Timeline for Differential equation with some constraints
Current License: CC BY-SA 3.0
8 events
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| Aug 29, 2011 at 17:55 | comment | added | Michael Hardy | @Noam: Well, now I've started looking at this answer while awake. First I wondered how you got $S_1$ from the first constraint. Then I saw how it can be done, if not how you did it, which I suspect is different. To be continued....... | |
| Aug 29, 2011 at 0:13 | history | edited | Noam D. Elkies | CC BY-SA 3.0 |
corrected several spelling errors
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| Aug 28, 2011 at 17:18 | comment | added | Noam D. Elkies | The arc-length parametrization is the inverse function of an arc-length integral, which involves a square root and can only rarely have an elementary formula (already for an ellipse we famously get elliptic integral). I see that Robert Bryant already worked out what happens here, and verified that there's almost never an elementary formula; so it won't get any more nice or neat than inverting the integral of the square root of a rational function. Did you have a reason to expect or hope for a particularly nice form here? | |
| Aug 28, 2011 at 3:35 | comment | added | Michael Hardy | The hour is late and I will look at this tomorrow. But even before digesting everything above, I've up-voted it because the identity following "$S_1$" looks just like what I wrote in another stackexchange posting (except for a factor of 2.....): math.stackexchange.com/questions/59508/trigonometric-identity And that one arose from a product of three sines, whereas this one arose from a sum of products of two sines, and there ought to be certain connections. So are you suggesting that there is a nice neat solution after all? Or only that there is one in a limiting case? | |
| Aug 28, 2011 at 3:20 | history | edited | Noam D. Elkies | CC BY-SA 3.0 |
(missed a couple of $S_1$'s...)
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| Aug 28, 2011 at 3:10 | history | edited | Noam D. Elkies | CC BY-SA 3.0 |
Corrected a few instances of "$S$" to "$S_1$"
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| Aug 28, 2011 at 2:55 | history | edited | Noam D. Elkies | CC BY-SA 3.0 |
Only 1/4 of $S$ is relevant because the algebra can't tell $\alpha,\beta,\gamma$ from their negatives.
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| Aug 28, 2011 at 2:33 | history | answered | Noam D. Elkies | CC BY-SA 3.0 |