> 数学 >
对点(x,y)的一次操作变换记为P1(x,y).定义其变换法则如下:
P1(x,y)=(x+y,x-y);且规定Pn(x,y)=P1[Pn-1(x,y)] (n为大于1的整数)
例如,P1=(3,-1),P2(1,2)=P1[P1(1,2)]=P1(3,-1)=(2,4),
P3(1,2)=P1[P2(1,2)]=P1(2,4)=(6,-2).
(1)P1(1,-1)=( _ ,_ ),P2(1,-1)=P1[P1(1,-1)]=P1( _ ,_ )=( _ ,_ ),
P3(1,-1)=P1[P2(1,-1)]=P1( _ ,_ )=( _ ,_ ) ,
P4(1,-1)=P1[P3(1,-1)]=P1( _ ,_ )=( _ ,_ );
(2)试根据(1)的规律求P5(1,-1),P6(1,-1),P2013(1,-1).
人气:370 ℃ 时间:2019-08-21 06:52:25
解答
P1(1,-1)=(1-1,1+1)=(0,2)P2(1,-1)=P1(P1(1,-1))=P1(0,2)=(2,-2)P3(1,-1)=.=P1(2,-2)=(0,4)P4(1,-1)=.=P1(0,-4)=(4,-4)所以P5=(0,8)P6=(8,-8)显然n为奇数 Pn=(0,2^((n+1)/2))n为偶...
推荐
猜你喜欢
© 2024 79432.Com All Rights Reserved.
电脑版|手机版