作业帮已知函数f(x)满足f(3)=1/3,3f(x)f(y)=f(x+y)+f(x-y),求f(1812)
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已知函数f(x)满足f(3)=1/3,3f(x)f(y)=f(x+y)+f(x-y),求f(1812)
已知函数f(x)满足f(3)=1/3,3f(x)f(y)=f(x+y)+f(x-y),求f(1812)
已知函数f(x)满足f(3)=1/3,3f(x)f(y)=f(x+y)+f(x-y),求f(1812)
因为f(3)=1/3,3f(x)f(y)=f(x+y)+f(x-y)
令y=3,得f(x)=f(x+3)+f(x-3)
所以,f(x-3)=f(x)+f(x-6)
两式相加,得 f(x+3)+f(x-6)=0
即,f(x)=-f(x-9)
又,f(x-9)=-f(x-18)
所以,f(x)=f(x-18)
即,f(x)=f(x+18)
所以,f(x)的周期为18
1812＝18×100＋12
所以,f(1812)=f(12)
又,f(12)=-f(3)=-1/3
所以,f(1812)=-1/3
3f(x)f(y)=f(x+y)+f(x-y),
let y = 3, so f(x) = f(x+3) + f(x-3), so f(x+3) = f(x) - f(x-3),
let x=3, so f(y) = f(3+y) + f(3-y), so f(0) = 2/3,
1812= 3*604,
f(...
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3f(x)f(y)=f(x+y)+f(x-y),
let y = 3, so f(x) = f(x+3) + f(x-3), so f(x+3) = f(x) - f(x-3),
let x=3, so f(y) = f(3+y) + f(3-y), so f(0) = 2/3,
1812= 3*604,
f(1812) = f(1812-3) - f(1812-6)
=f(1812-6)-f(1812-9) - f(1812-6) = -f(1812-9)
=[(-1)^201] * f(1812-9*201)
=-f(3) = -1/3
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