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A-Level · Further Pure Mathematics

FP.CP11 Maclaurin Series

EDEXCEL 9FM0 · Calculator allowed · about 155 minutes

Total Marks

Name: _______________________________ Date: ____ / ____ / ______

Answer ALL questions. Show all your working.

The function f(x) = e^x is to be expanded as a Maclaurin series.

(a)By finding successive derivatives of f(x) = e^x, evaluate f(0), f'(0), f''(0), f'''(0) and f''''(0).(2)

(b)Hence find the Maclaurin series for e^x, in ascending powers of x, up to and including the term in x^4.(3)

(c)State the general term of the series, and give the range of values of x for which the Maclaurin series for e^x is valid.(2)

(Total for Question 1 is 7 marks)

The function f(x) = cos x is to be expanded as a Maclaurin series.

(a)By finding successive derivatives of f(x) = cos x, find the values of f(0), f'(0), f''(0), f'''(0), f''''(0), f'''''(0) and f''''''(0).(2)

(b)Hence show that the Maclaurin series for cos x, up to and including the term in x⁶, is 1 - x²/2 + x⁴/24 - x⁶/720.(3)

(c)Write down the general term of the series, and state the values of x for which the expansion is valid.(2)

(Total for Question 2 is 7 marks)

The function f(x) = ln(1+x) is to be expanded as a Maclaurin series.

(a)Given that f(x) = ln(1+x), show that f''(0) = -1 and find the value of f'''(0).(3)

(b)Given further that f''''(0) = -6, find the Maclaurin series for ln(1+x), in ascending powers of x, up to and including the term in x^4.(3)

(c)State the range of values of x for which this series is valid.(1)

(Total for Question 3 is 7 marks)

You may use, without proof, the standard Maclaurin series e^x = 1 + x + x²/2! + x³/3! + ... and sin x = x - x³/3! + x⁵/5! - ..., both valid for all real x.

(a)By substituting into the series for e^x, find the Maclaurin series for e^2x, in ascending powers of x, up to and including the term in x^3.(3)

(b)By substituting into the series for sin x, find the Maclaurin series for sin(3x), in ascending powers of x, up to and including the term in x⁵, giving each coefficient as an exact fraction.(4)

(Total for Question 4 is 7 marks)

Maclaurin series can be used to evaluate limits that are not directly computable by substitution.

(a)By using the Maclaurin series for e^x, find lim(x->0) of (e^x - 1 - x)/x^2.(3)

(b)By using the Maclaurin series for cos x, find lim(x->0) of (1 - cos x - x²/2)/x^4.(4)

(Total for Question 5 is 7 marks)

Consider the function f(x) = e^x sin x.

(a)Write down the standard Maclaurin series for e^x and for sin x, in each case up to and including the term in x^4.(2)

(b)By multiplying these two series together, find the Maclaurin series for e^x sin x, in ascending powers of x, up to and including the term in x⁴, giving each coefficient as an exact fraction (or integer).(5)

(Total for Question 6 is 7 marks)

Let f(x) = ln(1 + sin x).

(a)Show that f''(x) = -1/(1 + sin x).(4)

(b)Hence find the value of f'''(0).(2)

(c)Show that the Maclaurin series for ln(1+sin x), up to and including the term in x³, is x - x²/2 + x³/6.(3)

(Total for Question 7 is 9 marks)

Let g(x) = 1/(1+x²).

(a)Write down the series expansion of 1/(1+x²), as a series in ascending powers of x, up to and including the term in x⁶, and state the range of values of x for which the expansion is valid.(3)

(b)By integrating the series in part (a) term by term, and using the fact that arctan(0) = 0, show that arctan x = x - x³/3 + x⁵/5 - x⁷/7 + ... .(3)

(c)Using the first four non-zero terms of this series, with x = 1/√3, find an estimate for the value of π, giving your answer to 3 decimal places. Hence find the percentage error in this estimate compared with the value of π given by your calculator, giving your answer to 2 significant figures.(4)

(Total for Question 8 is 10 marks)

A curve satisfies the differential equation dy/dx = x + y² and passes through the point (0, 1).

(a)Show that dy/dx = 1 when x = 0, and find the value of d^2y/dx² when x = 0.(4)

(b)Find the value of d^3y/dx³ when x = 0.(3)

(c)Hence find the series solution for y, in ascending powers of x, up to and including the term in x^3.(2)

(Total for Question 9 is 9 marks)

Consider the function f(x) = e^x √1+x.

(a)Write down the binomial expansion of (1+x)^1/2, in ascending powers of x, up to and including the term in x³, simplifying each coefficient.(3)

(b)Using the standard Maclaurin series for e^x, find the Maclaurin series for f(x) = e^x √1+x, in ascending powers of x, up to and including the term in x³, giving each coefficient as a fraction in its simplest form.(5)

(Total for Question 10 is 8 marks)

Consider the function h(x) = sin(x²).

(a)Using the standard Maclaurin series for sin x, show that the Maclaurin series for sin(x²), up to and including the term in x¹⁰, is x² - x⁶/6 + x¹⁰/120.(3)

(b)By integrating this series term by term, find an estimate for the integral from 0 to 0.5 of sin(x²) dx, giving your answer to 5 decimal places.(4)

(c)By considering the size of the next term in the integrated series, explain why this estimate is accurate to at least 5 decimal places.(2)

(Total for Question 11 is 9 marks)

A function y(x) satisfies the differential equation d^2y/dx² = x(dy/dx) + 2y, with y = 1 and dy/dx = 0 when x = 0.

(a)By repeated differentiation of the given equation, find the values of d^3y/dx³, d^4y/dx⁴ and d^5y/dx⁵ when x = 0. Show your method clearly.(6)

(b)Hence find the Maclaurin series for y, in ascending powers of x, up to and including the term in x^4.(3)

(c)Use your series to find an estimate, to 4 decimal places, for the value of y when x = 0.3.(2)

(Total for Question 12 is 11 marks)

Let f(x) = sec x.

(a)Show that f'(x) = sec x tan x, and hence show that f''(x) = sec x tan² x + sec³ x. State the value of f''(0).(5)

(b)Given further that f'''(x) = sec x tan³ x + 5 sec³ x tan x, show that f''''(0) = 5.(4)

(c)Hence show that the Maclaurin series for sec x, up to and including the term in x⁴, is 1 + x²/2 + 5x⁴/24.(2)

(Total for Question 13 is 11 marks)

Mark scheme · FP.CP11 Maclaurin Series

Question 1

(a) B1 recognises fⁿ(x) = e^x for every derivative oe
(a) B1 all five values correctly stated as 1 cao
(a) Answer: f(0) = f'(0) = f''(0) = f'''(0) = f''''(0) = 1
(b) M1 uses f(x) = f(0) + f'(0)x + f''(0)x²/2! + f'''(0)x³/3! + f''''(0)x⁴/4! oe
(b) A1 correct series with the values from part (a) substituted, unsimplified
(b) A1 e^x = 1 + x + x²/2 + x³/6 + x⁴/24 cao
(b) Answer: e^x = 1 + x + x²/2 + x³/6 + x⁴/24
(c) B1 general term xⁿ/n! oe
(c) B1 valid for all real x cao
(c) Answer: General term xⁿ/n!; valid for all real x

Question 2

(a) B1 derivatives cycle cos x, -sin x, -cos x, sin x with correct signs oe
(a) B1 all seven values correctly stated as 1, 0, -1, 0, 1, 0, -1
(a) Answer: f(0)=1, f'(0)=0, f''(0)=-1, f'''(0)=0, f''''(0)=1, f'''''(0)=0, f''''''(0)=-1
(b) M1 uses f(x) = f(0) + f'(0)x + f''(0)x²/2! + ... + f''''''(0)x⁶/6! oe
(b) A1 correct series with values substituted, odd-power terms vanish
(b) A1 cso: 1 - x²/2 + x⁴/24 - x⁶/720 as required
(b) Answer: cos x = 1 - x²/2 + x⁴/24 - x⁶/720
(c) B1 general term (-1)ⁿ x²ⁿ/(2n)! oe
(c) B1 valid for all real x cao
(c) Answer: General term (-1)ⁿ x²ⁿ/(2n)!; valid for all real x

Question 3

(a) M1 f'(x) = (1+x)^-1, f''(x) = -(1+x)^-2, f'''(x) = 2(1+x)^-3 oe
(a) A1 cso: f''(0) = -1 as required
(a) A1 f'''(0) = 2 cao
(a) Answer: f''(0) = -1, f'''(0) = 2
(b) M1 uses f(x) = f(0) + f'(0)x + f''(0)x²/2! + f'''(0)x³/3! + f''''(0)x⁴/4! with f(0) = 0, f'(0) = 1
(b) A1 correct series with values substituted, unsimplified
(b) A1 ln(1+x) = x - x²/2 + x³/3 - x⁴/4 cao
(b) Answer: ln(1+x) = x - x²/2 + x³/3 - x⁴/4
(c) B1 -1 < x ≤ 1 oe cao
(c) Answer: -1 < x ≤ 1

Question 4

(a) M1 substitutes u = 2x into 1 + u + u²/2! + u³/3!
(a) A1 1 + 2x + 4x²/2 + 8x³/6 unsimplified
(a) A1 e^2x = 1 + 2x + 2x² + (4/3)x³ cao
(a) Answer: e^2x = 1 + 2x + 2x² + (4/3)x³
(b) M1 substitutes u = 3x into u - u³/3! + u⁵/5!
(b) A1 3x - 27x³/6 + 243x⁵/120 unsimplified
(b) A1 coefficient of x³ simplified to -(9/2) cao
(b) A1 sin(3x) = 3x - (9/2)x³ + (81/40)x⁵ cao
(b) Answer: sin(3x) = 3x - (9/2)x³ + (81/40)x⁵

Question 5

(a) M1 writes e^x - 1 - x = x²/2 + x³/6 + x⁴/24 + ... using the standard series
(a) A1 divides by x² to get 1/2 + x/6 + x²/24 + ...
(a) A1 cao: limit = 1/2
(a) Answer: 1/2
(b) M1 writes 1 - cos x = x²/2 - x⁴/24 + x⁶/720 - ... using the standard series
(b) M1 subtracts x²/2 to leave -x⁴/24 + x⁶/720 - ...
(b) A1 divides by x⁴ to get -1/24 + x²/720 - ...
(b) A1 cao: limit = -1/24
(b) Answer: -1/24

Question 6

(a) B1 e^x = 1 + x + x²/2 + x³/6 + x⁴/24
(a) B1 sin x = x - x³/6 (+ 0x⁴)
(a) Answer: e^x = 1+x+x²/2+x³/6+x⁴/24; sin x = x - x³/6
(b) M1 attempts to multiply the two series, collecting terms of equal power up to x⁴
(b) A1 coefficient of x is 1
(b) A1 coefficient of x² is 1
(b) A1 coefficient of x³ is 1/3
(b) A1 coefficient of x⁴ is 0, so e^x sin x = x + x² + x³/3 (+ 0x⁴) cao
(b) Answer: e^x sin x = x + x² + x³/3

Question 7

(a) M1 differentiates f(x) using the chain rule to get f'(x) = cos x/(1+sin x)
(a) M1 attempts to differentiate f'(x) using the quotient rule
(a) A1 numerator simplifies to -(sin x + sin² x + cos² x) = -(1 + sin x) using sin² x + cos² x = 1
(a) A1 cso: f''(x) = -(1+sin x)/(1+sin x)² = -1/(1+sin x) as required
(a) Answer: f''(x) = -1/(1+sin x)
(b) M1 differentiates -(1+sin x)^-1 using the chain rule to get f'''(x) = cos x/(1+sin x)²
(b) A1 f'''(0) = 1/1 = 1 cao
(b) Answer: f'''(0) = 1
(c) M1 substitutes f(0)=0, f'(0)=1, f''(0)=-1, f'''(0)=1 into the Maclaurin formula
(c) A1 0 + 1(x) + (-1)x²/2! + 1(x³)/3! unsimplified
(c) A1 cso: x - x²/2 + x³/6 as required
(c) Answer: ln(1+sin x) = x - x²/2 + x³/6

Question 8

(a) B1 recognises 1/(1+x²) as a geometric series with first term 1 and common ratio -x²
(a) B1 1 - x² + x⁴ - x⁶ cao
(a) B1 valid for -1 < x < 1 oe
(a) Answer: 1/(1+x²) = 1 - x² + x⁴ - x⁶; valid for -1 < x < 1
(b) M1 uses d/dx(arctan x) = 1/(1+x²) and integrates each term of the series in part (a)
(b) A1 obtains x - x³/3 + x⁵/5 - x⁷/7 (+ C)
(b) A1 cso: C = 0 since arctan(0) = 0, giving the printed result
(b) Answer: arctan x = x - x³/3 + x⁵/5 - x⁷/7
(c) M1 uses arctan(1/√3) = π/6 and substitutes x = 1/√3 into the series
(c) M1 evaluates the sum of the first four non-zero terms (awrt 0.523)
(c) A1 π is approximately 6 x 0.52298 = 3.138 (3 dp) awrt
(c) A1 ft: percentage error = |π(calc) - estimate|/π(calc) x 100 = 0.12% (2 sf) awrt
(c) Answer: π is approximately 3.138; percentage error is approximately 0.12%

Question 9

(a) B1 cso: dy/dx = 0 + 1² = 1 when x = 0
(a) M1 differentiates dy/dx = x + y² implicitly to get d^2y/dx² = 1 + 2y(dy/dx)
(a) A1 substitutes x=0, y=1, dy/dx=1
(a) A1 cao: d^2y/dx² = 1 + 2(1)(1) = 3
(a) Answer: dy/dx = 1 and d^2y/dx² = 3 at x = 0
(b) M1 differentiates d^2y/dx² = 1 + 2y(dy/dx) to get d^3y/dx³ = 2(dy/dx)² + 2y(d^2y/dx²)
(b) A1 substitutes dy/dx=1, y=1, d^2y/dx²=3
(b) A1 cao: d^3y/dx³ = 2(1)² + 2(1)(3) = 8
(b) Answer: d^3y/dx³ = 8 at x = 0
(c) M1 substitutes y(0)=1, y'(0)=1, y''(0)=3, y'''(0)=8 into the Maclaurin formula
(c) A1 cao: y = 1 + x + (3/2)x² + (4/3)x³
(c) Answer: y = 1 + x + (3/2)x² + (4/3)x³

Question 10

(a) M1 uses (1+x)ⁿ = 1 + nx + n(n-1)x²/2! + n(n-1)(n-2)x³/3! with n = 1/2
(a) A1 coefficient of x² simplifies to -1/8
(a) A1 coefficient of x³ simplifies to 1/16, giving 1 + x/2 - x²/8 + x³/16 cao
(a) Answer: (1+x)^1/2 = 1 + x/2 - x²/8 + x³/16
(b) M1 attempts to multiply the series for e^x and for (1+x)^1/2, collecting terms up to x³
(b) A1 constant term is 1
(b) A1 coefficient of x is 3/2
(b) A1 coefficient of x² is 7/8
(b) A1 coefficient of x³ is 17/48 cao
(b) Answer: e^x √1+x = 1 + (3/2)x + (7/8)x² + (17/48)x³

Question 11

(a) M1 substitutes u = x² into u - u³/3! + u⁵/5!
(a) A1 u³ = x⁶ term gives -x⁶/6
(a) A1 cso: u⁵ = x¹⁰ term gives +x¹⁰/120, confirming x² - x⁶/6 + x¹⁰/120
(a) Answer: sin(x²) = x² - x⁶/6 + x¹⁰/120
(b) M1 integrates term by term to get x³/3 - x⁷/42 + x¹¹/1320 (+ C, C = 0)
(b) M1 substitutes the limits x = 0.5 and x = 0
(b) A1 evaluates the three terms as awrt 0.04167, 0.00019, 0.0000004
(b) A1 cao: integral is approximately 0.04148 (5 dp)
(b) Answer: Integral from 0 to 0.5 of sin(x²) dx is approximately 0.04148
(c) B1 identifies that the next term involves x¹⁵/75600 (from integrating -x¹⁴/5040)
(c) B1 evaluates this term at x = 0.5 as approximately 4 x 10^-10, which is far smaller than 0.000005, so the truncation error cannot affect the 5th decimal place
(c) Answer: The next term is approximately 4 x 10^-10 at x = 0.5, far below the rounding threshold of 0.000005, so the estimate is accurate to at least 5 dp

Question 12

(a) M1 differentiates y'' = xy' + 2y using the product rule to get y''' = 3y' + xy''
(a) A1 substitutes x=0, y'=0, y''=2 to get y'''(0) = 0
(a) M1 differentiates again to get y'''' = 4y'' + xy'''
(a) A1 substitutes x=0, y''=2, y'''=0 to get y''''(0) = 8
(a) M1 differentiates again to get y''''' = 5y''' + xy''''
(a) A1 substitutes x=0, y'''=0, y''''=8 to get y'''''(0) = 0
(a) Answer: y'''(0) = 0, y''''(0) = 8, y'''''(0) = 0
(b) M1 substitutes y(0)=1, y'(0)=0, y''(0)=2, y'''(0)=0, y''''(0)=8 into the Maclaurin formula
(b) A1 1 + 0 + 2x²/2! + 0 + 8x⁴/4! unsimplified
(b) A1 cao: y = 1 + x² + x⁴/3
(b) Answer: y = 1 + x² + x⁴/3
(c) M1 substitutes x = 0.3 into y = 1 + x² + x⁴/3
(c) A1 cao: y is approximately 1.0927 (4 dp)
(c) Answer: y(0.3) is approximately 1.0927

Question 13

(a) M1 differentiates f(x) = (cos x)^-1 using the chain rule
(a) A1 cso: f'(x) = sin x (cos x)^-2 = sec x tan x
(a) M1 differentiates f'(x) = sec x tan x using the product rule: sec x tan x . tan x + sec x . sec² x
(a) A1 cso: f''(x) = sec x tan² x + sec³ x as required
(a) B1 f''(0) = 0 + 1 = 1 cao
(a) Answer: f''(x) = sec x tan² x + sec³ x, and f''(0) = 1
(b) M1 differentiates sec x tan³ x using the product rule to get sec x tan⁴ x + 3sec³ x tan² x
(b) M1 differentiates 5sec³ x tan x using the product rule to get 15sec³ x tan² x + 5sec⁵ x
(b) A1 combines to f''''(x) = sec x tan⁴ x + 18sec³ x tan² x + 5sec⁵ x
(b) A1 cso: substitutes x=0 (sec0=1, tan0=0) to get f''''(0) = 0 + 0 + 5 = 5 as required
(b) Answer: f''''(0) = 5
(c) M1 substitutes f(0)=1, f'(0)=0, f''(0)=1, f'''(0)=0, f''''(0)=5 into the Maclaurin formula
(c) A1 cso: 1 + 0 + x²/2! + 0 + 5x⁴/4! = 1 + x²/2 + 5x⁴/24 as required
(c) Answer: sec x = 1 + x²/2 + 5x⁴/24

Maclaurin Series - Worksheets, Questions and Revision