Equivalence of Definitions of Weierstrass E-Function

Theorem
Let $\mathbf y, \mathbf z, \mathbf w$ be $n$-dimensional vectors.

Let $\mathbf y$ be such that $\mathbf y \left({a}\right) = A$ and $\mathbf y \left({b}\right) = B$.

Let $J$ be a functional such that:


 * $\displaystyle J \left[{\mathbf y}\right] = \int_a^b F \left({x, \mathbf y, \mathbf y'}\right) \rd x$

Definition 1 implies Definition 2
By Definition 1:


 * $E \left({x, \mathbf y, \mathbf z, \mathbf w}\right) = F \left({x, \mathbf y, \mathbf w}\right) - F \left({x, \mathbf y, \mathbf z}\right) + \left({\mathbf w - \mathbf z}\right) F_{\mathbf y'} \left({x, \mathbf y, \mathbf z}\right)$

By Taylor's Theorem, where expansion is done around $\mathbf w = \mathbf z$ and Lagrange form of remainder is used:


 * $\displaystyle F \left({x, \mathbf y, \mathbf w}\right) = F \left({x, \mathbf y, \mathbf z}\right) + \dfrac {\partial F \left({x, \mathbf y, \mathbf z}\right)} {\partial \mathbf y'} \left({\mathbf w - \mathbf z}\right) + \dfrac 1 2 \sum_{i, j \mathop = 1}^n \left({w_i - z_i}\right) \left({w_j - z_j}\right) \dfrac {\partial^2 F \left({x, \mathbf y, \mathbf z + \theta \left({\mathbf z - \mathbf w}\right)}\right)} {\partial y_i' y_j'}$

where $\theta \in \R: 0 < \theta < 1$.

Insertion of this expansion into the definition for Weierstrass E-Function leads to the desired result.