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已知f(x)是定义在R上的函数,对于任意实数a,b都有f(ab)=af(b)+bf(a),且f(2)=1求f(1\2)的

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已知f(x)是定义在R上的函数,对于任意实数a,b都有f(ab)=af(b)+bf(a),且f(2)=1求f(1\2)的值
f(2) = f(1*2)
= f(2) +2f(1)
f(1) = 0
f(1)= f(2*1/2) = 2f(1/2)+1/2 * f(2)
2f(1/2) + 1/2 = 0
f(1/2) = - 1/4
再问: 求f(2^-n)的解析式
再答: f(2^-n) =f(1/2^n) f(1/2^-2)= f(1/4) = f(1/2 * 1/2) = 1/2 * f(1/2) + 1/2 * f( 1/2) = f(1/2) =- 1/4 =- 2/8 f(2^-3)=f(1/8) = f(1/2 * 1/4) = 1/2 * f(1/4) + 1/4* f(1/2) = 1/2 * (-1/4) + 1/4 * (-1/4) = 3/4 * (-1/4) =-3/16 f(1/2^-4)=f(1/6)=f(1/4 * 1/4) = 1/2* f(1/4) = -1/8 = - 4/32 f(1/2^4) = 1/2 * f(1/8) + 1/8 * f(1/2) = 1/2* (-3/16 )+ 1/8 *(-1/4) = - 3/32 - 1/32 = -1/8 ... f(1/2^-n) = - n/ 2^(n+2) 意思: 2的(n+2)次方 分之 (-n) 不清楚再问