Expanding a function in terms of a vector

In summary, the homework statement is trying to find an equation for a variable that expands in terms of powers of a scalar, but is not sure what to use to expand. They are using the Taylor expansion with terms only up to the first derivative.
  • #1
tissuejkl
8
0

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.

Homework Equations

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.
 
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  • #2
tissuejkl said:

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.

Homework Equations

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} ##?
 
  • #3
Mark44 said:
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.
 
  • #4
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
Mark44 said:
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?
 
  • #6
tissuejkl said:
Is there a reason why the derivative in the second term should be the partial derivative ## \frac{\partial L}{\partial v^2} ##, then?
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:

What does it mean to "expand a function in terms of a vector"?

Expanding a function in terms of a vector means expressing the function as a linear combination of the given vector. This allows for a simpler representation of the function and can make calculations and analysis easier.

What is the purpose of expanding a function in terms of a vector?

Expanding a function in terms of a vector allows for a more concise and efficient representation of the function. It also allows for easier calculations and analysis of the function.

What is the process of expanding a function in terms of a vector?

The process typically involves finding the coefficients that represent the linear combination of the given vector that make up the function. This can be done through various methods such as using the dot product or solving a system of equations.

Can any function be expanded in terms of a vector?

Not all functions can be expanded in terms of a vector. The function must be a linear combination of the given vector in order for it to be expanded in this way.

What are some real-world applications of expanding a function in terms of a vector?

Expanding a function in terms of a vector is commonly used in fields such as physics, engineering, and economics. It can also be applied in data analysis and signal processing to simplify complex functions and make calculations more efficient.

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