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How is 0^0 defined?

  1. Jul 12, 2007 #1
    i was wondering how 0^0 is defined? can anybody please help?

    thanks in advance.
     
  2. jcsd
  3. Jul 12, 2007 #2

    matt grime

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    Please can we not start a very long thread on this one? If people want to see the debates on it then search the forums.

    Simply put the most logical definition of 0^0 is that it is equal to 1. This makes 'everything work' without having to make any 'except for 0 when it FOO is equal to 1' statements: partitions, functions, combinatorics, taylor series etc.
     
  4. Jul 12, 2007 #3
    It isn't. It's an undefined statement like 0/0.

    However, a lot of mathematicians like to set it equal to one, but it's really a matter of convenience. So be careful of this one.
     
  5. Jul 12, 2007 #4
  6. Jul 12, 2007 #5
    I, and most people, define it to be [tex]0^0[/tex].
    Why? Because it is useful for dealing with infinite series.


    But some people do not define it. And what I hate is when a person tells me he does not define it but when he writes the power series he completely overlooks [tex]0^0[/tex]

    The same way we define [tex]0!=1[/tex] (but there is actually another reason there).
     
  7. Jul 12, 2007 #6
    We can also define in terms of the continuity of the function x^0 or x^x, or (x+x)^x or whatever you want.
     
  8. Jul 13, 2007 #7

    Gib Z

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    The reason we define 0!=1 has very little relation to this..How many ways can we arrange nothing Kummer? Or if you want, you could take the recursive definition of the factorial function, [itex] n!= n\cdot (n-1)![/itex] and substituting n=1 gives the desired result.

    The reason 0^0 remains undefined is because the limit that represents it does not actually converge. Of course we could somewhat cheat by making some assumptions, eg say that it is the limit:
    [tex]\lim_{x^{+}\to 0} x^x[/tex], and that is equal to 1, but we assume that the Base and the exponent approach zero at the same rate.

    The correct limit is actually:
    [tex]\lim_{x\to 0 , y\to 0} x^y[/tex], which is multi valued.
     
  9. Jul 13, 2007 #8

    HallsofIvy

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    What do you mean "define it to be [tex]0^0[/tex]"? Did you mean to say "define it to be 1"?
     
  10. Jul 13, 2007 #9
    now i am really confused. is [tex]0^0 = 1[/tex] or not?
     
  11. Jul 13, 2007 #10
    Thank you.

    Yes, that is true. But that does not constitute a formal mathematical proof. The problem is that there is no proof and it is a matter of taste. My preference along with most people is to define it as 1 because it is useful in power series.

    Another reason is that the Gamma function evaluated at 1 is equal to 1, and that is a generalization of a factorial. But that is another story.
     
  12. Jul 14, 2007 #11

    HallsofIvy

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    No, it is an "indeterminate"- like 0/0, if you replace the "x" value in a limit by, say, 0 and get 0^0 the limit itself might have several different values.

    To take two obvious examples, if f(x)= x0, then f(0)= 00. For any positive x, f(x)= x0= 1 so the limit as x goes to 0 is 1. If we want to make this a continuous function, we would have to "define" 00= 1.

    However, if f(x)= 0x we again have f(0)= 00 but for any positive x, f(x)= 0x= 0 which has limit 0 as x goes to 0. If we want to make this a continuous function, we would have to "define" 00= 0.
     
    Last edited: Jul 14, 2007
  13. Jul 14, 2007 #12
    thanks a lot, HallsofIvy. that made it pretty clear to me.
     
  14. Jul 14, 2007 #13

    HallsofIvy

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    It might be still clearer now that I have edited it to say what I meant!
     
  15. Jul 15, 2007 #14
    yeah it's clear. 00 cannot be equal to both 0 and 1. so that's why it is indeterminate. am i right?
     
  16. Jul 15, 2007 #15

    HallsofIvy

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    Yes. Actually, it is possible to alter the limits slightly so as to get ANY number.
     
  17. Jul 16, 2007 #16

    matt grime

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    Come on people. It is not that 0^0 can be '0 and 1', but that a certain limit, x^y as x and y tend to 0 can be made to be arbitrary. That doesn't say what 0^0 is, just that the function f(x,y)=x^y has a nasty singularity at (0,0). But the symbol 0^0 has a perfectly well understood commonly accepted value as 1 for many other uses.
     
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