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I need help with a simplification

  1. Sep 12, 2006 #1
    I'm trying to follow this work, and I can't figure out how they are simplifying the following expression.

    [tex] \frac{r}{r_1} = \left( 1 - \frac{d}{r} \cos \theta + \left( \frac{d}{2r} \right)^2 \right)^{-1/2} [/tex]

    [tex] \frac{r}{r_1} \approxeq 1 - \frac{1}{2}\left( -\frac{d}{r} \cos \theta + \frac{d^2}{4r^2} \right) + \frac{3/4}{2} \left(-\frac{d}{r} \cos \theta + \frac{d^2}{4r^2} \right)^2 [/tex]

    This is with the assumption [itex] r \gg d [/itex].

    Are they doing a taylor expansion?

    Another related question.
    The book performs a binomial expansion from,
    [tex] \left( R^2 - \vec R \cdot \vec d + \frac{d^2}{4} \right)^{-3/2} \approxeq R^{-3}\left(1-\frac{\vec R \cdot \vec d}{R^2} \right) [/tex]

    So a binomial expansion needs to be of the form,
    [tex] (X+Y)^n [/tex] right?

    So would X just be [itex] R^2 [/itex] and Y would be the scalar [itex] -\vec R \cdot \vec d + \frac{d^2}{/4} [/itex] ?

    So does X and Y each have to be a function of one variable respectively? Or can they be any algebraic term? for example, could
    [tex] X = R^2 \cos \theta [/tex]
    [tex] Y = \sin (\phi+R) [/tex]
     
    Last edited: Sep 12, 2006
  2. jcsd
  3. Sep 13, 2006 #2
    I figured it out.

    It's a binomial series where:
    [tex] (X+Y)^n = \ldots [/tex]
    [tex] X = 1 [/tex]
    [tex] Y = \frac{d}{r} \left(-\cos \theta + \frac{4}{dr} \right) [/tex]


    Follow up question. Why use the binomial expansion? Why not perform a taylor expansion?
     
  4. Sep 13, 2006 #3

    J77

    User Avatar

    Taylor series is just a special case of binomial where you have a "one plus" with a positive exponent.

    (What you have can be called a negative binomial series.)
     
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