IAL 2024 Jan Q4 | 國際數學站

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 $\left( x - \frac{\pi}{6} \right)$ up to and including the term in $\left( x - \frac{\pi}{6} \right)^3$ , the Taylor series expansion about $\frac{\pi}{6}$ of

\begin{align*} y = \tan \left( \frac{3x}{2} \right) \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)

解答

(a)

Given $y = \tan\left(\frac{3x}{2}\right)$ , we find its derivatives up to the third order:

\begin{align*} \frac{\mathrm{d}y}{\mathrm{d}x} = &\,\frac{3}{2}\sec^2\left(\frac{3x}{2}\right)\\[4mm] \frac{\mathrm{d}^2y}{\mathrm{d}x^2} = &\,2 \cdot \frac{3}{2} \sec\left(\frac{3x}{2}\right) \cdot \sec\left(\frac{3x}{2}\right)\tan\left(\frac{3x}{2}\right) \cdot \frac{3}{2}\\[4mm] = &\,\frac{9}{2}\sec^2\left(\frac{3x}{2}\right)\tan\left(\frac{3x}{2}\right)\\[4mm] \frac{\mathrm{d}^3y}{\mathrm{d}x^3} = &\,\frac{9}{2} \left[ 2\sec\left(\frac{3x}{2}\right)\cdot \sec\left(\frac{3x}{2}\right)\tan\left(\frac{3x}{2}\right)\cdot \frac{3}{2} \cdot \tan\left(\frac{3x}{2}\right) + \sec^2\left(\frac{3x}{2}\right) \cdot \sec^2\left(\frac{3x}{2}\right) \cdot \frac{3}{2} \right]\\[4mm] = &\,\frac{27}{2}\sec^2\left(\frac{3x}{2}\right)\tan^2\left(\frac{3x}{2}\right) + \frac{27}{4}\sec^4\left(\frac{3x}{2}\right) \end{align*}

Now we evaluate $y$ and its derivatives at $x = \frac{\pi}{6}$ :

\begin{align*} y\left(\frac{\pi}{6}\right) = &\,\tan\left(\frac{\pi}{4}\right) = 1\\[4mm] y'\left(\frac{\pi}{6}\right) = &\,\frac{3}{2}\sec^2\left(\frac{\pi}{4}\right) = \frac{3}{2}(\sqrt{2})^2 = 3\\[4mm] y''\left(\frac{\pi}{6}\right) = &\,\frac{9}{2}\sec^2\left(\frac{\pi}{4}\right)\tan\left(\frac{\pi}{4}\right) = \frac{9}{2}(2)(1) = 9\\[4mm] y'''\left(\frac{\pi}{6}\right) = &\,\frac{27}{2}\sec^2\left(\frac{\pi}{4}\right)\tan^2\left(\frac{\pi}{4}\right) + \frac{27}{4}\sec^4\left(\frac{\pi}{4}\right)\\[4mm] = &\,\frac{27}{2}(2)(1)^2 + \frac{27}{4}(2)^2\\[4mm] = &\,27 + 27 = 54 \end{align*}

Using the Taylor series expansion formula about $x = a$ : $y(x) \approx y(a) + y'(a)(x-a) + \frac{y''(a)}{2!}(x-a)^2 + \frac{y'''(a)}{3!}(x-a)^3$

\begin{align*} y \approx &\,1 + 3\left(x - \frac{\pi}{6}\right) + \frac{9}{2}\left(x - \frac{\pi}{6}\right)^2 + \frac{54}{6}\left(x - \frac{\pi}{6}\right)^3\\[4mm] \approx &\,1 + 3\left(x - \frac{\pi}{6}\right) + \frac{9}{2}\left(x - \frac{\pi}{6}\right)^2 + 9\left(x - \frac{\pi}{6}\right)^3 \end{align*}

(b)

We want to approximate $\tan\left(\frac{3\pi}{8}\right)$ . Let $\frac{3x}{2} = \frac{3\pi}{8}$ , which means $x = \frac{\pi}{4}$ . The difference $\left(x - \frac{\pi}{6}\right)$ is:

\begin{align*} \frac{\pi}{4} - \frac{\pi}{6} = \frac{\pi}{12} \end{align*}

Substituting this into our expansion:

\begin{align*} \tan\left(\frac{3\pi}{8}\right) \approx &\,1 + 3\left(\frac{\pi}{12}\right) + \frac{9}{2}\left(\frac{\pi}{12}\right)^2 + 9\left(\frac{\pi}{12}\right)^3\\[4mm] \approx &\,1 + \frac{\pi}{4} + \frac{9}{2}\left(\frac{\pi^2}{144}\right) + 9\left(\frac{\pi^3}{1728}\right)\\[4mm] \approx &\,1 + \frac{\pi}{4} + \frac{\pi^2}{32} + \frac{\pi^3}{192} \end{align*}

Comparing to the given form $1 + \frac{\pi}{4} + \frac{\pi^2}{A} + \frac{\pi^3}{B}$ , we have $A = 32$ and $B = 192$ .