Unit 30 Review Exercises
Maclaurin Series
These questions mix the ideas from the whole unit. All of them are HL only AHL. Paper 1 means no calculator; Paper 2 means a GDC is allowed.
1 Paper 1 AHL Let \(f(x) = \sin\left(x + \dfrac{\pi}{6}\right)\). Use the Maclaurin series formula to find the series for \(f(x)\) up to and including the term in \(x^3\).
\(f(0) = \sin\frac{\pi}{6} = \frac{1}{2}\), \(\;f'(0) = \cos\frac{\pi}{6} = \frac{\sqrt{3}}{2}\), \(\;f''(0) = -\sin\frac{\pi}{6} = -\frac{1}{2}\), \(\;f'''(0) = -\cos\frac{\pi}{6} = -\frac{\sqrt{3}}{2}\).
\[ f(x) = \frac{1}{2} + \frac{\sqrt{3}}{2}x - \frac{1}{4}x^2 - \frac{\sqrt{3}}{12}x^3 + \cdots \]
See 30.1.
2 Paper 1 AHL Given that \(y = \mathrm{e}^{\cos x}\), show that \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = -y\sin x\). By repeated differentiation, find the Maclaurin series for \(y\) up to and including the term in \(x^3\).
Chain rule: \(y' = -\sin x\,\mathrm{e}^{\cos x} = -y\sin x\).
\(y'' = -y'\sin x - y\cos x\), \(\quad y''' = -y''\sin x - 2y'\cos x + y\sin x\).
At \(0\): \(y = \mathrm{e}\), \(y' = 0\), \(y'' = -\mathrm{e}\), \(y''' = 0\).
\[ y = \mathrm{e} - \frac{\mathrm{e}}{2}x^2 + 0x^3 + \cdots \]
See 30.1.
3 Paper 1 AHL The function \(y = f(x)\) satisfies \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = 1 + xy\) and \(f(0) = 0\). Find the Maclaurin series for \(f(x)\) up to and including the term in \(x^3\).
\(y' = 1 + xy\), \(\;y'' = y + xy'\), \(\;y''' = 2y' + xy''\).
At \(0\): \(y = 0\), \(y' = 1\), \(y'' = 0\), \(y''' = 2\).
\[ f(x) = x + \frac{2}{3!}x^3 + \cdots = x + \frac{x^3}{3} + \cdots \]
See 30.1.
4 Paper 1 AHL Find the first three terms of the Maclaurin series for \(\ln\left(1 + \dfrac{x}{2}\right)\), and state the values of \(x\) for which the series is valid.
\(\dfrac{x}{2} - \dfrac{1}{2}\left(\dfrac{x}{2}\right)^2 + \dfrac{1}{3}\left(\dfrac{x}{2}\right)^3 = \dfrac{x}{2} - \dfrac{x^2}{8} + \dfrac{x^3}{24}\).
Valid for \(-1 < \frac{x}{2} \leqslant 1\), i.e. \(-2 < x \leqslant 2\). See 30.2.
5 Paper 1 AHL Find the Maclaurin series for \((1 + x)^{-\frac{1}{3}}\) up to the term in \(x^2\), and state its interval of convergence.
\(1 - \dfrac{1}{3}x + \dfrac{\left(-\frac{1}{3}\right)\left(-\frac{4}{3}\right)}{2}x^2 = 1 - \dfrac{x}{3} + \dfrac{2x^2}{9} + \cdots\), \(\;\) valid for \(|x| < 1\). See 30.2.
6 Paper 1 AHL Find the Maclaurin series for \(\mathrm{e}^x\sin x\) up to and including the term in \(x^4\).
\(\left(1 + x + \frac{x^2}{2} + \frac{x^3}{6}\right)\left(x - \frac{x^3}{6}\right)\):
\(x\): \(1\); \(\;x^2\): \(1\); \(\;x^3\): \(\frac{1}{2} - \frac{1}{6} = \frac{1}{3}\); \(\;x^4\): \(\frac{1}{6} - \frac{1}{6} = 0\).
\(\mathrm{e}^x\sin x = x + x^2 + \dfrac{x^3}{3} + 0x^4 + \cdots\) See 30.3.
7 Paper 1 AHL Use Maclaurin series to find \(\displaystyle\lim_{x \to 0}\frac{\cos x - 1 + \frac{x^2}{2}}{x^4}\).
\(\cos x - 1 + \dfrac{x^2}{2} = \dfrac{x^4}{24} - \dfrac{x^6}{720} + \cdots\), so the fraction is \(\dfrac{1}{24} - \dfrac{x^2}{720} + \cdots \to \dfrac{1}{24}\).
See 30.3.
8 Paper 1 AHL
(a) Show that \(\ln\left(\dfrac{1 + x}{1 - x}\right) = 2\left(x + \dfrac{x^3}{3} + \dfrac{x^5}{5} + \cdots\right)\) for \(|x| < 1\).
(b) Find the value of \(x\) for which \(\dfrac{1 + x}{1 - x} = 3\), and hence use three terms of the series to find an approximation for \(\ln 3\) as a fraction.
(a) \(\ln(1 + x) - \ln(1 - x) = \left(x - \frac{x^2}{2} + \frac{x^3}{3} - \cdots\right) - \left(-x - \frac{x^2}{2} - \frac{x^3}{3} - \cdots\right) = 2x + \frac{2x^3}{3} + \frac{2x^5}{5} + \cdots\)
Both series need \(|x| < 1\).
(b) \(1 + x = 3 - 3x \Rightarrow x = \frac{1}{2}\) (inside \(|x| < 1\)).
\(\ln 3 \approx 2\left(\frac{1}{2} + \frac{1}{24} + \frac{1}{160}\right) = 1 + \frac{1}{12} + \frac{1}{80} = \frac{263}{240} \approx 1.0958\) (true value \(1.0986\)).
9 Paper 2 AHL
(a) Find the first three non-zero terms of the Maclaurin series for \(\cos(x^2)\).
(b) Hence find an approximation for \(\displaystyle\int_0^1\cos(x^2)\,\mathrm{d}x\), and compare it with the value from your GDC.
(a) Replace \(x\) by \(x^2\): \(\;\cos(x^2) = 1 - \dfrac{x^4}{2} + \dfrac{x^8}{24} - \cdots\)
(b) \(\displaystyle\int_0^1 \cdots\,\mathrm{d}x = \left[x - \frac{x^5}{10} + \frac{x^9}{216}\right]_0^1 = 1 - \frac{1}{10} + \frac{1}{216} = 0.904630\) (6 d.p.).
GDC (Numerical Integral): \(0.904524\). The approximation is correct to 3 decimal places. See 30.3.
10 Paper 2 AHL Find \(M_1(1.5)\), \(M_2(1.5)\), \(M_3(1.5)\) and \(M_4(1.5)\) for \(f(x) = \ln(1 + x)\). Compare with \(\ln 2.5\) and explain what you see.
\(M_1 = 1.5\), \(\;M_2 = 1.5 - 1.125 = 0.375\), \(\;M_3 = 0.375 + 1.125 = 1.5\), \(\;M_4 = 1.5 - 1.265625 = 0.234375\).
\(\ln 2.5 = 0.916291\ldots\) The values swing up and down and do not settle towards \(\ln 2.5\), because the terms \(\frac{1.5^k}{k}\) grow. \(x = 1.5\) is outside the interval of convergence \(-1 < x \leqslant 1\). See 30.2.