Timeline for Description of all total recursive functions where operator is effective?
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
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| Jan 24, 2017 at 13:14 | review | Close votes | |||
| Jan 24, 2017 at 18:26 | |||||
| Jan 24, 2017 at 13:00 | comment | added | Carl Mummert | I've voted to close this as unclear. I don't understand the question, and it seems others also have points of confusion. @Andrew - you can easily edit the question to add more detail. | |
| Jan 4, 2017 at 23:44 | comment | added | Joel David Hamkins | And if you don't have that meaning for $\mu$, but want to take just the least $y$ with $f(x,y)=0$, then the only $g$ for which the operator will be effective is when $g$ is the empty function, which never converges, since otherwise the question of the value and existence of $\mu y(f(x,y)=0)$ is not semi-computable for that interpretation of $\mu$. | |
| Jan 4, 2017 at 21:57 | comment | added | Joel David Hamkins | Isn't this always effective? If I have an algorithm for $g$ and $f$, then I can compute $g(\mu y(f(x,y)=0)$ by just trying out all the $y$'s until I find one. I assume that $\mu y$ means the least one such that $f(x,y)=0$ and $f(x,y')$ converges for all $y'<y$. | |
| Jan 4, 2017 at 21:54 | comment | added | Noah Schweber | Could you define your terms and notation? What is $\mathcal{F}_i$ (set of total functions of $i$ variables, or partial ones, or something else), and what does it mean for an operator to be effective? | |
| Jan 4, 2017 at 21:40 | comment | added | Noah Schweber | Incidentally, this was asked at MSE: math.stackexchange.com/questions/2069149/…. | |
| Jan 4, 2017 at 21:20 | review | First posts | |||
| Jan 4, 2017 at 21:32 | |||||
| Jan 4, 2017 at 21:19 | history | asked | Andrew | CC BY-SA 3.0 |