先声明一声,第一句话有误,f(x)只与x有关,和n无关的.只需要f(0)=1即可.
由f(2x)-f(x)=x^2得：f(x)-f(x/2)=x^2/4,f(x/2)-f(x/4)=x^2/16,……,f(x/2^n)-f(x/2^(n+1))=x^2/4^(n+1)
把所有式子相加得：f(x)=f(x/2^(n+1))+(1/4+1/16+…+1/4^(n+1))x^2
等式两边取极限,当n→+∞时,limf(x)=limf(x/2^(n+1))+lim[(1/4+1/16+…+1/4^(n+1))x^2]
即limf(x)=f(x),limf(x/2^(n+1))=f(0),lim(1/4+1/16+…+1/4^(n+1))=1/3
故,f(x)=1+(x^2)/3