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(2010)=?
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f(x)满足f(1)=1/44f(x)f(y)=f(x+y)+f(x-y)(x,y属于R)则f(2010)=?f(x)满足f(1)=1/44f(x)f(y)=f(x+y)+f(x-y)(x,y属于R)
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(2010)=?
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(2010)=?
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(2010)=?
取y=1,则 4f(x)f(1)=f(x+1)+f(x-1)
即 f(x)=f(x+1)+f(x-1)
所以 f(x+1)=f(x+2)+f(x) (在上式中,以x+1代替x)
两式相加,得 f(x+2)+f(x-1)=0
所以 f(x+2)=-f(x-1)
因此,f(x+6)=f[(x+4)+2]=-f[(x+4)-1]=-f(x+3)=-f[(x+1)+2]=f[(x+1)-1]=f(x)
就是说,函数是以6为周期的周期函数.
f(2010)=f(335*6)=f(0)
在已知等式中取x=1,y=0,则可得 4f(1)f(0)=2f(1),所以 f(0)=1/2
因此,f(2010)=f(0)=1/2.
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