Timeline for Implicit function theorem without uniqueness?
Current License: CC BY-SA 4.0
13 events
| when toggle format | what | by | license | comment | |
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| Dec 15, 2023 at 14:07 | comment | added | Loïc Teyssier | @MichaelHardy: I meant that some people try to avoid using the operator-like notation $\sin x$ in favor of the evaluation form $\sin(x)$. Not so much a matter of ambiguity as notational preferences, I'd say. For instance, for teaching material I always use the evaluation form (the operator notation leads more easily to multiplicative simplifications or other nonsense that I hope to avoid this way), but I'm more relaxed in research papers. | |
| Dec 14, 2023 at 21:06 | comment | added | Michael Hardy | @LoïcTeyssier : Can you clarify your comment? Do you mean evaluating the sine function at $x$ and then squaring? If so, then $(\sin x)^2$ also means that and is unambiguous. | |
| Dec 13, 2023 at 13:58 | comment | added | Loïc Teyssier | @MichaelHardy: parentheses can also be used to indicate functional evaluation. | |
| Dec 13, 2023 at 10:48 | answer | added | Kostya_I | timeline score: 2 | |
| Dec 13, 2023 at 9:33 | answer | added | Willie Wong | timeline score: 3 | |
| Dec 10, 2023 at 20:43 | comment | added | Michael Hardy | Parentheses can disambiguate when you use them to distinguish between $(\sin x)^2$ and $\sin(x^2),$ but I wonder what their purpose is when one writes $\sin(x)^2. \qquad$ | |
| Dec 10, 2023 at 19:51 | comment | added | terceira | The function $y^2-x^2$ exhibits locally the same behaviour (and is of course locally diffeomorphic to your example). It is dealt with by the abstract theory of algebraic geometry (resolution of singularities).. | |
| Dec 10, 2023 at 18:36 | comment | added | Robert Israel | What comes to mind is that if at $(0,0)$ the gradient of $f(x,y)$ is $(0,0)$ and the Hessian matrix is indefinite (one positive and one negative eigenvalue), then you have a saddle point and there should be two solution curves passing through $(0,,0)$. | |
| Dec 10, 2023 at 18:23 | comment | added | LSpice | @RobertIsrael, re, or, of course, even just $x^2 + y^2 = 0$. | |
| Dec 10, 2023 at 18:22 | history | edited | LSpice | CC BY-SA 4.0 |
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| Dec 10, 2023 at 18:14 | comment | added | Robert Israel | Another relevant example is $y^2 + \sin(x)^2 = 0$, where $(0,0)$ again is a solution but there is no solution in a neighbourhood of $x=0$. So you need something more than just $f(0,0) = 0$ to get existence. | |
| S Dec 10, 2023 at 17:02 | review | First questions | |||
| Dec 10, 2023 at 20:04 | |||||
| S Dec 10, 2023 at 17:02 | history | asked | Daisy_Duck | CC BY-SA 4.0 |