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Homework Help: Two variable diricklet

  1. Oct 2, 2007 #1
    1. The problem statement, all variables and given/known data

    Does the function f(x,y) has any partial derivatives?

    f(x,y) = (x^2 + y^2)^2, if both x and y are rational

    = 0 , otherwise

    2. Relevant equations

    3. The attempt at a solution

    Not sure what to do. I know this function has only one limit at (0,0) and is also continuous there, and nowhere else. So it is not differentiable for (x,y) =/= (0,0). However, this doesn't say anything about existence of the partials. Thanks for any help.
  2. jcsd
  3. Oct 3, 2007 #2


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    Have you written out the expression for the partial derivatives? In particular, if you look of
    [tex]\frac{f(x_0+ h,y_0)- f(x0,y0)}{h}[/tex]
    [tex]\frac{f(x_0,y_0+ h)- f(x0,y0)}{h}[/tex]
    for irrational x0, y0, and h, what do you get? What about for rational values. I would recommend that you look at (0,0) especially carefully.
    Last edited by a moderator: Oct 3, 2007
  4. Oct 3, 2007 #3
    And you can do this because of irrationals and rationals being dense in R. OK but what is it so special about (0,0)? I'm not seeing it. thanks
  5. Oct 3, 2007 #4
    I mean after all, with some of x, y, h, irrational we get 0/h and when you take the limit as h goes to 0, then what now?
  6. Oct 3, 2007 #5
    Wow forgot l'hospital for a sec there. Anyways lim 0/h =0. Ok so I got for the irrationals the partials are 0, while for the rationals the partials are non-zero only for (x,y)=/= (0,0). So the conclusion is that the partials exist only at (0,0)? Thanks
  7. Oct 3, 2007 #6
    Oh wait now I checked my work and is giving me that the x partial exists not only at
    (0,0) but also at the y-axis. Is that right?
  8. Oct 3, 2007 #7


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    There are actually partial derivatives at a lot of places, if I read your problem correctly. Take the line y=sqrt(2). Then f(x,y)=0 for any x-value along this line since y is irrational. So an x partial exists anywhere along this line.
  9. Oct 3, 2007 #8
    True! I considered the partials through the rationals and through the irrationals, and looked at where these two were equal. What I got was that they were equal only (for the x partial) at the y-axis. So what I did must be insufficient, and I am missing a major method of finding the answer. Can anyone share it? Thanks
  10. Oct 3, 2007 #9


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    There no real 'method'. You just have to think about it. I think you are right that the only rational (x,y) pts where partials exist is (0,0). Now we know if y is irrational then there is an x partial. What about if x is irrational? What about if they are both irrational? Is that all of the cases?
  11. Oct 3, 2007 #10
    Oh, so by symmetry if x is irrational then the y partial exists symilarly. I said that the partials exist through at the y-axis (for the x partial) including (0,0), because I found that this was the only place where these two partials are equal. So would this be the answer for the both rational or irrational cases?
  12. Oct 3, 2007 #11
    Oh now I see that the "both irrational" case is included in the "one irrational " case. So now my only concern is about the "both rational" case.
  13. Oct 3, 2007 #12


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    Right, but I don't see anything special about the y-axis. And, why should the two partials be equal?
  14. Oct 3, 2007 #13
    Oh I see the mistake I was doing. I think I got it now. Check this out:
    Take x,y rational. Letting h -> 0 through the irrational we get

    f_x = lim (h -> 0) [-(x^2 + y^2)^2/h], which exists only for (x,y) = (0,0), where it equals 0.

    Letting h->0 through the rationals we get

    f_x = (some function with x and y, so that if x = 0, then f_x = 0)

    Since Q and R\Q are dense in R, these two limits should be equal if they exist. Since they both work only if (x,y) = (0,0) then the partial exists only at (0,0) as you say? Thanks
    Last edited: Oct 3, 2007
  15. Oct 3, 2007 #14


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    That's a little confusing, but if you understand it... BTW in the case (x,y) rational and not (0,0), the function is not even continuous at (x,y) even when approached along x constant or y constant lines. This being the case, there really isn't any need to write down a difference quotient.
  16. Oct 3, 2007 #15
    I'm done with this Diricklet! :) Great help! Thanks a lot.
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