Maclaurin series: TMJC 2024 Promo Q5

(a)

By using the standard series from List of Formulae (MF27), find the series expansion of $\mathrm{e}^{x^2}(1+2x)^{1/2}$, up to and including the term in $x^2$. [3]

(a)

$$\begin{aligned} \mathrm{e}^{x^2}(1+2x)^{1/2} &=\mathrm{e}^{x^2}(1+2x)^{1/2}\\ &=\left(1+x^2+\ldots\right) \left(1+\frac12(2x)+\frac{\frac12\left(-\frac12\right)}{2!}(2x)^2+\ldots\right)\\ &=\left(1+x^2+\ldots\right)\left(1+x-\frac12x^2+\ldots\right)\\ &=1+x+\frac12x^2+\ldots\quad\left(\text{up to }x^2\right) \end{aligned}$$

Use MF 27 Standard Series for $\mathrm{e}^x$ and replace $x$ by $x^2$.

Use MF 27 Standard Series for $(1+x)^n$ and replace $n$ by $\dfrac12$ and $x$ by $2x$.

(b)

Given that $y=\sin^{-1}x$, show that $\dfrac{\,\mathrm{d}^2y}{\,\mathrm{d} x^2}=x\left(\dfrac{\,\mathrm{d} y}{\,\mathrm{d} x}\right)^3$.

Hence, by further differentiation, find the Maclaurin series for $\sin^{-1}x$ in ascending powers of $x$ up to and including the term in $x^3$. [5]

(b)

$$\begin{aligned} y&=\sin^{-1}x\\ \frac{\,\mathrm{d} y}{\,\mathrm{d} x}&=\frac{1}{\sqrt{1-x^2}}=(1-x^2)^{-\frac12}\\ \frac{\,\mathrm{d}^2y}{\,\mathrm{d} x^2}&=\left(-\frac12\right)(1-x^2)^{-\frac32}(-2x)\\ &=x(1-x^2)^{-\frac32}\\ &=x\left(\frac{\,\mathrm{d} y}{\,\mathrm{d} x}\right)^3\qquad\text{(Shown)}\\ \frac{\,\mathrm{d}^3y}{\,\mathrm{d} x^3}&=\left(\frac{\,\mathrm{d} y}{\,\mathrm{d} x}\right)^3+ 3x\left(\frac{\,\mathrm{d} y}{\,\mathrm{d} x}\right)^2\left(\frac{\,\mathrm{d}^2y}{\,\mathrm{d} x^2}\right) \end{aligned}$$ Remember to apply product rule here. Also, it is highly recommended to further differentiate from the “Shown” result, rather than uinsg your own expression or the previous un-simplified expression.

$$\begin{aligned} \text{when }x=0,\qquad y&=0,\quad\frac{\,\mathrm{d} y}{\,\mathrm{d} x}=1,\quad \frac{\,\mathrm{d}^2y}{\,\mathrm{d} x^2}=0,\quad\frac{\,\mathrm{d}^3y}{\,\mathrm{d} x^3}=1.\\ \sin^{-1}x&=x+\frac{1}{3!}x^3+\ldots\\ &=x+\frac16x^3+\ldots\quad\left(\text{up to }x^3\right) \end{aligned}$$