Maclaurin series: CJC 2024 Promo Q6

(a)

$$\begin{aligned} y&=\sin(\mathrm{e}^x-1)\\ \frac{\,\mathrm{d} y}{\,\mathrm{d} x} &= \cos(\mathrm{e}^x - 1)\,\mathrm{e}^x. \end{aligned}$$ $$\begin{aligned} \frac{\,\mathrm{d}^2 y}{\,\mathrm{d} x^2} &= \mathrm{e}^x\left[-\sin(\mathrm{e}^x - 1)\,\mathrm{e}^x\right] + \cos(\mathrm{e}^x - 1)\,\mathrm{e}^x\\ &= -\mathrm{e}^{2x}\sin(\mathrm{e}^x - 1) + \frac{\,\mathrm{d} y}{\,\mathrm{d} x}\\ &= \frac{\,\mathrm{d} y}{\,\mathrm{d} x} - y\mathrm{e}^{2x}\quad\text{(shown)} \end{aligned}$$

(b)

When $x = 0$: $$\begin{aligned} y &= \sin(\mathrm{e}^0 - 1) = 0\\ \frac{\,\mathrm{d} y}{\,\mathrm{d} x} &= \cos(\mathrm{e}^0 - 1)\,\mathrm{e}^0 = 1\\ \frac{\,\mathrm{d}^2 y}{\,\mathrm{d} x^2} &= 1. \end{aligned}$$ Hence $$\begin{aligned} \mathrm{f}(x) &= 0 + x + \frac{1}{2!}x^2+\cdots\\ &= x + \frac{1}{2}x^2+\cdots. \end{aligned}$$

(c)

$$\begin{aligned} \sqrt{1 + \sin(\mathrm{e}^x - 1)} &= \left[1 + \sin(\mathrm{e}^x - 1)\right]^{1/2}\\ &\approx \left[1 + x + \frac{x^2}{2}\right]^{1/2}\\ &= 1 + \frac{1}{2}\left(x + \frac{x^2}{2}\right) + \frac{\tfrac{1}{2}\left(\tfrac{1}{2} - 1\right)}{2!}\left(x + \frac{x^2}{2}\right)^{\!2} + \cdots\\ &= 1 + \frac{1}{2}x + \frac{1}{4}x^2 - \frac{1}{8}(x^2 + \cdots) + \cdots\\ &= 1 + \frac{1}{2}x + \frac{1}{8}x^2 + \cdots \end{aligned}$$