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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.

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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.

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