Find the nth derivative of an arbitrary function

Question

I want to define a function $\mathcal D$: $\Bbb F$ $\mapsto$ $\Bbb F'$ where $\Bbb F$ are infinitely differentiable functions $\Bbb R$ $\mapsto$ $\Bbb R$ and $\Bbb F'$ are the functions $\Bbb N$ $\mapsto$ $\Bbb F$, such that, if f is a function in $\Bbb F$ and f $^n$ is the nth derivative of f, $\mathcal D$( f )=f $^n$. My question is: is there a function in $\Bbb F$ for which I cannot define its nth derivative? If so, which? and why?, of course.

For example,

• $\mathcal D(x^a)= \frac {a!}{(a-n)!} x^{a-n}$ (the $n$th derivative of $x^a$)
• $\mathcal D (\ln(x))= \frac{(n-1)!}{x^n}(-1)^{n+1}$ (the $n$th derivative of $\ln(x)$)
• $\mathcal D(\sin(x))= \sin(x+\frac {n\pi}{2})$ (the $n$th derivative of $\sin(x)$)

and so on.

Is there a function $g$ for which I cannot define $\mathcal D(g)$?

Answer 1

It is possible to have something differentiable once that is not differentiable twice. Consider, for example, $g(x) = \int |x| dx$ with $g(0) = 0$.

Note that $g'(x) = |x|$ which is clearly well-defined, but $g''(x)$ does not exist at $x = 0$...

UPDATE

I can give you a function as a graph, represented by a hand drawing with infinite precision. Then, you will see the picture but creating a formula to reproduce it with infinite precision is impossible.