Vector Taylor series From pg. 35 of Classical Electrodynamics 3rd edition, Jackson,
$$\begin{aligned} \nabla^{2} \Phi_{a}(\mathbf{x}) &=-\frac{1}{4 \pi \epsilon_{0}} \int \rho\left(\mathbf{x}^{\prime}\right)\left[\frac{3 a^{2}}{\left(r^{2}+a^{2}\right)^{5 / 2}}\right] d^{3} x^{\prime} \end{aligned}$$
"Choose R such that $\rho(\mathbf{x'})$ changes little over the interior of the sphere... With a Taylor series expansion of the well-behaved $\rho (\mathbf{x'})$ around $\mathbf{x'} = \mathbf{x}$ one finds ..."
\begin{align}
\nabla^{2} \Phi_{a}(\mathbf{x}) &=-\frac{1}{\epsilon_{0}} \int_{0}^{R} \frac{3 a^{2}}{\left(r^{2}+a^{2}\right)^{5 / 2}}\left[\rho(\mathbf{x})+\frac{r^{2}}{6} \nabla^{2} \rho+\cdots\right] r^{2} d r+O\left(a^{2}\right),
\end{align}
where $r = |\mathbf{x'} -\mathbf{x}|$.
Could someone explain how to derive this Taylor series for a function of a vector? I've never seen this before and am at a loss.
 A: 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.
A: 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.
