Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Multiple times differentiable

  1. Aug 4, 2007 #1
    Sometimes I've encountered functions [itex]f:\mathbb{R}^n\to\mathbb{R}^m[/itex] being called N-times differentiable. What does it mean, precisely?

    I know that for a function to be differentiable, if is not enough that the partial derivatives [itex]\partial_i f_j[/itex] exist, but instead the derivative matrix [itex]Df[/itex] must exist and satisfy the equation

    f(x+u)=f(x) + (Df)u + |u|\epsilon(u)

    which is more than only existence of the partial derivatives.

    Does the N-times differentiability mean something else than only all partial derivatives [itex]\partial_1^{k_1} \partial_2^{k_2} \cdots \partial_n^{k_n} f_j[/itex], where [itex]k_1+k_2+\cdots+k_n=N[/itex], existing?
  2. jcsd
  3. Aug 4, 2007 #2
    To check differentiability it would be easier to use a thm rather than definition saying
    if all the partial derivatives exists and continuous in a nhd of x then e say f is differentiable at x.
    (note:I am not sure that this is the exact statement.Check from a textbook)
  4. Aug 4, 2007 #3
    You can take it to mean that the N-th partials all exist and are continuous.
  5. Aug 4, 2007 #4


    User Avatar
    Science Advisor
    Homework Helper

    or if you know that the derivative of amap R^n-->R^n is a matrix bva;ued function, you can take it to mean that matrix valued function is differentiable, etc.....
  6. Aug 5, 2007 #5


    User Avatar
    Science Advisor

    You should be careful to distinguish between the "derivative at a given point" and the "derivative function".

    The derivative of a function from [itex]\mathbb{R}^n[/itex] to [itex]\mathbb{R}^m[/itex] is an n by m matrix, and so an object in [itex]\mathbb{R}^{n+m}[/itex] - for every point in [itex]\mathbb{R}^n[/itex]. The derivative function, then, is a function from [itex]\mathbb{R}^n[/itex] to [itex]\mathbb{R}^{n+m}[/itex] and so its derivative, the second derivative of the original function is an n by n+m matrix, an object in [itex]\mathbb{R}^{2n+m}[/itex]. As you take more and more, because of all those new mixed derivatives, it gets more and more complicated!

    (Strictly speaking, a derivative is a linear transformation- we should say it can be represented by a matrix.)
  7. Aug 5, 2007 #6
    But shouldn't the image of the derivative function be in [itex]\mathbb{R}^{nm}[/itex], and not in [itex]\mathbb{R}^{n+m}[/itex]?

    The idea that we can start taking more derivatives and the dimensions just increase when this is done sounds good. I think I even had such thoughts at some point, but I wasn't sure if that is what is usually meant by this terminology of function being multiple times differentiable.
  8. Aug 5, 2007 #7


    User Avatar
    Science Advisor

    Yes, of course. Don't know what I was thinking! (Too early probably.)
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook