Consider a function
$$
f:\mathbb{R}^n\rightarrow\mathbb{R}^m
$$
given by $m$ functions $f\_i:\mathbb{R}^n\rightarrow \mathbb{R}$ that we can assume to be polynomials in $x\_1,\dots,x\_n$.


Does there exist any formula expressing $f$ as a linear combination of $\frac{\partial f}{\partial x\_1}(x\_1,\dots,x\_n),\dots,\frac{\partial f}{\partial x\_n}(x\_1,\dots,x\_n)$, $f(x\_1,0,\dots,0), \frac{\partial f}{\partial x\_1}(x\_1,0,\dots,0),\dots,\frac{\partial f}{\partial x\_n}(x\_1,0,\dots,0)$, where the coefficients of the linear combination are functions of $x\_1,\dots,x\_n$? Here $\frac{\partial f}{\partial x\_i} = \left(\frac{\partial f\_1}{\partial x\_i},\dots,\frac{\partial f\_m}{\partial x\_i}\right)$.


Basically I am asking for an analogue of Euler's formula for homogeneous polynomials which says that if $f:\mathbb{R}^n\rightarrow\mathbb{R}$ is a homogeneous polynomial of degree $d$ then $d\cdot f = x\_1\frac{\partial f}{\partial x\_1}(x\_1,\dots,x\_n) + \dots + x\_n\frac{\partial f}{\partial x\_n}(x\_1,\dots,x\_n)$.


Thank you.



Expressing a vector valued function in terms of its derivatives
$\newcommand{\pa}{\partial}\newcommand{\R}{\mathbb R}$The answer is no. Indeed, suppose the contrary: that for each polynomial $f$ there are functions $a\_j,b,c\_j$ such that
\begin{equation}
 f(x\_1,\dots,x\_n)=\sum\_{j\in[n]}a\_j(x\_1,\dots,x\_n)(\pa\_j f)(x\_1,\dots,x\_n) \\ 
 +b(x\_1,\dots,x\_n)f(x\_1,0\dots,0) \\ 
 +\sum\_{j\in[n]}c\_j(x\_1,\dots,x\_n)(\pa\_j f)(x\_1,0\dots,0) \tag{1}\label{1}
\end{equation}
for all $(x\_1,\dots,x\_n)\in\R^n$, where $[n]:=\{1,\dots,n\}$ and $\pa\_j f$ denotes the partial derivative of $f$ with respect to its $j$th argument.
Let now
\begin{equation}
 f(x\_1,\dots,x\_n):=3x\_2^2-2x\_2^3. 
\end{equation}
Then $(\pa\_j f)(1,1,0,\dots,0)=f(1,0\dots,0)=(\pa\_j f)(1,0\dots,0)=0$ for all $j\in[n]$, so that the right-hand side of \eqref{1} for $(x\_1,\dots,x\_n)=(1,1,0,\dots,0)$ is $0$, whereas the left-hand side of \eqref{1} for $(x\_1,\dots,x\_n)=(1,1,0,\dots,0)$ is $f(1,1,0,\dots,0)=1\ne0$, which contradicts \eqref{1}. $\quad\Box$


