Question

已知函数f(x),g(x)同时满足：g(x-y)=g(x)g(y)+f(x)f(x); f(-1)=-1, f(0)=0, f(1)=1, 。。。

已知函数f(x),g(x)同时满足：g(x-y)=g(x)g(y)+f(x)f(y); f(-1)=-1, f(0)=0, f(1)=1,
求g(0), g(1), g(2)的值。
求解题过程，，

Answer 1

g(0)=g(1-1)=g(1)g(1)+f(1)f(1)　得：g(0)=g(1)g(1)+1``` ```````````````1
g(1)=g(1-0)=g(1)g(0)+f(1)f(0) 得：g(1)=g(1)g(0)````````````````````````2
g(2)=g[1-(-1)]=g(1)g(-1)+f(1)f(-1) 得：g(2)=g(1)g(-1)-1```````````````3
联立1、2、3得：
g(0)=1, g(1)=0, g(2)=-1

Answer 2

g(x-y) = g(x)g(y) + f(x)f(y) f(0)=0 f(1)=1
g(0-0) = g ² (0) + f ² (0) => g(0) = 0 或 g(0) = 1
g(1-1) = g ² (1) + f ² (1) => g(0) = g ² (1) + 1
于是 g(0) = 1, g(1) = 0
又 g(0) = g(-1- (-1)) = g ² (-1) + f ² (-1) ， f(-1) = -1
=> g(-1) = 0
g(2) = g(1- (-1)) = g(1)g(-1) + f(1)f (-1) = 0 + (-1) = -1
综上：g(0) = 1, g(1) = 0, g(2) = -1

Answer 3

令y=0，g(x)=g(x)g(0)+f(x)f(0)=g(x)g(0)得g(0)=1
令x=1，y=1，g(0)=g(1)g(1)+f(1)f(1)=(1)g(1)+1=1得g(1)=0
令x=1，y=-1，g(2)=g(1)g(-1)+f(1)f(-1)=0+（-1）=-1