# Expanding a function in terms of a vector

tissuejkl

## Homework Statement

## 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.

## 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.

Mentor

## Homework Statement

## 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.

## 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.
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} ##?

tissuejkl
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} ##?
Yes. Sorry, I should have made that clearer.

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.

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

Staff Emeritus
$$L(\mathbf v^2+2\pmb v\cdot\pmb\varepsilon+\pmb\varepsilon^2) = L(\pmb v^2)+\varepsilon_i\frac{\partial}{\partial \varepsilon_i}\bigg|_{\pmb 0} L(\pmb v^2+2\pmb v\cdot\pmb\varepsilon+\pmb\varepsilon^2).$$