求椭圆周长的积分式和无穷级数展开式

luyuanhong

永远

陆老师好，红圈部分应该是小于2π吧

永远

呼叫陆老师

luyuanhong

谢谢楼上 永远 指出我的笔误！现已更正。

永远

求椭圆周长的积分式和无穷级数展开式

一个长半轴为\(a\)、短半轴为\(b\)的椭圆，参数方程为\(\left\{ {\begin{array}{*{20}{c}}
  {x = a\cos \theta } \\
  {y = b\sin \theta }
\end{array}} \right.\)（\(0 \leqslant \theta  < 2\pi\)）。
由于对称性，我们求椭圆周长时，只要先求\(0 \leqslant \theta  < \frac{\pi }{2}\)的一段，再乘以\(4\)就可以了。
将\(\theta\)的取值范围\([0,\frac{\pi }{2})\)等分成\(n\)段，相应地，椭圆弧也分成了\(n\)段。椭圆弧上第\(k\)段两端的坐标为\({P_{k - 1}}(a\cos \frac{{(k - 1)\pi }}{{2n}},b\sin \frac{{(k - 1)\pi }}{{2n}})\)和\({P_k}(a\cos \frac{{k\pi }}{{2n}},b\sin \frac{{k\pi }}{{2n}})\)。
连接\({P_0}{P_1}\;,\;{P_1}{P_2}\;, \cdots ,\;{P_{n - 1}}{P_n}\)，成为一条内接于椭圆的折线，这段折线的全长为
\(\small\begin{gathered}
  \sum\limits_{k = 1}^n {\overline {{P_{k - 1}}{P_k}} }  = \sum\limits_{k = 1}^n {\sqrt {{a^2}{{[\cos \frac{{k\pi }}{{2n}} - \cos \frac{{(k - 1)\pi }}{{2n}}]}^2} + {b^2}{{[\sin \frac{{k\pi }}{{2n}} - \sin \frac{{(k - 1)\pi }}{{2n}}]}^2}} }  \hfill \\
  \quad \quad \quad \quad \quad \;\; = \sum\limits_{k = 1}^n {\sqrt {{a^2}{{[ - 2\sin \frac{{(2k - 1)\pi }}{{4n}}\sin \frac{\pi }{{4n}}]}^2} + {b^2}{{[2\cos \frac{{(2k - 1)\pi }}{{4n}}\sin \frac{\pi }{{4n}}]}^2}} }  \hfill \\
  \quad \quad \quad \quad \quad \quad \quad \quad \quad  = \frac{\pi }{{2n}}\sum\limits_{k = 1}^n {\sqrt {{a^2}{{[\sin \frac{{(2k - 1)\pi }}{{4n}}\sin \frac{\pi }{{4n}}/\frac{\pi }{{4n}}]}^2} + {b^2}{{[\cos \frac{{(2k - 1)\pi }}{{4n}}\sin \frac{\pi }{{4n}}/\frac{\pi }{{4n}}]}^2}} }  \hfill \\ \end{gathered} \)
令\(n \to \infty\)，这段内接于椭圆的折线长就会无限接近于椭圆的弧长\(C\)。
由\(\mathop {\lim }\limits_{n \to \infty } \sin \frac{\pi }{{4n}}/\frac{\pi }{{4n}} = 1\)及定积分定义\(\;\displaystyle\mathop {\lim }\limits_{n \to \infty } \frac{\pi }{{2n}}\sum\limits_{k = 1}^n {f(\frac{{(k - \frac{1}{2})\pi }}{{2n}})}  = \int_0^{\frac{\pi }{2}} {f(\theta )d\theta }\)可知有（为方便研究，这里我们令：\(\lambda  = \tfrac{{a - b}}{{a + b}}\)）：\[\;\displaystyle\begin{align}
C &= \mathop {\lim }\limits_{n \to \infty } \sum\limits_{k = 1}^n {\overline {{P_{k - 1}}{P_k}} }  \\
&= \mathop {\lim }\limits_{n \to \infty } \frac{\pi }{{2n}}\sum\limits_{k = 1}^n {\sqrt {{a^2}{{\sin }^2}\frac{{(2k - 1)\pi }}{{4n}} + {b^2}{{\cos }^2}\frac{{(2k - 1)\pi }}{{4n}}} }  \\
&= \int_0^{\frac{\pi }{2}} {\sqrt {{a^2}{{\sin }^2}\theta  + {b^2}{{\cos }^2}\theta } \;d\theta }  \\
&= \int_0^{\frac{\pi }{2}} {\sqrt {{a^2}\left( {1 - {{\cos }^2}\theta } \right) + {b^2}{{\cos }^2}\theta } } d\theta  \\
&= \int_0^{\frac{\pi }{2}} {\sqrt {{a^2} - {a^2}{{\cos }^2}\theta  + {b^2}{{\cos }^2}\theta } } d\theta  \\
&= \int_0^{\frac{\pi }{2}} {\sqrt {{a^2} - {a^2}\tfrac{{1 + \cos 2\theta }}{2} + {b^2}\tfrac{{1 + \cos 2\theta }}{2}} } d\theta  \\
&= \int_0^{\frac{\pi }{2}} {\sqrt {{a^2} - \tfrac{{{a^2}}}{2} - \tfrac{{{a^2}}}{2}\cos 2\theta  + \tfrac{{{b^2}}}{2} + \tfrac{{{b^2}}}{2}\cos 2\theta } } d\theta  \\
  &= \int_0^{\frac{\pi }{2}} {\sqrt {\tfrac{{{a^2}}}{2} - \tfrac{{{a^2}}}{2}\cos 2\theta  + \tfrac{{{b^2}}}{2} + \tfrac{{{b^2}}}{2}\cos 2\theta } } d\theta  \\
&= \int_0^{\frac{\pi }{2}} {\sqrt {\tfrac{{{a^2} + {b^2}}}{2} - \tfrac{{{a^2} - {b^2}}}{2}\cos 2\theta } } d\theta  \\
&= \int_0^{\frac{\pi }{2}} {\sqrt {\tfrac{{2{a^2} + 2{b^2}}}{4} - \tfrac{{2{a^2} - 2{b^2}}}{4}\cos 2\theta } } d\theta  \\
&= \frac{1}{2}\int_0^{\frac{\pi }{2}} {\sqrt {\tfrac{{{{\left( {a + b} \right)}^2} + {{\left( {a - b} \right)}^2}}}{4} - \tfrac{{2\left( {a - b} \right)\left( {a + b} \right)}}{4}\cos 2\theta } } d\theta  \\
&= \frac{1}{2}\int_0^{\frac{\pi }{2}} {\sqrt {{{\left( {a + b} \right)}^2} + {{\left( {a - b} \right)}^2} - 2\left( {a - b} \right)\left( {a + b} \right)\cos 2\theta } } d\theta  \\
&= \frac{1}{2}\int_0^{\frac{\pi }{2}} {\sqrt {{{\left( {a + b} \right)}^2}\left[ {1 + {{\left( {\tfrac{{a - b}}{{a + b}}} \right)}^2} - 2\tfrac{{a - b}}{{a + b}}\cos 2\theta } \right]} } d\theta  \\
&= \frac{1}{2}\left( {a + b} \right)\int_0^{\frac{\pi }{2}} {\sqrt {1 + {{\left( {\tfrac{{a - b}}{{a + b}}} \right)}^2} - 2\tfrac{{a - b}}{{a + b}}\cos 2\theta } } d\theta  \\
  &= \frac{1}{2}\left( {a + b} \right)\int_0^{\frac{\pi }{2}} {\sqrt {1 + {\lambda ^2} - 2\lambda \cos 2\theta } } d\theta  \\
&= \frac{1}{2}\left( {a + b} \right)\int_0^{\frac{\pi }{2}} {\sqrt {1 + {\lambda ^2} - \lambda \left( {{e^{2i\theta }} + {e^{ - 2i\theta }}} \right)} } d\theta  \\
&= \frac{1}{2}\left( {a + b} \right)\int_0^{\frac{\pi }{2}} {\sqrt {1 + {\lambda ^2} - \lambda {e^{2i\theta }} - \lambda {e^{ - 2i\theta }}} } d\theta  \\
&= \frac{1}{2}\left( {a + b} \right)\int_0^{\frac{\pi }{2}} {\sqrt {1 + {\lambda ^2}{e^{2i\theta  - 2i\theta }} - \lambda {e^{2i\theta }} - \lambda {e^{ - 2i\theta }}} } d\theta  \\
&= \frac{1}{2}\left( {a + b} \right)\int_0^{\frac{\pi }{2}} {\sqrt {1 + \lambda {e^{2i\theta }}\lambda {e^{ - 2i\theta }} - \lambda {e^{2i\theta }} - \lambda {e^{ - 2i\theta }}} } d\theta  \\
&= \frac{1}{2}\left( {a + b} \right)\int_0^{\frac{\pi }{2}} {\sqrt {\left( {1 - \lambda {e^{2i\theta }}} \right) + \left( {\lambda {e^{2i\theta }}\lambda {e^{ - 2i\theta }} - \lambda {e^{ - 2i\theta }}} \right)} } d\theta  \\
&= \frac{1}{2}\left( {a + b} \right)\int_0^{\frac{\pi }{2}} {\sqrt {\left( {1 - \lambda {e^{2i\theta }}} \right) + \lambda {e^{ - 2i\theta }}\left( {\lambda {e^{2i\theta }} - 1} \right)} } d\theta  \\
&= \frac{1}{2}\left( {a + b} \right)\int_0^{\frac{\pi }{2}} {\sqrt {\left( {1 - \lambda {e^{2i\theta }}} \right) - \lambda {e^{ - 2i\theta }}\left( {1 - \lambda {e^{2i\theta }}} \right)} } d\theta  \\
  &= \frac{1}{2}\left( {a + b} \right)\int_0^{\frac{\pi }{2}} {\sqrt {\left( {1 - \lambda {e^{2i\theta }}} \right)\left( {1 - \lambda {e^{ - 2i\theta }}} \right)} } d\theta  \\
&= \frac{1}{2}\left( {a + b} \right){\int_0^{\frac{\pi }{2}} {\left( {1 - \lambda {e^{2i\theta }}} \right)} ^{\tfrac{1}{2}}}{\left( {1 - \lambda {e^{ - 2i\theta }}} \right)^{\tfrac{1}{2}}}d\theta  \\
&= \frac{1}{2}\left( {a + b} \right)\int_0^{\frac{\pi }{2}} {\sum\limits_{k = 0}^{ + \infty } {C_{\tfrac{1}{2}}^{\;k}} } {\left( { - \lambda {e^{2i\theta }}} \right)^{\;k}}\sum\limits_{j = 0}^{ + \infty } {C_{\tfrac{1}{2}}^{\;j}} {\left( { - \lambda {e^{ - 2i\theta }}} \right)^{\;j}}d\theta  \\
&= \frac{1}{2}\left( {a + b} \right)\int_0^{\frac{\pi }{2}} {\sum\limits_{k = 0}^\infty  {(_{\;k}^{1/2}){{( - \lambda )}^k}{e^{2k{\kern 1pt} \theta {\kern 1pt} i}}} \sum\limits_{j = 0}^\infty  {(_{\;j}^{1/2}){{( - \lambda )}^j}{e^{ - 2j{\kern 1pt} \theta {\kern 1pt} i}}} d\theta }  \\
&= \frac{1}{2}\left( {a + b} \right)\int_0^{\frac{\pi }{2}} {\sum\limits_{k = 0}^\infty  {\sum\limits_{j = 0}^\infty  {(_{\;k}^{1/2})(_{\;j}^{1/2}){{( - \lambda )}^{k + j}}{e^{2k\theta i}}{e^{ - 2j\theta i}}} } d\theta }  \\
&= \frac{1}{2}\left( {a + b} \right)\int_0^{\frac{\pi }{2}} {\sum\limits_{n = 0}^\infty  {\sum\limits_{k = 0}^n {(_{\;k}^{1/2})(_{\;n - k}^{1/2}){{( - \lambda )}^n}\cos [2(2k - n)\theta ]} } d\theta }  \\
&= \frac{1}{2}\left( {a + b} \right)\sum\limits_{n = 0}^\infty  {\sum\limits_{k = 0}^n {(_{\;k}^{1/2})(_{\;n - k}^{1/2}){{( - \lambda )}^n}} } \int_0^{\frac{\pi }{2}} {\cos [2(2k - n)\theta ]d\theta }  \\
\end{align}\]
其中，因为已知\(0 < \lambda  < 1\)，所以\(\left| {\,{{( - \lambda )}^n}\,} \right| < 1\)，再加上有\(\left| {\,\cos [2(2k - n)]\,} \right| \leqslant 1\)，所以双重求和式\(\sum\limits_{n = 0}^\infty  {\sum\limits_{k = 0}^n {(_{\;\,k}^{1/2})(_{\;n - k}^{\;1/2}){{( - \lambda )}^n}\cos [2(2k - n)]} } \)一致收敛，因此求和号与积分号可以互相交换。
当\(2k \ne n\)时，\(\;\displaystyle\int_0^{\frac{\pi }{2}} {\cos [2(2k - n)]d\theta }  = \left. {\frac{1}{{2(2k - n)}}\sin [2(2k - n)\theta ]} \right|_0^{\frac{\pi }{2}} = 0\)。

当\(2k = n\)时，\(\;\displaystyle\int_0^{\frac{\pi }{2}} {\cos [2(2k - n)]d\theta }  = \int_0^{\frac{\pi }{2}} {\cos 0d\theta }  = \int_0^{\frac{\pi }{2}} {1d\theta }  = \frac{\pi }{2}\)。
由此可见，在求和式\(\sum\limits_{n = 0}^\infty  {\sum\limits_{k = 0}^n {(_{\;\,k}^{1/2})(_{\;n - k}^{\;1/2}){{( - \lambda )}^n}\cos [2(2k - n)]} } \)中，\(2k = n\)的项都是\(0\)，只有\(2k = n = m\)的项不为\(0\)，所以双重求和式可以变成一重求和式，有
\(\begin{gathered}
  \frac{1}{2}\left( {a + b} \right)\sum\limits_{n = 0}^\infty  {\sum\limits_{k = 0}^n {(_{\;k}^{1/2})(_{\;n - k}^{{\kern 1pt} \;1/2}){{( - \lambda )}^n}} } \int_0^{\frac{\pi }{2}} {\cos [2(2k - n)]d\theta }  = \frac{\pi }{4}\left( {a + b} \right)\sum\limits_{m = 0}^\infty  {(_{\,{\kern 1pt} m}^{1/2})(_{\,{\kern 1pt} m}^{1/2}){{( - \lambda )}^{m + m}}}  \hfill \\
  \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad  = \frac{\pi }{4}\left( {a + b} \right)\sum\limits_{m = 0}^\infty  {{{(_{\,{\kern 1pt} m}^{1/2})}^2}{\lambda ^{2m}}}  \hfill \\
\end{gathered} \)
这就是求\(\frac{1}{4}\)椭圆周长的积分公式。

永远

对楼上陆教授贴子的整理