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

• gionole
In summary: Then ##L(v^2) = f(v)## and so ##L'(v^2) = f'(v)##. You could also introduce the function ##g## as the composition: ##g(v) = v^2##. Then ##f(v) = L(g(v))## and so ##f'(v) = L'(g(v))g'(v)##. Then ##L'(v^2) = L'(g(v))g'(v) = f'(v)2v = 2vL'(v^2)##. So we don't have to use partial derivatives at all.Just a note: the problem with using ##L'## to denote the derivative is that it is
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?

Niel said:
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 $$\begin{split} L(v^2 + 2\epsilon v + \epsilon^2) &= L((v + \epsilon)^2) \\ &= L(v^2) + \epsilon \left.\frac{d}{dz}L(z^2)\right|_{z=v} + O(\epsilon^2) \\ &= L(v^2) + 2\epsilon v L'(v^2) + O(\epsilon^2) \end{split}$$ where a prime denotes differentiation of a function of a single variable with respect to its argument. I suppose you could write $L'(v^2)$ 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.

fresh_42
PeroK said:
... 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 ?

Niel said:
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)##.

gionole

## What is a Taylor series and how is it used in the context of a Lagrangian?

A Taylor series is an infinite sum of terms calculated from the values of a function's derivatives at a single point. In the context of a Lagrangian, the Taylor series is used to approximate the function around a specific point, often to simplify complex expressions and facilitate analysis.

## How do I start deriving the Taylor series for a given Lagrangian?

To derive the Taylor series for a given Lagrangian, start by identifying the function and the point around which you will expand. Then, compute the derivatives of the Lagrangian with respect to the relevant variables at that point and use the Taylor series formula to express the Lagrangian as an infinite sum.

## What are the common mistakes to avoid when deriving a Taylor series for a Lagrangian?

Common mistakes include neglecting higher-order terms, incorrectly calculating derivatives, and not properly evaluating the derivatives at the expansion point. It's also important to ensure that the series converges within the desired range of approximation.

## Why is it important to show the Taylor series for a Lagrangian in physics?

Showing the Taylor series for a Lagrangian is important in physics because it allows for a simplified, approximate analysis of the system's behavior near a specific point. This can be particularly useful in perturbation theory, stability analysis, and when dealing with small oscillations around equilibrium points.

## Can you provide an example of deriving the Taylor series for a simple Lagrangian?

Sure! Consider a simple Lagrangian $$L = \frac{1}{2}mv^2 - \frac{1}{2}kx^2$$. To find the Taylor series around $$x = 0$$, compute the derivatives: $$L^{(0)} = L(0)$$, $$L^{(1)} = \left.\frac{dL}{dx}\right|_{x=0}$$, $$L^{(2)} = \left.\frac{d^2L}{dx^2}\right|_{x=0}$$, and so on. For this example, the higher-order terms vanish, and the series is simply $$L \approx L(0) + \frac{1}{2}\left.\frac{d^2L}{dx^2}\right|_{x=0}x^2$$, which simplifies to $$L \approx \frac{1}{2}mv^2 - \frac{1}{2}kx^2$$.

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