tan(nθ)表为tan(θ)

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$n$为正整数时，$\tan(nθ)$可表为有限连分数（n阶）
$\displaystyle \tan(nx) = \cfrac{n\tan x}{1 -\cfrac{(n^{2} - 1^{2})\tan^{2}x}{3 -\cfrac{(n^{2} - 2^{2})\tan^{2}x}{5 -\cfrac{(n^{2} - 3^{2})\tan^{2}x}{7 -\lower2ex\ddots\lower4ex\dfrac{(n^{2} - (n - 1)^{2})\tan^{2}x}{(2n - 1)} }}}}$
如$\displaystyle \tan 3x = \cfrac{3\tan x}{1 -\cfrac{8\tan^{2}x}{3-\cfrac{5\tan^{2}x}{5}}}$

爪机专用

还能这样表示😃

PS、相关帖子：
kuing.cjhb.site/forum.php?mod=viewthread&tid=11782
kuing.cjhb.site/forum.php?mod=viewthread&tid=11974&page=1#pid58059

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睡神 发表于 2024-3-28 02:42
高斯连分数！牛13

谢谢您提到高斯超几何函数。我找到了一篇相关文章：
tandfonline.com/doi/10.1080/0035919X.2012.727363

hbghlyj

$$\tan (n \arctan (t))=\frac{1}{i}\cdot\frac{(1+i t)^n-(1-i t)^n}{(1+i t)^n+(1-i t)^n}$$
等价于
$$\tan (n x)=\frac{\sum_{0 \leq k \leq n / 2}(-1)^k\binom{n}{2 k+1} \tan ^{2 k+1}(x)}{\sum_{0 \leq k<n / 2}(-1)^k\binom{n}{2 k} \tan ^{2 k}(x)}$$
除以$\tan(x)$，
$$\frac{\tan (n x)}{\tan (x)}=\frac{\sum_{0 \leq k \leq n / 2}(-1)^k\binom{n}{2 k+1} \tan ^{2 k}(x)}{\sum_{0 \leq k<n / 2}(-1)^k\binom{n}{2 k} \tan ^{2 k}(x)}$$
设 $z=-\tan ^2(x)$，则分子和分母写为 $f(x)=\sum_{k \geq 0}\binom{n}{2 k+1} z^k$ 和 $g(z)=\sum_{k \geq 0}\binom{n}{2 k} z^k$，上述方程变为
$$\frac{\tan (n x)}{\tan (x)}=\frac{f(z)}{g(z)}$$最后运用高斯连分数公式即可。

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hbghlyj 发表于 2024-9-24 15:13
最后运用高斯连分数公式即可。

下面是高斯连分数公式的证明：

需要用到₂F₁(a,b,c;z)函数的 contiguous relation：\begin{equation}\label{c1}
F\left(\left.\begin{array}{c}a, b \\ c\end{array} \right\rvert\, z\right)=F\left(\left.\begin{array}{c}a, b+1 \\ c+1\end{array} \right\rvert\, z\right)-\frac{a(c-b)}{c(c+1)} z F\left(\left.\begin{array}{c}a+1, b+1 \\ c+2\end{array} \right\rvert\, z\right)
\end{equation}和它的变体\begin{equation}\label{c2}
F\left(\left.\begin{array}{c}
a, b+1 \\
c+1
\end{array}\right| z\right)=  F\left(\left.\begin{array}{c}
a+1, b+1 \\
c+2
\end{array} \right\rvert\, z\right)  -\frac{(b+1)(c+1-a)}{(c+1)(c+2)} z F\left(\left.\begin{array}{c}
a+1, b+2 \\
c+3
\end{array} \right\rvert\, z\right)
\end{equation}用$F\left(\left.\begin{array}{c}a, b+1 \\ c+1\end{array} \right\rvert\, z\right)$除以\eqref{c1}，
\begin{equation}\label3\frac{F\left(\left.\begin{array}{c}a, b+1 \\ c+1\end{array} \right\rvert\, z\right)}{F\left(\left.\begin{array}{c}a, b \\ c\end{array} \right\rvert\, z\right)}=\cfrac1{1-\cfrac{a(c-b)}{c(c+1)} z \cfrac{F\left(\left.\begin{array}{c}a+1, b+1 \\ c+2\end{array} \right\rvert\, z\right)}{F\left(\left.\begin{array}{c}a, b+1 \\ c+1\end{array} \right\rvert\, z\right)}}
\end{equation}用$F\left(\left.\begin{array}{c}a+1, b+1 \\ c+2\end{array} \right\rvert\, z\right)$除以\eqref{c2}，
\begin{equation}\label4
\frac{F\left(\left.\begin{array}{c}
a+1, b+1 \\
c+2
\end{array} \right\rvert\, z\right)}{F\left(\left.\begin{array}{c}
a, b+1 \\
c+1
\end{array} \right\rvert\, z\right)}=\cfrac{1}{1-\cfrac{(b+1)(c-a+1)}{(c+1)(c+2)} z \cfrac{F\left(\left.\begin{array}{c}
a+1, b+2 \\
c+3
\end{array} \right\rvert\, z\right)}{F\left(\left.\begin{array}{c}
a+1, b+1 \\
c+2
\end{array} \right\rvert\, z\right)}}
\end{equation}
\eqref{4}代入\eqref{3}
$$
\frac{F\left(\left.\begin{array}{c}
a, b+1 \\
c+1
\end{array} \right\rvert\, z\right)}{F\left(\left.\begin{array}{c}
a, b \\
c
\end{array} \right\rvert\, z\right)}=\cfrac{1}{1-\cfrac{\cfrac{a(c-b)}{c(c+1)} z}{1-\cfrac{(b+1)(c-a+1)}{(c+1)(c+2)} z \cfrac{F\left(\left.\begin{array}{c}
a+1, b+2 \\
c+3
\end{array} \right\rvert\, z\right)}{F\left(\left.\begin{array}{c}
a+1, b+1 \\
c+2
\end{array} \right\rvert\, z\right)}}}
$$
可以再次使用\eqref{3}，但将 $a$、$b$ 和 $c$ 替换为 $a+1$、$b+1$ 和 $c+2$：
$$
\frac{F\left(\left.\begin{array}{c}
a, b+1 \\
c+1
\end{array} \right\rvert\, z\right)}{F\left(\left.\begin{array}{c}
a, b \\
c
\end{array} \right\rvert\, z\right)}=\cfrac{1}{1-\cfrac{\cfrac{a(c-b)}{c(c+1)} z}{1-\cfrac{\cfrac{(b+1)(c-a+1)}{(c+1)(c+2)} z }{
1-\cfrac{(a+1)(c-b+1)}{(c+2)(c+3)} z \cfrac{F\left(\left.\begin{array}{c}a+2, b+2 \\ c+4\end{array} \right\rvert\, z\right)}{F\left(\left.\begin{array}{c}a+1, b+2 \\ c+3\end{array} \right\rvert\, z\right)}
}}}
$$
继续交替使用\eqref{4}\eqref{3}得连分式
$$
\frac{F\left(\left.\begin{array}{c}
a, b+1 \\
c+1
\end{array} \right\rvert\, z\right)}{F\left(\left.\begin{array}{c}
a, b \\
c
\end{array} \right\rvert\, z\right)}
=\cfrac{1}{1-\cfrac{\cfrac{a(c-b)}{c(c+1)}z}{1-\cfrac{\cfrac{(b+1)(c-a+1)}{(c+1)(c+2)}z}{1-\cfrac{\cfrac{(a+1)(c-b+1)}{(c+2)(c+3)}z}{1-\cfrac{\cfrac{(b+2)(c-a+2)}{(c+3)(c+4)}z}{1-\ddots}}}}}$$

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hbghlyj 发表于 2024-9-24 16:41
需要用到₂F₁(a,b,c;z)函数的 contiguous relation：

\eqref{c1}的证明：
\begin{aligned}
\frac{a^{\bar{n}}(b+1)^{\bar{n}}}{(c+1)^{\bar{n}} n!} & -\frac{a(c-b)}{c(c+1)} \frac{(a+1)^{\overline{n-1}}(b+1)^{\overline{n-1}}}{(c+2)^{\overline{n-1}}(n-1)!} \\
& =\frac{(a+1)^{\overline{n-1}}(b+1)^{\overline{n-1}}}{(c+2)^{\overline{n-1}} n!}\left[\frac{a(b+n)}{c+1}-\frac{a(c-b) n}{c(c+1)}\right] \\
& =\frac{(a+1)^{\overline{n-1}}(b+1)^{\overline{n-1}}}{(c+2)^{\overline{n-1}} n!} \frac{a b(c+n)}{c(c+1)}=\frac{a^{\bar{n}} b^{\bar{n}}}{c^{\bar{n}} n!}
\end{aligned}对 $n\ge0$ 求和得到\eqref{c1}.

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文章第 3 部分是对 tan(nx) 的连分式展开式的另一种推导，没有使用超几何函数。这种技术已经产生过许多其他漂亮的展开式 [引用3、引用4]。