Show that the Taylor series for this Lagrangian is the following...

[gionole]

Homework Statement

Show Taylor series for the lagrangian is the following

Relevant Equations

..

We have $L(v^2 + 2v\epsilon + \epsilon^2)$. Then, the book proceeds to mention that we need to expand this in powers of $\epsilon$ and then neglect the terms above first order, we obtain:

$L(v^2) + \frac{\partial L}{\partial v^2}2v\epsilon$ (This is what I don't get).

We know taylor is given by: $p(x) = f(a) + f'(a)(x-a) + ....$

I tried doing this in respect to $\epsilon$. We know $\epsilon$ is super small, so our taylor is $f(0) + f'(0)(x-0) = L(v^2) + \frac{\partial L}{\partial \epsilon}(2v+2\epsilon)|(\epsilon=0)(\epsilon-0) = L(v^2) + \frac{\partial L}{\partial \epsilon}2v\epsilon$ (note that I don't have $\frac{\partial L}{\partial v^2}$)

If I try doing taylor with respect to $v^2$, then I don't get $L(v^2)$ as the first part because $f(a) != L(v^2)$

Where am I making a mistake?

 

[pasmith]

Homework Statement: Show Taylor series for the lagrangian is the following
Relevant Equations: ..

We have $L(v^2 + 2v\epsilon + \epsilon^2)$. Then, the book proceeds to mention that we need to expand this in powers of $\epsilon$ and then neglect the terms above first order, we obtain:

$L(v^2) + \frac{\partial L}{\partial v^2}2v\epsilon$ (This is what I don't get).

I'm not sure that's correct, or at least the notation is confusing.

We know taylor is given by: $p(x) = f(a) + f'(a)(x-a) + ....$

I tried doing this in respect to $\epsilon$. We know $\epsilon$ is super small, so our taylor is $f(0) + f'(0)(x-0) = L(v^2) + \frac{\partial L}{\partial \epsilon}(2v+2\epsilon)|(\epsilon=0)(\epsilon-0) = L(v^2) + \frac{\partial L}{\partial \epsilon}2v\epsilon$ (note that I don't have $\frac{\partial L}{\partial v^2}$)

This is incorrect. \epsilon here is playing the role of x and v is playing the role of a, but the role of f is played by (v \mapsto L(v^2)) rather than L. So we should find \begin{split}<br /> L(v^2 + 2\epsilon v + \epsilon^2) &amp;= <br /> L((v + \epsilon)^2) \\<br /> &amp;= L(v^2) + \epsilon \left.\frac{d}{dz}L(z^2)\right|_{z=v} + O(\epsilon^2) \\<br /> &amp;= L(v^2) + 2\epsilon v L&#039;(v^2) + O(\epsilon^2) \end{split} where a prime denotes differentiation of a function of a single variable with respect to its argument. I suppose you could write L&#039;(v^2) as \dfrac{\partial L}{\partial v^2} but if you want to use Leibnitz notation then \left.\dfrac{dL}{dz}\right|_{z = v^2} would be more usual.

 

[PeroK]

... the way I would look at it is this. Let $f(v) = L(v^2)$, so that $f'(v) = 2vL'(v^2)$ (by the chain rule). Then:
$$L((v+\epsilon)^2) = f(v+\epsilon) = f(v) + \epsilon f'(v) + O(\epsilon^2) = L(v^2) + 2\epsilon vL'(v)$$I don't like the notation $\frac{\partial L}{\partial v^2}$. Instead, as above, using $L'$ as the derivative of $L$ seems much neater to me.

 

[gionole]

... the way I would look at it is this. Let $f(v) = L(v^2)$, so that $f'(v) = 2vL'(v^2)$ (by the chain rule). Then:
$$L((v+\epsilon)^2) = f(v+\epsilon) = f(v) + \epsilon f'(v) + O(\epsilon^2) = L(v^2) + 2\epsilon vL'(v)$$I don't like the notation $\frac{\partial L}{\partial v^2}$. Instead, as above, using $L'$ as the derivative of $L$ seems much neater to me.

To me, what always seemed simpler in taylor was always find the taylor for simple function and then in terms of x, plug in the new value. Here, the idea is to write $L(v^2)$ in terms of taylor series and not the $L(v^2 + 2v\epsilon + \epsilon^2)$ and when we do taylor around point $v$ for this, we will just plug in $v+\epsilon$ and it will be great approximation since $\epsilon$ is super small and $v+\epsilon$ super close to point $v$.

Taylor for $L(v^2)$ is $f(a) + f'(a)(x-a) = f(v) + f'(v)(x-a) = L(v^2) + L'(v^2)(x-v)$ At this point, I don't know what the $L'(v^2)$ means - simply put, by what we derivate it with ? If derivative here means by $v$, then the saying that we do taylor around $v$ seems confusing. The around point for taylor and derivation should be by different variables I believe.

If we say that $L(v^2) + L'(v^2)(x-v) = L(v^2) + 2vL'(v^2)(x-v)$ and plugging in $L(v^2) + L'(v^2)(v+\epsilon-v) = L(v^2) + 2v\epsilon L'(v^2)$.. Thoughts ?

 

[PeroK]

Thoughts ?

$L(v^2)$ is a composition of the function $L$ with the function $v^2$. Once you see that, all should be clear. That's why formally we can introduce the function $f$ as this composition: $f(v) = L(v^2)$.