Maclaurin and Taylor series — WFM02 IAL Further Mathematics practice questions · Study-IAL

Maclaurin and Taylor series

WFM02Topic 2.63 questions

Series expansions of functions and their use in approximations and solving differential equations.

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Question 16 marks
$f(x) = \ln(1 + \sin x)$ , $-\dfrac{\pi}{2} < x < \dfrac{\pi}{2}$ .

(a) Show that $f''(x) = -\dfrac{1}{1 + \sin x}$ .

(b) Hence find the Maclaurin series of $f(x)$ up to and including the term in $x^3$ .
Mark scheme
- B1 $f'(x) = \dfrac{\cos x}{1 + \sin x}$ (1)
Question 25 marks
$y$ satisfies the differential equation $\frac{dy}{dx} = x + y^2, \qquad y = 1 \text{ at } x = 0$ Use the Taylor series method to find a series solution for in ascending powers of , up to and including the term in .

M1 Quotient rule:

f''(x) = \dfrac{-\sin x(1 + \sin x) - \cos^2 x}{(1 + \sin x)^2}

(1)

A1 Uses

\sin^2 x + \cos^2 x = 1

to get

\dfrac{-(1 + \sin x)}{(1 + \sin x)^2} = -\dfrac{1}{1 + \sin x}

(1)

f'''(x) = \dfrac{\cos x}{(1 + \sin x)^2}

(1)

f(0) = 0

f'(0) = 1

f''(0) = -1

f'''(0) = 1

(1)

f(x) \approx x - \dfrac{x^2}{2} + \dfrac{x^3}{6}

(1)

y2,y=

1 at x=

y

x

x^3

Mark scheme

B1 $\left(\dfrac{dy}{dx}\right)_0 = 0 + 1 = 1$ (1)
M1 Differentiates implicitly: $\dfrac{d^2y}{dx^2} = 1 + 2y\dfrac{dy}{dx}$ , so (1)
M1 $\dfrac{d^3y}{dx^3} = 2\left(\dfrac{dy}{dx}\right)^2 + 2y\dfrac{d^2y}{dx^2}$ (1)
A1 $\left(\dfrac{d^3y}{dx^3}\right)_0 = 2 + 6 = 8$ (1)
A1 $y = 1 + x + \dfrac{3}{2}x^2 + \dfrac{4}{3}x^3 + \dots$ (1)

2ydxdy

\left(\dfrac{d^2y}{dx^2}\right)_0 = 3

2(dxdy)2+

2ydx2d2y

34x3+

…