Homogeneous polynomial identity

Question

Suppose we have a homogeneous polynomial $P$ of degree $k$ in $n$ complex variables. The following formula seems to be true (can be proven by straightforward computations for $2$ or $3$): $$P(z_1,...,z_n)=\frac{1}{k!}\sum_{\epsilon_l \in \{0,1\}, l \in \{1,...,k \}} (-1)^{k-\sum_{j=1}^k \epsilon_j} P(a+(\sum_{j=1}^k \epsilon_j)z)$$ for any given $a \in \mathbb{C}^n$. So for $k=2$ it would be: $$P(z_1,...,z_n)=\frac{1}{2}(P(a)-2P(a+z)+P(a+2z)).$$Yet I have no idea for the proof.

Answer 1

Let me first reformulate the problem in a much more convenient way : I define $$Q(a,z) = \frac{1}{k!} \sum_{m=0}^k (-1)^{k-m}\binom{k}{m}P(a+mz)$$ and we have to show $Q(a,z) = P(z)$.

First we notice that $Q(0,z) = P(z)$ because $\sum_{m=0}^k (-1)^{k-m}\binom{k}{m} m^k = k!$.

Then it remains to show that $Q(a,z)$ is independent of $a$. We do that by induction on $k$. If $k=1$ this is clear because $Q(a,z) = P(a+z)-P(a) = P(z)$. So assume this is true for all homogeneous polynomials of degree $k-1$.

We have $$\begin{eqnarray*} \frac{\partial Q}{\partial a_i}(a,z) & = & \frac{1}{k!} \sum_{m=0}^k (-1)^{k-m}\binom{k}{m}(\partial_i P)(a+mz) \\ & = & \frac{1}{k!} \sum_{m=0}^k (-1)^{k-m}\binom{k-1}{m}(\partial_i P)(a+mz) +\frac{1}{k!} \sum_{m=0}^k (-1)^{k-m}\binom{k-1}{m-1}(\partial_i P)(a+mz)\\ & = & -\frac{1}{k!}\sum_{m=0}^{k-1} (-1)^{k-1-m}\binom{k-1}{m}(\partial_i P)(a+mz) + \frac{1}{k!} \sum_{m=0}^{k-1} (-1)^{k-1-m}\binom{k-1}{m}(\partial_i P)(a+z+mz)\\ & = & -(\partial_i P)(z) + (\partial_i P)(z) \\ & = & 0 \end{eqnarray*}$$ where we use that $\partial_i P$ is homogeneous of degree $k-1$ so we use twice the induction hypothesis (once with $a$ and once with $a+z$).

So the induction works and $Q(a,z) = P(z)$ for all $a$ in all degree.