Timeline for Iterative Root Finding
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 18, 2015 at 10:08 | vote | accept | MthQ | ||
| Nov 17, 2015 at 1:33 | history | edited | Pietro Majer | CC BY-SA 3.0 |
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| Nov 17, 2015 at 0:47 | comment | added | Pietro Majer | The preceding statement was wrong, so I edited and changed it. Now is correct, though more trivial. | |
| Nov 17, 2015 at 0:43 | history | undeleted | Pietro Majer | ||
| Nov 17, 2015 at 0:43 | history | edited | Pietro Majer | CC BY-SA 3.0 |
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| Nov 17, 2015 at 0:17 | history | deleted | Pietro Majer | via Vote | |
| Nov 16, 2015 at 20:02 | comment | added | Pietro Majer | In this case $G$ does not, because for any $M$ the zero set of $G$ meets the region $\{x>M, y>M\}$. | |
| Nov 16, 2015 at 15:39 | comment | added | MthQ | Not sure if I understand your answer. Suppose $G(x,y)=y^2-2x+1$, then $x_{i+1}=\frac{1}{2}(x_i^2+1)$, which does not converge to $x^*=1$ for $x_0=2$ (but $G$ does satisfy your property?) | |
| Nov 16, 2015 at 15:20 | history | answered | Pietro Majer | CC BY-SA 3.0 |