# Can't understand an expansion

## Homework Statement

Let
$$r_{\pm} = \sqrt{r^2 \mp rdcos\theta + (d/2)^2}$$
in my book it is said that for a approximation where $$d << r$$ (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 $$d$$. Then its given the following result

$$r_{\pm} \approx r(1 \mp \frac{d}{2r}cos\theta)$$
and it follows that

$$\frac{1}{r_{\pm}} \approx \frac{1}{r}(1 \pm \frac{d}{2r}cos\theta)$$

2. The attempt at a solution
So... he doesn't mention it, but I guess it is a Taylor expansion, right?
Rewriting the equation... $$r_{\pm} \approx r\sqrt{1 \mp \frac{dcos\theta}{r} + {\frac{d}{2r}}^2}.$$

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... $$r_{\pm} \approx r\left\{\sqrt{1 \mp \frac{d'cos\theta}{r} + {\frac{d'}{2r}}^2} + \frac{1}{2}*\frac{cos\theta/r + d'/2r^2}{\sqrt{1 \mp \frac{d'cos\theta}{r} + {\frac{d'}{2r}}^2}}\right\}...$$

3. Relevant equations
Taylor series
$$f(a) + \frac{f'(a)}{1!} (x - a) + \frac{f''(a)}{2!}(x - a)^2 + ...$$

LCKurtz
Homework Helper
Gold Member

## Homework Statement

Let
$$r_{\pm} = \sqrt{r^2 \mp rdcos\theta + (d/2)^2}$$
in my book it is said that for a approximation where $$d << r$$ (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 $$d$$. Then its given the following result

$$r_{\pm} \approx r(1 \mp \frac{d}{2r}cos\theta)$$
and it follows that

$$\frac{1}{r_{\pm}} \approx \frac{1}{r}(1 \pm \frac{d}{2r}cos\theta)$$

2. The attempt at a solution
So... he doesn't mention it, but I guess it is a Taylor expansion, right?

I think it is simpler than that. Just factor an r2 out under the radical:

$$\sqrt{r^2(1 \pm \frac d r \cos\theta +\frac{d^2}{r^2})}$$

If d << r you can ignore the last term in the parentheses; it's close to zero.

 On closer examination I'm not sure...
[Edit more] Yes, so ignoring that term you have

$$r\sqrt{1 \pm \frac d r \cos\theta}$$

Use the first two terms of the binomial expansion for that square root and it works.

Last edited:
Oh, thank you so much. Binomial series it was!

(Could you please answer me a non-related doubt? =D Sometimes my tex codes don't work: when I see a mistake and try to correct it, it doesn't change from what it was. ~.~ Sorry, I'm new to this forum, sorry about the poor english too. =X)

LCKurtz