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0^0 = 1?

  1. Apr 14, 2005 #1
    I've been told that any number raised to the zeroeth power is equal to 1. What about zero raised to the zeroeth power? Is that 1 or 0?

    [tex]0^0=1?[/tex]

    The Rev
     
  2. jcsd
  3. Apr 14, 2005 #2
    undefined!
     
  4. Apr 14, 2005 #3
    ...and indeterminate. In some cases when a generalization is desired, it is profitable to define it as 1, in others, as 0.
     
  5. Apr 14, 2005 #4

    Galileo

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    In series, it is always taken to be 1, so that you can write a series expansion compactly:

    [tex]f(x)=\sum_{n=0}^{\infty}a_nx^n[/tex]
    so that [itex]f(0)=a_0[/itex]

    For example, the geometric series:

    [tex]\sum_{n=0}^{\infty}x^n =\frac{1}{1-x}[/tex]
    is true at x=0 only if we define 0^0=1.
     
  6. Apr 14, 2005 #5

    CRGreathouse

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    I don't think that it's undefined; I think that it's 1. 0^0 is an empty product, and empty products are necessarily equal to 1. As Galileo points out above, we need 0^0=1 for series to have compact formulas.

    Everyone agrees that [tex]x^0=1[/tex] for [tex]x\neq0[/tex], but there's no reason to think that it should be different at 0 -- [tex]0^x[/tex] is only 0 for x > 0, since it's not defined for negative x.

    There's no problem accepting 0^0 as 1, and there are many good reasons to think it shouldn't be undefined or 0.
     
  7. Apr 14, 2005 #6
    Well, I think that the formula a^0=1 appears when you try to divide a^m by itself:

    (a^m)/(a^m) = a^(m-m) = a^0

    Since the first part of this equation equals 1, we have a^0=1

    But if a=0 we can't do the division (0^m)/(0^m)
     
  8. Apr 14, 2005 #7

    Zurtex

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    That's not mathematical at all. I've always had a simple way of looking at it:

    [tex]x^0 = \frac{x}{x}[/tex].
     
  9. Apr 14, 2005 #8

    shmoe

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    Everyone also agrees that [tex]0^x=0[/tex] for all [tex]x>0[/tex], so there should be no reason to think it's different at 0, so [tex]0^0=0[/tex] right? We have a problem with the limit of [tex]x^y[/tex] as [tex](x,y)\rightarrow (0,0)[/tex], it doesn't exist (different values depending on how x and y are approaching zero) so there is no obvious or natural choice of a value for [tex]0^0[/tex], so it's usually left as undefined (barring notational convenience).

    The reasons to call the symbol 0^0 1 is usually for notational convenience, like empty products, binomial theorem, power series, etc.
     
  10. Apr 14, 2005 #9
    Most of the mathematicians I know seem to take the position that 0^0 is technically undefined, but there's nothing wrong with letting it equal 1 for notational convenience.

    It's very rare for it to be convenient for 0^0 to be defined as any other value. The standard reason it needs to be something other than 1 is to extend 0^x to x=0, but in practice that doesn't seem to happen very often.


    Of course, if you have a limit of the form f(x)^g(x) with f(x),g(x) -> 0 you have to be careful; it won't always converge to one.
     
  11. Apr 14, 2005 #10
    What about [tex]x^x[/tex] as [tex]x \rightarrow 0[/tex]? I have always found [tex]x^x[/tex] fascinating for some obscure reason.
     
  12. Apr 14, 2005 #11
    x^x -> 1 as x -> 0, of course.

    In fact, I believe there's a result that says that if f(x),g(x) -> 0 as x approaches some limit, then f(x)^g(x) -> 1 as long as f and g are analytic.
     
  13. Apr 14, 2005 #12

    dextercioby

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    COUNTEREXAMPLE:


    The functions [itex] \frac{1}{3x+5} [/itex] and [itex] \frac{1}{x^{2}+4} [/itex] are analytical in every point from their domain...

    However,

    [tex] \lim_{x\rightarrow +\infty}\left(\frac{1}{x^{2}+4}\right)^{\frac{1}{3x+5}} =-\infty [/tex]

    Daniel.
     
  14. Apr 14, 2005 #13

    Hurkyl

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    My calculator symbolically evaluates that limit to 1...

    There's a more obvious counterexample:

    [tex]
    \lim_{x \rightarrow 0} 0^x = 0
    [/tex].

    :biggrin:


    The whole thing about 0^0 depends precisely what ^ means. As an operation on real numbers, 0^0 is undefined. The reason is that (0, 0) is not in the domain of ^.

    Okay, so you want the "philosophical" reason. :tongue: A crucial property about real operations is that they're continuous within their domain.

    However, ^ cannot be continuous at (0, 0) -- the classic examples demonstrating this fact are 0^x --> 0 and x^0 --> 1 as x --> 0.


    However, there are other meanings to ^. For example, there's a definition of ^ that means repeated multiplication. The empty product is, by definition, 1, so anything to the zero-th power is equal to 1. In polynomials and power series, this definition is what ^ "really" means -- that's why one uses 0^0 = 1.
     
  15. Apr 14, 2005 #14

    Zurtex

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    It is a very nice function though, when I first was thinking about it I tried to think about it in 4D in complex space and was pleasantly surprised when I started graphing it on mathematica recently I had quite a good idea oh how it looked.
     
  16. Apr 14, 2005 #15

    CRGreathouse

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    Empty products aren't mathematical?
     
  17. Apr 14, 2005 #16

    Hurkyl

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    A less trivial example is:

    [tex]
    \lim_{x \rightarrow 0} \left( e^{-1/x^2} \right)^{x^2} = \frac{1}{e}
    [/tex]

    Both the base and the exponent approach 0 from the positive side, but as we can see, the limit is not 1.

    (Note the base is not an analytic function of x, despite its infinite differentiability at 0!)
     
  18. Apr 14, 2005 #17

    Zurtex

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    No, I meant your approch to the problem.
     
  19. Apr 14, 2005 #18

    CRGreathouse

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    My approach was just stating that it's an empty product, and that empty products are always 1. The rest of my post was descriptive/informative (and perhaps poorly worded).
     
  20. Apr 14, 2005 #19
    take log both side...
    u get
    0 = (log 1)/log(x)
    log 1 = 0
    0/log(x) as x > 0
    = 0/inf
    interminate^_^
     
  21. Apr 14, 2005 #20
    That's what I get for quoting a theorem from memory. :redface:


    I was way too general it seems. After looking it up, it seems that the theorem actually requires that:

    1) f and g are non-zero
    2) f and g are analytic at zero
    3) f(x) -> 0 and g(x) -> 0 as x -> 0 from the right
    4) f(x) is positive from all positive x <= some value close to zero

    Then you get that f(x)^g(x) -> 1 as x -> 0 from the right.


    Of course, that result is much less impressive. But I'm pretty sure this one is true.



    Edited to remove the counterexample that wasn't really a counterexample.
     
    Last edited: Apr 15, 2005
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