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Let $0\neq y\in \mathbb{R^3}$ define a function $f$ on $\mathbb{R^3}$ as $$ f(x) = \frac1{\| x-y\|} $$ What are derivatives of $f$ in zero? Or equivalently, what is the Taylor series of $f$ at zero?


I was thinking about using Clifford algebra. We consider $x = x_i e_i$ and $y=y_i e_i$. Where $e_i$ are generators of Clifford algebra and $x_i,y_i$ are real numbers. Then $$ f(x) = \left\| \frac1{x-y} \right\| $$ The problem is that $\| \cdot \|$ is not linear, Thus expanding $\frac1{x-y}$ in terms of $x$ in useless.


Another approach I tried: I first expanded $f(x)$ in the first component $x_1$ and I tried to expand again in the next component $x_2$, but It become messy quickly and I got lost.


Based on the Michael Hoppe's comment you can derive $$ D_p\left( \frac1{\|x\|^k} \right) = -k \frac{\langle x, p \rangle}{\|x\|^{k+2}} $$ So $$ D_p\left( \frac1{\|x\|} \right) = - \frac{\langle x, p \rangle}{\|x\|^{3}} $$ $$ D_q D_p\left( \frac1{\|x\|} \right) = - \frac{\langle q, p \rangle}{\|x\|^{3}} + 3 \frac{\langle x, p \rangle \langle x, q \rangle}{\|x\|^{5}} $$ $$ D_r D_q D_p\left( \frac1{\|x\|} \right) = 3 \frac{\langle x, r \rangle \langle q, p \rangle + \langle x, q \rangle \langle r, p \rangle+ \langle x, p \rangle \langle r, q \rangle}{\|x\|^{5} } - (3 \cdot 5) \frac{\langle x, p \rangle \langle x, q \rangle \langle x, r \rangle}{\|x\|^{7}} $$

Hmm there is some kind of pattern, but I don't know if it can be written in a compact way.

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    $\begingroup$ Hint: $D_p(\|x\|)=\frac{\langle x,p\rangle}{\|x\|}$. $\endgroup$ – Michael Hoppe Apr 23 '16 at 13:27
  • $\begingroup$ @MichaelHoppe Not sure how is the hint helpful. Could you elaborate a little bit more, please. $\endgroup$ – tom Apr 23 '16 at 13:59
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Power of Legendre polynomials:

$$\frac{1}{\| \mathbf{x-y} \|}=\frac{1}{\sqrt{r^{2}-2rr'\cos \theta+r'^{2}}}= \frac{1}{r} \sum_{n=0}^{\infty} \left( \frac{r'}{r} \right)^{n} P_{n} (\cos \theta)$$

where $\displaystyle \, \cos \theta = \frac{\mathbf{x} \cdot \mathbf{y}}{\| \mathbf{y} \| \| \mathbf{x} \|} \,$ and $\displaystyle \, \begin{pmatrix} \mathbf{r'} \\ \mathbf{r} \end{pmatrix}= \left \{ \begin{array}{ccc} \begin{pmatrix} \mathbf{x} \\ \mathbf{y} \end{pmatrix} & , & \| \mathbf{x} \|<\| \mathbf{y} \| \\[5pt] \begin{pmatrix} \mathbf{y} \\ \mathbf{x} \end{pmatrix} & , & \| \mathbf{y} \|<\| \mathbf{x} \| \end{array} \right. \, $

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  • $\begingroup$ Surely usefull expansion but it is not a Taylor series in $x$, so I do not see how it answer my question. $\endgroup$ – tom Apr 24 '16 at 8:59
  • $\begingroup$ @tom Actually, the first few terms are quite similar to what you've got. See eqn (6) in physicspages.com/2012/02/20/… $\endgroup$ – Ng Chung Tak Apr 24 '16 at 9:36

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