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About Recursive Formula

  1. Oct 3, 2012 #1

    dyh

    User Avatar

    Hey guys,

    I would like to get some general form of recursive formula

    let f(s) = (a+bs)/(c+ds)

    given this I would like to get nth composite function of f

    i.e the general form of f^n(s)= fofofofofo........of(s) (nth composite)

    I can conjecture that the form of f^n(s) would be the same form with f(s) like

    f^n(s) = (α+βs)/(γ+δs)

    and α,β,γ,δ are would be expressed as function of a,b,c,d and n

    So, I would like to get the general from of α,β,γ,δ with the function of a,b,c,d and n.

    Thanks
     
  2. jcsd
  3. Oct 3, 2012 #2

    Mute

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    Homework Helper

    Your function is called a Mobius transformation. By looking up references about this function you may be able to find some information about the composition of functions that you want to find. A brief skim of the wikipedia article didn't yield a discussion of composition of the function, but I did see a section about fixed points of the transformation, which is related to composing the function infinitely many times.
     
  4. Oct 3, 2012 #3

    jasonRF

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    Science Advisor
    Gold Member

    dyh,

    Your conjecture is correct. The solution ends up consisting of matrix multiplies. For example, if
    [tex]
    z_1 = \frac{a z + b}{c z + d}
    [/tex]
    and
    [tex]
    z_2 = \frac{a_1 z_1 + b_1}{c_1 z_1 + d_1}
    [/tex]
    then if we want to write
    [tex]
    z_2 = \frac{a_2 z + b_2}{c_2 z + d_2}
    [/tex]
    we have
    [tex]
    \left( \begin{array}[cc] \\ a_2 & b_2 \\ c_2 & d_2 \end{array} \right)
    =
    \left( \begin{array}[cc] \\ a_1 & b_1 \\ c_1 & d_1 \end{array} \right)
    \left( \begin{array}[cc] \\ a & b \\ c & d \end{array} \right) .
    [/tex]

    You should work through the algebra yourself to verify that I didn't make a mistake. Composition of more than two of these leads to simply more matrices to multiply. In your case, [itex]a_1=a, b_1=b,[/itex] etc. so you would simply have an nth power of a matrix. If your matrix can be diagonalized then this nth power is relatively simple.

    By the way, this functional form is often called a Mobius (or bilinear) transformation. If you have ever learned about conformal mapping, you may recall that it maps circles/lines in the complex plane to other circles/lines.

    jason
     
    Last edited: Oct 3, 2012
  5. Oct 4, 2012 #4

    dyh

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    Thanks buddy! It is really helpful.
    I will try to get the specific form!
     
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