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.

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.