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第一種完全楕円積分の微分を導け

Last updated at Posted at 2020-07-22

それはさておき

\begin{align*}

\eqalign{
  & p = \sqrt {1 - k^2 \sin ^2 \phi }   \cr 
  & k^2 \sin ^2 \phi  = 1 - p^2 ,\;\;\;\;\sin ^2 \phi  = k^{ - 2}  - k^{ - 2} p^2   \cr 
  & \frac{d}
{{d\phi }}p^m  =  - mk^2 p^{m - 2} \sin \phi \cos \phi  \cr} 

\end{align*} 
\begin{align*}

\eqalign{
  & \frac{d}
{{d\phi }}p^m \sin \phi \cos \phi  = \left( { - mk^2 p^{m - 2} \sin \phi \cos \phi } \right)\sin \phi \cos \phi  + p^m \left( {1 - 2\sin ^2 \phi } \right)  \cr 
  &  = \left( {1 - 2\sin ^2 \phi } \right)p^m  - mp^{m - 2} k^2 \sin ^2 \phi \cos ^2 \phi   \cr 
  &  = \left( {1 - 2k^{ - 2}  + 2k^{ - 2} p^2 } \right)p^m  - m\left( {1 - p^2 } \right)\left( {1 - k^{ - 2}  + k^{ - 2} p^2 } \right)p^{m - 2}   \cr 
  &  = \left( {1 - 2k^{ - 2} } \right)p^m  + 2k^{ - 2} p^{m + 2}  - m\left( {1 - k^{ - 2}  + k^{ - 2} p^2 } \right)p^{m - 2}  + m\left( {1 - k^{ - 2}  + k^{ - 2} p^2 } \right)p^m   \cr 
  &  = \left( {1 - 2k^{ - 2} } \right)p^m  + 2k^{ - 2} p^{m + 2}  - m\left( {1 - k^{ - 2} } \right)p^{m - 2}  - mk^{ - 2} p^m  + m\left( {1 - k^{ - 2} } \right)p^m  + mk^{ - 2} p^{m + 2}   \cr 
  &  =  - m\left( {1 - k^{ - 2} } \right)p^{m - 2}  + \left\{ {1 - 2k^{ - 2}  + m\left( {1 - 2k^{ - 2} } \right)} \right\}p^m  + \left( {m + 2} \right)k^{ - 2} p^{m + 2}   \cr 
  &  = m\left( {k^{ - 2}  - 1} \right)p^{m - 2}  + \left( {m + 1} \right)\left( {1 - 2k^{ - 2} } \right)p^m  + \left( {m + 2} \right)k^{ - 2} p^{m + 2}  \cr} 

\end{align*} 

$$
m\left( {1 - k^2 } \right)\int {p^{m - 2} d\phi } - \left( {m + 1} \right)\left( {2 - k^2 } \right)\int {p^m d\phi } + \left( {m + 2} \right)\int {p^{m + 2} d\phi } = p^m \sin \phi \cos \phi
$$

$$
m\left( {1 - k^2 } \right)\int_0^{\frac{\pi }
{2}} {p^{m - 2} d\phi } - \left( {m + 1} \right)\left( {2 - k^2 } \right)\int_0^{\frac{\pi }
{2}} {p^m d\phi } + \left( {m + 2} \right)\int_0^{\frac{\pi }
{2}} {p^{m + 2} d\phi } = 0
$$

これを次数の上げ下げに使う

\begin{align*}

\eqalign{
  &  - \left( {1 - k^2 } \right)\int_0^{\frac{\pi }
{2}} {p^{ - 3} d\phi }  + \int_0^{\frac{\pi }
{2}} {pd\phi }  = 0  \cr 
  & \int_0^{\frac{\pi }
{2}} {p^{ - 3} d\phi }  = \frac{{E\left( k \right)}}
{{1 - k^2 }}  \cr 
  & \frac{d}
{{dk}}p^{ - 1}  = k^{ - 1} \left( {p^{ - 3}  - p^{ - 1} } \right)  \cr 
  & \frac{d}
{{dk}}K\left( k \right) = \frac{1}
{k}\left\{ {\frac{{E\left( k \right)}}
{{1 - k^2 }} - K\left( k \right)} \right\} =  - \frac{{K\left( k \right)}}
{k} + \frac{{E\left( k \right)}}
{{k\left( {1 - k^2 } \right)}} \cr} 

\end{align*} 
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