(adsbygoogle = window.adsbygoogle || []).push({}); 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

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# Taylor Expansion for rational function

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