# taylor expansion of function with a vector as variable

I know how to do a taylor expansion of a function from R to R. I dont know how to do taylor expansion of functions which have 3D vectors as variable. How can I do this? I would appreciate it if someone also worked out an example. Thank you!

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The title says "vector valued" and your question mentions "vectors as variable", which is it ? –  Joel Cohen Nov 28 '12 at 19:06
@JoelCohen sorry for the confusion, I am not good at the terminology. I meant a function which has a vector as a variable –  Badshah Nov 28 '12 at 19:07
Do you have an example in mind? –  Muphrid Nov 28 '12 at 19:14
@Muphrid f(x)=|a-x|^3 for example with x and a 3D vector and the vertical lines mean the length (here a is constant). –  Badshah Nov 28 '12 at 19:18
So $x$ and $a$ are both vectors, just $a$ is a constant vector? –  Muphrid Nov 28 '12 at 19:19

Then what you're looking for is the Taylor expansion of a scalar field--a function $f$ that maps $\mathbb R^n$ to $\mathbb R$.

An easy way to build up intuition about this is to do the expansion only in one direction. Let $\hat n$ be a unit vector and $t$ a scalar parameter. Let $x_0$, the point you want to expand around, be given by $x_0 = x - t \hat n$ or $x = x_0 + t \hat n$. There is only one direction connecting $x$ and $x_0$, and the magnitude can always be calculated (which fixes $t$). Then you can say

$$f(x) = f(x_0 + t \hat n) = f(x_0) + \left. \frac{\partial f}{\partial t} \right|_{x_0} t + \frac{1}{2} \left. \frac{\partial^2 f}{\partial t^2} \right|_{x_0} t^2 + \ldots$$

Now, identify $\partial f/\partial t$ as $\hat n \cdot \nabla f$. In addition, see that $t\hat n = x - x_0$. Some clever recombining of terms gives

$$f(x) = f(x_0) + (x-x_0) \cdot \nabla f|_{x_0} + \frac{1}{2} ([x - x_0] \cdot \nabla)^2 f|_{x_0} + \ldots$$

This is suitably general to cover any point $x$.

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I dont understand why f(x)=f(x_0 +tn). could you explain? and what does de upside down delta f mean? –  Badshah Nov 28 '12 at 20:28
I've added some additional language to clarify why $x = x_0 + t \hat n$. The symbol $\nabla$ (pronounced del) represents the vector derivative operator. Acting on a scalar field, it produces the gradient. –  Muphrid Nov 28 '12 at 21:52
@Muphrid How does the second-order term $t^2\partial_{tt}$ become $(x-x_0)\cdot \nabla)^2$? –  BillyJean Jan 1 '14 at 15:51
@BillyJean: $([x-x_0] \cdot \nabla)^2 = t^2 (\hat n \cdot \nabla)^2 = t^2 \partial_{tt}$. Remember I said to identify $\hat n \cdot \nabla = \partial_t$. –  Muphrid Jan 1 '14 at 21:54