4
$\begingroup$

I was wondering if such a function exist. I'm comfortable with derivatives of polynomial functions, and some other basic functions, but I'm wondering if there could exist a very complicated function that doesn't have a derivative.

$\endgroup$
  • 8
    $\begingroup$ Google "Weierstrass function". $\endgroup$ – David Mitra Oct 3 '15 at 11:49
  • 5
    $\begingroup$ Or, more simply, the absolute value function $|x|$ at $x=0$. $\endgroup$ – joriki Oct 3 '15 at 11:53
  • $\begingroup$ Related: Are Continuous Functions Always Differentiable? (And other posts linked there might be also of interest.) $\endgroup$ – Martin Sleziak Oct 3 '15 at 13:01
  • $\begingroup$ @mick, if my answer was useful, consider accepting it by clicking the gray tick on the left of it. $\endgroup$ – dbanet Oct 19 '15 at 18:08
10
$\begingroup$

The commonly known example is the Weierstrass Function $f$, defined as $$f(x)=\sum\limits_{k=1}^\infty\frac{\sin\left(\pi{k}^2x\right)}{\pi{k}^2}.$$

The $y=f(x)$ graph looks like this (and intuitively shows why it is differentiable nowhere):

Graph of the Weierstrass Function

Another example would be the Dirichlet Function $D$, defined as $$D(x)=\begin{cases}1,\;\;x\in\mathbb{Q},\\0,\;\;x\in\mathbb{I}.\end{cases}$$

Its graph $y=D(x)$ would look like a pair of lines $\displaystyle{y=\frac{1}{2}\pm\frac{1}{2}}$ (which of cousre is not a graph of any function), so is uninteresting to show. The interesting part is that $D$ is actually discontinuous everywhere, and therefore differentiable nowhere.

$\endgroup$
  • 15
    $\begingroup$ It's worth mentioning that the Weierstrass function is continuous everywhere. $\endgroup$ – joriki Oct 3 '15 at 12:10
  • $\begingroup$ does the Weierstrass function pass the vertical line test? $\endgroup$ – set5 Oct 3 '15 at 13:05
  • 1
    $\begingroup$ @mick Hint: It's a function. $\endgroup$ – Workaholic Oct 3 '15 at 13:18
  • 2
    $\begingroup$ @mick, any function passes the vertical line test. If you meant to ask about the Dirichlet Function $D$ instead, the source of confusion is clear: its graph only looks like a pair of parallel lines, but actually for every $x_0$ the line $x=x_0$ crosses only one of the “lines” of the graph (and “goes through a hole in the second one”, if that adds intuition). $\endgroup$ – dbanet Oct 3 '15 at 13:18

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.