Given that&nbsp;$y={{({{\sin }^{-1}}x)}^{2}}$, show that
\[\left( 1-{{x}^{2}} \right){{\left( \frac{\mathrm{d}y}{\mathrm{d}x} \right)}^{2}}=4y\]
and $\left( 1-{{x}^{2}} \right)\frac{{{\mathrm{d}}^{\text{2}}}y}{\mathrm{d}{{x}^{2}}}-x\frac{\mathrm{d}y}{\mathrm{d}x}=2$.[3]
By further differentiation of these results, find the Maclaurin series of $y$ up to including the term in ${{x}^{4}}$.[3]
The series found above is to be used to estimate a value for&nbsp;${{\pi }^{2}}$. Explain with the aid of a sketch why using $x=\frac{1}{2}$ in the series gives a better approximation than $x=1$, and hence find an approximation to ${{\pi }^{2}}$, giving your answer in fraction form. [4]

Given that $y={{({{\sin }^{-1}}x)}^{2}}$, show that

\[\left( 1-{{x}^{2}} \right){{\left( \frac{\mathrm{d}y}{\mathrm{d}x} \right)}^{2}}=4y\]

and $\left( 1-{{x}^{2}} \right)\frac{{{\mathrm{d}}^{\text{2}}}y}{\mathrm{d}{{x}^{2}}}-x\frac{\mathrm{d}y}{\mathrm{d}x}=2$.

By further differentiation of these results, find the Maclaurin series of $y$ up to including the term in ${{x}^{4}}$.

The series found above is to be used to estimate a value for ${{\pi }^{2}}$. Explain with the aid of a sketch why using $x=\frac{1}{2}$ in the series gives a better approximation than $x=1$, and hence find an approximation to ${{\pi }^{2}}$, giving your answer in fraction form. 

Maclaurin Series and Power Series - General

Maclaurin Series and Power Series

H2 Math

International Baccalaureate Higher Level Mathematics

IB HL Math

2013 ACJC P1 Q12

**Show that $(1-x^2)\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^2 = 4y$:**

$y = (\sin^{-1}x)^2$

$\frac{\mathrm{d}y}{\mathrm{d}x} = 2(\sin^{-1}x) \cdot \frac{1}{\sqrt{1-x^2}}$

$\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^2 = 4(\sin^{-1}x)^2 \cdot \frac{1}{1-x^2} = \frac{4y}{1-x^2}$

$\therefore (1-x^2)\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^2 = 4y$ (shown)

**Finding derivatives at x = 0:**

At $x = 0$: $y = 0$, $f'(0) = 0$

Differentiate w.r.t. x:
$(1-x^2)\frac{\mathrm{d}^2y}{dx^2} - x\frac{\mathrm{d}y}{\mathrm{d}x} = 2$

At $x = 0$: $f''(0) = 2$

Differentiate again:
$(1-x^2)\frac{\mathrm{d}^3y}{dx^3} - 2x\frac{\mathrm{d}^2y}{dx^2} - x\frac{\mathrm{d}^2y}{dx^2} - \frac{\mathrm{d}y}{\mathrm{d}x} = 0$

At $x = 0$: $f'''(0) = 0$

Differentiate again:
$(1-x^2)\frac{\mathrm{d}^4y}{dx^4} + \frac{\mathrm{d}^3y}{dx^3}(-2x) - 3\left[x\frac{\mathrm{d}^3y}{dx^3} + \frac{\mathrm{d}^2y}{dx^2} \cdot 1\right] - \frac{\mathrm{d}^2y}{dx^2} = 0$

At $x = 0$: $\frac{\mathrm{d}^4y}{dx^4} + 0 - 3(0 + 2) - 2 = 0$

$\therefore \frac{\mathrm{d}^4y}{dx^4} = 8$

**Maclaurin series:**

$\mathrm{f}(x) = f(0) + xf'(0) + \frac{x^2}{2!}f''(0) + \ldots + \frac{x^n}{n!}f^{(n)}(0) + \ldots$

$(\sin^{-1}x)^2 = 0 + x(0) + \frac{x^2}{2!}(2) + \frac{x^3}{3!}(0) + \frac{x^4}{4!}(8) + \ldots$

$= x^2 + \frac{1}{3}x^4 + \ldots$

**Using $x = \frac{1}{2}$:**
$(\sin^{-1}\frac{1}{2})^2 \approx \left(\frac{1}{2}\right)^2 + \frac{1}{3}\left(\frac{1}{2}\right)^4$
$\left(\frac{\pi}{6}\right)^2 \approx \frac{1}{4} + \frac{1}{48} = \frac{13}{48}$
$\frac{\pi^2}{36} \approx \frac{13}{48}$
$\therefore \pi^2 \approx \frac{39}{4}$
% error = 1.21%

**Using $x = 1$:**
$(\sin^{-1}1)^2 \approx 1^2 + \frac{1}{3}(1)^4$
$\left(\frac{\pi}{2}\right)^2 \approx \frac{4}{3}$
$\pi^2 \approx \frac{16}{3}$
% error = 45.9%

An approximation using $x = \frac{1}{2}$ is better than using $x = 1$ as the graph $(\sin^{-1}x)^2$ is closer to $y = x^2 + \frac{1}{3}x^4$ at $x = \frac{1}{2}$.

$\therefore \pi^2 \approx \frac{39}{4}$

2013 ACJC P1 Q12

2013 ACJC P1 Q12

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