Question

已知f(x)满足f(x+y)+f(x-y)=2f(x)f(y),x,y属于R，且f(0)不等于0，试判定f(x)的奇偶性

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Answer 1

2f(x)f(0)=f(x-0)+f(x+0)=2f(x),
0=2f(x)[f(0)-1],
0=2f(0)[f(0)-1],
0=f(0)-1, f(0)=1. f(x)非奇.
f(0+y)+f(0-y)=2f(0)f(y),
f(y)+f(-y)=2f(y),
f(-y)=f(y),
f(x)偶.