This equation comes from a physics script on electrodynamics, saying that this equation comes from a Taylor series expansion. enter image description here

I understand the first equality, but not the second one. It is really not clear for me how the second equality relates to Taylor series.

I try to make sense out of this but I'm stuck. I saw this post exactly about this problem, but there is no explanation on how we find this Taylor expansion.


If $\theta$ is the angle between $\vec{r}$ and $\vec{x}$, then $$ \frac{1}{\sqrt{r^2+x^2-2\vec{r}\cdot\vec{x}}} = (r^2+x^2-2rx\cos\theta)^{-1/2} =\frac{1}{r}\left(1+\frac{x^2}{r^2} - 2\frac{x}{r}\cos\theta \right)^{-1/2} $$ Use the Taylor series $(1+z)^{-1/2} = 1 -\frac{1}{2}z + O(z^2)$ as $z \to 0$.
Then as ${x/r} \to 0$ we get $$ \frac{1}{\sqrt{r^2+x^2-2\vec{r}\cdot\vec{x}}} = \frac{1}{r}\left(1 - \frac{1}{2}\left(- 2\frac{x}{r}\cos\theta\right) +O\left(\frac{x^2}{r^2}\right)\right) = \frac{1}{r} + \frac{x}{r^2}\cos\theta +O\left(\frac{x^2}{r^3}\right) \\= \frac{1}{r} + \frac{\vec{x}\cdot\vec{r}}{r^3} +O\left(\frac{x^2}{r^3}\right) $$

The OP did not say whether $x/r \to 0$ or $x/r \to \infty$. But if $\infty$, then we get a different answer, so it must mean $0$.

  • $\begingroup$ Thank you! it was easier than what I expected $\endgroup$ – Nicolas Schmid May 2 at 18:26

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