Timeline for the implicit function theorem in subsets of R$^2$
Current License: CC BY-SA 2.5
8 events
| when toggle format | what | by | license | comment | |
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| Aug 14, 2010 at 23:33 | comment | added | DarkLight | As for x$^2+y^2$=1,we can divide the space into two pieces,y>o and y<0,then there uniquely exists the function y=f(x) defined in the interval （1-$\alpha$,1],this is the key point. | |
| Aug 14, 2010 at 18:05 | comment | added | Piero D'Ancona | And, what about $F(x,y)=x^2+y^2$? Or $F\equiv0$ identically? Or $F=dist((x,y),K)^2$ where dist is the euclidean distance and $K is any closed set? | |
| Aug 14, 2010 at 15:45 | comment | added | Deane Yang | In the first statement of the implicit function theorem, don't (iii) and (iv) contradict each other? The implicit function theorem I know has condition (iv) only and no zero condition such as (iii). | |
| Aug 14, 2010 at 15:31 | comment | added | Thierry Zell | The condition (iii) $F_y(x,y)=0$ in the region would simply imply that F is a function of x only. Maybe what you wanted instead to replace (iii) by $F_y(x_0,y_0)=0$? Anyway, it's easy to see that there's no way to generalize the IFT in that case (try $x^2+y^2=1$). | |
| Aug 14, 2010 at 14:53 | comment | added | DarkLight | U stands for neighborhood，and $P_0=(x_0,y_0)$ | |
| Aug 14, 2010 at 14:32 | comment | added | Thierry Zell | What are U and P_0? | |
| Aug 14, 2010 at 14:20 | history | edited | Akhil Mathew |
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| Aug 14, 2010 at 13:24 | history | asked | DarkLight | CC BY-SA 2.5 |