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On page 33 of Jackson's Classical Electrodynamics (third edition), after obtaining the potential $$\Phi(x) = \frac{1}{4 \pi \epsilon_0} \int_S \frac{\sigma(x')}{|x-x'|} \mathrm da' - \frac{1}{4 \pi \epsilon_0} \int_S \frac{\sigma(x')}{|x-x' +nd|}\mathrm da''$$ he goes on and says:

For small $d$ we can expand $|x-x+nd|^{-1}$. Consider the general expression $|x+a|^{-1}$, where $|a|\ll|x|$. We write a Taylor series expansion in three dimensions: $$\frac{1}{x+a} = \frac{1}{x} + a \cdot \nabla \bigg( \frac{1}{x} \bigg) + \cdots$$

How does he get this expansion? Around what "point" he expands? I've been trying to find a way to get this expansion but no luck so far. Thank you a lot.

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See e.g. Wikipedia for Taylor's theorem for multivariate functions for a more rigorous explanation.

In particular, for a function $f:\mathbb{R}^n \to \mathbb{R}$, the first order approximation around point $\vec{\xi}_0$ is:

$$ f(\vec{\xi}) \approx f(\vec{\xi}_0) + (\vec{\xi}-\vec{\xi}_0) \cdot (\nabla f) (\vec{\xi}_0). $$

The intuition behind this is simply the fact that if $f$ is of 1st order (that is, $f(\vec{\xi}) = a + \vec{b} \cdot \vec{\xi}$ for some $a\in \mathbb{R}$ and $\vec{b} \in \mathbb{R}^n$), this approximation is exact, and close to a point, the function can be approximated as a 1st order function.

In your example, $f(\vec{\xi}) = 1/|\vec{\xi}|$ and since we want to approximate the values of that close to $\vec{x}$, we choose $\vec{\xi}_0 = \vec{x}$.

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  • $\begingroup$ Thank you very much for the explanation. Are there other sources where I can read about it other then wikipedia? Even the title sounds shady in that article "multivariate functions" ... Thank you anyway! $\endgroup$ – Xsnac Oct 14 '17 at 6:31

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