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.
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.