Maclaurin series: NYJC 2024 Promo Q8

(a)

$$\begin{aligned} y=\sqrt{\ln(x+\mathrm{e})}&\Rightarrow y^2=\ln(x+\mathrm{e})\Rightarrow \mathrm{e}^{y^2}=x+\mathrm{e}\\ 2y\mathrm{e}^{y^2}\frac{\,\mathrm{d} y}{\,\mathrm{d} x}&=1\Rightarrow 2y\frac{\,\mathrm{d} y}{\,\mathrm{d} x}=\mathrm{e}^{-y^2}.\quad\text{(shown)} \end{aligned}$$ OR

Differentiate w.r.t. $x$ gives $$2y\frac{\,\mathrm{d} y}{\,\mathrm{d} x}=\frac1{x+\mathrm{e}}=\frac1{\mathrm{e}^{y^2}} \quad\text{since }y^2=\ln(x+\mathrm{e})\Rightarrow x+\mathrm{e}=\mathrm{e}^{y^2}$$ That is, $2y\dfrac{\,\mathrm{d} y}{\,\mathrm{d} x}=\mathrm{e}^{-y^2}$.

Differentiate w.r.t. $x$ gives $$\begin{aligned} 2\left[y\frac{\,\mathrm{d}^2y}{\,\mathrm{d} x^2}+\left(\frac{\,\mathrm{d} y}{\,\mathrm{d} x}\right)^2\right] &=-2y\frac{\,\mathrm{d} y}{\,\mathrm{d} x}\mathrm{e}^{-y^2}\\ y\frac{\,\mathrm{d}^2y}{\,\mathrm{d} x^2}+\left(\frac{\,\mathrm{d} y}{\,\mathrm{d} x}\right)^2 &=-y\frac{\,\mathrm{d} y}{\,\mathrm{d} x}\mathrm{e}^{-y^2} \end{aligned}$$ When $x=0$, $y=1$, $\dfrac{\,\mathrm{d} y}{\,\mathrm{d} x}=\dfrac1{2\mathrm{e}}$, $\dfrac{\,\mathrm{d}^2y}{\,\mathrm{d} x^2}=-\dfrac3{4\mathrm{e}^2}$. $$y=1+\frac1{2\mathrm{e}}x+\frac1{2!}\left(-\frac3{4\mathrm{e}^2}\right)x^2+\cdots =1+\frac1{2\mathrm{e}}x-\frac3{8\mathrm{e}^2}x^2+\cdots.$$

(b)

$$\begin{aligned} \ln(x+\mathrm{e})&=\ln\mathrm{e}\left(1+\frac{x}{\mathrm{e}}\right)\\ &=1+\ln\left(1+\frac{x}{\mathrm{e}}\right)\\ &=1+\frac{x}{\mathrm{e}}-\frac12\left(\frac{x}{\mathrm{e}}\right)^2+\cdots\\ &=1+\left(\frac{x}{\mathrm{e}}-\frac{x^2}{2\mathrm{e}^2}+\cdots\right) \end{aligned}$$ $$\begin{aligned} y&=\sqrt{\ln(x+\mathrm{e})}\\&=\left[\ln(x+\mathrm{e})\right]^{1/2}\\ &=\left[1+\left(\frac{x}{\mathrm{e}}-\frac{x^2}{2\mathrm{e}^2}+\cdots\right)\right]^{1/2}\\ &=1+\frac12\left(\frac{x}{\mathrm{e}}-\frac{x^2}{2\mathrm{e}^2}+\cdots\right) +\frac{\left(\frac12\right)\left(-\frac12\right)}{2!} \left(\frac{x}{\mathrm{e}}-\frac{x^2}{2\mathrm{e}^2}+\cdots\right)^2+\cdots\\ &=1+\frac1{2\mathrm{e}}x-\frac3{8\mathrm{e}^2}x^2+\cdots \end{aligned}$$ which agrees with the expansion in (a).

(c)

$$\begin{aligned} \sqrt{\ln\left(\frac{1+10\mathrm{e}}{10}\right)}&=\sqrt{\ln\left(\frac1{10}+\mathrm{e}\right)}\\ &\approx1+\frac1{2\mathrm{e}}\left(\frac1{10}\right)-\frac3{8\mathrm{e}^2}\left(\frac1{10}\right)^2\\ &=1+\frac1{20\mathrm{e}}-\frac3{800\mathrm{e}^2}\\&=\frac{800\mathrm{e}^2+40\mathrm{e}-3}{800\mathrm{e}^2} \end{aligned}$$