# Expressing a vector valued function in terms of its derivatives

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.

• Consider the constant functions. May 15 at 16:57
• are the $f_i$'s homogeneous polynomials? May 15 at 16:59
• 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.
– R_O
May 15 at 17:19
• Missed that inclusion given the title. Sorry. May 15 at 18:12

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