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CTK Exchange
oniwaka
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Dec-13-06, 03:16 PM (EST) |
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"a = sin(f(x) * n); n = ???"
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I'm being haunted by this problem! My only solution at this point is the dirtiest of programmers tricks(the lookup table), which is workable for the level of precision I need. However, I can't help but think there's a better solution. a = sin(f(x) * n)
or
a = (exp(complex(0, f(x) * n)) - exp(-complex(0, f(x) * n))) / 2j a and x are known Thanks in advance for any help you might offer. |
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alexb
Charter Member
1930 posts |
Dec-13-06, 03:23 PM (EST) |
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1. "RE: a = sin(f(x) * n); n = ???"
In response to message #0
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Even assuming f(x) is a known function, the problem does not have a unique solution. Regardless, it requires a more or less standard application of arcsin: f(x)*n = arcsin(a) or
f(x)*n = π - arcsin(a). There may be a question of efficiency of course, in which case a lookup table may be a viable alternative. |
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oniwaka
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Dec-13-06, 09:19 AM (EST) |
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2. "RE: a = sin(f(x) * n); n = ???"
In response to message #1
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Thank you. I'll be able to rest easily now. Maybe I should take a math refresher course... I compiled a quick benchmark, and the lookup table is indeed faster in C. But since I'm prototyping in python, which has non-negotiable float precision (requiring type conversion for each lookup), arcsin(a)/f(x) is actually faster. Thanks again. |
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