MWresearch Posted July 10, 2015 Posted July 10, 2015 I'm running into a problem where I want to represent a series of radians as A*arccos(Bx+C)+D, like for instance pi/2, 3*pi/4, pi...18*pi/12.. or pi/2, pi, 3pi/2, 2pi, pi/2, pi, 3pi/2....ect. But, because the domain of arccos is limited and then becomes some imaginary hyperbolic trig function outside of the domain 0<x<pi, I'm having trouble figuring out how to do that. Like normally, you could represent a series of cyclical numbers like 1,-1,1,-1 as cos(pi*x). I want to do the same thing with radians using arccos(x) where it just keeps repeating periodically as x grows arbitrarily large
mathematic Posted July 11, 2015 Posted July 11, 2015 |Bx+C|≤1 will give you angles. > 1 leads to hyperbolic function.
MWresearch Posted July 11, 2015 Author Posted July 11, 2015 Yeah...that's more or less what I said within the first post except the exact domain of arcos(x) is [0, pi].
Endy0816 Posted July 11, 2015 Posted July 11, 2015 Can you do: Y = Your value DY = number of digits in Y Arccos(Y/10^DY) ?
MWresearch Posted July 11, 2015 Author Posted July 11, 2015 (edited) I don't know what you mean exactly, but it could lead to the right idea. If you increase by a smaller and smaller number that never increases past pi, then arccos should grow indefinitely large, but it does so logarithmically according to the hyperbolic trig function so it's still not cyclical. the only thing I could think that works is arccos(cos(pi*x))*pi or something like that, but I never see anything like that anywhere. Edited July 11, 2015 by MWresearch
mathematic Posted July 11, 2015 Posted July 11, 2015 Yeah...that's more or less what I said within the first post except the exact domain of arcos(x) is [0, pi]. The domain is [-1,1]. The range is [0,pi].
Endy0816 Posted July 12, 2015 Posted July 12, 2015 (edited) I don't know what you mean exactly, but it could lead to the right idea. If you increase by a smaller and smaller number that never increases past pi, then arccos should grow indefinitely large, but it does so logarithmically according to the hyperbolic trig function so it's still not cyclical. the only thing I could think that works is arccos(cos(pi*x))*pi or something like that, but I never see anything like that anywhere. It will convert arbitrary values into the range arccos is looking for. 123456 / 106 = 0.123456 -123 / 103 = -0.123 # of Digits to left of decimal = ⌊log10(|x|)⌋+1 arccos(x / 10⌊log10(|x|)⌋+1 ) Edited July 12, 2015 by Endy0816
MWresearch Posted July 15, 2015 Author Posted July 15, 2015 Yeah still not working for what I'm trying to do, maybe there's no way.
Endy0816 Posted July 16, 2015 Posted July 16, 2015 (edited) Are you wanting to keep your result normal via eliminating the imaginary bit? That's the only other option really. Edited July 16, 2015 by Endy0816
MWresearch Posted July 16, 2015 Author Posted July 16, 2015 I don't know, I just want to represent an indefinite series of radians as some function of arccos x. If there's some way to do that with archosh(x) then that's fine.
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