In what manner is the Taylor expansion used in the principle of stationary action?

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I have a little familiarity with some basic functions that can be expressed as Taylor series but I am stuck on the concept of using Taylor expansions on the "derivation" the principle of stationary action.



The action is equal to the integral of the Lagrangian and what is fed into the action is the points of the actual path and a deviation from the actual path. In all cases the end point of course are the same.



This is just equal to a polynomial with a squared term. i.e. standard equation for the variation in the action.



All is well so far but all derivations take this exact equation and begin to add higher order terms!



I do not understand where the higher order terms are coming from in the formula for stationary action.




taylor-expansion classical-mechanics




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    I have a little familiarity with some basic functions that can be expressed as Taylor series but I am stuck on the concept of using Taylor expansions on the "derivation" the principle of stationary action.



    The action is equal to the integral of the Lagrangian and what is fed into the action is the points of the actual path and a deviation from the actual path. In all cases the end point of course are the same.



    This is just equal to a polynomial with a squared term. i.e. standard equation for the variation in the action.



    All is well so far but all derivations take this exact equation and begin to add higher order terms!



    I do not understand where the higher order terms are coming from in the formula for stationary action.




    taylor-expansion classical-mechanics




    share|cite|improve this question























      up vote
      0
      down vote

      favorite









      up vote
      0
      down vote

      favorite











      I have a little familiarity with some basic functions that can be expressed as Taylor series but I am stuck on the concept of using Taylor expansions on the "derivation" the principle of stationary action.



      The action is equal to the integral of the Lagrangian and what is fed into the action is the points of the actual path and a deviation from the actual path. In all cases the end point of course are the same.



      This is just equal to a polynomial with a squared term. i.e. standard equation for the variation in the action.



      All is well so far but all derivations take this exact equation and begin to add higher order terms!



      I do not understand where the higher order terms are coming from in the formula for stationary action.




      taylor-expansion classical-mechanics




      share|cite|improve this question













      I have a little familiarity with some basic functions that can be expressed as Taylor series but I am stuck on the concept of using Taylor expansions on the "derivation" the principle of stationary action.



      The action is equal to the integral of the Lagrangian and what is fed into the action is the points of the actual path and a deviation from the actual path. In all cases the end point of course are the same.



      This is just equal to a polynomial with a squared term. i.e. standard equation for the variation in the action.



      All is well so far but all derivations take this exact equation and begin to add higher order terms!



      I do not understand where the higher order terms are coming from in the formula for stationary action.





      taylor-expansion classical-mechanics





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      share|cite|improve this question




      share|cite|improve this question








      edited Aug 9 at 14:30
























      asked Aug 6 at 14:41









      Sedumjoy

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