Taylor Expand Lagrangian to Second Order....

[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

Homework Equations

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.

 

[stevendaryl]

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

 

[sa1988]

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