Is x^0 Differentiable at x=0 Despite 0^0 Being Undefined?

[Shawn Garsed]

If the derivative of x^n equals nx^(n-1), then the derivative of x or x^1 equals x^0, but 0^0 is undefined. Does that mean x is not differentiable at zero?

 

[AdrianZ]

If the derivative of x^n equals nx^(n-1), then the derivative of x or x^1 equals x^0, but 0^0 is undefined. Does that mean x is not differentiable at zero?

Why do you think 0^0 is undefined? 0^0 = 1.

 

[Shawn Garsed]

Why do you think 0^0 is undefined? 0^0 = 1.

That's what it says in my book. That may not be the best answer, but that's where my question comes from.

 

[arildno]

If the derivative of x^n equals nx^(n-1), then the derivative of x or x^1 equals x^0, but 0^0 is undefined. Does that mean x is not differentiable at zero?

No.
But it means that that formula is only strictly valid for ALL n if you CHOOSE 0*0^(-1) (n=0) and 1*0^0 (n=1) in that particular context to mean 0 and 1, respectively.

Since those partial results are derivable by wholly independent ways, that local definition is utterly unproblematic.

 

[I like Serena]

As you can see here on wiki, your formula for the derivative of a power of x is indeed not generally valid.

Exceptions are specifically made for f(x)=c and for f(x)=x, in which cases f does have a proper derivative at x=0, but where the formula $nx^{n-1}$ is not properly defined.

 

[Shawn Garsed]

Thanks, I had a feeling that it had something to do with nx^(n-1) not being universally valid.

 

[HallsofIvy]

When ever you see a "general formula" that appears to have a problem for specific values, it is best to go back to the basic formulas.

Here, to find the derivative of f(x)= x, we can write
$$\frac{f(x+h)- f(x)}{h}= \frac{x+h- x}{h}= \frac{h}{h}$$
Now, we can say that is equal to 1 as long as h is not 0!

Fortunately, for problems like this (and, after all, the derivative always involves a fraction in which both numerator and denominator go to 0) we have the very important limit property:
If f(x)= g(x) for all x except x= a, then $\lim_{x\to a} f(x)= \lim_{x\to a}g(x)$.
That allows us, as long as we are taking limits to handle things like "0/0".

 

[Edgewood11]

0^0= 1

 

[HallsofIvy]

Please don't simply assert things (especially those others have already said were false) without giving some argument. What reason do you have to say that "0^0= 1"?

 

[LikeMath]

It seems to me that case should be eliminated from the general formula, so the formula is written like follows
If $f(x)=x^n$, $n\in \mathbb{N}$
then

$f^{\prime}(x)=\begin{cases} 1 & \text{ if } x= 0 \text{ and } n=1 \\ n x^{n-1} & \text{ otherwise.} \end{cases}$

 

[I like Serena]

@LikeMath: there is another special case for n=0.
However the formula also holds true for any $n \in \mathbb{R}$ if x>0.

 

[AdrianZ]

That's what it says in my book. That may not be the best answer, but that's where my question comes from.

hmm, sounds strange to me. I'm not a math expert like others on this forum, but from what I know from abstract algebra, if we take ordinary multiplication as a binary operation, then for any x we have x^0 = 1 (the identity element), or generally in a group we have g^0=e for any g in G.

I think your book says 0^0 is undefined because one can interpret it in two different ways and concludes that 0^0 = 0 and also 0^0=1. because:
1 - 0 times anything is equal to zero, hence 0^0 = 0
2 - x^0 = 1 for any x, hence 0^0=1
so your book considers 0^0 undefined because we can associate two different values to it (hence, it's not well-defined), but I think the first argument doesn't apply, because when we say 0^0 it means we're multiplying 0 zero times by itself, which means actually we don't multiply anything by 0. so I think the first case isn't relevant here or at least we can define 0^0=1, just like we define 0! = 1.

Please don't simply assert things (especially those others have already said were false) without giving some argument. What reason do you have to say that "0^0= 1"?

Do we really need to give any argument? Don't we define x^0 = 1 for any given x?



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Well, after having spent some time on this, I realized that 0^0 might really be ill-defined, because I found no example in group theory that could explain 0^0=1, but still when I analyzed the function f(x)=x^x I found out that x^x = 1 as x approaches 0. Is it really ill-defined? or does it cause any contradiction if we define 0^0=1?

 

[I like Serena]

hmm, sounds strange to me. I'm not a math expert like others on this forum, but from what I know from abstract algebra, if we take ordinary multiplication as a binary operation, then for any x we have x^0 = 1 (the identity element), or generally in a group we have g^0=e for any g in G.

R is not a group with multiplication, since 0 does not have an inverse.

Do we really need to give any argument? Don't we define x^0 = 1 for any given x?

You will need to specify a context for that definition, because contradictions are lurking around the corner.
EDIT: What should we do with 0^n for any given n?

 

[AdrianZ]

R is not a group with multiplication, since 0 does not have an inverse.

You will need to specify a context for that definition, because contradictions are lurking around the corner.
EDIT: What should we do with 0^n for any given n?

That is exactly what I thought of. Please refresh. (I edited my previous post)

I can't say whether 0^0 is undefined or not, it seems that if we define 0^0=1 no contradiction would arise, but on the other hand as you rightly pointed out (and I had already thought of it) R is not a group under multiplication (R-{0} is), so my group theory argument is obviously wrong, but I'm not quite sure whether we could say 0^0=1 or not, because f(x)=x^x suggests us that it's possible to define 0^0=1.

 

[D H]

The problem is by choosing different paths toward the origin, $\lim_{x,y\to 0} x^y$ can be made to have any value whatsoever. In other words, 0^0 is indeterminate (rather than undefined).

In many contexts such as the binomial expansion and the problem at hand, it is convenient to define 0^0 to be 1 as an abuse of notation. Extremely convenient, but still an abuse of notation.

 

[I like Serena]

That is exactly what I thought of. Please refresh. (I edited my previous post)

I can't say whether 0^0 is undefined or not, it seems that if we define 0^0=1 no contradiction would arise, but on the other hand as you rightly pointed out (and I had already thought of it) R is not a group under multiplication (R-{0} is), so my group theory argument is obviously wrong, but I'm not quite sure whether we could say 0^0=1 or not, because f(x)=x^x suggests us that it's possible to define 0^0=1.

For f(x)=0^x you might define 0^0=0.
For f(x)=x^0 you might define 0^0=1.

In other words, you would define 0^0 depending on which function f you're talking about.
The only reason you would do something like that, is to get out from under the burden to meticulously define f(0) in each case.

 

[AdrianZ]

The problem is by choosing different paths toward the origin, $\lim_{x,y\to 0} x^y$ can be made to have any value whatsoever. In other words, 0^0 is indeterminate (rather than undefined).

In many contexts such as the binomial expansion and the problem at hand, it is convenient to define 0^0 to be 1 as an abuse of notation. Extremely convenient, but still an abuse of notation.

Yup. That's an insightful post.

For f(x)=0^x you would define 0^0=0.
For f(x)=x^0 you would define 0^0=1.

In other words, you would define 0^0 depending on which function f you're talking about.
The only reason you would do something like that, is to get out from under the burden to meticulously define f(0) in each case.

You're right.

Actually I was thinking of a similar kind of abuse of notation that DH said. It's not immediately clear why we should define 0! = 1, but it turns out to be very useful in infinite series and binomial expansions, it would be a very useful abuse of notation, but still an abuse of notation. lol.
Thanks.

 

[D H]

0! = 1 is not an abuse of notation.

 

[I like Serena]

Actually I was thinking of a similar kind of abuse of notation that DH said. It's not immediately clear why we should define 0! = 1, but it turns out to be very useful in infinite series and binomial expansions, it would be a very useful abuse of notation, but still an abuse of notation. lol.
Thanks.

Check out http://en.wikipedia.org/wiki/Empty_product.
As DH said, 0!=1 is well defined. ;)

 

[AdrianZ]

0! = 1 is not an abuse of notation.

Why not? Can you disprove that 0! = 0?

Check out http://en.wikipedia.org/wiki/Empty_product.
As DH said, 0!=1 is well defined. ;)

Ok, I'm reading it. Thanks

 

[AdrianZ]

well, I read the page that ILS had given, I'm still not convinced why 0!=1 is well-defined (we define it this way!) but 0^0=1 is ill-defined.

I found this interesting explanation on wikipedia:

http://en.wikipedia.org/wiki/Exponentiation#Zero_to_the_zero_power

History of differing points of view

Different authors interpret the situation above in different ways:

Some argue that the best value for 0^0 depends on context, and hence that defining it once and for all is problematic.[13] According to Benson (1999), "The choice whether to define 0^0 is based on convenience, not on correctness."[14]
Others argue that 0^0 is 1. According to p. 408 of Knuth (1992), it "has to be 1", although he goes on to say that "Cauchy had good reason to consider 0^0 as an undefined limiting form" and that "in this much stronger sense, the value of 0^0 is less defined than, say, the value of 0 + 0" (emphases in original).[15]

The debate has been going on at least since the early 19th century. At that time, most mathematicians agreed that 0^0 = 1, until in 1821 Cauchy[16] listed 0^0 along with expressions like 0⁄0 in a table of undefined forms. In the 1830s Libri[17][18] published an unconvincing argument for 0^0 = 1, and Möbius[19] sided with him, erroneously claiming that

whenever

. A commentator who signed his name simply as "S" provided the counterexample of (e−1/t)t, and this quieted the debate for some time, with the apparent conclusion of this episode being that 00 should be undefined. More details can be found in Knuth (1992).[15]

It seems that we should blame Cauchy for this uninteresting discussion lol

 

[I like Serena]

well, I read the page that ILS had given, I'm still not convinced why 0!=1 is well-defined (we define it this way!) but 0^0=1 is ill-defined.

The difference is that 0^0 has different limits depending on the path you take to get there.
For 0! there is no such ambiguity: we're talking about integers only and only about multiplications of a whole number of factors.

It seems that we should blame Cauchy for this uninteresting discussion lol

Blame?
No, we should applaud his work, since he showed that the mathematicians before him hadn't done their job properly.
And now we have a consistent math framework (which leaves some things undefined LOL).

 

[AdrianZ]

The difference is that 0^0 has different limits depending on the path you take to get there.
For 0! there is no such ambiguity: we're talking about integers only and only about multiplications of a whole number of factors.

Well, I'm convinced that defining 0^0 is ambiguous, but I'm not convinced why 0!=1 is well-defined yet. I read the page you said, I have no problem with it, it says that in general they define prod({})=1. It's a very reasonable definition, but my question is, is it the ONLY way that we can define it? I mean do we get contradictions if we define it in any other way? It's more like an 'uninteresting' question in logic rather than mathematics I think.

Blame?
No, we should applaud his work, since he showed that the mathematicians before him hadn't done their job properly.
And now we have a consistent math framework (which leaves some things undefined LOL).

Well, I dare to disagree with you on this. Euler didn't have a clear understanding of what he did with infinite series, but he's by far the most prolific mathematician in history and his results influenced many great mathematicians like Lagrange, Gauss, Cauchy, Riemann, Ramanujan and many others and his ideas are still influential in almost every area of mathematics, so he did his job properly and efficiently in my opinion and probably he never asked whether 0^0=1 or not. lol maybe I'm saying that because I think rigorists are ruining the beauty of mathematics and are making mathematics a sub area of logic forcefully but I personally prefer mathematical beauty to rigor and that's why I said we should blame Cauchy for this 'uninteresting' question. It's just a matter of taste and personal opinions I think xD

 

[LikeMath]

@LikeMath: there is another special case for n=0.
However the formula also holds true for any $n \in \mathbb{R}$ if x>0.

Yes, thank you, it is just a slip of the pen.

 

[LikeMath]

I undo my previous opinion, now I am convinced that there is absolutely no problem in the formula.

If
$f(x)=x^n,\:n\in\mathbb{N}$, then (without any exceptions) we have
$f^{\prime}(x)=nx^{n-1}.$

For the choked piont when $n=1,x=0$
we have: $n=1$
$f^{\prime}(x)=1x^{1-1} =1x^0=1$, now we substitute $x=0$ to get that $f^{\prime}(0)=1$. So the argument that we have: First we must substitute n and then it becomes a new function of x, and hence substitute the value of x. We do not say x and n admit their values at the same time. In other words you have the constant n and then you have the variable x.
The same thing can be said for n=0.
I hope that the idea is written clearly, and sorry for my bad writing.

 

[I like Serena]

I disagree.
You can't state that 1x^0=1 for all x in R, without additional specifications.
You imply that for x=0 you take the limit and define the function in that point as being equal to that limit.
But such implications need to be stated explicitly to make your statement accurate.

Btw, for a correct statement, you also need to specify what kind of function f is, and in particular its domain.

 

[LikeMath]

Since we already know $f(x)=x$ is continuously differentiable, we do not need to define the function at that point, since it already equals the limit. So
$\lim_{x\to0} f^\prime(x)=\lim_{x\to0}x^0=1$, so $f^\prime(x)=1$ without any exception, so the formula is right, and no changes should be done!

On the other hand,
The function $f(x)=x^0$ in its own self (out side our problem) is defined on $x\in\mathbb{R}\backslash\{0\}$, and it is not defined at $x=0.$
No one knows what is $0^0$. They just know some limit behaviors which is not necessarily the value of $0^0$.