Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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!

share|improve this question
    
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

1 Answer 1

up vote 2 down vote accepted

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 $x$.

share|improve this answer
    
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 at 15:51
1  
@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 at 21:54

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.