# X not differentiable at x=0

1. Mar 24, 2012

### 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?

2. Mar 24, 2012

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

3. Mar 24, 2012

### Shawn Garsed

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

4. Mar 24, 2012

### arildno

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.

5. Mar 24, 2012

### 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.

6. Mar 24, 2012

### Shawn Garsed

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

7. Mar 25, 2012

### 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".

8. Mar 25, 2012

### Edgewood11

0^0= 1

9. Mar 25, 2012

### 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"?

10. Mar 27, 2012

### 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

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

11. Mar 27, 2012

### 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.

12. Mar 27, 2012

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.

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

--------------------------------------------------------------------------------------

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?

Last edited: Mar 27, 2012
13. Mar 27, 2012

### I like Serena

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?

14. Mar 27, 2012

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.

15. Mar 27, 2012

### D H

Staff Emeritus
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.

16. Mar 27, 2012

### I like Serena

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.

17. Mar 27, 2012

Yup. That's an insightful post.

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.

18. Mar 27, 2012

### D H

Staff Emeritus
0! = 1 is not an abuse of notation.

19. Mar 27, 2012

### I like Serena

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

20. Mar 27, 2012

21. Mar 27, 2012

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

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

Last edited: Mar 27, 2012
22. Mar 27, 2012

### I like Serena

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.

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).

23. Mar 27, 2012

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.

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

24. Mar 27, 2012

### LikeMath

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

25. Mar 28, 2012

### 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.

Last edited: Mar 28, 2012