Timeline for Can we decide whenever a function is the derivate of another function in this Language?
Current License: CC BY-SA 3.0
8 events
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| Jun 15, 2020 at 7:27 | history | edited | CommunityBot |
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| Nov 27, 2016 at 10:06 | comment | added | Emil Jeřábek | Wikipedia gives you a reference to the original paper. It is there for a reason. | |
| Nov 26, 2016 at 23:32 | comment | added | CoffeDeveloper | My feeling is that someone just "says" is because "other says". How to be sure someone is right, if he cannot show why is right (not speaking of you I'm in general sayings)?. Also your claim "to any set of E" is quite generic and assumes that you defined E as Richardson, if E in example is just a polynomial your claim do not hold anymore. People pretends me to be precise in the question, but then do generic answers without contextualization while it is obvious to everyone what I did not understand, no one cared to explain ^^. | |
| Nov 26, 2016 at 23:25 | comment | added | CoffeDeveloper | That's because you know what the theorem is about, if you read it on Wikipedia it is not clear at all. Everyone is good to criticize and downvote when someone do not understand because someone else wrote that in a bad way. Math should not be elitarie in any way. Everyone here just sad "that's obvious", but math is not at all obviouse, you have to show it is obvious. Math is made by many steps, if it is not clear enough you should explain what is missing step. Not insisting in "it is that way." show that way ^^. | |
| Nov 26, 2016 at 2:24 | comment | added | Pat Devlin | @DarioOO Richardson's theorem is very general. Note that it implies your question immediately. It applies to any set $E$. Your collection of functions is one such set. | |
| Nov 26, 2016 at 2:00 | history | bounty ended | CoffeDeveloper | ||
| Nov 26, 2016 at 1:59 | comment | added | CoffeDeveloper | Now I understand but I think it is more clear in the following way: "Since expressions mentioned in Richardson theorem are a subset of mine expressions, then mine expressions have a subset for which some expressions are not decidable on the condition." No one still exclude then that expressions using at least one time the "square root" function may be trivially decidable then. About the usage of $|x|$ I still have some doubts but Thanks for the spotlight. ^^ | |
| Nov 25, 2016 at 16:43 | history | answered | Emil Jeřábek | CC BY-SA 3.0 |