A=2aR+R(aos叭 由 R0=有限→bn=0 两=2 (9) 由(8)得 由 R=-HoRcos0 所以 aRP=-HoRCOS0 a1=-H0an=0 ks0-含致 (10) 由(6)(7)式得 含Rga0+2 w..p...cop 0R+2 [刊当n=0时有 =bo ao Ro a0=0 → 0= -bo bo=0 R [2]当n=1时有 aRRe-,R出+0P R 4=二3H。 4+246 P=-HP2P Ro A=-twRH。 4+24 P (cos ) R b a R n 0 n 1 n n 1  n  = +       = + n 由 1 R→0 = 有限bn = 0 n n n anR P  =  = 0 1 (9) 由(8)得 n n n n n P R b a R       = + 2 +1  由 2 R→ = −H0Rcos 所以 a1RP1 = −H0Rcos a1 = −H0 an = 0 n n 0 n 1 n 2 0 P R b H Rcos   = +  = −  + (10) 由 (6) (7)式得 ( )        + = − + = − +      =  = + −  = +  = n 0 n 0 n 2 n 0 n n 0 0 0 n 1 n 0 n 0 n 1 n 0 n n 0 n 0 R 0 n P R n 1 b na R P H cos P R b a R P H Rcos      1 当 n=0 时有        − = = 2 0 0 0 0 0 R b 0 R b a    = =  b 0 a 0 0 0 2 当 n=1 时有        − = − + = − + 3 1 0 0 1 1 0 0 1 2 1 0 1 1 0 1 0 0 1 R 2b a P H P P R b a R P H R P P    ( )        + − + = + − =  0 0 3 0 0 1 0 0 0 1 2 2 3 H R b a H       