题目

Problem

In this question you must show all stages of your working.

Solutions relying entirely on calculator technology are not acceptable.

(a) Determine, in ascending powers of $\bigg( x - \frac{\pi}{6} \bigg)$ up to and including the term in $\bigg( x - \frac{\pi}{6} \bigg)^3$ , the Taylor series expansion about $\frac{\pi}{6}$ of

\begin{align*} y = \tan \bigg( \frac{3x}{2} \bigg) \end{align*}

giving each coefficient in simplest form.

(7)

(b) Hence show that

\begin{align*} \tan \frac{3\pi}{8} \approx&\, 1+\frac{\pi}{4}+\frac{\pi^2}{A}+\frac{\pi^3}{B} \end{align*}

where $A$ and $B$ are integers to be determined.

(2)

题目中文翻译

本题必须展示所有运算步骤。

完全依赖计算器技术的解答不予接受。

(a) 求

\begin{align*} y=&\,\tan\bigg(\frac{3x}{2}\bigg) \end{align*}

在 $\frac{\pi}{6}$ 处的 Taylor 级数展开式，按 $\big(x-\frac{\pi}{6}\big)$ 的升幂排列，直至并包括 $\big(x-\frac{\pi}{6}\big)^3$ 项，各项系数须为最简形式。

(7)

(b) 由此证明

\begin{align*} \tan\frac{3\pi}{8}\approx&\,1+\frac{\pi}{4} +\frac{\pi^2}{A}+\frac{\pi^3}{B}, \end{align*}

其中 $A$ 、 $B$ 为待求的整数。

(2)

解答

(a)

解法一

思路

展开

Taylor series 到三次项需要 $y,y',y'',y'''$ 在 $x=\frac{\pi}{6}$ 的值。这里 $\frac{3x}{2}$ 在 $x=\frac{\pi}{6}$ 时等于 $\frac{\pi}{4}$ ，所以 $\tan\frac{\pi}{4}=1$ ， $\sec^2\frac{\pi}{4}=2$ ，数值会比较整齐。

答题过程

展开

Let

\begin{align*} y=\tan\bigg(\frac{3x}{2}\bigg) \end{align*}

Then

\begin{align*} y' =&\, \frac32\sec^2\bigg(\frac{3x}{2}\bigg) \end{align*}

Differentiate again:

\begin{align*} y'' =&\, \frac32\cdot 2\sec\bigg(\frac{3x}{2}\bigg) \bigg(\sec\bigg(\frac{3x}{2}\bigg) \tan\bigg(\frac{3x}{2}\bigg)\cdot\frac32\bigg)\\[4mm] =&\, \frac92\sec^2\bigg(\frac{3x}{2}\bigg) \tan\bigg(\frac{3x}{2}\bigg) \end{align*}

For the third derivative, use the product rule on $\sec^2\bigg(\frac{3x}{2}\bigg)\tan\bigg(\frac{3x}{2}\bigg)$ :

\begin{align*} y''' =&\, \frac92\bigg[ 3\sec^2\bigg(\frac{3x}{2}\bigg) \tan^2\bigg(\frac{3x}{2}\bigg)\\[2mm] &\,\hspace{2pt}+\frac32\sec^4\bigg(\frac{3x}{2}\bigg) \bigg]\\[4mm] =&\, \frac{27}{2}\sec^2\bigg(\frac{3x}{2}\bigg) \tan^2\bigg(\frac{3x}{2}\bigg)\\[2mm] &\,\hspace{2pt}+\frac{27}{4}\sec^4\bigg(\frac{3x}{2}\bigg) \end{align*}

At $x=\frac{\pi}{6}$ , we have $\frac{3x}{2}=\frac{\pi}{4}$ . Therefore

\begin{align*} y\bigg(\frac{\pi}{6}\bigg) =&\,1\\[2mm] y'\bigg(\frac{\pi}{6}\bigg) =&\, \frac32\cdot 2=3\\[2mm] y''\bigg(\frac{\pi}{6}\bigg) =&\, \frac92\cdot 2\cdot 1=9\\[2mm] y'''\bigg(\frac{\pi}{6}\bigg) =&\, \frac{27}{2}\cdot 2\cdot 1^2 +\frac{27}{4}\cdot 2^2\\[2mm] =&\,54 \end{align*}

Using Taylor’s formula about $x=\frac{\pi}{6}$ ,

\begin{align*} y =&\, 1+3\bigg(x-\frac{\pi}{6}\bigg) +\frac{9}{2!}\bigg(x-\frac{\pi}{6}\bigg)^2\\[4mm] &\,\hspace{2pt}+\frac{54}{3!}\bigg(x-\frac{\pi}{6}\bigg)^3+\cdots\\[4mm] =&\, 1+3\bigg(x-\frac{\pi}{6}\bigg) +\frac92\bigg(x-\frac{\pi}{6}\bigg)^2\\[2mm] &\,\hspace{2pt}+9\bigg(x-\frac{\pi}{6}\bigg)^3+\cdots \end{align*}

(b)

解法一

思路

展开

因为 $y=\tan\bigg(\frac{3x}{2}\bigg)$ ，要得到 $\tan\frac{3\pi}{8}$ ，令 $x=\frac{\pi}{4}$ 。这时 $x-\frac{\pi}{6}=\frac{\pi}{12}$ 。

答题过程

展开

Put $x=\frac{\pi}{4}$ in the expansion from part (a). Then

\begin{align*} x-\frac{\pi}{6} =&\, \frac{\pi}{4}-\frac{\pi}{6}\\[4mm] =&\, \frac{\pi}{12} \end{align*}

\begin{align*} \tan\frac{3\pi}{8} \approx&\, 1+3\bigg(\frac{\pi}{12}\bigg) +\frac92\bigg(\frac{\pi}{12}\bigg)^2 +9\bigg(\frac{\pi}{12}\bigg)^3\\[4mm] =&\, 1+\frac{\pi}{4} +\frac{9\pi^2}{288} +\frac{9\pi^3}{1728}\\[4mm] =&\, 1+\frac{\pi}{4} +\frac{\pi^2}{32} +\frac{\pi^3}{192} \end{align*}

Hence

\begin{align*} \boxed{A=32,\qquad B=192} \end{align*}

FP2 2024 Jan Q4

题目

解答

(a)

解法一

思路

答题过程

(b)

解法一

思路

答题过程