f(x)=(1+(1+x^2)^0.5)^0.5 x在0处的四阶泰勒展开 如果我设u=(1+x^2)^0.5,那么是不是u在1处的四阶泰勒展开?是否还需将u展开?f(u)=2^0.5+(2^0.5)/4*(u-1)-(2^0.5)/32*(u-1)^2u=1+(x^2)/2-(x^4)/8代入后得 f(x)=2^0.5+(2^0.5)/8*x^2

f(x)=x/(1+x^2)^0.5求f(f(f(x))) f(x)满足:2f(x)-f(1/x)=x+1,求f(x) 若f(x)满足f(x)-2f(1/x)=x,则f(x)=? 若F(x)满足f(x)+2f(1/x)=x,则f(x)= 设函数f(x)满足f(x)+2f(1/x)=x,求f(x) 已知f(x)满足2f(x)+f(1/x)=3x,求f(x) F(X)满足F(x)+2f(x分之1)=3X,求f(x) 已知f(x)满足2f(x)+f(1/x)=3x,求f(x)? 已知f(x)满足2f(x)+f(-x)=-3x+1,求f(x) 已知f(x)满足f(x)+2f(1/x)=3x,求f(x) , f(x)+f((x-1)/x)=2x; x!=0,1; 求f(x) f(x)+f[(x-1)/x]=2x x不等于0,1.求f(x). max{f(x),g(x)}=1/2(f(x)+g(x)+|f(x)-g(x)| f(x)={2x+1,x f(x)={1+x/2,x f(x)=(x-1)/(x+2) f(x)是奇函数f(1)=0.5 f(x+2)=f(x)+f(2) 求f(5) f(x-1)=2x/x-1求f(x)=?f(x+1)=?