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Equality of mixed partial derivatives of order >2

  1. Jan 3, 2012 #1
    I know that for any C2 function, the mixed second-order partials are equal, and I see that this should extend inductively to a statement about the kth partials of a Ck function, but I am having trouble figuring out exactly how this works.

    For example, take f:ℝ2 → ℝ .

    fxxy=fxyy is not true, right?

    Folland's Advanced Calculus says that in Taylor's theorem we can use the notation [itex]\partial[/itex][itex]\alpha[/itex]f to refer to "the generic partial derivative of order [itex]|\alpha|[/itex], since the order of differentiation doesnt matter". ([itex]\alpha[/itex] is a multi-index)

    But this doesnt make sense to me, considering that there are multiple possibilities for the order>2 partials, even when restricting it to mixed partials. Is this an oversight in Folland's wording, or am I missing something?

    Thanks in advance for your replies :)
  2. jcsd
  3. Jan 4, 2012 #2


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    Yes, that's right- but not really what you are asking. After all, fxx is not the samer as fyy. What fxy= fyx says is that, as long as the derivatives are continuous, the order of the derivatives does not matter.

    Apparently you are missing the fact that "order" does not mean changing the number of derivatives with respect to the different variables. "yxx" and "xyx" are different orders of the same thing, "yxx" and "xyy" are NOT.
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