I am trying to prove the following identity for symmetric multilinear functions. Let $V,W$ be vector spaces, and let $F : V^k \to W$ be $k$-linear and symmetric. Let $x,y \in V$, and suppose $x_0,x_1,...,x_N \in V$ are such that $x_0 = x, x_N = y$. I would like to know if $${1 \over k!} F[(y-x)^k] = \sum\limits_{i=1}^N \sum\limits_{j=1}^k {1 \over j! (k-j)!} F[(x_{i-1}-x)^{k-j}(x_i-x_{i-1})^j].$$ In the case $k=1$, this is straightforward. In the case $k=2$, the right hand side is $$\sum\limits_{i=1}^N F[(x_{i-1}-x)(x_i - x_{i-1})] + {1 \over 2} F[(x_i - x_{i-1})^2] $$ $$\sum\limits_{i=1}^N (\sum\limits_{j=1}^{i-1} F[(x_j - x_{j-1})(x_i - x_{i-1})]) + {1 \over 2} F[(x_i - x_{i-1})^2]$$ $$= \sum\limits_{i=1}^N ({1 \over 2} \sum\limits_{j \neq i} F[(x_j - x_{j-1})(x_i - x_{i-1})]) + {1 \over 2} F[(x_i - x_{i-1})^2]$$ $$= {1 \over 2} F[(y-x)^2].$$ Is there some straightforward argument for the general case?
Edit: I forgot that I want $F$ to be symmetric.
Just a hint toward the complete argument. Notice that the symmetry of $F$ implies that we can freely exchange two arguments without any difference in the outcome: $$F(b,b,\dots,A,\dots,b) = F(A,b,\dots,b,\dots,b)$$ so that the sum $$F(A,b,\dots,b,\dots,b) + F(b,A,\dots,b,\dots,b) + \cdots + F(b,b,\dots,A,\dots,b) + \cdots + F(b,b,\dots,b,\dots,A) $$ becomes $$k\cdot F(A,b,\dots,b,\dots,b) $$ and similarly the sum of all $F(b,\dots,A,\dots,A,\dots,b)$ over all possible changes in the order of the arguments amounts to $$\binom{k}{2} \cdot F(A,A,\dots,b) $$ So we have some combinatorial action going on. If we sum over all possible choices of how many $A$'s we feed into $F$, we have (in your notation) $$\sum_{j=1}^k \binom{k}{j} F[A^j b^{k-j}] $$ Does this make things clearer to you?