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 $\map{\mathbf y} a=A$ and $\map{\mathbf y} b=B$.

Let $J$ be a functional such that:


 * $\ds J \sqbrk {\mathbf y} = \int_a^b \map F {x, \mathbf y, \mathbf y'} \rd x$

Definition 1 implies Definition 2
By Definition 1:


 * $\map E {x, \mathbf y, \mathbf z, \mathbf w} = \map F {x, \mathbf y, \mathbf w} - \map F {x, \mathbf y, \mathbf z} + \paren {\mathbf w - \mathbf z} \map {F_{\mathbf y'} } {x, \mathbf y, \mathbf z}$

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


 * $\ds \map F {x, \mathbf y, \mathbf w} = \map F {x, \mathbf y, \mathbf z} + \frac {\partial \map F {x, \mathbf y, \mathbf z} } {\partial \mathbf y'} \paren {\mathbf w - \mathbf z} + \frac 1 2 \sum_{i, j \mathop = 1}^n \paren {w_i - z_i} \paren {w_j - z_j} \frac {\partial^2 \map F {x, \mathbf y, \mathbf z + \theta \paren {\mathbf z - \mathbf w} } } {\partial y_i'y_j'}$

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

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