題目 7. (10 points) Derive the complete Taylor series expansion for ln(1+2x1−2x)\ln\left(\frac{1+2x}{1-2x}\right)ln(1−2x1+2x) about x=0x=0x=0. (In the form ∑k=1∞akx2k−1\sum_{k=1}^{\infty} a_k x^{2k-1}∑k=1∞akx2k−1 with a general formula for aka_kak.) 解答 解法一 ln1+2x1−2x=ln(1+2x)−ln(1−2x)=2∑k=1∞(2x)2k−12k−1.\ln\frac{1+2x}{1-2x} =\ln(1+2x)-\ln(1-2x) =2\sum_{k=1}^\infty\frac{(2x)^{2k-1}}{2k-1}.ln1−2x1+2x=ln(1+2x)−ln(1−2x)=2k=1∑∞2k−1(2x)2k−1. 故 ak=22k2k−1,∣x∣<12.\boxed{a_k=\frac{2^{2k}}{2k-1},\qquad |x|<\frac12}.ak=2k−122k,∣x∣<21.
