从而有C4=0,c1=0 即y或→次多项式 1若l=k=2n,n=0,1,2 则a2=C2=Cn12=0 n+4 n+6 =..=0 yo(x) C +cx+.+C, x n n =cn+c,x2+…+cx→>次多项式 y,(x) 2n+1 2n+3 Cx+cx+.+ 2n+1 + C →无穷级数4 6 0, 0... k k c c 从而有 + + = = 0 1 即y 或y l → 项 次多 式 1.若l = k = = 2n n, 0,1,2,... 2 2 2 2 0 k l n c c c 则 + = + + = = 2 4 2 6 ... 0 n n c c \ + + = = = ( ) 2 2 0 0 2 2 ... n n y x = c + c x + + c x 2 0 2 ... l l = c + c x + + ® c x l次多项式 ( ) 3 2 1 2 3 1 1 3 2 1 2 3 ... ... n n n n y x c x c x c x c x + + = + + + + + + + → 无穷级数