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:


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

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:


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

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

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