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在允许取值范围内赋变量予特殊值,从而使问题获解的方法叫“特取法”,下面谈谈特取法解有关函数方程的几个问题。一、证明函数f(x)的周期性例1设函数f(x)定义在整数集,且满足f(0)=1,f(1)=0,f(x_1+x_2)+f(x_1-x_2)=2f(x_1)f(x_2),证明f(x)为周期函数。证明特取x_2=1,可得f(x_1+1)+f(x_1-1)=2f(x_1)f(1)=0 再用x_1+2代入x_1且特取x_2=1,可得f(x_1+3)+f(x_1+1)=2f(x_1+2)f(1)=0 由上述两式得f(x_1 +3)=f(x_1-1) 令x_1=x+1得f(x+4)=f(x) 故f(x)是以4为周期的函数。二、证明函数f(x)的奇偶性例2已知f(x+y)+f(x-y)=2f(x)·f(y)对于一切实数X、y都成立,且f(0)≠0,
The method of assigning variables to special values within the range of allowable values, so as to solve the problem is called “special method”. The following is a discussion of special methods for solving some problems related to function equations. First, prove the periodicity of the function f (x) Example 1 Let function f (x) be defined in the set of integers, and satisfy f (0) = 1, f (1) = 0, f (x_1 + x_2) + f (x_1) -x_2) = 2f(x_1)f(x_2), proving that f(x) is a periodic function. Prove that special x2 = 1, can be obtained f (x_1 +1) + f (x_1-1) = 2f (x_1) f (1) = 0 Then x_1 + 2 into x_1 and special take x_2 = 1, can be obtained f (x_1+3)+f(x_1+1)=2f(x_1+2)f(1)=0 By the above two formulae f(x_1 +3)=f(x_1-1) Let x_1=x+1 be f(x+4)=f(x) So f(x) is a function of 4 cycles. Second, prove the parity of the function f (x) Example 2 Know that f (x + y) + f (xy) = 2f (x) · f (y) for all real numbers X, y are true, and f (0 ) 0,