# Euler Lagrange Derivation (Taylor Series)

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1. Jan 15, 2016

### bananabandana

Mod note: Moved from Homework section
1. The problem statement, all variables and given/known data

Understand most of the derivation of the E-L just fine, but am confused about the fact that we can somehow Taylor expand $L$ in this way:

$$L\bigg[ y+\alpha\eta(x),y'+\alpha \eta^{'}(x),x\bigg] = L \bigg[ y, y',x\bigg] + \frac{\partial L}{\partial y} \alpha \eta(x) + \frac{\partial L}{\partial y'} \alpha \eta^{'} + O^{2}(\alpha \eta)$$

I'm really uncomfortable with the idea that you can treat whole functions as independent of each other (when they are clearly not). I've looked up a lot of derivations online & from textbooks and none of them seem to bother to explain this. Am I missing something obvious? Surely the fact that $y=f(x,t,\alpha )$ should be taken into account somewhere?

Thanks in advance for the help!

Last edited by a moderator: Jan 15, 2016
2. Jan 15, 2016

### mathman

The expansion looks ok. What dependence are you concerned about?

3. Jan 16, 2016

### Ssnow

Yes, it is exactly the Taylor expansion of $L\left[y+\alpha\eta(x),y'+\alpha\eta'(x),x\right]$ at the point $(y,y',x)$, obviously the term with the partial derivative of $x$ there isn't because $\frac{\partial L}{\partial x}\cdot(x-x)=0$...

4. Jan 20, 2016

### bananabandana

Yes, I know, but why are we allowed to treat $y(x)$ and $\dot{y}(x)$ as independent variables? (as we are required to do for the Taylor series?) They clearly aren't independent.

5. Jan 20, 2016

### PeroK

What's happening is that the symbols $y, \dot{y}, x$ are being used for two purposes. I'm only on my phone so it's difficult to type a lot of latex. Consider $L$ as any function of three variables. To be clear we will define $L$ using dummy variables $X,Y,Z$

E.g we could define $L(X,Y,Z) = X^2 + XYZ$

Now, we have three more functions by taking the partial derivatives of $L$:

$L_X, L_Y, L_Z$

Using this notation for simplicity.

Finally, we can do the following Taylor expansion for $L$

$L(X_0 + a, Y_0 + b, Z_0 + c) = L(X_0, Y_0, Z_0) + aL_X + bL_Y + cL_Z + \dots$

And that hold for any numerical values we choose for $X_0, Y_0, Z_0, a, b, c$. In other words, that equation holds for whatever expressions we plug in to replace the dummy variables we've used.

In this case, the expressions happen to be expressions in $x,y, \dot{y}, \eta$.

6. Jan 20, 2016

### PeroK

PS I forgot to mention:
$\frac{\partial L}{\partial y}$

Think of this not as the partial derivative of $L$ with respect to $y$, but the function formed by taking the partial derivative of $L$ with respect to its first argument.

7. Jan 22, 2016

### Ssnow

Geometrically $y(x)$ is a function of $x$ and you can think this as subset of the euclidean space, as the graph of a function on the plane for example. The derivative $\dot{y}(x)$ represent the tangent direction of the graph of the function at the point $x$. So $\dot{y}(x)$ generate the tangent space of $y(x)$ at $x$. In general they lives in two different spaces...