What does it mean for a function to be N-times differentiable?

[jostpuur]

Sometimes I've encountered functions $f:\mathbb{R}^n\to\mathbb{R}^m$ 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 $\partial_i f_j$ exist, but instead the derivative matrix $Df$ 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 $\partial_1^{k_1} \partial_2^{k_2} \cdots \partial_n^{k_n} f_j$, where $k_1+k_2+\cdots+k_n=N$, existing?

 

[matness]

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)

 

[kharranger]

You can take it to mean that the N-th partials all exist and are continuous.

 

[mathwonk]

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...

 

[HallsofIvy]

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

The derivative of a function from $\mathbb{R}^n$ to $\mathbb{R}^m$ is an n by m matrix, and so an object in $\mathbb{R}^{n+m}$ - for every point in $\mathbb{R}^n$. The derivative function, then, is a function from $\mathbb{R}^n$ to $\mathbb{R}^{n+m}$ and so its derivative, the second derivative of the original function is an n by n+m matrix, an object in $\mathbb{R}^{2n+m}$. 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.)

 

[jostpuur]

But shouldn't the image of the derivative function be in $\mathbb{R}^{nm}$, and not in $\mathbb{R}^{n+m}$?

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.

 

[HallsofIvy]

Yes, of course. Don't know what I was thinking! (Too early probably.)