Edexcel A-Level Mathematics A2 Fp2 6 Maclaurin and Taylor Series Questions

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 ]

$y=\tan^2x\Rightarrow \frac{dy}{dx}=2\tan x\sec^2x$ $\frac{dy}{dx}=2\tan x\sec^2x \Rightarrow \frac{d^2y}{dx^2}=2\sec^4x+4\sec^2x\tan^2x$ $\begin{aligned} \frac{d^2y}{dx^2} &=2\sec^4x+4\sec^2x\tan^2x\\ \Rightarrow\frac{d^3y}{dx^3} &=8\sec^4x\tan x+8\sec^2x\tan^3x+8\sec^4x\tan x \end{aligned}$
or
$\frac{d^2y}{dx^2}=6\sec^4x-4\sec^2x \Rightarrow \frac{d^3y}{dx^3}=24\sec^4x\tan x-8\sec^2x\tan x$ $8\sec^4x\tan x+8\sec^2x\tan x(\sec^2x-1)+8\sec^4x\tan x =8\sec^2x\tan x(3\sec^2x-1)$
(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 ]

$(y)_{\pi/3}=3,\quad (y')_{\pi/3}=8\sqrt3,\quad (y'')_{\pi/3}=80,\quad (y''')_{\pi/3}=352\sqrt3$ $y=3+8\sqrt3\left(x-\frac{\pi}{3}\right) +\frac{80}{2!}\left(x-\frac{\pi}{3}\right)^2 +\frac{352\sqrt3}{3!}\left(x-\frac{\pi}{3}\right)^3+\cdots$ $y=3+8\sqrt3\left(x-\frac{\pi}{3}\right) +40\left(x-\frac{\pi}{3}\right)^2 +\frac{176\sqrt3}{3}\left(x-\frac{\pi}{3}\right)^3+\cdots$
(3)

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Covered syllabus points

Question 1

Question (a)

Question (b)

Question 2

Question (a)

Question (b)