# Thread: Neglecting terms (and binomial expansion)

1. ## Neglecting terms (and binomial expansion)

From David J. Griffiths "Introduction to Electrodynamics 2. ed":
We're interested in the regime $\displaystyle r\gg s$ (...), so the third term is necligible, and the binomial expansion yields
$\displaystyle (...)\cong\frac{1}{r}\left(1\mp\frac{s}{r}\cos\the ta\right)^{-1/2}\cong\frac{1}{r}\left(1\pm\frac{s}{2r}\cos\theta \right)$
"The third term" is further up the page.

What is really going on here? I've been trying to understand it, but I can't figure it out. How does he get $\displaystyle \frac{1}{r}\left(1\pm\frac{s}{2r}\cos\theta\right)$?

Edit/context:
This is a step in deriving the formula
$\displaystyle V(P)\cong\frac{1}{4\pi\epsilon_{0}}\frac{qs\cos\th eta}{r^{2}}$
which in turn leads to
$\displaystyle V_{dip}(P)=\frac{1}{4\pi\epsilon_{0}}\frac{\mathbf {P}\cdot\mathbf{\hat{r}}}{r^{2}}$

2. Originally Posted by Matnater
From David J. Griffiths "Introduction to Electrodynamics 2. ed":

"The third term" is further up the page.

What is really going on here? I've been trying to understand it, but I can't figure it out. How does he get $\displaystyle \frac{1}{r}\left(1\pm\frac{s}{2r}\cos\theta\right)$?

Edit/context:
This is a step in deriving the formula
$\displaystyle V(P)\cong\frac{1}{4\pi\epsilon_{0}}\frac{qs\cos\th eta}{r^{2}}$
which in turn leads to
$\displaystyle V_{dip}(P)=\frac{1}{4\pi\epsilon_{0}}\frac{\mathbf {P}\cdot\mathbf{\hat{r}}}{r^{2}}$

You can expand the definition of the binomial theorem to non-integer exponents, defining $\displaystyle \binom {\alpha}{k}=\frac{\alpha(\alpha -1)(\alpha -2)\cdot\ldots\cdot (\alpha -k+1)}{k}\,,\,\,k\in\mathbb{N}$.

Thus, $\displaystyle \left(1\mp \frac{s}{r}\cos\theta\right)^{-1/2}$ $\displaystyle =1\mp \left(-\frac{1}{2}\right)\frac{s}{r}\cos\theta+\left(\fra c{3}{8}\right)\left(\frac{s}{r}\right)^2\cos^2\the ta+\ldots$ $\displaystyle \cong 1\pm\frac{s}{2r}\cos\theta$ , which apaprently is justified by the book since r is much larger than s thus from the third summand and on the coefficients are extremely low.

Tonio

3. Thank you! This really solved my problem!

### griffiths EM binomial expansion

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