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本帖最后由 hbghlyj 于 2024-11-15 09:09 编辑 Edward Waring, Miscellanea analytica, 1762; Meditationes Algebraicae, 1770, p. 225, 3d ed., 1782, pp. 1-4. No hint is given as to how Waring found (17); his proof was in effect by mathematical induction, being a verification that $s_k, s_{k-i},\dots, s_i$ satisfy Newton's formulae.
But (17) had been given earlier by Albert Girard, Invention Nouvelle en L'Algèbre, Amsterdam, 1629.
《Elementary theory of equations》第73页
9. Elementary Proof of Waring's Formula Divide each member of (16) into the negative of its derivative; we get
(19) $\quad\displaystyle \frac{-p_1-2 p_2 y-\cdots-n p_n y^{n-1}}{1+p_1 y+\cdots+p_n y^n} \equiv \frac{\alpha_1}{1-\alpha_1 y}+\cdots+\frac{\alpha_n}{1-\alpha_n y}$.
In the identity
(20) $\quad\displaystyle\frac{1}{1-Q} \equiv 1+Q+Q^2+\cdots+Q^{k-1}+\frac{Q^k}{1-Q}$
set $Q=\alpha_a y$ and multiply the resulting terms by $\alpha_g$. Hence the second member of (19) equals
(21) $\displaystyle\quad s_1+s_2 y+\cdots+s_k y^{k-1}+\frac{y^k \phi(y)}{1+p_1 y+\cdots+p_n y^n}$
the polynomial $\phi(y)$ being introduced in bringing the fractional terms
$$
\alpha_1^{k+1} /\left(1-\alpha_1 y\right)
$$
etc., to the common denominator (16).
In (20), we now set $Q=-p_1 y-\cdots-p_n y^n$. Thus
$$
\frac{1}{1+p_1 y+\cdots+p_n y^n} \equiv \sum_{r=0}^{k-1}(-1)^r\left(p_1 y+\cdots+p_n y^n\right)^r+\frac{y^k \psi(y)}{1+p_1 y+\cdots},
$$
where $\psi(y)$ is a polynomial. Expanding this $r$ th power by the multinomial theorem, we see that the left member of (19) equals
$$
\begin{array}{r}
d \sum(-1)^{r_1+\cdots+r_n+1} \frac{\left(r_1+\cdots+r_n\right)!}{r_{1}!\cdots r_{n}!} p_1^{r_1} \cdots p_n^{r_n} y^{r_1+2 r_2+\cdots+n r_n}+E \\
\left(d=p_1+2 p_2 y+\cdots\right),
\end{array}
$$
the sum extending over all integral values $\geqq 0$ of $r_1, r_2, \ldots, r_n$ such that $r_1+\cdots+r_n<k$, while $E$ is a fraction whose denominator is $1+p_1 y+\cdots$ and whose numerator is the product of $y^k$ by a polynomial in $y$. In the expansion of the part preceding $E$, the terms with the factor $y^k$ may be combined with $E$ after they are reduced to the same denominator as $E$. The resulting expression* is now of the same general form as (21), so that the coefficient of $y^{k-1}$ must equal $s_k$. This coefficient is the sum of
$$
\begin{aligned}
& \sum(-1)^{r_1+\cdots+r_n+1} \frac{\left(r_1+\cdots+r_n\right)!}{r_{1}!\ldots r_{n}!} p_1^{r_1+1} p_2^{r_2} \ldots p_n^{r_n} \\
& \left(r_1+2 r_2+\cdots+n r_n=k-1\right) \text {, } \\
& 2 \sum(-1)^{r_2+\cdots+r_n+1} \frac{\left(r_1+\cdots+r_n\right)!}{r_{1}!\ldots r_{n}!} p_1^{r_1} p_2{ }^{r_2+1} \ldots p_n{ }^{r_n} \\
& \left(r_1+2 r_2+\cdots+n r_n=k-2\right), \\
& 3 \sum(-1)^{r_1+\cdots+r_n+1} \frac{\left(r_1+\cdots+r_n\right)!}{r_{1}!\ldots r_{n}!} p_1^{r_1} p_2{ }^{r_2} p_3^{r_3+1} \ldots p_n^{r_n} \\
& \left(r_1+2 r_2+\cdots+n r_n=k-3\right) \text {, }
\end{aligned}
$$
In the first sum employ the summation index $r_1+1$ instead of $r_1$; in the second sum, $r_2+1$ instead of $r_2$; etc. We get
$$
\begin{aligned}
& \sum(-1)^{r_1+\cdots+r_n} \frac{\left(r_1+\cdots+r_n-1\right)!}{\left(r_1-1\right)!r_{2}!\ldots r_{n}!} p_1^{r_1} \ldots p_n^{r_n} \\
& 2 \sum(-1)^{r_1+\cdots+r_n} \frac{\left(r_1+\cdots+r_n-1\right)!}{r_{1}!\left(r_2-1\right)!\ldots r_{n}!} p_1^{r_1} \ldots p_n^{r_n} \\
& 3 \sum(-1)^{r_1+\cdots+r_n} \frac{\left(r_1+\cdots+r_n-1\right)!}{r_{1}!r_{2}!\left(r_3-1\right)!\ldots r_{n}!} p_1^{r_1} \ldots p_n^{r_n}
\end{aligned}
$$
where now (18) holds for each sum. Adding these sums, we evidently get the second member of (17). |
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tommywong
发表于 2024-11-14 14:12
