Wikipedia:Reference desk/Archives/Mathematics/2015 August 7

= August 7 =

Differentiables functions without an explicite derivative?
Hi everywhere. My question concerns the existence of functions from R (or a part of R) about which we could know that they are differentiable BUT without a known derivative. May be it seems obvious and the answer is NO! but in the other way round, only a extrem minority of the integrable functions do have a known antiderivative.

This could be a function having a derivative number (nombre dérivé in French) everywhere but these numbers would not enable an explicit definition of a function. I thank you for your thinkings about. Sent from France.--Jojodesbatignoles (talk) 07:52, 7 August 2015 (UTC)


 * The fact that an antiderivative ∫f(x)dx may not be expressed in terms of elementary functions, even if the function f is expressed in terms of elementary functions, does not make the antiderivative unknown. The derivatives of elementary functions are elementary, but the antiderivatives are often not. Bo Jacoby (talk) 11:38, 7 August 2015 (UTC).


 * Objection. That would mean saying that every solution to differential equations that are known to exist are known. YohanN7 (talk) 12:07, 7 August 2015 (UTC)
 * Many antiderivatives of elementary functions are known although they are not elementary functions. The error function is such an example. Bessel functions are defined as solutions of certain differential equations, not elementary but also by no means unknown. —Kusma (t·c) 12:26, 7 August 2015 (UTC)

Thanks. I'm the OP. So you think it's impossible to create a counterexample ; I think of using the Cantor set.--Jojodesbatignoles (talk) 12:53, 7 August 2015 (UTC)
 * I expect that if you "know" the function well enough to prove that it is differentiable, you kind of "know" its derivative as well. But the real question here is what "know" means. Any continuous function f on R has the antiderivative $$g(x)=\int_0^x f(t) dt$$. If f is analytic, you can even write down a power series for g. Doesn't that mean you "know" g as well as you know other transcendental analytic functions? —Kusma (t·c) 13:26, 7 August 2015 (UTC)


 * I think the answer would depend on what it means to "know" a function (or its derivative). Obviously, an answer like "if you know the function then you also know the derivative, since it is a limit of function values" is vacuous.  The class of all elementary functions, algebraic functions and, more generally, transcendental solutions of differential equations with elementary or algebraic coefficients, is closed under differentiation, and it is possible to write down the derivative explicitly in terms of such functions.  Another approach, in computational complexity theory, is that the notion of "knowing" a function is related to its computability.  Then the derivative of the function is also "knowable" in this sense (at least, if there is a uniform Lipschitz estimate known a priori).  See, for example, Ko, "Complexity theory of real functions".  A counterexample to at least one precise statement that you could formulate along these lines appears in Myhill (1971) A recursive function defined on a compact interval and having a continuous derivative that is not recursive, Michigan Math. J. 18, 97-98.   S ławomir  Biały  13:15, 9 August 2015 (UTC)