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Maclaurin Series: binomial expansion, composite exponential and series integration — NJC 2025 H2 Maths Prelim Paper 1
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Question
In this question, you may use expansions from the List of Formulae and Results (MF27).
It is given that \(a>0\).
(i) Find, in terms of \(a\), the series expansion of \(\dfrac{a}{a-x}-1\), in ascending powers of \(x\), up to and including the term in \(x^2\). State, in terms of \(a\), the range of \(x\) for which the expansion is valid.
(ii) Hence, find the Maclaurin expansion of \(\mathrm{e}^{\frac{a}{a-x}-1}\) in ascending powers of \(x\), up to and including the term in \(x^2\).
(iii) Use the expansion in part (ii) to approximate \(\displaystyle\int_0^{a/2}\mathrm{e}^{\frac{a}{a-x}-1}\,\mathrm{d}x\). Explain why this approximation is an under-estimation.
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(i) Binomial expansion and validity
Rewrite the fraction as \((1-\tfrac{x}{a})^{-1}\) and apply the binomial series, then subtract \(1\).
\[ \begin{aligned} \frac{a}{a-x} - 1 &= a(a-x)^{-1} - 1 \\ &= \left(1-\frac{x}{a}\right)^{-1} - 1 \\ &= \left[1+\left(-\frac{x}{a}\right)+\left(-\frac{x}{a}\right)^{2}+\cdots\right] - 1 \\ &= \frac{x}{a}+\frac{x^2}{a^2}+\cdots \end{aligned} \]Validity: \(\left|-\dfrac{x}{a}\right|<1 \Rightarrow |x|<a\) (since \(a>0\)), i.e. \(-a<x<a\).
\(\dfrac{a}{a-x}-1 \approx \dfrac{x}{a}+\dfrac{x^2}{a^2}+\cdots\), valid for \(-a<x<a\).
(ii) Composite exponential expansion
Substitute the series from (i) into \(\mathrm{e}^{u}=1+u+\tfrac{1}{2}u^2+\cdots\), keeping terms up to \(x^2\).
\[ \begin{aligned} \mathrm{e}^{\frac{a}{a-x}-1} &= \mathrm{e}^{\frac{x}{a}+\frac{x^2}{a^2}+\cdots} \\ &\approx 1 + \left(\frac{x}{a}+\frac{x^2}{a^2}\right) + \frac{1}{2}\left(\frac{x}{a}\right)^{2} + \cdots \\ &= 1 + \frac{x}{a} + \frac{x^2}{a^2} + \frac{x^2}{2a^2} + \cdots \end{aligned} \]\(\mathrm{e}^{\frac{a}{a-x}-1} \approx 1 + \dfrac{x}{a} + \dfrac{3x^2}{2a^2} + \cdots\)
(iii) Approximating the integral
Integrate the polynomial approximation term by term from \(0\) to \(a/2\).
\[ \begin{aligned} \int_0^{a/2} \mathrm{e}^{\frac{a}{a-x}-1}\,\mathrm{d}x &\approx \int_0^{a/2} \left(1+\frac{x}{a}+\frac{3x^2}{2a^2}\right)\,\mathrm{d}x \\ &= \left[x+\frac{x^2}{2a}+\frac{x^3}{2a^2}\right]_0^{a/2} \\ &= \frac{a}{2}+\frac{(a/2)^2}{2a}+\frac{(a/2)^3}{2a^2} \\ &= \frac{a}{2}+\frac{a}{8}+\frac{a}{16} = \frac{11a}{16} \end{aligned} \]Neglected terms for \(0<x<\dfrac{a}{2}\) are all positive, so the approximation under-estimates the integral.