How to show that $T_k(x_1,\ldots ,x_n)$ is the only polynomial of degree $k$ with the specific properties? Let $U\subset \mathbb{R}^n$ be an open set and $f:U\rightarrow \mathbb{R}$ is a $k$-times continusouly differentiable function.
Let $x_0\in U$ be fixed.
The $k$-th Taylor polynomial of $f$ in $x_0$ is $$T_k(x_1,\ldots ,x_n)=\sum_{m=0}^k\frac{1}{m!}\sum_{i_1=1}^n \ldots \sum_{i_m=1}^n \frac{\partial}{\partial{x_{i_1}}}\ldots \frac{\partial}{\partial{x_{i_m}}}f(x_0)\cdot x_{i_1}\cdot \ldots x_{i_m}$$
Show that $T_k(x_1,\ldots ,x_n)$ is the only polynomial of degree $k$ such that $$T_k(0)=f(x_0) \\ \frac{\partial}{\partial{x_{i_1}}}\ldots \frac{\partial}{\partial{x_{i_m}}}T_k(0)=\frac{\partial}{\partial{x_{i_1}}}\ldots \frac{\partial}{\partial{x_{i_m}}}f(x_0)$$ for all $m\in\{1,\ldots , k\}$, $i_\ell\in \{1, \ldots ,n\}$ and $\ell \in \{1, \ldots , m\}$.
To show that $T_k$ satisfies these properties we just have to replace $x_i=0$ for all $i$ for the first property and for the second one we have to calculate these partial derivatives, right?
But how can we show the uniqueness?
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EDIT :
I am trying to show that $$\frac{\partial}{\partial{x_{i_1}}}\ldots \frac{\partial}{\partial{x_{i_m}}}T_k(0)=\frac{\partial}{\partial{x_{i_1}}}\ldots \frac{\partial}{\partial{x_{i_m}}}f(x_0)$$
Is it maybe as follows? \begin{align*}\frac{\partial}{\partial{x_{i_1}}}T_k(x_1,\ldots ,x_n)&=\frac{\partial}{\partial{x_{i_1}}}\left [\sum_{m=0}^k\frac{1}{m!}\sum_{i_1=1}^n \ldots \sum_{i_m=1}^n \frac{\partial}{\partial{x_{i_1}}}\ldots \frac{\partial}{\partial{x_{i_m}}}f(x_0)\cdot x_{i_1}\cdot \ldots x_{i_m}\right ]\\ & =\sum_{m=0}^k\frac{1}{m!}\sum_{i_1=1}^n \ldots \sum_{i_m=1}^n \frac{\partial}{\partial{x_{i_1}}}\ldots \frac{\partial}{\partial{x_{i_m}}}f(x_0)\cdot \frac{\partial}{\partial{x_{i_1}}}\left (x_{i_1}\cdot \ldots x_{i_m}\right )\\  & =\sum_{m=0}^k\frac{1}{m!}\sum_{i_1=1}^n \ldots \sum_{i_m=1}^n \frac{\partial}{\partial{x_{i_1}}}\ldots \frac{\partial}{\partial{x_{i_m}}}f(x_0)\cdot x_{i_2}\cdot \ldots x_{i_m}\end{align*} Or how do we calculate the partial derivatives?
 A: I looks a bit difficult to show from the given form of the Taylor series polynomial. I think it is preferable to regroup powers of distinct $x_i$'s. A multinomial (polynomial in $n$ variables) may be written in a unique way as:
$$ P(x_1,...,x_n) = \sum_{k_1\geq 0,...,k_n\geq 0} c_{k_1,...,k_n} x_1^{k_1}\cdots x_n^{k_n} =: \sum_{{\bf k}\geq {\bf 0}} c_{\bf k} {\bf x}^{\bf k},$$
with some obvious definitions of multi-indices and where only finitely many $c_{\bf k}$'s are non-zero.
From an algebraic point of view this defines the $c_{\bf k}$'s uniquely. But to see it is unique in an analytic fashion you may take derivatives, using Schwarz' theorem showing that the relevant partial derivatives commute.
For ${\bf q}=(q_1,...,q_n)\geq {\bf 0}$ you may  verify by inspection that:
$$ \left(\frac{\partial}{\partial {\bf x}} \right)^{\bf q} P({\bf x})_{|{\bf x} = {\bf 0}} =  
\left(\left(\frac{\partial}{\partial x_1}\right)^{q_1} \cdots  \left(\frac{\partial}{\partial x_n}\right)^{q_n}  P \right)(0,...,0)  = (q_1)! \cdots (q_n)! c_{\bf q}.$$
Writing ${\bf q !}=  (q_1)! \cdots (q_n)! $ this yields (in compact notation):
$$c_{\bf q} = \frac{1}{\bf q !} \left(\frac{\partial}{\partial {\bf x}} \right)^{\bf q} P({\bf x})_{|{\bf x} = {\bf 0}} $$
which defines $c_{\bf q}$ uniquely. A Taylor polynomial is obtained by calculating the $c_{\bf q}$'s using $f$ instead of $P$ and up to some fixed total order, again defining the $c_{\bf q}$'s uniquely.
