How to expand r_{\pm} and \frac{1}{r_{\pm}} around d = 0

[emol1414]

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

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

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

 

[LCKurtz]

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.

[Edit] 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.

 

[emol1414]

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]

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)

Your English is fine, no need for apology. The TeX problem is a bug that has been reported many months ago and they can't seem to fix it. When the page displays old TeX, press F5 to refresh the screen in the browser which causes it to display the corrected code. That way you can see what is actually going to post. It is a nuisance but it works.