題目 3. (10 points) Write down the first three terms (three lowest order terms) of the Taylor series of tan−1(2x)1−x\dfrac{\tan^{-1}(2x)}{1-x}1−xtan−1(2x) at 000. (Hint: tan−1u=∫? du\tan^{-1}u=\int ?\,dutan−1u=∫?du) 解答 解法一 tan−1(2x)=2x−83x3+更高次項,11−x=1+x+x2+更高次項.\tan^{-1}(2x)=2x-\frac83x^3+\text{更高次項},\qquad \frac1{1-x}=1+x+x^2+\text{更高次項}.tan−1(2x)=2x−38x3+更高次項,1−x1=1+x+x2+更高次項. 相乘後的三個最低次項為 2x+2x2−23x3.\boxed{2x+2x^2-\frac23x^3}.2x+2x2−32x3.
