# Taylor expansion

1. Dec 26, 2009

### vertices

[I posted this in the Classical Mechanics sub forum - but I have not received any responses as yet -I think it is better placed here!]

I have an expression that I need to approximate. I Taylor expanded a bracket twice over and got the result I wanted to get, however I am not sure if what I have done is mathematically correct or not.

This is the expression:

$$\int{dt.(1+\epsilon\dot{\xi(t)}).L[q(t+\epsilon\xi(t))+{\delta}q(t+\epsilon\xi(t))]} - \int{dt.L[q(t)+{\delta}q(t)]}$$

I start with $$L[q(t+\epsilon\xi(t))]$$

If we first Taylor expand the bit in parenthesis (ie. $$q(t+\epsilon\xi(t))$$) we get:

$$q(t+\epsilon\xi(t))=q(t)+\frac{{\partial}q(t)}{{\partial}t}\epsilon\xi(t)+...$$

Therefore$$L[q(t+\epsilon\xi(t))=L[q(t)+\frac{{\partial}q(t)}{{\partial}t}\epsilon\xi(t)+...]$$

This is equal to:

$$L(q)+\frac{{\partial}L}{{\partial}q}\frac{{\partial}q(t)}{{\partial}t}\epsilon\xi(t)+...$$

but

$$\frac{{\partial}L}{{\partial}q}\frac{{\partial}q(t)}{{\partial}t} = \frac{{\partial}L}{{\partial}t}$$ ('CANCELLING' the 'dq's - I know this will have the mathematicians up in arms)

So:

$$L[q(t+\epsilon\xi(t))]=L(q)+\frac{{\partial}L}{{\partial}t}\epsilon\xi(t)+smaller terms]$$

Is it okay to Taylor expand in this way?

Thanks.

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