已知函数f（x）满足：f（1）=1/4,4f（x）f（y）=f（x+y）+f（x-y）（x、y属于R）,求f（2010）值.
问题描述：
已知函数f（x）满足：f（1）=1/4,4f（x）f（y）=f（x+y）+f（x-y）（x、y属于R）,求f（2010）值.
答
将y = 0 带入， 4f(x)f(0)=2f(x)
因为f(x)不可能永远=0，比如x=1时，f(x)不等于0，所以f(0) = 1 /2
将y=1代入， f(x+1) = f(x) - f(x-1)
推出f(2) = f(1)-f(0) = -1/4
f(3) = f(2)-f(1) = -1/2
f(4) = -1/4
f(5) = 1/4
f(6) = 1/2
f(7) = 1/4
看出来f(0)到f(5)为6个数1组，一次类推f(6)到f(11)循环，之后也一直循环下去
2010除6 = 335 正好整除
所以f(2010) = f(6) = f(0) = 1/2
答
4 f (0) f (0) = 2 f (0)f (0) = 1/24 f (0) f (x) = f (x) + f (x) = 2 f (x)4 f (x) f (1) = f (x + 1) + f (x - 1)f (x + 1) = f (x) - f (x - 1)f (x) = f (x - 1) - f (x - 2)事实上可以求得通项公式(1/4)*(1/2...