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(2010)=?f(x+2)=-f(x-1)即f(x)=f(x+6)
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![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(2010)=?f(x+2)=-f(x-1)即f(x)=f(x+6)](/uploads/image/z/6975700-52-0.jpg?t=%E5%B7%B2%E7%9F%A5%E5%87%BD%E6%95%B0f+%28x+%29%E6%BB%A1%E8%B6%B3f%281%29%3D1%2F4%2C4f%28x%29f%28y%29%3Df%28x%2By%29%2Bf%28x-y%29%28x%2Cy%E5%B1%9E%E4%BA%8ER%29%2C%E5%88%99f%282010%29%3D%3Ff%28x%2B2%29%3D-f%28x-1%29%E5%8D%B3f%28x%29%3Df%28x%2B6%29)
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(2010)=?f(x+2)=-f(x-1)即f(x)=f(x+6)
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(2010)=?
f(x+2)=-f(x-1)
即f(x)=f(x+6)
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(2010)=?f(x+2)=-f(x-1)即f(x)=f(x+6)
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.
f(x+6)=f[(x+4)+2]=-f[(x+4)-1]=-f(x+3)=-f[(x+1)+2]=f[(x+1)-1]=f(x)
是神马?不懂
f(0)=4f(1)f(0)=f(1+0)+f(1-0)=2f(1)=1/2.
f(x)=4f(x)f(1)=f(x+1)+f(x-1),
f(x)=f(x+1)+f(x-1).
f(x+1)=f(x+1+1)+f(x+1-1)=f(x+2)+f(x),
f(x)=f(x+1)+f(x-1)=f(x+2)+f(x)+f(x-1),
f(x+2)=-f(x-...
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f(0)=4f(1)f(0)=f(1+0)+f(1-0)=2f(1)=1/2.
f(x)=4f(x)f(1)=f(x+1)+f(x-1),
f(x)=f(x+1)+f(x-1).
f(x+1)=f(x+1+1)+f(x+1-1)=f(x+2)+f(x),
f(x)=f(x+1)+f(x-1)=f(x+2)+f(x)+f(x-1),
f(x+2)=-f(x-1).
f(x+6)=f(x+4+2)=-f(x+4-1)=-f(x+3)=-f(x+1+2)=f(x+1-1)=f(x).
6是f(x)的一个正周期.
f(2010)=f[6*350+0]=f(0)=1/2
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