Taylor approximation¶ first-order¶ f(x)≈f(a)+f′(a)(x−a) so the approximation of f(x)=log(1+x) around x=0 is log(1+x)≈x second-order¶ f(x)≈f(a)+f′(a)(x−a)+12f″(x)(x−a)2 example¶ given g(X) & μ=E(X) first-order g(X)≈g(μ)+g′(μ)(X−μ) Var(g(X))=[g′(μ)]2Var(X) second-order g(X)≈g(μ)+g′(μ)(X−μ)+12g″(X)(X−μ)2 E[g(X)]≈g(μ)+12g″(μ)Var(X)=g(μ)+12g″(μ)[g′(μ)]2Var(g(X)) when g(X)=log(X) E[log(X)]≈log(μ)−12Var(log(X)) logE(X)=E(logX)+12Var(logX) E(X)=eE(logX)+12Var(logX)