Skip to content
CalcGospel 國際數學圖譜
返回

CIE 9231 2022 November Paper 21 Q1

A Level / CIE / FP2

CIE 9231 2022 Nov Paper 21 Paper · Question 1

题目

Problem

Find the Maclaurin’s series for ln⁡(1+ex)\ln(1+\mathrm{e}^x) up to and including the term in x2x^2.

(5)
题目中文翻译

求 ln⁡(1+ex)\ln(1+\mathrm{e}^x) 的麦克劳林级数,展开至并包括 x2x^2 项。

(5)
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

展开

先求对数的导数,再用商法则求第二阶导数。代入 x=0x=0 得出展开系数。

答题过程

展开

For f(x)=ln⁡(1+ex)f(x)=\ln(1+e^x),

f′(x)= ex1+ex,f′′(x)= ex(1+ex)−e2x(1+ex)2= ex(1+ex)2.\begin{align*} f'(x)=&\,\frac{e^x}{1+e^x},\\ f''(x)=&\,\frac{e^x(1+e^x)-e^{2x}}{(1+e^x)^2}\\ =&\,\frac{e^x}{(1+e^x)^2}. \end{align*}

Thus f(0)=ln⁡2f(0)=\ln2, f′(0)=1/2f'(0)=1/2 and f′′(0)=1/4f''(0)=1/4. Using f(x)=f(0)+f′(0)x+12f′′(0)x2+O(x3)f(x)=f(0)+f'(0)x+\frac12f''(0)x^2+O(x^3) gives

ln⁡(1+ex)=ln⁡2+x2+x28+O(x3).\boxed{\ln(1+e^x)=\ln2+\frac x2+\frac{x^2}{8}+O(x^3)}.

解法二

思路

展开

也可先展开 exe^x,提出 1+ex1+e^x 的常数因子 22,再用 ln⁡(1+u)\ln(1+u) 的展开。

答题过程

展开

Using the standard expansion ex=1+x+x2/2+O(x3)e^x=1+x+x^2/2+O(x^3) gives 1+ex=2(1+x2+x24+O(x3))1+e^x=2(1+\frac x2+\frac{x^2}{4}+O(x^3)). Put u=x2+x24+O(x3)u=\frac x2+\frac{x^2}{4}+O(x^3). The standard expansion is ln⁡(1+u)=u−u2/2+O(u3)\ln(1+u)=u-u^2/2+O(u^3) and u2=x2/4+O(x3)u^2=x^2/4+O(x^3). Hence

ln⁡(1+ex)= ln⁡2+u−u22+O(u3)= ln⁡2+x2+x24−x28+O(x3)= ln⁡2+x2+x28+O(x3).\begin{align*} \ln(1+e^x)=&\,\ln2+u-\frac{u^2}{2}+O(u^3)\\ =&\,\ln2+\frac x2+\frac{x^2}{4}-\frac{x^2}{8}+O(x^3)\\ =&\,\ln2+\frac x2+\frac{x^2}{8}+O(x^3). \end{align*}