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Complex Partway Functions

  1. Nov 12, 2012 #1
    What I mean by a partway function is this:
    ff(x)=6x now as you probably know that f(x)=√6(x) or you could argue f(x)=-√6(x), with that function that you have just found being the partway function between x and f(x)=6x-Do you understand?
    But what about more complex partway functions like ff(x)=sin(x) so what is f(x)= to?, which is the same as saying what is the partway function between x and sin(x).
     
  2. jcsd
  3. Nov 12, 2012 #2

    Mark44

    Staff: Mentor

    No. Please provide a definition for this term - partway function.

     
  4. Nov 12, 2012 #3

    Mute

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    Homework Helper

    If I understand correctly, you are asking, if

    $$f(f(x)) = \sin x,$$
    then what is ##f(x)##?

    This particular equation has been studied, but I'm afraid I can't remember the name of the function or the wikipedia article.

    Edit: Ah! Here's the wikipedia article I was thinking of:

    Schroder's equation

    In particular, it discusses the "functional square root", a function such that ##h_{1/2}(h_{1/2}(x)) = h(x)##, which is relevant to the question of ##f(f(x)) = \sin x##.
     
  5. Nov 12, 2012 #4
    I would avoid using ff(x) to represent (f(x))^2 since it looks like the composite f(f(x)).

    Also, I'm a bit confused about what you mean by complex (complex numbers or complicated).

    That is, for y^2=6x, x>=0 if x is real or all x if we're using complex numbers.

    In the case of y^2=sin(x), assuming we're working with real numbers, sin(x)>=0 so x€U(n)[2nπ,(2n+1)π] for n€Z.
    So, y=+-√sin(x) for the same x in that set. (These are two distinct functions.)

    If we work with complex functions, then y=+-√sin(x) for all complex numbers x.

    Is this what you're looking for?
     
  6. Nov 13, 2012 #5
    You are on the right lines but your link doesn't include any information on that specific equation.
     
  7. Nov 13, 2012 #6

    Mute

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    Yes, it doesn't discuss f(f(x)) = sin x in particular, but rather a more general problem. If you can solve the functional equation ##\Psi(\sin x) = s \Psi(x)## for ##\Psi(x)## (which you may not be able to do analytically), then the "half" sine function would be given by ##h_{1/2}(x) = \Psi^{-1}(s^{1/2} \Psi(x))##, such that ##h_{1/2}(h_{1/2}(x)) = \sin x##. By searching some papers about solving Schroder's equation perhaps you can find one which discusses the case of sin x.
     
  8. Nov 13, 2012 #7
    If you click on the link, "functional square root", you would find that rin(rin(x))=sin(x) where rin(x) is the function you supposedly wanted.

    @dalek, in you first post where you said ff(x)=6x, and you also said f(x)=√6(x) which could mean either √(6x) or (√6)x. This was not clear as (f(x))^2 = 6x, and f(f(x))= 6x for the corresponding choice of f(x), respectively.
     
  9. Nov 13, 2012 #8
    Can you go through the notation because I don't understand it and is there a calculator on the internet for them, with degrees and radians?One of the reasons, I have asked this is that I have discovered some nice theorems to approximate such functions where x is in degrees,sin[2]x=sin sin x, sin[0.5]x roughly=(xsinx)^0.5, sin[0.25]roughly=x*(sinx/x)^0.25, sinx roughly=x*(sinx/x)^i - you should be able to see a nice pattern going on there.
     
  10. Nov 13, 2012 #9
    Is there a rin(x) calculator on the internet?
     
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