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Classical Electrodynamics by Jackson says

"With a Taylor series expansion of the well-behaved $\rho (\mathbf{x'})$ around $\mathbf{x'} = \mathbf{x}$ one finds ..."

and then he says basically that

$\rho (\mathbf{x'}) = \rho (\mathbf{x}) + \frac{r^2}{6}\nabla^2\rho + \ldots$

above, note that $ r = |\mathbf{x'} -\mathbf{x}|$ and we are in $3$ dimensions

Could someone explain how to derive this Taylor series result for a function of a vector? I've never seen this before and am at a loss.

UPDATE:

Perhaps the trick is to notice $\mathbf{x'} = \mathbf{x} -\mathbf{r}$ and then do some sort of expansion about $r = 0$?

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    The problem isn't somebody doing work for you (that's what we're here for), the problem is you not doing *your* part of the work, namely stating the question properly. You failed to mention that this is all inside an integral, multiplied by a spherically symmetric factor, and that there's a whole story about integrating over a small sphere in the text near it. If you pose the question properly, I'll be happy to answer it.2012-07-31

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enzotib has already provided the expansion of a real-valued function of a vector up to second order. Now we can make use of the fact that the function is being integrated over a spherical volume, multiplied by a spherically symmetric factor. The integral containing the linear term vanishes by symmetry. For the quadratic term, the Hessian can be split into a component proportional to the identity and a traceless part:

$\def\H{\mathbf H}\H=\def\tr{\operatorname{tr}}\frac{\tr\H}3\mathbf I+\left(\mathbf H-\frac{\tr\H}3\mathbf I\right)\;.$

The integral containing the traceless part vanishes by symmetry, and the integral containing the identity yields your quadratic term, since the trace of the Hessian is the Laplacian.

Please update the question to reflect the context that I used in the answer. Thanks.

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    @joriki I still don't see why does the term with the first order derivatives disappear in this integral? what symmetry do you use here?2017-04-06
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For function of several variables, the first few terms of the Taylor series assume the following form $f(\mathbf{x})=f(\mathbf{x}_0)+\bar\nabla^T{f}(\mathbf{x}_0)(\mathbf{x}-\mathbf{x}_0)+\frac{1}{2}(\mathbf{x}-\mathbf{x}_0)^T\mathbf{H}_f(\mathbf{x}_0)(\mathbf{x}-\mathbf{x}_0)+\ldots$ where $\mathbf{H}f(\mathbf{x})$ is the Hessian matrix.

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    @Ben: see third formula of this paragraph http://en.wikipedia.org/wiki/Taylor_series#Taylor_series_in_several_variables2013-02-22