Derivation of the prolongation formula for finding symmetries of diff equations from Olver I am having a problem with the derivation of the prolongation formula from PJ Olver's text :"Applications of Lie groups to differential equations" Page 105,106.
Considering a differential equation with independent variable(x) and one dependent variable(u). 
(x,u) $\subset$ $X \times U$
The first jet space $M^{(1)}$ has the coordinates (x,$u^1$) = (x,u,$u_j$).
Suppose u = f(x) is any function with $u_j = \frac{\partial u}{\partial x_j} $ 
First prolongation of a group action on M is given as:
$pr^{1} g_\epsilon . (x,u^{(1)}) = (\tilde{x},\tilde{u}^{(1)})$
The dependent variable is unchanged here so the following:
$\tilde{u} = \tilde{f}_\epsilon(\tilde{x}) = f[\Xi^{-1}_\epsilon(\tilde{x})] = f[\Xi_{-\epsilon} (\tilde{x})] $
Here I don't understand how $\tilde{f}$ has transformed to f. 
Now with some further calculations, following expression arises involving a pull back where I don't understand how an interchange in the order of differentiation has been done.
$\frac{\partial}{\partial\tilde{x}^j} [\frac{d\Xi^k_{-\epsilon}}{d\epsilon} ] (\Xi_{\epsilon}(x)) |_{\epsilon = 0} = \frac{\partial}{\partial x^j} [\frac{d\Xi^k_{\epsilon}}{d\epsilon} ] |_{\epsilon = 0} = - \frac{\partial \xi^k}{\partial x^j}(x)$  
Please anybody help, if any further details are needed I will add in.
 A: I also find Olver's proof confusing, but the prolongation formula is actually quite simple.
Let $x^i$ be independent variables and $u^{\alpha}$ be dependent variables. Then we have the jet bundle whose coordinates are $x^i,u_J^{\alpha}$. We want to compute the symmetries of the jet bundle which preserve the jets of graphs $u^{\alpha} = f(x^i)$. The 1-forms
$$ \theta_J^{\alpha} = d u_J^{\alpha} - u_{Ji}^{\alpha} dx^i$$
vanish on these jets, so if
$$ V = f^i \frac{\partial}{\partial x^i} + \phi^{\alpha}_j \frac{\partial}{\partial u^{\alpha}_J}$$
is an infinitesimal symmetry, then the 1-form $\mathcal{L}_V \theta_J^{\alpha}$ must also vanish on jets of graphs. If we evaluate this Lie derivative, it follows that
$$ \phi_{Ji}^{\alpha} = 
\frac{\partial \phi^{\alpha}_J}{\partial x^i} + \frac{\partial \phi^{\alpha}_J}{\partial u_K^{\beta}} u_{Ki}^{\beta} - u_{Jj}^{\alpha} \left( \frac{\partial f^j}{\partial x^i} + \frac{\partial f^j}{\partial u_K^{\beta}} u_{Ki}^{\beta} \right)$$
Here I am making heavy use of the Einstein summation convention. This is exactly the prolongation formula which Olver is proving.
