Was getting back into physics and reading a chapter on electrostatics which sets up the following situation. We have a configuration of point charges – one $-q$ at the point ($-d,0,0$) and one $+q$ at the point ($d,0,0$). The potential of this configuration is then just $$V(x) = \frac{1}{4\pi\epsilon_0}\left(\frac{q}{\lvert x-d \rvert} – \frac{q}{\lvert x + d\rvert}\right)$$ Now the book then says "Taylor expanding this expression for $x\gt\gt d$ tells you approximately how the field behaves far away". It's my understanding that performing a Taylor expansion at a point gives one the ability to approximate how a function behaves at that point when simply calculating the value might be too cumbersome or difficult. However, it's been a while since I've actually done a Taylor Expansion and was looking for some help in doing this for myself.
So, I know the general formula for a Taylor Expansion is the following $$f(x)\vert_a = \sum \frac{f^{(n)}(a)}{n!}(x-a)^n$$ Here in this case I'm a bit confused as to how to approach this given the information that we are expanding about a point $x$ such that $x\gt\gt d$ I get that it means we are investigating how the function behaves far away from the point $(d,0,0$) but not sure how this condition meaningfully changes out function. It's my first instinct to say that since $x \gt\gt d$ that adding or subtracting $d$ from $x$ doesn't really make a difference and thus $x + d = x – d =x$ but then that leaves us with $$V(x) = \frac{1}{4\pi\epsilon_0}\left(\frac{q}{x} – \frac{q}{x}\right) = 0$$ Which I know isn't what the example is prompting.
TL;DR
I want to know what the condition $x \gt\gt d$ meaningfully changes about how we proceed with the Taylor expansion of the potential function and then would also like some guidance in carrying out said expansion.
Best Answer
(Finite) Taylor expansion needs a small parameter that is sometimes obtained as the reciprocal of a large parameter. Here that small parameter is $d/x$, and you get that to appear in the formulas by dividing top and bottom by $x$. So you deal with $(q/x)/(1 \mp d/x)$ which is $q/x \pm qd/x^2$ to first order in $d/x$. (Here I am assuming you are doing all this on the line containing the charges.) As you noticed, first order expansion is the simplest approximation that actually says anything interesting about $V$.
In a more general situation in which $x$ and $d$ are just arbitrary vectors, you can write
$$|x \pm d|^{-1}=((x \pm d) \cdot (x \pm d))^{-1/2}=(|x|^2 \pm 2 (x \cdot d) + |d|^2)^{-1/2} \\=|x|^{-1} (1 \pm 2 (x/|x| \cdot d/|d|) (|d|/|x|) + |d|^2/|x|^2)^{-1/2}$$
and expand that.