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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 $\mathcal D(g)$?

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    $\begingroup$ What do you mean by "cannot define $D(g)$"? Do you mean that it does not exist or if we cannot write down an explicit formula for it? $\endgroup$
    – Winther
    Jan 22, 2018 at 1:46
  • $\begingroup$ Paraphrasing the question: Can you give me a function g such that I can't express its nth derivative as a formula (like ln, sin, etc) The derivative may exist, but it can't be written as a formula, the only way to express it the would be "g derived n times" $\endgroup$
    – Santiago Arranz
    Jan 22, 2018 at 1:51
  • $\begingroup$ If $g$ is an elementary function then $D(g)$ will be an elementary function ( math.stackexchange.com/questions/2194769/… ) $\endgroup$
    – Winther
    Jan 22, 2018 at 1:54
  • $\begingroup$ Does that mean I can't express it's nth derivative as a function $\Bbb N$ $\mapsto$ $\Bbb F$? $\endgroup$
    – Santiago Arranz
    Jan 22, 2018 at 2:05
  • $\begingroup$ see the update in my answer... $\endgroup$
    – gt6989b
    Jan 22, 2018 at 2:38

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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.

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  • $\begingroup$ Didn't think about that one! I was thinking about infinitely differentiable functions, but I didn't write it in the question since couldn't think of a reason that it would make a difference, but hey! this is one. I'll edit the question, but this is certainly interesting. $\endgroup$
    – Santiago Arranz
    Jan 22, 2018 at 2:03

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