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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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perhaps it has something to do w/ this math.stackexchange.com/questions/24817/… ? –  Timtam Jul 31 '12 at 9:52
    
I think it can be computed with the formula at the bottom of this page: mathworld.wolfram.com/TaylorSeries.html I'll try it out tomorrow if nobody wants to do the work for me (which is understabable heh!) –  Timtam Jul 31 '12 at 10:13
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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. –  joriki Jul 31 '12 at 10:17
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2 Answers

up vote 3 down vote accepted

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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you are right i will add the context and study your response than you –  Timtam Aug 1 '12 at 1:46
    
@Timtam: Are you still going to add the context? –  joriki Aug 4 '12 at 9:05
    
i will definitely do it I am just circling back to this in a moment... I have to relearn some of this... but I will change the context and then work on it –  Timtam Aug 4 '12 at 9:08
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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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Alright, so it turns out I'm having the exact same problem here, and am stumped as to how the term in the integrand takes the form given in Jackson. I've not seen this formula for a Taylor series for a function of several variables, I'm wondering what your source is for this, and where I can see a proof of it? Thanks –  user63499 Feb 22 '13 at 11:31
    
@Ben: see third formula of this paragraph en.wikipedia.org/wiki/… –  enzotib Feb 22 '13 at 21:21
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