已知实函数f(x)满足:f(x+y)+f(x-y)=2f(x)*cosy,求f(x) Solution 1: f '(y)-f '(-y)=-2f(0)siny f '(y)+f '(-y)=2f '(0)cosy f '(y)=-f(0)siny+f '(0)cosy f(y)=f(0)cosy+f '(0)siny + constwww.ddhw.com The const should be 0 to satisfy f(x+y)+f(x-y)=2f(x)*cosy. So f(x) = Asin(x+x0) where A and x0 are any real numbers Solution 2
Any function f(x) can be written as f(x) = F(x) + G(x) where F(-x) = F(x) and G(-x) = -G(x). So 1) f(y)+f(-y) = 2f(0)cosy or F(x) = f0cosx, where f0 = f(0). 2) f(2x) + f0 = 2f(x)cosx f0cos(2x) + G(2x) + f0 = 2f0(cosx)^2 + 2G(x)cosx G(2x) = 2G(x)cosx so G(x) = f1sinx where f1 is constant. So f(x) = f0cosx + f1sinx = Asin(x+x0) where A and x0 are any real constantwww.ddhw.com
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