dE( k )/dk
= ( E( k )-K( k ) )/k
dK( k )/dk
= ( E( k )/k’^2-K( k ) )/k
⇒
E( k )
= ( 1+k’ )E( ( 1-k’ )/( 1+k’ ) )-k’K( k )
⇒
J( ( a+b )/2 , √ab )
= ( J( a , b )+abI( a , b ) )/2
これが肝!
因みに
K( k )
= ( 2/( 1+k’ ) )K( ( 1-k’ )/( 1+k’ ) ) : Landen
でした・・・