当X=1,由已知条件可知f(y)=f(1+y)+f(1-y),->f(y+1)=f(y+2)+f(y)->f(y)=f(2+y)+f(y)+f(1-y)同理得:f(x)+f(x+3)=0.
由此判断函数为周期函数T=6,f(2013)=f(6*335+3)=f(3).
由f(1)知f(3)=-1/2.f(2013)=-1/2
已知函数f(x)满足:f(1)=1/4,4f(x)f(y)=f(x+y)+f(x-y).x,y是实数,则f(2013)=?
已知函数f(x)满足:f(1)=1/4,4f(x)f(y)=f(x+y)+f(x-y).x,y是实数,则f(2013)=?
数学人气:727 ℃时间:2019-08-21 15:02:44
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