$$ x \cdot y = \frac{1}{2 \cdot 2 !} \left( (x + y)^2 - (x - y)^2 \right) $$ $$ \begin{eqnarray} x \cdot y \cdot z &=& \frac{1}{2^2 \cdot 3 !} ((x + y + z)^3 - (x + y - z)^3 \nonumber \\ &-& (x - y + z)^3 + (x - y - z)^3 ), \end{eqnarray} $$ $$ \begin{eqnarray} x \cdot y \cdot z \cdot w &=& \frac{1}{2^3 \cdot 4 !} ( (x + y + z + w)^4 \nonumber \\ &-& (x + y + z - w)^4 - (x + y - z + w)^4 \nonumber \\ &+& (x + y - z - w)^4 - (x - y + z + w)^4 \nonumber \\ &+& (x - y + z - w)^4 + (x - y - z + w)^4 \nonumber \\ &-& (x - y - z - w)^4 ). \end{eqnarray} $$ 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 $r(\alpha, \beta)$ $=$ $\alpha - \beta \left\lfloor \frac{\alpha}{\beta}\right\rfloor$

  • 5
    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
up vote 33 down vote accepted

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).

  • 7
    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
  • @HideakiOkazaki You are welcome! Very interesting to note that you discovered some of these 22 years ago. I don't know though how old these polarization identities actually are.... – Suvrit Oct 11 '15 at 23:46
  • 1
    @mostafa You can see this polarization identity for instance in arxiv.org/pdf/math/0504397.pdf --- it is mentioned in other places too; I don't know the "oldest / canonical" citation for this though; maybe Alexandrov... – Suvrit Apr 22 '16 at 14:17

Your Answer

 
discard

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.