IAL 2025 June A Q3 | 國際數學站

Given that $y = \arcsin 2x$

(a) show that

\begin{align*} \frac{\mathrm{d}^3y}{\mathrm{d}x^3} = \frac{Ax^2 + 8}{\left(1 - 4x^2\right)^{\frac{5}{2}}} \end{align*}

where $A$ is a constant to be determined.

(3)

(b) Hence determine the Maclaurin series expansion for $\arcsin 2x$ in ascending powers of $x$ up to and including the term in $x^3$

(2)

The Maclaurin series expansion for $e^x$ is given by

\begin{align*} e^x = 1 + x + \frac{x^2}{2} + \ldots + \frac{x^r}{r!} + \ldots \end{align*}

\begin{align*} e^{3x}\arcsin 2x \approx Cx + Dx^2 + Ex^3 \end{align*}

where $C$ , $D$ and $E$ are constants to be determined.

(3)

解答

(a)

Given $y = \arcsin(2x)$ :

\begin{align*} \frac{\mathrm{d}y}{\mathrm{d}x} = &\,\frac{1}{\sqrt{1 - (2x)^2}} \cdot 2\\[4mm] = &\,\frac{2}{\sqrt{1 - 4x^2}}\\[4mm] = &\,2(1 - 4x^2)^{-\frac{1}{2}} \end{align*}

Differentiating again using the chain rule:

\begin{align*} \frac{\mathrm{d}^2y}{\mathrm{d}x^2} = &\,2\left(-\frac{1}{2}\right)(1 - 4x^2)^{-\frac{3}{2}} \cdot (-8x)\\[4mm] = &\,8x(1 - 4x^2)^{-\frac{3}{2}} \end{align*}

Differentiating a third time using the product rule:

\begin{align*} \frac{\mathrm{d}^3y}{\mathrm{d}x^3} = &\,8(1 - 4x^2)^{-\frac{3}{2}} + 8x\left(-\frac{3}{2}\right)(1 - 4x^2)^{-\frac{5}{2}} \cdot (-8x)\\[4mm] = &\,8(1 - 4x^2)^{-\frac{3}{2}} + 96x^2(1 - 4x^2)^{-\frac{5}{2}}\\[4mm] = &\,\frac{8(1 - 4x^2) + 96x^2}{(1 - 4x^2)^{\frac{5}{2}}}\\[4mm] = &\,\frac{8 - 32x^2 + 96x^2}{(1 - 4x^2)^{\frac{5}{2}}}\\[4mm] = &\,\frac{64x^2 + 8}{(1 - 4x^2)^{\frac{5}{2}}} \end{align*}

This is in the form $\frac{Ax^2+8}{(1-4x^2)^{\frac{5}{2}}}$ , where $A = 64$ .

(b)

To find the Maclaurin series expansion for $y = \arcsin(2x)$ , we evaluate the derivatives at $x = 0$ :

\begin{align*} y(0) = &\,\arcsin(0) = 0\\[4mm] y'(0) = &\,2(1 - 0)^{-\frac{1}{2}} = 2\\[4mm] y''(0) = &\,8(0)(1 - 0)^{-\frac{3}{2}} = 0\\[4mm] y'''(0) = &\,\frac{64(0) + 8}{(1 - 0)^{\frac{5}{2}}} = 8 \end{align*}

Using the Maclaurin series formula $f(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \dots$ :

\begin{align*} \arcsin(2x) \approx &\,0 + 2x + \frac{0}{2}x^2 + \frac{8}{6}x^3\\[4mm] \approx &\,2x + \frac{4}{3}x^3 \end{align*}

(c)

Using the given Maclaurin series for $\mathrm{e}^x$ :

\begin{align*} \mathrm{e}^{3x} \approx &\,1 + (3x) + \frac{(3x)^2}{2!} + \dots\\[4mm] \approx &\,1 + 3x + \frac{9}{2}x^2 \end{align*}

We multiply the series for $\mathrm{e}^{3x}$ and $\arcsin(2x)$ , keeping terms up to $x^3$ :

\begin{align*} \mathrm{e}^{3x} \arcsin(2x) \approx &\,\left(1 + 3x + \frac{9}{2}x^2\right) \left(2x + \frac{4}{3}x^3\right)\\[4mm] \approx &\,1\left(2x + \frac{4}{3}x^3\right) + 3x(2x) + \frac{9}{2}x^2(2x)\\[4mm] \approx &\,2x + \frac{4}{3}x^3 + 6x^2 + 9x^3\\[4mm] \approx &\,2x + 6x^2 + \frac{31}{3}x^3 \end{align*}

This matches the form $Cx + Dx^2 + Ex^3$ , with constants $C=2$ , $D=6$ , and $E=\frac{31}{3}$ .