# Taylor Expand Lagrangian to Second Order....

• sa1988
In summary, the conversation discusses a problem with understanding a Lagrangian equation and a solution that was provided. The solution involved taking the Sin and Cos parts up to their x2 terms and replacing them in the equation, while the person asking the question had made a mistake by not taking into account the 4a2x2 term and the Sin term, which were both zero.
sa1988

## Homework Statement

NOTE - When I post the thread my embedded images aren't showing up on my web browser, but they do show up when I bring it up to edit, so I don't know if other users can see the pictures or not... If not, they're here:
Problem outline: http://tinypic.com/r/34jeihj/9
Solution: http://tinypic.com/r/algpqh/9

## The Attempt at a Solution

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Given that x and x2 in the Lagrangian don't need expanding, I simply took the Sin and Cos parts up to their x2 terms and replaced them accordingly in the Lagrangian. However the given solution (below) seems to have done things rather differently. In particular I notice the 4a2x2 term and the Sin term have both become zero.

Can anyone explain why this is? Some obvious algebra error on my part, or a fundamental aspect of Taylor expansions that I haven't properly grasped? Many thanks.

Well, the term proportional to $x \dot{x} \dot{\theta} sin(\theta)$ is 4th order (treating $x$ and $\theta$ as the same order).

sa1988
stevendaryl said:
Well, the term proportional to $x \dot{x} \dot{\theta} sin(\theta)$ is 4th order (treating $x$ and $\theta$ as the same order).

Ahhh, so yes it was a fundamental error in understanding on my part.

Back to basics for me...

Thanks

## 1. What is the Taylor expansion of Lagrangian to second order?

The Taylor expansion of Lagrangian to second order is a mathematical technique used to approximate the Lagrangian function to a higher degree of precision. It involves expanding the function into a series of terms, with the first term being the original function and subsequent terms being the derivatives of the function evaluated at a specific point.

## 2. Why is it important to expand the Lagrangian to second order?

Expanding the Lagrangian to second order allows for a more accurate representation of the function, especially when dealing with complex systems. It also enables us to better approximate the behavior of the system and make more precise predictions.

## 3. How is the Taylor expansion of Lagrangian to second order calculated?

The Taylor expansion of Lagrangian to second order is calculated by using the Taylor series formula, which involves taking the derivatives of the function and evaluating them at a specific point. These derivatives are then multiplied by the corresponding powers of the difference between the point and the center of expansion. The resulting terms are then added together to get the expanded function.

## 4. What is the difference between first and second order expansions of the Lagrangian?

The first order expansion of the Lagrangian only takes into account the first derivative of the function, while the second order expansion considers both the first and second derivatives. This means that the second order expansion is a more accurate representation of the function, but it also involves more calculations.

## 5. In which situations would we use the Taylor expansion of Lagrangian to second order?

The Taylor expansion of Lagrangian to second order is commonly used in physics and engineering, particularly in the field of mechanics. It is used to approximate the behavior of complex systems, such as in the study of fluid dynamics and in the analysis of mechanical systems with multiple degrees of freedom.

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