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Math Help - Special function

  1. #1
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    Special function

    Is it possible to find a function that is continuous on neighborhood of (0,0), not differentiable at 0, where directional derivative is 0 for all directions u, and the directional derivative can be expressed as gradf(0)*unit vector, even though f is not diffable at 0?

    I have been trying functions like x/(x^2 + y^2), but they have all turned out to be differentiable.
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  2. #2
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    Actually, I just tried f(x,y) = x^3/(x^2 + y^2), and it is continuous along the path (x, x^1.5) (which is continuous), however, using the definition of differentiability,

    lim (x, y) -> (0,0) {f(0 + x, 0 + y) - f(0, 0) - grad(0)*(x, y)}/(x^2 + y^2)^0.5

    I am letting f approach along the path y = x^1.5
    f(x, x^1.5) = x^4.5/(x^2 + x^3)

    lim (x, x^1.5) -> (0, 0) x^4.5/[(x^2 + x^3)(x^2 + x^3)^0.5]

    By using L'hopital's rule several times, I still find that this limit is undefined, so I conclude that the function is not differentiable when approached along x, x^1.5,

    however, function is continuous when approached along this path because

    lim (x, x^1.5) -> (0,) x^4.5/(x^2 + x^3) = 0, so it is continuous along all paths I have tried so far.

    Plus, the directional derivative, I let u = (x,y), and then, Duf(0) = lim t -> 0 [f(0 + tu) - f(0)]/t = lim t -> 0 (ty)^3/((tx)2 + (ty)^2)) = 0, since t is left on numerator with no t's on the bottom.

    Does this function work?
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  3. #3
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    Quote Originally Posted by Rozaline View Post
    I just tried f(x,y) = x^3/(x^2 + y^2).
    .
    .
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    Does this function work?
    No, that function doesn't work. If you approach the origin along the x-axis (so that y=0) then f(x,0) = x^3/x^2 = x, whose derivative is 1. So that directional derivative is not 0.

    The simplest example I can find of a function that is continuous at 0, with directional derivative 0 in every direction, and yet not differentiable at 0, is f(x,y) = \frac{x^3y}{x^4+y^2} (with f(0,0)=0). Along the path y=kx the function is given by f(x,kx) = kx^2/(k^2+x^2), with derivative 0 at x=0. But if you approach 0 along the parabolic path y=x^2 then you get f(x,x^2) = x^5/(2x^4) = \tfrac12x, with derivative 1/2. So the function cannot be differentiable.
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