Edexcel IAL Maths FP2.6 Maclaurin and Taylor

[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)

[Maximum number: 4]

Given that

y \frac{\mathrm{~d}^{2} y}{\mathrm{~d} x^{2}}-4\left(\frac{\mathrm{~d} y}{\mathrm{~d} x}\right)^{2}+3 y=0

find a series solution for y in ascending powers of x, up to and including the term in $x^{3}$ , simplifying the coefficients where possible.

$\begin{aligned} \text { At } x=0 \quad \frac{\mathrm{~d}^{2} y}{\mathrm{~d} x^{2}}=\frac{4}{8}(1)^{2}-3=-\frac{5}{2} \quad \text { oe } \frac{\mathrm{d}^{3} y}{\mathrm{~d} x^{3}}=\frac{28}{64} \times 1^{3}-\frac{24}{8} \times 1=-\frac{41}{16} y=8+x-\frac{5}{2} \times \frac{x^{2}}{2!}-\frac{41}{16} \times \frac{x^{3}}{3!}+\ldots y=8+x-\frac{5}{4} x^{2}-\frac{41}{96} x^{3}+\ldots \end{aligned}$

B1

M1

M1

A1 (4)

[9]

Figure for Answer to Question 3 — Edexcel A-Level Mathematics A2

[Maximum number: 4]

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

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.

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.

[Maximum number: 3]

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

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

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: 3]

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

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.

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

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

Exam points

Covered syllabus points

Covered syllabus points

Question 1

Question (a)

Question (b)

Question 2

Question (a)

Question (b)

Question 3

Question 4

Question 5

Question 6