MathLabs

第6题

函数 f(x,y)f(x,y) 对所有非负整数 x,yx,y 满足 f(0,y)=y+1,f(0,y) = y+1, f(x+1,0)=f(x,1),f(x+1,0) = f(x,1), f(x+1,y+1)=f(x,f(x+1,y))f(x+1,y+1) = f\big(x, f(x+1,y)\big) 求 f(4,1981)f(4,1981)。
第 3/7 步：求解 x=2x=2 层
上一步下一步
f(2,0)=f(1,1)=3,f(2,y+1)=f(1,f(2,y))=f(2,y)+2  ⟹  f(2,y)=2y+3f(2,0)=f(1,1)=3,\quad f(2,y+1)=f(1,f(2,y))=f(2,y)+2 \implies f(2,y)=2y+3
详细分析

由规则 (2) 及第 2 步，f(2,0)=f(1,1)=1+2=3f(2,0)=f(1,1)=1+2=3。由规则 (3) 再次利用第 2 步，f(2,y+1)=f(1,f(2,y))=f(2,y)+2f(2,y+1)=f(1,f(2,y))=f(2,y)+2。以初值 33 求解此递推得 f(2,y)=2y+3f(2,y)=2y+3。