题目
Problem
Find the first three terms in the Maclaurin’s series for tanh−1(21ex) in the form 21lna+bx+cx2, giving the exact values of the constants a, b and c.
[6]
题目中文翻译
求 tanh−1(21ex) 的麦克劳林级数的前三项,并将其写成 21lna+bx+cx2 的形式,给出常数 a、b 和 c 的准确值。
(6)
原卷通用作答要求
Answer all questions. You should use a calculator where appropriate.
You must show all necessary working clearly; no marks will be given for unsupported answers from a calculator.
Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place for angles in degrees, unless a different level of accuracy is specified in the question.
请回答所有题目。适当时应使用计算器。必须清楚展示所有必要的计算过程;仅给出计算器答案而没有过程,不予评分。除非题目另有精度要求,非精确数值答案保留 3 位有效数字,以度为单位的角度保留 1 位小数。
解答
解法一
思路
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令 u=ex/2,利用 u′=u 求前两阶导数。常数项用反双曲正切的对数形式求出。
答题过程
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Let f(x)=tanh−1(u) with u=ex/2. Since u′=u,
f′(x)=f′′(x)==1−u2u,(1−u2)2u(1−u2)−u(−2u2)(1−u2)2u(1+u2).
At x=0, u=21. The logarithmic form gives
f(0)=21ln1−1/21+1/2=21ln3.
Also f′(0)=32 and f′′(0)=910. Substituting into f(0)+xf′(0)+2x2f′′(0) gives
f(x)=21ln3+32x+95x2+O(x3).
Therefore (a,b,c)=(3,32,95).
解法二
思路
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也可先写成两个对数之差,再分别代入 ex 和 ln(1+w) 的展开。
答题过程
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Use ex=1+x+21x2+O(x3) and ln(1+w)=w−21w2+O(w3). For u=ex/2=21+2x+4x2+O(x3),
ln(1+u)==ln(1−u)==ln23+ln(1+3x+6x2+O(x3))ln23+3x+9x2+O(x3),ln21+ln(1−x−2x2+O(x3))ln21−x−x2+O(x3).
Since tanh−1u=21(ln(1+u)−ln(1−u)), taking half the difference gives
f(x)==21ln3+21(31+1)x+21(91+1)x2+O(x3)21ln3+32x+95x2+O(x3).
Thus a=3, b=32, and c=95.