Maclaurin Series: binomial expansion and term-by-term integration — ACJC 2025 H2 Math Prelim P2
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Question
It is given that \(\mathrm{f}(x) = \dfrac{1}{\sqrt{1-4x^2}}\) where \(-\dfrac{1}{2} < x < \dfrac{1}{2}\).
(a) Using standard series from the List of Formulae (MF27), find the Maclaurin's expansion of \(\mathrm{f}(x)\), up to and including the term in \(x^6\).
(b) Hence find the first four non-zero terms of the Maclaurin series for \(\sin^{-1} 2x\).
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(a) Binomial expansion of \((1-4x^2)^{-1/2}\)
Write \(\mathrm{f}(x)\) as a power and apply the \((1+u)^n\) binomial series with \(u=-4x^2\).
\[ \begin{aligned} \mathrm{f}(x) &= \frac{1}{\sqrt{1-4x^2}} = \left(1-4x^2\right)^{-\frac{1}{2}}\\ &= 1 + \left(-\tfrac{1}{2}\right)(-4x^2) + \frac{(-\tfrac{1}{2})(-\tfrac{3}{2})}{2!}(-4x^2)^2 + \frac{(-\tfrac{1}{2})(-\tfrac{3}{2})(-\tfrac{5}{2})}{3!}(-4x^2)^3 + \cdots\\ &= 1 + 2x^2 + 6x^4 + 20x^6 + \cdots \end{aligned} \](b) Integrate the series term by term
Since differentiating \(\sin^{-1}2x\) gives \(\dfrac{2}{\sqrt{1-4x^2}}\), integrating \(\mathrm{f}(x)\) recovers \(\tfrac{1}{2}\sin^{-1}2x\), so integrate the part (a) series and fix the constant.
Since \(\dfrac{\mathrm{d}}{\mathrm{d}x}\left(\sin^{-1}2x\right) = \dfrac{2}{\sqrt{1-4x^2}}\), we have
\[ \int \frac{1}{\sqrt{1-4x^2}}\,\mathrm{d}x = \frac{1}{2}\sin^{-1}2x + c. \]Integrating the series from (a) term by term:
\[ \begin{aligned} \frac{1}{2}\sin^{-1}2x + c &= \int 1 + 2x^2 + 6x^4 + 20x^6\,\mathrm{d}x\\ &= x + \frac{2}{3}x^3 + \frac{6}{5}x^5 + \frac{20}{7}x^7 + c \end{aligned} \]so
\[ \sin^{-1}2x + d = 2x + \frac{4}{3}x^3 + \frac{12}{5}x^5 + \frac{40}{7}x^7, \]where \(d = 2c\). Substituting \(x = 0\): \(\sin^{-1}0 = 0 \Rightarrow d = 0\).