Timeline for Is there a continuous function on $f:\mathbb{R} \rightarrow \mathbb{R}$ with uncountably infinite turning points?
Current License: CC BY-SA 2.5
13 events
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| Mar 30, 2011 at 23:56 | comment | added | Richard Stanley | Not exactly what you're asking, but there are functions $\mathbb{R} \to \mathbb{R}$ that are nonconstant and continuous, and for all $x_0\in\mathbb{R}$ and every deleted neighborhood $N(x_0)$ of $x_0$ there is an $x\in N(x_0)$ for which $f(x)=f(x_0)$. Such functions are called locally recurrent. See for instance jstor.org/pss/2312661. | |
| Mar 30, 2011 at 15:47 | vote | accept | Kate | ||
| Mar 30, 2011 at 12:39 | vote | accept | Kate | ||
| Mar 30, 2011 at 15:47 | |||||
| Mar 30, 2011 at 12:34 | answer | added | Jon Bannon | timeline score: 4 | |
| Mar 30, 2011 at 12:26 | comment | added | Kate | sorry again, I fixed that up when I re-read it. I wrote it in a hurry and didn't write what I was thinking. The function is on $\mathbb{R}$ | |
| Mar 30, 2011 at 12:25 | comment | added | Gerry Myerson | Ah - you answered my question while I was typing it in. Thanks. | |
| Mar 30, 2011 at 12:24 | comment | added | Gerry Myerson | Still confusing. You ask about functions on ${\bf R}^2$ but your examples are of functions on $\bf R$ - only the graph is in ${\bf R}^2$. So what do you mean? | |
| Mar 30, 2011 at 12:20 | comment | added | Kate | Sorry for being so late to edit and for being vague, I think the definition of turning point I use above makes it impossible. | |
| Mar 30, 2011 at 12:19 | history | edited | Kate | CC BY-SA 2.5 |
added 332 characters in body; added 20 characters in body; edited title
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| Mar 30, 2011 at 11:20 | answer | added | ght | timeline score: 1 | |
| Mar 30, 2011 at 11:11 | comment | added | S. Carnahan♦ | Please use the "edit" link below the question, and describe the definition of "turning point" that you are using. | |
| Mar 30, 2011 at 5:55 | comment | added | Gerhard Paseman | You need to be more specific. The (graph of the) real-valued function f(x,y) = x - x^2 has uncountably many points (x,y) with a partial derivative of 0 and second partial negative. It is likely there are 2-D versions of Brownian motion which might come closer to what you actually intend to visualize. Gerhard "Ask Me About System Design" Paseman, 2011.03.29 | |
| Mar 30, 2011 at 5:07 | history | asked | Kate | CC BY-SA 2.5 |