f(x+y)+f(x-y)=2f(x)乘f(y) 且f(0)不等于0 f(c/2)=0 求证f(x+2c)=f(x)

y=0
f(x)+f(x)=2f(x)f(0)
f(0)不等于0, f(0)=1.

y=c/2
f(x+c/2)+f(x-c/2)=2f(x)f(c/2)=0
f(x+c/2)=-f(x-c/2)

x=x+c/2
f(x+c)=-f(x)
x=x+c
f(x+2c)=-f(x+c)=f(x)
f(x+2c)=f(x)，得证。

证明：令y=c/2得：f(x+c/2)+f(x-c/2)=0
∴f(x+c/2)=-f(x-c/2)
∴f(x+2c)=f((x+3*c/2)+c/2)=-f(x+3*c/2-c/2)=-f(x+c)=-f((x+c/2)+c/2)
=f(x+c/2-c/2)=f(x)
故证得原命题