函数方程f[(x-1)/(x+1)]+f(-1/x)+f[(1+x)/(1-x)]=cosx该解为无数个但可以用一个表达式表示

Question

函数方程f[(x-1)/(x+1)]+f(-1/x)+f[(1+x)/(1-x)]=cosx
该解为无数个 但可以用一个表达式表示

Answer 1

f[(x-1)/(x+1)]+f(-1/x)+f[(1+x)/(1-x)]=cosx①；将x代换为(1+x)/(1-x),①式变为f[(x-1)/(x+1)]+f(x)+f(-1/x)=cos[(1+x)/(1-x)]②；将x代换为(x-1)/(x+1),①式变为f(-1/x)+f(x)+f[(1+x)/(1-x)]=cos[(x-1)/(x+1)]③；将x代换为-1/x,①式变为f[(x-1)/(x+1)]+f(x)+f[(1+x)/(1-x)]=cos(1/x)④；①减去②得f[(1+x)/(1-x)]-f(x)=cosx-cos[(1+x)/(1-x)]⑤；①减去③得f[(x-1)/(x+1)]-f(x)=cosx-cos[(x-1)/(x+1)]⑥；①减去④得f(-1/x)-f(x)=cosx-cos(1/x)⑦；⑤、⑥、⑦相加得f[(1+x)/(1-x)]+f[(x-1)/(x+1)]+f(-1/x)-3f(x)=3cosx-cos[(1+x)/(1-x)]-cos[(x-1)/(x+1)]-cos(1/x),即cosx-3f(x)=3cosx-cos[(1+x)/(1-x)]-cos[(x-1)/(x+1)]-cos(1/x),所以f(x)=(-2/3)cosx+(1/3)cos[(1+x)/(1-x)]+(1/3)cos[(x-1)/(x+1)]+(1/3)cos(1/3)(毕).