ProofOfWaringsFormula2 2005-11-10 02:41:57 2005-11-10 02:55:07 Proof Waring's formula 0 % this is the default PlanetMath preamble. as your knowledge % of TeX increases, you will probably want to edit this, but % it should be fine as is for beginners. % almost certainly you want these \usepackage{amssymb} \usepackage{amsmath} \usepackage{amsfonts} % used for TeXing text within eps files %\usepackage{psfrag} % need this for including graphics (\includegraphics) %\usepackage{graphicx} % for neatly defining theorems and propositions %\usepackage{amsthm} % making logically defined graphics %\usepackage{xypic} % there are many more packages, add them here as you need them % define commands here The following is a proof of the Waring's formula using formal power series. We will work with formal power series in indeterminate $z$ with coefficients in the ring $\mathbb{Q}[x_1,\ldots,x_n]$. We also need the following equality \[ -\log(1-z) = \sum_{j=1}^\infty \frac{z^j}{j}. \] Taking log on both sides of \[ 1 - \sigma_1z+\ldots + (-1)^n \sigma_n z^n = \prod_{m=1}^n(1-x_mz), \] we get \begin{equation} \log(1 - \sigma_1z+\ldots + (-1)^n \sigma_n z^n) = \sum_{m=1}^n \log(1-x_mz), \label{eq} \end{equation} Waring's formula will follow by comparing the coefficients on both sides. The right hand side of the above equation equals \[ \sum_{m=1}^n \sum_{j=1}^\infty \frac{x_m^j}{j}z^j \] or \[ \sum_{j=1}^\infty \left( \sum_{m=1}^n x_m^j \right) \frac{z^j}{j} \] The coefficient of $z^k$ is equal to $S_k/k$. On the other hand, the left hand side of \eqref{eq} can be written as \[ \sum_{j=1}^\infty \frac{1}{j}(\sigma_1z-\sigma_2z^2+\ldots+(-1)^{n-1} \sigma_n z^n)^j. \] For each $j$, the coefficient of $z^k$ in \[(\sigma_1z-\sigma_2z^2+\ldots+(-1)^{n-1} \sigma_n z^n)^j \] is \[\sum_{i_1,\ldots,i_n} (-1)^{i_2+i_4+i_6+\ldots} \frac{j!}{i_1!\cdots i_n!}\sigma_1^{i_1} \cdots \sigma_n^{i_n}, \] where the summation is extended over all $n$-tuple $(i_1,\ldots,i_n)$ whose entries are non-negative integers, such that \begin{gather*} i_1+i_2+\ldots+i_n = j \\ i_1+2i_2+\ldots +ni_n = k. \end{gather*} So the coefficient of $z^k$ in the left hand side of \eqref{eq} is \[ \sum_{j=1}^\infty \sum_{i_1,\ldots,i_n} (-1)^{i_2+i_4+i_6+\ldots} \frac{(j-1)!}{i_1!\cdots i_n!}\sigma_1^{i_1} \cdots \sigma_n^{i_n}, \] or \[\sum (-1)^{i_2+i_4+i_6+\ldots} \frac{(i_1+\ldots+i_n-1)!}{i_1!\cdots i_n!}\sigma_1^{i_1} \cdots \sigma_n^{i_n}. \] The last summation is over all $(i_1,\ldots, i_n)\in \mathbb{Z}^n$ with non-negative entries such that $i_1+2i_2+\ldots+ni_n=k$.