“…(44)) b11 ( 1 ) = a 11 (s 0 ) + A 2,00 ( 1 − 2 1 )= 1 − 1 2p |α m | p−2 α m α m (p − 1 + 1 ) + (p − 1)( 1 − 1)α 2 m +A 2,00 ( 1 − 2 1 ) − |B(2m)|(45)and, for j ≥ 2,b 1j ( 1 ) = a 1j (s 0 ) − A 2,00 (1 − j )(1 − 2 1 ) = ( j − 1) |α m | p−2 2p α m α m (2 1 + p − 2) denoting by B k,1j the coefficient of k 1 in b 1j , j ≥ 1, we have: -Constant term in b 11 : This is B 0,11 = b 11 (0) = a 11 (s 0 ) m | p−2 α m α m (p − 1) − (p − 1Coefficient of 1 in b 11 : From (45) we obtain b 11 ( 1 ) = |α m | p−2 2p α m α m (2 1 + p − 2) + (p − 1)α 2 m (2 1 − 2) + A 2,00 (1 − 2 1 ) (47)and therefore the coefficient isB 1,11 = b 11 (0) = |α m | p−2 2p (p − 2)α m α m − 2(p − 1)α 2 m + A 2,00 .-Coefficient of 2 1 in b11 : From (47),B 2,11 = 1 2 b 11 (0) = |α m | p−2 2pα m α m + (p − 1)α 2 m − A 2,00 = 0.…”