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.

  • 1
    $\begingroup$ Consider the constant functions. $\endgroup$ May 15, 2022 at 16:57
  • $\begingroup$ are the $f_i$'s homogeneous polynomials? $\endgroup$ May 15, 2022 at 16:59
  • $\begingroup$ I think the constant function is ok since we can use also $f(x_1,0,\dots,0)$. No, in general the polynomials are not homogeneous. $\endgroup$
    – R_O
    May 15, 2022 at 17:19
  • $\begingroup$ Missed that inclusion given the title. Sorry. $\endgroup$ May 15, 2022 at 18:12

1 Answer 1


$\newcommand{\pa}{\partial}\newcommand{\R}{\mathbb R}$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}. $\quad\Box$


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