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Continuity of d/dx [x^(1/3)]

  1. Jul 9, 2012 #1
    Hey everyone,

    I was just curious about the nature of the cube root function [itex]f(x)=x^{1/3}[/itex]. I know that its derivative is obviously [itex]\frac{1}{3}x^{-2/3}[/itex] which has a discontinuity at [itex]x=0[/itex]. However, in the non-mathematical sense, the graph of [itex]y=f(x)[/itex] looks smooth - I don't see any angles or cusps like [itex]y=|x|[/itex] or [itex]y=x^{2/3}[/itex].

    Now, granted, the discontinuity in [itex]f'(x)[/itex] is a vertical asymptote where [itex]\delta(M)[/itex] can always be chosen such that:
    [tex]|x|<\delta \Rightarrow f'(x)>M[/tex]
    So unlike the asymptotes of [itex]\frac{1}{x}[/itex] for example, or the derivative of [itex]y=x^{2/3}[/itex], the function both sides of the discontinuity "goes to positive infinity" so the tangent line to [itex]x^{1/3}[/itex] approaches the same slope on either side of [itex]x=0[/itex].

    My question is: Is this a good enough explanation as to why an apparently smooth function has a discontinuous derivative? If not, why doesn't this violate the definition of a (mathematically) smooth function? (This may border on a philosophical question, so my apologies in advance if so)
     
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  3. Jul 10, 2012 #2

    haruspex

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    Yes to the first, but I don't understand your reasoning in the second. How do you think it would lead to such? Mathematically smooth is very much more demanding than apparently smooth.
     
  4. Jul 10, 2012 #3
    My reasoning is best described mathematically as follows (and I admit, rather poorly at that):
    Suppose we rotate the coordinate axes by [itex]\frac{\pi}{6}[/itex] (i.e., counterclockwise) and label the new grid [itex](x^*,y^*)[/itex]. We could technically have in this new coordinate system a function [itex]y^*=f^*(x)[/itex] that, although not expressible in terms of elementary functions, describes the graph of [itex]y=x^{1/3}[/itex].

    My claim is that
    [tex] \frac{\mathrm{d}y^*}{\mathrm{d}x} [/tex]
    is continuous. For other functions with "clearly" discontinuous derivatives, i.e., [itex]y_2=|x|, y_3=x^{2/3},\ldots [/itex] this is not the case.

    Just wanted to point this out because it's different from most other piecewise-smooth functions in that respect. Any thoughts?
     
  5. Jul 10, 2012 #4
    Your question is kind of a non-question - your f(x) is not mathematically smooth, that is, it is not in C^inf. It is only in C^0 or D^1.
     
  6. Jul 10, 2012 #5

    haruspex

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    OK, so you're thinking of it as a smooth curve in the plane, in more of a co-ordinate free sense.
    That's true, but that is rather different from a function. E.g. you could rotate it so that there are multiple y values for some x values, so now it is not even a function in the standard definition.
    One way around this is to parameterise, making x and y functions of some other variable that changes monotonically along the curve.
     
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