Waring公式

hbghlyj

青青子衿 发表于 2015-8-28 09:25
回复 5# tommywong
华林公式(Waring Formula)
fq.math.ca/Scanned/4-2/draim.pdf

但这篇只证明了Waring Formula的2元情形！文中的证明是对n归纳。使用\begin{aligned}
r_1^{k+1}+r_2^{k+1} & =\left(r_1^{k+1}+r_2^{k+1}+r_1 r_2^{k}+r_1^{k} r_2\right)-\left(r_1 r_2^{k}+r_1^{k} r_2\right) \\
& =\left(r_1^{k}+r_2^{k}\right)\left(r_1+r_2\right)-r_1 r_2\left(r_1^{k-1}+r_2^{k-1}\right) \\
& =p\left(r_1^{k}+r_2^{k}\right)+q\left(r_1^{k-1}+r_2^{k-1}\right)
\end{aligned}一般情形也能类似地证明吗？

hbghlyj

hbghlyj

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).

tommywong

維基條目Newton's identities有這兩條公式：

en.wikipedia.org/wiki/Newton%27s_identities#E … ymmetric_polynomials

Expressing power sums in terms of elementary symmetric polynomials

$\displaystyle p_m = (-1)^m m \sum_{r_1 + 2r_2 + \cdots + mr_m = m \atop r_1\ge 0, \ldots, r_m\ge 0} \frac{(r_1 + r_2 + \cdots + r_m - 1)!}{r_1!\,r_2! \cdots r_m!} \prod_{i=1}^m (-e_i)^{r_i}$

en.wikipedia.org/wiki/Newton%27s_identities#E … _terms_of_power_sums

Expressing elementary symmetric polynomials in terms of power sums

$\displaystyle e_n=(-1)^n  \sum_{m_1 + 2m_2 + \cdots + nm_n = n \atop m_1 \ge 0, \ldots, m_n \ge 0} \prod_{i=1}^n \frac{(-p_i)^{m_i}}{m_i ! \, i^{m_i}}$

我記得是我把inductive step寫上去的，不過其實我沒有在學校讀過專門研究多項式根和組合數的相關學科。我也忘記了什麼是Waring Formula，於是我又重新認識了一遍。

hbghlyj

tommywong 发表于 2024-11-14 06:12
維基條目Newton's identities有這兩條公式：

请问：Waring公式可以推广到分数幂吗？

如$\frac12$次幂
\[\sqrt x+\sqrt y=\sum _{i=0}^{\infty } 2^i \binom{\frac{1}{2}}{i} (x+y)^{\frac{1}{2}-i} (x y)^{i/2}\]

hbghlyj

发到 MSE 问问。

hbghlyj

$\frac13$次幂：
$$\root3\of x+\root3\of y=\sum _{i=0}^{\infty } \frac{  3^i }{1+i} \binom{\frac{i+1}{3}}{i}(x+y)^{\frac13-\frac{2 i}{3}} (x y)^{\frac i3}$$
那$\frac14$次幂呢？