IAL 2023 Jan Q1 | 國際數學站

Given that $y = \ln(5 + 3x)$

(a) determine, in simplest form, $\frac{\mathrm{d}^3y}{\mathrm{d}x^3}$

(3)

(b) Hence determine the Maclaurin series expansion of $\ln(5 + 3x)$ , in ascending powers of $x$ up to and including the term in $x^3$ , giving each coefficient in simplest form.

(2)

(c) Hence write down the Maclaurin series expansion of $\ln(5 - 3x)$ in ascending powers of $x$ up to and including the term in $x^3$ giving each coefficient in simplest form.

(1)

(d) Use the answers to parts (b) and (c) to determine the first $2$ non-zero terms, in ascending powers of $x$ , of the Maclaurin series expansion of

\begin{align*} \ln\left(\frac{5 + 3x}{5 - 3x}\right)\\[2mm] \end{align*}

(2)

解答

(a)

Given $y = \ln(5+3x)$ , we differentiate with respect to $x$ to find the first three derivatives:

\begin{align*} \frac{\mathrm{d}y}{\mathrm{d}x} = &\,\frac{3}{5+3x} = 3(5+3x)^{-1}\\[4mm] \frac{\mathrm{d}^2y}{\mathrm{d}x^2} = &\,-3(5+3x)^{-2} \cdot 3\\[4mm] = &\,-9(5+3x)^{-2}\\[4mm] = &\,-\frac{9}{(5+3x)^2}\\[4mm] \frac{\mathrm{d}^3y}{\mathrm{d}x^3} = &\,-9(-2)(5+3x)^{-3} \cdot 3\\[4mm] = &\,54(5+3x)^{-3}\\[4mm] = &\,\frac{54}{(5+3x)^3} \end{align*}

(b)

To find the Maclaurin series expansion of $\ln(5+3x)$ , we evaluate $y$ and its derivatives at $x=0$ :

\begin{align*} y(0) = &\,\ln 5\\[4mm] y'(0) = &\,\frac{3}{5+0} = \frac{3}{5}\\[4mm] y''(0) = &\,-\frac{9}{(5+0)^2} = -\frac{9}{25}\\[4mm] y'''(0) = &\,\frac{54}{(5+0)^3} = \frac{54}{125} \end{align*}

Using the Maclaurin series formula $f(x) \approx f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3$ :

\begin{align*} \ln(5+3x) \approx &\,\ln 5 + \frac{3}{5}x - \frac{9}{25}\frac{x^2}{2!} + \frac{54}{125}\frac{x^3}{3!}\\[4mm] \approx &\,\ln 5 + \frac{3}{5}x - \frac{9}{50}x^2 + \frac{9}{125}x^3 \end{align*}

(c)

To find the Maclaurin series for $\ln(5-3x)$ , replace $x$ with $-x$ in the expansion from part (b):

\begin{align*} \ln(5-3x) \approx &\,\ln 5 + \frac{3}{5}(-x) - \frac{9}{50}(-x)^2 + \frac{9}{125}(-x)^3\\[4mm] \approx &\,\ln 5 - \frac{3}{5}x - \frac{9}{50}x^2 - \frac{9}{125}x^3 \end{align*}

(d)

Using laws of logarithms, $\ln\left(\frac{5+3x}{5-3x}\right) = \ln(5+3x) - \ln(5-3x)$ . Subtracting the series from (c) from the series from (b):

\begin{align*} \ln\left(\frac{5+3x}{5-3x}\right) \approx &\,\left( \ln 5 + \frac{3}{5}x - \frac{9}{50}x^2 + \frac{9}{125}x^3 \right)\\[4mm] &\,\hspace{2pt}- \left( \ln 5 - \frac{3}{5}x - \frac{9}{50}x^2 - \frac{9}{125}x^3 \right)\\[4mm] \approx &\,\left(\frac{3}{5}x - \left(-\frac{3}{5}x\right)\right) + \left(\frac{9}{125}x^3 - \left(-\frac{9}{125}x^3\right)\right)\\[4mm] \approx &\,\frac{6}{5}x + \frac{18}{125}x^3 \end{align*}

These are the first 2 non-zero terms in the expansion.