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Taylor series

  1. Oct 12, 2009 #1
    I can't work out how to calculate the Taylor series for

    [tex] \frac{1}{|R-r|} [/tex]

    when R>>r, but they are both vectors. We were told to expand in r/R but I did the step below and I'm not sure where to go from there

    I got to

    [tex] \frac{1}{R \sqrt{1 - (2R.r)/R^2 + (r^2)/(R^2)}} [/tex]

    I also know the result in first order is

    [tex] \frac{1}{R} + \frac{R.r}{R^3} [/tex]
     
  2. jcsd
  3. Oct 12, 2009 #2

    kuruman

    User Avatar
    Homework Helper
    Gold Member

    Simplify and rewrite your expression as
    [tex]
    \frac{1}{R \sqrt{1 - 2r/R + (r/R)^2}}
    [/tex]

    Now define a (small) quantity ε =r/R in terms of which you have

    [tex]
    \frac{1}{R \sqrt{1 - 2r/R + (r/R)^2}} =\frac{1}{R}f(\epsilon)
    [/tex]

    Where
    [tex]f(\epsilon)=(1-2\epsilon+\epsilon^2)^{-1/2}[/tex]

    Expand f(ε) in Taylor series the usual way, then replace ε with r/R.
     
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