Question

高斯函数问题解方程[3-2x]+[(x+1)/2]=2

Answer 1

好久没有做这种题了,只想出来一个很笨的解法.令 t = (x+1)/2, 设t = n+s, n为t的整数部分,0 ≤ s < 1, [s]=0.原方程化为[5-4t] + [t] = 2[5-4n-4s]+[t] =2[1-4n+4(1-s)] +[t] =21-4n +[4(1-s)] +n+[s] =2[4(1-s)] = 3n+1 0 < 4(1-s) ≤4, 所以n只能为 0 或 1n=0时, [4(1-s)] = 1,1≤ [4(1-s)] <2, 1/2< t≤ 3/4, 即0