n倍角公式

tommywong

$\displaystyle cos n\theta =\frac{1}{2} \sum_{r=0}^{\lfloor \frac{n}{2} \rfloor} \frac{nC_{n-r}^r}{n-r} (-1)^r (2cos\theta)^{n-2r}$

$sin n\theta=?$

Tesla35

回复 1# tommywong

先确定$\sin n\theta$符号,
    $$\sin n\theta=\sqrt{1-\cos^2n\theta}$$
把上式代入即可

isee

复数三角形式……

战巡

回复 1# tommywong


en.wikipedia.org/wiki/Chebyshev_polynomials
第二类切比雪夫多项式，自己看吧，不多解释.........

tommywong

证明：

$\displaystyle \frac{a^{n+1}-b^{n+1}}{a-b}=\sum_{k=0}^{[\frac{n}{2}]} C_{n-k}^k (a+b)^{n-2k} (-ab)^k$

青青子衿

回复 5# tommywong

证明：
\(\displaystyle \frac{a^{n+1}-b^{n+1}}{a-b}=\sum_{k=0}^{[\frac{n}{2}]} C_{n-k}^k (a+b)^{n-2k} (-ab)^k\) ...
tommywong 发表于 2014-4-1 22:06

华林公式(Waring Formula)
fq.math.ca/Scanned/4-2/draim.pdf

hbghlyj

PlanetMath

Let $x_1,\ldots, x_n$ be $n$ indeterminates. For $k\geq 1$, let $\sigma_k$ be the $k$th elementary symmetric polynomials in $x_1,
\ldots, x_n$, and $S_k$ be the $k$th power sum defined as
\[
  S_k = \sum_{i=1}^n x_i^k.
\]
Like the Newton's formula, the Waring formula is a relation between $\sigma_k$ and $S_k$:
\[S_k = \sum (-1)^{(i_2+i_4+i_6+\ldots)} \frac{(i_1+i_2+\ldots+i_n-1)!k}{i_1!i_2!\cdots i_n!}
\sigma_1^{i_1} \sigma_2^{i_2} \cdots \sigma_n^{i_n},
\]
where the summation is over all $n$-tuples $(i_1,\ldots, i_n)\in\mathbb{Z}^n$ with non-negative components, such that
\[
i_1+2i_2+\ldots+ni_n = k.
\]
In particular, when there are two indeterminates, i.e. $n=2$, the Waring formula reads
\[
  x_1^k + x_2^k = \sum_{i=0}^{\lfloor k/2 \rfloor}
  (-1)^i\frac{k}{k-i}\binom{k-i}{i}(x_1+x_2)^{k-2i}(x_1x_2)^i.
\]

---

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 $i_1+2i_2+\ldots+ni_n=k$.