Timeline for certain trigonometric homeomorphisms
Current License: CC BY-SA 3.0
19 events
| when toggle format | what | by | license | comment | |
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| Jul 24, 2013 at 6:02 | history | edited | Michael Hardy | CC BY-SA 3.0 |
added 2 characters in body
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| Jul 24, 2013 at 0:05 | comment | added | Igor Rivin | @MichaelHardy see the last edit. | |
| Jul 24, 2013 at 0:04 | history | edited | Igor Rivin | CC BY-SA 3.0 |
fixed minor bug
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| Jul 23, 2013 at 18:06 | history | edited | Igor Rivin | CC BY-SA 3.0 |
added explanation
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| Jul 23, 2013 at 17:53 | comment | added | Michael Hardy | I'm not sure this question's been answered yet, but it's been reduced to a problem not involving trigonometric functions. Neither of the two functions $x\mapsto (1+x^2)/(1-x^2)$ and $x\mapsto 2x/(1-x^2)$ is a homeomorphism from $\mathbb R\cup\{\infty\}$ to itself, but their sum is such a homeomorphism. Just how addition and subtraction behave with respect to this class of functions could bear examination. | |
| Jul 21, 2013 at 23:38 | comment | added | Michael Hardy | OK, I've decided to "accept" this incomplete answer, since I think I can work out the rest of it. | |
| Jul 21, 2013 at 23:37 | vote | accept | Michael Hardy | ||
| Jul 21, 2013 at 20:18 | comment | added | Malik Younsi | More generally, any rational function of the form $R(z)=\sum_{j=1}^{n}a_j/(z-p_j)$ ,where every $p_j$ is real and the $a_j$'s are real and all of the same sign, will be an $n$-to-$1$ mapping of $\mathbb{R} \cup \{\infty\}$ onto itself whose derivative does not change sign. | |
| Jul 21, 2013 at 19:27 | comment | added | Michael Hardy | . . . . . and now I've looked at Alexandre Eremenko's example. What he says is true. His function is a two-to-one continuous mapping from the circle $\mathbb R\cup\{\infty\}$ to itself, and it's locally one-to-one at every point and its derivative is everywhere positive. | |
| Jul 21, 2013 at 19:01 | vote | accept | Michael Hardy | ||
| Jul 21, 2013 at 19:25 | |||||
| Jul 21, 2013 at 18:13 | comment | added | Michael Hardy | @Qfwfq : I think Igor Rivin's later edit to his answer tacitly tells you how to construct an algorithm that would write $(x-a)^2+(x-b)^2+(x-c)^2$ as a sum of two squares of polynomials. I was going to ask essentially the same question, but you beat me to it. | |
| Jul 21, 2013 at 18:08 | comment | added | Michael Hardy | . . . . . so in other words, your answer amounts to showing that it's all reducible to just one such function. | |
| Jul 21, 2013 at 18:08 | comment | added | Michael Hardy | @IgorRivin : Your later edit certainly clarifies things. If the only property of $\alpha\mapsto\tan\frac\alpha2$ that is relied on is that it is itself a homeomorphism from $\mathbb R\bmod2\pi$ to $\mathbb R\cup\{\infty\}$ then $\alpha\mapsto\sec\alpha+\tan\alpha$ also works (and no surprise there, since it is itself a tangent half-angle function, being equal to $\tan(\frac\pi4+\frac\alpha2)$), but also every function $\alpha\mapsto\text{some rational function of sine and cosine}$ that is itself such a homeomorphism will work just as well. | |
| Jul 21, 2013 at 16:50 | comment | added | Igor Rivin | @AlexandreEremenko You are right, of course -- my answer is not quit complete, but this is not hard to fix... | |
| Jul 21, 2013 at 16:46 | comment | added | Igor Rivin | @Qfwfq see the edit, though your case falls into the completing the square category ( notice that you have to take square roots to do this) | |
| Jul 21, 2013 at 16:45 | history | edited | Igor Rivin | CC BY-SA 3.0 |
Added explanation.
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| Jul 21, 2013 at 15:34 | comment | added | Qfwfq | How can one express, say, $(x-a)^2+(x-b)^2+(x-c)^2$ as a sum of two squares? (just curiosity) | |
| Jul 21, 2013 at 8:10 | comment | added | Alexandre Eremenko | Derivative of $-1/x-1/(x-1)$ does not change sign but this is not a homeomorphism. | |
| Jul 21, 2013 at 0:14 | history | answered | Igor Rivin | CC BY-SA 3.0 |