answer is only 1 for other intergers to the power of 0 becuse say take 3
3to the 3 = 27 3 to the 2=9 3 to the 1= 3 you are dividing by three each time so it makes sense for the nxt to be 0 .
you do not get this with 0

 

[masudr]

The reason

$$n^{-a} = 1/n^a$$

makes sense is because the rule of multiplication of powers is the same as adding the exponents (if the powers are of the same base) and the definition that $n^0 \equiv 1$ for all nonzero $n$.

Since:

$$n^0 \equiv 1$$

$$\Rightarrow n^{-a} n^a \equiv 1$$

$$\Rightarrow n^{-a} \equiv 1/n^a.$$

 

[NEWO]

huh? if you are doing the three times tables then yeah but 0^0 can not be applied in this way. x^0 = 1 for non zero values of x, this can not be strictly said exactly for 0^0 (my opinion changed with more reading). I think if some one is to prove that 0^0= 1 and 0/0 is infinite you would both win the nobel prize and also become a billionaire.

 

[pivoxa15]

no, I am only at secondary school but it makes more sense for the answer to be 0.
the answer is only 1 for other intergers to the power of 0 becuse say take 3
3to the 3 = 27 3 to the 2=9 3 to the 1= 3 you are dividing by three each time so it makes sense for the nxt to be 0 .
you do not get this with 0

Your argument with base 3 is good but If you take 0 as your base than it dosen't work because you can't divide by 0. Going from 0^5 to 0^4 doesn't work by dividing by 0.

But if you are trying to say that since 0^n where n is any number except 0 is 0 than 0^0 ought also be 0. That's not a bad one but Universe_Man's argument although wrong seems more convincing if I had to choose between the two.

 

[Universe_Man]

Wait it would be 0. I apologize if this is repetition, but I would like to give my own interpretation, just to see how I did.

$1/.01=100$ right?

So what if you were to put more and more zeros in the decimal, and always having it end in 1? You could make your decimal incredibly mind numbingly small and your solution would get closer and closer to infinity. So 1/0 could be defined as infinity couldn't it? If so:

given $0^-1=1/0= /infty$ and $0^1=0$

and $(0^1)(0^-1)=(/infty)(0)=0^0$

Then $0^0=0$

Because it doesen't matter what you multiply 0 by, you will still get zero.
plus if you put $0^0$ into the calculator on windows, it gives you 0 which confirms it. And since 0 is a definable value, $0^0$ is defined.

 

[masudr]

Infinity is defined, in fact, as a limit:

$$\lim_{x\rightarrow0}\frac{1}{x}\equiv\infty.$$

You have just carefully defined a process of getting to infinity. That's what infinity is: the result of some kind of process, it's not a number. And then you have gone on to disregard your careful definition of the concept of infinity and manipulated it as a real number. The reason infinity is not a proper number is because it leads to contradictions unless treated correctly as a limit.

Don't worry about what $0^0$ is; it's not a problem. The value of $0^0$ does not logically follow from the idea of exponentiation as being repeated multiplication. If we define it as $1$ and we can still be consistent with the rest of the rules of the game, then that's what we do.

And surely you're joking (Mr. Man): some people who use this forum are much, much smarter than the people who made Windows Calculator. I'd trust them much more than I'd trust that software.

 

[d_leet]

plus if you put $0^0$ into the calculator on windows, it gives you 0 which confirms it.

Uhh.. Not that the value that the windows calculator confirms anything at all, but when I try that I get 1 not 0, and why on Earth would what a calculator returns as the value of some expression validate whether or not 0⁰ is defined?

 

[shmoe]

plus if you put $0^0$ into the calculator on windows, it gives you 0 which confirms it. And since 0 is a definable value, $0^0$ is defined.

Here's a hint- don't use software for mathematical definitions, and certainly not a built in cheapo like windows calculator. It's sometimes easier to arbitrarily assign values to otherwise undefined things when it comes to programs, it gives a less likely chance of your program exploding, but the user has to be aware of potentially funky results.

Infinity is defined, in fact, as a limit:

$$\lim_{x\rightarrow0}\frac{1}{x}\equiv\infty.$$

First it should be:

$$\lim_{x\rightarrow0^{+}}\frac{1}{x}=\infty$$

The left hand limit will give -infinity. Secondly I wouldn't call this "defining infinity", rather we can define the use of the infinity symbol to mean a specific kind of divergence when it comes to limits (or a kind of convergence in some version of the extended reals if you prefer), but this is maybe just my preference for how to think about this notation.