4.递推公式 dJ,(x)、d 2k 2k 2k dx dx k=0 k!r(v+k+1)2 k!r(v+k+1)2 k=l+1 ∑ l+1 +2+12+1 =0 (+1)!r(v+l+2)2 ()+1+21x+21=-21(x) l!(v+1++1)2 x dx 对一v、m诺依曼函数、汉克尔函数满足同样关系。写作2(x) d rz,(x Z+1(x) Z-1Z./x=-Z Z-Z=2Z x [x"z,(x)=x"z(x)2+1n/x=2 Z-21z,/x+Z1=04. 递推公式 k k k k x dx k k d x J x dx d 2 2 0 ) 2 1 ( ! ( 1) 1 ) ( 1) ( ) ( +  =  + + =  −     2 2 1 1 ) 2 1 ( ! ( 1) 2 ( 1) + −  =  + + =  − k k k k x k k k   2 1 2 1 0 1 ) 2 1 ( ( 1)! ( 2) 1 ( 1) + + +  = + +  + + + =  − l l l l x l l l   k = l +1 l l l l x x l l 1 2 1 2 0 ) 2 1 ( ! ( 1 1) 1 ( 1) 1 + + + +  =  + + + = −  −       x J (x) +1 = − [ ( )] ( ) 1 x J x x J x dx d =  −    对 − 、m 诺依曼函数、汉克尔函数满足同样关系。 写作 Z (x)      x Z x x Z x dx d ( ) ] ( ) [ +1 = − [ ( )] ( ) 1 x Z x x Z x dx d =  −    1 ' /  −  = −Z + Z Z x 1 ' /  +  = Z − Z Z x 2 ' Z −1 − Z +1 = Z Z +1 − 2Z / x + Z −1 = 0