Taylor formula uniqueness for multivariate maps

Let $f: U \to \mathbb{R}$ a function $k$ times differentiable in $U$ an K+1 times differentiable in $a \in U$, where $U \subset \mathbb{R}^n$ is open. I'm interested in showing the uniqueness of Taylor's formula in several variables. Then, let $\phi_i: \mathbb{R}^n \times \cdots \times \mathbb{R}^n \to \mathbb{R}$ the $i$-linear transformation such that $$f(a+v) = f(a) + \phi_1 \cdot v + \phi_2 \cdot v^2 + \ldots + \phi_{k+1} \cdot v^{k+1} + r(v),$$ where $\lim_{v \to 0} \frac{r(v)}{|v|^{k+1}} = 0$. Denoting $$d^pf(a) \cdot v = \sum_{i_1, \ldots, i_p = 1}^n \dfrac{\partial^n f}{\partial x_{i_1} \cdots \partial x_{i_p}}(a) \alpha_{i_1} \cdots \alpha_{i_p}$$, where $v= (\alpha_{i_1}, \ldots, \alpha_{i_p})$. I need to show that $$\phi_i \cdot v^i = \frac{1}{i!}d^if(a)\cdot v^i$$ for all $i$. I need to use Taylor's formula, but I don't see how, in the case of real functions I just derive the polynomial and apply it to 0, but here I don't know how to do that.

Thank you for your help.

Solution 1:

What you're looking to prove is equivalent to prove that $$0=\phi_1 \cdot v + \phi_2 \cdot v^2 + \ldots + \phi_{k+1} \cdot v^{k+1} + r(v),$$ where $\lim\limits_{v \to 0} \frac{r(v)}{|v|^{k+1}} = 0$ implies that $\phi_1, \dots, \phi_{k+1}$ are all identically vanishing. And you can do that by induction on $k$.

For example when $k=0$, you have to prove that if $$0=\phi_1 \cdot v + r(v),$$ with $\lim\limits_{v \to 0} \frac{r(v)}{|v|} = 0$ then $\phi_1$ is the zero linear form on $\mathbb R^n$. And this is clear as if it exists $u \neq 0$ such that $\phi_1 \cdot u \neq 0$, then

$$0 = \phi_1 \cdot u + \frac{r(\lambda u)}{\lambda}$$ for $0 \neq \lambda \in \mathbb R$ which leads to the contradiction $\lim\limits_{\lambda \to 0} \frac{r(\lambda u)}{\lambda} = 0$ while $\phi_1 \cdot u \neq 0$.

And you can proceed in a similar way for the induction step.