0
$\begingroup$

For example, we can define a function

$$f(x) = \begin{cases} x^2 & \text{if } x < 2 \\ 4 \cdot(x - 1) & \text{if } x \ge 2 \end{cases}$$

This function seems to have a derivative.

$$f'(x) = \begin{cases} 2x & \text{if } x < 2 \\ 4 & \text{if } x \ge 2 \end{cases}$$

However, $f$ has no second derivative at $2$. Contrast this to $\sin$, which has an infinite number of derivatives. It seems that $\sin$ is somehow "smoother" than $f$.

Is there any significance to this?

$\endgroup$
7
  • $\begingroup$ My understanding is that you're asking for some kind of intuition behind the difference between a $k$-differentiable map and an $m$-differentiable map. Could you make it more clear if this is not the case? $\endgroup$ Nov 29 '19 at 13:14
  • $\begingroup$ What do you mean by $k$-differentiable maps? $\endgroup$ Nov 29 '19 at 13:15
  • $\begingroup$ Diff $k$ times but not $k+1$. $\endgroup$ Nov 29 '19 at 13:16
  • $\begingroup$ Yes, I was asking whether or not $k$-differentiability is important anywhere, and if $\infty$-differentiable maps are somehow a special class of functions. In fact, "$k$-differentiability" may actually be an answer, since having a name for it means it is at least somewhat significant $\endgroup$ Nov 29 '19 at 13:17
  • 1
    $\begingroup$ Smooth functions are nice because they form the the largest class of functions on which differentiation is an operation into the same class. If you take a $k$ times differentiable function, its derivative needn't be $k$ times differentiable. But a derivative of an infinitely differentiable function is infinitely differentiable. $\endgroup$
    – Wojowu
    Nov 29 '19 at 13:27
1
$\begingroup$

wojowu, Hans Lundmark and Arnaud Mortier answered the question in the comments: This is called smoothness, and $\sin$ is an infinitely smooth function.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.