# Expanding a function in terms of a vector

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1. Jul 9, 2015

### tissuejkl

1. The problem statement, all variables and given/known data
$L (v^2 + 2 \pmb{v} \cdot \pmb{ \epsilon } ~ + \pmb{ \epsilon} ^2)$, where $\pmb{\epsilon}$ is infinitesimal and $\pmb{v}$ is a constant vector ($v^2$ here means $\pmb{v} \cdot \pmb{v}$ ), must be expanded in terms of powers of $\pmb{\epsilon}$ to give $$L(v^2)+\frac{\partial L}{\partial v^2}\ 2 \ \pmb{v} \cdot \pmb{\epsilon}$$
where terms above the first order have been neglected.
2. Relevant equations

3. The attempt at a solution
I thought about using the Taylor expansion to expand L in terms of $\pmb{\epsilon}$, but that seems to work only if $\epsilon$ were a scalar, so I'm not quite sure what to use to expand L.

2. Jul 9, 2015

### Staff: Mentor

What is your question? Are you trying to show that $L (v^2 + 2 \pmb{v} \cdot \pmb{ \epsilon } ~ + \pmb{ \epsilon} ^2) \approx L(v^2)+\frac{\partial L}{\partial v^2}\ 2 \ \pmb{v} \cdot \pmb{\epsilon}$?

3. Jul 9, 2015

### tissuejkl

Yes. Sorry, I should have made that clearer.

4. Jul 9, 2015

### Staff: Mentor

It looks to me like they're doing something like this:
$f(x + \Delta x) \approx f(x) + f'(x) \cdot \Delta x$
IOW, it's the Taylor expansion with terms only up to the first derivative.

5. Jul 9, 2015

### tissuejkl

Is there a reason why the derivative in the second term should be the partial derivative $\frac{\partial L}{\partial v^2}$, then?

6. Jul 9, 2015

### Fredrik

Staff Emeritus
The domain of L is a subset of $\mathbb R$, not a subset of $\mathbb R^n$ with n>1, so L doesn't have partial derivatives. Maybe that's just a terrible notation for $L'(\pmb v^2)$.

I would start like this and then just use the chain rule:
$$L(\mathbf v^2+2\pmb v\cdot\pmb\varepsilon+\pmb\varepsilon^2) = L(\pmb v^2)+\varepsilon_i\frac{\partial}{\partial \varepsilon_i}\bigg|_{\pmb 0} L(\pmb v^2+2\pmb v\cdot\pmb\varepsilon+\pmb\varepsilon^2).$$

Last edited: Jul 9, 2015