Question

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。

Answer 1

4 f (0) f (0) = 2 f (0)
f (0) = 1/2
4 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)
事实上可以求得通橘森模项春拿公圆缓式
(1/4)*(1/2+(1/2*I)*sqrt(3))^n+(1/4)*(1/2-(1/2*I)*sqrt(3))^n
解得
f(2010) = 1/2

Answer 2

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，弊清清所以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，一次正拍类推f(6)到f(11)循环，之后也一租前直循环下去

2010除6 = 335 正好整除

所以f(2010) = f(6) = f(0) = 1/2