注1：若 lim B=0 I 则 k-→o (I-B)1=I+B+B2+…十Bk+… 事实上 (I-B)(I+B+B2+…+Bk)=I-Bk+1 注2：X=BX-)+f=B(BXk-2)+)+f=… =BkX0)+（I+B++Bk-1)f ≈(I-B)1f 66 : 若 lim  0 则  k k B ( I - B) -1 = I + B + B2 + ······ + Bk + ······ 事实上 ( I - B)( I + B + B2 + ······ + Bk ) =I – Bk+1 : X(k) =B X(k-1) + f = B(B X(k-2) + f) + f =···· = Bk X(0) + ( I + B + ····+ Bk-1)f ≈ ( I – B ) -1 f