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B Difference between these functions .

  1. Aug 29, 2016 #1
    What's the difference between [tex]f(x)=3[/tex] and [tex]f(x)=3x^0[/tex] ? and why Limit of the second function when [tex]x\rightarrow0[/tex] exists ? and is the second function continuous at [tex]x=0[/tex] ?
     
  2. jcsd
  3. Aug 29, 2016 #2
    You can't present a function by just giving a rule, like you do. You need to give the domain and the codomain too. So what are they?

    Also, what is your convention for ##0^0##? Yes, mathematicians disagree on what it should be.
     
  4. Aug 29, 2016 #3

    mfb

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    Assuming R->R:
    00 does not have a single common definition, so f(0) needs a separate definition in the second case. If you define f(0)=3, then both functions are identical, and the second function is continuous, otherwise it is not.
     
  5. Aug 29, 2016 #4

    jedishrfu

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  6. Aug 29, 2016 #5
  7. Aug 29, 2016 #6

    fresh_42

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    Maybe it's not. But I also think it's natural, simply for the reason that taking something to a power is something multiplicative. And therefore it makes sense to define it as the unit 1.
     
  8. Aug 29, 2016 #7
    Sure, there are plenty of reasons why it should be ##1##. The best are set theoretic and category theoretic. But there is no consensus.
     
  9. Aug 29, 2016 #8

    jedishrfu

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    At least its a very binary decision and mathematicians can wear the shirt:

    "There are 10 different answers to the equation y = 0^0"
     
  10. Aug 29, 2016 #9

    mfb

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    Binary? What about defining 00 as the limit $$\lim_{x \to 0} x^{1/\log(x)} = e$$? You can also get every other number of course.
     
  11. Aug 29, 2016 #10
    Yes, the function ##y^x## has an essential singularity around ##(0,0)##.
     
  12. Aug 29, 2016 #11

    jedishrfu

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    It all reduces to 1's and 0's somehow. :-)

    That's a good example, I didn't see it before. It appears then that there's a whole class of possibilities too with other variations.
     
  13. Aug 29, 2016 #12

    fresh_42

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    The vast, boring, senseless, stupid, Sisyphus-like, unprofitable and endless job to rewrite formerly beautiful and nice formulas like, e.g.
    $$\exp(x) = \sum_{n = 0}^{\infty} \frac{x^n}{n!}$$
    is justification enough for the choice of ##1##.
     
  14. Aug 29, 2016 #13
    The thing is though that ##0^0 = 1## in this case only when the exponent is already seen as an integer. If the exponent is seen as a real number then it's better to leave ##0^0## undefined. So we have this weird context-dependent rule: ##y^x = 1## if ##x## is an integer and undefined when it can take on a continuous range of variables. Such a definition would be the most interesting one, but have no idea how to formalize that in a neat way. Maybe some kind of typed logic or something.
     
  15. Aug 29, 2016 #14
    Computer languanges can deal with this very well if you take as input ##0^0##. Then it would find the type of the exponent. If the exponent is an INT, then it's ##1##, if it's a double then it's undefined. But how to do this in mathematical context?
     
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