√(ρ^2+z^2) = z√(1+ρ^2/z^2) 以ρ为自变数,z为参变量,将f(ρ)=√(1+ρ^2/z^2)在0点展成Taylor级数
为此,首先计算出：f(0)＝1；f '(0)=0；f ''(0)=1/z^2；f '''(0)=0; f ''''(0)=-3/z^4.
于是
z f(z) = z [f(0) + f '(0)ρ/z + f ''(0)ρ^2/(2z^2) +f '''(0)ρ^3/(3!z^3)+f ''''(0)ρ^4/(4!z^4)+ . ]
= z + ρ^2/(2z) - ρ^4/(8z^3) + ...
因此：
√(ρ^2+z^2) = z f(z) = z + ρ^2/(2z) - ρ^4/(8z^3) + ...