network profile
| bio | website | |
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| age | ||
| visits | member for | 1 year, 2 months |
| seen | Dec 3 '12 at 23:51 | |
| stats | profile views | 7 |
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Oct 5 |
accepted | Simple test if point is above or below sine curve |
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Oct 2 |
comment |
Simple test if point is above or below sine curve I can follow it fine, I'm disputing whether or not it's correct. Specifically, I'm arguing that it's inconclusive. For instance, lets substitute logs for sine, so we want to know, given some test point $(x,y)$ and an arbitrary test parameter $\epsilon > 0$, is the point below the log curve: $\ln(x) - y < \epsilon$. By your argument, because there's a log in the inequality, there's no way to test it without computing the log. But of course we can rewrite it: $e^x / e^y < e^{\epsilon}$, and perform out test without actually calculating the function in question. |
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Oct 2 |
comment |
Simple test if point is above or below sine curve Agreed, which is why that particular test isn't useful. But it's a counter example to the idea that there's no way to test it without calculating $\sin(x)$. So I guess the question would be, is it provable that there's no way to perform the test I'm looking for which is easier than calculating $\sin(x)$. This particular example is not easier, but that doesn't mean there aren't any. |
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Oct 2 |
comment |
Simple test if point is above or below sine curve But that's my point, you don't have to: $y < sqrt(1-\cos^2x)$ implies that $y < \sin(x)$. |
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Oct 1 |
comment |
Cosine of average of angles @AméricoTavares: I actually liked you answer better than the accepted one: sign ambiguity is not an issue since all of my angles are first quadrant. Thanks for the help. |
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Oct 1 |
comment |
Cosine of average of angles Thanks, @did. I'll concede that I can't escape using square roots for this, so I guess I'll have to practice Newton's method. |
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Oct 1 |
accepted | Cosine of average of angles |
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Oct 1 |
comment |
Cosine of average of angles @robjohn: Thanks, that's a good explanation for what seems to be the running theme in most of these responses =). I'll concede that there's not likely a better solution that doesn't involve square roots. |
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Oct 1 |
awarded | Supporter |
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Oct 1 |
comment |
Simple test if point is above or below sine curve Hi, Tony. Is that more than just a gut instinct? I know primality is a lot different than a continuous function, but just as an illustration you can definitively test if a number is prime, but those tests won't actually produce a prime number on their own. And for sine, for instance, it is always the case the $\sin^2x + \cos^2x = 1$, so that is a definitive test for sine which just happens to not be useful because it requires me to know the cosine a priori. |
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Sep 28 |
awarded | Custodian |
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Sep 28 |
reviewed | Approve suggested edit on Simple test if point is above or below sine curve |
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Sep 28 |
revised |
Simple test if point is above or below sine curve Added context for the problem and hopefully for solutions. |
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Sep 28 |
comment |
Simple test if point is above or below sine curve I'm not familiar with that third point: can you describe where this comes from and how I would choose an appropriate alpha? (Or give me the name of a theorem, for instance). Thanks! |
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Sep 28 |
comment |
Simple test if point is above or below sine curve Because I'm doing this by hand, without a calculator or look up table. I know the sines of certain landmark angles (the multiples of 10-degrees) and I'm trying to figure out how to use these to find the sine of other angles by hand. |
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Sep 28 |
comment |
Cosine of average of angles Agreed. This is a nice attempt, but you've changed the problem from finding the cosine of the average to finding the cosine of half the difference. The point is really that I only the cosines (and sines) for A and B, and need to find the cosine of their average without relying on the sines or cosines or any other angles. |
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Sep 28 |
comment |
Cosine of average of angles @AméricoTavares: Well I can't see how either, that's why I asked =). It's not that it's ugly, it's that it's hard to compute by hand. I don't mind doing one or two 3 or 4 digit multiplications by hand, but finding the square root of numbers in [0, 1] is not really practical. |
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Sep 28 |
awarded | Commentator |
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Sep 28 |
comment |
Cosine of average of angles @MarkBennet: I'm not necessarily expecting a simpler answer, just hoping for one. I'm trying to work out an iterative approach to refining an estimation of the sine of an angle which can be done relatively quickly by hand. The slope of the secant that connects the lower reference angle (A) to x (the point in question) is somewhere between cos(A) and cos(B), so if I can find the cos of a particular angle between A and B, I can narrow down the interval and repeat, until I have two points close to x, and their cosines. |
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Sep 28 |
asked | Simple test if point is above or below sine curve |
