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),则f(2011)=?
人气:227 ℃ 时间:2019-08-22 07:07:33
解答
由题意得,任意数带进去都满足原式.此类题目要先代入几组数找出规律
4f(1)f(0)=f(1)+f(1)即f(0)=2f(1)=1/2
找他的一般规律,设y=1
则4f(x)f(1)=f(x+1)+f(x-1)
即f(x)=f(x+1)+f(x-1)
由已知的f(1)=1/4,f(0)=1/2算出f(1)=f(1+1)+f(0)
即f(1)=f(2)+f(0)
即f(2)=f(1)-f(0)=-1/4
由以上原理可推知当x=0,1,2,3,4,5,6,7,8……时
f(x)=1/2, 1/4,-1/4, -1/2,-1/4, 1/4, 1/2, 1/4,-1/4
寻其中规律为分母以2,4,4为一周期循环,正负号除x=0,1外以-,-,-,+,+,+为一周期循环
这样可推知f(2011)=1/4
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