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Differentiability of a function

  1. Nov 14, 2013 #1
    1. The problem statement, all variables and given/known data
    I have the function f, defined as follows:
    f=0 if xy=0
    f= ##xysin(\frac{1}{xy})## if ##xy \neq 0##

    Study the differentiability of this function.

    3. The attempt at a solution

    there are no problems in differentiating the function where ##xy\neq0##.
    the partials in (0,0) are both zero.
    If i consider xy as a single variable i get from the definition of differential that:
    ##\lim_{xy \to 0} \frac{f(xy)-f(0)-\nabla f(0)(xy)}{\sqrt{(xy)^2-0}}##=##\lim_{xy \to 0} \frac{xy sin(1/xy)}{xy}## doesn't exist

    On the other hand, if i consider x,y separately i get:
    ##\lim_{(x,y) \to (0,0)} \frac{f(x,y)-f(0,0)-\nabla f(0,0)(x,y)}{\sqrt{x^2+y^2}}##=##\lim_{xy \to 0} \frac{xy sin(1/xy)}{\sqrt{x^2+y^2}}=0## and thus the function is differentiable even when x,y are 0.

    these two results are conflicting. what am i doing wrong?
     
  2. jcsd
  3. Nov 14, 2013 #2

    mfb

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    The results are not conflicting. The existence of the partial derivatives is not sufficient to have a differentiable function.
     
  4. Nov 14, 2013 #3
    yes, i know that, but ny problem is that the differential doesn't exist if i consider xy as a single variable, but it does if i consider x and y separately, please look at the two different applications of the definition of differentiability i wrote
     
  5. Nov 14, 2013 #4

    PeroK

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    I don't think that the partial derivatives exist (except trivially for x = 0 and y = 0). E.g.

    [itex]\frac{\partial{f}}{\partial{x}}|_{(0,y)} \ = \lim_{x \rightarrow 0}\frac{xysin(\frac{1}{xy})}{x} = \lim_{x \rightarrow 0}ysin(\frac{1}{xy})[/itex]

    Which doesn't exist.
     
  6. Nov 15, 2013 #5
    should I deduce than that it is differentiable only in (0,0)?
     
  7. Nov 15, 2013 #6

    PeroK

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    Yes, it's differentiable at (0, 0) but nowhere else on the x- and y-axes.
     
  8. Nov 15, 2013 #7

    mfb

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    You did not use the definition of differentiability in (x,y) at all. If the function would be differentiable at (0,0), both derivative attempts would have to work. They do not, showing your function is not differentiable at that point.
     
    Last edited: Nov 15, 2013
  9. Nov 15, 2013 #8

    PeroK

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    Why is the function not differentiable at (0,0)?
     
  10. Nov 15, 2013 #9

    mfb

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    This post (and #11 and #13) is wrong, see post 15

    It is not even continuous.

    Consider (x,y)=(h,h) with h≠0. Then ## f(h,h)=h^2 \sin(\frac{1}{h^2}) > h^2 \frac{1}{2h^2} = \frac{1}{2}## using sin(x)>x/2 for small x>0.
    Clearly the limit for ##h \to 0## cannot be 0, but f(0,0) is zero.
     
    Last edited: Nov 15, 2013
  11. Nov 15, 2013 #10

    PeroK

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    I'm not sure about [itex]sin(x) > 1/(2x)[/itex] given that [itex]|sin(x)| \le 1[/itex]

    As (x, y) -> (0,0), [itex]\frac{xy}{\sqrt{x^2+y^2}} → 0[/itex]
     
  12. Nov 15, 2013 #11

    mfb

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    I don't see the connection of your global upper bound to my local lower bound (valid at least up to pi/2).
    Plot

    I should have specified x>0 in "sin(x)>1/(2x) for small x", but that is not an issue as h^2 is always positive (I edited it in my previous post now).

    Where does the denominator come from?
     
  13. Nov 15, 2013 #12

    PeroK

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    Ok, so take x = 0.1:

    1/2x = 1/0.2 = 5

    And sin(anything) can't possibly be > 5
     
  14. Nov 15, 2013 #13

    mfb

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    Sorry, my mistake. I meant sin(x)>x/2 (for 0<x<pi/2). The plot and the application to calculate the limit were correct.
     
  15. Nov 15, 2013 #14

    PeroK

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    [itex]\lim_{h→0}h^2 \sin(\frac{1}{h^2}) = 0[/itex]

    If h is getting small and |sin| is bounded by 1, then that limit must be zero, surely?

    [itex]1/h^2[/itex] is getting very large, not very small
     
  16. Nov 15, 2013 #15

    mfb

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    You are right, the function is differentiable. Sorry for the confusion.

    I really need that free weekend now...
     
  17. Nov 16, 2013 #16
    thank you very much :)
     
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