把(1)式改写为 F=F(h)+kh+k2h2+k3h+、(3) 用h/2代替(3)式中的h得 hh F=F(=)+k+k2+k3+ 4 用2乘(4)式再减去(3)式消去含h的项得 h F=[F()+(F()-F(h)+k2(-h2)+k3(-h3) 2 2 2 令F(h)=F(h)且记 F2(h)=F1()+[F(=)-F1(h)把(1)式改写为 (3) 用h/2代替(3)式中的h,得 (4) 用2乘(4)式再减去(3)式,消去含h的项,得 (5) 令 ,且记 * 2 3 1 2 3 F F h k h k h k h = + + + + ( ) ... 2 3 * 1 2 3 ( ) ... 2 2 4 8 h h h h F F k k k = + + + + 2 3 * 2 3 2 3 [ ( ) ( ( ) ( ))] ( ) ( ) ... 2 2 2 2 h h h h F F F F h k h k h = + − + − + − + 1 F h F h ( ) ( ) = 2 1 1 1 ( ) ( ) [ ( ) ( )] 2 2 h h F h F F F h = + −