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Taylor Expansion for rational function

  1. Feb 16, 2010 #1
    1. The problem statement, all variables and given/known data
    Find the taylor expansion of the following formula in the case where [tex] r > > d[/tex] to the first order in [tex]\epsilon = \frac{d}{r}[/tex]

    [tex] \frac{1}{r_{+}} = \frac{1}{\sqrt{r^{2} + (\frac{d}{2})^{2} - rdcos\theta}}[/tex]



    2. Relevant equations

    [tex](1 + \epsilon)^{m} = 1+m\epsilon[/tex], where [tex]\epsilon << 1[/tex] (First order Taylor expansion)


    3. The attempt at a solution
    [tex]\frac{1}{r_{+}} = \frac{1}{\sqrt{r^{2}(1 + \frac{d^{2}}{4r^{2}} - \frac{d}{r}cos\theta)}}

    =\frac{1}{r \sqrt{1 + \frac{\epsilon^{2}}{4} - \epsilon cos\theta}}

    =\frac{1}{r} \left( 1 + \frac{\epsilon^{2}}{4} - \epsilon cos\theta \right)^{\frac{-1}{2}}
    = \frac{1}{r} \left(1 + \left(\frac{-1}{2} \right) \frac{\epsilon^{2}}{4} + \frac{1}{2}\epsilon cos\theta \right)

    = \frac{1}{r} \left(1 - \frac{1}{8} \left( \frac{d}{r}\right)^{2} + \frac{d}{2r} cos\theta \right)[/tex]

    but the answer my instructor gives is

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

    Can someone please point out where I made a mistake? Can I just assume that [tex] \left(\frac{d}{r} \right)^{2} [/tex] must be extremely close to zero because it is being squared? I also am somewhat confused on how to apply the relevant equation for an expression that has more that just 1 and epsilon being raised to the m.




    Thanks in advance,

    KEØM
     
    Last edited: Feb 16, 2010
  2. jcsd
  3. Feb 17, 2010 #2
    yes you may neglect the term you mentioned for r>>d. it's contribution will be small. shouldn't write strictly equal but approximately. note this is part of the formulation of an electric dipole. higher order moments become less significant at far distances more significant at short distances (with respect to the charge configuration).
     
  4. Feb 17, 2010 #3

    HallsofIvy

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    For this particular problem, it is not just that "r>> d". The important point is that the problem said "to first order". That means dropping squares and higher powers.
     
  5. Feb 17, 2010 #4
    Thank you both for replying. That really helps!

    KEØM
     
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