Question

定义域在R上的函数满足f(x+y)+f(x-y)=2f(x)f(y) f(0)≠0 f(1/2)=0 求证f(x)为偶函数 f(x)为周期函数

定义域在R上的函数满足f(x+y)+f(x-y)=2f(x)f(y) f(0)≠0 f(1/2)=0 求证f(x)为偶函数 f(x)为周期函数 若函数在[0，1]内单调 求f(1/3)=? f(1/6)=?

Answer 1

1令x，y都等于零
可以得出f(0)=1
当x=0时
f(y)+f(-y)=2f(y)
即f(-y)=f(y)
所以f(x)为偶函数
2令y=1/2
则f(x+1/2)+f(x-1/2)=0
即f(x)=-f(x+1)
所以f(x-1/2)+f(x-3/2）=0
两式子相减
得出f(x+1/2)=f(x-3/2）
即f（x）=f(x+2)
f(x)是以2为周期的周期函数
3令x=1/3,y=1/6
f(1/2)+f(1/6)=2f(1/3)f(1/6)
即f(1/6)=2f(1/3)f(1/6)且f(1/6)不等于0
解得2f(1/3)=1/2
在令x=1/6,y=1/6
3/2=f(1/3)+f(0)=2f(1/6)f(1/6)
得f(1/6)=sqrt3/2

Answer 2

f(0+0)+f(0-0)=2f(0)f(0)
2f(0)=2f(0)*f(0)
f(0)(f(0)-1)=0
因f(0)≠0
所以：f(0)=1

f(0+x)+f(0-x)=2f(0)f(-x)
f(x)+f(-x)=2f(-x)
f(-x)=f(x)
f(x)为偶函数

f(x+(1/2))+f(x-(1/2))=2f(x)f(1/2)=0
f(x+(1/2))=-f(x-(1/2))
f(x+(1/2))=-f((x-1)+(1/2))=f((x-1)-(1/2))=f(x-(3/2))=f((x+(1/2))-2)
f(x)=f(x-2)
f(x)为周期函数,且周期为2

f((1/2)+(1/2))+f((1/2)-(1/2))=2f(1/2)f(1/2)
f(1)+f(0)=0
f(1)=-f(0)=-1
f((2/3)+(1/3))+f((2/3)-(1/3))=2f(2/3)f(1/3)
f(1)+f(1/3)=2f(2/3)f(1/3)
-1+f(1/3)=2f(2/3)f(1/3) ----------------------(1)
f((1/3)+(1/3))+f((1/3)-(1/3))=2f(1/3)f(1/3)
f(2/3)+f(0)=2f(1/3)f(1/3)
f(2/3)+1=2f(1/3)f(1/3) ---------------(2)
(1)+(2)得：
[f(1/3)+f(2/3)][1-2f(1/3)]=0
而函数在[0，1]内单调，f(0)=1,f(1/2)=0,显然f(1/3)>0, f(2/3)<0
所以：1-2f(1/3)=0, f(1/3)=1/2
或：f(1/3)+f(2/3)=0
f(1/3)=-f(2/3),带入(1)
2[f(1/3)]^2+f(1/3)-1=0，f(1/3)=1/2,或f(1/3)=-1(不合理，应该舍弃）

所以：f(1/3)=1/2