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Thread: Curious trig functions

  1. #1 Curious trig functions 
    Forum Radioactive Isotope MagiMaster's Avatar
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    Does anyone know anything about the trig functions:

    C(a) = b s.t. cos(a+b) = b
    S(a) = b s.t. sin(a+b) = b

    I can't find any mention of these anywhere. One way to get the values is to iterate the cos and sin functions. For example, C(0) = cos(cos(cos(...))) ~= 0.73908513321516064165531208767387. (As good as Windows calculator can do.) S(0) = 0. Also, these functions are periodic with period 2*pi.

    Hmm... Thinking about it just now, it seems like you could work out a Taylor series for these functions using the series for cos, cos(cos), etc... but IIRC function composition covolutes the coefficients, right? Can an infinite convolution be worked out?


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  3. #2  
    Forum Professor river_rat's Avatar
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    are you interested on the whole domain or around some point?


    As is often the case with technical subjects we are presented with an unfortunate choice: an explanation that is accurate but incomprehensible, or comprehensible but wrong.
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  4. #3  
    Forum Radioactive Isotope MagiMaster's Avatar
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    Well, the whole domain. But, any information would be nice. The functions seem to be continuous for all reals. (But that's just based on very limited observations.)
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  5. #4  
    Forum Professor river_rat's Avatar
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    Well if you treat a as a small parameter then you could always just do a asymptotic expansion for your function - might give you some more information.

    im trying to remember fixed point results for continuity - fixed points of a smooth function depend continuously on the parameters? It looks differentiable (edit) actually - check if you have a well defined derivative. That will give you continuity
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  6. #5  
    Forum Radioactive Isotope MagiMaster's Avatar
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    How would you get the derivative of that function? Assuming f(x) = cos(x+b), would dC/dx be the same as lim(n->Inf) df^n(x)/db (where f^n is repeated composition)? There'd have to be some way to write that out without just requiring infinite uses of the chain rule, right?
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  7. #6  
    Forum Professor river_rat's Avatar
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    Implicit deriviatives should do the trick - just check if your result is defined everywhere and handle the points where it is not seperately
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  8. #7  
    Forum Professor river_rat's Avatar
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    Just checked - your function is differentiable everywhere except for a = 3/2 pi + 2 pi k but i think C(3/2 pi) = 0 so the argument for continuity there should not be too difficult.
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  9. #8  
    Forum Radioactive Isotope MagiMaster's Avatar
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    Yeah, C(3/2 pi) would be 0. cos(3/2 pi + x) = sin(x), so C(3/2 pi) = S(0).

    BTW, I was reading the wikipedia article on implicit differentiation, but it doesn't really say how you would apply it to a function defined as: f(x)=y s.t. g(y)=z. How did you do it?
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  10. #9  
    Forum Professor river_rat's Avatar
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    Okay, your equation is just Cos(a + b(a)) = b(a)

    Take derivatives on both sides to find that -Sin(a + b(a)) (1 + b'(a)) = b'(a) and then just solve for b'(a) to see that b'(a) = -Sin(a + b)/(1 + Sin(a + b))

    That is defined everwhere except when a+b = 3/2 pi + 2 pi k ...
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  11. #10  
    Forum Radioactive Isotope MagiMaster's Avatar
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    Ah right. I see. I suppose C(a) = cos(a + C(a)) is probably a better definition anyway. With the derivative, it should be possible to find the power series for the function, but what's the usual thing to do when you can't find an exact representation for one of the coefficients (the first one in this case)?
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  12. #11  
    Forum Professor river_rat's Avatar
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    The usual trick is to either use a decimal approx. for the first co-eff

    i.e. C(a) ~ 0.7 + ...

    or to call that number "beta" s.t. beta = Cos(beta) and then C(a) ~ beta + ...

    A power series is still going to be a real pain to work out though (as you have to invert it to solve for the C(a)). Good luck!
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  13. #12  
    Forum Radioactive Isotope MagiMaster's Avatar
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    Yeah. I checked around and couldn't find an existing constant that equals C(0) (although, I think you'd probably want to define the power series around pi/2).

    If you got b'(a) = -Sin(a + b)/(1 + Sin(a + b)) wouldn't this imply that C'(x) = -sin(x + C(x))/(1 + sin(x + C(x))? If so, I don't understand what you mean about needing to invert the power series...

    Actually, working the values out for C(pi/2) and C'(pi/2) gives the nice numbers 0 and 1/2. I think C''(x) = C(x)*(1 + 2*sin(x + C(x)))/(1 + sin(x + C(x)))^3. If so, then C''(pi/2) would be 0. Unfortunately, it looks unlikely that there won't be any real pattern to the derivatives. Can you really define a function as a power series without a closed form for the coefficients?
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