1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
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

    [tex]
    f(x+u)=f(x) + (Df)u + |u|\epsilon(u)
    [/tex]

    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

    mathwonk

    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

    HallsofIvy

    User Avatar
    Staff Emeritus
    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

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Yes, of course. Don't know what I was thinking! (Too early probably.)
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Multiple times differentiable
  1. Differentiable (Replies: 2)

Loading...