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## Homework Statement

Let

[tex]r_{\pm} = \sqrt{r^2 \mp rdcos\theta + (d/2)^2}[/tex]

in my book it is said that for a approximation where [tex]d << r[/tex] (we don't want d=0 cause the purpose is to find the potential, and with d=0 there would be no potential) so he does a expansion carried to

*first order*in [tex]d[/tex]. Then its given the following result

[tex]r_{\pm} \approx r(1 \mp \frac{d}{2r}cos\theta)[/tex]

and it follows that

[tex]\frac{1}{r_{\pm}} \approx \frac{1}{r}(1 \pm \frac{d}{2r}cos\theta)[/tex]

**2. The attempt at a solution**

So... he doesn't mention it, but I guess it is a Taylor expansion, right?

Rewriting the equation... [tex]r_{\pm} \approx r\sqrt{1 \mp \frac{dcos\theta}{r} + {\frac{d}{2r}}^2}.[/tex]

But I'm getting a bit confused about it... i can't figure out how exactly he comes up with that. Cause to me... the expansion, based on the equation given bellow for Taylor Series, would go like... [tex]r_{\pm} \approx r\left\{\sqrt{1 \mp \frac{d'cos\theta}{r} + {\frac{d'}{2r}}^2} + \frac{1}{2}*\frac{cos\theta/r + d'/2r^2}{\sqrt{1 \mp \frac{d'cos\theta}{r} + {\frac{d'}{2r}}^2}}\right\}...[/tex]

It seems so simple, but I can't get what I'm missing. =/ Anyone could please help?

**3. Relevant equations**

Taylor series

[tex]f(a) + \frac{f'(a)}{1!} (x - a) + \frac{f''(a)}{2!}(x - a)^2 + ...[/tex]