Theorem: Let $V$ and $W$ be finite dimensional Banach spaces, $G \subset V$ an open subset, $f: G \to W$ a $n$-times differentiable function and $p \in G$. Then, we have $$ f(x) = \bigg(\sum_{k = 0}^{n} \frac{1}{k!} D^k_p f(\underbrace{x - p, \ldots, x - p}_{k-\text{times}}) \bigg) + R(x) $$ so that $$ \lim_{x \to p} \frac{R(x)}{\| x - p \|^n} = 0, $$ where $D_p f(v) := \lim\limits_{t \to 0} \frac{f(p + tv) - f(p)}{t}$ is the first directional derivative of $f$ and the $k$-th derivative is defined inductively.
My question is regarding the following addendum: Is $f$ is $(n + 1)$-times differentiable and real valued (!) and $\overline{px} \subset G$, we have $$ \exists q \in \overline{px}: R(x) = \frac{1}{(n + 1)!} D_q^{n + 1}f (\underbrace{x - p, \ldots, x - p}_{(n + 1)-\text{times}}) $$
We prove the addendum by using the one dimensional Taylor theorem: Define $g: [0,1] \to \mathbb{R}, \ t \mapsto f(p + tv)$, then there exists a $\tau \in [0,1]$ so that $$f(x) = g(1) = \sum_{k = 0}^{n} \frac{g^(k)(0)}{k!} + \frac{1}{(n + 1)!}g^{(n + 1)}(\tau)$$
I want to find a example for a function which doesn't meet the requirements of the addendum and for which statement doesn't hold because I can't really grasp why exactly those requirements are necessary.
Any help is greatly appreciated.
This generalization of one dimensional Taylor’s formula with the remainder in Lagrange form fails because, in general, we cannot simultaneously provide the respective fraction $\tau\in [0,1]$ for different coordinate components. For instance, consider a function $f:R\to \Bbb R^2$, $x\mapsto (x^2,x^3)$ for each $x\in\Bbb R$ and $n=0$. For each $q,v\in\Bbb R$ we have $D_q(v)=vf’(q)=v(2q, 3q^2)$. If $p=0$, $0\ne x\in\Bbb R$ and $f(x)=f(p)+D_q(x-p)$ for some $q\in\Bbb R$ then $x^2=2qx$ and $x^3=3q^2x$, which implies $4q^2=3q^2$ and $q=0=x$, a contradiction.