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I already know that there are functions that have no anti-derivative (although I haven't seen a proof), but is there a function with no derivative? and I don't mean something like dirichlet function or floor/ceiling functions. In other words, is there some function that is defined in terms of algebraic operations that have no derivative?

(just for reference I've taken calc I,II,III,linear algebra and am currently in diff eq's)

edit: by algebraeic operations what i really meant is "closed-form expression"

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closed as unclear what you're asking by Alan, mrf, Daniel, uranix, user230715 Sep 11 '15 at 22:10

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    $\begingroup$ See en.wikipedia.org/wiki/Weierstrass_function $\endgroup$ – 1-___- Sep 11 '15 at 19:42
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    $\begingroup$ What is an algebraic operation for you? $\endgroup$ – Dominic Michaelis Sep 11 '15 at 19:43
  • $\begingroup$ There are functions like $f(x)= \left| x \right|$ that have no continuous derivative in $\mathbb R$, but they have in $(-\infty, 0)$ and $(0, +\infty)$. $\endgroup$ – mkspk Sep 11 '15 at 19:44
  • $\begingroup$ @user2770287 I think that taking $\sin$ is not an algebraic operation (as $\sin(x)$ for $x$ algebraic $\neq 0$ is always non algebraic. Furthermore taking an infinite sum doesn't seem to be algebraic for me either $\endgroup$ – Dominic Michaelis Sep 11 '15 at 19:44
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    $\begingroup$ $x^{2/3}, x \in \mathbb R,$ has no derivative at $0.$ $\endgroup$ – zhw. Sep 11 '15 at 19:51
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I understand that you mean a derivative with no closed form formula.

No, there aren't such functions, as there is an algorithm to compute the symbolic derivative.

Any closed form expression can be expressed as a syntax tree (f.i. $\cos(x^2+3)$, is described by a $\cos$ node with a son; the son is a sum ($+$) node with two sons; the left son is a square node with a son, which is the independent variable $x$; the right son is the constant $3$).

And every kind of node corresponds to a derivation rule, where every closed formula derives as another closed formula. $$(u+v)'=u'+v'$$ $$(uv)'=u'v+uv'$$ $$(\cos(u))'=-u'\sin(u)$$ $$(e^u)'=u'e^u$$ $$\cdots$$

By induction, the result is a closed formula.


If you allow special functions, such as $\Gamma(x)$, then you have to also allow their derivatives, which may not be expressible in terms of the original function.

Or said differently, if you allow the $\Gamma$ function alone, then, yes, there are functions without a derivative.

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  • $\begingroup$ Thank you, this answers my question very well 'closed form' was the term I was looking for. $\endgroup$ – Ryan Joseph Sep 11 '15 at 19:56
  • $\begingroup$ For more details about the "allowed" functions, see en.wikipedia.org/wiki/…. $\endgroup$ – Yves Daoust Sep 11 '15 at 19:59
  • $\begingroup$ There can be issues with domains though; the derivative has a symbolic expression by this algorithm, but that expression might not be well-defined on the entire domain of the original function. See the example of $x^{2/3}$ given in zhw.'s comment on the question, for instance. $\endgroup$ – Eric Wofsey Sep 11 '15 at 20:21
  • $\begingroup$ @EricWofsey: a symbolic expression does not have a domain, does it ? I can derive $\sqrt{-x^2-1}$ symbolically, even though the domain of the function is empty. $\endgroup$ – Yves Daoust Sep 11 '15 at 20:29

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