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X not differentiable at x=0

  1. Mar 24, 2012 #1
    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. jcsd
  3. Mar 24, 2012 #2
    Why do you think 0^0 is undefined? 0^0 = 1.
     
  4. Mar 24, 2012 #3
    That's what it says in my book. That may not be the best answer, but that's where my question comes from.
     
  5. Mar 24, 2012 #4

    arildno

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    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.
     
  6. Mar 24, 2012 #5

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    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.
     
  7. Mar 24, 2012 #6
    Thanks, I had a feeling that it had something to do with nx^(n-1) not being universally valid.
     
  8. Mar 25, 2012 #7

    HallsofIvy

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    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
    [tex]\frac{f(x+h)- f(x)}{h}= \frac{x+h- x}{h}= \frac{h}{h}[/tex]
    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 [itex]\lim_{x\to a} f(x)= \lim_{x\to a}g(x)[/itex].
    That allows us, as long as we are taking limits to handle things like "0/0".
     
  9. Mar 25, 2012 #8
  10. Mar 25, 2012 #9

    HallsofIvy

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    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"?
     
  11. Mar 27, 2012 #10
    It seems to me that case should be eliminated from the general formula, so the formula is written like follows
    If [itex]f(x)=x^n[/itex], [itex]n\in \mathbb{N}[/itex]
    then

    [itex]f^{\prime}(x)=\begin{cases}
    1 & \text{ if } x= 0 \text{ and } n=1 \\
    n x^{n-1} & \text{ otherwise.}
    \end{cases}[/itex]
     
  12. Mar 27, 2012 #11

    I like Serena

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    @LikeMath: there is another special case for n=0.
    However the formula also holds true for any ##n \in \mathbb{R}## if x>0.
     
  13. Mar 27, 2012 #12
    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
  14. Mar 27, 2012 #13

    I like Serena

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    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?
     
  15. Mar 27, 2012 #14
    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.
     
  16. Mar 27, 2012 #15

    D H

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    The problem is by choosing different paths toward the origin, [itex]\lim_{x,y\to 0} x^y[/itex] 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.
     
  17. Mar 27, 2012 #16

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    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.
     
  18. Mar 27, 2012 #17
    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.
     
  19. Mar 27, 2012 #18

    D H

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    0! = 1 is not an abuse of notation.
     
  20. Mar 27, 2012 #19

    I like Serena

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    Check out http://en.wikipedia.org/wiki/Empty_product.
    As DH said, 0!=1 is well defined. ;)
     
  21. Mar 27, 2012 #20
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