Maclaurin Series: logarithmic expansion, composition with \(\sin\) and \(\cos\), and integral approximation — VJC 2025 H2 Math Prelim P2
What this question tests
Question
It is given that \(\mathrm{f}(x) = \ln(a + x)\), \(x \in \mathbb{R}\), \(x > -a\), where \(a\) is a constant.
(a) Using the standard series from the List of Formulae (MF27), find the series expansion for \(\mathrm{f}(x)\), up to and including the term in \(x^3\).
It is given that \(a = 1\).
(b) Hence, or otherwise, show that the series expansion of \(\sin\!\left[\mathrm{f}(x)\right]\), up to and including the term in \(x^3\), is given by \(x - \dfrac{1}{2}x^2 + \dfrac{1}{6}x^3\).
(c) Deduce the Maclaurin series for \(\cos\!\left[\mathrm{f}(x)\right]\) up to and including the term in \(x^2\).
(d) Find \(\displaystyle\int_{1}^{3}\!\left(x - \tfrac{1}{2}x^2 + \tfrac{1}{6}x^3\right) \mathrm{d}x\). Without the use of a calculator or any further calculation, explain whether this value is a good approximation to the value of \(\displaystyle\int_{1}^{3}\sin\!\left[\mathrm{f}(x)\right]\mathrm{d}x\).
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(a) Expanding \(\ln(a+x)\)
Factor out \(a\) so the bracket has the form \(1+u\), then apply the standard logarithmic series with \(u = \dfrac{x}{a}\).
\[ \begin{aligned} \ln(a + x) &= \ln\!\left[a\!\left(1 + \dfrac{x}{a}\right)\right]\\ &= \ln a + \ln\!\left(1 + \dfrac{x}{a}\right)\\ &= \ln a + \left[\dfrac{x}{a} - \dfrac{1}{2}\!\left(\dfrac{x}{a}\right)^{2} + \dfrac{1}{3}\!\left(\dfrac{x}{a}\right)^{3} + \cdots\right]\\ &= \ln a + \dfrac{x}{a} - \dfrac{x^2}{2a^2} + \dfrac{x^3}{3a^3} + \cdots \end{aligned} \](b) Composing inside \(\sin\)
Put \(a=1\) to get \(\mathrm{f}(x)=\ln(1+x)\), substitute it into \(\sin u = u - \dfrac{u^3}{3!} + \cdots\), and keep terms only up to \(x^3\).
With \(a = 1\): \(\mathrm{f}(x) = \ln(1+x) = x - \dfrac{x^2}{2} + \dfrac{x^3}{3} + \cdots\)
\[ \begin{aligned} \sin\!\left[\ln(1+x)\right] &= \left(x - \dfrac{x^2}{2} + \dfrac{x^3}{3} + \cdots\right) - \dfrac{1}{6}\!\left(x - \dfrac{x^2}{2} + \cdots\right)^{3} + \cdots\\ &= x - \dfrac{x^2}{2} + \dfrac{x^3}{3} - \dfrac{x^3}{6} + \cdots\\ &= x - \dfrac{x^2}{2} + \dfrac{x^3}{6} + \cdots \quad \text{(shown)} \end{aligned} \](c) Deducing the series for \(\cos[\mathrm{f}(x)]\)
Differentiate the part (b) result: by the chain rule the left side becomes \(\dfrac{1}{1+x}\cos[\ln(1+x)]\), then multiply back through by \(1+x\).
\[ \begin{aligned} \dfrac{1}{1+x}\cos\!\left[\ln(1+x)\right] &= 1 - x + \dfrac{1}{2}x^2 + \cdots\\ \cos\!\left[\ln(1+x)\right] &= (1+x)\!\left(1 - x + \dfrac{1}{2}x^2 + \cdots\right)\\ &= 1 - \dfrac{1}{2}x^2 + \cdots \end{aligned} \](d) Evaluating the integral and judging the approximation
Integrate the polynomial directly, then check whether \(x\) over the interval of integration stays within the range where the series is valid.
\[ \begin{aligned} \int_{1}^{3}\!\left(x - \dfrac{x^2}{2} + \dfrac{x^3}{6}\right)\mathrm{d}x &= \left[\dfrac{x^2}{2} - \dfrac{x^3}{6} + \dfrac{x^4}{24}\right]_{1}^{3}\\ &= \left(\dfrac{9}{2} - \dfrac{27}{6} + \dfrac{81}{24}\right) - \left(\dfrac{1}{2} - \dfrac{1}{6} + \dfrac{1}{24}\right)\\ &= 3 \end{aligned} \]For \(x \in (1,3] \not\subseteq (-1,1]\), the series expansion \(\sin\!\left[\ln(1+x)\right] = \sin\!\left(x - \dfrac{x^2}{2} + \dfrac{x^3}{3} + \cdots\right)\) is not valid. Hence the value is not a good approximation.