Maclaurin series: SAJC 2024 Promo Q8

(i)

From $y = \mathrm{e}^{\tan^{-1}(x)}$, differentiating with respect to $x$: $$\begin{aligned} \frac{\,\mathrm{d} y}{\,\mathrm{d} x} &= \mathrm{e}^{\tan^{-1}(x)}\left(\frac{1}{1 + x^2}\right). \end{aligned}$$

Multiplying both sides by $(1 + x^2)$ and noting $y = \mathrm{e}^{\tan^{-1}(x)}$: $$\begin{aligned} (1 + x^2)\frac{\,\mathrm{d} y}{\,\mathrm{d} x} &= y. \quad\text{(1)} \end{aligned}$$ Differentiating $(1)$ with respect to $x$: $$\begin{aligned} 2x\frac{\,\mathrm{d} y}{\,\mathrm{d} x} + (1 + x^2)\frac{\,\mathrm{d}^2 y}{\,\mathrm{d} x^2} &= \frac{\,\mathrm{d} y}{\,\mathrm{d} x}\\ (1 + x^2)\frac{\,\mathrm{d}^2 y}{\,\mathrm{d} x^2} &= (1 - 2x)\frac{\,\mathrm{d} y}{\,\mathrm{d} x}. \quad\text{((2)\quad(shown))} \end{aligned}$$

Alternative Solution

From $y=\mathrm{e}^{\tan^{-1}(x)}$, $$\begin{aligned} \ln y&=\tan^{-1}(x),\\ \frac1y\frac{\,\mathrm{d} y}{\,\mathrm{d} x}&=\frac1{1+x^2},\\ (1+x^2)\frac{\,\mathrm{d} y}{\,\mathrm{d} x}&=y. \quad\text{(1)} \end{aligned}$$ Differentiate (1) with respect to $x$, $$\begin{aligned} 2x\frac{\,\mathrm{d} y}{\,\mathrm{d} x}+(1+x^2)\frac{\,\mathrm{d}^2y}{\,\mathrm{d} x^2}&=\frac{\,\mathrm{d} y}{\,\mathrm{d} x},\\ (1+x^2)\frac{\,\mathrm{d}^2y}{\,\mathrm{d} x^2}&=(1-2x)\frac{\,\mathrm{d} y}{\,\mathrm{d} x}. \quad\text{(2)} \end{aligned}$$

(ii)

Differentiating $(2)$ with respect to $x$: $$\begin{aligned} (1 + x^2)\frac{\,\mathrm{d}^3 y}{\,\mathrm{d} x^3} + 2x\frac{\,\mathrm{d}^2 y}{\,\mathrm{d} x^2} &= (1 - 2x)\frac{\,\mathrm{d}^2 y}{\,\mathrm{d} x^2} - 2\frac{\,\mathrm{d} y}{\,\mathrm{d} x}. \quad\text{(3)} \end{aligned}$$

At $x = 0$: $y = \mathrm{e}^{\tan^{-1}(0)} = 1$. From $(1)$: $\dfrac{\,\mathrm{d} y}{\,\mathrm{d} x} = 1$. From $(2)$: $\dfrac{\,\mathrm{d}^2 y}{\,\mathrm{d} x^2} = 1$. From $(3)$: $\dfrac{\,\mathrm{d}^3 y}{\,\mathrm{d} x^3} = 1 - 2 = -1$. Hence $$\begin{aligned} y &\approx 1 + x(1) + \frac{x^2}{2}(1) + \frac{x^3}{6}(-1)\\ &= 1 + x + \frac{1}{2}x^2 - \frac{1}{6}x^3. \end{aligned}$$

(iii)

Choosing $x = \dfrac{1}{\sqrt{3}}$ so that $\tan^{-1}\!\left(\dfrac{1}{\sqrt{3}}\right) = \dfrac{\pi}{6}$: $$\begin{aligned} \mathrm{e}^{\frac{\pi}{6}} &\approx 1 + \frac{1}{\sqrt{3}} + \frac{1}{2}\left(\frac{1}{\sqrt{3}}\right)^{\!2} - \frac{1}{6}\left(\frac{1}{\sqrt{3}}\right)^{\!3}\\ &= 1 + \frac{1}{6} + \frac{1}{\sqrt{3}} - \frac{1}{18}\left(\frac{1}{\sqrt{3}}\right)\\ &= \frac{7}{6} + \left(1 - \frac{1}{18}\right)\frac{1}{\sqrt{3}}\\ &= \frac{7}{6} + \frac{17}{18}\left(\frac{\sqrt{3}}{3}\right)\\ &= \frac{7}{6} + \frac{17}{54}\sqrt{3}. \end{aligned}$$ Hence $p = 7$, $q = 6$, $r = 17$ and $s = 54$.