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Thread: Continued fraction expansion

  1. #1
    Sep 2011

    Continued fraction expansion

    Hello everyboby.

    I am very new here, so please excuse me, if I'm doing something wrong. But I need your help anyway.

    There is a dissertation:

    The author has following equation there (page 41):

    from which he derives the next equation with the aid of the "continued fraction expansion" (on the same page):

    He mentions, that he used a MATLAB program for that, but he doesn't describe it anywhere in detail.

    I tried to obtain the same result in MATLAB and in Mathcad, but it looks completely different:

    The reason is, that MATLAB and Mathcad have a built-in fuction, which realizes the continued fraction expansion in another form.

    But I need exactly the same form as the author has in his dissertation!

    Could you help me please to get it?

    P.S. I must tell you, that my knowledge in this part of the mathematics is not very big. So could you please provide me just a "ready-to-use" formula for the required expansion?
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  2. #2
    MHF Contributor
    Opalg's Avatar
    Aug 2007
    Leeds, UK

    Re: Continued fraction expansion

    The way that the calculation goes in the dissertation is first to take the given expression for Z_{jc}(s) and (using the "simplify" function in Mathcad) add the four fractions together to form a single fraction

    Z_{jc}(s) = \frac{0.56882s^3 + 65.659s^2 + 766.16s + 912.28} {s^4 + 364.39s^3 + 10582s^2 + 77404s + 70175}.

    The procedure for turning this into a continued fraction is then to repeatedly invert it and divide the numerator by the denominator. The first steps will go like this:

    \frac1{Z_{jc}(s)} = \frac{s^4 + 364.39s^3 + 10582s^2 + 77404s + 70175} {0.56882s^3 + 65.659s^2 + 766.16s + 912.28} = 1.758s + \frac{\text{some remainder}} {0.56882s^3 + 65.659s^2 + 766.16s + 912.28}

    (notice that 1.758 = 1/0.56882).

    The next step will be to invert that last fraction so that the "remainder" becomes the denominator. This time, the numerator and denominator of the new fraction will both be cubic polynomials in s, so the quotient will be a constant (in fact 0.0023 according to the solution given in the dissertation), and the remainder will be of the form

    \frac{\text{quadratic polynomial}}{\text{cubic polynomial}}.

    Again, invert that fraction, divide top by bottom and get the remainder, and so on.
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  3. #3
    Sep 2011

    Re: Continued fraction expansion

    Thank you very much! It works!
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