Pearson Edexcel IAL Mathematics FP2.6 Maclaurin & Taylor series Question Bank

Practise Maclaurin and Taylor series by finding derivatives, using recurrence from differential equations and writing finite expansions.

Exam points

calculate higher derivatives and evaluate them at x=0 or a given expansion point
use derivative relationships to obtain missing coefficients in a series solution
write expansions in ascending powers of x, x-1 or x-π/3 with simplified coefficients

Covered syllabus points

FP2.6.1 - Third and higher order derivatives FP2.6.2 - Derivation FP2.6.3 - Derivation FP2.6.4 - Taylor series method for differential equations

[Maximum number: 7]

Given that $y=e^x\sin x$

(a)

show that

\frac{\mathrm{d}^{6} y}{\mathrm{~d} x^{6}}=k \frac{\mathrm{~d}^{2} y}{\mathrm{~d} x^{2}}

where k is a constant to be determined.

[ 4 ]

y=e^x\sin x,

y'=e^x\sin x+e^x\cos x,

y''=2e^x\cos x,

y'''=2e^x\cos x-2e^x\sin x,

y^{(4)}=-4e^x\sin x=-4y.

Hence

y^{(6)}=-4y''

so k=-4.

(b)

Hence determine the first 5 non-zero terms in the Maclaurin series expansion for y, giving each coefficient in simplest form.

[ 3 ]

At x=0,

y=0,\quad y'=1,\quad y''=2,\quad y'''=2,\quad y^{(4)}=0,\quad y^{(5)}=-4,\quad y^{(6)}=-8

Therefore

y=0+x+\frac{x^2}{2}\cdot2+\frac{x^3}{3!}\cdot2-\frac{x^5}{5!}\cdot4-\frac{x^6}{6!}\cdot8+\cdots

= =x+x^2+\frac{x^3}{3}-\frac{x^5}{30}-\frac{x^6}{90}+\cdots.

[Maximum number: 8]

Given that $y=\tan ^{2} x$

(a)

show that

\frac{\mathrm{d}^{3} y}{\mathrm{~d} x^{3}}=8 \tan x \sec ^{2} x\left(p \sec ^{2} x+q\right)

where p and q are integers to be determined.

[ 5 ]

$y=\tan^2x\Rightarrow \frac{dy}{dx}=2\tan x\sec^2x$
B1

$\frac{dy}{dx}=2\tan x\sec^2x \Rightarrow \frac{d^2y}{dx^2}=2\sec^4x+4\sec^2x\tan^2x$
M1 A1

$\begin{aligned} \frac{d^2y}{dx^2} &=2\sec^4x+4\sec^2x\tan^2x\\ \Rightarrow\frac{d^3y}{dx^3} &=8\sec^4x\tan x+8\sec^2x\tan^3x+8\sec^4x\tan x \end{aligned}$
or
$\frac{d^2y}{dx^2}=6\sec^4x-4\sec^2x \Rightarrow \frac{d^3y}{dx^3}=24\sec^4x\tan x-8\sec^2x\tan x$
M1

$8\sec^4x\tan x+8\sec^2x\tan x(\sec^2x-1)+8\sec^4x\tan x =8\sec^2x\tan x(3\sec^2x-1)$
A1
(5)

(b)

Hence determine the Taylor series expansion about $\frac{\pi}{3}$ of $\tan ^2 x$ in ascending powers of $x-\frac{\pi}{3}$ up to and including the term in $\left(x-\frac{\pi}{3}\right)^3$ , giving each coefficient in simplest form.

[ 3 ]

$(y)_{\pi/3}=3,\quad (y')_{\pi/3}=8\sqrt3,\quad (y'')_{\pi/3}=80,\quad (y''')_{\pi/3}=352\sqrt3$
M1

$y=3+8\sqrt3\left(x-\frac{\pi}{3}\right) +\frac{80}{2!}\left(x-\frac{\pi}{3}\right)^2 +\frac{352\sqrt3}{3!}\left(x-\frac{\pi}{3}\right)^3+\cdots$
M1

$y=3+8\sqrt3\left(x-\frac{\pi}{3}\right) +40\left(x-\frac{\pi}{3}\right)^2 +\frac{176\sqrt3}{3}\left(x-\frac{\pi}{3}\right)^3+\cdots$
A1
(3)

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Covered syllabus points

FP2.6 - Maclaurin and Taylor series question 1

Question (a)

Question (b)

FP2.6 - Maclaurin and Taylor series question 2

Question (a)

Question (b)