# I with a simplification

I'm trying to follow this work, and I can't figure out how they are simplifying the following expression.

$$\frac{r}{r_1} = \left( 1 - \frac{d}{r} \cos \theta + \left( \frac{d}{2r} \right)^2 \right)^{-1/2}$$

$$\frac{r}{r_1} \approxeq 1 - \frac{1}{2}\left( -\frac{d}{r} \cos \theta + \frac{d^2}{4r^2} \right) + \frac{3/4}{2} \left(-\frac{d}{r} \cos \theta + \frac{d^2}{4r^2} \right)^2$$

This is with the assumption $r \gg d$.

Are they doing a taylor expansion?

Another related question.
The book performs a binomial expansion from,
$$\left( R^2 - \vec R \cdot \vec d + \frac{d^2}{4} \right)^{-3/2} \approxeq R^{-3}\left(1-\frac{\vec R \cdot \vec d}{R^2} \right)$$

So a binomial expansion needs to be of the form,
$$(X+Y)^n$$ right?

So would X just be $R^2$ and Y would be the scalar $-\vec R \cdot \vec d + \frac{d^2}{/4}$ ?

So does X and Y each have to be a function of one variable respectively? Or can they be any algebraic term? for example, could
$$X = R^2 \cos \theta$$
$$Y = \sin (\phi+R)$$

Last edited:

I figured it out.

It's a binomial series where:
$$(X+Y)^n = \ldots$$
$$X = 1$$
$$Y = \frac{d}{r} \left(-\cos \theta + \frac{4}{dr} \right)$$

Follow up question. Why use the binomial expansion? Why not perform a taylor expansion?

J77
Taylor series is just a special case of binomial where you have a "one plus" with a positive exponent.

(What you have can be called a negative binomial series.)