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Nov 26, 2016 at 2:00 vote accept CoffeDeveloper
S Nov 26, 2016 at 2:00 history bounty ended CoffeDeveloper
S Nov 26, 2016 at 2:00 history notice removed CoffeDeveloper
Nov 25, 2016 at 16:43 answer added Emil Jeřábek timeline score: 3
Nov 24, 2016 at 14:59 history edited CoffeDeveloper CC BY-SA 3.0
minor clarifications
Nov 24, 2016 at 14:48 history edited CoffeDeveloper CC BY-SA 3.0
added 261 characters in body
Nov 24, 2016 at 14:37 history edited CoffeDeveloper CC BY-SA 3.0
fixed power stuff to have only real exponent
Nov 24, 2016 at 14:33 history rollback CoffeDeveloper
Rollback to Revision 9
Nov 24, 2016 at 14:32 history edited CoffeDeveloper CC BY-SA 3.0
removed pow function which is not trivial to derivate nor is taught in high-school
Nov 24, 2016 at 14:20 history edited CoffeDeveloper CC BY-SA 3.0
more clear and added a punctualization
S Nov 24, 2016 at 14:16 history bounty started CoffeDeveloper
S Nov 24, 2016 at 14:16 history notice added CoffeDeveloper Authoritative reference needed
Nov 24, 2016 at 14:13 history edited CoffeDeveloper CC BY-SA 3.0
added subset of derivation rules for my language and fixed a few points in the bullet list
Nov 23, 2016 at 11:05 comment added CoffeDeveloper I do not understan why, Richardson use "abs" and don't have "sqrt and pow" functions.. Maybe you can link the theorem in an answer @EmilJeřábek ? (at least according to wikipedia page) I'm not going to trust a theorem without explaination if I'm not working under the assumptions of the theorem
Nov 23, 2016 at 10:49 comment added Emil Jeřábek All right. So the question is indeed answered negatively by Richardson's theorem.
Nov 23, 2016 at 9:58 history edited CoffeDeveloper CC BY-SA 3.0
derivating sqrt in 0 may be a problem
Nov 23, 2016 at 9:53 history edited CoffeDeveloper CC BY-SA 3.0
added 18 characters in body
Nov 23, 2016 at 9:49 comment added CoffeDeveloper I edited to made it very formal so that everyone is happy and can understand now ^^
Nov 23, 2016 at 9:47 history edited CoffeDeveloper CC BY-SA 3.0
Made it more formal, and removed a statement that would made the set decidable by definition
Nov 22, 2016 at 22:00 comment added Gerry Myerson You ask whether an elementary function can be integrated. You ask about integrability in closed form. You ask about integration in finite time. You ask about an elementary function having an elementary anti-derivative. You ask about an elementary function having an anti-derivative. No one can figure out what it is that you actually want to know, when you ask so many different things, as if they were the same thing. Please, take some time to think through what it is that you really want to know, then come back and write a question that asks that in a clear and cohesive way.
Nov 22, 2016 at 19:05 comment added CoffeDeveloper Let us continue this discussion in chat.
Nov 22, 2016 at 16:52 history edited CoffeDeveloper CC BY-SA 3.0
added 209 characters in body
Nov 22, 2016 at 16:44 comment added CoffeDeveloper help me make that consistent then :P. Elementary antiderivative. Both input and output should be something a highschool student can write and understand.. Elementary Watson
Nov 22, 2016 at 16:42 comment added Emil Jeřábek And now, after another update, the question is again inconsistent. Are you asking about “antiderivative” or about “elementary antiderivative”? Choose one or the other. It makes a huge difference.
Nov 22, 2016 at 16:38 comment added Emil Jeřábek Richardson’s theorem states that, under some natural conditions on the definition of “elementary”, it is undecidable if a given elementary function has an elementary antiderivative. It does not involve finding the antiderivative. Any way I look at it, this is exactly what you want in the question.
Nov 22, 2016 at 16:36 comment added CoffeDeveloper No it does not contradicts. One fact is claiming that "Not all antiderivatives of elementary functions, are elementary functions" (Richardson theorem). Another fact is asking does there exist an algorithm that can tell if there exist an elementary antiderivative for a given (but arbitrary ) elementary function. I think I did not told that enough well to get me understanded :)
Nov 22, 2016 at 16:32 history edited CoffeDeveloper CC BY-SA 3.0
added 255 characters in body
Nov 22, 2016 at 16:30 comment added CoffeDeveloper No the question is about the existence of an algorithm that tell us if a function has an antiderivative (otherwise my algorithm should just return a "False" value which is very trivial) @EmilJeřábek The theorem just states that we cannot always find the anti-derivative. Which is a general statement.
Nov 22, 2016 at 15:38 answer added Pat Devlin timeline score: 2
Nov 22, 2016 at 15:26 comment added Emil Jeřábek I may be missing something, but after the clarification that the question is about the existence of an elementary antiderivative, it is answered negatively by Richardson's theorem as stated in the Wikipedia article, isn’t it?
Nov 22, 2016 at 15:03 comment added Pat Devlin @JoelDavidHamkins I like your idea. Perhaps we tried to integrate $(a -b)^2$ which integrates to $0$ iff $a=b$ a.e. (and hence everywhere by continuity). So asking if an integral equals $0$ is undecidable.
Nov 22, 2016 at 14:44 comment added CoffeDeveloper @JoelDavidHamkins edited to reflect that.
Nov 22, 2016 at 14:43 history edited CoffeDeveloper CC BY-SA 3.0
added 262 characters in body
Nov 22, 2016 at 14:28 comment added Joel David Hamkins Could you clarify whether you want to know just yes/no whether the given function is integrable, or do you want to know yes/no whether the integral is itself represented by an elementary function, or do you want to compute to find an expression that is equal to the integral? You ask at first about the decision problem, but your algorithm seems aimed at the latter.
Nov 22, 2016 at 13:27 vote accept CoffeDeveloper
Nov 23, 2016 at 9:51
Nov 22, 2016 at 13:26 comment added CoffeDeveloper @PatDevlin probably :P sorry
Nov 22, 2016 at 13:25 comment added Pat Devlin @DarioOO I think you maybe misread what I wrote.
Nov 22, 2016 at 13:25 comment added Joel David Hamkins Can we reduce Richardson's problem to the integrability problem? Fix a sufficiently bad $f(x)$, and then given expressions $a,b$ ask whether $f(x)[a(x)-b(x)]$ is integrable. The idea is that if $a(x)=b(x)$ for all $x$, then the answer will be yes, since the function is constant $0$; but if not, then since $f$ is sufficiently bad, it won't be. If this can work, then it will be undecidable, since we can't decide whether $a(x)=b(x)$ for all $x$.
Nov 22, 2016 at 13:23 answer added Pat Devlin timeline score: 1
Nov 22, 2016 at 13:13 comment added CoffeDeveloper No you can't @PatDevlin there are simple expressions that are proovably not integrable, I'm asking if integrability can be proved for all expressions with an algorithm. IF you speak about numerical integrability you are right, I'm speaking of integrability in closed form.
Nov 22, 2016 at 13:12 comment added Pat Devlin It feels like anything you randomly write down can't be integrated even if it's a short and simple expression. Which always amazed me because we can take the derivative of ANYTHING a high schooler can think of.
Nov 22, 2016 at 13:06 comment added Joel David Hamkins Probably relevant: Richardson's theorem, en.wikipedia.org/wiki/Richardson's_theorem.
Nov 22, 2016 at 12:35 review First posts
Nov 22, 2016 at 12:42
Nov 22, 2016 at 12:32 history asked CoffeDeveloper CC BY-SA 3.0