# Are the following identities well known?

$$x \cdot y = \frac{1}{2 \cdot 2 !} \left( (x + y)^2 - (x - y)^2 \right)$$ $$\begin{eqnarray} x \cdot y \cdot z &=& \frac{1}{2^2 \cdot 3 !} ((x + y + z)^3 - (x + y - z)^3 \nonumber \\ &-& (x - y + z)^3 + (x - y - z)^3 ), \end{eqnarray}$$ $$\begin{eqnarray} x \cdot y \cdot z \cdot w &=& \frac{1}{2^3 \cdot 4 !} ( (x + y + z + w)^4 \nonumber \\ &-& (x + y + z - w)^4 - (x + y - z + w)^4 \nonumber \\ &+& (x + y - z - w)^4 - (x - y + z + w)^4 \nonumber \\ &+& (x - y + z - w)^4 + (x - y - z + w)^4 \nonumber \\ &-& (x - y - z - w)^4 ). \end{eqnarray}$$ The identity that rewrites a product of $n$ variables $( n \ge 2$, $n \in \boldsymbol{\mathbb{Z}_+})$ as additions of $n$ th power functions is as given below: $$\begin{eqnarray} & &x_0 \cdots x_1 \cdot x_{n-1} = \frac{1}{2^{n - 1} \cdot n !} \cdot \sum_{j = 0}^{2^{ n - 1} -1} ( - 1 )^{\sum_{m = 1}^{n - 1} \sigma_m(j)} \times ( x_0 + (-1)^{\sigma_1(j)} x_1 + \cdots + (- 1)^{\sigma_{n - 1}(j)} x_{n - 1})^n, \\ & &\sigma_m(j) = r(\left\lfloor \frac{j}{2^{m - 1}}\right\rfloor, 2), m (\ge 1), j (\ge 0) \in \mathbb{Z}_+, \; \left\lfloor x \right\rfloor = \max \{ n \in \mathbb{Z}_+ ; n \le x, x \in \mathbb{R} \} \end{eqnarray}$$ where $r(\alpha, \beta)$, $\alpha, \beta (\ge 1) \in \mathbb{Z}_+$ means the remainder of the division of $\alpha$ by $\beta$ such that $r(\alpha, \beta)$ $=$ $\alpha - \beta \left\lfloor \frac{\alpha}{\beta}\right\rfloor$

• Why do you need to recode the subsets of $\left\{1,2,\ldots,n-1\right\}$ as numbers in $\left\{0,1,\ldots,2^{n-1}-1\right\}$ ? – darij grinberg Oct 9 '15 at 2:55
• I intend to constructively make an one sided approximation method of a continuous function of many variables by using the general type identity above. Therefore, I summarized the general type identity above as a constructive form. I need to use $\{0,1,...,2^{n-1}\}$ type. – Hideaki Okazaki Oct 12 '15 at 0:11

Although not the exactly the same due to $2^{n-1}$ instead of $2^n$ terms, the OP's formula seems to be essentially the well-known polarization formula for homogeneous polynomials, which is stated as following:
Any polynomial $f$, homogeneous of degree $n$ can be written as $f(x)=H(x,\ldots,x)$ for a specific multilinear form $H$. One has the following polarization formula for $H$ (see also this MO post): \begin{equation*} H(x_1,\ldots,x_n) = \frac{1}{2^n n!}\sum_{s \in \{\pm 1\}^n}s_1\ldots s_n f\Bigl(\sum\nolimits_{j=1}^n s_jx_j\Bigr) \end{equation*}
In your case, $f=x^n$, so $H(x_1,\ldots,x_n) = x_1\cdots x_n$ (please note off by 1 indexing).
• The OP's formula is obtained upon division by $2$. – darij grinberg Oct 9 '15 at 2:56
• @darijgrinberg: thanks! I think I got confused due to the extra $\lfloor . \rfloor$ notation in the OP, and did not feel like verifying :-) – Suvrit Oct 9 '15 at 3:22
• Thank you very much. I found these identies from n = 2 until n= 6, 22 years ago. Then 3 years ago, I summarized the general type identity above, by using a remainder form. I also proved that general type identity above, by using a recurrence formula. Now I understood that the general type identity above, is a special form of polarization formula in the case of $f =x^n$, and H(x_0, \ldots, x_{n-1})\$. – Hideaki Okazaki Oct 11 '15 at 23:11