Using the Binomial Theorem to expand the magnitude of the difference of two vectors

Question

I have the following expression that I need to expand using the Binomial Theorem: $$\frac{1}{\mid\vec{r}-\vec{d}\mid}$$ Now the Binomial Theorem is the following: $$(x+y)^r = \sum^{\infty}_{k=0}\binom{r}{k}x^{r-k}y^{k}$$ I believe I start this by rewriting the initial equation above as: $$\frac{1}{\mid\vec{r}-\vec{d}\mid} = \left(\left(\vec{r}-\vec{d}\right)^T\left(\vec{r}-\vec{d}\right)\right)^{-1/2}=\left(\vec{r}^T\vec{r}-2\vec{r}^T\vec{d}+\vec{d}^T\vec{d}\right)^{-1/2}$$ Thus, I have an expression made up of scalars instead of vectors which is desireble for the Binomial Theorem. However, I have three terms instead of 2. I could just make $x = \vec{r}^T\vec{r}$ and y = $-2\vec{r}^T\vec{d}+\vec{d}^T\vec{d}$ I guess, but I'm not sure. I guess I am wondering if I am on the right track or if there is a specific formula expanding the magnitude of the difference of vectors. Thanks in advance.

Answer 1

Note that: $$|\mathbf{r}-\mathbf{d}|=√(\mathbf{r}-\mathbf{d})\cdot(\mathbf{r}-\mathbf{d})=√(r^2+d^2-2\mathbf{r}\cdot\mathbf{d})$$

You can now expand using the binomial theorem in the form of:

$$√(r^2+X),\quad X=d^2-2\mathbf{r}\cdot\mathbf{d}$$

Hope that helps.

Answer 2

Presumably, $\vec d$ is a perturbation of $\vec r$ and you want a development in terms of $\vec d$.

WLOG we can consider that $r$ is unit (otherwise, normalize it). Then

$$\|\vec r-\vec d\|^2=1-2\vec r\vec d+\vec d^2=1-2\cos\theta\,d+d^2.$$

Alpha gives you a Taylor development, which is quite tedious by hand:

https://www.wolframalpha.com/input/?i=Taylor+1%2Fsqrt(1-2cos(theta)+d%2Bd%5E2)

You can also keep the terms $t:=d^2-2\cos\theta d$ together and consider the Taylor development of

$$\frac1{\sqrt{1+t}}=1-\frac t2+\frac{3t^2}8-\cdots$$