## Variational Operator/First Variation - Taylor Expansion

1. The problem statement, all variables and given/known data
Folks, how is the following expansion obtained for the following function

##F(x,u,u')## where x is the independent variable.

The change ##\epsilon v## in ##u## where ##\epsilon## is a constant and ##v## is a function is called the variation of ##u## and denoted by ##\delta u \equiv \epsilon v##

2. Relevant equations
3. The attempt at a solution

##\Delta F=F(x,u+\epsilon v, u'+\epsilon v')-F(x,u,u')##.

Expanding in powers of ##\epsilon## (treating ##u+\epsilon v## and ##u'+\epsilon v'## as dependent functions)

##\displaystyle \Delta F= F(x,u,u')+ \epsilon v \frac {\partial F}{\partial u}+ \epsilon v' \frac{\partial F}{\partial u'}+ \frac{(\epsilon v)^2}{2!} \frac{\partial^2 F}{\partial u^2}+\frac{(\epsilon v)(\epsilon v')}{2!} \frac{\partial^2 F}{\partial u \partial u'}+\frac{(\epsilon v')^2}{2!} \frac{\partial^2 F}{\partial u'^2}+....-F(x,u,u')##

How is the above line obtained...I looked at Taylor expansion but could not recognize its application to this function. Any links or tips will be appreciated.

Regards

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It's basically Taylor expansion on functions of several variables.

On the other hand, the Euler-Lagrangian equation can be derived by not using Taylor expansion which may be more straightforward to some readers(using partial derivative and total differential).

 Quote by bugatti79 How is the above line obtained...I looked at Taylor expansion but could not recognize its application to this function. Any links or tips will be appreciated.

 Quote by klondike It's basically Taylor expansion on functions of several variables. On the other hand, the Euler-Lagrangian equation can be derived by not using Taylor expansion which may be more straightforward to some readers(using partial derivative and total differential).
Where can I learn more about Taylor expansion of several variables? In this case are the variables u and u' so we are dealing with 2 variables since x is an independent variable...?

## Variational Operator/First Variation - Taylor Expansion

 Quote by bugatti79 1. The problem statement, all variables and given/known data Folks, how is the following expansion obtained for the following function ##F(x,u,u')## where x is the independent variable. The change ##\epsilon v## in ##u## where ##\epsilon## is a constant and ##v## is a function is called the variation of ##u## and denoted by ##\delta u \equiv \epsilon v## 2. Relevant equations 3. The attempt at a solution ##\Delta F=F(x,u+\epsilon v, u'+\epsilon v')-F(x,u,u')##. Expanding in powers of ##\epsilon## (treating ##u+\epsilon v## and ##u'+\epsilon v'## as dependent functions) ##\displaystyle \Delta F= F(x,u,u')+ \epsilon v \frac {\partial F}{\partial u}+ \epsilon v' \frac{\partial F}{\partial u'}+ \frac{(\epsilon v)^2}{2!} \frac{\partial^2 F}{\partial u^2}+\frac{(\epsilon v)(\epsilon v')}{2!} \frac{\partial^2 F}{\partial u \partial u'}+\frac{(\epsilon v')^2}{2!} \frac{\partial^2 F}{\partial u'^2}+....-F(x,u,u')## How is the above line obtained...I looked at Taylor expansion but could not recognize its application to this function. Any links or tips will be appreciated. Regards
 Quote by klondike It's basically Taylor expansion on functions of several variables.
I have identified the expansion as that of 2 variables ##u## and ##u'## based on this wolfram link eqns 33/34 http://mathworld.wolfram.com/TaylorSeries.html

1) I am not sure why the author threw in the ##F(x,u,u')## at the end to balance the first term in last equation of original post...?

2) The book continues on from this last equation to write
##\displaystyle \epsilon v \frac{\partial F}{\partial u}+\epsilon v' \frac{\partial F}{\partial u'} +\epsilon R_1(\epsilon)##

Does ##\epsilon R_1(\epsilon)## represent the higher order terms that can be neglected since they are small? This approaches 0 as ##\epsilon## approaches 0.

 Quote by bugatti79 1) I am not sure why the author threw in the ##F(x,u,u')## at the end to balance the first term in last equation of original post...?
Perhaps it helps to look at the Taylor expansion on single variable function where
f(x)=f(a)+f'(a)(x-a)+0.5f''(a)(x-a)2+o(x-2). Ignoring 2nd and higher order terms, we have f(x)=f(a)+f'(a)(x-a). It says f changed from f(a) to f(x) due to independent valuable changed from a to x, where f(a) is the initial point, f(x) is the final point, and (x-a) is the change in the independent variable. Hence the net change Δf = f(x)-f(a)=f(a)+f'(a)(x-a)-f(a)

 Quote by bugatti79 2) The book continues on from this last equation to write ##\displaystyle \epsilon v \frac{\partial F}{\partial u}+\epsilon v' \frac{\partial F}{\partial u'} +\epsilon R_1(\epsilon)## Does ##\epsilon R_1(\epsilon)## represent the higher order terms that can be neglected since they are small? This approaches 0 as ##\epsilon## approaches 0.
They are higher order infinitesimals than those of εη and εη'. Sorry, my η is your v. When taking limit as ε->0 (in the step of finding dS/dε), they vanished.

 Quote by klondike Perhaps it helps to look at the Taylor expansion on single variable function where f(x)=f(a)+f'(a)(x-a)+0.5f''(a)(x-a)2+o(x-2). Ignoring 2nd and higher order terms, we have f(x)=f(a)+f'(a)(x-a). It says f changed from f(a) to f(x) due to independent valuable changed from a to x, where f(a) is the initial point, f(x) is the final point, and (x-a) is the change in the independent variable. Hence the net change Δf = f(x)-f(a)=f(a)+f'(a)(x-a)-f(a) They are higher order infinitesimals than those of εη and εη'. Sorry, my η is your v. When taking limit as ε->0 (in the step of finding dS/dε), they vanished.
Thanks for the insight, that helped!