# Taylor series of a modulus argument

What is the definition of a Taylor series of the function $F(|\vec a -\vec x|)$ about the point $\vec a$ in $\vec x$?

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As filmor notes, the Taylor series isn't well-defined if you consider this as a function of the full vector $\vec x$. If you don't mind a singularity at $\vec a$ (or if the odd derivatives of $F$ at $\vec a$ vanish), you can use the one-dimensional Taylor series for $F$ to write

$$\sum_{n=0}^\infty\frac{F^{(n)}(|\vec a-\vec x|)}{n!}|\vec a-\vec x|^n\;.$$

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This definition is the same as Taylor series of function $F_2(\vec x)=F(|\vec x|)$. I'm not sure if you are familiar with vector function Taylor series. In general case the formula is kind of long, so I'll just give and example for bivariate function $f(x,y)$ about the point $(a,b)$ $$f(x,y)=f(a,b) + \frac{\partial f}{\partial x} \bigg|_{(a,b)} (x-a) + \frac{\partial f}{\partial y}\bigg|_{(a,b)} (y-b) + \frac{\partial^2 f}{\partial x^2}\bigg|_{(a,b)} (x-a)^2 + 2\frac{\partial^2 f}{\partial x \partial y}\bigg|_{(a,b)} (x-a)(y-b)+\frac{\partial^2 f}{\partial y^2}\bigg|_{(a,b)} (y-b)^2 + \dots$$ But there is still one thing, you see, the derivative of modulus ($|\cdot|$) does not exist around zero vector. So if you could give us a little more detail on why are you trying to solve a problem like that, them maybe you will get a good answer.

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There is no proper definition for this (if you take $F(\lvert \vec a - \vec x\rvert)$ as a function of $\vec x$), as $\lvert\vec v\rvert$ is not differentiable at $\vec v = 0$ (in this case: $\vec x = \vec a$).

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