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Taylor Expand Lagrangian to Second Order...

  1. Dec 16, 2016 #1
    1. The problem statement, all variables and given/known data

    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

    2zxtiro.png

    2. Relevant equations


    3. The attempt at a solution

    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.

    algpqh.png
     
  2. jcsd
  3. Dec 16, 2016 #2

    stevendaryl

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    Staff Emeritus
    Science Advisor

    Well, the term proportional to [itex]x \dot{x} \dot{\theta} sin(\theta)[/itex] is 4th order (treating [itex]x[/itex] and [itex]\theta[/itex] as the same order).
     
  4. Dec 19, 2016 #3
    Ahhh, so yes it was a fundamental error in understanding on my part.

    Back to basics for me...

    Thanks
     
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