Notices
Results 1 to 14 of 14

Thread: 0 to the power of zero?

  1. #1 0 to the power of zero? 
    Forum Freshman
    Join Date
    Jul 2006
    Location
    usa
    Posts
    34
    What is 0 to the power of zero?[/img]


    "When you eliminate the inpossible, whatever remains, no matter how improbbable, must be the truth"- Sherlock Holmes
    Reply With Quote  
     

  2.  
     

  3. #2  
    Forum Sophomore Matt Lacey's Avatar
    Join Date
    Sep 2006
    Location
    Southampton, England
    Posts
    119
    1.


    Reply With Quote  
     

  4. #3  
    Forum Sophomore
    Join Date
    Jul 2005
    Posts
    121
    0 to the power 0 is often left undefined. The usual reason to define it is for notational convenience, e.g. to make power series a little simpler to write. In this case it will usually be defined as 1.
    Reply With Quote  
     

  5. #4  
    Forum Freshman
    Join Date
    Jul 2006
    Location
    usa
    Posts
    34
    Thanks. That is what I thought. It is an odd situation though. Can someone please describe why any # to the power of 0 is one?
    "When you eliminate the inpossible, whatever remains, no matter how improbbable, must be the truth"- Sherlock Holmes
    Reply With Quote  
     

  6. #5  
    Forum Sophomore
    Join Date
    Jul 2005
    Posts
    121
    That's how it's defined, x^0=1 when x is non-zero. There's really nothing more to it than that.

    Why is this a good or sensible or useful definition? This is another question. The most basic reason is it's required if we want the power laws x^(n+m)=(x^n)*(x^m) to be true when n and m are any integers.
    Reply With Quote  
     

  7. #6  
    Forum Isotope Zelos's Avatar
    Join Date
    Jun 2006
    Posts
    2,755
    it is 1 since anything divided withitself (not 0) is allways 1 x/x=1=x^1/x^1=x^(1-1)=x^0
    I am zelos. Destroyer of planets, exterminator of life, conquerer of worlds. I have come to rule this uiniverse. And there is nothing u pathetic biengs can do to stop me

    On the eighth day Zelos said: 'Let there be darkness,' and the light was never again seen.

    The king of posting
    Reply With Quote  
     

  8. #7  
    Guest
    Quote Originally Posted by Zelos
    it is 1 since anything divided withitself (not 0) is allways 1 x/x=1=x^1/x^1=x^(1-1)=x^0
    Brilliant! Zelos you have just answered an age old question!

    U0^U0 = 1U

    where U = universe! 8)

    All that money wasted on Quantum Theory...
    Reply With Quote  
     

  9. #8  
    Forum Radioactive Isotope mitchellmckain's Avatar
    Join Date
    Oct 2005
    Location
    Salt Lake City, UTAH, USA
    Posts
    3,112
    Shmoe and I just finished a discussion of this in "power of 4"

    Quote Originally Posted by shmoe
    What would you think if I suggested 0^0 should be 0 based on:

    0^1=0
    0^0.5=0
    0^0.1=0
    0^0.001=0

    and then went on to suggest 0^0 should be 1 based on:

    1^0=1
    0.5^0=1
    0.1^0=1
    0.001^0=1
    Shmoe is giving an example of function of two variables which has two different limits at the origin from two different directions, f(x,y) = x^y at (x,y)=(0,0). But I don't see that algebra or consistency with the logarithm will work on this one either.

    taking the logarithm of both sides of 0^0 = a
    0 log 0 = log a
    But log 0 is undefined except as the limit of negative infinity.

    Also the other approach of (log a)/log 0 = 0 doesn't help, even when you use the limit of log 0 -> negative infinity for any a solves this.

    Therefore I think we must conclude that 0^0 is undefined

    Hmmm.... what about a limit from a different angle
    like the 45 degree angle using the limit of x^x as x approaches 0

    emacs lisp gives me the following progression

    (expt .1 .1) 0.7943282347242815
    (expt .01 .01) 0.954992586021436
    (expt .001 .001) 0.9931160484209338
    (expt .0001 .0001) 0.9990793899844618
    (expt .00001 .00001) 0.9998848773724686
    (expt .000001 .000001) 0.9999861845848758
    (expt .0000001 .0000001) 0.9999983881917339

    which suggests that the limit is 1.0

    I think this means that the limit is 1.0 from all directions except along x=0 where f(x,y) = 0^y = 0

    I don't know that might be a case for 1.0 being the best answer if there is any answer at all, other than undefined that is.
    Quote Originally Posted by shmoe
    Quote Originally Posted by mitchellmckain
    I think this means that the limit is 1.0 from all directions except along x=0 where f(x,y) = 0^y = 0
    What if you approach along the curve (x,log(a)/log(x)), where a>0 and x>0 and x->0?
    Shmoe is suggesting this because on this curve our two variable function f(x,y) = x^y becomes x^(log base x of a) = a, which means that the limit as x goes to 0 on this curve is a.

    For this limit to work (without resorting to analytic continuations of our function) I think that a must not only be greater than 0 but also less than 1.

    Now the question I immediately had was at what angle does this approach the origin and as you would expect the tangent to this curve at the origin is the line x=0.

    As a result I was correct in my conclusion that approaching the origin at any angle except the line x=0 does give the limit of 1. It is only that when you approach the origin from the angle of the line x=0 that you can get any limit from 0.0 to 1.0 depending on how you take the limit (that is what curve you follow).
    See my physics of spaceflight simulator at http://www.relspace.astahost.com

    I now have a blog too: http://astahost.blogspot.com/
    Reply With Quote  
     

  10. #9  
    Forum Ph.D. william's Avatar
    Join Date
    Jun 2006
    Location
    Wherever I go, there I am
    Posts
    935
    This smells like a branch cut is involved....?
    "... the polhode rolls without slipping on the herpolhode lying in the invariable plane."
    ~Footnote in Goldstein's Mechanics, 3rd ed. p. 202
    Reply With Quote  
     

  11. #10  
    Forum Professor river_rat's Avatar
    Join Date
    Jun 2006
    Location
    South Africa
    Posts
    1,497
    Quote Originally Posted by william
    This smells like a branch cut is involved....?
    Branch cuts with real valued functions is quite a novel idea

    Not sure how Mitch got tied into analytic continuations, the function f(x, y) = x<sup>y</sup> has no limit as x, y -> 0. You cant even give it a continous extension

    f(x, x^2) -> 1 as x -> 0 but f(2<sup>-1/x</sup>, x) -> 1/2 as x -> 0
    As is often the case with technical subjects we are presented with an unfortunate choice: an explanation that is accurate but incomprehensible, or comprehensible but wrong.
    Reply With Quote  
     

  12. #11  
    New Member
    Join Date
    Oct 2006
    Posts
    4
    0^0 = 0, nothing or all by itself.

    0 is the barriar between the real number and the imaginary number. If 0 x 0 or 0^0 equals any real or imaginary number, 0 cannot exist as zero.
    Reply With Quote  
     

  13. #12  
    M
    M is offline
    Forum Junior
    Join Date
    Sep 2006
    Posts
    282
    0^0 = 0^k/0^k for any integer k except zero

    0^k = 0, hence 0^0 = 0/0,

    which is generally undefined, but can be defined as a limit for a particular problem, like the limit of f(x)/g(x) as x-> 0, with f(0)=g(0)=0. To arbitrarily define it as 1 is dubious.
    Reply With Quote  
     

  14. #13  
    M
    M is offline
    Forum Junior
    Join Date
    Sep 2006
    Posts
    282
    not to confuse you:

    the equal sign in 0^0 = 0/0 should not be taken literally, but as saying both sides are equivalent in the sense that they are both undefined.

    Reply With Quote  
     

  15. #14  
    New Member
    Join Date
    Nov 2006
    Posts
    4
    Quote Originally Posted by shmoe
    0 to the power 0 is often left undefined. The usual reason to define it is for notational convenience, e.g. to make power series a little simpler to write. In this case it will usually be defined as 1.
    0^0 is undefined. I agree with Zelos. But I never knew it could also be considered equals to 1. Any example in what equation 0^0 = 1?
    Reply With Quote  
     

Bookmarks
Bookmarks
Posting Permissions
  • You may not post new threads
  • You may not post replies
  • You may not post attachments
  • You may not edit your posts
  •