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Question about 0^0 ?

  1. Jun 30, 2011 #1
    can I write it like this [itex] 0^{1-1} [/itex]
    which is [itex] 0^{1}0^{-1} [/itex]
    which would eventually by 0/0 .
    And I thought we defined 0^0=1
    so then 0/0 could equal 1 .
    I read some of the recent 0^0 thread but wasn't sure if this is in there.
     
  2. jcsd
  3. Jun 30, 2011 #2

    pwsnafu

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    00 is what is called http://en.wikipedia.org/wiki/Indeterminate_form" [Broken]. This label means that in different contexts 00 can be equal to different values because there is not enough information in the notation.

    In discrete mathematics we usually define 00=1 as a convention, and we also do this with Taylor series as well. However, it's possible to justify different values for the expression in other contexts.
     
    Last edited by a moderator: May 5, 2017
  4. Jun 30, 2011 #3
    ok, I thought it would be something like that.
     
  5. Jul 3, 2011 #4
    In my opinion 0^0 depends on how you approach zero on the Real number line. You have 4 cases. Two ways to approach the base, two ways to approach the exponent. 0^0 also depends on how 'fast' the base approaches 0 and how 'fast' the exponent approaches 0 They don't have to approach 0 at the same rate. Let us only consider RATIONAL approximations to zero for example (1/10)^(1/1000) can be computed, how would you compute (1/10)^(1/e)? I would like to entertain any suggestions about that.

    Anyway... what i want to point out is the following. Approaching the base 0 FROM THE LEFT is going to give you problems when you have a radical with AN EVEN INDEX because that is going to produce an IMAGINARY value. I sort of have this vision that approaching the base zero from the right the function is sort of 'well behaved' in the sense that the values remain in the real plane, but as soon as you 'pass' zero going from right to left you get this 'explosion' of an uncountable number of DISCONTINUOUS tentacles elevating from the REAL PLANE into a COMPLEX 3D PLANE. I don't have enough info to know what that looks like.

    Just some thoughts, I welcome criticism and suggestions, after all, this is a way to learn.

    :smile:
     
    Last edited by a moderator: May 5, 2017
  6. Jul 3, 2011 #5

    Hurkyl

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    What you mean to say is something like
    Defining 00 as a limit of xy as (x,y) goes (0,0) doesn't work, because the limit depends on how you approach (0,0)​
     
  7. Jul 3, 2011 #6
  8. Jul 3, 2011 #7
    I read all the posts from the other thread. No one mentioned that you can use RATIONAL approximations and get as close to zero as you want from right to left. What you get is DISCONTINUOUS of course but do the plotted points follow a pattern? As a graph do the plotted points 'appear' to approach a value or do they zig-zag all over the place. I don't have a graphing calculator so please somebody put y = x^x in a TI89 and zoom in on zero.

    I am aware that for sensitive issues like this the calculator should not be trusted but it can plot much faster than i can. If i had TI89 i would write a small program that would plot y = x^x for RATIONAL x only using a small step size, probably .001

    I can't read the document cause my browser does not decode, can you cut and paste and send it to me as a file attachment? I don't know if this will work, don't know how cut and paste affect scribd file.

    Like I said before in, my opinion, 0^0 is complicated but having said that, I don't think we should abandon our quest to understand it.

    Question: Can we use a set of 3 mutually perpendicular axes, (x, y, iz) to graph functions of two Real numbers (x, y) that can give Complex values?

    Example: If f(x,y) = Real then the graph stays in x-y plane where iz is zero, if f(x, y) = i(Real) then the graph elevates or lowers from the Real plane.

    I have not seen this done so i am thinking there might be irreconcilable problems with this approach.

    :smile:
     
  9. Jul 4, 2011 #8
    Yes I agree but plotting many points close to zero maybe can provide some insight?

    If you vary the parameters by which you take the limit, for example x^2 goes to zero much faster than x and x^4 goes to zero much faster than x^2 etc. We must be very carefull to account for every possibility.

    Also, calculating (1/10)^(1/e) EXACTLY may not be possible, however 1/e can be calculated to any degree of accuracy using a rational approximation. If you can do it this way then you can define x^x for all diminishing values of x to the immediate right of zero so the function becomes 'continuous' provided you 'fix' the rate of change of the exponent and base.

    Years ago i read a math text on numerical analysis in which exponents like 1/e were approximated using fractions and the author claimed it was a valid extension. This book had some amazing techniques...long ago...can't remember...but i digress.

    Finally, I have taken many limits and am aware of the numerous rules associated with them but this investigation requires much more than that.:smile:
     
  10. Jul 4, 2011 #9

    Hurkyl

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    You're posts sound a little confused to it's hard to offer much.


    I assume by "compute" you mean "find a decimal approximation". In this sense, computing real exponentiation [itex]a^b[/itex] is not a problem. It can be done straightforwardly from

    [tex]a^b = e^{b \ln a}[/tex]​

    The logarithm is easy to compute because
    [tex]\ln (m \cdot 2^n) = \ln m + n \ln 2[/tex]​
    Nothing in the right hand side poses difficulty.

    The taylor series for [itex]y = e^x[/itex] is really good, so it's only a problem to compute if x has large magnitude. (And if it's negative, it's only a problem if "zero" is not accurate enough of an answer)

    But Newton's method comes to our rescue: we use it to find a root of [itex]f(y) = \ln x - y[/itex] (this root will satisfy [itex]y = e^x[/itex]).

    This works by starting with, say, [itex]y_0=1[/itex] and iterating:
    [tex]y_{n+1} = y_n (1 + x - \ln y_n)[/tex]​
    For negative x, though, you need to start with a [itex]y_0[/itex] near zero.



    People have done a lot of work on methods for quickly and accurately computing classical functions like logarithms and exponentials. I assume it shouldn't be too hard to find materials on the topic via google search....
     
  11. Jul 4, 2011 #10
    I appreciate any offer, thanx. Just sharing ideas, i like that. I have done all these things you mention. Exponential functions, Newtons Method, limits. Got honors in Cal 2 and linalg, tutored them in math center for 8 years. That was 4 years ago. Now not as sharp as i was but still thirsty for knowledge.
     
  12. Jul 4, 2011 #11
    The word 'compute' i used in my first post only and i did not mean decimal approximation. That came in second post. To clarify i meant exact value of (1/10)^(1/e) so necessarilly there is a discontinuity at that exact value of y as a function of x. Now, using the rational approximations, that discontinuity is 'supposedly' squeezed into non-nexistence from both sides. (Bet you like that idea...lol)

    However, knowing what i know now about transcendental numbers, uncountable, un-nameable etc.
    I am beginning to wonder if the rational approximations for the exponent accurately describe the behavior of exponential functions and whether or not any exponential GRAPH is continuous at all.

    But even after all that, something WORSE happens to the left of zero. That was my point. To make these ideas more precise i would need to write pages for post so i gotta choose what i think is more important for the argument.:biggrin:
     
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