Free worked solutions · 2026 Edition

Chapter 17: Maclaurin series worked solutions

Worked solutions to the practice questions in Chapter 17 of The Ultimate Guide for the IB: Mathematics: Analysis and Approaches HL (2026 Edition). Each solution shows every step with the mark it earns, the way an IB mark scheme does.

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Basics

One idea at a time, in the order the chapter teaches them.

Question 1

[8 marks]

Let f(x) = \ln(1 + x), for x > -1.

(a) Find f'(x), f''(x), f'''(x) and f^{(4)}(x).

[2]

(b) Hence find the Maclaurin series for f(x) up to and including the term in x^4.

[3]

(c) You may assume that f^{(n)}(x) = \dfrac{(-1)^{n-1}(n - 1)!}{(1 + x)^{n}} for n \in \mathbb{Z}^{+}. Show that the coefficient of x^n in the Maclaurin series for f(x) is \dfrac{(-1)^{n-1}}{n}.

[2]

(d) Explain why the function y = \ln x does not have a Maclaurin series.

[1]

Show solutionHide solution

(a) f'(x) = \dfrac{1}{1 + x}, f''(x) = -\dfrac{1}{(1 + x)^2}, f'''(x) = \dfrac{2}{(1 + x)^3}, f^{(4)}(x) = -\dfrac{6}{(1 + x)^4}

Note: award A1 for two or three correct derivatives.

(b) f(0) = 0, f'(0) = 1, f''(0) = -1, f'''(0) = 2, f^{(4)}(0) = -6

f(x) = 0 + x - \dfrac{1}{2!}x^2 + \dfrac{2}{3!}x^3 - \dfrac{6}{4!}x^4 + \ldots

= x - \dfrac{x^2}{2} + \dfrac{x^3}{3} - \dfrac{x^4}{4} + \ldots

(c) f^{(n)}(0) = (-1)^{n-1}(n - 1)!

coefficient = \dfrac{f^{(n)}(0)}{n!} = \dfrac{(-1)^{n-1}(n - 1)!}{n!} = \dfrac{(-1)^{n-1}}{n} A1AG

(d) \ln x is not defined at x = 0 (nor are its derivatives), so f(0), f'(0), \ldots do not exist

Final answer

(a) \frac{1}{1+x}, -\frac{1}{(1+x)^2}, \frac{2}{(1+x)^3}, -\frac{6}{(1+x)^4}
(b) x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}
(c) shown
(d) \ln x is not defined at x=0

Question 2

[6 marks]

The Maclaurin series for (1 + px)e^{qx}, where p and q are constants and p > 0, begins

1 + 5x + \dfrac{21}{2}x^2 + cx^3 + \ldots

(a) Find the value of p and the value of q.

[4]

(b) Find the value of c.

[2]

Show solutionHide solution

(a) (1 + px)\left( 1 + qx + \dfrac{q^2x^2}{2} + \dfrac{q^3x^3}{6} + \ldots \right) = 1 + (p + q)x + \left( pq + \dfrac{q^2}{2} \right)x^2 + \ldots

p + q = 5 and pq + \dfrac{q^2}{2} = \dfrac{21}{2}

q(5 - q) + \dfrac{q^2}{2} = \dfrac{21}{2} \Rightarrow q^2 - 10q + 21 = 0 \Rightarrow q = 3 or q = 7

q = 7 gives p = -2 < 0, rejected; so p = 2, q = 3

(b) c = \dfrac{pq^2}{2} + \dfrac{q^3}{6} = \dfrac{2 \times 9}{2} + \dfrac{27}{6}

= \dfrac{27}{2}

Final answer

(a) p=2, q=3
(b) c=\frac{27}{2}

Question 3

[7 marks]

(a) Write down the expansion of (1 - x)^{-1} as a geometric series, up to and including the term in x^4, and state the values of x for which it is valid.

[1]

(b) By differentiating both sides of your answer to part (a), show that

(1 - x)^{-2} = 1 + 2x + 3x^2 + 4x^3 + \ldots

[2]

(c) Write down the coefficient of x^{n-1} in the series in part (b), and check this coefficient for n = 3 by using the extended binomial theorem.

[1]

(d) Hence find the exact value of \displaystyle\sum_{n=1}^{\infty} \dfrac{n}{4^{n}}.

[3]

Show solutionHide solution

(a) (1 - x)^{-1} = 1 + x + x^2 + x^3 + x^4 + \ldots, valid for -1 < x < 1

(b) \dfrac{d}{dx}(1 - x)^{-1} = (1 - x)^{-2}

differentiating term by term: 0 + 1 + 2x + 3x^2 + 4x^3 + \ldots A1AG

(c) coefficient of x^{n-1} is n; for n = 3: \dfrac{(-2)(-3)}{2!}(-x)^2 = 3x^2, coefficient 3

(d) \displaystyle\sum_{n=1}^{\infty} \dfrac{n}{4^{n}} = \dfrac{1}{4}\displaystyle\sum_{n=1}^{\infty} n\left( \dfrac{1}{4} \right)^{n-1}

= \dfrac{1}{4}\left( 1 - \dfrac{1}{4} \right)^{-2} with x = \dfrac{1}{4} inside -1 < x < 1

= \dfrac{1}{4} \times \dfrac{16}{9} = \dfrac{4}{9}

Final answer

(a) 1+x+x^2+x^3+x^4, valid for |x|<1
(b) shown
(c) n
(d) \frac49

Question 4

[7 marks]

Use Maclaurin series to find each limit.

(a) \displaystyle\lim_{x \to 0} \dfrac{\ln(1 + 3x) - 3x}{x^2}

[2]

(b) \displaystyle\lim_{x \to 0} \dfrac{\sin 2x - 2x\cos x}{x^3}

[3]

(c) \displaystyle\lim_{x \to 0} \dfrac{e^{2x^2} - \cos x}{x^2}

[2]

Show solutionHide solution

(a) \ln(1 + 3x) - 3x = -\dfrac{9x^2}{2} + 9x^3 - \ldots

\displaystyle\lim_{x \to 0}\left( -\dfrac{9}{2} + 9x - \ldots \right) = -\dfrac{9}{2}

(b) \sin 2x = 2x - \dfrac{4x^3}{3} + \ldots and 2x\cos x = 2x - x^3 + \ldots

\sin 2x - 2x\cos x = -\dfrac{x^3}{3} + \ldots

limit = -\dfrac{1}{3}

(c) e^{2x^2} - \cos x = \left( 1 + 2x^2 + \ldots \right) - \left( 1 - \dfrac{x^2}{2} + \ldots \right) = \dfrac{5x^2}{2} + \ldots

limit = \dfrac{5}{2}

Note: l'Hopital's rule is an acceptable alternative method if the form \dfrac{0}{0} is checked at every stage; award the marks for a correct limit found that way.

Final answer

(a) -\frac92
(b) -\frac13
(c) \frac52

Question 5

[9 marks]

The function y = f(x) satisfies the differential equation

\dfrac{dy}{dx} = 1 + xy, \quad \text{with } y = 2 \text{ when } x = 0.

(a) State the gradient of the curve y = f(x) at the point (0, 2).

[1]

(b) Show that \dfrac{d^2y}{dx^2} = y + x\dfrac{dy}{dx}.

[2]

(c) Find expressions for \dfrac{d^3y}{dx^3} and \dfrac{d^4y}{dx^4}, and their values when x = 0.

[3]

(d) Hence find the Maclaurin series for y up to and including the term in x^4.

[2]

(e) Use the terms of your series up to x^3 to estimate f(0.2), giving your answer as a fraction in its lowest terms.

[1]

Show solutionHide solution

(a) 1 + 0 \times 2 = 1

(b) differentiating 1 + xy with the product rule: \dfrac{d^2y}{dx^2} = 1 \times y + x\dfrac{dy}{dx} M1A1AG

(c) \dfrac{d^3y}{dx^3} = \dfrac{dy}{dx} + \dfrac{dy}{dx} + x\dfrac{d^2y}{dx^2} = 2\dfrac{dy}{dx} + x\dfrac{d^2y}{dx^2}

\dfrac{d^4y}{dx^4} = 3\dfrac{d^2y}{dx^2} + x\dfrac{d^3y}{dx^3}

at x = 0: \dfrac{d^2y}{dx^2} = 2, \dfrac{d^3y}{dx^3} = 2, \dfrac{d^4y}{dx^4} = 6

(d) y = 2 + x + \dfrac{2}{2!}x^2 + \dfrac{2}{3!}x^3 + \dfrac{6}{4!}x^4 + \ldots

= 2 + x + x^2 + \dfrac{x^3}{3} + \dfrac{x^4}{4} + \ldots

(e) 2 + 0.2 + 0.04 + \dfrac{0.008}{3} = \dfrac{841}{375}

Note: FT from their series in (d).

Final answer

(a) 1
(b) shown
(c) y'''=2y'+xy'', y^{(4)}=3y''+xy'''; values 2 and 6
(d) 2+x+x^2+\frac{x^3}{3}+\frac{x^4}{4}
(e) \frac{841}{375}

Tutorial

Exam-style questions on each question type, with the marks the IB would give.

Question 6

[7 marks]

(a) Find the Maclaurin series of (1+x)^{1/2} and of (1-x)^{-1/2}, each up to and including the term in x^3.

[2]

(b) Hence find the Maclaurin series of \sqrt{\dfrac{1+x}{1-x}} up to and including the term in x^3.

[3]

(c) By writing \sqrt{\dfrac{1+x}{1-x}} = (1+x)(1-x^2)^{-1/2}, check your answer to part (b).

[2]

Show solutionHide solution

(a) (1+x)^{1/2} = 1 + \dfrac x2 - \dfrac{x^2}{8} + \dfrac{x^3}{16} + \cdots

[1]

(1-x)^{-1/2} = 1 + \dfrac x2 + \dfrac{3x^2}{8} + \dfrac{5x^3}{16} + \cdots

[1]

(b) Multiply the two series, keeping powers up to x^3.

[1]

x: \dfrac12 + \dfrac12 = 1; x^2: \dfrac38 + \dfrac14 - \dfrac18 = \dfrac12.

[1]

x^3: \dfrac{5}{16} + \dfrac{3}{16} - \dfrac1{16} + \dfrac1{16} = \dfrac12, so the series is 1 + x + \dfrac{x^2}{2} + \dfrac{x^3}{2} + \cdots

[1]

(c) \sqrt{\dfrac{1+x}{1-x}} = \dfrac{1+x}{\sqrt{(1-x)(1+x)}} = (1+x)(1-x^2)^{-1/2}.

[1]

(1-x^2)^{-1/2} = 1 + \dfrac{x^2}{2} + \cdots, so (1+x)\left(1 + \dfrac{x^2}{2}\right) = 1 + x + \dfrac{x^2}{2} + \dfrac{x^3}{2} + \cdots, as before.

[1]

Final answer

(a) 1+\frac x2-\frac{x^2}{8}+\frac{x^3}{16} and 1+\frac x2+\frac{3x^2}{8}+\frac{5x^3}{16}
(b) 1+x+\frac{x^2}{2}+\frac{x^3}{2}
(c) agrees

Question 7

[5 marks]

(a) Find the first four terms of the Maclaurin series of \dfrac{1-e^{-x^2}}{x^2}.

[2]

(b) Hence find an approximation for \displaystyle\int_0^1\frac{1-e^{-x^2}}{x^2}\,dx, giving your answer as a fraction in its lowest terms.

[3]

Show solutionHide solution

(a) In e^u = 1 + u + \dfrac{u^2}{2} + \dfrac{u^3}{6} + \dfrac{u^4}{24} + \cdots put u = -x^2: e^{-x^2} = 1 - x^2 + \dfrac{x^4}{2} - \dfrac{x^6}{6} + \dfrac{x^8}{24} - \cdots, so 1 - e^{-x^2} = x^2 - \dfrac{x^4}{2} + \dfrac{x^6}{6} - \dfrac{x^8}{24} + \cdots

[1]

\dfrac{1-e^{-x^2}}{x^2} = 1 - \dfrac{x^2}{2} + \dfrac{x^4}{6} - \dfrac{x^6}{24} + \cdots

[1]

(b) \displaystyle\int_0^1\left(1 - \dfrac{x^2}{2} + \dfrac{x^4}{6} - \dfrac{x^6}{24}\right)dx = \left[x - \dfrac{x^3}{6} + \dfrac{x^5}{30} - \dfrac{x^7}{168}\right]_0^1.

[1]

= 1 - \dfrac16 + \dfrac1{30} - \dfrac1{168}.

[1]

= \dfrac{840 - 140 + 28 - 5}{840} = \dfrac{723}{840} = \dfrac{241}{280}.

[1]

Final answer

(a) 1-\frac{x^2}{2}+\frac{x^4}{6}-\frac{x^6}{24}
(b) \frac{241}{280}

Question 8

[6 marks]

Consider \displaystyle\lim_{x\to0}\frac{\ln(1+x)-x\,e^{-ax}}{x^3}, where a\in\mathbb{R}.

(a) Use Maclaurin series to show that a finite limit exists only if a = \tfrac12.

[3]

(b) Find the value of the limit when a = \tfrac12.

[2]

(c) State how many times l'Hôpital's rule would have to be applied to find this limit.

[1]

Show solutionHide solution

(a) \ln(1+x) = x - \dfrac{x^2}{2} + \dfrac{x^3}{3} - \cdots and x\,e^{-ax} = x\left(1 - ax + \dfrac{a^2x^2}{2} - \cdots\right) = x - ax^2 + \dfrac{a^2x^3}{2} - \cdots

[1]

Numerator = \left(a - \dfrac12\right)x^2 + \left(\dfrac13 - \dfrac{a^2}{2}\right)x^3 + \cdots

[1]

Dividing by x^3 gives \dfrac{a-\frac12}{x} + \left(\dfrac13 - \dfrac{a^2}{2}\right) + \cdots, which has a finite limit only if a - \dfrac12 = 0, i.e. a = \dfrac12.

[1]

(b) With a = \dfrac12 the limit is \dfrac13 - \dfrac{1}{2}\cdot\dfrac14.

[1]

= \dfrac{5}{24}.

[1]

(c) Three times (with a = \tfrac12 the numerator and its first two derivatives are all 0 at x = 0, and the third derivative is \tfrac54 \ne 0).

[1]

Final answer

(a) shown
(b) \frac{5}{24}
(c) three times

Question 9

[8 marks]

A function y = f(x) satisfies \dfrac{dy}{dx} = x + e^y with y = 0 when x = 0.

(a) Find the value of \dfrac{dy}{dx} when x = 0. Show that \dfrac{d^2y}{dx^2} = 1 + e^y\dfrac{dy}{dx}, and find its value when x = 0.

[3]

(b) Find an expression for \dfrac{d^3y}{dx^3} in terms of y, \dfrac{dy}{dx} and \dfrac{d^2y}{dx^2}, and find its value when x = 0.

[2]

(c) It is given that \dfrac{d^4y}{dx^4} = e^y\left[\left(\dfrac{dy}{dx}\right)^3 + 3\dfrac{dy}{dx}\dfrac{d^2y}{dx^2} + \dfrac{d^3y}{dx^3}\right]. Find its value when x = 0.

[1]

(d) Hence find the Maclaurin series of y up to and including the term in x^4.

[2]

Show solutionHide solution

(a) y'(0) = 0 + e^0 = 1.

[1]

Differentiating y' = x + e^y (chain rule on e^y): y'' = 1 + e^y y'.

[1]

y''(0) = 1 + 1\times1 = 2.

[1]

(b) Differentiate y'' = 1 + e^yy' with the product rule: y''' = e^y(y')^2 + e^yy''.

[1]

y'''(0) = 1 + 2 = 3.

[1]

(c) y^{(4)}(0) = 1\times\left(1^3 + 3(1)(2) + 3\right) = 10.

[1]

(d) y = 0 + x + \dfrac{2}{2!}x^2 + \dfrac{3}{3!}x^3 + \dfrac{10}{4!}x^4 + \cdots

[1]

= x + x^2 + \dfrac12x^3 + \dfrac5{12}x^4 + \cdots

[1]

Final answer

(a) 1; shown; 2
(b) y'''=e^y(y')^2+e^yy''; 3
(c) 10
(d) x+x^2+\frac{x^3}{2}+\frac{5x^4}{12}

Question 10

[10 marks]

The function y = f(x) is defined implicitly by e^y + y = x + 1.

(a) Show that y = 0 when x = 0.

[1]

(b) By differentiating the equation with respect to x, show that \left(e^y+1\right)\dfrac{dy}{dx} = 1, and find the value of \dfrac{dy}{dx} when x = 0.

[2]

(c) Show that \left(e^y+1\right)\dfrac{d^2y}{dx^2} + e^y\left(\dfrac{dy}{dx}\right)^2 = 0, and find the value of \dfrac{d^2y}{dx^2} when x = 0.

[3]

(d) Find the value of \dfrac{d^3y}{dx^3} when x = 0.

[3]

(e) Hence write down the Maclaurin series of y up to and including the term in x^3.

[1]

Show solutionHide solution

(a) At x = 0: e^y + y = 1, which is satisfied by y = 0 (e^0 + 0 = 1); the left side is increasing in y, so this is the only solution.

[1]

(b) Differentiate: e^y\dfrac{dy}{dx} + \dfrac{dy}{dx} = 1, so (e^y+1)y' = 1.

[1]

At x = 0: (1+1)y' = 1, so y'(0) = \dfrac12.

[1]

(c) Differentiate (e^y+1)y' = 1 using the product rule: (e^y+1)y'' + e^yy'\cdot y' = 0.

[1]

(e^y+1)y'' + e^y(y')^2 = 0.

At x = 0: 2y'' + \dfrac14 = 0, so y''(0) = -\dfrac18.

[2]

(d) Differentiate again: (e^y+1)y''' + e^yy'y'' + e^y\left(2y'y''\right) + e^y(y')^3 = 0, that is (e^y+1)y''' + 3e^yy'y'' + e^y(y')^3 = 0.

[1]

At x = 0: 2y''' + 3\left(\dfrac12\right)\left(-\dfrac18\right) + \dfrac18 = 0.

[1]

2y''' = \dfrac{3}{16} - \dfrac{2}{16} = \dfrac1{16}, so y'''(0) = \dfrac{1}{32}.

[1]

(e) y = \dfrac12x + \dfrac{-1/8}{2}x^2 + \dfrac{1/32}{6}x^3 + \cdots = \dfrac x2 - \dfrac{x^2}{16} + \dfrac{x^3}{192} + \cdots

[1]

Final answer

(a) shown
(b) shown; \frac12
(c) shown; -\frac18
(d) \frac{1}{32}
(e) \frac x2-\frac{x^2}{16}+\frac{x^3}{192}

Mastery

Full exam questions that mix question types, at Paper 1 and Paper 2 standard.

Question 11

[6 marks]

(a) By composing two standard series, show that \arctan(\sin x) = x - \dfrac{x^3}{2} + \cdots

[4]

(b) Deduce the value of \displaystyle\lim_{x\to0}\dfrac{\arctan(\sin x)-x}{x^3}.

[2]

Show solutionHide solution

(a) Let u = \sin x = x - \dfrac{x^3}{6} + \cdots and use \arctan u = u - \dfrac{u^3}{3} + \cdots

[1]

Up to x^3: u = x - \dfrac{x^3}{6} and u^3 = x^3 + \cdots.

[2]

\arctan(\sin x) = x - \dfrac{x^3}{6} - \dfrac{x^3}{3} + \cdots = x - \dfrac{x^3}{2} + \cdots

[1]

(b) \dfrac{\arctan(\sin x)-x}{x^3} = -\dfrac12 + \cdots

[1]

The limit is -\dfrac12.

[1]

Final answer

(a) shown
(b) -\frac12

Question 12

[9 marks]

For small values of x, \cos x can be approximated by a rational function

R(x) = \dfrac{12 + px^2}{12 + qx^2},

where p and q are constants, chosen so that (12 + qx^2)\cos x and 12 + px^2 have the same Maclaurin series up to and including the term in x^4.

(a) Find the Maclaurin series for (12 + qx^2)\cos x up to and including the term in x^4.

[2]

(b) Hence show that q = 1, and find the value of p.

[3]

(c) Find R(1), and the sum of the first three non-zero terms of the Maclaurin series for \cos x at x = 1. Hence find the percentage error of each approximation compared with \cos 1.

[4]

Show solutionHide solution

(a) (12 + qx^2)\left( 1 - \dfrac{x^2}{2} + \dfrac{x^4}{24} - \ldots \right)

= 12 + (q - 6)x^2 + \left( \dfrac{1}{2} - \dfrac{q}{2} \right)x^4 + \ldots

(b) 12 + px^2 has no x^4 term, so \dfrac{1}{2} - \dfrac{q}{2} = 0

q = 1 A1AG

p = q - 6 = -5

(c) R(1) = \dfrac{7}{13} = 0.538\ (0.538461\ldots); series value 1 - \dfrac{1}{2} + \dfrac{1}{24} = \dfrac{13}{24} = 0.542\ (0.541666\ldots) A1A1

\cos 1 = 0.540302\ldots (M1)

percentage errors: R(1): -0.341\%; series: 0.253\%

Note: accept the size of each error without a sign. Accept 0.340\% to 0.341\% and 0.252\% to 0.253\%.

Final answer

(a) 12+(q-6)x^2+\left(\frac12-\frac q2\right)x^4
(b) q=1, p=-5
(c) R(1)=\frac{7}{13}, series sum \frac{13}{24}; errors -0.341\% and 0.253\%

Question 13

[8 marks]

Let F(x) = \displaystyle\int_0^x \frac{e^t - 1}{t}\,dt, for x > 0.

(a) Find the Maclaurin series of \dfrac{e^t - 1}{t} up to and including the term in t^3.

[2]

(b) Hence find the Maclaurin series of F(x) up to and including the term in x^4.

[2]

(c) Use your answer to part (b) to find an approximate value for F\!\left(\tfrac12\right), giving your answer as a fraction.

[2]

(d) Show that the coefficient of x^n in the series of F(x) is \dfrac{1}{n \cdot n!}.

[2]

Show solutionHide solution

(a) e^t - 1 = t + \dfrac{t^2}{2} + \dfrac{t^3}{6} + \dfrac{t^4}{24} + \cdots

[1]

\dfrac{e^t - 1}{t} = 1 + \dfrac{t}{2} + \dfrac{t^2}{6} + \dfrac{t^3}{24} + \cdots

[1]

(b) F(x) = \displaystyle\int_0^x\left(1 + \dfrac t2 + \dfrac{t^2}{6} + \dfrac{t^3}{24}\right)dt.

[1]

F(x) = x + \dfrac{x^2}{4} + \dfrac{x^3}{18} + \dfrac{x^4}{96} + \cdots

[1]

(c) F\!\left(\tfrac12\right) \approx \dfrac12 + \dfrac{1}{16} + \dfrac{1}{144} + \dfrac{1}{1536}.

[1]

Common denominator 4608: \dfrac{2304 + 288 + 32 + 3}{4608} = \dfrac{2627}{4608}.

[1]

(d) The term in t^{n-1} of \dfrac{e^t-1}{t} is \dfrac{t^{n-1}}{n!} (dividing the \dfrac{t^n}{n!} term of e^t - 1 by t).

[1]

Integrating gives \dfrac{x^n}{n\cdot n!}.

[1]

So the coefficient of x^n is \dfrac{1}{n\cdot n!}.

Final answer

(a) 1+\frac t2+\frac{t^2}{6}+\frac{t^3}{24}
(b) x+\frac{x^2}{4}+\frac{x^3}{18}+\frac{x^4}{96}
(c) \frac{2627}{4608}
(d) shown

Question 14

[14 marks]

Let g(x) = (1+x)^{-3}.

(a) Use mathematical induction to prove that g^{(n)}(x) = \dfrac{(-1)^n(n+2)!}{2(1+x)^{n+3}} for n\in\mathbb{Z}^+.

[6]

(b) Hence show that the coefficient of x^n in the Maclaurin series of g(x) is (-1)^n\dfrac{(n+1)(n+2)}{2}.

[2]

(c) Find the exact value of \displaystyle\sum_{n=0}^{\infty}\dfrac{(n+1)(n+2)}{3^n}.

[3]

(d) Use part (b) and the series of (1+x)^{-2} to find the coefficient of x^n in the Maclaurin series of \dfrac{x}{(1+x)^3}.

[3]

Show solutionHide solution

; (a) n = 1: g'(x) = -3(1+x)^{-4}, and the formula gives \dfrac{(-1)\,3!}{2(1+x)^4} = \dfrac{-3}{(1+x)^4}. True for n = 1.

[1]

Assume the result is true for n = k: g^{(k)}(x) = \dfrac{(-1)^k(k+2)!}{2}(1+x)^{-(k+3)}.

[1]

Differentiate: g^{(k+1)}(x) = \dfrac{(-1)^k(k+2)!}{2}\cdot\big(-(k+3)\big)(1+x)^{-(k+4)}.

[2]

= \dfrac{(-1)^{k+1}(k+3)!}{2(1+x)^{k+4}}, which is the result for n = k+1.

[1]

True for n = 1, and true for n = k\Rightarrow true for n = k+1; so true for all n\in\mathbb{Z}^+ by induction.

[1]

(b) g^{(n)}(0) = \dfrac{(-1)^n(n+2)!}{2}, so the coefficient is \dfrac{g^{(n)}(0)}{n!} = \dfrac{(-1)^n(n+2)!}{2\,n!}.

[1]

Since \dfrac{(n+2)!}{n!} = (n+1)(n+2),

[1]

the coefficient is (-1)^n\dfrac{(n+1)(n+2)}{2}.

(This also holds at n = 0, where g(0) = 1.)

(c) g(x) = \sum(-1)^n\dfrac{(n+1)(n+2)}{2}x^n. Put x = -\dfrac13 (valid, |x|<1): g\!\left(-\tfrac13\right) = \sum\dfrac{(n+1)(n+2)}{2}\cdot\dfrac{1}{3^n}.

[1]

g\!\left(-\tfrac13\right) = \left(\dfrac23\right)^{-3} = \dfrac{27}{8}.

[1]

So \sum\dfrac{(n+1)(n+2)}{3^n} = 2\times\dfrac{27}{8} = \dfrac{27}{4}.

[1]

(d) \dfrac{1}{(1+x)^2} has coefficient (-1)^n(n+1) and \dfrac{1}{(1+x)^3} - \dfrac{1}{(1+x)^2} = \dfrac{1-(1+x)}{(1+x)^3} = \dfrac{-x}{(1+x)^3}.

[1]

Coefficient of x^n in \dfrac{-x}{(1+x)^3}: (-1)^n\left[\dfrac{(n+1)(n+2)}{2} - (n+1)\right] = (-1)^n\dfrac{n(n+1)}{2}.

[1]

So the coefficient of x^n in \dfrac{x}{(1+x)^3} is (-1)^{n+1}\dfrac{n(n+1)}{2}.

[1]

Final answer

(a) proof
(b) shown
(c) \frac{27}{4}
(d) (-1)^{n+1}\frac{n(n+1)}{2}

Question 15

[16 marks]

A function y = f(x) satisfies (3+\sin x)\,y = 1.

(a) Show that (3+\sin x)\,y' + y\cos x = 0, and find y(0) and y'(0).

[3]

(b) By differentiating further, show that

(3+\sin x)\,y'' + 2y'\cos x - y\sin x = 0,

and

(3+\sin x)\,y''' + 3y''\cos x - 3y'\sin x - y\cos x = 0.

[4]

(c) Find the values of y''(0) and y'''(0).

[3]

(d) Hence write down the Maclaurin series of y up to and including the term in x^3.

[1]

(e) Check your answer to part (d) by writing y = \dfrac13\left(1 + \dfrac{\sin x}{3}\right)^{-1} and using a standard series.

[4]

(f) Hence find \displaystyle\lim_{x\to0}\dfrac{y-\frac13+\frac x9}{x^2}.

[1]

Show solutionHide solution

(a) Differentiating (3+\sin x)y = 1 with the product rule: (3+\sin x)y' + (\cos x)\,y = 0.

[1]

At x = 0: 3y(0) = 1, so y(0) = \dfrac13.

[1]

Then 3y'(0) + \dfrac13 = 0, so y'(0) = -\dfrac19.

[1]

(b) Differentiate (3+\sin x)y' + y\cos x = 0: \cos x\,y' + (3+\sin x)y'' + y'\cos x - y\sin x = 0.

[2]

(3+\sin x)y'' + 2y'\cos x - y\sin x = 0.

Differentiate again: \cos x\,y'' + (3+\sin x)y''' + 2y''\cos x - 2y'\sin x - y'\sin x - y\cos x = 0.

[2]

(3+\sin x)y''' + 3y''\cos x - 3y'\sin x - y\cos x = 0.

(c) At x = 0 the second relation gives 3y'' + 2\left(-\dfrac19\right) = 0, so y''(0) = \dfrac{2}{27}.

[2]

The third gives 3y''' + 3\left(\dfrac2{27}\right) - \dfrac13 = 0, so 3y''' = \dfrac13 - \dfrac{2}{9} = \dfrac19 and y'''(0) = \dfrac1{27}.

[1]

(d) y = \dfrac13 - \dfrac x9 + \dfrac{2/27}{2}x^2 + \dfrac{1/27}{6}x^3 + \cdots = \dfrac13 - \dfrac x9 + \dfrac{x^2}{27} + \dfrac{x^3}{162} + \cdots

[1]

(e) Let u = \dfrac{\sin x}{3} = \dfrac x3 - \dfrac{x^3}{18} + \cdots and use (1+u)^{-1} = 1 - u + u^2 - u^3 + \cdots

[1]

Up to x^3: u^2 = \dfrac{x^2}{9} and u^3 = \dfrac{x^3}{27}.

[1]

(1+u)^{-1} = 1 - \dfrac x3 + \dfrac{x^3}{18} + \dfrac{x^2}{9} - \dfrac{x^3}{27} = 1 - \dfrac x3 + \dfrac{x^2}{9} + \dfrac{x^3}{54}.

[1]

Multiplying by \dfrac13: \dfrac13 - \dfrac x9 + \dfrac{x^2}{27} + \dfrac{x^3}{162}, as in part (d).

[1]

(f) \dfrac{y - \frac13 + \frac x9}{x^2} = \dfrac{1}{27} + \dfrac{x}{162} + \cdots\to\dfrac1{27}.

[1]

Final answer

(a) shown; y(0)=\frac13, y'(0)=-\frac19
(b) shown
(c) \frac{2}{27}, \frac{1}{27}
(d) \frac13-\frac x9+\frac{x^2}{27}+\frac{x^3}{162}
(e) agrees
(f) \frac{1}{27}

Question 16

[17 marks]

Let f(x) = \dfrac{e^{x}}{1 + x}, for x > -1.

(a) Write down the first four terms of the expansion of (1 + x)^{-1} in ascending powers of x, and state the values of x for which it is valid.

[2]

(b) Hence show that the Maclaurin series for f(x), up to and including the term in x^3, is 1 + \dfrac{x^2}{2} - \dfrac{x^3}{3}.

[3]

(c) Use the series to write down the values of f'(0) and f''(0). Hence show that the graph of y = f(x) has a stationary point at (0, 1) and determine its nature.

[4]

(d) Show that f'(x) = \dfrac{xe^{x}}{(1 + x)^2}, and hence confirm that x = 0 is the only stationary point of f.

[2]

(e) Use the first three non-zero terms of the series to find an approximate value for \displaystyle\int_{0}^{0.5} \dfrac{e^{x}}{1 + x}\,dx. Find the value of this integral given by your GDC, and find the percentage error of the approximation.

[4]

(f) Find \displaystyle\lim_{x \to 0} \dfrac{2f(x) - 2 - x^2}{x^3}.

[2]

Show solutionHide solution

(a) 1 - x + x^2 - x^3

valid for -1 < x < 1

(b) \left( 1 + x + \dfrac{x^2}{2} + \dfrac{x^3}{6} \right)\left( 1 - x + x^2 - x^3 \right), terms up to x^3

x: 1 - 1 = 0; x^2: \dfrac{1}{2} - 1 + 1 = \dfrac{1}{2}

x^3: \dfrac{1}{6} - \dfrac{1}{2} + 1 - 1 = -\dfrac{1}{3}, so f(x) = 1 + \dfrac{x^2}{2} - \dfrac{x^3}{3} + \ldots A1AG

(c) f'(0) = 0 (coefficient of x) and \dfrac{f''(0)}{2!} = \dfrac{1}{2} \Rightarrow f''(0) = 1 A1A1

f(0) = 1 and f'(0) = 0, so (0, 1) is a stationary point

f''(0) = 1 > 0, so it is a local minimum point

(d) f'(x) = \dfrac{(1 + x)e^{x} - e^{x}}{(1 + x)^2} = \dfrac{xe^{x}}{(1 + x)^2} A1AG

e^{x} > 0 and (1 + x)^2 > 0 for x > -1, so f'(x) = 0 only when x = 0

(e) \displaystyle\int_{0}^{0.5} \left( 1 + \dfrac{x^2}{2} - \dfrac{x^3}{3} \right)dx = \left[ x + \dfrac{x^3}{6} - \dfrac{x^4}{12} \right]_{0}^{0.5}

= 0.5 + 0.0208333\ldots - 0.0052083\ldots = 0.515625

GDC: \displaystyle\int_{0}^{0.5} \dfrac{e^{x}}{1 + x}\,dx = 0.517300\ldots

percentage error = \dfrac{0.515625 - 0.517300\ldots}{0.517300\ldots} \times 100\% = -0.324\%

Note: accept 0.324\% without a sign; accept 0.323\% to 0.324\%.

(f) 2f(x) - 2 - x^2 = -\dfrac{2x^3}{3} + \ldots

limit = -\dfrac{2}{3}

Note: l'Hopital's rule is an acceptable alternative method if the form \dfrac{0}{0} is checked at every stage; award the marks for a correct limit found that way.

Final answer

(a) 1-x+x^2-x^3, valid for |x|<1
(b) shown
(c) f'(0)=0, f''(0)=1; minimum at (0,1)
(d) shown
(e) \frac{33}{64}=0.515625; GDC 0.517300; error -0.324\%
(f) -\frac23

Going further

Beyond the syllabus: the worked examples and investigations of the chapter's Going further pages.

G1 e is irrational

G1 Worked example

[9 marks]

For n\in\mathbb{Z}^+, let s_n=\displaystyle\sum_{k=0}^{n}\frac{1}{k!} and t_n=e-s_n, where e=\displaystyle\sum_{k=0}^{\infty}\frac{1}{k!}. You may use the result 0<n!\,t_n<\dfrac1n.

(a) Show that n!\,s_n is an integer for every n\in\mathbb{Z}^+.

[2]

(b) Hence, without using a value of e, find the integer part of 720e.

[2]

(c) Find the least value of n for which the result guarantees that s_n differs from e by less than 10^{-6}.

[2]

(d) Show that n!\,t_n>\dfrac{1}{n+1}, and hence find \displaystyle\lim_{n\to\infty}n\cdot n!\,t_n.

[3]

Show solutionHide solution

(a) n!\,s_n=\displaystyle\sum_{k=0}^{n}\frac{n!}{k!}

for k\le n, \dfrac{n!}{k!}=(k+1)(k+2)\cdots n, a product of integers (equal to 1 when k=n), so every term and the sum are integers

(b) 720=6! and 6!\,s_6=720+720+360+120+30+6+1=1957

720e=6!\,s_6+6!\,t_6 with 0<6!\,t_6<\dfrac16, so the integer part of 720e is 1957

(c) t_n<\dfrac{1}{n\cdot n!}<10^{-6}\iff n\cdot n!>10^6

8\times8!=322\,560 and 9\times9!=3\,265\,920, so n=9

(d) n!\,t_n=\dfrac{n!}{(n+1)!}+\dfrac{n!}{(n+2)!}+\cdots>\dfrac{n!}{(n+1)!}=\dfrac1{n+1}, since every later term is positive

multiply both bounds by n: \dfrac{n}{n+1}<n\cdot n!\,t_n<1

\dfrac{n}{n+1}\to1, so n\cdot n!\,t_n\to1

Final answer

(b) 1957
(c) n=9
(d) 1

G1 Investigation

[12 marks]

For n\in\mathbb{Z}^+, let s_n=\displaystyle\sum_{k=0}^{n}\frac{1}{k!} and t_n=e-s_n, where e=\displaystyle\sum_{k=0}^{\infty}\frac{1}{k!}. You may use the result 0<n!\,t_n<\dfrac1n. You may also use the fact that n!\,s_n is an integer.

(a) Suppose that e=\dfrac pq, where p,q\in\mathbb{Z}^+. Show that q!\,t_q would then be an integer, and use the result above to obtain a contradiction.

[3]

(b) Putting x=\dfrac12 in the Maclaurin series for e^x gives \sqrt e=\displaystyle\sum_{k=0}^{\infty}\frac{1}{2^kk!}. Let u_n=\sqrt e-\displaystyle\sum_{k=0}^{n}\frac{1}{2^kk!}. Use your GDC to find 2^nn!\,u_n for n=1, 2 and 3, and compare each value with \dfrac{1}{2n+1}.

[2]

(c) Show that 2^nn!\,u_n=\dfrac{1}{2(n+1)}+\dfrac{1}{2^2(n+1)(n+2)}+\dfrac{1}{2^3(n+1)(n+2)(n+3)}+\cdots, and hence that 0<2^nn!\,u_n<\dfrac{1}{2n+1}.

[3]

(d) Prove that \sqrt e is irrational.

[2]

(e) The Maclaurin series gives e^2=\displaystyle\sum_{k=0}^{\infty}\frac{2^k}{k!}. Explain why the argument in (a), applied to this series, does not show that e^2 is irrational.

[2]

Show solutionHide solution

(a) q!\,e=q!\cdot\dfrac pq=(q-1)!\,p, an integer; q!\,s_q is an integer

so q!\,t_q=q!\,e-q!\,s_q is an integer

but 0<q!\,t_q<\dfrac1q\le1, and no integer lies strictly between 0 and 1: contradiction, so e is irrational

(b) n=1: 0.297; n=2: 0.190; n=3: 0.139; each is less than \dfrac13, \dfrac15, \dfrac17=0.143 respectively

Note: award A1 for two correct values with the comparison.

(c) 2^nn!\,u_n=\displaystyle\sum_{k=n+1}^{\infty}\frac{2^nn!}{2^kk!}, and the term with k=n+j is \dfrac{1}{2^j(n+1)(n+2)\cdots(n+j)}

each of these is at most \dfrac{1}{\bigl(2(n+1)\bigr)^j}, with strict inequality for j\ge2, and every term is positive

so 0<2^nn!\,u_n<\dfrac{\frac1{2(n+1)}}{1-\frac1{2(n+1)}}=\dfrac{1}{2n+1}

(d) if \sqrt e=\dfrac pq, then 2^qq!\sqrt e=2^q(q-1)!\,p and 2^qq!\displaystyle\sum_{k=0}^{q}\frac{1}{2^kk!}=\sum_{k=0}^{q}2^{q-k}\frac{q!}{k!} are integers, so 2^qq!\,u_q is an integer

but 0<2^qq!\,u_q<\dfrac{1}{2q+1}<1: contradiction, so \sqrt e is irrational

(e) multiplying the tail by q! gives \displaystyle\sum_{k=q+1}^{\infty}\frac{2^kq!}{k!}, whose first term alone is \dfrac{2^{q+1}}{q+1}\ge2

the tail is not trapped between 0 and 1, so being an integer gives no contradiction

Final answer

(a) q!\,t_q is an integer in (0,1): impossible
(b) 0.297, 0.190, 0.139, each below \frac1{2n+1}
(c), (d) proved in the solutions
(e) the tail is at least \frac{2^{q+1}}{q+1}\ge2

G2 Exponentials beat every power

G2 Worked example

[8 marks]

(a) Use the Maclaurin series for e^x to explain why e^x>\dfrac{x^4}{24} for every x>0.

[2]

(b) Hence show that 0<x^3e^{-x}<\dfrac{24}{x} for x>0, and deduce the value of \displaystyle\lim_{x\to\infty}x^3e^{-x}.

[3]

(c) By using the substitution t=e^{-x}, find \displaystyle\lim_{t\to0^+}t(\ln t)^3.

[3]

Show solutionHide solution

(a) e^x=\displaystyle\sum_{k=0}^{\infty}\frac{x^k}{k!}, and for x>0 every term \dfrac{x^k}{k!} is positive

so e^x is greater than any one of its terms, in particular \dfrac{x^4}{4!}=\dfrac{x^4}{24}

(b) x^3e^{-x}=\dfrac{x^3}{e^x}, which is positive, and \dfrac{x^3}{e^x}<\dfrac{x^3}{x^4/24}

=\dfrac{24}{x}

\dfrac{24}x\to0 as x\to\infty, so, squeezed between 0 and \dfrac{24}x, x^3e^{-x}\to0

(c) t=e^{-x} gives \ln t=-x, and t\to0^+ exactly when x\to\infty

t(\ln t)^3=e^{-x}(-x)^3=-x^3e^{-x}

\to0 by (b), so \displaystyle\lim_{t\to0^+}t(\ln t)^3=0

Final answer

(b) 0
(c) 0

G2 Investigation

[11 marks]

This investigation compares x^{10} with e^x for x>0.

(a) Use your GDC to find the two positive solutions of x^{10}=e^x.

[2]

(b) Use the Maclaurin series for e^x to show that 0<x^{10}e^{-x}<\dfrac{11!}{x} for x>0, and deduce \displaystyle\lim_{x\to\infty}x^{10}e^{-x}. Explain why this inequality on its own gives a poor estimate of the larger solution in (a).

[3]

(c) Show that, for x>0, x^{10}<e^x exactly when x<e^{x/10}. By comparing e^{x/10} with a single term of its Maclaurin series, show that x^{10}<e^x whenever x\ge10\left(10\cdot k!\right)^{\frac{1}{k-1}}, for each integer k\ge2.

[3]

(d) Find the value of k that gives the smallest bound in (c), and state this bound. Compare it with your answer to (a).

[3]

Show solutionHide solution

(a) x=1.12 and x=35.8 (1.1183\ldots and 35.771\ldots)

A1A1

(b) every term of e^x is positive for x>0, so e^x>\dfrac{x^{11}}{11!} and 0<\dfrac{x^{10}}{e^x}<\dfrac{x^{10}\cdot11!}{x^{11}}=\dfrac{11!}{x}

\dfrac{11!}{x}\to0, so x^{10}e^{-x}\to0

the bound is below 1 only when x>11!=39\,916\,800, so it shows x^{10}<e^x only far beyond the actual crossing at 35.8

(c) both sides are positive, so taking the tenth root, x^{10}<e^x\iff x<\left(e^x\right)^{1/10}=e^{x/10}

for x>0, e^{x/10}>\dfrac{(x/10)^k}{k!}

\dfrac{(x/10)^k}{k!}\ge x\iff x^{k-1}\ge10^k\,k!\iff x\ge10\left(10\cdot k!\right)^{\frac1{k-1}}; for such x, e^{x/10}>x

(d) k=2: 200; k=3: 77.5; k=4: 62.1; k=5: 58.9; k=6: 59.1; k=7: 60.8

the least is at k=5: x^{10}<e^x for every x\ge58.9

the true crossing is 35.8: the bound is safe but about 23 too large, because one term of the series is much smaller than the whole sum

Final answer

(a) 1.12, 35.8
(b) 0; the bound needs x>11!
(d) k=5, x\ge58.9 (true value 35.8)

TI-84 Plus CE keystrokes

The GDC panels in the book give keystrokes for the TI-Nspire CX II and the Casio fx-CG50. These are the TI-84 Plus CE keystrokes for the same calculator jobs. The full list of TI-84 keystrokes is in Chapter 19.

Job	TI-84 Plus CE keystrokes
Gradient at a point	`MATH, 8:nDeriv(expression, X, value)`
Definite integral	`MATH, 9:fnInt(expression, X, lower, upper)`
Solve an equation	`MATH, Numeric Solver (MATH, 0:Solver on older OS versions): enter the equation, then SOLVE. Or graph both sides and use 2nd TRACE (CALC), 5:intersect`

From the book

The Ultimate Guide for the IB: Mathematics: Analysis and Approaches HL

Chapter 17 teaches maclaurin series from the basics to full exam questions: how the IB asks it, every question type with a marked example, and these practice questions in three steps.

About the book