1.已知f（x）=ax²+bx+c,f(0)=0,且f(x+1)=f(x)+x+1,试求f(x)的表达式。

Question

2.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(1)g(2)g(3)的值。 帮我解下这两道题，步骤写详细点，我是初三升高中，还没正式上课，尽量写得通俗易懂。

Answer 1

1. 在f(x) = ax²+bx+c中取x = 0得c = f(0) = 0, 故f(x) = ax²+bx.
于是f(x+1)-f(x) = a(x+1)²+b(x+1)-(ax²+bx) = a((x+1)²-x²)+b = a(2x+1)+b = 2ax+(a+b).
而由条件f(x+1)-f(x) = x+1, 得x+1 = 2ax+(a+b), 即(2a-1)x+(a+b-1) = 0.
因为对任意x都成立, 有2a-1 = 0, a+b-1 = 0, 解得a = b = 1/2.
因此f(x) = 1/2·x²+1/2·x = x(x+1)/2.

2. 代入y = 0得g(x) = g(x)g(0)+f(x)f(0) = g(x)g(0).
若g(x)恒等于0, 有0 = g(x-y) = g(x)g(y)+f(x)f(y) = f(x)f(y).
但代入x = y = 1得0 = 1, 矛盾. 因此存在a使g(a) ≠ 0.
于是由g(a) = g(a)g(0)得g(0) = 1.

代入x = y = 1得1 = g(0) = g(1)²+f(-1)² = g(1)²+1, 即g(1)² = 0, 故g(1) = 0.
代入x = 0, y = 1得g(-1) = g(0)g(1)+f(0)f(1) = 0.
代入y = -1得g(x+1) = g(x)g(-1)+f(x)f(-1) = -f(x), 即有g(x) = -f(x-1).
代入y = 1得g(x-1) = g(x)g(1)+f(x)f(1) = f(x), 即有g(x-2) = f(x-1).
于是g(x) = -g(x-2).
代入x = 2得g(2) = -g(0) = -1, 而代入x = 3得g(3) = -g(1) = 0.
因此g(1) = 0, g(2) = -1, g(3) = 0.

注: 实际上g(x) = cos(πx/2), f(x) = sin(πx/2)是一组满足条件的函数.

Answer 2

a=0.5 b=0.5 c=o,,,,把原式子代进去算就可以了