Timeline for Implicit function theorem with singularities of any order
Current License: CC BY-SA 4.0
11 events
| when toggle format | what | by | license | comment | |
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| Apr 17, 2023 at 12:45 | vote | accept | Lorenzo Q | ||
| Apr 15, 2023 at 2:24 | answer | added | Lorenzo Q | timeline score: 1 | |
| Mar 24, 2023 at 16:29 | history | edited | Lorenzo Q | CC BY-SA 4.0 |
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| Mar 16, 2023 at 20:53 | history | edited | Lorenzo Q | CC BY-SA 4.0 |
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| Mar 15, 2023 at 11:41 | history | edited | Lorenzo Q | CC BY-SA 4.0 |
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| Mar 15, 2023 at 11:28 | history | edited | Lorenzo Q | CC BY-SA 4.0 |
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| Mar 15, 2023 at 11:26 | comment | added | Lorenzo Q | @cs89 Thank you for your remark, I've made it more explicit. There can be higher order terms in $x$ or $z$. Take for instance $f(x,z)= z^2+x^2 + x^3$, clearly this doesn't have the same zeroes as $z^2+x^2$ because we can't obtain a bound like $|x^3|\leq \varepsilon |z^2+x^2|$ for $x,z$ small (the latter can be zero even when $x$ is non-zero). | |
| Mar 15, 2023 at 11:23 | history | edited | Lorenzo Q | CC BY-SA 4.0 |
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| Mar 15, 2023 at 9:25 | comment | added | cs89 | I suspect there is a problem with your notation $f(x,z) \sim z^k + h^k$. Indeed, if $\sim$ takes the usual meaning, in a neighborhood of your reference point, we have $\frac 12 |z^k+x^h| \leq |f(x,z)| \leq 2 |z^k+x^h|$, so the zero-set of $f$ and of $z^k+x^h$ are the same. So your conclusion follows from the polynomial case which you explain. | |
| Mar 15, 2023 at 9:23 | comment | added | cs89 | By translation (on $x$ and $z$), and scaling (on $f$ and $z$), you can assume that $\bar x = 0$, $\bar z = 0$ and $C_1 = C_2 = 1$. This might simplify the exposition. | |
| Mar 14, 2023 at 20:26 | history | asked | Lorenzo Q | CC BY-SA 4.0 |