问答题 设[x] =x 0 .x 1 x 2 …x n ,求证:x=-x 0 +
【正确答案】正确答案:当x≥0时,x 0 =0, [x] =0.x 1 x 2 …x n = x i 2 -i =x 当x<0时,x 0 =1, [x] =1.x 1 x 2 …x n =2+x x=1.x 1 x 2 …=x n -2=-1+0.x 1 x 2 …x n =-1+ x i 2 i-1 综合上述两种情况,可得出x=-x 0 +
【答案解析】