Mathematics for the international student: mathematics HL (Option): Calculus HL Topic 9 FM Topic 5 1921972335, 9781921972331
Mathematics HL Option Calculus has been written as a companion book to the Mathematics HL (Core) textbook. Together they
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- Catherine Quinn
- Chris Sangwin
- Robert Haese
- Michael Haese
- Commentary
- From IBDocs.org
Table of contents :
Mathematics HL Option: Calculus
Foreword
Acknowledgements
Using the interactive CD
Table of contents
Symbols and notation
A - Number properties
Exercise A
Answers
B - Limits
Exercise B.1
Answers
Exercise B.2
Answers
Exercise B.3
Answers
C - Continuity of functions
Exercise C.1
Answers
Exercise C.2
Answers
D - Differentiable functions
Discussion
Exercise D
Answers
E - L'Hôpitals rule
Exercise E
Answers
F - Rolle's theorem and the mean value theorem (MVT)
Exercise F
Answers
G - Riemann sums
Exercise G.1
Answers
Exercise G.2
Answers
H - The fundamental theorem of calculus
Discussion
Exercise H
Answers
I - Improper integrals
Exercise I.1
Answers
Exercise I.2
Answers
J - Sequences
Exercise J.1
Answers
Exercise J.2
Answers
K - Infinite series
Exercise K.1
Answers
Exercise K.2
Answers
Exercise K.3
Answers
Exercise K.4
Answers
L - Taylor and Maclaurin series
Exercise L.1
Answers
Exercise L.2
Answers
Exercise L.3
Answers
Exercise L.4
Answers
Historical note
M - Differential equations
Exercise M
Answers
N - Separable differential equations
Exercise N
Answers
O - The integrating factor method
Exercise O
Answers
Activity - Laplace transforms
P - Taylor or Maclaurin series developed from a differential equation
Exercise P
Answers
Review set A
Answers
Review set B
Answers
Review set C
Answers
Review set D
Answers
Theory of knowledge - Torricellis trumpet
Appendix A - Methods of proof
Theory of knowledge - Axioms and occam's razor
Appendix B - Formal definition of a limit
Worked Solutions
Citation preview
HAESE MATHEMATICS Specialists in mathematics publishing
Mathematics
for the international student
Mathematics HL (Option): Calculus
HL Topic 9 FM Topic 5 Catherine Quinn Chris Sangwin Robert Haese Michael Haese
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for use with IB Diploma Programme black
Y:\HAESE\IB_HL_OPT-Calculus\IB_HL_OPT-Calculus_00\001IB_HL_OPT-Calculus_00.cdr Tuesday, 19 March 2013 2:00:41 PM BRIAN
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MATHEMATICS FOR THE INTERNATIONAL STUDENT Mathematics HL (Option): Calculus Catherine Quinn Chris Sangwin Robert Haese Michael Haese
B.Sc.(Hons), Grad.Dip.Ed., Ph.D. M.A., M.Sc., Ph.D. B.Sc. B.Sc.(Hons.),Ph.D.
Haese Mathematics 152 Richmond Road, Marleston, SA 5033, AUSTRALIA Telephone: +61 8 8210 4666, Fax: +61 8 8354 1238 Email: [email protected] www.haesemathematics.com.au Web: National Library of Australia Card Number & ISBN 978-1-921972-33-1 © Haese & Harris Publications 2013 Published by Haese Mathematics. 152 Richmond Road, Marleston, SA 5033, AUSTRALIA First Edition
2013
Artwork by Brian Houston and Gregory Olesinski. Cover design by Piotr Poturaj. Typeset in Australia by Deanne Gallasch. Typeset in Times Roman 10 \Qw_ . Printed in Hong Kong through Prolong Press Limited. The textbook and its accompanying CD have been developed independently of the International Baccalaureate Organization (IBO). The textbook and CD are in no way connected with, or endorsed by, the IBO. This book is copyright. Except as permitted by the Copyright Act (any fair dealing for the purposes of private study, research, criticism or review), no part of this publication may be reproduced, stored in a retrieval system, or transmitted in any form or by any means, electronic, mechanical, photocopying, recording or otherwise, without the prior permission of the publisher. Enquiries to be made to Haese Mathematics. Copying for educational purposes: Where copies of part or the whole of the book are made under Part VB of the Copyright Act, the law requires that the educational institution or the body that administers it has given a remuneration notice to Copyright Agency Limited (CAL). For information, contact the Copyright Agency Limited. Acknowledgements: While every attempt has been made to trace and acknowledge copyright, the authors and publishers apologise for any accidental infringement where copyright has proved untraceable. They would be pleased to come to a suitable agreement with the rightful owner.
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Disclaimer: All the internet addresses (URLs) given in this book were valid at the time of printing. While the authors and publisher regret any inconvenience that changes of address may cause readers, no responsibility for any such changes can be accepted by either the authors or the publisher.
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FOREWORD Mathematics HL (Option): Calculus has been written as a companion book to the Mathematics HL (Core) textbook. Together, they aim to provide students and teachers with appropriate coverage of the two-year Mathematics HL Course, to be first examined in 2014. This book covers all sub-topics set out in Mathematics HL Option Topic 9 and Further Mathematics HL Topic 5, Calculus. The aim of this topic is to introduce students to the basic concepts and techniques of differential and integral calculus and their applications. Detailed explanations and key facts are highlighted throughout the text. Each sub-topic contains numerous Worked Examples, highlighting each step necessary to reach the answer for that example. Theory of Knowledge is a core requirement in the International Baccalaureate Diploma Programme, whereby students are encouraged to think critically and challenge the assumptions of knowledge. Discussion topics for Theory of Knowledge have been included on pages 129 and 140. These aim to help students discover and express their views on knowledge issues. The accompanying student CD includes a PDF of the full text and access to specially designed graphing software. Graphics calculator instructions for Casio fx-9860G Plus, Casio fx-CG20, TI-84 Plus and TI-nspire are available from icons located throughout the book. Fully worked solutions are provided at the back of the text, however students are encouraged to attempt each question before referring to the solution. It is not our intention to define the course. Teachers are encouraged to use other resources. We have developed this book independently of the International Baccalaureate Organization (IBO) in consultation with experienced teachers of IB Mathematics. The Text is not endorsed by the IBO. In this changing world of mathematics education, we believe that the contextual approach shown in this book, with associated use of technology, will enhance the students understanding, knowledge and appreciation of mathematics and its universal applications. We welcome your feedback. Email:
[email protected]
Web:
www.haesemathematics.com.au
CTQ
CS
RCH PMH
ACKNOWLEDGEMENTS
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The authors and publishers would like to thank all those teachers who offered advice and encouragement on this book, with particular mention to Peter Blythe.
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USING THE INTERACTIVE STUDENT CD The interactive CD is ideal for independent study. Students can revisit concepts taught in class and undertake their own revision and practice. The CD also has the text of the book, allowing students to leave the textbook at school and keep the CD at home. By clicking on the relevant icon, a range of interactive features can be accessed: w Graphics calculator instructions for the Casio fx-9860G Plus, Casio fxCG20, TI-84 Plus and the TI-nspire w Interactive links to graphing software
INTERACTIVE LINK
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GRAPHICS CALCUL ATOR INSTRUCTIONS
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TABLE OF CONTENTS
5
TABLE OF CONTENTS
SYMBOLS AND NOTATION USED IN THIS BOOK A B C D E F G H
6
Number properties Limits Continuity of functions Differentiable functions l’Hôpital’s Rule Rolle’s theorem and the Mean Value Theorem (MVT) Riemann sums The Fundamental Theorem of Calculus Z 1 f (x) dx Improper integrals of the form
I
9 12 20 25 29 34 38 44 50
a
Sequences Infinite series Taylor and Maclaurin series Differential equations Separable differential equations The integrating factor method Taylor or Maclaurin series developed from a differential equation Review set A Review set B Review set C Review set D
57 66 89 105 112 117 120 123 124 125 126
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INDEX
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WORKED SOLUTIONS
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APPENDIX B (Formal definition of a limit)
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THEORY OF KNOWLEDGE (Axioms and Occam’s razor)
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APPENDIX A (Methods of proof)
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THEORY OF KNOWLEDGE (Torricelli’s trumpet)
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J K L M N O P
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6
SYMBOLS AND NOTATION USED IN THIS BOOK
¼ > > < 6
is is is is is
f......g 2 2 = N Z Q R Z+ µ ½ ) ) Á f : A!B f : x 7! y
approximately equal to greater than greater than or equal to less than less than or equal to
the set of all elements ...... is an element of is not an element of the set of all natural numbers the set of integers the set of rational numbers the set of real numbers the set of positive integers is a subset of is a proper subset of implies that does not imply that f is a function under which each element of set A has an image in set B f is a function under which x is mapped to y
f (x)
the image of x under the function f
f ±g
or f (g(x)) the composite function of f and g
jxj
the modulus or absolute value of x
[a, b]
the closed interval a 6 x 6 b
] a, b [
the open interval a < x < b the nth term of a sequence or series
un fun g
the sequence with nth term un
Sn S1 n P ui
the sum of the first n terms of a sequence the sum to infinity of a series u1 + u2 + u3 + :::: + un
i=1 n Q
u1 £ u2 £ u3 £ :::: £ un
ui
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IB HL OPT 2ed Calculus
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7
lim f(x)
the limit of f(x) as x tends to a
lim f (x)
the limit of f(x) as x tends to a from the positive side of a
lim f (x)
the limit of f(x) as x tends to a from the negative side of a
maxfa, bg 1 P cn xn
the maximum value of a or b
x!a
x!a+
x!a¡
the power series whose terms have form cn xn
n=0
dy dx f 0 (x)
the derivative of y with respect to x the derivative of f (x) with respect to x
2
d y dx2 f 00 (x) dn y dxn f (n) (x) R y dx Rb y dx a
the second derivative of y with respect to x the second derivative of f (x) with respect to x the nth derivative of y with respect to x the nth derivative of f (x) with respect to x the indefinite integral of y with respect to x
the definite integral of y with respect to x between the limits x = a and x = b ex exponential function of x ln x the natural logarithm of x sin, cos, tan the circular functions csc, sec, cot the reciprocal circular functions arcsin, arccos, arctan the inverse circular functions A(x, y) the point A in the plane with Cartesian coordinates x and y [AB]
the line segment with end points A and B
AB
the length of [AB]
b A
the angle at A
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is parallel to
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is perpendicular to
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the angle between [CA] and [AB] the triangle whose vertices are A, B, and C
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b or ]CAB CAB 4ABC k ?
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IB HL OPT 2ed Calculus
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9
CALCULUS
A
NUMBER PROPERTIES
IMPORTANT NUMBER SETS You should be familiar with the following important number sets: ² Z + = f1, 2, 3, .... g is the set of positive integers. ² N = f0, 1, 2, 3, .... g is the set of natural numbers. ² Z = f...., ¡2, ¡1, 0, 1, 2, .... g is the set of integers. p
² Q is the set of rational numbers. These are numbers which can be expressed in the form where q p, q 2 Z , q 6= 0. ² R is the set of real numbers comprising the rational numbers Q , and the irrational numbers which lie on the number line but cannot be expressed as a ratio of integers. The number sets follow the hierarchy Z + ½ N ½ Z ½ Q ½ R . In this option topic we will be principally concerned with the set R . Rigorous treatments of the algebraic and set theoretic properties of R , such as the fact that R is a continuous set, are available in a variety of calculus and analysis books. However, we will outline here only those results of most immediate relevance to our work with limits, sequences, and series. A closed interval consisting of all real numbers from a to b inclusive is denoted [a, b]. [a, b] is fx j a 6 x 6 bg. An open interval consisting of all real numbers between a and b is denoted ] a, b [ . ] a, b [ is fx j a < x < bg.
a
b
a
b
THE ABSOLUTE VALUE FUNCTION y
For any a 2 R , the absolute value of a, denoted by jaj, is defined by: ½ a if a > 0 jaj = ¡a if a < 0
y = |x| x
The absolute value jaj of a number a 2 R is the distance from a to the origin on the real number line. More generally, the distance between two numbers a, b 2 R on the number line is given by ja ¡ bj. The absolute value function has the following properties: 1 jaj > 0 for all a 2 R . 2 j¡aj = jaj for all a 2 R . 3 jabj = jaj jbj for all a, b 2 R . 4 a = jaj or ¡jaj, and hence ¡jaj 6 a 6 jaj for all a 2 R .
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5 If c > 0 then jaj 6 c if and only if ¡c 6 a 6 c.
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IB HL OPT 2ed Calculus
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CALCULUS
We will also need the property of real numbers that: If a 6 c and b 6 d, then a + b 6 c + d, for a, b, c, d 2 R . Proof of Property 5: ) Suppose that jaj 6 c. Since a 6 jaj and ¡a 6 jaj, we find a 6 jaj 6 c and ¡a 6 jaj 6 c ) a 6 c and ¡a 6 c.
For a proof “if and only if ” we prove it one way ) and then the other (.
But ¡a 6 c is equivalent to ¡c 6 a, so combining the two inequalities we have ¡c 6 a 6 c. ( If ¡c 6 a 6 c, then a 6 c and ¡c 6 a: ) ¡a 6 c: ) since jaj = a or ¡a, jaj 6 c.
THE TRIANGLE INEQUALITY The Triangle Inequality states: For any a, b 2 R , ja + bj 6 jaj + jbj. Proof: From Property 4 we have ¡jaj 6 a 6 jaj and ¡jbj 6 b 6 jbj for all a, b 2 R . Adding these inequalities gives ¡(jaj + jbj) 6 a + b 6 jaj + jbj. Using Property 5 with c = jaj + jbj, this is equivalent to ja + bj 6 jaj + jbj. Corollaries: 1 ja ¡ bj 6 jaj + jbj for all a, b 2 R . 2 jaj ¡ jbj 6 ja + bj for all a, b 2 R . 3 jaj ¡ jbj 6 ja ¡ bj for all a, b 2 R . Proofs: 1 By the Triangle Inequality, we have ja + cj 6 jaj + jcj for all a, c 2 R . ) letting c = ¡b, we get ja ¡ bj 6 jaj + j¡bj for all a, b 2 R . ) ja ¡ bj 6 jaj + jbj jaj = j(a + b) + (¡b)j
2 )
jaj 6 ja + bj + j¡bj )
jaj 6 ja ¡ bj + jbj
for all a, b 2 R
by the Triangle Inequality.
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jaj ¡ jbj 6 ja ¡ bj
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jaj = j(a ¡ b) + bj
3
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jaj ¡ jbj 6 ja + bj
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IB HL OPT 2ed Calculus
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CALCULUS
EXERCISE A 1 Prove that jaj > 0 for all a 2 R . 2 Prove that j¡aj = jaj for all a 2 R . 3 Prove that ja1 + a2 + .... + an j 6 ja1 j + ja2 j + .... + jan j for any a1 , a2 , ...., an 2 R . 4 If a < x < b and a < y < b show that jx ¡ yj < b ¡ a. Interpret this result geometrically. 5 Prove that ja ¡ bj 6 ja ¡ cj + jc ¡ bj. 6 Prove that if jx ¡ aj
.
7 If jx ¡ aj < " and jy ¡ bj < " show that j(x + y) ¡ (a + b)j < 2". 8 The Archimedean Property states that for each pair of positive real numbers a and b, there is a natural number n such that na > b. Use the Archimedean Property to prove that for each positive number " there is a natural number n such that
1 < ". n
The properties of R in questions 8 and 9 are needed later in the course.
9 Prove the Bernoulli Inequality by mathematical induction: If x > ¡1 then (1 + x)n > 1 + nx for all n 2 Z + .
10 The Well-Ordering Principle states that every non-empty subset of Z + has a least element. Show that the Well-Ordering Principle does not apply to R + , the set of positive real numbers. 11 If r 6= 0 is rational and x is irrational, prove that r + x and rx are irrational.
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For questions 10 and 11 you will need to write a proof by contradiction. For help with this, consult Appendix A: Methods of proof.
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IB HL OPT 2ed Calculus
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11
12
CALCULUS
B
LIMITS
Consider a real function f(x) whose domain D is a subset of R . We can write f : D ! R .
1 is not a real number. x ! 1 refers to positive values of x becoming increasingly large.
We wish to examine the behaviour of the function:
² as x approaches some particular finite value a 2 R , so x ! a ² as x tends to infinity or negative infinity, so x ! 1 or x ! ¡1, when D = R .
We shall work with the following informal definition of a limit of a function. For more information, consult Appendix B: Formal definition of a limit. Let a 2 R be a fixed real number. Let f be a function defined in an open interval about x = a, except f (a) need not be defined. We say the number l is the limit of f as x approaches a, provided f(x) becomes as close as we like to l by choosing values of x close enough, but not equal to, the number a. We write lim f(x) = l. x!a
From this definition, we see that f (x) gets closer and closer to l as x gets closer and closer to a, from either side of a. If the function does not approach a finite value l, we say the limit does not exist (DNE). If f (x) gets closer and closer to l as x gets closer and closer to a from the right of a (where x > a), we say “x approaches a from the right”, and write lim f(x) = l. x!a+
If f (x) gets closer and closer to l as x gets closer and closer to a from the left of a (where x < a), we say “x approaches a from the left”, and write lim f (x) = l. x!a¡
For a 2 R , lim f (x) exists if and only if both
lim f(x) and
x!a
x!a¡
In this case lim f (x) = lim f (x) = lim f (x). x!a
x!a¡
lim f (x) exist and are equal.
x!a+
x!a+
For example: lim (x + 1) = 2
y
x!1
y=x+1
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In this case f(1) can be evaluated directly, but we do not do this. Instead, when calculating limits we consider the behaviour of f(x) for x close to 1.
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IB HL OPT 2ed Calculus
Y:\HAESE\IB_HL_OPT-Calculus\IB_HL_OPT-Calculus_01\012IB_HL_OPT-Calculus_01.cdr Tuesday, 19 February 2013 3:11:41 PM GR8GREG
CALCULUS
x2 ¡ 1 x¡1
² y =
is not defined at x = 1. This is not a
y
problem, since in determining the limit as x approaches 1, we never let x actually reach the value 1. But as we make x very, very close to value 1, we can make f (x) as close as we like to value 2. x2 ¡ 1 (x ¡ 1)(x + 1) = lim x!1 x ¡ 1 x!1 (x ¡ 1)
y=
2
x2 ¡ 1 x¡1
1
fsince x 6= 1g
lim
13
1
x
= lim x + 1 x!1
=2 ²
y y=
As x ! 1¡ , f(x) ! ¡1. As x ! 1+ , f(x) ! 1.
1 x¡1
There is no finite real number that f (x) approaches as x approaches value 1, so the limit of f as x approaches 1 does not exist (DNE).
x
-1
½ ² Consider f(x) =
x!1
1 x¡1
DNE.
x + 1, x 6 1 x > 1. x2 ,
As x ! 1¡ , f(x) ! 2 and so
y
lim f(x) = 2.
y = f(x)
x!1¡
As x ! 1+ , f(x) ! 1 and so Since
lim
So,
x=1
lim f (x) = 1.
x!1+
2 1
lim f (x) 6= lim f(x), there is no unique
x!1¡
x!1+
value which the function approaches as x ! 1. ) lim f (x) DNE.
1
x
x!1
We can make f(x) as close as we like to value 5 by making x large enough. As x ! 1, f(x) ! 5 from below. We write lim f(x) = 5¡ .
y
5
y =5-
x!¡1
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Similarly, as x ! ¡1, f (x) ! 5 from above. We write lim f(x) = 5+ .
1 x
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IB HL OPT 2ed Calculus
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14
CALCULUS
Example 1 By examining the graphs of y = x, y = a
lim x DNE
a
lim
b
x!1
x!1
1 , x
1 , establish the following limits: x2 1 1 c lim 2 = 0 d lim DNE x!1 x x!0 x
and y =
1 =0 x
b
y
y y=
y=x
1 x
x x
As x ! 1, y = x ! 1 ) lim x DNE. x!1
) y
c
lim
x!1
1 = 0. x
d y=
1 ! 0+ x
As x ! 1,
y
1 x2
y=
1 x
x
x
As x ! 1, )
1 ! 0+ x2
As x ! 0+ ,
1 = 0. x!1 x2
As x ! 0¡ ,
lim
)
1 x
lim
x!0
1 !1 x 1 ! ¡1 x
DNE.
EXERCISE B.1 1
1
1 By examining the graphs of y = ¡x, y = ¡ , and y = ¡ 2 , establish the following limits: x x ³ ´ ³ ´ ³ ´ 1 1 1 a lim (¡x) DNE b lim ¡ =0 c lim ¡ 2 = 0 d lim ¡ DNE x!1
x!1
x!1
x
x!0
x
x
2 Sketch each function and determine the existence or otherwise of lim f (x). x!a
x2 + x ¡ 2 b f (x) = , x+2
a f (x) = 3x + 2, a = ¡1
( a Sketch the function f (x) =
lim f (x) and ¼¡
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x! 2
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b Hence determine
2x , ¼
100
3
sin x, x >
a = ¡2
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IB HL OPT 2ed Calculus
Y:\HAESE\IB_HL_OPT-Calculus\IB_HL_OPT-Calculus_01\014IB_HL_OPT-Calculus_01.cdr Tuesday, 19 February 2013 3:16:22 PM GR8GREG
CALCULUS
15
c Determine the existence or otherwise of lim f (x): ¼
x! 2
a Sketch the function f(x) =
4
x+1 . x¡1
b Determine the existence or otherwise of: i lim f(x) ii lim f (x)
iii
x!1
x!¡1
8 2 < x , x>3 5 Let f (x) = 5, x=3 : 3x, x < 3. a Sketch the graph of y = f(x).
b Evaluate
lim f (x)
x!1
lim f (x) and
x!3¡
lim f (x).
x!3+
c Does lim f(x) exist? Explain your answer. x!3
p 6 Does lim x exist? Explain your answer. x!0
7 For each of the following functions, discuss the limits: lim f (x)
i
½ a f(x) =
lim f(x)
ii
x!0¡
3
(x ¡ 1) , x 6 0 1, x>0
c f(x) = sin
iii
x!0+
b f(x) =
³ ´ 1 , x 6= 0
8 < sin x,
lim f (x)
x!0
x0
x
THE LIMIT LAWS We often define new functions using a sum, composition, or some other combination of simpler functions. The limit laws help us calculate limits for these new functions using what we already know about the simpler functions. For information about proving these laws, consult Appendix B: Formal definition of a limit. If f (x) = c a constant, where c 2 R , then
lim f (x) = lim c = c, for all a 2 R .
x!a
x!a
We can state the limit laws as follows: Consider real functions f(x) and g(x) for which lim f (x) = l and lim g(x) = m, x!a x!a where a, l, m 2 R .
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m
for all n 2 Z + lim f (x) = l p p lim n f (x) = n l for all n 2 Z + provided l > 0 n
n
x!a
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for any constant c 2 R
x!a
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lim cf (x) = cl
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IB HL OPT 2ed Calculus
Y:\HAESE\IB_HL_OPT-Calculus\IB_HL_OPT-Calculus_01\015IB_HL_OPT-Calculus_01.cdr Thursday, 7 February 2013 3:34:49 PM BRIAN
16
CALCULUS
From the first limit law we see that multiplying by the constant 1 will not change the value or existence of a limit. When evaluating many limits, it often helps to multiply by 1 in a well-chosen form. We will use this technique for rational functions.
RATIONAL FUNCTIONS f(x) is a rational function if it can be written in the form f(x) =
p(x) , where p, q are real polynomials. q(x)
Suppose xm is the highest power of x present in either p or q. To examine lim
x!1
p(x) p(x) or lim , we divide all terms in the numerator and denominator by xm . q(x) x!¡1 q(x) 1
m p(x) p(x) = lim £ x1 x!1 q(x) x!1 q(x)
lim
1 xm 1 xm
since
xm
= 1 for all x 6= 0 and multiplying by the constant 1
does not change the value of the limit. For rational functions, this allows us to make use of the known limits
lim
x!§1
1 = 0 for all m 2 Z + . xm
Example 2 Determine the existence or otherwise of the following limits: a
x+5 x!1 ¡2x2 + x + 1
lim
lim
a
x!1
b
x+5 ¡2x2 + x + 1
10x2 ¡ 5 x!1 3x2 + x + 2
lim
c
lim
b
x!1
1
x!1
=
0
+
¡2 +
+
= lim
1 x2
x!1
fas x ! 1,
¡2
x2
5 x2
1 x
1 x
1
= lim
x2
= lim
10x2 ¡ 5 3x2 + x + 2
2 10x2 ¡ 5 £ x1 2 x!1 3x + x + 2
2 x+5 £ x1 2 x!1 ¡2x + x + 1
= lim
1 x
x2 + x + 1 x!1 x¡2
lim
! 0 and
1 x2
=
! 0g
10 3
10 ¡
5 x2
1 3+ x + x22
=0 c
1 1 1+ x + x12 2 x2 + x + 1 x2 + x + 1 = lim £ x1 = lim 1 x!1 x¡2 x!1 x¡2 x!1 ¡ x22 x x2
lim
As x ! 1, the numerator ! 1, but the denominator ! 0. Hence as x ! 1, )
x2 + x + 1 x!1 x¡2
x2 + x + 1 ! 1. x¡2
lim
DNE.
EXERCISE B.2 1 Evaluate the following limits, where possible:
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µ
x2 + 3x ¡ 4 x!0 x¡1
c lim
1 f lim y !5 y ¡ 5
100
x+1 ¡ 2x ¡ 3
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x!¡1 x2
1 e lim y !2 y ¡ 5
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5
x2 + 3x ¡ 4 d lim x!1 x¡1
lim
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x+1 ¡ 2x ¡ 3
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x!1 x2
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a lim
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µ
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¶
IB HL OPT 2ed Calculus
Y:\HAESE\IB_HL_OPT-Calculus\IB_HL_OPT-Calculus_01\016IB_HL_OPT-Calculus_01.cdr Thursday, 7 February 2013 3:35:22 PM BRIAN
CALCULUS
17
2 Evaluate the following limits, where possible: a
ln x 2¡x
lim p
x!2¡
d lim
µ!0
lim
b
cos µ µ
x!0
sin x ex
¼¡ x! 2
3 Evaluate the following limits, where possible:
r
4x2 ¡ 5x + 1 b lim x!1 x2 + x + 1
3+x a lim x!1 x + 5
lim
c
x!1
p
x2 + x + x
p
4 By first multiplying by
x2
+x+x
lim
x!¼ ¡
sin x 1 ¡ cos x
tan x sec x
lim
e
c
lim
, find
x!1
x2 + 3 x
d
lim
x!1 x2
1+x +x+1
p x2 + x ¡ x. p
x2 + x + x
p
x2 + x + x
5 Let f be a function with domain R , and let a, l 2 R lim f(x) = l
Use the limit laws to prove that
be constants. Suppose lim f (x) exists. x!a
if and only if
x!a
= 1, for x 6= 0.
lim (f(x) ¡ l) = 0.
x!a
THE SQUEEZE THEOREM The Squeeze Theorem shows us that inequalities between functions are preserved when we take limits. Let f, g, h be real functions and let a, l 2 R . Suppose that f (x) 6 g(x) 6 h(x) for all x 6= a in some open interval containing a. If
lim f (x) = l = lim h(x) then
x!a
lim g(x) = l .
x!a
x!a
So, a function g(x) is forced to have the same limit as f and h if g is squeezed between them. Example 3 Use the Squeeze Theorem to evaluate
lim x2 cos
x!0
³ ´ 9 . x
³ ´ 9 x
6 1 for all x 2 R , ³ ´ 9 6 x2 ¡x2 6 x2 cos
Since ¡1 6 cos
x
2
lim ¡x = 0 = lim x2 .
Now
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x!0
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x!0
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5
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³ ´ 9 x
=0
IB HL OPT 2ed Calculus
Y:\HAESE\IB_HL_OPT-Calculus\IB_HL_OPT-Calculus_01\017IB_HL_OPT-Calculus_01.cdr Thursday, 7 February 2013 3:38:49 PM BRIAN
18
CALCULUS
Consider the limit lim
x!0
sin x . Since lim sin x = 0 and lim x = 0, we say this limit has indeterminate x!0 x!0 x
form 00 . We cannot use the limit laws to determine whether or not this limit exists, or what its value might be. However, one way to evaluate the limit is to use the Squeeze Theorem. lim
The Fundamental Trigonometric Limit is
Proof:
Consider the unit circle with angle 0 < µ < ¼2 , and also points P(cos µ, sin µ), Q(cos µ, 0), R(1, 0), and S(1, tan µ).
y 1
S (1, tan µ)
P
µ! 0
sin µ = 1. µ
µ Q R1
x
Clearly, area of 4OPQ < area of sector OPR < area of 4ORS 1 2
)
cos µ sin µ < 12 µ < 1 2
Since 0 < µ < ¼2 ,
1 2
sin µ cos µ
sin µ > 0
µ 1 < sin µ cos µ 1 sin µ > > cos µ ) cos µ µ sin µ 1 < ) cos µ < µ cos µ
) cos µ
0 ¡1, x < 0 y
1 x¡5
y = f(x)
1
x
x -1 x=5
f is not defined at x = 5, and lim f(x) DNE.
f is defined for all x, but there is a ‘jump’ discontinuity at x = 0.
x!5
lim f (x) = ¡1 and
f is continuous for all x 2 R , x 6= 5. f has an essential discontinuity at x = 5.
x!0¡
)
lim f (x) = 1
x!0+
lim f(x) DNE.
x!0
f is continuous for all x 2 R , x 6= 0. f has an essential discontinuity at x = 0. Example 5 Discuss the continuity of the following functions. If there is a removable discontinuity, describe how this could be removed. ( sin x ½ 2 , x 6= 0 x , x 6= 2 x a f (x) = b f (x) = 6, x = 2 0, x=0 a
f is defined for all x, but there is a discontinuity at x=2 lim f (x) = 4 and lim f(x) = 4
y 6
y = f(x)
x!2¡
4
x!2+
lim f(x) = 4
) But
lim f(x) 6= f (2), so there is a removable
x!2
discontinuity when x = 2.
x
2
x!2
f is continuous for all x 2 R , x 6= 2.
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The discontinuity can be removed by defining a new function based on f, but which is continuous at x = 2. ½ 2 x , x= 6 2 This is g(x) = which is actually just g(x) = x2 . 4, x=2
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IB HL OPT 2ed Calculus
Y:\HAESE\IB_HL_OPT-Calculus\IB_HL_OPT-Calculus_01\021IB_HL_OPT-Calculus_01.cdr Tuesday, 19 February 2013 3:21:20 PM GR8GREG
22
CALCULUS
b
f is defined for all x, but there is a discontinuity at x = 0. lim f (x) = 1 and lim f (x) = 1
y 1 y = f(x)
x!0¡
x!0+
lim f (x) = 1
) x
But
x!0
lim f(x) 6= f (0), so there is a removable
x!0
discontinuity when x = 0. f is continuous for all x 2 R , x 6= 0. The discontinuity can be removed by defining a new function based on f , but which is ( sin x , x= 6 0 x continuous at x = 0. This is g(x) = 1, x = 0.
EXERCISE C.1 1 Suppose f and g are functions which are continuous at x = a. Use the limit laws to prove that the following functions are also continuous at x = a: a f (x) g(x)
b f (x) § g(x)
f (x) , g(x)
c
for g(a) 6= 0
e [f (x)]n , n 2 Z +
d c f (x), for c 2 R a constant 2 Consider the function f with graph:
y 6
b For which values of 8 x, > > > > x2 ¡ 6, > > > > < 5, 3 Let f (x) = 6, > > > > 5 , > > > 4¡x > : 2x,
y = f(x)
4
a Complete the following: i f has an essential discontinuity at x = :::::: ii f has a removable discontinuity at x = ::::::
1
2
x
5 x=6
x 2 R is f continuous? x < ¡2 ¡2 6 x < 0 06x2 ½ ½ kx, x > 0 k + x, x>2 e f(x) = f f(x) = 0, x60 jk + 2j , x < 2 6 For a real function f with domain D, an alternative (equivalent) definition of continuity is as follows: “f is continuous at a 2 D if, for any sequence fxn g of values in D such that lim xn = a, then lim f(xn ) = f(a).” n!1 n!1 ½ 1, x 2 Q The Dirichlet function is defined by g(x) = 0, x 2 =Q. a Consider the sequence xn = a +
p 2 , a 2 Q , n 2 Z + . Use the given definition of continuity n
to show that g(x) is discontinuous at x = a. b Now suppose a 2 = Q , and that xn is the decimal expansion of a to n decimal places, n 2 Z + . Use the given definition of continuity to show that g(x) is discontinuous at x = a. c Hence discuss the continuity of the Dirichlet function. 7 Consider two functions f, g and two values a, l 2 R . Suppose f is continuous on R . If l = g(a), you may assume the result: “If g is continuous at a, and f is continuous at g(a), then f ± g is continuous at a.” lim f (g(x)) = f(l). ´ b Show that lim f (g(x)) = f lim g(x) whenever lim g(x) exists. x!a x!a x!a ³ ´ c Show that lim f(g(x)) = f lim g(x) whenever lim g(x) exists. a If
lim g(x) = l, show that
x!a
³
x!1
x!1
1
a Show that x n = eln x
8
x!a
1 n
x!1
for x > 0, x 2 R .
b Use the fact that f(x) = ex
is continuous on R to prove that
x 2 R , n 2 Z +.
1
lim x n = 1 for x > 0,
n!1
THE INTERMEDIATE VALUE THEOREM (IVT) A function f is continuous on a closed interval [a, b], a < b, if f is continuous at x for all x 2 ] a, b [ , and also lim f (x) = f(a) and lim f(x) = f(b). x!a+
x!b¡
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The following theorem formalises the intuitive property that a function f which is continuous on a closed interval [a, b] has no ‘breaks’ or ‘holes’ between x = a and x = b, and will in fact take every value between f(a) and f (b) as we increase x from a to b.
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IB HL OPT 2ed Calculus
Y:\HAESE\IB_HL_OPT-Calculus\IB_HL_OPT-Calculus_01\023IB_HL_OPT-Calculus_01.cdr Tuesday, 19 February 2013 3:24:42 PM GR8GREG
24
CALCULUS
The Intermediate Value Theorem (IVT) states that: Suppose a function f is continuous on a closed interval [a, b]. If k is any value between f (a) and f (b), then there exists c 2 [a, b] such that f (c) = k. y
y
f(a)
f(b) k
k
or f(a)
f(b)
c
a
x
b
a
c
b
x
BOUNDED FUNCTIONS A function f is bounded on [a, b] if, for all x 2 [a, b], jf(x)j 6 M for some M 2 R . In other words, a function is bounded on [a, b] if it does not tend to infinity or negative infinity on the interval [a, b]. If a function f is continuous on [a, b], then f is bounded on [a, b]. It follows that if f is continuous on [a, b], a < b, then: ² f has a maximum value M on the interval [a, b] where M = f(xM ) for some xM 2 [a, b] ² f has a minimum value m on the interval [a, b] where m = f (xm ) for some xm 2 [a, b].
EXERCISE C.2 1 Using the IVT, explain why: a f (x) = x3 + x ¡ 3 has a real zero in the interval [0, 2] b f (x) =
4 x¡2
does not have a real zero in the interval [1, 3] even though f(1) < 0 and
f (3) > 0.
2 Consider the function f(x) = x3 ¡ 9x2 + 24x ¡ 10 on the interval [1, 5]. Find: a the maximum value M of f on [1, 5], and the values xM 2 [1, 5] such that f (xM ) = M b the minimum value m of f on [1, 5], and the values xm 2 [1, 5] such that f(xm ) = m. 3 Suppose f is continuous on [a, b], a < b, and suppose f(a), f (b) have opposite signs. Prove that f has at least one zero between a and b. a Prove that for each constant r 2 R , r > 0, there exists a real value of x such that sin x = 1 ¡ rx. Hint: Let f (x) = sin x + rx ¡ 1 and apply the IVT to a suitable interval.
4
b Suppose r = 1. Use your calculator to solve sin x = 1 ¡ x on the domain x 2 ] ¡2, 2 [ . a Use the IVT to prove that f (x) = x15 +
5
1 1 + sin2 x + (r + 1)2
has at least one zero on the
interval ] ¡1, 1 [ for any constant r 2 R .
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Explain your answer.
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] ¡1, 1 [ of g(x) = rx17 +
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b Let r 2 R , r > 0 be any constant. Can the IVT be used to prove the existence of a zero in
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IB HL OPT 2ed Calculus
Y:\HAESE\IB_HL_OPT-Calculus\IB_HL_OPT-Calculus_01\024IB_HL_OPT-Calculus_01.cdr Friday, 15 March 2013 3:12:00 PM BRIAN
CALCULUS
D
25
DIFFERENTIABLE FUNCTIONS
Suppose f is a real function with domain D containing an open interval about x = a. f (a + h) ¡ f (a) exists. If this limit exists we denote it by h f (a + h) ¡ f (a) f 0 (a) = lim is the derivative of f at x = a. h h!0
f is differentiable at x = a if f 0 (a), and
lim
h!0
If f is differentiable at a for all a 2 D, then we say f is a differentiable function. By letting x = a + h in the definition of the derivative, and noting that as h ! 0, x = a + h ! a, we obtain the following alternative form for the derivative of f at x = a: f 0 (x) = lim
x!a
f (x) ¡ f (a) . x¡a
Example 6 lim
a Prove that
h!0
cos h ¡ 1 = 0. h
b Using the limit definition of the derivative, prove that if f(x) = sin x then f 0 (x) = cos x. cos h ¡ 1 cos h ¡ 1 cos h + 1 = lim £ h h cos h + 1 h!0
a lim
h!0
cos2 h ¡ 1 h!0 h(cos h + 1)
= lim = lim
h!0
= lim
h!0
= lim
h!0
=1£ =0
¡ sin2 h h(cos h + 1) sin h ¡ sin h £ h cos h + 1 sin h ¡ sin h £ lim h h!0 cos h + 1
fby the limit laws since both limits existg
0 2
b Let f (x) = sin x. If f 0 (x) exists, then it is given by sin(x + h) ¡ sin x h sin x cos h + sin h cos x ¡ sin x = lim h h!0 h ³ ´ i cos h ¡ 1 sin h = lim sin x + £ cos x h h h!0 ³ ´³ ´ ³ ´³ ´ cos h ¡ 1 sin h = lim sin x lim + lim cos x lim h h!0 h!0 h!0 h!0 h
f 0 (x) = lim
h!0
fby the limit laws, since each of these limits existsg = sin x £ 0 + cos x £ 1 fsince x is independent of hg = cos x
DIFFERENTIABILITY AND CONTINUITY
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is differentiable at x = a, then f is continuous at x = a.
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IB HL OPT 2ed Calculus
Y:\HAESE\IB_HL_OPT-Calculus\IB_HL_OPT-Calculus_01\025IB_HL_OPT-Calculus_01.cdr Thursday, 7 February 2013 5:47:07 PM BRIAN
26
CALCULUS
lim f(a + h) ¡ f (a)
Proof:
h!0
f (a + h) ¡ f (a) £h h f (a + h) ¡ f (a) £ lim h = lim h h!0 h!0
= lim
h!0
fby the limit laws since both limits existg
= f 0 (a) £ 0 =0 )
lim f (a + h) = f (a)
h!0
By letting x = a + h, and since h can take both positive and negative values, this is equivalent to lim f (x) = f(a). x!a
) f is continuous at x = a. It follows that if a function is not continuous at x = a, then it is not differentiable at x = a. The converse of this theorem is not true, however. If a function is continuous at x = a, it is not necessarily differentiable there. The set of all differentiable functions is therefore a proper subset of the set of all continuous functions.
TESTING FOR DIFFERENTIABILITY For functions which are defined by different expressions on separate intervals, we need a formal test to see whether the function is differentiable. To do this we first need to define: f (a + h) ¡ f (a) h
0 (a) = lim ² the left-hand derivative of f at x = a is f¡
h!0¡
0 (a) = lim ² the right-hand derivative of f at x = a is f+
h!0+
f (a + h) ¡ f (a) . h
A function f : D ! R is differentiable at x = a, a 2 D, if: 1 f is continuous at x = a, and 0 (a) = lim 2 f¡
h!0¡
f (a + h) ¡ f (a) h
Example 7
½
Prove that f (x) = jxj =
h!0+
f (a + h) ¡ f (a) h
x!0¡
y
x!0¡
lim f (x) = lim x = 0
and
x!0+
x!0+
y = jxj
lim f (x) = f(0) = 0
x!0
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) f is continuous at x = 0.
5
both exist and are equal.
x, x>0 is continuous but not differentiable at x = 0. ¡x, x < 0
lim f (x) = lim (¡x) = 0,
f (0) = 0,
)
0 (a) = lim and f+
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IB HL OPT 2ed Calculus
Y:\HAESE\IB_HL_OPT-Calculus\IB_HL_OPT-Calculus_01\026IB_HL_OPT-Calculus_01.cdr Tuesday, 19 February 2013 3:26:36 PM GR8GREG
CALCULUS
½ Now
f 0 (x)
=
1, x>0 ¡1, x < 0
27
since the derivative of f exists on the open interval ] ¡1, 0 [
and on the open inteval ] 0, 1 [ . )
0 0 (0) = ¡1 and f+ (0) = 1 f¡
)
0 0 f¡ (0) 6= f+ (0)
) f is not differentiable at x = 0.
Example 8
½
Consider f(x) =
sin x, x>0 with domain R . 2 x + 5x, x < 0
a Prove that f is continuous but not differentiable at x = 0. b Write down f 0 (x) as a piecewise defined function. a f (0) = 0 is defined.
y
lim f (x) = lim (x2 + 5x) = 0
x!0¡
and )
x!0¡
x
lim f (x) = lim sin x = 0
x!0+
x!0+
y = f(x)
lim f (x) = f(0) = 0
x!0
) f(x) is continuous at x = 0. Using known derivatives for open intevals of R , ½ cos x, x>0 we have f 0 (x) = 2x + 5, x < 0 0 (0) = lim (2x + 5) = 2 £ 0 + 5 = 5 f¡
)
x!0¡
0
and f+ (0) = lim cos x = cos 0 = 1 0 f¡ (0)
Since
x!0+
0 6= f+ (0), f is not differentiable at x = 0.
We observe on the graph of y = f (x) that the curve does not have a unique tangent at x = 0. ½ cos x, x>0 0 b From a we have f (x) = 2x + 5, x < 0.
DISCUSSION There are functions which are continuous everywhere in their domain but which are differentiable nowhere!
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Research and discuss the Weierstrass function.
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IB HL OPT 2ed Calculus
Y:\HAESE\IB_HL_OPT-Calculus\IB_HL_OPT-Calculus_01\027IB_HL_OPT-Calculus_01.cdr Tuesday, 19 February 2013 3:27:11 PM GR8GREG
28
CALCULUS
EXERCISE D 1 Let f (x) = cos x. Use the definition of the derivative to prove that f 0 (x) = ¡ sin x. ½ x ¡ 5, x > 5 2 Prove that f(x) = j x ¡ 5 j = is continuous but not differentiable at x = 5. 5 ¡ x, x < 5 ½ y x + 2, x>0 3 Let f (x) = 2 x + 3x, x < 0. y = f(x) Explain why f (x) is not differentiable at x = 0.
x
-3
½ 4 Let f (x) =
¡x2 + 5x + 6, x > 1 3x + 10, x < 1.
a Sketch the function y = f (x). 0 (1) i f¡
b Calculate:
0 (1) ii f+
c Is f differentiable at x = 1? Explain your answer. 5 For each of the following functions and the given value of a, determine whether the function is differentiable at x = a. ½ ½ 1 + sin x, x>0 cos x, x > 0 a f (x) = b f (x) = , a=0 , a=0 x2 + x + 1, x < 0 x3 , x2 c f (x) = , a=2 3 x + 2x + 1, x < 2 6 Investigate the continuity and differentiability of f at x = 0 if ½ k sin x, x > 0 f(x) = , where k 2 R is any constant. tan x, x < 0 7 Find constants c, d 2 R so that the given function is differentiable at x = 1. ½ 2 ½ sin(x ¡ 1) + cx, x > 1 x , x61 a f (x) = b f (x) = x 1 x2 ¡ x + d, ³ ´ ½ 3 1 x sin , x 6= 0 8 Let f (x) = x 0, x = 0. a Prove that f is continuous and differentiable at x = 0. b Write down f 0 (x) as a piecewise defined function.
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c Is f 0 (x) continuous at x = 0? Explain your answer.
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IB HL OPT 2ed Calculus
Y:\HAESE\IB_HL_OPT-Calculus\IB_HL_OPT-Calculus_01\028IB_HL_OPT-Calculus_01.cdr Tuesday, 19 February 2013 3:28:01 PM GR8GREG
CALCULUS
E
29
ˆ L’HOPITAL’S RULE
The limit laws do not help us to deal with limits which have indeterminate forms. These include: Type
Description
0 0
lim
x!a
1 1
f (x) g(x)
lim
x!a
0£1
lim f (x) = 0 and
where
x!a
lim g(x) = 0
x!a
where f(x) ! §1 and g(x) ! §1 when x ! a
lim [f(x) g(x)] where
x!a
For example, consider
lim f(x) = 0 and g(x) ! §1 when x ! a
x!a
2x ¡ 1 . x!0 x
lim
lim (2x ¡ 1) = 0 and
Since
f (x) g(x)
lim (x) = 0, we have the indeterminate form
x!0
x!0
0 0.
To address these types of limits, we use l’Hˆopital’s Rule: Suppose f (x) and g(x) are differentiable and g0 (x) 6= 0 on an open interval that contains the point x = a. If
lim f (x) = 0 and
lim g(x) = 0, or, if as x ! a, f (x) ! §1 and g(x) ! §1,
x!a
then
x!a
f (x) f 0 (x) = lim 0 x!a g(x) x!a g (x)
lim
provided the limit on the right exists.
Proof of one case of l’Hˆopital’s Rule: The derivative of a function f (x) at a point x = a, denoted by f 0 (a), is given by f 0 (a) = lim
x!a
f (x) ¡ f (a) . x¡a
Using this definition of the derivative, we prove the case of l’Hˆopital’s Rule in which f (a) = g(a) = 0, f 0 (x) and g0 (x) are continuous, and g 0 (a) 6= 0. Under these conditions, f (x) f (x) ¡ f (a) = lim x!a g(x) x!a g(x) ¡ g(a)
lim
f (x)¡f (a) x¡a x!a g(x)¡g(a) x¡a
fmultiplying by
= lim
f (x)¡f (a) x¡a
lim
g(x)¡g(a) x¡a
lim f 0 (x)
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x!a
1 x¡a 1 x¡a
fsince f 0 (x) and g 0 (x) are continuousg
lim g 0 (x)
100
x!a
50
=
75
f 0 (a) g 0 (a)
25
=
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x!a
75
=
lim
x!a
fsince f(a) = g(a) = 0g
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IB HL OPT 2ed Calculus
Y:\HAESE\IB_HL_OPT-Calculus\IB_HL_OPT-Calculus_01\029IB_HL_OPT-Calculus_01.cdr Monday, 18 March 2013 10:11:52 AM BRIAN
30
CALCULUS
Example 9 Use l’Hˆopital’s Rule to evaluate: 2x ¡ 1 x!0 x
a lim
lim xe¡x
b
c
x!1
lim
x!1
ln x x
a lim (2x ¡ 1) = 0 and lim x = 0, so we can use l’Hˆopital’s Rule if the resulting limit exists. x!0
) =
x!0
2x ¡ 1 lim x!0 x
2x ¡ 1 x
lim
x!0
d lim dx (2x ¡ 1)
x!0
Pp .
has indeterminate form
fl’Hˆopital’s Ruleg
d lim dx (x)
x!0
=
lim 2x ln 2
x!0
lim 1
x!0
=
ln 2 1
= ln 2
lim xe¡x has
b As x ! 1, e¡x ! 0, so we can use l’Hˆopital’s Rule.
x!1
lim xe¡x
)
indeterminate form 1 £ 0.
x!1
= lim
x!1
=
x ex
fconvert limit to a quotient with the form
lim d (x) x!1 dx lim d (ex ) x!1 dx
1 g 1
fl’Hˆopital’s Ruleg
lim 1
x!1
=
lim (ex )
x!1
=0
fsince
lim 1 = 1 and as x ! 1, ex ! 1g
x!1
c As x ! 1, ln x ! 1 and x ! 1, so we can use l’Hˆopital’s Rule. )
lim
x!1
=
=
ln x x
lim d (ln x) x!1 dx lim d (x) x!1 dx lim
fl’Hˆopital’s Ruleg
ln x 1 . has indeterminate form x 1
³ ´ 1 x
x!1
lim 1
x!1
=
lim
x!1
0 1
fsince
lim
³ ´ 1 x
x!1
=0
= 0g
Important: A common error is to attempt to evaluate the Fundamental Trigonometric Limit lim
x!0
sin x x
using l’Hˆopital’s Rule. We cannot use l’Hˆopital’s Rule in this case as the derivative of sin x from first principles itself requires the use of
lim
x!0
sin x = 1, x
as shown in
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Example 6.
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IB HL OPT 2ed Calculus
Y:\HAESE\IB_HL_OPT-Calculus\IB_HL_OPT-Calculus_01\030IB_HL_OPT-Calculus_01.cdr Tuesday, 19 February 2013 3:38:58 PM GR8GREG
CALCULUS
EXERCISE E 1 Evaluate, if possible, the following limits using l’Hˆopital’s Rule. 1 ¡ cos x x!0 x2
b
ex ¡ 1 ¡ x x!0 x2
ln x x!1 x ¡ 1
d
ex x!1 x
a lim c lim e
lim x ln x
f
x2 + x x!0 sin 2x
h
x!0+
g lim i lim
x + sin x x ¡ sin x
k lim
ax ¡ bx , sin x
x!0 x!0
2 Attempt to find
lim
3 Show that
lim
lim
lim
x!0
Some of these limits were evaluated in Exercises B.2 and B.3 using other methods.
arctan x x
lim
x!0+
sin x p x
lim x2 ln x
j
x!0+
where a, b > 0 are real constants. lim
¼¡ x! 2
¼ 2
tan x sec x
using l’Hˆopital’s Rule.
¡ arccos x ¡ x
= 16 .
x3
x!0
Example 10 Find
lim
x!0+
ln(cos 3x) . ln(cos 2x)
lim ln(cos 3x) = 0 and
x!0+
ln(cos 3x) ) lim = lim x!0+ ln(cos 2x) x!0+
lim ln(cos 2x) = 0, so we can use l’Hˆopital’s Rule.
x!0+
à ¡3 sin 3x ! cos 3x
fl’Hˆopital’s Ruleg
¡2 sin 2x
³
cos 2x
´ = lim x!0+ µ ¶ µ ¶ sin 3x 3 cos 2x = lim £ lim 3 sin 3x cos 2x 2 sin 2x cos 3x
x!0+
sin 2x
lim
sin 3x sin 2x
µ
lim sin 3x = 0 and
£
3 2
lim sin 2x = 0,
x!0+
x!0+
so we can use l’Hˆopital’s Rule again. µ ¶ ln(cos 3x) 3 cos 3x ) lim = lim £ x!0+
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fl’Hˆopital’s Ruleg
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5
=
3 2 9 4
75
=
2 cos 2x
100
ln(cos 2x)
25
x!0+
At step (¤) we could have alternatively used the Fundamental Trigonometric Limit.
(¤)
75
Now
x!0+
25
=
2 cos 3x
x!0+
¶
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IB HL OPT 2ed Calculus
Y:\HAESE\IB_HL_OPT-Calculus\IB_HL_OPT-Calculus_01\031IB_HL_OPT-Calculus_01.cdr Thursday, 7 February 2013 6:05:35 PM BRIAN
31
32
CALCULUS
4 Evaluate: ln(cos 5x) ln(cos 3x)
lim
a
x!0+
lim
b
¼¡ x! 2
ln(sin 2x) ln(sin 3x)
Example 11 lim (sec x ¡ tan x).
Evaluate
¼¡
x! 2
As x !
¼¡ , 2
both sec x ! 1 and tan x ! 1.
We therefore need to convert the difference sec x ¡ tan x into a quotient, and then apply l’Hˆopital’s Rule. Now sec x ¡ tan x = )
1 sin x 1 ¡ sin x ¡ = cos x cos x cos x
³
lim (sec x ¡ tan x) = lim ¼¡
¼¡
x! 2
x! 2
1 ¡ sin x cos x
´
lim (1 ¡ sin x) = 0 and
where
lim cos x = 0
¼¡
¼¡
x! 2
lim
)
lim (sec x ¡ tan x) = ¼¡
x! 2
x!
¼¡ 2
lim
¼¡ x! 2
=
0 1
x
(¡ cos x)
fl’Hˆopital’s Ruleg
(¡ sin x)
=0
5 Evaluate, if possible: ³ ´ 1 1 a lim ¡ x!0+
x! 2
b
sin x
lim
³
x!0+
1 1 ¡ x sin 2x
´ c
lim (sec2 x ¡ tan x) ¼¡
x! 2
Example 12 ex x!1 xn
where n 2 Z + .
lim
Evaluate, if possible:
For all n 2 Z + , as x ! 1, ex ! 1 and xn ! 1, so we can use l’Hˆopital’s Rule. )
ex ex = lim n x!1 x x!1 nxn¡1 ex = lim x!1 n(n ¡ 1)xn¡2
lim
We use l’Hˆopital’s Rule n times.
.. . ex x!1 n! 1 = lim ex n! x!1
= lim
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As x ! 1, ex ! 1, so the given limit DNE.
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IB HL OPT 2ed Calculus
Y:\HAESE\IB_HL_OPT-Calculus\IB_HL_OPT-Calculus_01\032IB_HL_OPT-Calculus_01.cdr Tuesday, 19 February 2013 3:43:09 PM GR8GREG
CALCULUS
6
xk , x!1 ex
k 2 Z +.
lim
a Evaluate, if possible:
33
b Hence explain why, as x ! 1, the exponential function ex increases more rapidly than any fixed positive power of x. 7 Prove that as x ! 1, ln x increases more slowly than any fixed positive power of x. ³ ´ 1 8 a Prove that lim x ln 1 + = 1. x!1 x ¡ ¢ ³ ´ 1 1 x x ln 1+ x b By writing 1+ =e and using the fact that f (x) = ex is continuous on R , x ³ ´ 1 x = e. prove that lim 1 + x!1 x ³ ´ a x c Prove that for a 6= 0, lim 1 + = ea . x!1
x
9 By writing xsin x = esin x ln x and using the fact that f (x) = ln x is continuous on x 2 R , x > 0, prove that lim xsin x = 1. x!0+
1
cyan
lim x x = 1.
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10 Prove that
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IB HL OPT 2ed Calculus
Y:\HAESE\IB_HL_OPT-Calculus\IB_HL_OPT-Calculus_01\033IB_HL_OPT-Calculus_01.cdr Tuesday, 19 February 2013 4:43:40 PM GR8GREG
34
CALCULUS
F
ROLLE’S THEOREM AND THE MEAN VALUE THEOREM (MVT)
Suppose a real function f is defined on domain D. f : D ! R is continuous on a closed interval [a, b], a < b, if: 1 f is continuous at x = c for all c 2 ] a, b [ , and 2
lim f(x) = f (a) and
x!a+
lim f (x) = f (b)
x!b¡
For a function f : D ! R we define: f is differentiable on an open interval ] a, b [ , a < b if f is differentiable at x = c for all c 2 ] a, b [ . f is differentiable on a closed interval [a, b] , a < b, if: 1 f is differentiable on ] a, b [ , and 0 (a) and f 0 (b) both exist. 2 f+ ¡
ROLLE’S THEOREM Suppose function f : D ! R is continuous on the closed interval [a, b], and differentiable on the open interval ] a, b [.
y y = f(x)
If f (a) = f (b) = 0, then there exists a value c 2 ] a, b [ such that f 0 (c) = 0.
a
c
b
x
Rolle’s theorem guarantees that between any two zeros of a differentiable function f there is at least one point at which the tangent line to the graph y = f(x) is horizontal. Proof of Rolle’s theorem: Since f is continuous on [a, b], it attains both a maximum and minimum value on [a, b]. If f takes positive values on [a, b], let f(c) be the maximum of these. Now f (a) = 0 = f(b) and f (c) > 0, so c 2 ] a, b [ . Since f is differentiable at c, f must have a local maximum at x = c. ) f 0 (c) = 0. Similarly, if f takes negative values on [a, b], let the minimum of these be f (c). It follows that f has a local minimum at x = c, and therefore f 0 (c) = 0. Finally, if f (x) = 0 for all x 2 [a, b] then clearly f 0 (c) = 0 for all c 2 ] a, b [ . Rolle’s theorem is a lemma used to prove the following theorem.
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A lemma is a proven proposition which leads on to a larger result.
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IB HL OPT 2ed Calculus
Y:\HAESE\IB_HL_OPT-Calculus\IB_HL_OPT-Calculus_01\034IB_HL_OPT-Calculus_01.cdr Tuesday, 19 February 2013 4:45:46 PM GR8GREG
CALCULUS
35
THE MEAN VALUE THEOREM (MVT) (THE LAGRANGE FORM) If f is a function continuous on [a, b] and differentiable on ] a, b [, then f (b) ¡ f (a) = f 0 (c) £ (b ¡ a) for some number c 2 ] a, b [ . The MVT tells us that for such a function f there exists at least one value c 2 ] a, b [ such that the tangent to the curve y = f (x) at x = c has gradient equal to the gradient of the chord through points (a, f(a)) and (b, f (b)).
y
y = f(x) gradient = f(b) ¡ f(a) b¡a gradient = f 0 (c)
a
c
b
x
Proof of the MVT: Let h(x) = f (x) ¡
h
i
f (b) ¡ f (a) (x ¡ a) ¡ f (a) b¡a
for x 2 [a, b].
Since f is continuous on [a, b] and differentiable on ] a, b [ , so is the function h. Now, h(a) = h(b) = 0, so by Rolle’s theorem there exists c 2 ] a, b [ such that h0 (c) = 0. ³ ´ f (b) ¡ f (a) for x 2 ] a, b [ But h0 (x) = f 0 (x) ¡ b¡a ³ ´ f (b) ¡ f (a) ) h0 (c) = f 0 (c) ¡ =0 b¡a
f (b) ¡ f (a) . ) f 0 (c) = b¡a
We have seen that the derivative of a constant function is zero. We can now prove the converse of this result, that if a continuous function f has derivative equal to zero, then f is a constant function. FIRST COROLLARY OF THE MVT Suppose the function f is continuous on [a, b] and differentiable on ] a, b [ . If f 0 (x) = 0 for all x 2 ] a, b [ , then f (x) is a constant function on [a, b]. Proof: Let x 2 ] a, b ] and apply the MVT to f on the interval [a; x].
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Then f (x) ¡ f (a) = f 0 (c)(x ¡ a) for some c 2 ] a, x [ = 0 £ (x ¡ a) =0 ) f (x) = f (a) for all x 2 ] a, b ] ) f (x) = f (a) for all x 2 [a, b]
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A corollary is a result which follows directly from a theorem, and which is worth stating in its own right.
IB HL OPT 2ed Calculus
Y:\HAESE\IB_HL_OPT-Calculus\IB_HL_OPT-Calculus_01\035IB_HL_OPT-Calculus_01.cdr Tuesday, 19 February 2013 4:46:46 PM GR8GREG
36
CALCULUS
SECOND COROLLARY OF THE MVT Suppose F and G are two functions continuous on [a, b] and differentiable on ] a, b [ . If F 0 (x) = G0 (x) for all x 2 ] a, b [ , then F (x) = G(x) + C for some constant C, for all x 2 [a, b]. Let H(x) = F (x) ¡ G(x), for x 2 [a, b].
Proof:
) H is differentiable on ] a, b [ and H 0 (x) = 0 on ] a, b [ . By the first corollary of the MVT, H(x) is a constant for all x 2 [a, b]. Hence F (x) = G(x) + C
as required.
ANTIDERIVATIVES A function f has an antiderivative G on [a, b] if there exists a function G continuous on [a, b] such that G0 (x) = f (x) for all x 2 ] a, b [ . If a function f has an antiderivative G, then G is unique up to the addition of a constant. Suppose f has two antiderivatives F , G on [a, b].
Proof:
By definition, F 0 (x) = G0 (x) = f (x) for all x 2 ] a, b [ . By the second corollary to the MVT, F (x) = G(x) + C, where C is a constant.
Z f (x) dx = G(x)+C
We define
to be the indefinite integral of function f with respect to x. The
function f is called the integrand and the indefinite integral of f is thus the set of all antiderivatives of f on [a, b].
EXERCISE F 1 Determine whether or not Rolle’s theorem applies to the function f on the given interval [a, b]. If Rolle’s theorem does apply, find all values c 2 ] a, b [ for which f 0 (c) = 0. a f (x) = 3x3 + 5x2 ¡ 43x + 35, [a, b] = [¡5, 2 13 ] b f (x) = jxj ¡ 5, [a, b] = [¡5, 5] 1
c f (x) = 2 ¡ , [a, b] = [¡ 12 , 7] x+1 ½ ¡2x ¡ 5, x < ¡1 d f (x) = , [a, b] = [¡2 12 , 2] x > ¡1 x2 ¡ 4, 2 For each of the following functions, find: i the number of real zeros of the derivative f 0 (x), as guaranteed by Rolle’s theorem ii the exact number of real zeros of the derivative f 0 (x). b f (x) = (x ¡ 1)2 (x2 ¡ 9)(x ¡ 2)
a f (x) = (x ¡ 1)(x ¡ 2)(x ¡ 4)(x ¡ 5)
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c f (x) = (x ¡ 1)2 (x2 + 9)(x ¡ 2)
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IB HL OPT 2ed Calculus
Y:\HAESE\IB_HL_OPT-Calculus\IB_HL_OPT-Calculus_01\036IB_HL_OPT-Calculus_01.cdr Tuesday, 19 February 2013 3:47:35 PM GR8GREG
CALCULUS
37
3 Show that the given function f satisfies the MVT on the given interval [a, b]. Find all values of c such that f(b) ¡ f (a) = f 0 (c) £ (b ¡ a). p a f(x) = x3 , [a, b] = [¡2, 2] b f(x) = x ¡ 2, [a, b] = [3, 6] 1 x
c f(x) = x + , [a, b] = [1, 3] 4 Let f (x) =
p x.
a Use the MVT to show there exists c 2 ] 49, 51 [ such that
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1 8
c Hence prove
c
on the interval [49, 64]. p < 51 ¡ 7 < 17 without using decimal expansions.
5
b Graph
1 y= p x
p 1 51 ¡ 7 = p .
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IB HL OPT 2ed Calculus
Y:\HAESE\IB_HL_OPT-Calculus\IB_HL_OPT-Calculus_01\037IB_HL_OPT-Calculus_01.cdr Thursday, 7 February 2013 6:15:51 PM BRIAN
38
CALCULUS
G
RIEMANN SUMS
Let y = f (x) be a function which is non-negative and continuous on the interval [a, b], a < b. y y = f(x) A
a
x
b
Let A denote the area under the curve y = f (x) and above the x-axis for a 6 x 6 b. Archimedes of Syracuse (c. 287 - 212 BC) developed the following method for approximating the area A using the sum of areas of rectangles: A partition of the interval [a, b] is a set of n + 1 points in [a, b] which divide the interval into n subintervals [x0 , x1 ], [x1 , x2 ], ...., [xn¡1 , xn ], where x0 = a, xn = b, and x0 < x1 < :::: < xn¡1 < xn . We write partition P = fa = x0 , x1 , ...., xn = bg. The length of the ith subinterval [xi¡1 , xi ], i = 1, ...., n is ¢xi = xi ¡ xi¡1 . y y = f(x)
a = x0
xn = b x
xn-1
x2
x1
Let xi¤ 2 [xi¡1 , xi ] be an arbitrarily chosen value in the ith subinterval, i = 1, ...., n. Let Ai = the area of the rectangle whose base is the width of the ith subinterval and with height equal to f (xi¤ ) ) Ai = f (xi¤ ) £ ¢xi y y = f(x)
f(x2*) A2 x0
x1 x2* x2
The area A can be approximated by A ¼
n P i=1
xn
x
Ai = f(x1¤ )¢x1 + :::: + f (xn¤ )¢xn =
n P
f (xi¤ )¢xi
i=1
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and this is called a Riemann sum.
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IB HL OPT 2ed Calculus
Y:\HAESE\IB_HL_OPT-Calculus\IB_HL_OPT-Calculus_01\038IB_HL_OPT-Calculus_01.cdr Tuesday, 19 February 2013 3:49:52 PM GR8GREG
CALCULUS
39
LOWER AND UPPER RIEMANN SUMS Suppose f is non-negative and continuous on [a, b], and let partition P = fa = x0 , x1 , ...., xn = bg. If we choose the values xi¤ 2 [xi¡1 , xi ], i = 1, ...., n so that f (xi¤ ) is the minimum value of f (x) on the interval [xi¡1 , xi ], then the Riemann Sum is called a lower Riemann sum and is denoted n P f (xi¤ )¢xi . Ln = i=1
If we choose the values xi¤ 2 [xi¡1 , xi ], i = 1, ...., n so that f(xi¤ ) is the maximum value of f (x) on the interval [xi¡1 , xi ], then the Riemann Sum is called a upper Riemann sum and is denoted n P Un = f (xi¤ )¢xi . i=1
REGULAR PARTITIONS Partition P is regular if the n subintervals have equal length. In this case: b¡a = ¢xi n³ ´ b¡a i ² xi = a + n
² ¢x =
for all i = 1, ...., n
for all i = 0, ...., n ³ ´ n b¡a P f (xi¤ ). ² the Riemann sum is n
i=1
When evaluating Riemann sums, the following identities may be useful: ² ² ²
n P i=1 n P i=1 n P
c = cn for c a constant i=
n(n + 1) 2
i2 =
i=1
n(n + 1)(2n + 1) 2n3 + 3n2 + n = 6 6
CALCULATING EXACT AREAS For the function f(x), the rectangles for the lower sum lie under the curve y = f (x) and the rectangles for the upper sum lie above the curve. The exact area A under the curve and above the x-axis on [a, b] therefore satisfies Ln 6 A 6 Un for all partitions of [a, b] into n subintervals, n 2 Z + . For a regular partition, in the limit as n ! 1, the width of each subinterval ! 0. lim ¢x = lim
n!1
n!1
b¡a = 0. n
At the same time, the approximation for the area A is squeezed between the lower and upper Riemann sums. lim Ln = lim A = lim Un n!1
n!1
n!1
For f a non-negative continuous function on [a, b], a < b, the exact area A under the curve y = f (x) and above the x-axis on [a, b] is the unique value which satisfies Ln 6 A 6 Un for all regular partitions of [a, b].
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lim Ln = A = lim Un .
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In particular,
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IB HL OPT 2ed Calculus
Y:\HAESE\IB_HL_OPT-Calculus\IB_HL_OPT-Calculus_01\039IB_HL_OPT-Calculus_01.cdr Friday, 8 February 2013 9:58:07 AM BRIAN
40
CALCULUS
Example 13 Suppose f (x) = x2 on [0, 1]. a Let P be a regular partition of [0, 1] into n subintervals. Find ¢x, and write xi in terms of i and n. 1 1 1 ¡ + 2. 3 2n 6n 1 1 1 Un = + + 2. 3 2n 6n
b Show that the lower Riemann sum is given by Ln = c Show that the upper Riemann sum is given by
d Hence find the area A under the curve y = x2 and above the x-axis on [0, 1]. a There are n subintervals from x0 = 0 to xn = 1.
y
1¡0 1 = ) ¢x = n n
y = x2
) xi = x0 + i¢x =
i , n
M3
i = 1, ...., n.
m3 1 n
2 n
3 n
x
1
b The minimum value of f(x) on [xi¡1 , xi ] c The maximum value of f (x) on [xi¡1 , xi ] ³ ´ ³ ´2 i¡1 2 i , i = 1, ...., n. , i = 1, ...., n. is mi = f (xi¡1 ) = is Mi = f (xi ) = n
n
) the lower Riemann sum is n n 1 P 1 P (i ¡ 1)2 Ln = mi = 2 n i=1
n i=1
=
n 1 P
n3
) the upper Riemann sum is n n 1 P 1 P i2 Mi = Un = 2
n
n i=1
(i ¡ 1)2
=
i=1 nP ¡1
1 i2 n3 i=0 ³ ´ 1 (n ¡ 1)(n)(2n ¡ 1) = 3 n 6
=
1 = 3 n
=
n i=1 n
µ
2n3 ¡ 3n2 + n 6
n 1 P i2 3 n i=1
1 = 3 n
=
¶
µ
2n3 + 3n2 + n 6
¶
1 1 1 + + 2 3 2n 6n
1 1 1 ¡ + 2 3 2n 6n
d The rectangles for the lower sum lie under the curve y = x2 and the rectangles for the upper sum lie above the curve. ) the exact area A under the curve y = x2 and above the x-axis on [0, 1] satisfies Ln 6 A 6 Un for all regular partitions of [0, 1] into n subintervals, n 2 Z + . 1 1 1 1 1 1 ¡ + 2 6A6 + + 2 for all n 2 Z + . 3 2n 6n 3 2n 6n ³ ´ ³ ´ 1 1 1 1 1 1 1 Now lim ¡ + 2 = = lim + + 2 n!1 n!1 3 2n 6n 3 3 2n 6n
)
lim A = A, since A is a constant.
and
n!1
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) by the Squeeze Theorem, A =
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IB HL OPT 2ed Calculus
Y:\HAESE\IB_HL_OPT-Calculus\IB_HL_OPT-Calculus_01\040IB_HL_OPT-Calculus_01.cdr Tuesday, 19 February 2013 4:48:00 PM GR8GREG
CALCULUS
41
EXERCISE G.1 1 Let y = sin x on [0, ¼]. Let P = f0, ¼4 , ¼2 , 3¼ 4 , ¼g be a regular partition of [0, ¼]. Calculate L4 and U4 exactly, and sketch the corresponding areas on graphs of the function. 2 Suppose f (x) = 2x on [1, 4]. a Let P be a regular partition of [1, 4] into n subintervals. Find ¢x, and write xi in terms of i and n. b Write expressions for Ln and Un . c Hence find the area A under the curve f (x) = 2x and above the x-axis on [1, 4]. 3 Suppose f (x) = x2 on [1, 2]. a Let P be a regular partition of [1, 2] into n subintervals. Find ¢x, and write xi in terms of i and n. b Write expressions for Ln and Un . c Hence show that the area A under the curve f (x) = x2 and above the x-axis on [1, 2] is 7 2 3 units .
DEFINITE INTEGRALS Let f , not necessarily non-negative or continuous, be any function defined on the interval [a, b]. Let P = fa = x0 , x1 , ...., xn = bg be any partition of [a, b]. If the values xi¤ 2 [xi¡1 , xi ], i = 1, ...., n are arbitrarily chosen then R =
n P
f (xi¤ )¢xi
where
i=1
¢xi = xi ¡ xi¡1 , i = 1, ...., n is called the Riemann sum of f on [a, b] for partition P and selection fx1¤ , x2¤ , ...., xn¤ g. b¡a = ¢xi n
If P is a regular partition then ¢x =
In the limit as n ! 1, ¢x ! 0 and
for all i = 1, ...., n, and R =
n (b ¡ a) P f (xi¤ ). n i=1
n (b ¡ a) P f(xi¤ ). n!1 n i=1
lim R = lim
n!1
If this limit exists, we say f is integrable on [a, b]. We denote the value of this limit by Z b n (b ¡ a) P f (x) dx = lim f (xi ) and call this the definite integral of function f from a to b. n!1
a
n
i=1
The function f is the integrand. If a function f is continuous on [a, b], then f is integrable on [a, b].
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The interval [a, b] is called the domain of integration. The values a and b are also called, respectively, the lower and upper limits of the integral.
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IB HL OPT 2ed Calculus
Y:\HAESE\IB_HL_OPT-Calculus\IB_HL_OPT-Calculus_01\041IB_HL_OPT-Calculus_01.cdr Wednesday, 20 February 2013 11:29:01 AM GR8GREG
42
CALCULUS
INTERPRETING DEFINITE INTEGRALS AS AREAS If f is an integrable function which takes negative values on [a, b], we must take care when interpreting definite integrals as areas. We observe that: ² If f(x) > 0 on [a, b], a < b, then Z b f (x) dx = A = the area under the curve.
y
y = f(x)
a
A
² If f(x) 6 0 on [a, b] with a < b, then Z b f (x) dx = ¡A = ¡(the area between the curve a and the x-axis) In this case the values f (xi¤ ) in the corresponding Riemann sum will be negative.
Z
a
b
a
b
y
y = f(x)
Z
b
a
function on [a, b] with a < b.
x
A
f (x) dx = ¡
If we reverse the direction of integration we obtain
x
a
f (x) dx for f an integrable b
By interpreting definite integrals in terms of areas, the following results can be demonstrated: For f(x), g(x) integrable functions on [a, b]: ² If f takes positive and negative values on [a, b] then the area bounded by y = f(x), the x-axis, and the lines x = a and x = b is given by Z b jf(x)j dx.
y
y = f(x)
a
b
x
a
In this case, Z b f(x) dx = (enclosed area above the x-axis) a
a
y
f(x) dx = 0
Z
c
f (x) dx + a
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g(x) dx
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f (x) dx§
a
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b
25
for all c 2 [a, b] Z b Z (f(x)§g(x)) dx = ² a
f(x) dx c
0
a
y = f(x)
b
5
f (x) dx =
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Z
b
25
Z ²
100
a
5
²
75
Z
¡ (enclosed area below the x-axis)
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IB HL OPT 2ed Calculus
Y:\HAESE\IB_HL_OPT-Calculus\IB_HL_OPT-Calculus_01\042IB_HL_OPT-Calculus_01.cdr Tuesday, 19 February 2013 3:51:50 PM GR8GREG
CALCULUS
EXERCISE G.2
Z
1 Suppose f is a function continuous on R , and that Z 5 Z 6 1 f(x) dx = ¡ 2 , and f (x) dx = ¡2. 2
3
Z
7
f (x) dx = 2,
1
2
43
f (x) dx = 12 ,
3
a Find: Z i
6
Z f (x) dx
Z
5
f (x) dx
ii
1
v
Z
7
Z f (x) dx ¡
3
5
iii
5
Z f (x) dx
7
Z
2
vi
f(x) dx 1
2
3
4
iv
Z f (x) dx +
f (x) dx
4 7
f(x) dx 6
b Sketch a possible graph for y = f(x) on [1, 7]. 8 06x 0 be such that x + h < b.
y
y = f(t)
x x+h
a
Now
F (x + h) ¡ F (x) 1 = h h
=
1 h
µZ
Z
x+h
x
f(t) dt ¡ a x+h
Z
b
t
¶ f (t) dt
a
f (t) dt x
1 = (x + h) ¡ x
Z
x+h
f (t) dt x
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= f (c) for some c 2 [x, x + h], since f is continuous on [a, b].
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IB HL OPT 2ed Calculus
Y:\HAESE\IB_HL_OPT-Calculus\IB_HL_OPT-Calculus_01\045IB_HL_OPT-Calculus_01.cdr Tuesday, 19 February 2013 3:52:17 PM GR8GREG
46
CALCULUS
)
lim
h!0+
F (x + h) ¡ F (x) = lim f(c) = f(x) h h!0+
since x 6 c 6 x + h and f is continuous on [a, b].
In the same way for h < 0, x + h > a, we find lim
h!0¡
F (x + h) ¡ F (x) = lim f(c) = f(x) h h!0¡
since x + h 6 c 6 x and f is continuous on [a, b].
) F 0 (x) = f(x) as required. Since F is differentiable on ] a, b [ , F is continuous on ] a, b [ . We still need to show that F is continuous on [a, b]. Since f is continuous on [a, b], f is integrable on [a, b] Z a Z b ) F (a) = f (t) dt = 0 and F (b) = f(t) dt a
Now
Z
are both defined.
a
Z
x
f (t) dt lim F (x) = lim x!a+ x!a+ a Z a f (t) dt =
and
lim F (x) = lim
x!b¡
x!b¡
Z
f(t) dt a
b
f (t) dt
=
a
x
a
=0
= F (b).
= F (a) Thus F (x) is continuous on [a, b]. The FTOC (Part 1) can also be written as:
If f is continuous on [a, b] then
d dx
µZ
x
¶ f (t) dt = f(x), for x 2 ] a, b [ .
a
The FTOC (Part 1) guarantees that any function f continuous on [a, b] has the antiderivative Z x f(t) dt on [a, b], even if we cannot write it down. F (x) = a
Z
x
For example, f (x) = sin(x3 ) has an antiderivative F (x) =
sin(t3 ) dt on [ 0, 1 [ even though
0
we cannot write down an antiderivative of sin(x3 ) explicitly as a function.
THE FUNDAMENTAL THEOREM OF CALCULUS (PART 2) If f is any function integrable on [a, b] which has an antiderivative G on [a, b], Z b f (x) dx = [G(x)]ba = G(b) ¡ G(a).
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IB HL OPT 2ed Calculus
Y:\HAESE\IB_HL_OPT-Calculus\IB_HL_OPT-Calculus_01\046IB_HL_OPT-Calculus_01.cdr Friday, 8 February 2013 10:07:56 AM BRIAN
CALCULUS
47
Proof: Case: If f is continuous on [a, b] then f is integrable on [a, b]. Z x By FTOC (Part 1) the function F (x) = f(t) dt is an antiderivative of f on [a, b]. a
By the second corollary to the MVT, F (x) = G(x) + C ) F (a) = G(a) + C
Z
where C is a constant.
a
f (t) dt = G(a) + C
) a
) 0 = G(a) + C ) C = ¡G(a).
Hence F (x) = G(x) ¡ G(a) ) F (b) = G(b) ¡ G(a) Z b f(t) dt = G(b) ¡ G(a) as required. ) a
Z
Case: Suppose f is not continuous on [a, b]. By the premise of the theorem,
b
f(x) dx a
exists, and G is a function continuous on [a, b] such that G0 (x) = f (x) for x 2 ] a, b [ . Let P = fa = x0 , x1 , ...., xn = bg be a regular partition of [a, b] with n 2 Z + . Consider the ith subinterval [xi¡1 , xi ], i = 1, 2, ...., n. By the Mean Value Theorem,
G(xi ) ¡ G(xi¡1 ) = G0 (xi¤ ) xi ¡ xi¡1
Thus G(xi ) ¡ G(xi¡1 ) = G0 (xi¤ )¢x, where ¢x =
for some xi¤ 2 [xi¡1 , xi ].
b¡a = xi ¡ xi¡1 . n
Summing over all subintervals in the partition we obtain n P
G0 (xi¤ )¢x =
i=1
n P
(G(xi ) ¡ G(xi¡1 ))
i=1
= [G(x1 ) ¡ G(x0 )] + [G(x2 ) ¡ G(x1 )] + :::: + [G(xn ) ¡ G(xn¡1 )] = G(xn ) ¡ G(x0 ) )
n P
f(xi¤ )¢x = G(b) ¡ G(a), since G0 (xi¤ ) = f(xi¤ ), xn = b, and x0 = a.
i=1
)
·
lim
n!1
n P
¸ f(xi¤ )¢x = G(b) ¡ G(a) which is a constant.
i=1
Z
b
)
f (x) dx = G(b) ¡ G(a) as required. a
If a function is integrable and has an explicitly defined antiderivative, then the FTOC (Part 2) allows us to calculate definite integrals easily. · ¸1 Z 1 3 3 x3 x3 and therefore x2 dx = = 13 ¡ 03 = 13 . For example, if f (x) = x2 then G(x) =
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IB HL OPT 2ed Calculus
Y:\HAESE\IB_HL_OPT-Calculus\IB_HL_OPT-Calculus_01\047IB_HL_OPT-Calculus_01.cdr Friday, 8 February 2013 10:09:20 AM BRIAN
48
CALCULUS
DISCUSSION Does the existence of an antiderivative guarantee that a function defined on [a, b] is integrable on [a, b]?
Example 14 Find
F 0 (x)
Z
x
a F (x) =
if:
Z
3
cos t dt
x
¶ cos t dt
µZ
¶ et dt
5
x2
à Z d = ¡
3
1
dx
3
= cos x
d = dx
This is valid since f(t) = cos3 t is a continuous function.
d = du
!
x2
et dt
5
à Z ¡ µ Z ¡
u
¶ du e dt £ t
dx
5
= ¡2xex
Z
x2
if F (x) = x
d = du
c
ÃZ
u
µZ
c u
1 dt t2 + 3
d + dx
1 1 £ 2x ¡ 2 u2 + 3 x +3 2x 1 ¡ 2 = 4 x +3 x +3
95
50
c
¶
1 dt t2 + 3
fletting u(x) = x2 g fChain ruleg
fvalid since f (t) =
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x
where c is a constant
x +3
dx
=
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µ Z ¡
¶ 1 1 dt ¡ 2 t2 + 3 x +3 ¶ 1 du 1 dt £ ¡ 2 2 t +3
c
1 dt t2 + 3
!
x2
c
µZ
!
c
yellow
50
d dx
1 dt + t2 + 3 x
75
=
Z
x2
25
d = dx
1 dt t2 + 3
x
ÃZ
fvalid since f(t) = et is continuousg
2
!
x2
0
d = dx
ÃZ
5
d F 0 (x) = dx
fChain ruleg
1 dt. t2 + 3
100
Find
F 0 (x)
where u(x) = x2
et dt
5
= ¡eu £ 2x
Example 15
!
u(x)
95
d = dx
d dx
b F 0 (x) =
1 is continuousg t2 + 3
100
F 0 (x) µZ
et dt x2
1
a
5
b F (x) =
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IB HL OPT 2ed Calculus
Y:\HAESE\IB_HL_OPT-Calculus\IB_HL_OPT-Calculus_01\048IB_HL_OPT-Calculus_01.cdr Friday, 8 February 2013 10:09:49 AM BRIAN
49
CALCULUS
EXERCISE H 1 Find F 0 (x) if: Z x a F (x) = sin4 t dt
Z
1
Z
dt (t2 + 2)15
0
4
c F (x) =
x
b F (x) =
t2
e
Z
x2
esin t dt
d F (x) =
dt
x
¡1
Z
2
1 dx = 2 x ¡1
2 Explain what is wrong with the argument:
h
i2
¡1 x ¡1
fFTOCg
= ¡ 12 ¡ (1) 3
d dx
a Can
µZ
x
= ¡ 32
¶ 1 dt for x 2 R
be calculated using the Fundamental Theorem of Calculus?
¡1 t
Explain your answer. b For which values of x can the derivative in a be calculated? 4 Calculate the following derivatives: ÃZ ! 2x d t a dt b p dx
1 + t4
1
Z
10
5 Find ¡3
Z
5
6 Find 0
d dx
ÃZ
!
x2
sin(e ) dt t
d dx
c
x
µZ
sin x
x3
¶
1 dt 1 + t2
j x j dx.
8 sin x, > < 2x, f(x) dx for f (x) = > : 1, x
0 6 x 6 ¼2 1 and diverges if p 6 1.
From Example 16 we conclude the important result:
Z
1
1 dx xp
1
converges if p > 1 and diverges if p 6 1.
THE COMPARISON TEST FOR IMPROPER INTEGRALS Suppose 0 6 f (x) 6 g(x) for all x > a. Z 1 Z g(x) dx is convergent, then so is ² If a
Z
1
² If
Z f (x) dx is divergent, then so is
a
Z
1
Determine whether 2
Now we know that 0 6
1 dx p x¡1
1
2
1 x¡1
1
Z
2
1 1 p dx ¡ p dx x x 1
Z
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1 dx p x¡1
50
1
75
Z
1
2
0
95
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1
exists, but from Example 16
is divergent, and so
100
1 p dx x
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Z )
1
for all x > 2.
2
1 p dx x 1
where
Z
5
Z
1 p dx = x
g(x) dx.
p p x ¡ 1 6 x for all x > 2 1 x
Z
1
is convergent or divergent.
so it follows that 0 6 p 6 p Now
f(x) dx.
a
a
100
Example 17
1
black
1 p dx x
is divergent.
is divergent by the Comparison Test.
IB HL OPT 2ed Calculus
Y:\HAESE\IB_HL_OPT-Calculus\IB_HL_OPT-Calculus_01\051IB_HL_OPT-Calculus_01.cdr Friday, 8 February 2013 10:17:07 AM BRIAN
51
52
CALCULUS
Z
1
If
Z
1
j f (x) j dx converges then
a
f (x) dx converges.
a
Proof: By definition, ¡ j f (x) j 6 f(x) 6 j f(x) j ) 0 6 f(x) + j f(x) j 6 2 j f (x) j Z 1 Z f (x) + j f (x) j dx 6 2 ) 06 a
Z
1
a
Z
1
then
Z
1
f(x) + j f (x) j dx.
a
Z
1
j f (x) j dx = A and
Supposing
Z
j f (x) j dx is convergent then so is
a
1
j f(x) j dx
a
) by the Comparison Test, if
Z
1
f(x) + j f(x) j dx = B
where A, B 2 R
a
f (x) dx = B ¡ A.
a
1
Hence
f (x) dx is convergent.
a
Example 18
Z
1
Using integration by parts and the Comparison Test, prove that 1
Z
1
sin x dx = lim x b!1
1
Z
b
sin x dx x
is convergent.
sin x dx x
1
Z b h i cos x b cos x = lim ¡ ¡ lim dx fintegration by partsg x x2 b!1 b !1 1 1 Z ³ ´ 1 cos b cos x + cos 1 ¡ dx = lim ¡ b x2 b!1 1 Z 1 cos x dx = cos 1 ¡ 2
¯ ¯ 1 ¯ cos x ¯ Now 0 6 ¯ 2 ¯ 6 2 x
x
1
x
for all x > 1,
Z
1
1
dx is convergent. and we also know from Example 16 that x2 1 Z 1¯ Z 1 ¯ cos x ¯ cos x ¯ dx. ) ¯ 2 ¯ dx is also convergent, and hence so is 2
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x
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)
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IB HL OPT 2ed Calculus
Y:\HAESE\IB_HL_OPT-Calculus\IB_HL_OPT-Calculus_01\052IB_HL_OPT-Calculus_01.cdr Friday, 8 February 2013 10:17:46 AM BRIAN
CALCULUS
53
EVALUATING IMPROPER INTEGRALS When an improper integral is convergent, we may be able to evaluate it using a variety of techniques. These include use of the limit laws, l’Hˆopital’s Rule, integration by parts, and integration by substitution. Example 19 Z Evaluate
1
xe¡x dx, for any constant a 2 R .
a
Z
1
Z
¡x
b
dx = lim
xe
b!1
a
xe¡x dx
a
µ ¶ Z b £ ¤b ¡xe¡x a ¡ ¡e¡x dx b!1 a ³ £ ¤ b´ ¡b ¡a = lim ¡be + ae ¡ e¡x a b!1 ¡ ¢ = lim ¡be¡b + ae¡a ¡ e¡b + e¡a b!1 ¡ ¢ = e¡a (a + 1) + lim e¡b (¡1 ¡ b) b!1 µ ¶ ¡1 ¡ b ¡a = e (a + 1) + lim b = lim
b!1
fintegration by partsg
e
Now as b ! 1, ¡1 ¡ b ! ¡1 and eb ! 1. Z 1 ¡1 ) xe¡x dx = e¡a (a + 1) + lim b fl’Hˆopital’s Ruleg a
b!1
= e¡a (a + 1)
e
EXERCISE I.1 1 Use the Comparison Test for improper integrals to test for convergence: Z 1 Z 1 2 x x ¡1 a dx b dx p 5 2 2x + 3x + 1
1
Z
1
2 Determine whether 1
3 Test for convergence: Z 1 2 x +1 a dx 4 1
Z
1
4 Show that
sin x dx x3
Z
1
b
x +1
x7 + 1
2
is convergent.
Z
2
e¡x dx
1
c
0
1
Z
ln x dx x
1
d
e¡x ln x dx
1
e¡x cos x dx is convergent.
0
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using the substitution u = ex .
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a
dx ex + e¡x
25
1
6 Evaluate
0
Z
b
5
a
Z
dx x2 + a2
100
5 Evaluate: Z 1 a
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IB HL OPT 2ed Calculus
Y:\HAESE\IB_HL_OPT-Calculus\IB_HL_OPT-Calculus_01\053IB_HL_OPT-Calculus_01.cdr Wednesday, 20 February 2013 11:30:12 AM GR8GREG
54
CALCULUS
Z
1
µ
7 Evaluate 1
¶
1 1 p ¡p x x+3
dx. 1
8 Find the area in the first quadrant under the curve y = 2 . x + 6x + 10 Z 1 ln x 9 Prove that dx is divergent for p 6 1. p x
e
10
Z
1
a Evaluate the integral
Z
xn e¡x dx for n = 0, 1, 2, 3.
0
1
xn e¡x dx for n 2 Z + .
b Predict the value of 0
c Prove your prediction using mathematical induction.
APPROXIMATION TO THE IMPROPER INTEGRAL
Z
1
a
f (x) dx, a 2 Z Z
Suppose f is continuous and positive on [a, 1 [ , a 2 Z . We consider the integral
1
f(x) dx in a
terms of area under the curve for x > a. Consider a regular partition P = fa, a + 1, a + 2, ....g of [ a, 1 [ with subintervals of length ¢x = 1. y
f(a)
f(a+1) f(a+2) f(a+3) ...
Z 1 Area = f (x)dx a
. e tc .
y = ¦(x)
a
x
a+1 a+2 a+3
For each interval of length one along the x-axis, we can draw a rectangle of height equal to the value of the function on one side of the rectangle. For example, using the left side of the rectangle, the rectangle from x = a to x = a + 1 would have height f (a), the rectangle from x = a + 1 to x = a + 2 would have height f (a + 1), and so on. y
f(a) f(a+1)
f(a+2) f(a+3) y = ¦(x)
a
x
a+1 a+2 a+3
Z 1 f (x) dx may As with definite integrals approximated by Riemann sums, the improper integral a be approximated by the corresponding sum of areas of these rectangles. Z 1 1 b 1 P P P f (x) dx ¼ f (i) = lim f (i) where f(i) is also called an infinite series.
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IB HL OPT 2ed Calculus
Y:\HAESE\IB_HL_OPT-Calculus\IB_HL_OPT-Calculus_01\054IB_HL_OPT-Calculus_01.cdr Wednesday, 20 February 2013 9:26:20 AM GR8GREG
CALCULUS
55
DECREASING AND INCREASING FUNCTIONS Suppose the function f(x) is decreasing for all x > a. y
f(a)
f(a+1) f(a+2) f(a+3) y = ¦(x)
x
a+1 a+2 a+3
a
By taking the height of each rectangle to be the value of the function at the left endpoint of each subinterval, we obtain an upper sum U where Z 1 1 P f (x) dx 6 f (i) = U . i=a
a
By taking the height of each rectangle to be the value of the function at the right endpoint of each subinterval, we obtain a lower sum L where Z 1 1 P L= f(i + 1) 6 f (x) dx. i=a
a
Hence, for a function which is decreasing on [a, 1[ , Z 1 1 1 P P f(i + 1) 6 f (x) dx 6 f (i) = U . L= i=a
i=a
a
Similarly, for any continuous function which is increasing on [a, 1[ , Z 1 1 1 P P f(i) 6 f (x) dx 6 f (i + 1) = U . L= i=a
i=a
a
Example 20
Z
1
Write down a series which approximates
2
e¡x dx.
0
Z
1
2
e¡x dx ¼
0
1 P
2
e¡i
i=0
EXERCISE I.2 1 Write down a series which approximates: Z 1 1 a dx p
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e¡x dx
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IB HL OPT 2ed Calculus
Y:\HAESE\IB_HL_OPT-Calculus\IB_HL_OPT-Calculus_01\055IB_HL_OPT-Calculus_01.cdr Wednesday, 20 February 2013 9:29:36 AM GR8GREG
56
CALCULUS 2
2 Consider the function f(x) = e¡x . a Show that f (x) is decreasing for all x > 0.
Z
1
b Write upper and lower sums that approximate
f(x) dx. 0
c Write an inequality that relates the sums in b to the integral. 3 Consider the function f(x) = ¡
1 . x2
a Show that f (x) is increasing for all x > 0.
Z
1
b Write upper and lower sums that approximate
¡
1
1 dx. x2
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c Write an inequality that relates the sums in b to the integral.
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IB HL OPT 2ed Calculus
Y:\HAESE\IB_HL_OPT-Calculus\IB_HL_OPT-Calculus_01\056IB_HL_OPT-Calculus_01.cdr Friday, 8 February 2013 10:24:13 AM BRIAN
CALCULUS
J
57
SEQUENCES A sequence is a list of numbers, called terms, in a definite order.
We write the sequence a1 , a2 , a3 , .... as fan g, where an is the nth term of the sequence. The sequence is infinite if it contains an infinite number of terms, so n 2 Z + . The sequence is finite if it contains a finite number of terms, so n 2 Z + , n 6 N . For example, the sequence fan g, where an = f 12 ,
2 3,
3 4,
4 5,
5 6,
....g in this order.
n , n 2 Z + , denotes the infinite set of discrete values n+1
an 1
We can plot an against n to give:
Qw
n
Consider the function f (x) =
x , x+1
x 2 R , x 6= ¡1.
1
x £ x1 lim f (x) = lim x!1 x!1 x + 1 x
= lim
x!1
=
1 1 1+ x
1 1+0
fsince
1 = 0g x!1 x
n =1 n+1
for n 2 Z + .
lim
=1 It follows that
lim
n!1
By considering n sufficiently large, the terms of the sequence will be as close as we like to value 1. We say that L = 1 is the limit of the sequence fan g, where an =
n , n 2 Z +. n+1
This is true even though the terms never actually reach the limit value 1. A sequence fan g has a limit L if for each " > 0 there exists a positive integer N such that jan ¡ Lj < " for all terms an with n > N. We write
lim an = L.
n!1
If the limit of a sequence exists, then we say the sequence converges. Otherwise, the sequence diverges.
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If a sequence converges, then its limit is unique.
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IB HL OPT 2ed Calculus
Y:\HAESE\IB_HL_OPT-Calculus\IB_HL_OPT-Calculus_01\057IB_HL_OPT-Calculus_01.cdr Wednesday, 20 February 2013 10:25:30 AM GR8GREG
58
CALCULUS
Proof: Suppose fan g is a convergent sequence with two limits L1 , L2 where L1 6= L2 . If " = jL1 ¡ L2 j, then " > 0 since L1 6= L2 . Since Since
" 2
lim an = L1 , there exists N1 2 Z + such that n > N1 ensures jan ¡ L1 j < .
n!1
" 2
lim an = L2 , there exists N2 2 Z + such that n > N2 ensures jan ¡ L2 j < .
n!1
Suppose n = max(N1 , N2 ) is the maximum of N1 and N2 . ) jL1 ¡ L2 j = jL1 ¡ an + an ¡ L2 j 6 jL1 ¡ an j + jan ¡ L2 j
fTriangle Inequalityg
6 jan ¡ L1 j + jan ¡ L2 j
0 and for all n > 0 we have jan ¡ cj < ". ) by the definition of a limit of a sequence fan g, lim c = c. n!1
1 = 0. n!1 n
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lim
Result 2:
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IB HL OPT 2ed Calculus
Y:\HAESE\IB_HL_OPT-Calculus\IB_HL_OPT-Calculus_01\058IB_HL_OPT-Calculus_01.cdr Friday, 15 March 2013 3:09:15 PM BRIAN
CALCULUS
1 for n 2 Z + . n
Let an =
Proof:
59
¯ ¯ ¯1 ¯ 1 ) jan ¡ 0j = ¯ ¡ 0¯ = . n
n
For any " > 0 there exists N 2 Z + such that N " > 1
fArchimedean propertyg
1 < ". N 1 1 < . n N
) For n > N ,
) for any " > 0, there exists N 2 Z + such that for n > N, jan ¡ 0j =
1 1 < < ". n N
) by the definition of a limit of a sequence,
Result 3:
If p > 0, then
Proof:
Let an =
lim
n!1
1 np
lim an = lim
n!1
n!1
1 = 0. n
1 = 0. np
for n 2 Z + . 1
Suppose " > 0 is given. Then " p > 0 and by the Archimedean property there exists 1
N 2 Z + such that N " p > 1. )
1 1 < "p . N
) since y = xp , p > 0, is an increasing function,
1 1 p p < (" ) = ". Np
) for any " > 0, there exists N 2 Z + such that for n > N, ¯ ¯ 1 1 ¯1 ¯ jan ¡ 0j = ¯ p ¡ 0¯ = p < p < ". n
n
N
) by the definition of a limit of a sequence, lim an = lim
n!1
n!1
1 =0 np
for all p > 0.
Consider the sequence fcn g, c 2 R .
Result 4:
If 0 6 jcj < 1, the sequence fcn g converges to 0, so
lim cn = 0.
n!1
If jcj > 1, the sequence fcn g diverges. As n ! 1, jcjn ! 1. If c = 0 then
Proof:
lim cn = 0.
fResult 1g
n!1
Let an = cn for n 2 Z + , where c is a constant such that 0 < jcj < 1. If we let jcj =
1 , 1+d
then d =
1 ¡ 1 > 0. jcj
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By the Bernoulli Inequality (see Exercise A question 9), since d > 0, (1 + d)n > 1 + nd > 0 for all n 2 Z + .
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IB HL OPT 2ed Calculus
Y:\HAESE\IB_HL_OPT-Calculus\IB_HL_OPT-Calculus_01\059IB_HL_OPT-Calculus_01.cdr Friday, 8 February 2013 10:27:40 AM BRIAN
60
CALCULUS
) jcjn =
1 1 1 6 < (1 + d)n 1 + nd nd
for all n 2 Z + .
Given " > 0, "d > 0, so by the Archimedean property there exists N 2 Z + such that N"d > 1. )
1 < ". Nd
) jan ¡ 0j = jcn ¡ 0j = jcn j = jcjn
N. Hence
lim bn = L.
n!1
The Squeeze Theorem still holds even if the condition an 6 bn 6 cn only applies for every natural number from some point n > k. The finite number of sequence terms from n = 1 to n = k does not affect the ultimate convergence (or divergence) of the sequence.
BOUNDED SEQUENCES A sequence of real numbers fan g is said to be bounded if there exists a real number M > 0 such that jan j 6 M for all n 2 Z + . We can deduce that:
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Every convergent sequence is bounded.
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IB HL OPT 2ed Calculus
Y:\HAESE\IB_HL_OPT-Calculus\IB_HL_OPT-Calculus_01\060IB_HL_OPT-Calculus_01.cdr Tuesday, 19 February 2013 3:56:41 PM GR8GREG
CALCULUS
Let fan g be a sequence where
Proof:
61
lim an = a.
n!1
If we let " = 1, then by the definition of convergence there exists a natural number N such that jan ¡ aj < 1 for all n > N. But from Corollary 3 of the Triangle Inequality, jan j ¡ jaj 6 jan ¡ aj < 1 for all n > N . ) jan j 6 1 + jaj for all n > N. If we define M as the maximum value in the set f1+jaj, ja1 j, ...., jaN ¡1 jg then jan j 6 M for all n 2 Z + . ) the sequence fan g is bounded.
COMPARISON TEST FOR SEQUENCES Consider sequences fan g, fbn g such that 0 6 an 6 bn . If fan g diverges, then fbn g diverges. If fbn g converges, then fan g converges.
ALGEBRA OF LIMITS THEOREMS Suppose fan g converges to a real number a and fbn g converges to a real number b. 1
lim (an + bn ) = lim an + lim bn = a + b
n!1
n!1
n!1
2 The sequence fan bn g converges and
µ 3 If b 6= 0 then
lim
n!1
an bn
¶ =
lim (an bn ) =
³
n!1
lim an
n!1
lim bn
=
n!1
lim an
n!1
´³
´ lim bn = ab.
n!1
a . b
These results can be extended to finite sums and products of limits using mathematical induction. Proof of 2: For n 2 Z + , we have an bn ¡ ab = an bn ¡ an b + an b ¡ ab = an (bn ¡ b) + b(an ¡ a). By the Triangle Inequality, jan bn ¡ abj 6 jan (bn ¡ b)j + jb(an ¡ a)j = jan j jbn ¡ bj + jbj jan ¡ aj Since fan g and fbn g are convergent sequences, they are bounded and there exists M1 , M2 > 0 such that jan j 6 M1 and jbn j 6 M2 for all n 2 Z + . If we let M be the greater of M1 and M2 , then jan bn ¡ abj 6 M jbn ¡ bj + M jan ¡ aj for all n 2 Z + . lim an = a and
Since
n!1
such that jan ¡ aj
N1 , and jbn ¡ bj
N2 .
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IB HL OPT 2ed Calculus
Y:\HAESE\IB_HL_OPT-Calculus\IB_HL_OPT-Calculus_01\061IB_HL_OPT-Calculus_01.cdr Friday, 8 February 2013 10:29:41 AM BRIAN
62
CALCULUS
We have applied the formal definition of the limit of a sequence to rigorously establish some key results for sequences that can now be used to deal very efficiently with more general problems. Example 21
³ ´n 4 5
Suppose an =
+
3 ¡9 n
for all n 2 Z + . Find
lim an .
n!1
By the generalised version of 1 of the Algebra of Limits Theorems, h³ ´n i ³ ´n 4 3 4 3 lim + ¡ 9 = lim + lim + lim (¡9) 5
n!1
5
n!1
n
n!1
n
n!1
provided each of these limits exist. ¡ ¢n Now lim 45 = 0 fsince 0 < 45 < 1g n!1 ³ ´ 3 1 = lim 3 £ lim =3£0 =0 lim n!1
n!1
n
n!1
n
lim (¡9) = ¡9
n!1
)
h³ ´n 4 5
lim
n!1
+
i
3 ¡ 9 = 0 + 0 ¡ 9 = ¡9 n
Example 22 Let an =
2n2 + 4n ¡ 3 n2 ¡ 4 ln n
for all n 2 Z + . Find
lim an .
n!1
4
3
2+ ¡ 2 2n2 + 4n ¡ 3 n n Dividing the numerator and denominator by n , = 4 ln n n2 ¡ 4 ln n 1¡ 2 n ³ ´ 4 3 lim 2 + ¡ 2 n!1 n n ³ ´ ) lim an = 4 ln n n!1 lim 1 ¡ 2 2
n!1
)
lim
n!1
1 =0 n2
lim
Using the limit laws,
n
n!1
³ ´ ³ ´ ³ ´ 4 3 1 1 2 + ¡ 2 = lim (2) + 4 lim ¡ 3 lim 2 n
n!1
n
n!1
n!1
n
n
GRAPHICS CALCUL ATOR INSTRUCTIONS
You can use your calculator to check your answer.
=2+0+0 =2 Now 0 < ln n < n for all n > 1 ln n 1 < n2 n 4 ln n 4 0< 2 < n n
0
an
or an+1 6 an
for all n.
To show that a sequence is monotone we show that either an+1 ¡ an > 0 or that an+1 ¡ an 6 0 for all n 2 Z + . Alternatively, suppose an can be represented by a differentiable function f (x), x 2 R , x > 1 such that an = f(n) for all n 2 Z + . If f 0 (x) > 0 for all x > 1, or if f 0 (x) 6 0 for all x > 1, then fan g is monotone. THE MONOTONE CONVERGENCE THEOREM A monotone sequence of real numbers is convergent if and only if it is bounded. For example: ² The sequence fan g where an = n, n 2 Z + , is a monotone increasing sequence which is unbounded. Clearly fan g diverges. 1
² The sequence fan g where an = 1 ¡ , n 2 Z + , is a monotone increasing sequence which is n ¯ ¯ 1¯ ¯ bounded since jan j = ¯1 ¡ ¯ < 1 for all n 2 Z + . n
We have shown previously that fan g is convergent and
lim an = 1.
n!1
EXERCISE J.2 a Prove that the sequence fun g with nth term un =
1
2n ¡ 7 3n + 2
is:
i monotone increasing ii bounded. b Determine whether the following sequences are monotone and bounded, and calculate their limits if they exist: n n o n o 3 1 i ii n n ¡n
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IB HL OPT 2ed Calculus
Y:\HAESE\IB_HL_OPT-Calculus\IB_HL_OPT-Calculus_01\064IB_HL_OPT-Calculus_01.cdr Tuesday, 19 February 2013 3:57:38 PM GR8GREG
65
CALCULUS
½ 2 Prove that the sequence
¾
1 £ 3 £ 5 £ :::: £ (2n ¡ 1) 2n n!
3 The sequence fxn g is defined by x1 = 0, xn =
is convergent.
p
4 + 3xn¡1 .
a Use mathematical induction to show that fxn g is monotone increasing. b Evaluate x1 , x2 , x3 , ...., x11 , and hence suggest an upper bound for fxn g. c Use mathematical induction to prove that fxn g is bounded. d Hence find
lim xn .
n!1
a Find the values of 1 + 11 , 1 +
4
1 1 , 1+ 1 1 + 11 1+
, 1+
1 + 11
1 1
1+
1+
1 1+
1 1
b Give a recursive definition for the sequence above in terms of un . c Show that fun g is bounded but not monotone. d Given that fun g converges, find the exact value of
lim un .
n!1
a Consider the sequence fun g defined by u1 = a (a positive constant) and ³ ´ 2 for all n 2 Z + . un+1 = 12 un +
5
un
Use your calculator to find u1 , u2 , u3 , ...., u8 when: i a=5 ii a = 1
iii a of your choice.
b Is fun g monotone for all a > 0? c Do your experimental results in a suggest the existence of a limit L such that If so, find L. d If u1 = a where a > 0 and un+1 =
2!
3!
n
n
1
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ii Show that en < 1 + 1 + + + :::: + < 1 + 1 + + 2 + :::: + n¡1 for all 2! 3! n! 2 2 2 + n2Z . iii Hence show that fen g is convergent. ³ ´ ³ ´ 1 n 1 n i Given that lim 1 + = e ¼ 2:718, show that lim 1 ¡ = e¡1 . n!1 n!1 n n ³ ´ n! ii Hence show that lim = 0 using the Squeeze Theorem. n
5
d
lim un
n!1
for all n 2 Z + .
i Show that 2 6 en < en+1
c
n
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50
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un
75
f
³ ´ 3 un + , find
given that it exists. ³ ´ k State lim un given that fun g is defined by u1 = 6 and un+1 = 12 un + . n!1 un p Suggest a recurrence relationship for generating values of 3 k. ³ ´ 1 n Expand 1+ , n 2 Z + , using the Binomial Theorem. n ³ ´ 1 n Define fen g by en = 1 + and show that en equals: n ³ ´ ³ ´³ ´ h³ ´³ ´ ³ ´i 1 1 1 1 2 1 1 2 n¡1 1¡ + 1¡ 1¡ + :::: + 1¡ 1¡ :::: 1 ¡ 1+1+
25
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lim un = L.
n!1
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IB HL OPT 2ed Calculus
Y:\HAESE\IB_HL_OPT-Calculus\IB_HL_OPT-Calculus_01\065IB_HL_OPT-Calculus_01.cdr Tuesday, 19 February 2013 3:58:01 PM GR8GREG
66
CALCULUS
K
INFINITE SERIES
Let fu1 , u2 , u3 , ....g be an infinite sequence. We can form a new sequence fSn g = fS1 , S2 , S3 , ....g by letting S1 = u1 S2 = u1 + u2 .. .
n P
Sn = u1 + u2 + :::: + un =
ui .
i=1
The value Sn , which is the sum of the first n terms of fun g, is called the nth partial sum. Each term of fSn g is a finite series. If
1 P
lim Sn =
n!1
un
exists and equals some finite value S, then the infinite series is convergent.
n=1
Otherwise it is divergent. Example 24 Let fun g be defined by un = rn¡1 where r 2 R , r 6= 0, n 2 Z + . Find an expression for Sn , the nth partial sum of fun g, which does not involve a summation. Sn =
n P
n P
ui =
i=1
ri¡1 = 1 + r + r2 + :::: + rn¡1
i=1
rSn = r + r2 + r3 + :::: + rn
) )
rSn ¡ Sn = rn ¡ 1 )
rn ¡ 1 r¡1
Sn =
It is often important to know when
lim Sn =
n!1
1 P
un
exists, and if so, what its value is. In general
n=1
it is not possible to write Sn as an explicit expression as we did in Example 24. However, we shall see that more difficult functions can often be expressed as simpler infinite series. Great mathematicians such as Euler and Newton did much of their foundation work using infinite series representations of functions, though it was not until much later that other mathematicians such as Cauchy and Lagrange rigorously established when such representations were valid. Since convergence of a series is in effect convergence of a sequence of partial sums, many of the sequence results apply. For example: an
and
magenta
an §
1 P
bn
are also both convergent.
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1 P
0
can = c
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are convergent series, then
100
1 P
50
²
bn
n=1
5
n=1
1 P
75
1 P
If
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IB HL OPT 2ed Calculus
Y:\HAESE\IB_HL_OPT-Calculus\IB_HL_OPT-Calculus_01\066IB_HL_OPT-Calculus_01.cdr Friday, 8 February 2013 10:36:55 AM BRIAN
CALCULUS
67
However, because the form of the sequence of partial sums is generally too unwieldy to deal with using our earlier methods, we need a special set of tests and conditions for determining when the limits of these partial sums exist. We start with a useful result which can tell us something either about a series
1 P
an or its associated
n=1
sequence of general terms fan g: 1 P
If the series
an
is convergent then
n=1
lim an = 0.
n!1
Let Sn = a1 + a2 + :::: + an ) an = Sn ¡ Sn¡1
Proof:
Now
1 P
an is convergent, so fSn g is convergent (by definition).
n=1
Letting
lim Sn = S,
lim Sn¡1 = S
n!1
n!1
lim an = lim (Sn ¡ Sn¡1 ) = S ¡ S = 0
)
n!1
n!1
1 = 0, n!1 n
lim
We shall show later that even though
1 P 1 n=1
n
diverges extremely slowly.
Therefore, the converse of the above theorem is not true. However, we may establish the following Test for Divergence: If
lim an
n!1
lim an 6= 0, then the series
does not exist, or if
n!1
1 P
an
is divergent.
n=1
Example 25 1 P
n2 2+4 5n n=1
Show that the series
The nth term of the series is an = n2 n!1 5n2 + 4 1
diverges. n2 . 5n2 + 4
lim an = lim
)
n!1
= lim
n!1
= Since
5 + n42
1 5
lim an 6= 0, the series diverges.
n!1
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The Test for Divergence puts no sign restriction on each term of fan g. However, all of the following series tests only apply to series with positive terms.
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IB HL OPT 2ed Calculus
Y:\HAESE\IB_HL_OPT-Calculus\IB_HL_OPT-Calculus_01\067IB_HL_OPT-Calculus_01.cdr Friday, 8 February 2013 10:37:25 AM BRIAN
68
CALCULUS
THE COMPARISON TEST FOR SERIES Let fan g be a positive sequence, so an > 0 for all n. 1 1 P P bn such that an 6 bn , then an is also convergent. If there exists a convergent series n=1
1 P
Conversely, if an > bn and
n=1
1 P
bn diverges, then so does
n=1
an .
n=1
Proof of the first part: Let fAn g and fBn g be the sequences of partial sums associated with an and bn respectively. Since an , bn > 0, fAn g and fBn g are monotonic increasing. Suppose an 6 bn , and that fBn g is convergent with
lim Bn = B.
n!1
) 0 6 An 6 Bn 6 B. ) An is also a bounded monotonic sequence, and therefore converges by the Monotone Convergence Theorem. With a minor adjustment to the proof, the result can be shown to hold if an > 0 for all n. 1 P However, the difficulty in using the Comparison Test is in finding a suitable bn . n=1
An appropriate geometric series often tends to work. Indeed, convergent geometric series are used in the proofs of some of the most general and important convergence tests. GEOMETRIC SERIES 1 P A series ak is a geometric series if there exists a constant r, called the common ratio of the series, k=1
such that ak+1 = rak for all k 2 Z + , and a1 = a is a constant. 1 1 P P In this case ak = a + ar + ar2 + ar3 + :::: = ark¡1 . k=1
k=1
n P
= a(1 + r + r2 + :::: + rn¡1 ). 1 1 P P rk¡1 , or equivalently rk . Since a 2 R is a constant it suffices to examine
The nth partial sum Sn =
ar
k¡1
k=1
k=1
If jrj < 1, then the geometric series
1 P
k=0
rk converges with sum S =
k=0
1 P k=0
rk =
1 . 1¡r
If jrj > 1 then the geometric series diverges. Proof: 1 P
If r = 1 then
1k¡1 = 1 + 1 + 1 + 1 + ...., and the nth partial sum is
k=1
Sn =
n P
1k¡1 = 1 + 1 + :::: + 1 = n. | {z } k=1 n times
1 P
Now lim Sn = lim n DNE, so
cyan
1k diverges.
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IB HL OPT 2ed Calculus
Y:\HAESE\IB_HL_OPT-Calculus\IB_HL_OPT-Calculus_01\068IB_HL_OPT-Calculus_01.cdr Friday, 8 February 2013 10:37:51 AM BRIAN
69
CALCULUS
If r = ¡1 then
½
1 P
k¡1
(¡1)
= 1 ¡ 1 + 1 ¡ 1 :::: and the nth partial sum is Sn =
k=1
1, n odd 0, n even.
DNE as Sn alternates between constants 1 and 0 as n ! 1.
lim Sn
n!1
n P
If jrj 6= 1 then Sn =
rk¡1 = 1 + r + :::: + rn¡1
rn ¡ 1 r¡1 rn ¡ 1 lim Sn = lim n!1 n!1 r ¡ 1 k=1
)
=
ffrom Example 24g
lim rn = 0 for jrj < 1
By Result 4 of the limit theorems for sequences,
n!1
lim rn DNE for jrj > 1.
and Hence for jrj < 1 For jrj > 1,
lim Sn =
n!1
lim Sn
1 , 1¡r
1 P
and so
rk =
k=0
1 P
DNE and so
n!1
n!1
rk
1 1¡r
is convergent.
diverges.
k=0
Example 26 Test the series
1 P
1 n+1 2 n=1
for convergence.
Now 2n is positive for all n, and 2n + 1 > 2n . ¡ ¢n 1 1 < n = 12 for all n 2 Z + . ) 0< n 2 +1
But
2
1 ¡ ¢n P 1 n=1
is a convergent geometric series and therefore,
2
1 P
1 n+1 2 n=1
by the Comparison Test,
also converges.
We cannot use the Comparison Test in the same way as in Example 26 to test the series
1 P n=1
2n
1 ¡1
for convergence. However, the following test may be useful when the Comparison Test cannot be applied directly.
THE LIMIT COMPARISON TEST Suppose that
1 P
an
1 P
and
n=1
cyan
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and
1 P
bn
diverges, then
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50
1 P
an
diverges.
n=1
75
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0
n=1
5
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50
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25
0
an ! 1 as n ! 1 bn
5
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5
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100
50
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5
3 If
an = c, c > 0, then the series either both converge or both diverge. bn 1 1 P P a lim n = 0 and bn converges, then an converges. n!1 bn n=1 n=1
lim
95
2 If
are series with positive terms.
n!1
100
1 If
bn
n=1
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IB HL OPT 2ed Calculus
Y:\HAESE\IB_HL_OPT-Calculus\IB_HL_OPT-Calculus_01\069IB_HL_OPT-Calculus_01.cdr Wednesday, 20 February 2013 11:31:59 AM GR8GREG
70
CALCULUS
Proof of 1: c 2
Since c > 0, 0 < " = . Since
an = c, bn
lim
n!1
using the definition of a limit there exists N 2 Z + such that
¯ ¯ ¯ an ¯ c ¡c¯ < ¯
)
)
for all n > N
bn 2 c an c ¡ < ¡c< 2 bn 2 c a 3c ) < n < 2 b n ³ ´ ³2 ´ c 3c bn < an < bn 2 2 1 P
Now if
bn
for all n > N
n=1
an
bn
1 ³ ´ P c
diverges then so does
n=1
n=1
1 P
Hence by the Comparison Test,
bn .
also converges.
n=1
1 P
2
n=1
1 P
Hence by the Comparison Test, However, if
1 ³ ´ P 3c
converges then so does
an
2
bn .
also diverges.
n=1
Example 27 Test the series
Let an = )
1 P
1 n¡1 2 n=1
1 2n ¡ 1
for convergence or divergence.
and bn =
1 . 2n
an 2n = lim n n!1 bn n!1 2 ¡ 1 1
lim
= lim
n!1
=1 Since
1 P
1 n 2 n=1
1¡
³ ´n 1
n 2 since
lim
¡ 1 ¢n
n!1
2
o =0
is a convergent geometric series,
1 P
1 n¡1 2 n=1
fLimit Comparison Test 1g
converges also.
THE INTEGRAL TEST The Integral Test links the sum of a series to the integral of a positive function. Z 1 1 P We have seen that if a is an integer, f (i) ¼ f (x) dx i=a
f (i) ¼
magenta
f (x) dx
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1
1
i=1
0
In particular, when a = 1,
a
Z
5
1 P
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IB HL OPT 2ed Calculus
Y:\HAESE\IB_HL_OPT-Calculus\IB_HL_OPT-Calculus_01\070IB_HL_OPT-Calculus_01.cdr Tuesday, 19 February 2013 3:58:35 PM GR8GREG
CALCULUS
71
The Integral Test is: Suppose that f is a continuous, positive, decreasing function on [ 1, 1 [ . Let an = f(n), n 2 Z + , so that the terms in fan g are all positive. Z 1 1 P f (x) dx is convergent, then an is convergent. 1 If 1
Z
n=1
1
2 If
f (x) dx is divergent, then
1
1 P
an
is divergent.
n=1
Z
1
Clearly this test is only of practical use if
f (x) dx can be evaluated relatively easily.
1
Proof of 1: If f(x) is a continuous, positive, decreasing function, then we can approximate the integral Z 1 f (x) dx using lower and upper sums: 1
a1 a2
a2
y = f(x)
a3
y = f(x) an-1
an 1
2
3
n
The lower sum
Z
a2 + a3 + :::: + an + :::: 6
1
n=1
1
Hence, 1
Z
1
If
f (x) dx 6
1 P
2
3
n
The upper sum
Z 1 f (x) dx a1 + a2 + :::: + an¡1 + :::: > 1 Z 1 1 P ) f (x) dx 6 an
1
f (x) dx 1 Z 1 1 P an 6 a1 + f (x) dx
)
Z
1
Z
an 6 a1 +
1
1
n=1
f (x) dx
1
n=1
f (x) dx converges then the sequence of partial sums fSn g of
1
1 P
an
is bounded and
n=1
monotonic increasing, and hence convergent also. From the proof we also gain the useful result: If f(x) is a continuous, positive, decreasing function on [1, 1 [ , then Z 1 Z 1 1 P f(x) dx 6 an 6 a1 + f (x) dx.
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IB HL OPT 2ed Calculus
Y:\HAESE\IB_HL_OPT-Calculus\IB_HL_OPT-Calculus_01\071IB_HL_OPT-Calculus_01.cdr Wednesday, 20 February 2013 11:35:41 AM GR8GREG
72
CALCULUS
Example 28 1 P
1 2+1 n n=1
Test
f (x) =
1 x2 + 1
for convergence.
is continuous, positive, and decreasing for x > 1.
) the conditions to use the Integral Test are satisfied. Z 1 Z b 1 1 dx = lim dx Now 2 2 x +1
1
b!1
GRAPHICS CALCUL ATOR INSTRUCTIONS
x +1
1
= lim [arctan x]b1 b!1 ¡ ¢ = lim arctan(b) ¡ ¼4 b!1 ¼ ¼ 2 ¡ 4
=
Z )
1
=
You can use your calculator to estimate
Z
¼ 4
1 P
f (x) dx is convergent, and so is
1
n=1
1
f (x) dx
1
1 . n2 + 1
Example 29 1 P 1
For what values of p is the series
n=1
If p < 0 then
lim
n!1
If p > 0 then
1 = lim njpj n!1 np
lim
In each of these cases,
n!1
which diverges, and if p = 0 then
1 6= 0, np
1 = 0. np
lim
n!1
convergent?
np
lim
n!1
1 = 1. np
so by the Test for Divergence, the series diverges.
Since the function f(x) =
1 xp
is continuous, positive, and
decreasing for x > 1, we can apply the Integral Test: Z 1 Z b 1 1 dx = lim dx p p 1
b!1
x
·
1
x
¸b
1 x1¡p for p 6= 1 = lim b!1 1 ¡ p 1 1 1 lim b1¡p ¡ = 1 ¡ p b!1 1¡p
(
=
¡
1 1¡p
if p > 1
if 0 < p < 1 Z b 1 1 dx = lim dx = lim [ln jxj]b1
DNE
cyan
magenta
= lim (ln b) which DNE b!1
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75
converges if p > 1 and diverges if p 6 1.
25
95
1 p n n=1
0
1 P
100
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5
95
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5
) by the Integral Test, the series
b!1
x
1
5
b!1
x
1
50
1
75
Z For p = 1,
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IB HL OPT 2ed Calculus
Y:\HAESE\IB_HL_OPT-Calculus\IB_HL_OPT-Calculus_01\072IB_HL_OPT-Calculus_01.cdr Wednesday, 20 February 2013 11:37:07 AM GR8GREG
73
CALCULUS
p-SERIES
1 P 1
The series
n=1
this form.
is called the p-series, and can be used to rapidly test the convergence of series of
np
As shown in Example 29: 1 P 1
The p-series
converges if p > 1 and diverges if p 6 1.
np
n=1
1 P 1
1 P 1 p = 0:5 n n n=1 n=1
For example, the series
1 P 1
The p-series with p = 1 is
n=1
n
1 2
=1+
is divergent since p = 1 3
+
1 4
+
1 P
Test for convergence or divergence
sin
lim an = lim
)
lim
1 . n sin
n!1
1 = 0. n
n
³ ´ 1 n
1 n
n!1
n!1
³ ´ 1 .
n=1
Let x =
< 1.
+ :::: and is called the harmonic series.
This is an example of a series which is divergent even though Example 30
1 2
= lim
x!0
sin(x) =1 x
by the Fundamental Trigonometric Limit.
1 P
) by the Limit Comparison Test
sin
³ ´ 1 n
n=1
diverge.
1 P 1
Since the harmonic series
n=1 n
n=1
1 P
diverges,
1 P 1
and sin
n
either both converge or both
³ ´
n=1
1 n
also diverges.
EXERCISE K.1 1 Determine whether the following series converge or diverge using an appropriate geometric series, the Comparison Test, or the Test for Divergence. ´ 1 1 1 1 ³ P P P P 1 n2 3n + 2n 1 1 a b c d ¡ 2n n 2 n=1
e
n=1
3(n + 1)(n + 2)
n=1
2
2 Use the Limit Comparison Test with bn = p 3 Determine whether
1 n n n=1
1 P
1 n! n=1
and
n=1
to show that the series
n3
1 P
6
n
n
1 P 2n2 + 3n is convergent. p n=1
5 + n7
are convergent using the Comparison Test.
4 Determine whether the following series converge or diverge using the Comparison Test or Limit Comparison Test. 1 1 1 P P P 1 1 sin2 n a b c p p p 3
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n n=1 1 + 3
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n(n + 1)(n ¡ 1)
n=2
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n(n + 1)(n + 2) p 1 P n d n ¡ 1 n=2 n=1
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n n
1 P
1 ln n n=2
IB HL OPT 2ed Calculus
Y:\HAESE\IB_HL_OPT-Calculus\IB_HL_OPT-Calculus_01\073IB_HL_OPT-Calculus_01.cdr Friday, 8 February 2013 10:56:53 AM BRIAN
74
CALCULUS 1 P
5 Find all the values of x 2 [0, 2¼] for which the series
2n jsinn xj converges.
n=0
1 P
6 Find c if
(1 + c)¡n = 2.
n=2
7 Use the Integral Test to determine whether the following series converge: 1 1 1 P P P n ln n ¡n2 a b ne c 2 n=1 n + 1 ¼ 4
8 Show that
n=1 n
n=1
p + p + :::: + p = p . c Hence prove that the sequence fSn g of partial sums diverges. 1 P 1 d Does converge or diverge? Explain your answer. p n
n=1
12 Consider the series
1 P n2 n=1
3n
. n P i2
a Use induction to prove that the nth partial sum is Sn =
i=1
b Find
3i
=
3 ¡ 3¡n (n2 + 3n + 3) . 2
lim Sn .
n!1
c Does the series series.
1 P n2 n=1
3n
converge? If not, explain why. If so, find the value of the sum of the
APPROXIMATING BY TRUNCATING AN INFINITE SERIES The tests for infinite series can help us determine whether a series is convergent or divergent. However, even if we prove an infinite series is convergent, we do not in general have techniques to 1 P determine the value of the limit S = an . It is only in special cases such as geometric series that n=1
we can do this.
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Instead, we can approximate the sum S of a convergent infinite series using a partial sum Sk .
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IB HL OPT 2ed Calculus
Y:\HAESE\IB_HL_OPT-Calculus\IB_HL_OPT-Calculus_01\074IB_HL_OPT-Calculus_01.cdr Tuesday, 19 February 2013 4:01:58 PM GR8GREG
CALCULUS
Suppose we approximate the sum of a convergent series S = 1 P
an ¼
n=1
k P
1 P
75
an by the sum Sk of its first k terms:
n=1
for some k 2 Z + .
an = a1 + a2 + :::: + ak
n=1
If f is a continuous, positive, decreasing function on [ k, 1 [ , then we can apply the Integral Test. 1 P an and the error satisfies The error in the approximation is Rk = S ¡ Sk = n=k+1 Z 1 Z 1 f(x) dx < Rk < f (x) dx. k+1
k
Example 31 1 P
Suppose we can use the Integral Test to show that
an
is convergent, where an = f (n).
n=1
1 P a Show that the error Rk in approximating an n=1 Z 1 Z 1 satisfies f(x) dx < Rk < f (x) dx. k+1
by a1 +a2 +::::+ak
k
1 P 1
b Hence determine the number of terms necessary to approximate decimal places. a The error Rk = S ¡ Sk =
1 P
k P
an ¡
n=1
n=1
n3
for some k 2 Z +
correct to two
y
an
a1
n=1
= ak+1 + ak+2 + ak+3 + ::::
a2
y = f(x)
a3
From the areas of lower rectangles, we deduce Z 1 f (x) dx. Rk = ak+1 + ak+2 + ak+3 + ::::
f(x) dx
Z
1
Hence
Z f (x) dx < Rk
10 fas k > 0g
n3
to 2 decimal places.
IB HL OPT 2ed Calculus
Y:\HAESE\IB_HL_OPT-Calculus\IB_HL_OPT-Calculus_01\075IB_HL_OPT-Calculus_01.cdr Tuesday, 26 February 2013 12:40:51 PM BRIAN
76
CALCULUS
EXERCISE K.2 1
1 P
1 2 5n n=1
a Estimate the error when
is approximated by its first 12 terms. 1 P 1
b How many terms are necessary to approximate
n=1
1 P
2 The nth partial sum of a series
is Sn =
an
n=1
a Find an .
b Write
1 P
correct to 6 decimal places?
n4
n¡1 . n+1
an
in expanded form and find its sum.
n=1
3
a Use your calculator to evaluate the partial sums S1 , S2 , S3 , S4 , 1 P n S5 , and S6 for . Give your answers in rational form. n=1
(n + 1)!
b Conjecture a formula for Sn . c Use mathematical induction to prove your conjecture. 1 P n d Hence find . n=1
GRAPHICS CALCUL ATOR INSTRUCTIONS
(n + 1)!
1 P 1
4 The harmonic series is defined by
n=1 n
=1+
1 2
+
1 3
+
1 4
+ :::: .
Consider the following sequence of partial sums for the harmonic series: S1 = 1 In Example 29 we saw the S2 = 1 + 12 most efficient proof that the ¡ ¢ Harmonic series is divergent. S4 = 1 + 12 + 13 + 14 ¡1 1¢ This is an alternative proof. 1 2 >1+ 2 + 4 + 4 =1+ 2 ¡ ¢ ¡ ¢ S8 = 1 + 12 + 13 + 14 + 15 + 16 + 17 + 18 ¡ ¢ ¡ ¢ > 1 + 12 + 14 + 14 + 18 + 18 + 18 + 18 = 1 + 32 a Use the same method to find an inequality involving S16 . b Conjecture an inequality involving S2m , m 2 Z + . Use mathematical induction to prove your conjecture. c Show that S2m ! 1 as m ! 1 and hence prove that fSn g is divergent.
ALTERNATING SERIES Thus far, we have dealt with series with only positive terms. An alternating series is one whose terms are alternately positive and negative. For example, 1 ¡
1 2
+
1 4
¡
1 8
+
1 16
¡
1 32
+ ::::
THE ALTERNATING SERIES TEST If the alternating series
1 P
(¡1)n¡1 bn = b1 ¡ b2 + b3 ¡ :::: satisfies 0 6 bn+1 6 bn
n=1
cyan
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lim bn = 0, then the series is convergent.
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IB HL OPT 2ed Calculus
Y:\HAESE\IB_HL_OPT-Calculus\IB_HL_OPT-Calculus_01\076IB_HL_OPT-Calculus_01.cdr Tuesday, 19 February 2013 4:04:04 PM GR8GREG
CALCULUS
77
The theorem also applies if the first term is negative, since we can multiply through by constant ¡1 and proceed as above. Proof: The (2n + 2)th partial sum of the series is S2n+2 = b1 ¡ b2 + :::: ¡ b2n + b2n+1 ¡ b2n+2 , where the bi are all non-negative and non-increasing. We therefore find that S2n+1 = S2n + b2n+1 S2n+2 = S2n + b2n+1 ¡ b2n+2 S2n+3 = S2n+1 ¡ b2n+2 + b2n+3 = S2n+2 + b2n+3 Since b2n+1 > b2n+2 > b2n+3 , we find S2n+1 > S2n+3 > S2n+2 > S2n . Also, S2n+2 = (b1 ¡ b2 ) + (b3 ¡ b4 ) + :::: + (b2n+1 ¡ b2n+2 ). Because the bi are non-increasing, each expression in brackets is > 0. Hence Sn > 0 for any even n, and since S2n+1 > S2n+2 , Sn > 0 for all n. Finally, since S2n+1 6 b1 , we conclude that b1 > S2n+1 > S2n+3 > S2n+2 > S2n > 0. Hence the even partial sums S2n and the odd partial sums S2n+1 are bounded. The S2n are monotonically non-decreasing, while the odd sums S2n+1 are monotonically non-increasing. Thus the even and odd series both converge. Since S2n+1 ¡ S2n = b2n+1 , the sums converge to the same limit if and only if
lim bn = 0.
n!1
The convergence process is illustrated in the following diagram. 0
b1
0
b1
b 1- b 2
0
b 1- b 2 + b3
0 b 1- b 2 + b3 - b4
b1 b1
even partial sums odd partial sums 0
b1
lim Sn
n®1
If 0 6 bn+1 6 bn
for all n 2 Z + but
between the two values
lim S2n
n!1
lim bn 6= 0, then the series will eventually oscillate
n!1
lim S2n+1 .
and
n!1
even partial sums
odd partial sums
0
b1 lim S2 n
lim S2 n + 1
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IB HL OPT 2ed Calculus
Y:\HAESE\IB_HL_OPT-Calculus\IB_HL_OPT-Calculus_01\077IB_HL_OPT-Calculus_01.cdr Tuesday, 19 February 2013 4:07:23 PM GR8GREG
78
CALCULUS
Example 32 Show that 1 ¡ 12 + 13 ¡ 14 + :::: =
1 P (¡1)n¡1
1 P
This is an alternating series of the form where bn =
Compare this with the harmonic series which diverges!
(¡1)n¡1 bn ,
n=1
1 . n
1 1 < , n+1 n
Since
converges.
n
n=1
the series satisfies 0 < bn+1 < bn
for all n 2 Z + . lim bn = lim
Also,
n!1
n!1
1 P (¡1)n¡1
)
n
n=1
1 =0 n
converges by the Alternating Series Test.
Suppose a convergent infinite series converges to a sum S. If the nth partial sum Sn is used to estimate the sum S, the truncation error is defined by Rn = jS ¡ Sn j. THE ALTERNATING SERIES ESTIMATION THEOREM Suppose S =
1 P
(¡1)n¡1 bn
is the sum of a convergent alternating series,
n=1
for all n 2 Z + and
so 0 6 bn+1 6 bn
lim bn = 0.
n!1
The truncation error Rn = jS ¡ Sn j 6 bn+1 . Proof: S ¡ Sn =
1 P
(¡1)k¡1 bk ¡
k=1
n P
(¡1)k¡1 bk
k=1 n+1
= (¡1)n bn+1 + (¡1)
bn+2 + ::::
= (¡1) [(bn+1 ¡ bn+2 ) + (bn+3 ¡ bn+4 ) + ::::] n
Since br > br+1
for all r 2 Z + , bn+r > bn+r+1
for all r 2 Z + .
) (bn+1 ¡ bn+2 ) + (bn+3 ¡ bn+4 ) + :::: > 0 ) Rn = jS ¡ Sn j = (bn+1 ¡ bn+2 ) + (bn+3 ¡ bn+4 ) + :::: = bn+1 ¡ (bn+2 ¡ bn+3 ) ¡ (bn+4 ¡ bn+5 ) ¡ :::: = bn+1 ¡ [(bn+2 ¡ bn+3 ) + (bn+4 ¡ bn+5 ) + ::::]
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since
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IB HL OPT 2ed Calculus
Y:\HAESE\IB_HL_OPT-Calculus\IB_HL_OPT-Calculus_01\078IB_HL_OPT-Calculus_01.cdr Friday, 8 February 2013 11:44:03 AM BRIAN
CALCULUS
79
Example 33 1 P (¡1)n¡1
Find the sum of
correct to 3 decimal places.
n!
n=1
1 P
This is an alternating series of the form
(¡1)n¡1 bn , where bn =
n=1
1 1 0< < , (n + 1)! n!
Now
Also, 0 < ) since
1 1 < n! n 1 lim = 0, n!1 n
so 0 < bn+1 < bn
lim
n!1
1 . n!
for all n 2 Z + .
1 = lim bn = 0 n!1 n!
fSqueeze Theoremg
) the series converges by the Alternating Series Test. Now S = 1 ¡
1 2
+
Notice that b7 =
1 6
¡
1 5040
and S6 = 1 ¡
1 24
+
1 120
¡
1 720
1 2000 = 0:0005 1 1 1 1 2 + 6 ¡ 24 + 120
+
1 5040
¡
1 720
+ ::::
0 for all n, absolute convergence is the same as convergence. 1 P (¡1)n¡1
A series such as
n
n=1
which is convergent but not absolutely convergent, is called
conditionally convergent. So what is important about absolute and conditional convergence? We all know that the addition of scalars is commutative, so a + b = b + a. Furthermore, if we have a N P an , then we can also reorder the terms without affecting the total sum. Infinite series finite sum n=1
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which are absolutely convergent behave like finite series, in that we can reorder the terms of the series without affecting the sum. However, the same is not true for conditionally convergent series!
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IB HL OPT 2ed Calculus
Y:\HAESE\IB_HL_OPT-Calculus\IB_HL_OPT-Calculus_01\079IB_HL_OPT-Calculus_01.cdr Friday, 8 February 2013 11:44:10 AM BRIAN
80
CALCULUS
S =1¡
For example, let
1 2S 1 2S 3 2S 3 2S
) ) Adding (1) and (2) gives )
=
1 2
¡
=0+ =1+ =1+
1 2
1 3
+
¡
1 4
+
1 5
¡
1 6
+
1 7
1 8
¡
+ :::: (1)
1 1 1 4 + 6 ¡ 8 + :::: 1 1 1 1 2 + 0 ¡ 4 + 0 + 6 + 0 ¡ 8 + :::: 0 + 13 ¡ 12 + 15 + 0 + 17 ¡ 14 + :::: 1 1 1 1 1 3 ¡ 2 + 5 + 7 ¡ 4 + ::::
(2)
We thus obtain a rearrangement of the original series, but with a different sum! In fact, Riemann showed that by taking groups of sufficiently large numbers of negative or positive terms, it is possible to rearrange a conditionally convergent series so it adds up to any arbitrary real value. THEOREM OF ABSOLUTE CONVERGENCE 1 P
If a series
an
is absolutely convergent then it is convergent.
n=1
By the definition of absolute value, ¡ jan j 6 an 6 jan j
Proof:
) 1 P
Now if
0 6 an + jan j 6 2 jan j
is absolutely convergent then 2
an
n=1
jan j is convergent.
n=1
1 P
) by the Comparison Test,
1 P
(an + jan j) is convergent.
n=1
1 P
But
1 P
an =
n=1
n=1
1 P
) since
(an + jan j) ¡
1 P
jan j since the series is absolutely convergent.
n=1
1 P
(an + jan j) and
n=1
jan j are both convergent,
n=1
1 P
an
is convergent.
n=1
Example 34 1 P cos n
cos 1 cos 2 + 2 + :::: has terms with different signs, but is not an alternating series. 12 2
¯ ¯ 1 ¯ cos n ¯ However, ¯ 2 ¯ 6 2 n
n
for all n 2 R , and
1 P 1 n=1
) by the Comparison Test,
¯ 1 ¯ P ¯ cos n ¯ ¯ 2 ¯ is convergent. n
n=1
) by the Theorem of Absolute Convergence,
1 P cos n
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n2
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95
=
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1 P cos n
is convergent.
n2
n=1
75
Show that
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IB HL OPT 2ed Calculus
Y:\HAESE\IB_HL_OPT-Calculus\IB_HL_OPT-Calculus_01\080IB_HL_OPT-Calculus_01.cdr Friday, 8 February 2013 11:44:28 AM BRIAN
CALCULUS
81
THE RATIO TEST The Ratio Test can be used to determine whether a general series is absolutely convergent, and hence convergent:
¯ ¯ 1 P ¯a ¯ lim ¯ n+1 ¯ < 1, then an is absolutely convergent. n!1 an n=1 ¯ ¯ 1 P ¯a ¯ 2 If lim ¯ n+1 ¯ > 1, then an is divergent. n!1 an n=1 ¯ ¯ ¯a ¯ 3 If lim ¯ n+1 ¯ = 1, the Ratio Test is inconclusive. 1 If
n!1
an
Proof of 1: Let un = jan j, with an 6= 0 for all n 2 Z + . u
Suppose that lim n+1 = L < 1, so given " > 0 there exists a positive integer N ¯ n!1 ¯un ¯u ¯ such that ¯ n+1 ¡ L¯ < " for all n > N. un
In particular, as L < 1 we can choose r such that L < r < 1 and let " = r ¡ L > 0. ¯ ¯ ¯u ¯ Now ¯ n+1 ¡ L¯ < " )
)
)
un un+1 ¡L 1, we let " = L ¡ 1. n!1 an ¯ ¯ ¯ jan+1 j ¯ + ¯ such that ¯ ¡ L¯¯ < " for all n > N ) there exists N 2 Z If
jan j
jan+1 j ¡L 0. b Show that S4 > S2 . Hence prove that in general, S2n > S2n¡2 and 0 6 S2 6 S4 6 :::: 6 S2n 6 :::: . c Show that S2n = b1 ¡ (b2 ¡ b3 ) ¡ (b4 ¡ b5 ) :::: ¡ (b2n¡2 ¡ b2n ) ¡ b2n and S2n 6 b1 . d Hence prove that S2n is convergent. Let lim S2n = S. n!1
e Show that S2n+1 = S2n + b2n+1 . lim bn = 0 then
f Show that if
lim S2n+1 = S
n!1
n!1
and hence
lim Sn = S.
n!1
7 Determine whether these series are absolutely convergent, conditionally convergent, or divergent: 1 1 1 P P P (¡3)n 2n arctan n a b (¡1)n 2 c (¡1)n 3 n=1
d
n=1
g
n!
1 P
n +1
n=1
´ 1 ³ P 1 ¡ 3n n
1 P (¡1)n¡1 ln n
e
3 + 4n
f
n
n=1
n
n=1
1 P (¡1)n n=2
n ln n
¡p p ¢ (¡1)r¡1 r+1¡ r
r=0
8
1 P xn
a Show that
n=0
b Deduce that
n!
lim
n!1
converges for all x 2 R . xn =0 n!
for all x 2 R .
9 Test these series for convergence or divergence:
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n(n + 1)
d
n=1
³ ´
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IB HL OPT 2ed Calculus
Y:\HAESE\IB_HL_OPT-Calculus\IB_HL_OPT-Calculus_01\083IB_HL_OPT-Calculus_01.cdr Tuesday, 19 February 2013 4:11:00 PM GR8GREG
n 2
n2 + 4n
n! 2 £ 5 £ 8 £ :::: £ (3n + 2) n=0
100
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25
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2n c 8n ¡5 n=1
1 cos P
1 P
75
f
4 n=2 n ¡ 1
50
p
n=1
1 P n3 + 1
1 P
1
0
n!
1 P
5
n=0
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10n
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1 P
84
CALCULUS 1 P 1
10 Test the series
1 P 1
for absolute convergence using the Ratio Test. ¯ ¯ ¯a ¯ Hence explain why the Ratio Test is inconclusive when lim ¯ n+1 ¯ = 1. n=1
and
n2
n=1
n
n!1
an
POWER SERIES 1 P
A power series is a series of the form 1 P
or more generally
cn xn = c0 + c1 x + c2 x2 + ::::
n=0
cn (x ¡ a)n = c0 + c1 (x ¡ a) + c2 (x ¡ a)2 + :::: .
n=0
The convergence of a power series will often, but not always, depend on the value of x. 1 P
For example, consider the power series
where cn = 1 for all n. This is in fact the
cn xn
n=0
geometric series 1 + x + x2 + x3 + x4 + :::: , which converges for all jxj < 1. We use the Ratio Test to determine the convergence of a power series. Example 36 1 P (x ¡ 3)n
For what values of x is
n
n=1
(x ¡ 3)n an = n
If
convergent?
¯ ¯ ¯ ¯ ¯ n ¯ an+1 ¯ ¯¯ (x ¡ 3)n+1 ¯ then ¯ £ ¯=¯ n an n+1 (x ¡ 3) ¯ ¯ ¯ ¯ (x ¡ 3)n ¯ =¯ ¯ n+1 ¯ ¯ ¯ ¯ ¯ (x ¡ 3) ¯ =¯ ¯ ¯ 1+ 1 ¯ n ¯ ¯ ¯a ¯ ) lim ¯ n+1 ¯ = jx ¡ 3j n!1
1 P
By the Ratio Test,
an
is divergent if jx ¡ 3j > 1, but is absolutely convergent and hence
an
n=1
jx ¡ 3j < 1
convergent if
)
¡1 < x ¡ 3 < 1 )
2 R. If the series converges only when x = a, we say R = 0. If the series converges for all x 2 R , we say R is infinite. The interval of convergence I is the set of all points x for which the power series converges. Most of the interval of convergence may be deduced from the radius of convergence. However, we need to consider convergence for the cases jx ¡ aj = R separately. 1 P (x ¡ 3)n
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For instance, we saw in Example 36, that
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IB HL OPT 2ed Calculus
Y:\HAESE\IB_HL_OPT-Calculus\IB_HL_OPT-Calculus_01\085IB_HL_OPT-Calculus_01.cdr Friday, 8 February 2013 12:15:18 PM BRIAN
86
CALCULUS
Example 37 1 P (¡3)n xn
Find the radius and interval of convergence for
. p n+1
n=0
If
(¡3)n xn an = p n+1
¯ p ¯ ¯ ¯ n + 1 ¯¯ ¯ an+1 ¯ ¯¯ (¡3)n+1 xn+1 £ ¯ ¯=¯ p an (¡3)n xn ¯ n+2 r n+1 = 3 jxj
then
r
n+2
= 3 jxj
2 1+ n
¯ ¯ ¯a ¯ lim ¯ n+1 ¯ = 3 jxj
)
n!1
1 P
By the Ratio Test,
1 1+ n
an
1 3
converges if jxj
13 .
) the radius of convergence is R = 13 . To determine the interval of convergence, we must consider what happens when x = § 13 . ³ ´n 1 1 1 (¡3)n ¡ 3 1 P P P 1 an = = If x = ¡ 13 , p p n=0
n+1
n=0
1 P
Letting r = n + 1,
an =
n=0
1 P
If x = 13 ,
an =
n=0
1 (¡3)n P
³ ´n
which is a divergent p-series.
1 3
1 P (¡1)n n=0
1 0:5 r r=1
p n+1
n=0
=
n+1
n=0
1 P
p n+1
So, the interval of convergence of
which converges by the Alternating Series Test. 1 P
is ] ¡ 13 ,
an
n=0
1 3
].
Example 38 a Find the radius of convergence and interval of convergence for lim
b Hence show that a If
n!1
xn n!
1 P xn n=0
n!
, where x 2 R .
= 0 for all x 2 R .
¯ ¯ ¯ ¯ ¯ xn+1 n! ¯¯ ¯ an+1 ¯ ¯ then lim ¯ £ n¯ ¯ = lim ¯ n!1 n!1 (n + 1)! an x
xn an = n!
= lim
n!1
jxj n+1
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= 0 for any constant x 2 R ) by the Ratio Test the series has an infinite radius of convergence, and its interval of convergence is R . 1 P xn xn b Since converges for all x 2 R , lim = 0 for all x 2 R .
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IB HL OPT 2ed Calculus
Y:\HAESE\IB_HL_OPT-Calculus\IB_HL_OPT-Calculus_01\086IB_HL_OPT-Calculus_01.cdr Tuesday, 19 February 2013 4:12:21 PM GR8GREG
CALCULUS
87
DIFFERENTIATION AND INTEGRATION OF POWER SERIES A power series can be differentiated or integrated term by term over an interval contained entirely within its interval of convergence. In the case of differentiation, an open interval is required. If f (x) =
1 P
n
cn x
f 0 (x)
then
=
n=0
1 P
Z
n¡1
ncn x
f (x) dx =
and
n=1
1 P
cn xn+1 . n +1 n=0
Example 39 ¶ Z 0:1 µ 1 P (¡3)n xn p dx. Find 0
n+1
n=0
1 P (¡3)n xn
has interval of convergence ] ¡ 13 ,
p n+1 n=0
From Example 37, the series
1 3
].
) since [0, 0:1] lies entirely within the interval of convergence, ¶ µZ 0:1 ¶ Z 0:1 µ 1 1 P (¡3)n xn P (¡3)n xn p dx = p dx 0
n+1
n=0
0
n=0
·
1 P (¡3)n
=
n=0
p n+1
n+1
¸ 0:1
xn+1 n+1 0
1 P (¡3)n (0:1)n+1
=
n=0
(n + 1)
3 2
EXERCISE K.4 1 Use the geometric series to write a formula for each of the following. In each case find the corresponding radius of convergence and interval of convergence. 1 1 1 P P P a x3n b (2 ¡ x)n c (¡1)n x4n n=0
n=0
n=0
2 Find the interval of convergence of
1 P n=0
1 . x2n
3 Find the radius and interval of convergence for each of the following series: 1 1 1 P P P 3n xn (¡1)n x2n¡1 a n5n xn b c d 2 n=0 (n + 1)
n=1
n=1
4 Find the radius and interval of convergence of
(2n ¡ 1)!
1 P 2 £ 4 £ 6 £ :::: £ (2n)xn n=1
1 P
1 £ 3 £ 5 £ :::: £ (2n ¡ 1)
(¡1)n
n=2
(2x + 3)n n ln n
.
5 A function f is defined by f (x) = 1 + 2x + x2 + 2x3 + x4 + ::::, so f is a power series with c2n¡1 = 2 and c2n = 1 for all n 2 Z + . a Find the interval of convergence for the series.
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b Write an explicit formula for f (x).
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IB HL OPT 2ed Calculus
Y:\HAESE\IB_HL_OPT-Calculus\IB_HL_OPT-Calculus_01\087IB_HL_OPT-Calculus_01.cdr Wednesday, 20 February 2013 11:39:19 AM GR8GREG
88
CALCULUS 1 P
6 Suppose that the radius of convergence of a power series
cn xn
is R.
n=0
1 P
Find the radius of convergence of the power series
cn x2n .
n=0
1 P
7 Suppose the series
1 P
has radius of convergence 2 and the series
cn xn
n=0
dn xn
has radius
n=0
of convergence 3. What can you say about the radius of convergence of the series
1 P
(cn + dn )xn ?
n=0
1 P xn
8 Show that the power series
n=1
1 P nxn¡1
and the series of derivatives
n2 3n
n2 3n
n=1
have the same
radius of convergence, but not the same interval of convergence. µ1 ¶ Z xµ 1 n¶ P xn P t d 9 Find and dt. For what values of x do these series converge? dx
µ1 P
cyan
Z b
dx
x
(¡1)n 2n n=0 x
¡2
¶
2n
¡1:5 µ P 1
¶
Z dx
c 0
1 2
µ
1 P
¶ x2n
dx
n=0
dx defined? Explain your answer.
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0:1
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10 Find: Z a
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IB HL OPT 2ed Calculus
Y:\HAESE\IB_HL_OPT-Calculus\IB_HL_OPT-Calculus_01\088IB_HL_OPT-Calculus_01.cdr Friday, 8 February 2013 12:15:44 PM BRIAN
CALCULUS
L
89
TAYLOR AND MACLAURIN SERIES 1 P
Let
cn (x ¡ a)n
be a power series with radius of convergence
R > 0,
and interval of
n=0
convergence I. 1 P
For each x with jx ¡ aj < R, the series 1 P
therefore define a function f (x) =
cn (x ¡ a)n
converges to a finite value. The series may
n=0
cn (x ¡ a)n
with domain I.
n=0
Functions defined in this way may look awkward. However, as we have seen, convergent power series can be added, differentiated, and integrated just like ordinary polynomials. They are very useful because we can express many different functions as power series expansions.
TAYLOR SERIES EXPANSIONS Suppose f (x) =
1 P
cn (x ¡ a)n
n=0
= c0 + c1 (x ¡ a) + c2 (x ¡ a)2 + :::: where
jx ¡ aj < R.
Since we can differentiate the power series on an open interval contained in I, f 0 (x) = c1 + 2c2 (x ¡ a) + 3c3 (x ¡ a)2 + :::: f 00 (x) = 2c2 + 6c3 (x ¡ a) + :::: .. . f (n) (x) = n! cn + (n + 1)! cn+1 (x ¡ a) + :::: f (a) = c0
Using the above formulae, we find that:
f 0 (a) = c1 f 00 (a) = 2c2 = 2!c2 .. . f (n) (a) = n!cn Hence cn =
f (n) (a) n!
where 0! = 1 and f (0) (x) = f (x).
We can use this information to reconstruct the function f in terms of the values of its derivatives evaluated at x = a: For such a function f, the Taylor series expansion of f (x) about x = a is 1 P f 0 (a) f 00 (a) f 000 (a) f (k) (a) f (x) = f(a) + (x ¡ a) + (x ¡ a)2 + (x ¡ a)3 :::: = (x ¡ a)k . 1!
2!
3!
k=0
k!
When a = 0, this is also called the Maclaurin series expansion of f (x). This is 1 P x2 x3 f (k) (0)xk f (x) = f(0) + xf 0 (0) + f 00 (0) + f 000 (0) + :::: =
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IB HL OPT 2ed Calculus
Y:\HAESE\IB_HL_OPT-Calculus\IB_HL_OPT-Calculus_01\089IB_HL_OPT-Calculus_01.cdr Friday, 8 February 2013 5:20:19 PM BRIAN
90
CALCULUS
We have shown for a function f (x) with domain I, that if: 1 f has a convergent power series representation for x 2 I, and 2 the derivatives of f of all orders exist on I, or I with its endpoints removed if I is closed, then f has a Taylor series representation and this is the only power series representation of f. However, we still need to consider the conditions under which the power series representation of f will converge.
TAYLOR POLYNOMIALS If f (k) (x) exists at x = a for k = 0, 1, ...., n, then the nth degree Taylor polynomial approximation to f (x) about x = a is Tn (x) = f (a) + f 0 (a)(x ¡ a) + :::: + =
n P (x ¡ a)k
k!
k=0
f (n) (a) (x ¡ a)n n!
f (k) (a)
When a = 0, this is also called the nth degree Maclaurin polynomial
n P xk k=0
k!
f (k) (0).
Consider the function f(x) = ex . f (n) (x) = ex
exists for all n 2 Z + and x 2 R , and f (n) (0) = e0 = 1 for all n 2 Z + .
The nth degree Taylor approximation to ex about 0 is Tn (x) = 1 +
x x2 xn + + :::: + . 1! 2! n!
Graphs of f(x) = ex , T1 (x) = 1 + x, T2 (x) = 1 + x +
x2 , 2!
T5 (x) = 1 + x +
x2 x3 x4 x5 + + + 2! 3! 4! 5!
y
T2 (x) = 1 +
and
x x2 + 1! 2!
2 1
f(x) = e
x
are shown alongside:
-4
-3
-2
-1
1
2
x
-1
T5 (x) = 1 + x +
x2 x3 x4 x5 + + + 2! 3! 4! 5!
T1 (x) = 1 + x -2
Note that Tn (x) ¼ f (x) when x is close to 0. As n increases, Tn (x) approximates f (x) = ex closely for an increasingly large subset of I = R . If we denote Rn (x : a) to be the error involved in using Tn (x) to approximate f(x) about x = a on I, then f (x) = Tn (x) + Rn (x : a).
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IB HL OPT 2ed Calculus
Y:\HAESE\IB_HL_OPT-Calculus\IB_HL_OPT-Calculus_01\090IB_HL_OPT-Calculus_01.cdr Tuesday, 19 February 2013 4:13:59 PM GR8GREG
CALCULUS
91
Consider a function f with domain I containing x = a (not at an endpoint of I). Suppose f (n+1) (x) exists on I, or I with its endpoints removed if I is closed. For x 2 I, Taylor’s formula with remainder is f (x) = Tn (x) + Rn (x : a) n P f (k) (a)(x ¡ a)k where Tn (x) = k!
k=0
and Rn (x : a) =
f (n+1) (c)(x ¡ a)n+1 (n + 1)!
where c is between x and a, so either c 2 ] a, x [ or c 2 ] x, a [ , depending on the values of a and x. This formula for Rn (x : a) is called the Lagrange form of the error term. Proof of Taylor’s Formula: Consider an interval [a, b] µ I. We define a new function f 00 (x) f (n) (x) (b ¡ x)2 ¡ :::: ¡ (b ¡ x)n ¡ K(b ¡ x)n+1 2! n!
h(x) = f (b) ¡ f (x) ¡ f 0 (x)(b ¡ x) ¡
where K is the constant found by solving h(a) = 0. Note that h(b) = 0. Since f and its n + 1 derivatives exist on [a, b], h is continuous on [a, b] and differentiable on ] a, b [ . Since h(a) = h(b) = 0 we can apply Rolle’s theorem to h on [a, b]. Thus there exists c 2 ] a, b [ such that h0 (c) = 0. Using the product rule n times we obtain for x 2 I and x > a: · ¸ £ 0 ¤ f (3) (x)(b ¡ x)2 0 0 00 00 h (x) = ¡f (x) + f (x) ¡ f (x)(b ¡ x) + f (x)(b ¡ x) ¡ 2! · ¸ · ¸ (3) 2 (4) 3 (n) f (x)(b ¡ x) f (x)(b ¡ x) f (x)(b ¡ x)n¡1 f (n+1) (x)(b ¡ x)n + ¡ + :::: + ¡ 2!
3!
(n ¡ 1)!
n!
+ (n + 1)K(b ¡ x)
n
) h0 (x) = ¡
f (n+1) (x)(b ¡ x)n + (n + 1)K(b ¡ x)n n!
Since h0 (c) = 0, (n + 1)K(b ¡ c)n = )
K=
f (n+1) (c) (b ¡ c)n n! f (n+1) (c) (n + 1)!
We can repeat the argument with interval [b, a] µ I, for x 2 I and x < a. It follows that Rn (x : a) =
f (n+1) (c)(x ¡ a)n+1 (n + 1)!
as required.
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Before the invention of sophisticated calculators, calculations involving irrational numbers like e were difficult. Approximations using Taylor polynomials and rational values of x were often used.
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IB HL OPT 2ed Calculus
Y:\HAESE\IB_HL_OPT-Calculus\IB_HL_OPT-Calculus_01\091IB_HL_OPT-Calculus_01.cdr Tuesday, 19 February 2013 4:16:29 PM GR8GREG
92
CALCULUS
Example 40 a Find the 3rd degree Taylor polynomial of sin x about ¡ ¢ b Hence estimate sin ¼5 .
¼ 4.
c Use the error term to show your approximation is accurate to at least 3 decimal places. f(x) = sin x
)
f 0 (x) = cos x
)
f 00 (x) = ¡ sin x
)
f 000 (x) = ¡ cos x
)
a
f
¡¼¢
= sin
4
¡ ¢ 0 ¼
¡¼¢ 4
¡¼¢
=
p1
2
p1 4 4 = 2 ¡ ¢ ¡ ¢ f 00 ¼4 = ¡ sin ¼4 = ¡ p12 ¡ ¢ ¡ ¢ f 000 ¼4 = ¡ cos ¼4 = ¡ p12
f
= cos
³ ´³ ´2 ³ ´³ ´3 f 00 ¼4 x ¡ ¼4 f 000 ¼4 x ¡ ¼4 ¡ ¢ ¡ ¢ T3 (x) = f 4 + f 0 ¼4 x ¡ ¼4 + + 2! 3! ³ ´2 ³ ´3 ¼ ¼ x¡ 4 x¡ 4 ¡ ¢ = p12 + p12 x ¡ ¼4 ¡ p12 ¡ p12
¡¼¢
)
2!
b T3
¡¼¢ 5
=
p1
2
+
p1
¡¼ 5
2
¡
¢ ¼ 4
³ ¡
p1
2
¼ 0:5878 ¯ ³ ´ ¯ ¯ sin c ¼ ¡ ¼ 4 ¯ ¯ ¯ ¡ ¢¯ ¯ 5 4 ¯ c ¯R3 ¼5 : ¼4 ¯ = ¯¯ ¯ 4! ¯ ¯ ³ ´4 6
)
¼ ¡ 20
¡ ¼4
³ ¡
2!
where
¼ 5
p1
¼ 5
´3 ¡¼ 4
2
3!
6
) T6 (0:4) ¼ 1:491 824 356 approximates e0:4 with an error less than 0:000 01.
EXERCISE L.1 a Find the 3rd degree Maclaurin polynomial T3 (x) of sin x. ¡ ¢ b Hence estimate sin ¼5 correct to 4 decimal places. ¯ ¡ ¢ ¡ ¢¯ c Show that ¯T3 ¼5 ¡ sin ¼5 ¯ < 0:0065 . Hence show the approximation is accurate to 1 decimal place.
1
2 Find an upper bound for the error in using the approximation sin x ¼ x ¡ ¡0:3 6 x 6 0:3 .
x3 x5 + 3! 5!
on the interval
3 Use the Maclaurin series for sin x to compute sin 3± correct to 5 decimal places. ¡ ¢ 4 Find a Taylor polynomial approximation to cos ¼6 + 0:2 with error < 0:001 . 5 Using the power series expansion of
2 e¡x ,
Z evaluate
1
2
e¡x dx to 3 decimal places.
0
Z
2
1
6 Using the power series expansion of ex , evaluate
2
ex dx to 3 decimal places.
0
7 Estimate the value of e¡1 to 6 decimal places using the Alternating Series Estimation Theorem. 8 Use a Taylor polynomial to estimate each value with accuracy indicated.
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a e3 with error < 0:0001 (You may assume e < 2:72 .) ¡¼ ¢ b sin 2 + 0:1 with error < 0:0001
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IB HL OPT 2ed Calculus
Y:\HAESE\IB_HL_OPT-Calculus\IB_HL_OPT-Calculus_01\093IB_HL_OPT-Calculus_01.cdr Thursday, 14 March 2013 3:35:13 PM BRIAN
94
CALCULUS
CONVERGENCE OF THE TAYLOR SERIES From Example 41, the error term for the Maclaurin polynomial approximation of f (x) = ex is f (n+1) (c)xn+1 ec xn+1 = , (n + 1)! (n + 1)!
Rn (x : 0) = By Taylor’s formula, ex =
n P xk k=0 k!
+
ec xn+1 (n + 1)!
where c is between x and 0.
where c is some value between x and 0.
Consider what happens as n ! 1. 1 P xk
By the Ratio test,
k=0
converges for all x 2 R . (See Example 38.)
k!
1 P xk
But before we can conclude that ex =
k=0
ec jxjn+1 n!1 (n + 1)!
lim jRn (x : 0)j = lim
Now
n!1
jxjn+1 =0 n!1 (n + 1)!
For any x 2 R , )
lim
k!
, we need to find
lim Rn (x : 0).
n!1
where ec is a constant dependent on x.
(See Example 38.)
ec jxjn+1 jxjn+1 = ec lim =0 n!1 (n + 1)! n!1 (n + 1)!
lim jRn (x : 0)j = lim
n!1
)
lim Rn (x : 0) = 0
n!1
We therefore conclude that ex =
1 P xk k=0
for all x 2 R .
k!
TAYLOR’S THEOREM If f (x) has derivatives of all orders on interval I which contains a, then 1 f (x) = Tn (x) + Rn (x : a) for all x 2 I, and 1 P f (n) (a)(x ¡ a)n
2 f (x) =
Rn (x : a) =
lim Rn (x : a) = 0 where
if and only if
n!
n=0
f (n+1) (c)(x ¡ a)n+1 (n + 1)!
n!1
for some constant c between x and a, where c depends on x.
1 Proved above as Taylor’s formula.
Proof:
2 For a general function f and for any n 2 Z + , f(x) = Tn (x) + Rn (x : a) for all x 2 I 1 P f (n) (a)(x ¡ a)n exists if and only if this series converges. Now lim Tn (x) = n!1
n!
n=0
This occurs if and only if
lim Tn (x) = lim (f(x) ¡ Rn (x : a))
n!1
n!1
= f (x) ¡ lim Rn (x : a) exists.
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if and only if
n!
n=0
25
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n!1
1 P f (n) (a)(x ¡ a)n
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IB HL OPT 2ed Calculus
Y:\HAESE\IB_HL_OPT-Calculus\IB_HL_OPT-Calculus_01\094IB_HL_OPT-Calculus_01.cdr Monday, 11 February 2013 9:01:08 AM BRIAN
CALCULUS
95
Taylor’s Theorem tells us that the Taylor series of a function f does not necessarily converge, but when it does converge, it equals f(x) if and only if lim Rn (x : a) = 0. n!1
ADDITION OF SERIES 1 P
Suppose
an xn
1 P
has radius of convergence Ra and
n=0
bn xn
has radius of convergence Rb .
n=0
1 P
The sum of the two series is
an xn +
n=0
1 P
bn xn =
n=0
1 P
(an + bn )xn
which has radius of
n=0
convergence being the minimum of Ra and Rb .
Example 42 a Show that f (x) = cos x is equal to its Maclaurin series expansion, and find the radius of convergence. b Hence find Maclaurin series expansions for the following, including their radii of convergence: i cos(2x) ii sin x f(x) = cos x
)
f (0) = 1
f (x) = ¡ sin x
)
f 0 (0) = 0
f 00 (x) = ¡ cos x
)
f 00 (0) = ¡1
f 000 (x) = sin x
)
f 000 (0) = 0
f (4) (x) = cos x
)
f (4) (0) = 1
a
0
By Taylor’s Theorem, f (x) = cos x = f(0) + =1¡
f 0 (0) f 00 (0) 2 f 000 (0) 3 f (n) (0)xn x+ x + x + :::: + + Rn (x : 0) 1! 2! 3! n!
x2 x4 x6 x2k + ¡ + :::: + (¡1)k + R2k+1 (x : 0), 2! 4! 6! (2k)!
where R2k+1 (x : 0) =
f (2k+1) (c)x2k+1 (2k + 1)!
k 2Z+
for c a constant between 0 and x
¯ (2k+1) ¯ ¯f (c)¯ = jsin(c)j 6 1, ¯ ¯ ¯ f (2k+1) (c)x2k+1 ¯ ¯ lim jRn (x : 0)j = lim ¯¯ ¯ n!1 (2k + 1)! k!1
Since
jxj2k+1 =0 k!1 (2k + 1)!
6 lim
) by the Squeeze Theorem,
lim jRn (x : 0)j = 0 for all x 2 R ,
n!1
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(2n)!
n=0
25
1 P (¡1)n x2n
lim Rn (x : 0) = 0 for all x 2 R
n!1
5
) ) f(x) = cos x =
fsee Example 38g
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IB HL OPT 2ed Calculus
Y:\HAESE\IB_HL_OPT-Calculus\IB_HL_OPT-Calculus_01\095IB_HL_OPT-Calculus_01.cdr Monday, 11 February 2013 9:01:15 AM BRIAN
96
CALCULUS
b
1 P (¡1)n x2n
cos x =
i Since
,
(2n)!
n=0
1 P (¡1)n (2x)2n
cos(2x) =
for all x 2 R .
(2n)!
n=0
The radius of convergence is infinite. ii Now since the Maclaurin series expansion of cos x is defined on R , it is both differentiable and integrable on its interval of convergence R . Since
cos x = 1 ¡
x2 x4 x6 + ¡ + :::: 2! 4! 6!
for all x 2 R
d 2x 4x3 6x5 (cos x) = ¡ sin x = ¡ + ¡ + :::: dx 2! 4! 6!
) Alternatively,
sin x = x ¡
x3 x5 + ¡ :::: 3! 5!
Z
sin x ¡ sin(0) =
for all x 2 R .
x
cos t dt ¶ Z xµ 1 P (¡1)n t2n 0
)
sin x =
0
=
n=0 µZ x 1 P 0
n=0
=
dt
¶
(¡1)n t2n dt (2n)!
·1 ¸ P (¡1)n t2n+1 x n=0
=
(2n)!
(2n + 1)!
0
1 P (¡1)n x2n+1
=x¡
for all x 2 R
(2n + 1)!
n=0
x3 x5 x7 + ¡ + :::: 3! 5! 7!
for all x 2 R
The radius of convergence is infinite.
EXERCISE L.2 a Show that f (x) = sin x is equal to its Maclaurin series expansion, and find the radius of convergence.
1
b Hence find the Maclaurin series expansions for the following, including their radii of convergence: i x sin x ii sin(3x) iii cos x a Show that f (x) = e¡x is equal to its Maclaurin series expansion, and find the radius of convergence.
2
2
b Hence find the Maclaurin series expansion for e¡x , including its radius of convergence. 3 Find the Taylor series expansion about x = 2 for f(x) = ln x and its associated radius of convergence. a Find the Maclaurin series expansion for f(x) = 2x and its associated interval of convergence. 1 P b Hence find an such that an = 7.
4
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IB HL OPT 2ed Calculus
Y:\HAESE\IB_HL_OPT-Calculus\IB_HL_OPT-Calculus_01\096IB_HL_OPT-Calculus_01.cdr Monday, 11 February 2013 9:01:25 AM BRIAN
CALCULUS
97
Example 43 Let f(x) = ln(1 + x). a
i Find the Maclaurin series for f(x) and its associated interval of convergence I. ii Show that f (x) = ln(1 + x) equals its Maclaurin series for 0 < x 6 1 by showing lim Rn (x : 0) = 0 for 0 < x 6 1. n!1
b Complete the proof that ln(1 + x) equals its Maclaurin series for ¡1 < x 6 1 in the following steps: 1
i Use the geometric series to write down the power series representation for , jxj < 1. 1+x Z x 1 ii Use the fact that ln(1 + x) ¡ ln(1 + 0) = dt to find the Maclaurin series for 1 P (¡1)k
c Hence find the exact value of f (x) = ln(1 + x)
i
f 0 (x) = f 00 (x) = f 000 (x) = .. . f (n) (x) = )
)
1 1+x ¡1 (1 + x)2 ¡1 £ ¡2 (1 + x)3 (¡1)n¡1 (n ¡ 1)! (1 + x)n
.
k
k=1
a
1+t
0
ln(1 + x), jxj < 1.
f(0) = ln 1 = 0
)
f 0 (0) = 1
)
f 00 (0) = ¡1
)
f 000 (0) = (¡1)2 2! .. .
)
f (n) (0) = (¡1)n¡1 (n ¡ 1)!
f (x) = Tn (x) + Rn (x : 0) =0+
1£x 1 £ x2 2!x3 3!x4 (¡1)n¡1 (n ¡ 1)!xn ¡ + ¡ + :::: + + Rn (x : 0) 1! 2! 3! 4! n!
=x¡
x2 x3 x4 (¡1)n¡1 xn + ¡ + :::: + + Rn (x : 0) 2 3 4 n
where Rn (x : 0) =
f (n+1) (c)xn+1 (¡1)n xn+1 = . (n + 1)! (1 + c)n+1 (n + 1) 1 P (¡1)k+1 xk
Thus the Maclaurin series for ln(1 + x) is
k=1
k
= x¡
x2 x3 x4 + ¡ + :::: 2 3 4
By the Ratio test this converges for jxj < 1. We ignore the case x = ¡1, since f (x) is undefined at this point. 1 P (¡1)k+1 If x = 1, we have which converges by the Alternating Series Test. k
k=1
Hence the Maclaurin series for ln(1 + x) has interval of convergence ] ¡1, 1 ] . ¯ ¯ n+1 ¯ ¯ x ¯ where c is between 0 and x. ii jRn (x : 0)j = ¯¯ n+1 (1 + c) (n + 1) ¯ If 0 < c < x 6 1 then xn+1 1 6 for 0 < x 6 1 n+1 n+1 1 = 0. fSqueeze Theoremg lim jRn (x : 0)j = lim n!1 n!1 n + 1
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0 6 jRn (x : 0)j
p. In these cases (1 + x)p =
p ¡ ¢ P p n n x ,
n=0
which is the binomial formula. This series is finite and therefore convergent for all x 2 R . For other values of p, the series has radius of convergence R = 1. Example 45 a Use the Maclaurin series expansion of (1 + x)p to write down the Maclaurin series for 1
¡2
(1 ¡ x2 )
.
Z
x
b Use the fact that arccos x ¡ arccos(0) = 0
p
¡1
1 ¡ t2
dt to find a Maclaurin series for
arccos x. For which values of x is this valid? c
i Use the series in b to find the first three non-zero terms of arccos(x2 ). Z 0:4 ii Hence find an approximate value for arccos(x2 ) dx. 0
1 P p(p ¡ 1) :::: (p ¡ k + 1)
k!
k=0
=1+
k=1
1 P
=1+
³
¡ 12
³
¡ 12
¡ 32
k!
´³
¡ 32
´
³ ::::
2k (k!)
k=1
1 P 1 £ 3 £ 5 £ :::: £ (2k ¡ 1)
=1+
2k k!
k=1
(¡1)k
x2k x2k
x2k £
2 £ 4 £ 6 £ :::: £ 2k £ x2k 2 £ 4 £ 6 £ :::: £ 2k
1 P
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xk for all jxj < 1 ´³ ´ ³ ´
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IB HL OPT 2ed Calculus
Y:\HAESE\IB_HL_OPT-Calculus\IB_HL_OPT-Calculus_01\100IB_HL_OPT-Calculus_01.cdr Monday, 11 February 2013 9:03:41 AM BRIAN
CALCULUS 1
¡2
) (1 ¡ x2 )
1 P (2k)!
=1+
4k (k!)2
k=1 Z x
b arccos x ¡ arccos(0) =
p
0
)
arccos x ¡
¡1
¯ ¯ for all ¯x2 ¯ < 1, which is for all jxj < 1.
dt
1 ¡ t2
Z
¼ 2
x2k
µ ¶ 1 P (2k)! 2k 1+ t dt k 2
x
=¡ 0
µ 1 P =¡ x+
k=1
4 (k!)
(2k)! x2k+1 k (k!)2 (2k + 1) 4 k=1
)
arccos x =
¡x¡
¼ 2
101
¶
1 P
(2k)! x2k+1 k (k!)2 (2k + 1) 4 k=1
This is valid provided jxj < 1. These values of x all lie in the domain of arccos, so the series is valid for ¡1 < x < 1. c
arccos x =
i
2
¼ 2 ¼ 2
¡ x ¡ 16 x3 2
fusing the first 3 termsg
¡ 16 x6 Z 0:4
arccos(x ) ¼ ¡ x Z 0:4 ¡ ¢ ii arccos x2 dx ¼ 0
· =
for jxj < 1 µ ¶ x6 2 ¼ ¡ x ¡ dx since j0:4j < 1 2 6
0
¸ 0:4
¼ 2x
¡
¼ 0:607
x3 x7 ¡ 3 7 0
EXERCISE L.3 1 Use the Maclaurin series expansion of (1 + x)p to find the Maclaurin series for each given function and find the associated radius of convergence. a f(x) =
1 1 ¡ x2
b f (x) =
1 1 + x3
c f (x) =
1 1 ¡ x3
a Use the Maclaurin series expansion of (1 + x)p to write down the Maclaurin series for (1 + x2 )¡1 and state the associated radius of convergence. Z x 1 b Use the fact that arctan x ¡ arctan(0) = dt to obtain the Maclaurin series for 2
2
1+t
0
arctan x. For which values of x is this valid? c
i Use the series in b to find the first four non-zero terms of arctan(x2 ). Z 1 ii Hence find an approximate value for arctan(x2 ) dx. 0
¡
0
expansion for arcsin x for jxj < R. 1 P ¼ 1 c If = + ak , find ak .
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a Use the (1 + x)p series to write a Maclaurin series for (1 ¡ x2 ) 2 and state the radius of convergence R. Z x 1 b Use the fact that arcsin x ¡ arcsin(0) = dx to obtain a Maclaurin series p
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IB HL OPT 2ed Calculus
Y:\HAESE\IB_HL_OPT-Calculus\IB_HL_OPT-Calculus_01\101IB_HL_OPT-Calculus_01.cdr Monday, 11 February 2013 9:04:13 AM BRIAN
102
CALCULUS
PRODUCT OF SERIES 1 P
Suppose
an xn
1 P
has radius of convergence Ra and
n=0
bn xn
has radius of convergence Rb .
n=0
The product of the series is µ1 ¶µ 1 ¶ 1 P P P an xn bn xn = cn xn n=0
n=0
n=0
= a0 b0 + (a0 b1 + a1 b0 )x + (a0 b2 + a1 b1 + a2 b0 )x2 + :::: + cn xn + :::: where cn = a0 bn + a1 bn¡1 + :::: + an b0 . The radius of convergence is the minimum of Ra and Rb . Example 46 1 P
Let
an xn =
n=0
x sin x
be the power series expansion of
x , x 6= k¼, k 2 Z . sin x
Use the Maclaurin series expansion of sin x to find a0 , a1 , a2
so that
for x near 0.
µ ¶ x3 x5 we find x = (a0 + a1 x + a2 x2 + ::::) x ¡ + ¡ :::: 3! 5! ³ ´ a0 2 3 = a0 x + a1 x + a2 ¡ x + ::::
1 P x = an xn , sin x n=0
Since
x ¼ a0 +a1 x+a2 x2 sin x
3!
Comparing LHS and RHS and equating coefficients, a0 = 1, a1 = 0, a2 =
1 6
x ¼ 1 + 16 x2 sin x
)
EXERCISE L.4
1 p a Find the first four terms of the power series representation for 1 + x = (1 + x) 2 , where 1 p p P 1+x= an xn and using ( 1 + x)2 = 1 + x. jxj < 1, by letting
1
n=0
b Check that your answer in a agrees with the (1 + x)p series with p = 12 . 2
a Use known Maclaurin series expansions to find the first four non-zero terms in the Maclaurin series expansion of ex arccos x.
For the Maclaurin series expansion of arccos x, see Example 45.
b For which values of x does the expansion in a equal the full Maclaurin series expansion?
3
a Use known Maclaurin series expansions to find the first three non-zero terms in the Maclaurin series expansion of
x . cos x
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IB HL OPT 2ed Calculus
Y:\HAESE\IB_HL_OPT-Calculus\IB_HL_OPT-Calculus_01\102IB_HL_OPT-Calculus_01.cdr Monday, 11 February 2013 9:07:07 AM BRIAN
CALCULUS
103
4 Use the Maclaurin series expansions of sin x and cos x to determine the first three non-zero terms sin x . cos x
of the Maclaurin series expansion for tan x =
5 It can be shown that the exponential function equals its Maclaurin series for all z 2 C , where C is the set of all complex numbers. 1 P zk ) ez = for all complex numbers z. k=0
k!
Let z = iµ where µ 2 R . a Use the above result to write down the simplified expanded series for eiµ . b Hence, using the known Maclaurin series for cos µ and sin µ, µ 2 R , prove Euler’s formula eiµ = cos µ + i sin µ. a Prove that 1 + x 6 ex for all x > 0. b Hence show that if uk > 0 for all k, then n Q (1 + uk ) = (1 + u1 )(1 + u2 ) :::: (1 + un ) 6 eu1 +u2 +::::+un .
6
k=1
1 P
c If n Q
uk
n Q
converges, deduce the behaviour of
n=1
uk = u1 £ u2 £ :::: £ un
k=1
(1 + uk ) as n ! 1.
k=1
7 In this question, use the following steps for Euler’s proof of
1 P 1 n=1
n2
=
¼2 . 6
x3 x5 x7 You may assume that sin x = x ¡ + ¡ + :::: for all x 2 R . 3! 5! 7! sin x a Find all the zeros of sin x and of for x 2 R . x sin x b Find the power series expansion for and its interval of convergence. x
c Find all the zeros of
³ ´³ ´³ ´³ ´ x x x x 1¡ 1+ 1¡ 1+ :::: ¼
2¼
¼
2¼
d Show that: µ ¶µ ¶µ ¶ ³ ´³ ´³ ´³ ´ x x x x x2 x2 x2 1¡ 1+ 1¡ 1+ :::: = 1 ¡ 2 1¡ 2 1 ¡ 2 :::: ¼
2¼
¼
2¼
4¼
¼
and comment on Euler’s claim that x2 x4 x6 + ¡ + :::: = 1¡ 3! 5! 7!
9¼
µ ¶µ ¶µ ¶ x2 x2 x2 1¡ 2 1¡ 2 1 ¡ 2 :::: 4¼
¼
9¼
e By equating the coefficients of x2 in this last equation, prove that: 1 P 1 1 1 1 1 ¼2 = 2 + 2 + 2 + 2 + :::: = . 2 n=1
1 P 1
f As
n=1
n2
1
n
2
3
4
6
is absolutely convergent, we can write
n=1
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=
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| {z } even n 1 P and
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{z } odd n
1 . (2n ¡ 1)2 n=1
IB HL OPT 2ed Calculus
Y:\HAESE\IB_HL_OPT-Calculus\IB_HL_OPT-Calculus_01\103IB_HL_OPT-Calculus_01.cdr Monday, 11 February 2013 9:13:17 AM BRIAN
104
CALCULUS
HISTORICAL NOTE Euler was able to derive a method to sum all series of the 1 P 1 form , k 2 Z +. 2k n
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IB HL OPT 2ed Calculus
Y:\HAESE\IB_HL_OPT-Calculus\IB_HL_OPT-Calculus_01\104IB_HL_OPT-Calculus_01.cdr Monday, 11 February 2013 9:15:15 AM BRIAN
CALCULUS
M
105
DIFFERENTIAL EQUATIONS
A differential equation is an equation involving the derivative(s) of an unknown function. Suppose y is a function of x, so y = y(x). Examples of differential equations for this function are: dy x2 = dx y
d2 y dy ¡3 + 4y = 0 dx2 dx
dy = ¡0:075y 3 dx
Such equations not only arise in pure mathematics, but are also used to model and solve problems in applied mathematics, physics, engineering, and the other sciences. For example: A parachutist
A falling object
Object on a spring
y
m
v d2 y = 9:8 dx2
m
dv = mg ¡ av2 dt
m
Water from a tank
Current in an RL Circuit
d2 y = ¡ky dt2
Dog pursuing cat y
curve of pursuit
R
(x, y)
E L
L
H
p dH = ¡a H dt
dI + RI = E dt
x
d2 y = dx2
r 1+
x
³
´ dy 2 dx
FIRST ORDER DIFFERENTIAL EQUATIONS In this course we will only deal with differential equations of the form f(x, y)
dy + g(x, y) = 0 dx
where y = y(x).
These are known as first order differential equations since there is only one derivative in the equation, and it is a first derivative.
SOLUTIONS OF DIFFERENTIAL EQUATIONS A function y(x) is said to be a solution of a differential equation if it satisfies the differential equation for all values of x in the domain of y. dy = x. dx
Z
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In this case we can use direct integration to find y =
x dx = 12 x2 + c.
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For example, a very simple differential equation is
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IB HL OPT 2ed Calculus
Y:\HAESE\IB_HL_OPT-Calculus\IB_HL_OPT-Calculus_01\105IB_HL_OPT-Calculus_01.cdr Tuesday, 19 February 2013 4:19:26 PM GR8GREG
106
CALCULUS
In solving the differential equation we obtain a constant of integration. We say that y = 12 x2 + c is a general solution to the differential equation. It describes a family of curves which all satisfy the differential equation. If we are given initial conditions for the problem, such as a value of y or
dy for a specific value of x, dx
then we can evaluate c. This gives us a particular solution to the problem, which is one particular curve from the family of curves described by the general solution. Example 47 Consider the differential equation
dy ¡ 3y = 3. dx
a Show that y = ce3x ¡ 1 is a solution to the differential equation for any constant c. b Sketch the solution curves for c = §1, §2, §3. c Find the particular solution which passes through (0, 2). d Find the equation of the tangent to the particular solution at (0, 2). a If y = ce3x ¡ 1 then )
dy = 3ce3x . dx
dy ¡ 3y = 3ce3x ¡ 3(ce3x ¡ 1) dx
= 3ce3x ¡ 3ce3x + 3 =3 ) the differential equation is satisfied for all x 2 R .
y
b The solution curves for c = §1, §2, §3 are shown alongside:
y = 2e 3x - 1
2 y = 3e 3x - 1 -1
-2
y = e 3x - 1 1 x
y = -3e
3x
-1 -4
-2 y = -e 3x - 1 y = -2e 3x - 1
c y = ce3x ¡ 1 is a general solution to the differential equation. The particular solution passes through (0, 2), so 2 = ce3£0 ¡ 1 ) c=3 ) the particular solution is y = 3e3x¡1 dy = 3 + 3y dx
d
y
) at the point (0, 2),
dy =3+3£2=9 dx
) the gradient of the tangent to the particular solution y = 3e3x¡1 at (0, 2), is 9. ) the equation of the tangent is
-2
2 y = 3e 3x - 1 -1
x
y¡2 =9 x¡0
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IB HL OPT 2ed Calculus
Y:\HAESE\IB_HL_OPT-Calculus\IB_HL_OPT-Calculus_01\106IB_HL_OPT-Calculus_01.cdr Tuesday, 19 February 2013 4:19:43 PM GR8GREG
CALCULUS
107
SLOPE FIELDS Given any first order differential equation of the form f (x, y) we can write
dy + g(x, y) = 0 dx
where y = y(x),
dy g(x, y) =¡ = h(x, y). dx f (x, y)
We may therefore deduce the gradient of the solution curves to the differential equation at any point (x, y) in the plane where h(x, y) is defined, and hence the equations of the tangents to the solution curves. The set of tangents at all points (x, y) is called the slope field of the differential equation. For example, the table alongside shows the values of dy = x(y ¡ 1) dx
for the integer grid points x, y 2 [¡2, 2]. y
¡2 ¡1 0 1 2
¡2 6 4 2 0 ¡2
x 0 0 0 0 0 0
¡1 3 2 1 0 ¡1
1 ¡3 ¡2 ¡1 0 1
2 ¡6 ¡4 ¡2 0 2
y
By representing the gradients at many different grid points as line segments, we obtain a slope field of the tangents to the solution curves as shown.
4 2
-4
-2
2
4 x
2
4 x
-2 -4 y
Now the tangent to a curve approximates that curve at and near the points of tangency. Therefore, by following the direction given by the slope field at a given point, we can approximate solution curves of the differential equation. The horizontal line in the figure is the solution curve corresponding to the initial condition that y = 1 when x = 0.
4 2
-4
-2 -2 -4
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Although it is quite straightforward to obtain a few slope field points by hand, a larger or more refined field is best obtained using technology. You can click on the icon on your CD or else download software for your graphics calculator.
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IB HL OPT 2ed Calculus
Y:\HAESE\IB_HL_OPT-Calculus\IB_HL_OPT-Calculus_01\107IB_HL_OPT-Calculus_01.cdr Wednesday, 20 February 2013 9:52:24 AM GR8GREG
SLOPE FIELDS
108
CALCULUS
An isocline is the set of all points on a slope field at which the tangents to the solution curves have the same gradient. For a first order differential equation written in the form
Isoclines are not solutions to the differential equation. In general, they are not straight lines.
dy = h(x, y), an isocline is a curve given by h(x, y) = k, dx
where k is a particular constant. There may also be an isocline corresponding to points where the gradients of the tangents to the solution curve are undefined.
For example, suppose
dy 1 ¡ x2 ¡ y2 = . dx y¡x+2
dy is discontinuous when dx
²
y 2
y ¡ x + 2 = 0, which is
when y = x ¡ 2. This isocline is the red line on the slope field. dy is zero when dx
²
-2
1 ¡ x2 ¡ y 2 = 0,
2
which is when x2 + y 2 = 1. This isocline is the green circle on the slope field.
x
-2
Example 48 dy = xy dx
Consider the differential equation a Verify that y = ce
x2 2
where x, y > 0.
is a general solution for any constant c 2 R , c > 0.
b Construct the slope field using integer grid points for x, y 2 [0, 4]. c Sketch the particular solution curve through P(2, 1). 2
d Verify algebraically that the curve you have sketched is y = e Draw the isoclines corresponding to k = 1, 3, and 6. a If y = ce
x2 2
1 x2 e . e2
2
then
x dy x = 2 £ £ ce 2 dx 2 x2
= xce 2 µ x2 ¶ = x ce 2 = xy > 0 for all x
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IB HL OPT 2ed Calculus
Y:\HAESE\IB_HL_OPT-Calculus\IB_HL_OPT-Calculus_01\108IB_HL_OPT-Calculus_01.cdr Wednesday, 20 February 2013 9:54:59 AM GR8GREG
109
CALCULUS
b, c 0 0 0 0 0 0
0 1 2 3 4
y
x 2 0 2 4 6 8
1 0 1 2 3 4
3 0 3 6 9 12
y 4
4 0 4 8 12 16
2
y=
1 x2 e e2
3 2
k = 6, y =
x2 2
d The general solution is y = ce . 1 The particular solution passes through (2, 1)
k = 3, y =
1 = ce 1 = ce2
)
c=
3 x
k = 1, y =
22 2
) )
6 x
1
2
3
1 x
x
4
1 e2 2
Hence P(2, 1) lies on the particular solution curve y =
1 x2 e . e2
e The isoclines are shown in red.
EULER’S METHOD OF NUMERICAL INTEGRATION Euler’s Method uses the same principle as slope fields to find a numerical approximation to the solution of the differential equation Since the gradient
dy = f (x, y). dx
dy indicates the direction in which the solution curve goes at any point, we reconstruct dx
the graph of the solution as follows: We start at a point (x0 , y0 ) and move a small distance in the direction of the slope field to find a new point (x1 , y1 ). We then move a small distance in the direction of the slope field at this new point, and so on. If we step h units to the right each time, then x1 = x0 + h and y1 = y0 + h f (x0 , y0 ).
gradient¡=¡f(x0, y 0)
(x0, y 0)
More generally, xn+1 = xn + h and yn+1 = yn + h f (xn , yn ).
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Clearly, Euler’s Method only gives an approximate solution to an initial value problem. However, by decreasing the step size h and hence increasing the number of course corrections, we can usually improve the accuracy of the approximation.
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(x1, y 1)
h
Euler’s Method will be less accurate when the gradient f (x, y) is large.
IB HL OPT 2ed Calculus
Y:\HAESE\IB_HL_OPT-Calculus\IB_HL_OPT-Calculus_01\109IB_HL_OPT-Calculus_01.cdr Wednesday, 20 February 2013 9:58:29 AM GR8GREG
110
CALCULUS
Example 49 dy =x+y dx
Suppose
where y(0) = 1. Use Euler’s Method
with a step size of 0:2 to find an approximate value for y(1). Now xn+1 = xn + h ) given f (x, y) =
If we are given a differential equation and an initial point, we call it an initial value problem or IVP.
and yn+1 = yn + h f(xn , yn ) dy =x+y dx
xn+1 = xn + 0:2
and step size h = 0:2,
and yn+1 = yn + 0:2(xn + yn )
Using the initial conditions, y0 x0 = 0 y1 x1 = 0 + 0:2 = 0:2 y2 x2 = 0:2 + 0:2 = 0:4 y3 x3 = 0:4 + 0:2 = 0:6 y4 x4 = 0:6 + 0:2 = 0:8 y5 x5 = 0:8 + 0:2 = 1
=1 = 1 + 0:2(0 + 1) = 1:2 = 1:2 + 0:2(0:2 + 1:2) = 1:48 = 1:48 + 0:2(0:4 + 1:48) = 1:856 = 1:856 + 0:2(0:6 + 1:856) = 2:3472 = 2:3472 + 0:2(0:8 + 2:3472) = 2:9766
So, y(1) ¼ 2:98 .
EXERCISE M dy = 2x ¡ y. dx
1 Consider the differential equation
a Show that y = 2x ¡ 2 + ce¡x
is a solution to the differential equation for any constant c.
b Sketch the solution curves for c = 0, §1, §2. c Find the particular solution which passes through (0, 1). d Find the equation of the tangent to the particular solution at (0, 1). a Show that y =
2
p x2 + c is a general solution to the differential equation
for any c 2 R . b Hence solve the IVP
dy x = , dx y
dy x = , dx y
y > 0,
y(3) = 4.
3 Slope fields for two differential equations are plotted below for x, y 2 [¡2, 2]. Use the slope fields to graph the solution curves satisfying y(1) = 1. a
b
y
y
2
2
x 2 x
-2
-2
2
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IB HL OPT 2ed Calculus
Y:\HAESE\IB_HL_OPT-Calculus\IB_HL_OPT-Calculus_01\110IB_HL_OPT-Calculus_01.cdr Wednesday, 20 February 2013 10:01:53 AM GR8GREG
CALCULUS
dy = 10y tan x dx
4 Consider the differential equation
111
where x is measured in degrees. Draw the
slope field using integer grid points where x, y 2 [¡2, 2]. dy = x2 + y ¡ 1. dx
5 Consider the differential equation
Sketch isoclines for k = 0, 2, 4 on a grid for x, y 2 [¡5, 5]. Hence construct a slope field, and sketch the solution curve satisfying y(0) = 1. 6 The slope field for the differential equation dy ¡1 + x2 + 4y 2 = dx y ¡ 5x + 10
y
is shown alongside.
2
a Sketch the particular solution passing through the origin.
1 -3
b Sketch the isocline corresponding to: i
dy being undefined dx
1
dy = 0. dx
ii
-1 3
x
-1 -2
7 Use Euler’s Method with step size 0:2 to estimate y(1) for the initial value problem dy = 1 + 2x ¡ 3y, dx
y(0) = 1.
8 Use Euler’s Method with step size 0:1 to estimate y(0:5) for the initial value problem
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y(0) = 0:5. Assume x and y are in radians.
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dy = sin(x + y), dx
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IB HL OPT 2ed Calculus
Y:\HAESE\IB_HL_OPT-Calculus\IB_HL_OPT-Calculus_01\111IB_HL_OPT-Calculus_01.cdr Wednesday, 13 March 2013 11:41:54 AM BRIAN
112
CALCULUS
N
SEPARABLE DIFFERENTIAL EQUATIONS dy f (x) = , dx g(y)
Differential equations which can be written in the form
where y = y(x), are known as
separable differential equations. dy f (x) = dx g(y)
Notice that if
then g(y)
dy = f(x). dx
If we integrate both sides of this equation with respect to x we obtain Z Z dy g(y) dx = f (x) dx dx Z Z ) by the Chain Rule, g(y) dy = f (x) dx. The problem of solving the differential equation is hence reduced to the problem of finding two separate integrals. Example 50 Solve the initial value problem 2x
dy ¡ 1 = y2 , dx
dy ¡ 1 = y2 dx dy ) 2x = y2 + 1 dx 1 1 dy = y 2 + 1 dx 2x
y(1) = 1.
2x
)
Z
Integrating both sides with respect to x gives
1 dy dx = y 2 + 1 dx
Z
)
1 dy = y2 + 1
)
arctan y = )
¡1
¢ But y(1) = 1, so 1 = tan 2 ln 1 + c ) 1 = tan c ) c = ¼4
Z Z
1 dx 2x 1 dx 2x
1 2
ln jxj + c ¡ ¢ y = tan 12 ln jxj + c
) the particular solution of the differential equation is y = tan
¡1 2
ln jxj +
¼ 4
¢ .
Example 51 a Show that
2 1 1 = ¡ . x2 ¡ 1 x¡1 x+1
b Find the general solution of the differential equation
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1 1 x + 1 ¡ (x ¡ 1) 2 ¡ = = 2 as required. x¡1 x+1 (x ¡ 1)(x + 1) x ¡1
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dy x2 y + y = 2 . dx x ¡1
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IB HL OPT 2ed Calculus
Y:\HAESE\IB_HL_OPT-Calculus\IB_HL_OPT-Calculus_01\112IB_HL_OPT-Calculus_01.cdr Friday, 15 March 2013 5:30:35 PM BRIAN
CALCULUS
dy x2 y + y = 2 dx x ¡1
b
1 dy x2 + 1 = 2 y dx x ¡1 x2 ¡ 1 + 2 = x2 ¡ 1 2 =1+ 2 x ¡1 1 1 =1+ ¡ x¡1 x+1
)
y(x2 + 1) x2 ¡ 1
=
113
fusing ag
Integrating both sides with respect to x gives Z Z ³ ´ 1 dy 1 1 dx = 1+ ¡ dx y dx x¡1 x+1 Z 1 ) dy = x + ln jx ¡ 1j ¡ ln jx + 1j + c y ¯´ ³ ¯ ¯x ¡ 1¯ where ln A = c ln jyj = x + ln A ¯ ¯ x+1 ³ ´ x¡1 is the general solution of the differential equation. ) y = Aex x+1
Example 52 When an object travels through a resistive medium, the rate at which it loses speed at any given instant is given by kv m s¡2 , where v is the speed of the body at that instant and k is a positive constant. Suppose the initial speed is u m s¡1 . By formulating and solving an appropriate differential equation, show that the time taken for the body to decrease its speed to The rate of change of speed is given by
m s¡1 is
1 ln 2 seconds. k
dv . dt
Our differential equation must reflect that the body loses speed, so 1 dv = ¡k. v dt
Separating the variables, we find
1 2u
Z
Integrating both sides with respect to t gives )
1 dv = ¡k v
dv = ¡kv. dt
Z dt
ln jvj = ¡kt + c v = Ae¡kt
)
where A = ec
The initial speed (at t = 0) is u, so u = Ae¡k£0 = A. ) v = ue¡kt is the particular solution of the differential equation. When v = 12 u we have )
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IB HL OPT 2ed Calculus
Y:\HAESE\IB_HL_OPT-Calculus\IB_HL_OPT-Calculus_01\113IB_HL_OPT-Calculus_01.cdr Thursday, 14 March 2013 3:37:17 PM BRIAN
114
CALCULUS
Example 53 Consider a curve in the first quadrant. The tangent at any point P cuts the x-axis at Q. Given that OP = PQ, where O is the origin, and that the point (1, 4) lies on the curve, find the equation of the curve. Since OP = OQ, triangle OPQ is isosceles. Hence [PA] is the perpendicular bisector of [OQ].
y P(x, y)
The coordinates of A are (x, 0), so the coordinates of Q are (2x, 0). Since (PQ) is the tangent to the curve at P, the gradient of the curve at P is the same as the gradient of [PQ]. Hence )
dy y =¡ dx x 1 dy 1 =¡ y dx x
Z
Integrating both sides with respect to x gives ) )
O
A x
1 dy = ¡ y
Z
x
Q x
1 dx x
ln jyj = ¡ ln jxj + c
ln jxj + ln jyj = c )
ln jxyj = c xy = ec = k
)
where k is a constant.
Since the curve passes through (1, 4), 1 £ 4 = k ) the equation of the curve is xy = 4 or y =
4 , where x > 0. x
HOMOGENEOUS DIFFERENTIAL EQUATIONS ³ ´
Differential equations of the form
dy y = f , dx x
where y = y(x), are known as homogeneous
differential equations. They can be solved using the substitution y = vx where v is a function of x. The substitution will always reduce the differential equation to a separable differentiable equation as follows. If y = vx where v is a function of x, then
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dy dv = x + v fproduct ruleg dx dx ³ ´ dv vx x+v =f = f(v) ) dx x dv f (v) ¡ v = ) dx x 1 1 dv ) = which is a separable differential equation. f (v) ¡ v dx x
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IB HL OPT 2ed Calculus
Y:\HAESE\IB_HL_OPT-Calculus\IB_HL_OPT-Calculus_01\114IB_HL_OPT-Calculus_01.cdr Friday, 15 March 2013 5:31:35 PM BRIAN
CALCULUS
115
Example 54 dy x + 2y = . dx x
a Use the substitution y = vx, where v is a function of x, to solve b Find the particular solution if y =
3 2
when x = 3.
dy dv =v+x . dx dx dv x + 2vx = Comparing with the differential equation, we find v+x dx x dv ) v+x = 1 + 2v dx dv ) x =1+v dx 1 dv 1 ) = 1 + v dx x
a If y = vx, then using the product rule we obtain
Z
Integrating with respect to x gives ) ) But v =
y , so x
1 dv = v+1
Z
1 dx x
ln jv + 1j = ln jxj + c ln jv + 1j = ln jAxj where ln jAj = c ) v + 1 = Ax
y + 1 = Ax x
y = Ax2 ¡ x, where A is a constant.
) b Substituting y =
3 2
3 2
and x = 3 into the general solution, we find )
9A =
) ) the particular solution is y =
1 2 2x
= A £ 32 ¡ 3
A=
¡ x.
9 2 1 2
EXERCISE N 1 Solve the following initial value problems: a (2 ¡ x)
dy = 1, dx
y(4) = 3
c ey (2x2 + 4x + 1) d x 2
dy = cos2 y, dx
b
dy = (x + 1)(ey + 3), dx
y(e) =
y(1) = 0
y(0) = 2
¼ 4
3¡x 1 2 = ¡ . x2 ¡ 1 x¡1 x+1
a Show that
dy ¡ 3x sec y = 0, dx
b Solve
dy 3y ¡ xy = 2 , dx x ¡1
y(0) = 1.
3 According to Newton’s law of cooling, the rate at which a body loses temperature at time t is proportional to the amount by which the temperature T (t) of the body at that instant exceeds the temperature R of its surroundings.
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a Express this information as a differential equation in terms of t, T , and R. b A container of hot liquid is placed in a room with constant temperature 18± C. The liquid cools from 82± C to 50± C in 6 minutes. Show that it takes 12 minutes for the liquid to cool from 26± C to 20± C.
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IB HL OPT 2ed Calculus
Y:\HAESE\IB_HL_OPT-Calculus\IB_HL_OPT-Calculus_01\115IB_HL_OPT-Calculus_01.cdr Monday, 11 February 2013 1:09:39 PM BRIAN
116
CALCULUS
4 The tangent at any point P on a curve cuts the x-axis at the point Q. PQ = 90± , where O is the origin, and that the point (1, 2) lies on the curve, find Given that Ob the equation of the curve. 5 The tangent at any point P on a curve cuts the x-axis at A and the y-axis at B. Given that AP : PB = 2 : 1 and that the curve passes through (1, 1), find the equation of the curve. 6 A radioactive substance decays at a rate proportional to the mass m(t) remaining at the time. Suppose the initial mass is m0 . a Construct and solve the appropriate initial value problem and hence obtain a formula for m(t). b If the mass is reduced to 45 of its original value in 30 days, calculate the time required for the mass to decay to half its original value. 7 Solve the following homogeneous differential equations using the substitution y = vx, where v is a function of x. a
dy x¡y = dx x
b
dy x+y = dx x¡y
dx
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y dy x = y + ex dx
b Solve
0
dy y 2 ¡ x2 = dx 2xy
a Show that the substitution y = vx (where v is a function of x) will reduce all inhomogeneous ³ ´ dy y y = +f g(x) to separable form. differential equations of the form
8
5
c
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IB HL OPT 2ed Calculus
Y:\HAESE\IB_HL_OPT-Calculus\IB_HL_OPT-Calculus_01\116IB_HL_OPT-Calculus_01.cdr Monday, 11 February 2013 1:09:53 PM BRIAN
CALCULUS
O
THE INTEGRATING FACTOR METHOD
Suppose a first order linear differential equation is of the form
dy +P (x)y = Q(x), dx
where y = y(x).
Generally, but not always, this type of equation is not separable. However, suppose there is a function I(x), called an integrating factor, such that d dy (I(x)y) = I(x) + I(x) P (x)y dx dx
.... (¤)
= I(x) Q(x) Then integrating both sides with respect to x would give Z I(x)y = I(x) Q(x) dx Z 1 I(x) Q(x) dx and we could hence find a solution for y. ) y= I(x)
Now if such an integrating factor exists, then from (¤), I(x)
dy dy + I 0 (x)y = I(x) + I(x) P (x)y dx dx
)
I 0 (x) = I(x) P (x)
)
I 0 (x) = P (x) I(x)
Integrating both sides with respect to x, Z 0 Z I (x) dx = P (x) dx I(x) Z ln jIj + c = P (x) dx R ) I(x) = Ae P (x) dx where A = e¡c and is conventionally set as 1. R I(x) = e
Thus the integrating factor is
P (x) dx
.
THE INTEGRATION FACTOR METHOD dy + P (x) y = Q(x) dx
Suppose
where y = y(x).
R 1 Calculate the integrating factor I(x) = e integration.
P (x) dx
. You do not need a constant of
2 Multiply the differential equation through by I(x). Z 3 Simplify the LHS and hence obtain I(x) y = I(x) Q(x) dx + c, where c is a constant.
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4 Integrate to obtain the general solution.
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IB HL OPT 2ed Calculus
Y:\HAESE\IB_HL_OPT-Calculus\IB_HL_OPT-Calculus_01\117IB_HL_OPT-Calculus_01.cdr Tuesday, 19 February 2013 4:25:59 PM GR8GREG
118
CALCULUS
Example 55 Solve the differential equation
dy + 3x2 y = 6x2 . dx
R The integrating factor is I(x) = e
3x2 dx
= ex
3 3
Multiplying the differential equation through by ex gives 3
3 3 dy + 3x2 ex y = 6x2 ex dx ³ 3´ 3 d ) yex = 6x2 ex dx Z
ex
3
)
3
)
3
6x2 ex dx
yex =
3
yex = 2ex + c 3 ) y = 2 + ce¡x
Example 56 Solve the initial value problem cos x
dy = y sin x + sin(2x), dx
y(0) = 1.
dy sin x sin(2x) ¡ y= dx cos x cos x
We can rewrite the differential equation as
dy + (¡ tan x)y = 2 sin x dx
)
The differential equation is not separable, but is of a form such that we can use an integrating factor. R The integrating factor is I(x) = e ¡ tan x dx = eln(cos x) = cos x. Multiplying the equation through by the integrating factor gives cos x
dy + (¡ cos x tan x)y = sin(2x) dx d (y cos x) = sin(2x) ) dx Z
)
y cos x =
sin(2x) dx
= ¡ 12 cos(2x) + c But when x = 0, y = 1 ) 1 = ¡ 12 cos(0) + c ) c = 32
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) the solution of the initial value problem is y cos x =
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IB HL OPT 2ed Calculus
Y:\HAESE\IB_HL_OPT-Calculus\IB_HL_OPT-Calculus_01\118IB_HL_OPT-Calculus_01.cdr Monday, 11 February 2013 1:10:10 PM BRIAN
CALCULUS
119
EXERCISE O 1 Solve the following using the integrating factor method. a
dy + 4y = 12 dx
c
dy + y = x + ex , dx
b y(1) = 1
dy ¡ 3y = ex , dx
d x
2 Solve the differential equation (x + 1)y + x
y(0) = 2
dy + y = x cos x dx
dy = x ¡ x2 . dx
ACTIVITY
LAPLACE TRANSFORMS
Laplace transforms provide a useful link between improper integrals and differential equations. Z 1 e¡sx f (x) dx. The Laplace transform of a function f (x) is defined as F (s) = Lff(x)g = 0
What to do: 1 Show that: a Lfeax g =
1 , s¡a
c Lfsin(ax)g =
b Lfxg =
s>a
a , s2 + a2
1 , s2
s>0
s>0
2 Show that: a Lff 0 (x)g = sLff(x)g ¡ f (0)
b Lff 00 (x)g = s2 Lff(x)g ¡ sf(0) ¡ f 0 (0)
3 Consider the differential equation f 00 (x) + f (x) = x, f (0) = 0, f 0 (0) = 2. Assuming that Lfg(x)+h(x)g = Lfg(x)g + Lfh(x)g, show that Lff(x)g =
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Hence find a possible solution function f (x) and check your answer.
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IB HL OPT 2ed Calculus
Y:\HAESE\IB_HL_OPT-Calculus\IB_HL_OPT-Calculus_01\119IB_HL_OPT-Calculus_01.cdr Monday, 11 February 2013 1:10:20 PM BRIAN
120
CALCULUS
P
TAYLOR OR MACLAURIN SERIES DEVELOPED FROM A DIFFERENTIAL EQUATION
Consider a first order differential equation with initial condition: dy = h(x, y) dx
where y(x0 ) = y0 for some constants x0 , y0 2 R .
Depending on the form of h(x, y), it may or may not be possible to obtain an explicit particular solution y(x). However, provided the necessary derivatives exist, we can determine the nth degree Taylor polynomial approximation to the particular solution y(x) about x0 as follows: y(x0 ) = y0 y0 (x) = h(x, y) d (h(x, y)) dx ¢ d ¡ 00 y (x) :::: y000 (x) = dx
y00 (x) =
)
y 0 (x0 ) = h(x0 , y0 )
)
we can obtain y 00 (x0 ) using x0 , y0 , and y 0 (x0 ) found above.
Once y (x0 ), y 0 (x0 ), y00 (x0 ), ::::, y(n) (x0 ) have been determined, the nth degree Taylor polynomial approximation to y(x) about x0 can then be written as: Tn (x) = y (x0 ) + y0 (x0 ) (x ¡ x0 ) +
y00 (x0 ) (x ¡ x0 )2 y(n) (x0 ) (x ¡ x0 )n + :::: + . 2! n!
Example 57 Consider the initial value problem
dy = x ¡ 4xy, dx
y(0) = 1.
a Without solving the system, find the first three non-zero terms of the Maclaurin polynomial expansion for y. b Solve the system exactly to find the particular solution.
)
dy = x ¡ 4xy = x (1 ¡ 4y) dx d2 y d = (x (1 ¡ 4y)) dx2 dx ³ ´ dy = 1 ¡ 4y + x ¡4 dx
)
d3 y dy dy = ¡4 + 1 £ ¡4 dx3 dx dx
³
d4 y d2 y d2 y = ¡8 2 ¡ 4 2 + x 4 dx dx dx
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µ ¶ d2 y + x ¡4 2 dx
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= ¡12
d2 y
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)
dy ¡ 4x 2 dx dx
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´
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IB HL OPT 2ed Calculus
Y:\HAESE\IB_HL_OPT-Calculus\IB_HL_OPT-Calculus_01\120IB_HL_OPT-Calculus_01.cdr Monday, 11 February 2013 1:11:54 PM BRIAN
CALCULUS
121
dy =0 dx
Now y(0) = 1, so at (0, 1):
d2 y = 1 ¡ 4 + 0 = ¡3 dx2 d3 y =0 dx3 d4 y = 36 dx4
) y has Maclaurin polynomial 1 ¡
3x2 36x4 + :::: which is 1 ¡ 32 x2 + 32 x4 :::: 2! 4!
dy = x (1 ¡ 4y) dx 1 dy =x 1 ¡ 4y dx
b
Z
1 dy = 1 ¡ 4y
)
¡ 14 ln j1 ¡ 4yj =
Z
x dx But y(0) = 1
x2 +c 2
1 4
)
2
ln j1 ¡ 4yj = ¡2x + c
B = ¡ 34 µ ¶ 3 ) y = 14 1 + 2x2
)
¡2x2
1 ¡ 4y = Ae y=
1 4
¡B = 1
¡ Be¡2x
2
e
EXERCISE P dy e¡y = ¡ 1, dx 3
1 Consider the differential equation
where y(0) = 0.
a Find the first three non-zero terms in the Maclaurin polynomial expansion of y. µ ¶ 1 + 2e¡x b Verify that y = ln is a particular solution. 3
dy y = 2x + , dx x
2 Consider the differential equation
where
dy = 1 when x = 1. dx
a Find the first three non-zero terms in the Maclaurin polynomial expansion of y about x = 1. b Solve the system exactly to find the particular solution. dy 3x ¡ 2y = , dx x
3 Consider the initial value problem
y(1) = 0.
a Without solving the system, find the first three non-zero terms of the Taylor polynomial expansion for y about x = 1. b Solve the system exactly to find the particular solution. 4 Consider the initial value problem cos x
dy + y sin x = 1, dx
y(0) = 2.
a Find the first three non-zero terms in the Maclaurin polynomial expansion of y.
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b Solve the system exactly to find the particular solution.
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IB HL OPT 2ed Calculus
Y:\HAESE\IB_HL_OPT-Calculus\IB_HL_OPT-Calculus_01\121IB_HL_OPT-Calculus_01.cdr Monday, 11 February 2013 1:12:21 PM BRIAN
122
CALCULUS
Example 58 1 P xk
Given that
k!
k=0
steps.
is convergent for all x 2 R , show that ex =
1 k P x k=0
a Show that if y(x) =
1 P xk k=0 k!
c Hence prove ex =
1 P xk
k!
k=0
using the following
dy = y. dx
, then
b Solve the differential equation
k!
dy = y to obtain the general solution. dx
.
1 P xk
x2 x3 =1+x+ + + :::: k! 2! 3! k=0 dy 2x 3x2 4x3 =1+ + + + :::: dx 2! 3! 4!
a Let y(x) = )
=1+x+ =
1 P xk k=0
x2 x3 + + :::: 2! 3!
k!
=y b Consider ) Z )
dy =y dx 1 dy =1 y dx
Z
1 dy = y
1 dx
) ln jyj = x + c, where c is a constant ) y = Aex , where A = §ec is a constant. 1 P xk c Since y(x) = is a solution to the differential equation for all x 2 R , 1 P xk
y(x) =
k!
k=0
k=0
k!
= Aex for some constant A.
Now y(0) = 1, so A = 1 1 P xk for all x 2 R . Hence ex = k=0
5 Let y(x) =
k!
1 P p(p ¡ 1) :::: (p ¡ n + 1)xn
n!
n=0
for jxj < 1. From Exercise L.2 question 8, y is a well
defined function on jxj < 1 since the binomial series was shown to be convergent on the interval ¡1 < x < 1. a Show that for ¡1 < x < 1, y satisfies the differential equation (1 + x)
dy = py. dx
b Find the general solution to the differential equation in a.
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c Hence show that y(x) = (1 + x)p for all x 2 ] ¡1, 1 [ .
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IB HL OPT 2ed Calculus
Y:\HAESE\IB_HL_OPT-Calculus\IB_HL_OPT-Calculus_01\122IB_HL_OPT-Calculus_01.cdr Thursday, 14 March 2013 3:38:43 PM BRIAN
CALCULUS
123
REVIEW SET A lim
1 Prove that 2 Find
x!1
ln x = 0. x
ex sin x . x!0 x
lim
3 Find the limit, if it exists, of the sequence fun g as n tends to infinity, if un equals: 8 ¡ 2n ¡ 2n2 4 + 6n + 7n2
a
c p
b 3+
2n + 13
d arctan n
6n2 + 5n ¡ 7
4 Prove that the series
5
13
1 2 3 4 5 + 3 + 3 + 3 + 3 + :::: +1 2 +1 3 +1 4 +1 5 +1
´n 1 ³ P 3x
a For what values of x does b Find
in terms of x.
x¡2
1 P n(3x)n¡1
c Show that
n=0
converges.
converge?
x¡2
n=1
´n 1 ³ P 3x n=0
1 + n [1 + (¡1)n ] n
(x ¡ 2)n+1
=
1 4(x + 1)2
for the x values found in a.
6 Find the set of real numbers for which the series x +
Z
1
7 Using an appropriate Maclaurin series, evaluate
x2 x3 + + :::: 1¡x (1 ¡ x)2
converges.
sin(x2 ) dx to three decimal places.
0
8 Let X be a random variable such that P(X = x) = 1 P
Prove that
e¡¸ ¸x x!
for x = 0, 1, 2, ....
P(X = x) = 1.
x=0
a Draw the slope field using integer grid points for x, y 2 [¡3, 3] for the differential dy x = . dx y
b Draw on your slope field the isoclines dy xy = dx x¡1
10 Solve the differential equation
dy =k dx
p dy y ¡ = x, dx x
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11 By finding a suitable integrating factor, solve
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given that y = 2 when x = 2.
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IB HL OPT 2ed Calculus
Y:\HAESE\IB_HL_OPT-Calculus\IB_HL_OPT-Calculus_01\123IB_HL_OPT-Calculus_01.cdr Thursday, 14 March 2013 3:45:13 PM BRIAN
124
CALCULUS
12 On the slope field for
dy = 2x ¡ y 2 dx
y
shown,
sketch the solution curves through a (0, 0)
b (2, 3).
4
2
-2
2
4
x
-2
REVIEW SET B 1 Find the limit, if it exists, of the sequence fun g as n tends to infinity, if un equals: a c
(¡1)n (2n ¡ 1) n
p p n+5¡ n¡1
2 Prove that the series x +
x2 x3 x4 + + + :::: 2 3 4
b
0:9n 1 + 0:1n
d
n2 2n3 ¡ 2 3n + 1 6n + 1
is convergent for ¡1 < x < 1 and divergent
for jxj > 1. Determine the convergence or divergence of the series for x = §1. ³ ´ 1 P (k ¡ 1)¼ sin is convergent. 3 Determine whether or not the series 2k
k=1
1 P 1+r
4 Use the Comparison Test to prove that the series
1 + r2
r=1
a Show that the series Sn =
5
n P (¡1)k+1 k=3
ln(k ¡ 1)
diverges.
converges as n ! 1.
b Find the maximum possible error in using S10 to estimate
1 P (¡1)k+1 k=3
ln(k ¡ 1)
.
6 Find the Taylor series expansion of (x ¡ 1)ex¡1 about x = 1 up to the term (x ¡ 1)3 . 7 Find constants a and b such that y = ax + b is a solution of the differential equation dy = 4x ¡ 2y. dx
8 Find the general solution of the differential equation
dy = 2xy2 ¡ y 2 . dx
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9 Find the equation of the curve through (2, 1) given that for any point (x, y) on the curve, the y-intercept of the tangent to the curve is 3x2 y3 .
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IB HL OPT 2ed Calculus
Y:\HAESE\IB_HL_OPT-Calculus\IB_HL_OPT-Calculus_01\124IB_HL_OPT-Calculus_01.cdr Wednesday, 20 February 2013 10:09:26 AM GR8GREG
CALCULUS
125
10 Match the slope fields A, B, and C to the differential equations: dy =y+1 dx
a A
dy =x¡y dx
b B
y
c C
y
2
dy = x ¡ y2 dx
y 2
2
2 x
-2
2 x
-2 -2
-2
2 x
-2 -2
11 The population P of an island is currently 154. The population growth in the foreseeable future ³ ´ dP P = 0:2P 1 ¡ , t > 0. is given by 400
dt
1 1 + P 400 ¡ P
a Write
as a single fraction.
b Find P as a function of time t years. c Estimate the population in 20 years’ time. d Is there a limiting population size? If so, what is it? 12 Let f(x) =
1 P
x2 . (1 + x2 )i i=1
a Find f (0). b For x 6= 0, show that
1 P (1 + x2 ) 1 f (x) = ri where r = . 2 x 1 + x2 i=0
c For which values of x 2 R is f(x) convergent? Hence state the domain of f . d For each x for which f(x) is convergent, find the value of f (x). e Sketch y = f (x).
REVIEW SET C 1 Find the limit, if it exists, of the sequence fun g as n tends to infinity, if un equals: 1 p en b (3n + 2n ) n c d (¡1)n ne¡n a n ¡ n2 + n n!
1 P
2 Explain why the series
1
3r
is not convergent.
r=1
1 P
3 Determine whether
n=2
4 Suppose
1 P
a Prove that
is convergent or divergent.
is convergent, where an > 0 for all n 2 Z + .
an
n=1
1 ln(n2 )
1 P
an2
´ 1 ³ P 1 2 an ¡ are also convergent.
and
n=1
n
n=1
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b Would these results follow if an 2 R ?
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IB HL OPT 2ed Calculus
Y:\HAESE\IB_HL_OPT-Calculus\IB_HL_OPT-Calculus_01\125IB_HL_OPT-Calculus_01.cdr Wednesday, 20 February 2013 10:13:35 AM GR8GREG
126
CALCULUS
5
a Show that
1 1 1 = ¡ . x(x + 1) x x+1
1 P
b Use a and the Integral Test to prove that the series
n=1
1 n(n + 1)
converges.
n P
1 be the nth partial sum for the series in b. i(i + 1) i=1 1 i Use a to show that Sn = 1 ¡ . n+1
c Let Sn =
ii Hence find the value of the sum of the convergent series
1 P
1 . n(n + 1) n=1
6 Let Rn be the error term in approximating f(x) = ln(1 + x) for 0 6 x < 1 using the first n + 1 terms of its Maclaurin series. Prove that jRn j 6
1 n+1
for 0 6 x < 1.
7 Find a simplified expression for 1 ¡ x + x2 ¡ x3 + :::: where ¡1 < x < 1. Hence find a power series expansion for f (x) = ln(1 + x) for ¡1 < x < 1. 8 A curve passes through the point (1, 2) and satisfies the differential equation dy = x ¡ 2y. dx
Use Euler’s Method with step size 0:1 to estimate the value of y when x = 1:6 . 9 A water tank of height 1 m has a square base 2 m £ 2 m. When a tap at its base is opened, the water flows out at a rate proportional to the square root of the depth of the water at any given time. Suppose the depth of the water is h m, and V is the volume of water remaining in the tank after t minutes. a Write down a differential equation involving
dV and h. dt
b Explain why V = 4h m3 at time t. Hence write down a differential equation involving dh and h. dt
c Initially the tank is full. When the tap is opened, the water level drops by 19 cm in 2 minutes. Find the time it takes for the tank to empty. 10 Use the substitution y = vx where v is a function of x, to solve the differential equation dy x y = + . dx y x
11 Use an integrating factor to solve
dy = cos x ¡ y cot x, dx
y
¡¼¢ 2
= 0.
REVIEW SET D 1 Find the limit, if it exists, of the sequence fun g as n tends to infinity, if un equals: ³ ´ ³ ´ 3 £ 5 £ 7 £ :::: £ (2n + 1) 1 1 b n(2 cos ¡ sin ¡ 2) a 2 £ 5 £ 8 £ :::: £ (3n ¡ 1)
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n
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1 n(ln n)2 n=2
5
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1 P
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5
2 Prove that the series
n
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IB HL OPT 2ed Calculus
Y:\HAESE\IB_HL_OPT-Calculus\IB_HL_OPT-Calculus_01\126IB_HL_OPT-Calculus_01.cdr Wednesday, 20 February 2013 11:42:37 AM GR8GREG
CALCULUS 1 P (x ¡ 3)n
3 Determine the interval and radius of convergence of the series
converges or diverges.
n+5
n=0
.
n2
n=1
´n 1 ³ P n
4 Determine whether the series
3
127
5 Estimate e0:3 correct to three decimal places using the Maclaurin approximation: f 00 (0) x2 f (n) (0) xn f (n+1) (c) xn+1 + :::: + + 2! n! (n + 1)!
f(x) = f (0) + f 0 (0) x +
6 Use the substitution y = vx where v is a function of x, to solve the initial value problem xy 7
dy = 1 + x + y2 , dx
a Prove that e ¡
y(1) = 0. n P 1 k=0
k!
=
ec (n + 1)!
where 0 < c < 1.
b Using the fact that e < 3, show that for n > 3: n P 1 1 3 0, find the equation of the curve which passes through the point (1, 5). 3x x
10 The inside surface of y = f (x) is a mirror. Light is emitted from O(0, 0). All rays that strike the surface of the mirror are reflected so that they emerge parallel to the axis of symmetry (the x-axis).
y y = f(x)
P(x, y) ® q
a Explain why µ = 2®. b Explain why the gradient of the tangent at a general point P(x, y) on the mirror is given
x tangent at P
dy = tan ®. dx
by
c Use the identity
y = -f(x)
p
2 tan ® tan(2®) = 1 ¡ tan2 ®
x2 + y 2 ¡ x . y
to deduce that tan ® = p
d Find the general solution to the differential equation the substitution r2 = x2 + y2 .
dy = dx
x2 + y2 ¡ x y
by making
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e What is the nature of y = f (x)?
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IB HL OPT 2ed Calculus
Y:\HAESE\IB_HL_OPT-Calculus\IB_HL_OPT-Calculus_01\127IB_HL_OPT-Calculus_01.cdr Thursday, 14 March 2013 3:43:17 PM BRIAN
128
CALCULUS
11 Consider the differential equation
dy = y ln x, x > 0, where y(1) = 1. dx
a Find the first three non-zero terms in the Taylor polynomial expansion of y about x = 1. b Hence estimate the solution to the initial value problem
dy = y ln x, dx
y(1) = 1.
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c Solve the initial value problem in b exactly.
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IB HL OPT 2ed Calculus
Y:\HAESE\IB_HL_OPT-Calculus\IB_HL_OPT-Calculus_01\128IB_HL_OPT-Calculus_01.cdr Monday, 11 February 2013 1:23:20 PM BRIAN
CALCULUS
THEORY OF KNOWLEDGE Consider f (x) =
129
TORRICELLI’S TRUMPET
1 . An infinite area sweeps out a finite volume. Can this be reconciled with our x
intuition? What does this tell us about mathematical knowledge? Calculus provides us with a set of procedures for manipulating symbols, such as differentiation and integration. Justifying how, and importantly when, such manipulations lead to correct answers is the subject of analysis which underpins calculus. These justifications need to make use of arguments involving infinite processes such as infinitely small quantities and limits over an infinitely large range. These are very subtle concepts which mathematicians and philosophers have discussed and debated for many thousands of years, such as in Archimedes’ work in finding areas and volumes. Much of the debate has focused on paradoxes. A paradox is an argument that appears to produce an inconsistency, such as resulting in 1 = 0. Often the paradox enables a definition, technique, or logical argument to be refined. However there are still some things which appear to be paradoxical and Torricelli’s trumpet, also sometimes called Gabriel’s Horn, is one of these.
To create this shape we take the graph of y =
1 x
for x > 1 and rotate it around the x-axis to
create a solid of revolution. The two questions we wish to ask are: ² What is the volume of this shape? ² What is the surface area of this shape? To answer these questions we shall use the modern techniques of calculus. Torricelli’s original arguments, made in the early 1600s, predate the systematic development of calculus by Newton and Leibnitz later that century. His argument relied on Cavalieri’s Principle: Take two solids included between parallel planes, and cut the solids with another parallel plane. If the areas are always in the same ratio, then the volumes are also in the same ratio.
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Think of a solid, such as a cone, made up of a stack of cards. You can displace the cards without altering the volume. If you can change the shape of the cards without altering the areas then you also don’t change the volume. Cavalieri’s Principle is another extension to the case when the respective areas are in constant ratio.
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IB HL OPT 2ed Calculus
Y:\HAESE\IB_HL_OPT-Calculus\IB_HL_OPT-Calculus_01\129IB_HL_OPT-Calculus_01.cdr Monday, 11 February 2013 2:53:28 PM BRIAN
130
CALCULUS
1 What place does an “intuitive” rule such as Cavalieri’s Principle have in mathematics? 2 How can Cavalieri’s Principle be used to find the volume of a sphere from the known volume of a circular cone? Can this be regarded as a valid proof of the formula? In order to consider the volume and surface area we will look at a finite part of the trumpet from x = 1 to x = N and then let N ! 1. VOLUME y
We imagine a plane perpendicular to the x-axis, cutting the axis at x. 1 . x
The trumpet and the plane intersect at a circle of radius The area of the circle is simply ¼
y = Qv
1 . x2
x
To calculate the volume we sum the infinitesimal areas of these Z N h iN ³ ´ 1 1 1 discs as ¼ dx = ¼ ¡ = ¼ 1 ¡ . 2 1
x
x 1
N
As N ! 1 we see that the volume converges to ¼. It is somewhat surprising that an infinite shape can have a finite volume. However, this is similar to the idea that an infinite series of positive terms can sum to a finite length. SURFACE AREA
Z
r
N
To calculate the surface area we use the formula S = 2¼
y
1+
³
1
1 dy 1 we have = ¡ 2 and so the integral becomes Since y = x dx x
Noting that
1 x
Z S = 2¼
r 1 1 1 + 4 > , we find that x
´ dy 2 dx. dx
1
N
1 x
r 1 1 + 4 dx. x
x
Z
N
S = 2¼ 1
1 x
r Z 1 1 + 4 dx > 2¼ x
1
N
1 dx = 2¼ ln(N). x
As N ! 1, ln(N) diverges, and so the surface area S also diverges. Hence we have a finite volume with an infinite surface area. This is a strange and controversial result. Thomas Hobbes, the English Philosopher, said “to understand this for sense, it is not required that a man should be a geometrician or logician, but that he should be mad ”. 3 Imagine filling the trumpet with a liquid. While the trumpet holds a finite volume, what happens when a thin layer coats the surface?
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4 What do we do when a physical problem yields a mathematical result with no physical meaning?
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IB HL OPT 2ed Calculus
Y:\HAESE\IB_HL_OPT-Calculus\IB_HL_OPT-Calculus_01\130IB_HL_OPT-Calculus_01.cdr Tuesday, 26 February 2013 12:42:48 PM BRIAN
APPENDICES
APPENDIX A:
131
METHODS OF PROOF
Greek mathematicians more than 2000 years ago realised that progress in mathematical thinking could be brought about by conscious formulation of the methods of abstraction and proof. By considering a few examples, one might notice a certain common quality or pattern from which one could predict a rule or formula for the general case. In mathematics this prediction is known as a conjecture. Mathematicians love to find patterns, and try to understand why they occur. Experiments and further examples might help to convince you that the conjecture is true. However, problems will often contain extra information which can sometimes obscure the essential detail, particularly in applied mathematics. Stripping this away is the process of abstraction. For example, by considering the given table of values one may conjecture: “If a and b are real numbers then a < b implies that a2 < b2 .” However, on observing that ¡2 < 1 but (¡2)2 6< 12 we have a counter-example. In the light of this we reformulate and refine our conjecture:
a 1 3 4 5 6
b 2 5 5 7 9
a2 1 9 16 25 36
b2 4 25 25 49 81
“If a and b are positive real numbers then a < b implies a2 < b2 .” The difficulty is that this process might continue with reformulations, counter-examples, and revised conjectures indefinitely. At what point are we certain that the conjecture is true? A proof is a flawless logical argument which leaves no doubt that the conjecture is indeed a truth. If we have a proof then the conjecture can be called a theorem. Mathematics has evolved to accept certain types of arguments as valid proofs. They include a mixture of both logic and calculation. Generally mathematicians like elegant, efficient proofs. It is common not to write every minute detail. However, when you write a proof you should be prepared to expand and justify every step if asked to do so. We have already examined in the HL Core text, proof by the principle of mathematical induction. Now we consider other methods.
DIRECT PROOF In a direct proof we start with a known truth and by a succession of correct deductions finish with the required result. Example 1: Prove that if a, b 2 R then a < b ) a < a b < 2 2 a a a b + < + ) 2 2 2 2 a+b ) a< 2
Proof: a < b )
a+b 2
fas we are dividing by 2 which is > 0g fadding
a to both sidesg 2
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Sometimes it is not possible to give a direct proof of the full result and so the different possible cases (called exhaustive cases) need to be considered and proved separately.
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132
APPENDICES
Example 2: Prove the geometric progression: For n 2 Z , n > 0, 8 < rn+1 ¡ 1 , r 6= 1 1 2 n 1 + r + r + :::: + r = r¡1 : n + 1, r=1 1 + r1 + r2 + :::: + rn
Proof: Case r = 1:
= 1 + 1 + 1 + :::: + 1
fn + 1 timesg
=n+1 Let Sn = 1 + r1 + r2 + :::: + rn .
Case r 6= 1:
Then rSn = r1 + r2 + r3 + :::: + rn+1 ) rSn ¡ Sn = rn+1 ¡ 1 ) (r ¡ 1)Sn = r ) Sn =
n+1
fafter cancellation of termsg
¡1
¡1 r¡1
rn+1
fdividing by r ¡ 1 since r 6= 1g
Example 3: Alice looks at Bob and Bob looks at Clare. Alice is married, but Clare is not. Prove that a married person looks at an unmarried person. Proof: We do not know whether Bob is married or not, so we consider the different (exhaustive) cases: If Bob is married, then a married person (Bob) looks at an unmarried person (Clare).
Case: Bob is married.
Case: Bob is unmarried. If Bob is unmarried, then a married person (Alice) looks at an unmarried person (Bob). Since we have considered all possible cases, the full result is proved.
EXERCISE p I 1 Let I = 2, which is irrational. Consider I I and I I , and hence prove that an irrational number to the power of an irrational number can be rational.
PROOF BY CONTRADICTION (AN INDIRECT PROOF)
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In proof by contradiction we deliberately assume the opposite to what we are trying to prove. By a series of correct steps we show that this is impossible, our assumption is false, and hence its opposite is true.
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APPENDICES
133
Example 4: Consider Example 1 again but this time use proof by contradiction: Prove that if a, b 2 R then a < b ) a
2 2
For a < b, suppose that a >
fmultiplying both sides by 2g
) 2a > a + b ) a>b fsubtracting a from both sidesg which is false. Since the steps of the argument are correct, the supposition must be false and the alternative, a
0. )
x=
p q
where p, q 2 Z , q 6= 0 fand since x > 0, integers p, q > 0g
p q
3 =8 µ p ¶q ) 3q = 8q
)
) 3p = 8q which is impossible since for the given possible values of p and q, 3p is always odd and 8q is always even. Thus, the assumption is false and its opposite must be true. Hence x is irrational. Example 6: Prove that no positive integers x and y exist such that x2 ¡ y2 = 1. Proof (by contradiction): Suppose x, y 2 Z + exist such that x2 ¡ y 2 = 1. ) (x + y)(x ¡ y) = 1 ) x + y = 1 and x ¡ y = 1 or x + y = ¡1 and x ¡ y = ¡1 | {z } | {z } case 1 case 2 ) x = 1, y = 0 (from case 1) or x = ¡1, y = 0 (from case 2) Both cases provide a contradiction to x, y > 0. Thus, the supposition is false and its opposite is true. There do not exist positive integers x and y such that x2 ¡ y 2 = 1.
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Indirect proof often seems cleverly contrived, especially if no direct proof is forthcoming. It is perhaps more natural to seek a direct proof for the first attempt to prove a conjecture.
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134
APPENDICES
ERRORS IN PROOF One must be careful not to make errors in algebra or reasoning. Examine carefully the following examples. Example 7: Consider Example 5 again: Prove that the solution of 3x = 8 is irrational. 3x = 8
Invalid argument: )
log 3x = log 8
) x log 3 = log 8 log 8 ) x= log 3 ) x is irrational.
where both log 8 and log 3 are irrational.
The last step is not valid. The argument that an irrational divided by an irrational is always p p2
irrational is not correct. For example, Dividing by zero is not a valid operation.
2
= 1, and 1 is rational.
a is not defined for any a 2 R , in particular 00 = 6 1. 0
Example 8: Invalid “proof” that 5 = 2 0=0 )
0£5=0£2
)
0£5 0£2 = 0 0
)
fdividing through by 0g
5 = 2, which is clearly false.
This invalid step is not always obvious, as illustrated in the following example. Example 9: Invalid “proof” that 0 = 1: Suppose a = 1 )
a2 = a
)
a2 ¡ 1 = a ¡ 1
) (a + 1)(a ¡ 1) = a ¡ 1 )
a + 1 = 1 .... (¤)
)
a=0 So, 0 = 1
The invalid step in the argument is (¤) where we divide both sides by a ¡ 1. Since a = 1, a ¡ 1 = 0, and so we are dividing both sides by zero.
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Another trap to be avoided is to begin by assuming the result we wish to prove is true. This readily leads to invalid circular arguments.
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APPENDICES
Example 10: Prove without decimalisation that
p 3¡1 >
Invalid argument: p 3 ¡ 1 > p12 ³ ´2 p ) ( 3 ¡ 1)2 > p12 p ) 4 ¡ 2 3 > 12 p 7 ) 2 >2 3 p ) 7>4 3
p1
2
.
fboth sides are > 0, so we can square themg
)
72 > 48
fsquaring againg
)
49 > 48
which is true.
p Hence 3 ¡ 1 >
p Although 3 ¡ 1 >
p1
2
p1
135
is true.
2
is in fact true, the above argument is invalid because we began by
assuming the result. A valid method of proof for
p 3¡1>
p1
2
can be found by either:
² reversing the steps of the above argument, or by p ² using proof by contradiction (supposing 3 ¡ 1 6
p1
2
).
It is important to distinguish errors in proof from a false conjecture. Consider the table alongside, which shows values of n2 ¡ n + 41 for various values of n 2 N . From the many examples given, one might conjecture: “For all natural numbers n, n2 ¡ n + 41 is prime.” This conjecture is in fact false. For example, for n = 41, n2 ¡ n + 41 = 412 is clearly not prime.
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It takes only one counter-example to prove a conjecture is false.
n 1 2 3 4 5 6 7 8 9 10 11 12 13 ¢¢¢ 30 ¢¢¢ 99 ¢¢¢
n2 ¡ n + 41 41 43 47 53 61 71 83 97 113 131 151 173 197 ¢¢¢ 911 ¢¢¢ 9743 ¢¢¢
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136
APPENDICES
IMPLICATIONS AND THEIR CONVERSE If .... then .... Many statements in mathematics take the form of an implication “If A then B”, where A and B are themselves statements. The statement A is known as the hypothesis. The statement B is known as the conclusion. Implications can be written in many forms in addition to “If A then B”. For example, the following all have the same meaning:
8 9 implies > > > > > < so > = hence A B. > > > > thus > > : ; therefore Given a statement of the form “If A then B”, we can write a converse statement “If B then A”. If we know the truth, or otherwise, of a given statement, we can say nothing about the truth of the converse. It could be true or false. A statement and its converse are said to be (logically) independent. For example, suppose x is an integer. ² The statement odd” is false. ² The statement even” is true. ² The statement is also true. ² The statement is also false.
“If x is odd, then 2x is even” is true, but its converse “If 2x is even, then x is “If 2x is even, then x is odd” is false, but its converse “If x is odd, then 2x is “If x > 1, then ln x > 0” is true, and its converse “If ln x > 0, then x > 1” “If x = 5, then x2 = 16” is false, and its converse “If x2 = 16, then x = 5”
EXERCISE Prove or disprove: 1 If x is rational then 2x 6= 3. 2 If 2x 6= 3 then x is rational.
EQUIVALENCE Some conjectures with two statements A and B involve logical equivalence or simply equivalence. We say A is equivalent to B, or A is true if and only if B is true. The phrase “if and only if” is often written as “iff” or ,.
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A , B means A ) B and B ) A
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APPENDICES
In order to prove an equivalence, we need to prove both implications: A ) B
137
and B ) A.
For example: x2 = 9 , x = 3 is a false statement. x = 3 ) x2 = 9 is true but x2 = 9 6) x = 3 as x may be ¡3. Example 11: Prove that (n + 2)2 ¡ n2 is a multiple of 8 , n is odd. Proof: ())
(n + 2)2 ¡ n2
is a multiple of 8
2
) n + 4n + 4 ¡ n2 = 8a for some integer a ) 4n + 4 = 8a ) n + 1 = 2a ) n = 2a ¡ 1 ) n is odd. (()
n is odd ) n = 2a ¡ 1 for some integer a ) n + 1 = 2a ) 4n + 4 = 8a ) (n2 + 4n + 4) ¡ n2 = 8a ) (n + 2)2 ¡ n2
is a multiple of 8.
In the above example the ()) argument is clearly reversible to give the (() argument. However, this is not always the case. Example 12: Prove that for all x 2 Z + , x is not divisible by 3 , x2 ¡ 1 is divisible by 3. Proof: ())
x is not divisible by 3 ) either x = 3k + 1 or x = 3k + 2 for some k 2 Z + [ f0g ) x2 ¡ 1 = 9k2 + 6k or 9k2 + 12k + 3 = 3(3k2 + 2) or 3(3k 2 + 4k + 1) ) x2 ¡ 1 is divisible by 3.
(()
x2 ¡ 1 is divisible by 3 ) 3 j x2 ¡ 1 ) 3 j (x + 1)(x ¡ 1) ) 3 j (x + 1) or 3 j (x ¡ 1) fas 3 is a prime numberg ) 3-x
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or in other words, x is not divisible by 3.
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APPENDICES
NEGATION For any given statement A, we write not A or :A to represent the negation of the statement A. A x>0
:A x60
x is prime
x is not prime
x is an integer
x is not an integer
x is rational z is real
x is irrational z = a + bi, a, b 2 R , b 6= 0
x is a multiple of 3
x is not a multiple of 3 or x = 3k + 1 or 3k + 2 for k 2 Z + [ f0g
For example:
For x 2 R : For z 2 C : For x 2 Z + [ f0g:
PROOF OF THE CONTRAPOSITIVE To prove the statement “If A then B”, we can provide a direct proof, or we can prove the logically equivalent contrapositive statement “If not B, then not A” which we can also write as “If :B, then :A”. For example, the statement “If it is Jon’s bicycle, then it is blue” is logically equivalent to “If that bicycle is not blue, then it is not Jon’s”. Example 13: Prove that for a, b 2 R , “ab is irrational ) either a or b is irrational”. Proof using contrapositive: a and b are both rational ) a =
p r and b = where p, q, r, s 2 Z , q 6= 0, s 6= 0 q s
) ab =
µ ¶³ ´ p q
r s
=
pr fwhere qs 6= 0, since q, s 6= 0g qs
) ab is rational
fsince pr, qs 2 Z g
Thus ab is irrational ) either a or b is irrational. Example 14: Prove that if n is a positive integer of the form 3k + 2, k > 0, k 2 Z , then n is not a square. Proof using contrapositive: If n is a square then n has one of the forms (3a)2 , (3a + 1)2 or (3a + 2)2 , where a 2 Z + [ f0g. ) n = 9a2 , 9a2 + 6a + 1 or 9a2 + 12a + 4 ) n = 3(3a2 ), 3(3a2 + 2a) + 1 or 3(3a2 + 4a + 1) + 1 ) n has the form 3k or 3k + 1 only, where k 2 Z + [ f0g ) n does not have form 3k + 2.
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Thus if n is a positive integer of the form 3k + 2, k > 0, k 2 Z , then n is not a square.
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APPENDICES
139
USING PREVIOUS RESULTS In mathematics we build up collections of important and useful results, each depending on previously proven statements. Example 15: Prove the conjecture: “The recurring decimal 0:9 = 0:999 999 99:::: is exactly equal to 1”. Proof (by contradiction): Suppose 0:9 < 1 ) 0:9
1 is absurd, 0:9 = 1. Proof (Direct Proof): 0:9 = 0:999 999 99:::: = 0:9 + 0:09 + 0:009 + 0:0009 + :::: ¡ ¢ 1 1 1 = 0:9 1 + 10 + 100 + 1000 + :::: µ1 ¶ P ¡ 1 ¢i 9 = 10 10 i=0 Ã !
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10 9
0
9 = 10 £ =1
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fUsing the previously¯ proved Geometric Series ¯ 1 1 ¯ with r = 10 and ¯ 10 < 1g
1 1 ¡ 10
5
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1
9 10
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=
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APPENDICES
THEORY OF KNOWLEDGE
AXIOMS AND OCCAM’S RAZOR
In order to understand complicated concepts, we often try to break them down into simpler components. But when mathematicians try to understand the foundations of a particular branch of the subject, they consider the question “What is the minimal set of assumptions from which all other results can be deduced or proved?” The assumptions they make are called axioms. Whether the axioms accurately reflect properties observed in the physical world is less important to pure mathematicians than the theory which can be developed and deduced from the axioms. Occam’s razor is a principle of economy that among competing hypotheses, the one that makes the fewest assumptions should be selected. 1 What value does Occam’s razor have in understanding the long-held belief that the world was flat? 2 Is the simplest explanation to something always true? 3 Is it reasonable to construct a set of mathematical axioms under Occam’s razor? One of the most famous examples of a set of axioms is given by Euclid in his set of 13 books called Elements. He gives five axioms, which he calls “postulates”, as the basis for his study of Geometry: 1. Any two points can be joined by a straight line. 2. Any straight line segment can be extended indefinitely in a straight line. 3. Given any straight line segment, a circle can be drawn having the segment as radius and one endpoint as centre. 4. All right angles are congruent. 5. Parallel postulate: If two lines intersect a third in such a way that the sum of the inner angles on one side is less than two right angles, then the two lines inevitably must intersect each other on that side if extended far enough. 4 Is the parallel postulate genuinely an axiom, or can it be proved from the others? 5 What happens if you change the list of axioms or do not include the parallel postulate?
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6 What other areas of mathematics can we reduce to a concise list of axioms?
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141
APPENDICES
APPENDIX B:
FORMAL DEFINITION OF A LIMIT
The informal definition of a limit we were given in Section B is sufficient for our study here. However, for completeness we justify its use by considering and interpreting the formal definition: The formal definition of a limit of a function: Let f be a function defined in an open interval about x = a, except f (a) need not be defined. The number l is called the limit of f as x approaches a if, for any value " > 0, there exists a corresponding value ± > 0 such that 0 < jx ¡ aj < ± ensures jf(x) ¡ lj < ". We write lim f(x) = l to denote “the limit of f as x approaches a, is l”. x!a
To interpret this definition, note that l and a are constants, and x and f(x) are variables. The value " > 0 is any positive value, but in particular we consider " a very, very small positive value. In this case
y
jf(x) ¡ lj < " , ¡" < f (x) ¡ l < " , l ¡ " < f (x) < l + " , the value f (x) is very, very close to l
l+" l l-"
f (x) will lie in the shaded band of values shown.
x
Similarly, the value ± > 0 is a positive value, and in particular we consider ± a very, very small positive value. y 0 < jx ¡ aj < ± , x 6= a and ¡± < x ¡ a < ± , x 6= a and a ¡ ± < x < a + ± so that x is very, very close to, but not equal to, the value a.
In this case
a-± a a+±
x
Putting this all together, we have that f has limit l as x approaches a if for any choice of " > 0, there always exists a value ± > 0 such that the value f (x) will be within distance " from l, whenever x is within distance ± from a, but not equal to a. In particular, this must be true for " > 0 chosen as small as we like. y y = f(x) l+" l l-"
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Y:\HAESE\IB_HL_OPT-Stat-Prob\IB_HL_OPT-Stat-Prob_AA\141IB_HL_OPT-Stat-Prob_AA.cdr Tuesday, 26 February 2013 10:14:10 AM BRIAN
142
APPENDICES
These algebraic and geometric interpretations of the formal definition lead to the informal definition of the limit of a function given in Section B.
ESTABLISHING LIMITS By considering the identity function f (x) = x, we intuitively see that lim x = a for any constant a 2 R .
y
x!a
This is proved formally as follows: y = f(x) = x
l=a
Suppose f(x) = x and a 2 R is a constant. For any " > 0, there exists ± = " > 0 such that 0 < jx ¡ aj < ± ) jx ¡ aj < "
fsince ± = "g
) jf (x) ¡ aj < "
fsince f (x) = xg
a
x
Since an appropriate ± can be found for any choice of " > 0, by the formal definition of a limit, lim x = a, for a any constant.
x!a
Arguments such as this can be used to establish the limit laws presented in Section B. When we define new functions using a sum, composition, or some other combination of simpler functions, the limit laws help us calculate limits for these new functions using what we already know about the simpler functions. 3(x2 ¡ 1) . x!1 x¡1
Example: Determine the limit lim Using Limit Laws 3(x2 ¡ 1) x¡1 3(x ¡ 1)(x + 1) = lim x!1 (x ¡ 1)
lim
x!1
= lim 3(x + 1)
fusing Limit Laws with
x!1
x¡1 = 1 a constant, since x 6= 1g x¡1
= lim (3x + 3) x!1 ³ ´³ ´ ³ ´ 3 lim x + lim 3 fby the Limit Laws, since each limit existsg = xlim !1 x!1 x!1 fsince lim 3 = 3 from the Limit Laws
=3£1+3
x!1
and lim x = 1 from above using a = 1g
=6
x!1
Using the formal definition directly (not required for this course) From the graph of y =
3(x2 ¡ 1) we observe that for x near 1, f (x) is close to the value 6. x¡1
3(x2 ¡ 1) = 6, and must prove this formally. x!1 x¡1
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We conjecture that lim
IB HL OPT 2ed
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Y:\HAESE\IB_HL_OPT-Stat-Prob\IB_HL_OPT-Stat-Prob_AA\142IB_HL_OPT-Stat-Prob_AA.cdr Wednesday, 20 February 2013 1:35:11 PM GR8GREG
APPENDICES
143
¯ ¯ ¯ 3(x2 ¡ 1) ¯ ¯ For x 6= 1, jf (x) ¡ 6j < " , ¯ ¡ 6¯¯ < " x¡1 ¯ ¯ ¯ 3(x ¡ 1)(x + 1) ¯ ¯ ¡ 6¯¯ < " , ¯ (x ¡ 1) , , , ,
j3(x + 1) ¡ 6j < " j3x + 3 ¡ 6j < " j3x ¡ 3j < " 3 jx ¡ 1j < "
, jx ¡ 1j
0 such that 0 < jx ¡ 1j < ± implies jf (x) ¡ 6j < ". 3 3(x2 ¡ 1) = 6. Thus by the formal definition of the limit, lim x!1 x¡1
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Therefore, for all " > 0, there exists ± =
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Y:\HAESE\IB_HL_OPT-Stat-Prob\IB_HL_OPT-Stat-Prob_AA\143IB_HL_OPT-Stat-Prob_AA.cdr Tuesday, 19 February 2013 4:18:05 PM BRIAN
WORKED SOLUTIONS 6 jx ¡ aj
0, jaj = a ) jaj > 0 .... (1) If a < 0, jaj = ¡a where ¡a > 0 ) jaj > 0 .... (2) From (1) and (2), jaj > 0 for all a 2 R . if ¡a > 0 0 if ¡a = 0 ¡(¡a) if ¡a < 0
= =
n" > 1 ) " >
if a < 0 if a = 0 if a > 0
0 a
na
= jaj
(2) If Pk is true, then (1 + x)k > 1 + kx, k 2 Z + .
or j¡aj = j¡1 £ aj = j¡1j jaj = 1 jaj = jaj
Now x > ¡1 ) 1 + x > 0
f jabj = jaj jbj g
)
)
y
(1 + x)k+1 > 1 + (k + 1)x, as kx2 > 0
® lies in A, but is smaller than ® 2
) ) )
r + x is irrational. c Similarly suppose rx = , c, d 2 Z , d 6= 0 d ) x=
a
y
x
b
1
a
)
lim (¡x) DNE.
x!1
x y = -x
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As x ! 1, ¡x ! ¡1
y
fTriangle inequalityg
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c which 2 Q , a contradiction dr
) rx is irrational.
If two points lie in a particular interval [a, b] on the number line then the distance between them is less than the length of the interval.
0
a where a, b 2 Z , b 6= 0 b a x = ¡ r which 2 Q , a contradiction b
r+x =
EXERCISE B.1
b
5 ja ¡ bj = j(a ¡ c) + (c ¡ b)j 6 ja ¡ cj + jc ¡ bj
® _ "w
¼¡ , y ! 1 below 2
lim f (x) = 1 ¼¡
As x !
x
´ 1
lim f (x) = 1 ¼+
x! 2
c Since
lim f (x) = ¼¡
x! 2
lim f (x) = 1, ¼+
x! 2
lim f (x) = 1 ¼
x! 2
DNE.
x
¼+ , y ! 1 below 2
4
a
y
y = -Qv
2
a
x
x! 2
) d
2¼
x+1 y = ^^^% x - 1_ y=1
y -1 -1
y = 3x + 2
x
x
(-1, -1)
x=1
b ¡
As x ! ¡1 , As x ! ¡1+ , )
ii As x ! 1, y ! 1 above iii As x ! 1¡ , As x ! 1+ ,
lim (3x + 2) = ¡1
x!¡1
b f (x) = )
3x + 2 ! ¡1 below 3x + 2 ! ¡1 above
(x + 2)(x ¡ 1) x2 + x ¡ 2 = x+2 x+2
f (x) =
i As x ! ¡1, y ! 1 below
5
y ! ¡1 y!1
x2 + x - 2_ f(x) = ^^^^^^%^ x+2
(3, 5)
x
y = 3x, x < 3 3
y ! 9 below y ! 9 above
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c Yes the limit exists fas both LH and RH limits are 9g lim f (x) = 9
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x2 + x ¡ 2 = ¡3 x+2
5
lim
x!¡2
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b As x ! 3¡ , As x ! 3+ ,
y ! ¡3 below y ! ¡3 above
75
As x ! ¡2¡ , As x ! ¡2+ ,
lim f (x) DNE.
x!1
(3, 9) missing point
y
-1
)
lim f (x) = 1
x!1
y = x2, x > 3
if x 2 R , x 6= ¡2 undefined if x = ¡2
missing point (-2, -3)
)
lim f (x) = 1
x!¡1
y
a
nx ¡ 1
1
o
)
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Y:\HAESE\IB_HL_OPT-Calculus\IB_HL_OPT-Calculus_AN\146IB_HL_OPT-Calculus_an.cdr Friday, 8 March 2013 10:44:11 AM BRIAN
IB HL OPT 2ed Calculus
WORKED SOLUTIONS p x does not exist
6 For x < 0, )
lim
lim
x+1 x2 ¡ 2x ¡ 3 x+1 = lim x!¡1 (x + 1)(x ¡ 3) lim
x!¡1
DNE.
x!0¡
So, although 7
b
p
x!0+
x = 0, lim
p
x DNE.
x!0
a
= ¡ 14
y y = 1, x > 0
1
=
-1
d
=5
lim f (x) = ¡1
x!0¡
lim f (x) = 1
=
x!0+
=
iii Since the limits in i and ii are different, lim f (x) DNE. x!0
b
=
y
¡
1 1 ¡ 15 ¡3 2 3 ¡ 13 £ 10 1 ¡ 10
y!5
1 -"w
¢
µ
1 1 ¡ y 5
1 1 ¡ y 5
5¡y 5y(y ¡ 5)1 ¡1 = lim y!5 5y
x2 - 3x + 5 f(x) = ^^^%^^%^ , x > 0 5__ _ _ _ _ _ _ _ _ _ _ _ _ _ _ x
-2¼
µ
1 y¡5
lim
f
f(x) = sin_x, x < 0
-¼
1 lim y!2 y ¡ 5
e
ii As x ! 0+ , y = 1
-Es"
x2 + 3x ¡ 4 x¡1 (x ¡ 1)(x + 4) = lim x!1 x¡1 lim
x!1
i As x ! 0¡ , y ! ¡1 below
)
x!0 ¡4 ¡1
=4
x y = (x - 1)3, x < 0
fx 6= ¡1g
x2 + 3x ¡ 4 x¡1
lim
c
)
147
-1
= lim
¶
¶
fy 6= 5g
y!5
-1
fx 6= 1g
1 = ¡ 25
i As x ! 0¡ , y ! 0 ii As x !
0+ ,
)
y!1
)
lim f (x) = 0
x!0¡
lim f (x) = 1
2
x!0+
)
iii Since the limits in i and ii are different, lim f (x) DNE.
)
x!0
c
y
i As x ! 0¡ , y ! 0
)
y!0
)
lim f (x) = 0
=0
x!0¡
d
lim f (x) = 0
x!0+
)
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lim
µ!0¡
DNE
tan x sec x
lim
sin x cos x cos x 1
¼¡ x! 2
=
cos µ µ
¼¡
lim µ = 0
µ!0¡
lim
x! 2
f cos x 6= 0g
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=1
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0
x+1 x2 ¡ 2x ¡ 3
5
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5
=
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=
lim cos µ = 1 but
µ!0¡
e
lim
DNE
sin x 0 = =0 ex 1 sin x c lim x!¼¡ 1 ¡ cos x 0 = 1 ¡ (¡1)
EXERCISE B.2 x!1 2 ¡4 ¡ 12
p 2¡x =0
x!0
x!0
a
x!2¡
lim
x!2¡
b lim
iii lim f (x) = 0
1
x!2¡
ln x lim p 2¡x
)
x
ii As x !
lim ln x 6= 0 but
Now f(x) = sin(_Qv_), x 6 = 0
0+ ,
x0 p 2 ¡ x exists.
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IB HL OPT 2ed Calculus
148 3
WORKED SOLUTIONS 3+x x+5
lim
a
x!1
3 x
= lim
lim
b
x!1
+1
= lim
5 1+ x
x!1
=1
r
lim
c
x!1
r
= lim
lim
d
x!1 x2
)
= lim
x2 + x ¡ x
= lim
³p
x!1
= lim
x!1
= lim
x!1
x2 + x + x
cos x sin x ³ x ´ = lim cos x x!0 sin x = lim x x!0
=1£1 =1
!
µ
x!0
x!0
= lim
fas x > 0g
x!0
=
+1
g
1 2
lim
x!0+
x!a
, ,
x!a
³ sin x ´ p
x!a
x + sin x = lim a lim x!0 x ¡ sin x x!0
2
But as x ! 0, 1 ¡ )
sin 3µ b lim µ!0 µ ³ sin 3µ ´ = lim 3 µ!0 3µ ³ sin 3µ ´ = 3 lim 3µ!0 3µ
x
x
b lim
h!0
= lim
h!0
´
¡ sin2 h h(cos h + 1)
³ sin h ´2 h
h!0
= 0 £ 12 £
1 cos h + 1
1 2
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´³
= lim ¡h
=3
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cos h ¡ 1 cos h ¡ 1 cos h + 1 = lim h!0 h h cos h + 1 cos2 h ¡ 1 = lim h!0 h(cos h + 1)
= 3£1
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sin x x sin x 1¡ x 1+
sin x !1¡1=0 x
the limit DNE.
3µ ! 0 alsog
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sin 2x
£1
x!a
lim (f (x) ¡ l) = 0
fas µ ! 0,
50
sin 2x 1 2
³ 2x ´i
1 2
x!a
sin2 µ a lim µ!0 µ ³ sin µ ´ = lim sin µ µ!0 µ
75
2
+
lim f (x) ¡ lim l = 0
=1£0 =0
25
h x ³ 2x ´
¶
lim f (x) = lim l
x!a
EXERCISE B.3
0
x2 x + sin 2x sin 2x
=1£0 =0
5 If lim f (x) = l ,
5
¶
sin x p = lim x x!0+
1 2
=
µ
x2 + x sin 2x
=0£1+
1 = 1+1
1
lim
f
1
1+
£1
lim x cot x
= lim
1 1+ x +x 1 x
=1
=
7 4 7 4
x!0
x2 + x + x
q q
=
e
p
x
=1£1
µ!0
p
x2 + x + x x
cos µ
fSince lim
x2 + x ¡ x2
= lim
x!1
´
x2 + x ¡ x
sin µ
sin µ = 1, µ µ = 1g lim µ!0 sin µ
1+x +x+1
0+0 1+0+0
Ãp
sin 7x 4x ³ sin 7x ´ 7 = lim 4 x!0 7x ³ sin 7x ´ = 74 £ lim 7x!0 7x lim
x!0
³ µ ´
µ!0
p
x!1
d
cos µ
=0
the limit DNE.
lim
4
=
µ tan µ µ
µ!0 sin µ
= lim
3 !1 x
x+
= lim
1 1 +x x2 1 x!1 1+ x + x12
3 x
But as x ! 1,
q
5 4¡ x + x12
1 1+ x + x12 4¡0+0 = 1+0+0
+3 x
x+
x!1
µ!0
=4
x2
lim
c
x!1
0+1 1+0
=
4x2 ¡ 5x + 1 x2 + x + 1
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Y:\HAESE\IB_HL_OPT-Calculus\IB_HL_OPT-Calculus_an\148IB_HL_OPT-Calculus_an.cdr Thursday, 7 March 2013 9:04:48 AM BRIAN
IB HL OPT 2ed Calculus
WORKED SOLUTIONS
³
´³
1 ¡ cos x 1 ¡ cos x 1 + cos x = lim x!0 x2 x2 1 + cos x sin2 x = lim 2 x!0 x (1 + cos x)
c lim
x!0
³ sin x ´2
= lim 1
x
x!0
= 1 £ 12 £
´
Now 0 6 jxj 6 2x )
0 jxj 2x 6 6 1 + x4 1 + x4 1 + x4 0 2x ! =0 1 + x4 1
As x ! 0+ ,
1 1 + cos x
)
1 2
1 2
=
149
lim
x!0+
jxj = 0 .... (1) fSqueeze Theoremg 1 + x4
Consider x < 0
y = -2x
y y = |x|
3
³1´
a As ¡1 6 cos
y=0
³1´
¡x2 6 x2 cos
x
6 1, then
x2
x2
6 x2
But lim (¡x2 ) = lim (x2 ) = 0 x!0
³1´
lim x2 cos
)
Now 0 6 jxj 6 ¡2x
x!0
x2
x!0
³1´
b As ¡1 6 sin
³1´ x
6 x fif x > 0g
)
But lim (¡x) = lim (x) = 0 x!0
)
lim x sin
x!0
However, if x < 0 ³ ´ ³ ´ 1 1 ¡1 6 sin 6 1 ) x 6 x sin 6 ¡x x x
EXERCISE C.1 1
a Let F (x) = f (x)g(x) As f (x) and g(x) are defined at x = a, F (a) = f (a)g(a) is also defined at x = a.
and as lim (¡x) = lim (x) = 0, x!0
)
x!0
³1´
lim x sin
x
x!0
jxj = 0 .... (2) fSqueeze Theoremg 1 + x4
= 0 fSqueeze Theoremg
x
x!0
lim
x!0¡
Now lim F (x) = lim [f (x)g(x)]
= 0 fSqueeze Theoremg
x!a
x!a
= lim f (x) £ lim g(x)
1
x!a
c As ¡1 6 sin x 6 1 and e x > 0 for all x 2 R , ¡
1 1 ex
sin x
6
e
6
1 x
1
= F (a)
1 ex
) 1
à )
lim
¡
x!0
¡ )
lim e
!
1 e
1 ¡x
Ã
= lim
1 x
x!0
¢
x!0
1 e
As f (x) and g(x) are defined at x = a, F (a) = f (a) § g(a) is also defined at x = a.
=0
1 x
f (x) g(x) is continuous at x = a.
b Let F (x) = f (x) § g(x)
!
Now lim F (x) = lim (f (x) § g(x)) x!a
x!a
= lim f (x) § lim g(x)
sin x = 0 fSqueeze Theoremg
x!a
y = 2x
y
= F (a) )
y = |x| y=0
f (x) § g(x) is continuous at x = a.
c Let F (x) =
x
f (x) where g(x) 6= 0 g(x)
Now F (a) =
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F (a) is defined.
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)
5
x!a
= f (a) § g(a)
d Consider x > 0
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x!a
= f (a) £ g(a)
1 ! 1 and e x ! 1 x
But as x ! 0,
¡2x !0 1 + x4
From (1) and (2), lim g(x) = 0
x!0
³1´
0 jxj ¡2x 6 6 1 + x4 1 + x4 1 + x4
But, as x ! 0¡ ,
6 1, then
x
¡x 6 x sin
)
= 0 fSqueeze Theoremg
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Y:\HAESE\IB_HL_OPT-Calculus\IB_HL_OPT-Calculus_an\149IB_HL_OPT-Calculus_an.cdr Thursday, 7 March 2013 1:44:34 PM BRIAN
IB HL OPT 2ed Calculus
150
WORKED SOLUTIONS Now lim F (x) = lim x!a
x!a
f (x) g(x)
4
y
a 12
y = x2
lim f (x)
x!a
=
(3, 9)
9
lim g(x)
x!a
)
6
f (a) g(a)
=
3
= F (a) f (x) is continuous at x = a. g(x)
y = 3x
½
Now lim F (x) = lim c f (x)
f (x) =
x!a
= c lim f (x) x!a
b
= c f (a)
y=5
5
e Let F (x) = [f (x)]n , n 2 Z +
2
) F (a) = [f (a)]n is defined as f (a) is defined. Now lim F (x) = lim [f (x)]n
h
x!a
=
lim f (x)
[f (x)]n
3
-15
f (x) is continuous for all x 2 R , x 6= 3. At x = 3 we have a ‘jump’ discontinuity which is essential.
a
a
y
=1+1+1 =3 ) let k = 3
5__ _ _ _ _ y = ^^^^ 4-x
b
sin 3x x ³ sin 3x ´ = lim 3 x!0 3x ³ sin 3x ´ = 3 £ lim 3x!0 3x lim
x!0
y = 2x
8 y=5
4
x3 ¡ 1 x¡1 (x ¡ 1)(x2 + x + 1) = lim x!1 x¡1 lim
x!1
i f has an essential discontinuity at x = 6. ii f has a removable discontinuity at x = 5. b f is continuous on fx j x 2 R , x 6= 5 or 6g
=3£1
c f (2) = 22 = 4 ) let k = 4 1 d y = has an essential discontinuity at x = 0 which x cannot be removed ) no value for k can be found.
-4
x=4
cyan
magenta
yellow
50
25
0
e y = kx represents the infinite number of lines passing through O(0, 0). ) k2R
5
95
100
50
75
25
0
5
95
100
50
75
25
0
5
95
f is continuous at x = ¡2. f has a ‘jump’ discontinuity at x = 0. f has a removable discontinuity at x = 3. f has a ‘break’ discontinuity at x = 4 due to the vertical asymptote at x = 4.
100
50
75
25
0
5
i ii iii iv
fas x ! 0, 3x ! 0g
=3 ) let k = 3
x
95
4
100
2
75
-2
y = x2 - 6
b
x
x!a
= F (a) is continuous at x = a.
y=x
8
(3, -14)
a
-4
6
y = x2 - 10x + 7
-10
in
5 2
4
-5
= [f (a)]n )
x>3 x 0, then k > ¡2, and f (x) has graph: y
) x
1 n
1
= (eln x ) n
1
1
) xn = en
(2, -k - 2)
1
x
1
y=x+k (2, k + 2)
1
lim f (g(n)) = f
From 7 c,
)
a f (x) is continuous for all x 2 R and ) on [0, 2]. f (0) = ¡3 and f (2) = 7 Thus, with k = 0, f (0) 6 k 6 f (2) Hence, by the IVT, there exists c 2 [0, 2] such that f (c) = 0 ) f (x) has a zero on [0, 2]. y b
1 Z+
But g(a) = 1 since a 2 Q lim g(xn ) 6= g(a)
n!1
lim xn = a
n!1
)
(3, 4)
g(x) is discontinuous at x = a.
b If xn is the decimal expansion of a to n decimal places then xn 2 Q )
1
x
(1, -4)
lim g(xn ) = 1 fDirichlet functiong
x=2
f (x) has an essential discontinuity at x = 2 ) the IVT will not apply. Clearly for [1, 2[ , f (x) < 0 and for ]2, 3], f (x) > 0 ) f (x) is never zero on [1, 3].
lim g(xn ) 6= g(a)
n!1
lim xn = a
Also
n!1
)
3
n!1
But g(a) = 0 since a 2 =Q )
= f (ln 1)
EXERCISE C.2
lim g(xn ) = 0 fDirichlet functiong
But
n!1
n!1
= e0 =1
n!1
)
lim x
1 n
´
lim g(n)
= f (0)
p 2 , a2Q n
Since a 2 Q , xn 2 = Q for all n 2 )
³
n!1
) f (x) is discontinuous f (x) is continuous at x = 2 for all x 2 R So, if k > ¡2, k 2 R , f (x) is continuous and if k < ¡2, k 2 R , f (x) is continuous for all x 6= 2. )
a xn = a +
1 n
b From a, x n = f (g(n)) where f (n) = en , g(n) = ln x n
x
6
ln x
) x n = eln x
y = -k - 2
(2, k + 2)
x = eln x
a
y y=x+k
y=k+2
8
151
g(x) is discontinuous at x = a.
2 f (x) = x3 ¡ 9x2 + 24x ¡ 10 has graph:
c From a and b, the Dirichlet function is nowhere continuous.
y
7
(2, 10)
(5, 10)
(1, 6)
(4, 6)
a If g(x) is continuous at x = a, then lim g(x) = g(a) = l
.... (1)
x!a
If f (x) is continuous at g(a) then x
f (g(x)) = f (g(a)) = f (l)
f 0 (x) = 3x2 ¡ 18x + 24 ) (2, 10) is a local maximum (4, 6) is a local minimum. = 3(x2 ¡ 6x + 8) = 3(x ¡ 2)(x ¡ 4) a On [1, 5], m = 10 and xm = 2 or 5 + - + b On [1, 5], m = 6 and xm = 1 or 4 2 4
But g(x) ! g(a) when x ! a lim f (g(x)) = f (l)
x!a
b From a, if lim g(x) exists then, x!a
x!a
cyan
´
lim g(x)
magenta
yellow
25
0
5
95
x!1
100
95
50
75
25
0
5
95
100
50
75
25
0
5
x!1
³
50
lim f (g(x)) = f (l) = f
75
x!1
25
lim g(x) exists then
3 As f (a) and f (b) have opposite signs, then f (a) 6 f (b) or f (b) 6 f (a) By the IVT, there exists c 2 [a, b] such that f (a) 6 f (c) = 0 6 f (b) or f (b) 6 f (c) = 0 6 f (a) ) there exists a zero c 2 [a, b] ) there exists at least one zero of f (x).
ffrom (1)g
95
´
lim g(x)
0
c If
³
5
=f
100
x!a
50
lim f (g(x)) = f (l)
100
)
75
lim
g(x)!g(a)
black
Y:\HAESE\IB_HL_OPT-Calculus\IB_HL_OPT-Calculus_an\151IB_HL_OPT-Calculus_an.cdr Friday, 8 March 2013 10:48:05 AM BRIAN
IB HL OPT 2ed Calculus
152
WORKED SOLUTIONS a Let f (x) = sin x + rx ¡ 1, r > 0 for x 2 ]0,
4
Now f (0) = ¡1 < 0 and f Since f (0) and f exists c 2 [0,
³¼´ 2
³¼´ 2
n
¼ [. 2
h!0
¼ =r >0 2
2 f (5) = 0 is defined. lim f (x) = lim (5 ¡ x) = 0 and
x!5¡
are opposite in sign, by the IVT there
y y = sin_x 1
A
-2 2
1
lim f (x) = lim (x ¡ 5) = 0
)
a If f (x) =
5
x!5
Thus f (x) is continuous at x = 5.
) f (x) = 0 ) x ¼ 0:511
Now f (x) =
) f (x) is not differentiable at x = 5 3 f (x) =
4 f (x) =
= ¡1 + )
y
(1, 13)
(1, 10) y = 3x + 10, x1
¡1 6 f (¡1) 6 ¡0:4146
0 (1) = 3 i f¡
b
Since f is continuous for all x, r 2 R , f is also continuous on [¡1, 1] and since f (¡1) < 0 < f (1) )
x>1 x0
0 (0) 6= f 0 (0) f+ ¡
1 + 1 + sin2 (1) + (r + 1)2
and also
n x + 2,
x2 + 3x, x < 0 has an essential ‘jump’ discontinuity at x = 0 ) f (x) is not continuous at x = 0 ) f (x) is not differentiable at x = 0 0 (0) = 0 and f 0 (0) = 2(0) + 3 = 3 and Note: f+ ¡
for x = §1 and r 2 R . f (1) = (1)
x>5 x 0 for r 2 R 1 ) 06 6 0:5854 1 + sin2 x + (r + 1)2
15
n x ¡ 5,
0 0 ) f¡ (x) = ¡1 and f+ (x) = 1
x
1 + 1 + sin2 x + (r + 1)2
So, for r 2 R ,
x!5+
lim f (x) = 0 = f (5)
When r = 1, at A, sin x = 1 ¡ x
y=1-x
x15
x!5¡
x!5+
¼ ] such that f (c) = 0 2
b
o
sin h cos h ¡ 1 = 1 and lim = 0 fExample 6g h h!0 h
lim
0 (1) = ¡2(1) + 5 = 3 ii f+ 0 (1) = f 0 (1), f (x) is not continuous at c No, although f¡ + x = 1.
there exists an x 2 [¡1, 1] such that f (x) = 0.
1 , where r 2 R , r > 0, there is a 5x discontinuity at x = 0. f is not continuous on the interval [¡1, 1], so the IVT cannot be used to prove the existence of a zero.
b If g(x) = rx17 +
5
a
y 2 y = x2 + x + 1,
EXERCISE D 1 If f (x) = cos x
y = 1 + sin_x, x > 0
1
x1 x 2
0 f+ (2) 0 f¡ (2)
sin(x ¡ 1) + cx, x2 ¡ x + d,
f (1) = sin 0 + c = c is defined.
As f (x) is discontinuous at x = 0, it is not differentiable at x = 0.
)
d = ¡1
c = 2, d = ¡1
b f (x) =
c
x!1+
lim f (x) = f (1) , c + d = 1 .... (1)
0 (1) = 2(1) = 2 and f 0 (1) = c Now f¡ + 0 ) f (1) exists , c = 2
y = cos_x, x > 0
(0, 1)
x!1¡
lim f (x) = lim (cx + d) = c + d
and
) f (x) is differentiable at x = 0 b
lim f (x) = lim x2 = 1
Also,
0 0 ) f¡ (0) = f+ (0) = 1
) and
= 8(2) = 16
x!0
0 6= f+ (2)
x!0
) lim f (x) = 0
fSqueeze Theoremg
x!0
) f (x) is not differentiable at x = 2
) lim f (x) = f (0) x!0
) f (x) is continuous at x = 0
6 f (0) = k sin 0 = 0 ) f (0) is defined. Also,
f 0 (x) = 3x2 sin
lim f (x) = lim (k sin x) = k(0) = 0
where
x!0¡
and
x!0¡
x!0+
)
Consider x > 0
lim f (x) = lim (tan x) = tan 0 = 0 x!0+
¡1 6 sin
lim f (x) = 0 = f (0)
x!0
) f (x) is continuous at x = 0 for all k 2 R .
and
1
¡1 6 cos
0 Now f¡ (0) = sec2 (0) = 2 = 1 and 1
³1´ x
³1´ x
³ ´ 1 x
6 1 ) x > ¡x cos
7
Thus,
f (x) is differentiable at x = 0 , k = 1
a f (x) =
x2 , cx + d,
¡3x2 ¡ x 6 3x2 sin
But
x
¡ x cos
³1´
95
x
> ¡x
x ³ ´ 1 x
³1´ x
6 3x2
6x
6 3x2 + x
x!0+
0
lim f (x) = 0 .... (1) fSqueeze Theoremg
x!0+
100
50
75
25
0
5
95
50
75
25
0
100
yellow
³1´
lim (¡3x2 ¡ x) = 0 = lim (3x2 + x)
x!0+
)
5
95
50
75
25
0
5
95
100
50
75
25
0
5
100
magenta
³1´
or ¡3x2 ¡ x 6 f 0 (x) 6 3x2 + x
x61 x>1
f (1) = 12 = 1 is defined.
cyan
1 x
) ¡x 6 ¡x cos
0 0 ) f¡ (0) = f+ (0) , k = 1
½
³ ´
6 1 ) ¡3x2 6 3x2 sin
0 (0) = k cos(0) = k f+
)
¡ x cos
black
Y:\HAESE\IB_HL_OPT-Calculus\IB_HL_OPT-Calculus_an\153IB_HL_OPT-Calculus_an.cdr Thursday, 7 March 2013 1:50:59 PM BRIAN
IB HL OPT 2ed Calculus
154
WORKED SOLUTIONS )
Consider x < 0 as 3x2 > 0, 2
2
¡3x 6 3x sin and as ¡x > 0,
³1´ x
³1´
x 6 ¡x cos
x
Thus, ¡3x2 + x 6 3x2 sin
³1´ x
2
)
2
6 3x
lim
ex ex = lim which does not exist x x!1 1
lim
ex x
x!1
x!1
e As x ! 0+ , x ! 0, ln x ! ¡1 The limit has type 0 £ 1, so we can use l’Hˆopital’s Rule.
6 ¡x
¡ x cos
³1´ x
2
)
lim x ln x = lim
x!0+
x!0+
= lim
x!0+
lim (¡3x2 + x) = 0 = lim (3x2 ¡ x)
x!0¡
=0
x!0¡
f As x ! 0, arctan x ! 0
b
f 0 (x)
=
x!0
3x2 sin
³1´ x
0,
¡ x cos
³1´ x
The limit has type ,
x 6= 0
)
x=0
lim
x!0
x!0+
)
lim f 0 (x) = 0 = f 0 (0)
x!0¡
lim
)
lim
x!0+
a As x ! 0, 1 ¡ cos x ! 0 and x2 ! 0
x!0
= 0 , 0
Note:
x!0
=
The limit has type )
so we again use l’Hˆopital’s Rule.
0 , 0
lim
x!1
The limit has type 0 £ 1, so we can use l’Hˆopital’s Rule.
1 2
)
lim x2 ln x = lim
x!0+
= lim
x!0+
1 x
¡2x¡3
= lim ¡ 12 x2 x!0+
magenta
yellow
95
100
50
75
25
0
5
95
100
50
75
25
0
5
95
so we can use l’Hˆopital’s Rule.
100
50
1 , 1
ln x x¡2
=0
75
25
0
5
95
100
50
75
25
0
5
x!0+
so we can use l’Hˆopital’s Rule.
d As x ! 1, ex ! 1
cyan
so we can use l’Hˆopital’s Rule.
j As x ! 0+ , x2 ! 0 and ln x ! ¡1
ln x 1 x = lim = lim =1 x!1 x x ¡ 1 x!1 1
The limit has type
0 , 0
x + sin x 1 + cos x = lim lim x!0 x ¡ sin x x!0 1 ¡ cos x
1
)
´
which DNE as 1 + cos x ! 2 and 1 ¡ cos x ! 0
c As x ! 1, ln x ! 0 and x ¡ 1 ! 0 The limit has type
x!0+
³
p sin x sin x x p = lim + x x x!0
i As x ! 0, x + sin x ! 0 and x ¡ sin x ! 0
ex ¡ 1 ¡ x ex = lim 2 x!0 2 x
lim
lim
=0£1 =0
ex ¡ 1 ¡ x ex ¡ 1 = lim lim x!0 x!0 x2 2x )
so we can use l’Hˆopital’s Rule.
=0
£1
0 , 0
0 , 0
p x!0
x!0+
so we can use l’Hˆopital’s Rule.
This limit also has type
1 2
= 2(0)(1)
b As x ! 0, ex ¡ 1 ¡ x ! 1 ¡ 1 ¡ 0 = 0 and x2 ! 0 The limit has type
so we can use l’Hˆopital’s Rule.
p = lim 2 x cos x
1 ¡ cos x sin x = lim x!0 2x x2 sin x = 12 lim x!0 x 1 2 1 2
=1
sin x cos x p = lim 1 x x!0+ 1 ¡2 x 2
so we can use l’Hˆopital’s Rule.
=
¶
1
h As x ! 0+ , sin x ! 0 and
EXERCISE E
lim
1 1+x2
+x 2x + 1 = lim x!0 2 cos 2x sin 2x
x!0
The limit has type
)
0 , 0
=
) f 0 (x) is continuous for all x
0 , 0
µ
x2
lim f 0 (x) = f 0 (0)
x!0
The limit has type
so we can use l’Hˆopital’s Rule.
arctan x = lim x x!0
The limit has type
lim f 0 (x) = 0 = f 0 (0) and
1
0 , 0
g As x ! 0, x2 + x ! 0 and sin 2x ! 0
c From (1) and (2) above,
)
¡x¡2
x!0+
x!0¡
So, for x 6= 0, lim f 0 (x) = 0
(
1 x
= lim (¡x)
lim f 0 (x) = 0 .... (2) fSqueeze Theoremg
)
ln x x¡1
6 3x2 ¡ x
or ¡3x + x 6 f (x) 6 3x ¡ x But
DNE
black
Y:\HAESE\IB_HL_OPT-Calculus\IB_HL_OPT-Calculus_an\154IB_HL_OPT-Calculus_an.cdr Thursday, 7 March 2013 12:37:16 PM BRIAN
IB HL OPT 2ed Calculus
155
WORKED SOLUTIONS k As x ! 0, ax ¡ bx ! 0 and sin x ! 0 0 , 0
The limit has type )
ax
lim
¡ bx
The limit has type
ax ln a ¡ bx ln b
= lim
sin x
x!0
a As x ! 0+ , ln(cos 5x) ! 0 and ln(cos 3x) ! 0
4
so we can use l’Hˆopital’s Rule.
ln a ¡ ln b 1 ³a´ = ln provided a > 0, b > 0 b =
2 As x !
¼¡ , 2
)
sin 5x cos 3x sin 3x cos 5x ³ sin 5x ´ ³ 3x ´ ³ cos 3x ´ 1 = lim 53 5 5x sin 3x 3 cos 5x x!0+ ³ sin 5x ´ ³ 3x ´ = 25 £ lim £ lim £ 9 5x 5x!0+ 3x!0+ sin 3x
1 , 1
so we can use l’Hˆopital’s Rule.
tan x sec2 x = lim ¼ ¡ sec x tan x sec x x!
lim
¼¡ x! 2
lim
sec x tan x
lim
sec x tan x sec2 x
¼¡ x! 2
=
¼¡ x! 2
= b
lim
¼¡ x! 2
ln(sin 2x) is not defined when sin 3x < 0 ln(sin 3x) ) ¼ 2
and
tan x sec x
1 1
£1£1£1
that is, when ¼ 6 3x 6 2¼
fusing l’Hˆopital’s Rule againg =
25 9 25 9
=
2
=
5 3
lim
x!0+
tan x ! 1 and sec x ! 1
The limit has type
so we can use l’Hˆopital’s Rule.
¡5 sin 5x ln(cos 5x) cos 5x ) lim = lim x!0+ ln(cos 3x) x!0+ ¡3 sin 3x cos 3x which is ³ ´³ ´
cos x
x!0
0 , 0
)
¼ 3
6x6
y
2¼ 3
S
lies in this interval
lim
¼¡ x! 2
ln(sin 2x) ln(sin 3x)
A
¼
2¼ x T
DNE
C
which is back to where we started. So, l’Hˆopital’s Rule was no use here. sin x
tan x cos x = = sin x provided cos x 6= 0 1 sec x ¼ cos x that is, x 6= 2 tan x = lim sin x = 1 ¼¡ sec x x!
Note:
)
lim
¼¡
x! 2
5
1 1 sin x ¡ x ¡ = x sin x x sin x ³1 ´ ³ sin x ¡ x ´ 1 ) lim ¡ = lim sin x x sin x x!0+ x x!0+ a
The limit has type
2
) 3 As x ! 0,
¡ arccos x ¡ x !
¼ 2 ¼ 2
that is,
lim
¼ 2
0 , 0
x!0
= lim
x!0
= lim
1¡x2 3x2
x!0+
The limit also has type )
lim
p
3x2 3x2
1 6
1 ¡ x2
Ã
1 ¡ x2
p
p
3
=
x!0+
1+ 1+
p
1 ¡ x2
b
p
1 ¡ x2
1 ¡ (1 ¡ x2 )
1 ¡ x2 (1 + 1
1 ¡ x2 (1 +
¡
x!0+
0 , 0
´ 1
sin x
p
1 ¡ x2 )
¡ sin x cos x + [cos x + x(¡ sin x)] ¡ sin x = lim + 2 cos x ¡ x sin x x!0 0 = 2¡0 = lim
x!0+
1 1 sin 2x ¡ x ¡ = x sin 2x x sin 2x ³1 ´ 1 sin 2x ¡ x ) lim ¡ = lim sin 2x x!0+ x x!0+ x sin 2x The limit has type
p
)
1 ¡ x2 )
yellow
lim
x!0+
³1
x
¡
0 , 0
so we can use l’Hˆopital’s Rule.
1 sin 2x
´
= lim
x!0+
95
2 cos 2x ¡ 1 sin 2x + x(2 cos 2x)
2(1) ¡ 1 0+0
the limit DNE.
100
50
75
25
0
5
95
100
50
75
25
0
5
95
50
75
25
100
magenta
cos x ¡ 1 sin x + x cos x
so we again use l’Hˆopital’s Rule.
=
5
0
x
= lim
=0
!
)
cyan
³1
´
¡1
p
1¡
1 3(1)(2)
95
lim
3
so we can use l’Hˆopital’s Rule.
0+ p 1
=
100
50
75
25
0
5
x!0
so we can use l’Hˆopital’s Rule.
1 ¡ x sin x
x3
x!0
= lim
¡0
0 , 0
¡ arccos x ¡ x
x!0
= lim
¼ 2
¡ arccos x ¡ x ! 0 and x ! 0
The limit has type )
¡
¼ 2
³1
black
Y:\HAESE\IB_HL_OPT-Calculus\IB_HL_OPT-Calculus_an\155IB_HL_OPT-Calculus_an.cdr Friday, 8 March 2013 10:59:17 AM BRIAN
IB HL OPT 2ed Calculus
156
WORKED SOLUTIONS c sec2 x ¡ tan x =
1 sin x ¡ cos2 x cos x
=
1 ¡ sin x cos x cos2 x 1¡
=
1 2
x!1
x!1
1 ¡ 12 (1) 0
= lim
x!1
the limit DNE.
=1
1 , 1
The limit has type )
lim
x!1
1+
b
so we can use l’Hˆopital’s Rule.
xk kxk¡1 = lim x x!1 e ex
And, by using l’Hˆopital’s Rule repeatedly = lim
x!1
³
) )
k(k ¡ 1)xk¡2 ex
1+
lim
x!1
1 x
@ 1+x
1 1 + x¡1 1 fx¡1 = ! 0g x
1 ln =e x
1+
1
1 x
c
ex
lim
³
´x
¡
x!1
fk is fixed, e ! 1g x
1+
x!1
³
x
= lim
x !1 a
ffrom ag
20 1xa 3a 6@1 + ³ 1 ´ A 7 4 5 x
=e
9 x = eln x
x!1
ffrom bg
= esin x ln x ln x ¡1
= e [sin x]
1 kxk
As x ! 0+ , ln x ! ¡1 and [sin x]¡1 ! 1 The limit has type
) for large x, xk is very much greater than ln x ) ln x increases more slowly than any fixed power of x.
a As x ! 1,
³
ln 1 +
1 x
´
1 that is, lim ln 1 + x!1 x Consider x ln 1 +
³
´
ln x x lim = lim x!0+ [sin x]¡1 x!0+ ¡[sin x]¡2 cos x
)
= lim
x!0+
=0
The limit has type )
´
magenta
yellow
95
50
75
25
0
so we can use l’Hˆopital’s Rule.
=0
5
95
100
50
75
25
0
5
95
100
50
75
25
0
5
95
100 cyan
0 , 0
¡[sin x]2 x cos x
ln x ¡2 sin x cos x lim = lim x!0+ [sin x]¡1 x!0+ cos x + x(¡ sin x) ¡2(0)(1) = 1¡0
1 !0 x The limit has type 0 £ 1, so we can use l’Hˆopital’s Rule.
As x ! 1, ln 1 +
so we can use l’Hˆopital’s Rule. 1
´ 1
x
1 , 1
! ln 1 = 0
³
³
.... (1)
ln x Now consider as x ! 0+ [sin x]¡1
=0
50
a
) xsin x = (eln x )sin x
so we can use l’Hˆopital’s Rule.
1 ln x x lim = lim x!1 xk x!1 kxk¡1
75
¢´
x!1
100
1 , 1
= lim
25
1
lim
The limit has type
0
¡
a
xk
ln x , k 2Z+ x!1 xk As x ! 1, ln x ! 1 and xk ! 1
5
¢
lim x ln 1+ x
a
8
1
x ln 1+ x
= lim e
20 1xa 3a 1 6 7 = lim 4@1 + ³ ´ A 5 x!1 x
´ a x
b The result in a implies that for large x, is greater than for any fixed k 2 Z + . ) ex increases more rapidly than any fixed positive power of x.
)
¢
=e =e
ex
7 Consider
fa = eln a g
1
k!x0
=0
¡
¢
x ln 1+ x
=e
.. . x!1
1
1+ x
=e
k(k ¡ 1)(k ¡ 2)xk¡3 = lim x!1 ex
= lim
A
¡x¡2
´x
³
´
1
¡x¡2
¡
a For all k 2 Z + , as x ! 1, xk ! 1 and ex ! 1.
6
0
¡1
= lim
¼¡
1 x
ln(1 + x¡1 ) x¡1
= lim
cos2 x
x! 2
)
x!1
sin 2x
lim (sec2 x ¡ tan x) =
)
³
lim x ln 1 +
Now
black
Y:\HAESE\IB_HL_OPT-Calculus\IB_HL_OPT-Calculus_an\156IB_HL_OPT-Calculus_an.cdr Thursday, 7 March 2013 12:45:56 PM BRIAN
IB HL OPT 2ed Calculus
WORKED SOLUTIONS ln x ¡1
x!0+
x!0+
10 x
1
=
(eln x ) x
=e
2 x
= e0 =1 1 x
y
-2_Qw
lim x sin x = lim e [sin x]
Hence,
157
f(x) = -2x - 5, -2_Qw 6 x 6 -1
ln x x
f(x) = x2 - 4, -1 6 x 6 2
(-1, -3)
ln x . x As x ! 1, ln x ! 1 lim
Consider
x!1
The limit has type
1 , 1
½
so we can use l’Hˆopital’s Rule.
f 0 (x)
1
)
lim
x!1
)
ln x x = lim =0 x!1 1 x
lim e
ln x x
f (¡1 12 ) = f (2) = 0 and f (x) is continuous and differentiable on [¡2 12 , 2].
lim x x = 1
2
a
f (x) = 3x3 + 5x2 ¡ 43x + 35 on [¡5, 2 13 ]
a
Rolle’s Theorem applies and f 0 (c) = 0 , c = 0 which lies in ]¡2 12 , 2[ .
)
x!1
EXERCISE F 1
f (¡5) = f (2 13 ) = 0
f 0 (x) = 9x2 + 10x ¡ 43 for all x 2 R ) f (x) is continuous and differentiable for all x 2 [¡5, 2 13 ] and f (¡5) = f (2 13 ) = 0 ) Rolle’s Theorem applies p ¡5 § 2 103 f 0 (c) = 0 , c = 9
i f (x) = (x ¡ 1)(x ¡ 2)(x ¡ 4)(x ¡ 5) has zeros: 1, 2, 4, and 5 As f (x) is a real polynomial, continuous and differentiable for all x 2 R , by Rolle’s Theorem, there exist zeros of f 0 (x) in the intervals ]1, 2[ and ]2, 4[ and ]4, 5[ . So at least 3 real zeros exist for f 0 (x). ii f (x) has sign diagram and graph:
+
both of which lie in ]¡5, 2 13 [ b f (x) = jxj ¡ 5 on [¡5, 5]. y
-5
5
If x > 0, If x < 0,
x
0 f+ (x) = 1
b
0 f¡ (x) = ¡1
Thus f (x) is not differentiable for all x 2 [¡5, 5] Rolle’s Theorem does not apply. 1 on [¡ 12 , 7]. x+1
-
is continuous on [¡ 12 , 7] 1 f 0 (x) = exists for all x 2 R , x 6= ¡1 (x + 1)2
Rolle’s Theorem does not apply.
magenta
yellow
95
50
75
25
0
5
95
100
50
75
25
0
on [¡2 12 , 2]
5
95
x < ¡1 x > ¡1
100
50
75
25
0
5
95
100
50
75
25
0
5
cyan
¡2x ¡ 5, x2 ¡ 4,
+ 1
x
2
+ 3
x
i f (x) = (x ¡ 1)2 (x2 + 9)(x ¡ 2) has two real zeros: 1 and 2. As f (x) is a polynomial, continuous and differentiable for all x 2 R , by Rolle’s Theorem, there exists a zero of f (x) in ]1, 2[ . So at least one zero exists for f 0 (x).
100
c
f (¡ 12 ) = 0 and f (7) = 1 78 6= 0
½
+ 5
As y = f (x) has 4 turning points, f 0 (x) has 4 distinct real zeros, 3 guaranteed by Rolle’s Theorem and x = 1.
is differentiable on [¡ 12 , 7]
d f (x) =
+ -3
)
)
4
i f (x) = (x ¡ 1)2 (x2 ¡ 9)(x ¡ 2) = (x + 3)(x ¡ 1)2 (x ¡ 2)(x ¡ 3) which has zeros: ¡3, 1, 2, and 3. As f (x) is a polynomial, continuous and differentiable for all x 2 R , by Rolle’s Theorem, there exist zeros of f 0 (x) in the intervals ]¡3, 1[ and ]1, 2[ and ]2, 3[ . So at least 3 zeros exist for f 0 (x). ii f (x) has sign diagram and graph:
f (x) is continuous for all x 2 R , x 6= ¡1
)
+ 2
We see that there are 3 turning points ) there are exactly 3 real distinct zeros of f 0 (x).
-5
c f (x) = 2 ¡
1
f (x) is not differentiable at x = 0.
½
)
x < ¡1 x > ¡1
) f 0 (¡1) = ¡2
1
)
¡2, 2x,
0 0 ) f¡ (¡1) = f+ (¡1) = ¡2
= e0 = 1
x!1
=
black
Y:\HAESE\IB_HL_OPT-Calculus\IB_HL_OPT-Calculus_an\157IB_HL_OPT-Calculus_an.cdr Wednesday, 13 March 2013 10:26:40 AM BRIAN
IB HL OPT 2ed Calculus
158
WORKED SOLUTIONS f (x) is continuous for all x > 0 and f 0 (x) exists for all x>0 Thus, on [49, 51] f (x) is continuous and differentiable. Hence, by the MVT, there exists c 2 [49, 51] such that p p 51 ¡ 49 = f 0 (c)(51 ¡ 49) p 1 ) 51 ¡ 7 = 2 £ p
ii f (x) has sign diagram and graph:
-
-
+
1
f 0 (x) has exactly 2 real distinct zeros, the one guaranteed by Rolle’s Theorem and the one at x = 1.
)
3
2
)
a f (x) = x3 is continuous on [¡2, 2] and differentiable on ]¡2, 2[ .
p
2 c
1 51 ¡ 7 = p c
b
y
Thus, by the MVT, there exists c in [¡2, 2] such that f (2) ¡ f (¡2) = f 0 (c)(2 ¡ ¡2) ) 8 ¡ (¡8) = f 0 (c) £ 4 ) f 0 (c) = 4
1_ _ _ _ f(x) = %% ~`x_ _
, c = § p2 3 p f (x) = x ¡ 2 is continuous on [3, 6] and differentiable on ]3, 6[ .
y
(6, 2)
x
by the MVT, there exists c on [3, 6] such that f (6) ¡ f (3) = f 0 (c)(6 ¡ 3)
1 8
< p
1 xp
+1 dx converges x4 + 1
)
1 dx + x2
1 dx converge x4
x2
fComparison Testg
2
) e¡x 6 e¡x 2
) 0 6 e¡x 6 e¡x
Z
) 06
50
75
25
0
1
1
Z
1 dx and x2
1 1
Z
b For x > 1, x2 > x ) ¡x2 6 ¡x
5
95
100
yellow
x2 + 1 dx < x4 + 1
½Z
1
2
¡x
e 1
50
95
50
75
100
1
25
0
5
95
100
1
1
x dx 2x5 + 3x2 + 1
75
by the Comparison Test
cyan
1
where
¾
25
Z
converges.
50
Z
1 dx converges for all p > 1 xp
0
1
1
75
1
1
5
½Z
1 dx converges for p > 1 xp
x2 + 1 x2 + 1 1 1 < = 2 + 4 4 x +1 x4 x x
Z
x x 1 < < 4 for all x > 1 5 2 5 2x + 3x + 1 2x x x 1 < 4 for all x > 1 ) 0< 2x5 + 3x2 + 1 x Z 1 Z 1 x 1 dx < dx ) 0< 2x5 + 3x2 + 1 x4 1 1 1
¾
1
sin x dx converges x3
) 0
1.
Z
dx
1 dx converges x3
½Z
)
b Assumptions:
where
¾
x2 ¡ 1 dx is a positive constant p x7 + 1
fComparison Testg
= 5 f (2)
5
1
)
) F 0 (3) = f (g(3) ¡ 2) £ g 0 (3)
)
1
1
fChain ruleg
fComparison Testg
1
where
= f (g(x) ¡ 2) g 0 (x)
a
2
1
Z
where u = g(x)
¾
1 dx converges for p > 1 xp
¯ ¯ ¯ sin x ¯ 1 For all x > 1, 0 6 ¯ 3 ¯ 6 3 x x Z 1¯ Z 1 ¯ 1 ¯ sin x ¯ dx 6 ) 06 ¯ x3 ¯ x3
= f (u ¡ 2) g 0 (x)
1
½Z
1
g(x)
0
dx
x2 ¡ 1 dx converges p x7 + 1
2
f (t ¡ 2) dt 1
=
1
g(x)
F (x) =
a
1
x2 ¡ 1 dx converges p x7 + 1
)
=0
1 x1:5
1
1
1
1
g 0 (1) = f (1) = 3
g 00 (x)
½Z
)
fFTOCg
Z
10
Z
=1
0
95
)
= f (3)
f (t) dt
= f (x)
1
g (3)
x
1
dx
2g
Z
x2 ¡ 1 x7 + 1
where
¼ 4
1 x1:5
p
1
Z
£ 2 + 14 ¼(1)2
1
Z
1
dx 6
e¡x dx
¡1
100
=
e
=6+
g (x)
c
2
1
) 0
0 for all x2R,
1 dx x
du dx dx
=
1
1 1
1
=
¡e¡x dx
2 e
=
0
0
0
2
·
1
0
c Pn is:
Z
1
x3 e¡x dx = 6 = 3!
Z
= ¡x2 e¡x + 2
Z
x2 e¡x dx = 2 = 2!
(1) P0 was proved in a.
2 ¡x
x2 e¡x dx = ¡x2 e¡x ¡
)
´
x e
v=x v 0 = 2x
Z
1
0
2
If u = e , u = ¡e¡x ,
1
x1 e¡x dx = 1 = 1!
(2) Assuming the truth of Pk ,
1
If n = 2, we need to find 0
+1
¡eb
b!1
1
¤b 0
¶
x0 e¡x dx = 1 = 0!
So, we predict
¡e¡x (x + 1)
0
¡(b3 + 3b2 + 6b + 6) ¡ (¡6) eb
0
b
xe¡x dx = lim
b!1
1
0
v=x v0 = 1
¸b
=6
b
Z
)
µ
= lim
xe¡x dx
xe¡x dx = ¡xe¡x ¡
= ¡e
¡(x3 + 3x2 + 6x + 6) ex
b!1
Using Integration by Parts with u0 = e¡x , u = ¡e¡x ,
¡x
·
b!1
¡ (¡1)
0
Z
x3 e¡x dx
= lim
=1¡0 =1
Z
1
)
e¡x dx
b!1
0
Z
¶
b
e¡x dx = lim
black
Y:\HAESE\IB_HL_OPT-Calculus\IB_HL_OPT-Calculus_an\165IB_HL_OPT-Calculus_an.cdr Friday, 8 March 2013 12:20:41 PM BRIAN
¡2x ex2
IB HL OPT 2ed Calculus
166
WORKED SOLUTIONS 2
For x > 0, ¡2x < 0 and ex > 0 ) f 0 (x) < 0 )
f (x) is decreasing for x > 0. 1 P
b U=
f (i) =
i=0 1
P
L= 1 P
f (i + 1) =
Z
2
1 P
e an = ¡(i+1)2
=
e
1
2
e¡x dx
0 1 P
1 P i=1
¡1 < i2
1
)
EXERCISE J.1 As n ! 1, n + n3 ! 1 ) )
lim an = 0
´
) an =
lim an = 0 3¡ ¡ 5n = 2 5n2 + 2n ¡ 6 5+ n ¡
lim an =
n!1
=
6 n2
)
)
cyan
magenta
yellow
50
25
0
5
95
100
50
75
25
0
5
95
100
50
75
25
0
+
n2 + 1 + n + 1 ¡ n2
fdividing each term by
p ng
fan g converges to 1. cos2 n 2n
¯ ¯ ¯ cos2 n ¯ ¯ ¯ ¯ 2n ¯ 6
1 , n 2Z+ 2n
)
1 !0 2n jan j is convergent to 0 fSqueeze Theoremg
)
fan g converges to 0.
As n ! 1,
2 n2
+ n12 + n13
5
95
1+
1 n
n2 + 1 + n
n2
p
Now 0 6
+ n2
1+
p
n2 + 1 ¡ n
fan g is divergent.
d an =
+ 2n n3 + n2 + n + 1
100
50
75
25
0
5
=
!
n
(n2 + 2n)(n2 + 1) ¡ n3 (n + 1) (n + 1)(n2 + 1)
1 n
! Ãp
n2 + 1 + n
1
1 As n ! 1, p ! 0 n
n4 + 2n3 + n2 + 2n ¡ n4 ¡ n3 = n3 + n2 + n + 1 =
+1¡n
1
n(n + 2) n3 ¡ 2 n+1 n +1
n3
16 81
=
1¡ p p n¡1 n = c an = p 1 n+1 1+ p
3 5
75
d an =
3
n2 + 1 + n p and as n ! 1, n2 + 1 ! 1 =
5 2 6 ! 0, ! 0, and 2 ! 0 n n n
As n ! 1, )
p
n!1
c an =
n2
p
=
5 n
¡ 2 ¢4
1
p Ã
3n2
¶4
1 n! = (n + 3)! (n + 3)(n + 2)(n + 1)
an =
b
n+1 n ³ ´ 1 = ln 1 + n As n ! 1, an ! ln 1 )
3+
3 n 7 n
) fan g converges to 0.
b an = ln(1 + n) ¡ ln(n)
³
2¡
7 ¡3 ! 0, and !0 n n
lim an =
n!1
= ln
µ =
As n ! 1, (n + 3)(n + 2)(n + 1) ! 1 ) an ! 0
1 !0 n + n3
95
1
3n + 7
n!1
a an =
2
1 a an = n + n3
³ 2n ¡ 3 ´4
As n ! 1,
1 P ¡1 ¡1 dx < x2 (i + 1)2 i=1
1
lim an = 0
n!1
100
c
Z
p p n ! 1, n+1!1
f an =
1 P ¡1 ¡1 , L= 2 2 (i + 1) i=1 i
i=1
p ¢ n+1¡ n
i=0
) f 0 (x) > 0 for all x > 0
b U=
p ¶ n+1+ n p p n+1+ n
µp
¡p
n+1¡n p n+1+ n
1
0
p p n+1¡ n
= p
2
e¡i
f (x) = ¡ 2 = ¡x¡2 x
a
n!1
i=0
e¡(i+1)
0, 2 +
sin_µ µ
)
µ 1 x
1 n
)
jan j is convergent to 0 fSqueeze Theoremg
v u u 3 t
=
Thus j(an + bn ) ¡ (a + b)j < " for all n > N )
¡ n14 + n46 6
1 + n32 + n34 + n16 1
=0
fan g is convergent to 0.
)
1 2 3 n + 2 + 2 + :::: + 2 n2 n n n 1 + 2 + 3 + :::: + n = n2
3 an =
½
n(n+1) 2 = n2
= =
1 2 1 2
³
n
1 1+ n
As n ! 1,
n P
¾
) )
an !
1 2
)
n+1>n
magenta
provided
we require n ¡ 1 >
¯ ¯ 3n + 5 ¯ 7n ¡ 4 lim
n!1
¡
3 7
¯ ¯ N = 1 +
lim an = a and
1 "
lim bn = b,
n!1
lim ¯bn = ¯b
n!1
lim (®an + ¯bn )
Hence,
n!1
³ 1 ´n
= ® lim an + ¯ lim bn
fas ®, ¯ are constantsg
n!1
n!1
= ®a + ¯b lim (an ¡ bn ) = a ¡ b
n!1
yellow
50
25
0
=0
5
95
1 n+1
´n
For ® = 1, ¯ = ¡1,
100
lim
fLimit Lawg
n!1
1 nn ³
n!1
= lim ®an + lim ¯bn n!1
n
50
)
95
50
75
1 !0 nn
25
0
5
95
100
50
75
25
0
n+1
2n
there exists N such that jan ¡ aj
2 n
100
But
¯ ³ ´¯ ¯sin 1 ¯ 6 ¯ ¯ n
y 1
167
black
Y:\HAESE\IB_HL_OPT-Calculus\IB_HL_OPT-Calculus_an\167IB_HL_OPT-Calculus_an.cdr Friday, 8 March 2013 12:31:19 PM BRIAN
IB HL OPT 2ed Calculus
168
WORKED SOLUTIONS
EXERCISE J.2 1
a
ii Let cn =
i un =
2n ¡ 7 3n + 2
en
en+1
2(n + 1) ¡ 7 2n ¡ 7 = ¡ 3(n + 1) + 2 3n + 2 2n ¡ 5 2n ¡ 7 = ¡ 3n + 5 3n + 2 =
(2n ¡ 5)(3n + 2) ¡ (2n ¡ 7)(3n + 5) (3n + 5)(3n + 2)
=
6n2 ¡ 11n ¡ 10 ¡ [6n2 ¡ 11n ¡ 35] (3n + 5)(3n + 2)
=
25 (3n + 5)(3n + 2)
=
(1 ¡ e)(e3n+1 + en ) (e2n+2 ¡ 1)(e2n ¡ 1)
) fun g is monotonic (increasing)
Also, cn > 0 for all n 2 Z +
i If bn =
Now as n ! 1, en ! 1, and e¡n ! 0 1 ) !0 en ¡ e¡n 1 ) lim n =0 n!1 e ¡ e¡n
=
3n+1 + 32n+1 ¡ 3n ¡ 32n+1 (1 + 3n+1 )(1 + 3n )
=
3n (3 ¡ 1) (1 + 3n+1 )(1 + 3n )
1 £ 3 £ 5 £ 7 £ :::: £ (2n ¡ 1) 2n n! un+1 1 £ 3 £ 5 £ 7 £ :::: £ (2n ¡ 1)(2n + 1)2n n! = un 1 £ 3 £ 5 £ 7 £ :::: £ (2n ¡ 1)2n+1 (n + 1)!
Let un =
2 )
> 0 for all n 2 Z + +
for all n 2 Z + .
) fun g is bounded
1 !0 1 + 3n
Proof: (by induction)
p
(1) x2 = 4 + 3(0) = 2 ) x2 > x1 ) P1 is true.
cyan
magenta
yellow
50
75
25
0
5
95
100
50
75
25
0
5
95
100
50
75
25
0
n!1
5
95
lim bn = 1
100
)
a Pn : fxn g is monotonic increasing, that is, xn+1 > xn for all n 2 Z +
95
From (1), as n ! 1,
is its greatest lower bound.
Thus fun g is convergent. p 3 x1 = 0 and xn+1 = 4 + 3xn , n 2 Z +
6 bn < 1 for all n 2 Z +
) fbn g is bounded
1 2
Also un > 0 ) 0 < un < 1
) bn < 1 for all n 2 Z + 3 4
2n + 1 2n + 2
Thus u1 =
3n 1 + 3n ¡ 1 Also bn = = 1 + 3n 1 + 3n 1 ) bn = 1 ¡ .... (1) 1 + 3n So,
=
100
3 4
2n + 1 2(n + 1)
) un+1 < un for all n 2 Z + fun > 0g ) fun g is monotonic (decreasing)
3 4
Hence its smallest member is b1 = bn >
=
< 1 for all n 2 Z +
) fbn g is monotonic (increasing) )
e for all n 2 Z + e2 ¡ 1
As fcn g is monotonic and bounded ) fcn g is convergent.
3n+1 3n ¡ n+1 1+3 1 + 3n
) bn+1 > bn for all n 2 Z
fen > 0, e2n > 1g
) fcn g is bounded
2n ¡ 7 3n + 2 < 3n + 2 3n + 2
3n , 1 + 3n
bn+1 ¡ bn =
0 < cn 6
Thus
So ¡1 6 un < 1 ) fun g is bounded.
50
e3n+1 (1 ¡ e) + en (1 ¡ e) (e2n+2 ¡ 1)(e2n ¡ 1)
) fcn g is monotonic (decreasing) e is its least upper bound. Thus c1 = 2 e ¡1
Thus fun g has an upper bound of 1
75
=
cn+1 < cn for all n 2 Z +
)
) un < 1 for all n 2 Z +
25
e3n+1 ¡ en+1 ¡ e3n+2 + en (e2n+2 ¡ 1)(e2n ¡ 1)
> 0 for all n 2 Z + ) un+1 > un for all n 2 Z +
Also, as un =
0
=
Now 1 ¡ e < 0, e3n+1 + en > 0, e2n+2 > 1, and e2n > 1 ) cn+1 ¡ cn < 0 for all n 2 Z +
ii As fun g is monotonic increasing u1 = ¡1 is its greatest lower bound.
5
en
) cn+1 ¡ cn = 2n+2 ¡ 2n e ¡1 e ¡1
un+1 ¡ un
b
1 en = 2n ¡n ¡e e ¡1
black
Y:\HAESE\IB_HL_OPT-Calculus\IB_HL_OPT-Calculus_an\168IB_HL_OPT-Calculus_an.cdr Wednesday, 13 March 2013 4:07:29 PM BRIAN
IB HL OPT 2ed Calculus
169
WORKED SOLUTIONS (2) If Pk is true, then xk+1 > xk , k 2 Z + .... (¤) 2 2 Now xk+2 ¡ xk+1 = 4 + 3xk+1 ¡ (4 + 3xk ) = 3(xk+1 ¡ xk ) >0 ffrom ¤ g 2 2 > xk+1 ) xk+2 ) xk+2 > xk+1
d Given fun g converges, then lim un = lim un+1 = L, say n!1
n!1
) 2
L ¡L¡1=0
)
) Pn is true fPOMIg
5 un+1 = a
Proof: (by induction) P1 is true (2) If Pk is true then xk < 4 for k 2 Z + 2 Now xk+1 = 4 + 3xk )
fusing Pk g
< 16 fas xk+1 > 0g
)
d From a and c, fxn g is convergent.
r lim xn+1 =
Since
n!1
) )
) 4
lim xn+1 = L also.
n!1
4+3
³
p L = 4 + 3L L2 = 4 + 3L
´
u4 =
3 2 5 3 8 5
)
)
L=
for all n 2 Z + .
1 un
)
´
2 L
fas L > 0g
yellow
1 2
L+
) )
lim un =
n!1
If
p
fas L > 0g 3.
lim un = L then
n!1
)
95
´
3 . un
L
1 2
³
un +
k un
´
f Suppose u1 = a, a > 0, and un+1 =
50
un +
lim un+1 = L also.
then
n!1
75
³
´n!1 3
e If u1 = 6 and un+1 = p lim un = k.
25
0
1 2
³
L2 = 3 p L= 3
)
5
95
100
50
75
25
0
5
95
100
50
75
25
0
2 L
3 2L = L + L 3 L= L
) u1 = 2 and un+1 = 1 +
5
95
100
50
75
L+
lim un+1 = L also.
n!1
2 L
n!1
) fun g is bounded.
25
³
lim un = L then
If
b un is defined by:
magenta
=1 iii u1 = 7 u2 ¼ 3:642 90 = 1:5 u3 ¼ 2:095 94 ¼ 1:416 67 u4 ¼ 1:525 08 ¼ 1:414 22 u5 ¼ 1:418 24 ¼ 1:414 21 ¼ 1:414 21 u6 ¼ 1:414 21 u7 ¼ 1:414 21 ¼ 1:414 21 ¼ 1:414 21 u8 ¼ 1:414 21
d Suppose u1 = a, a > 0, and un+1 =
c fun g is not monotonic as u2 < u1 but u3 > u2 . All terms of the sequence fun g are clearly positive. 1 So, as un¡1 > 0 and un = 1 + , we deduce un¡1 un > 1 for all n 2 Z + . 1 < 1, for n ¡ 1 > 1 Thus un¡1 > 1 and un¡1 ) un 6 2 ) 1 < un 6 2 for all n 2 Z +
cyan
1 2
L2 = 2 p L= 2
)
n!1
u3 =
0
L=
n!1
lim xn = 4 2 1
ii u1 u2 u3 u4 u5 u6 u7 u8
lim un = L then
2L = L +
)
lim xn ,
L ¡ 3L ¡ 4 = 0 (L ¡ 4)(L + 1) = 0 ) L = 4 or ¡1 ) L = 4 fas xn > 0g
u2 =
5
L=
)
2
a u1 = 2 =
=5 = 2:7 ¼ 1:720 37 ¼ 1:441 46 ¼ 1:414 47 ¼ 1:414 21 ¼ 1:414 21 ¼ 1:414 21
L=
1 2
³
k L+ 2 L
1 2
³
un +
lim un+1 = L also.
´
k . un2
´ n!1
100
)
´
n!1
Hence, 0 6 xn < 4, so fxn g is bounded. lim xn = L say, then
2 un
c Yes, and if
) Pn is true fPOMIg
n!1
un +
p 1§ 5 2
b No, as in a where u1 = 1, u2 > u1 but u3 < u2 .
Thus P1 is true and Pk+1 is true whenever Pk is true
If
L=
³
1 2
i u1 u2 u3 u4 u5 u6 u7 u8
(1) If n = 1, 0 < 4 )
) xk+1 < 4
)
p 1+ 5 But L > 0, so lim un = n!1 2
=0 x5 ¼ 3:8752 x9 ¼ 3:9975 =2 x6 ¼ 3:9529 x10 ¼ 3:9991 ¼ 3:1623 x7 ¼ 3:9823 x11 ¼ 3:9996 ¼ 3:6724 x8 ¼ 3:9934 The experimentation suggests that 4 may be an upper bound. p c Pn : If x1 = 0 and xn+1 = 4 + 3xn for all n 2 Z + then xn < 4.
b x1 x2 x3 x4
2 xk+1
1 L
L2 = L + 1
)
fas xk+2 , xk+1 > 0g
Thus P1 is true and Pk+1 is true whenever Pk is true
2 ) xk+1 < 4 + 3(4)
L=1+
black
Y:\HAESE\IB_HL_OPT-Calculus\IB_HL_OPT-Calculus_an\169IB_HL_OPT-Calculus_an.cdr Wednesday, 13 March 2013 4:11:24 PM BRIAN
IB HL OPT 2ed Calculus
170
WORKED SOLUTIONS
)
) )
lim un =
p 3
1+
=1+
1 n
) g(x) > 0 Since f 0 (x) = f (x) g(x) where f (x) > 0, g(x) > 0 for all x > 0, we deduce f 0 (x) > 0 for all x > 0 )
³ ´1 n
+:::: +
³
n
+
³ ´ 1 n ´
n n¡1
+ n2
2
1
+
nn¡1
³1´
en = 1 + n
+
3
n3
n
nn
n
³
µ
::::
´
n ¡ [n ¡ 1] n
³
· 1 ln(1 + x¡1 ) + x
¶ + (x + 1)¡2
d
2
¡ 1 ¢n¡1
)
2
yellow
³
´
1 n =e n!1 n ³ n + 1 ´n lim =e n!1 n lim
i )
50
25
0
¡1 1 + x(x + 1) (x + 1)2
95
=
100
¡1 1 + x2 + x (x + 1)2
75
¡ 1 ¢n
fsum of a GSg
Thus 2 6 en < 3 ) fen g is bounded. So, en is monotonic (increasing) and bounded ) fen g is convergent.
=
magenta
1 ¡ 12
) en < 3 for all n 2 Z + , n > 3
x+1 x2 x2
¡ ¢n 1(1 ¡ 12 )
) en < 1 + 2(1 ¡ ) en < 3 ¡
50
µ
1 + x¡1
i 1
25
0
5
95
100
50
75
cyan
¡x¡2 1 + x¡1
iii Thus en < 1 +
95
= f (x) g(x), say.
¸
50
= f (x) ln(1 + x¡1 ) ¡
Now g 0 (x) =
1 1 1 1 + + + :::: + , 2! 3! 4! n! 1 1 1 en < 1 + 1 + 12 + 2 + 3 + :::: + n¡1 2 2 2
So, as en = 1 + 1 +
75
h
25
= 2k Thus P3 is true and Pk+1 is true whenever Pk is true. ) Pn is true fPOMIg
25
= f (x) ln(1 + x
x¡1 )¡ 1 + x¡1
¸
¡x¡2
0
¡1
0
> 2 £ 2k¡1
5
)
·
5
) (k + 1)! = (k + 1)k!
)
95
¡1
= ex ln(1+x
, n>0
Proof: (by induction)
´
100
f 0 (x)
)
¡1
´n
(1) If n = 3, 3! > 22 ) 6 > 4 ) P3 is true (2) If Pk is true, then k! > 2k¡1 , k 2 Z +
, x > 0, x 2 R .
Now f (x) = ex ln(1+x
³
Pn : n! > 2n¡1 for all n > 3, n 2 Z +
¶
´³
75
´x
5
f (x) =
1 1+ x
´
n+1 1 1 > 1+ n+1 n > en for all n > 1
ii Using b, since 2 3 n¡1 1 < 1, 1¡ < 1, 1¡ < 1, 1¡ < 1, ...., 1¡ n n n n 1 1 1 1 en < 1 + 1 + + + + :::: + 2! 3! 4! n! 1 1 < n¡1 for all n > 3, We now need to show that n! 2 n 2 Z + or n! > 2n¡1 .
i From b, en = 1 + 1+ many other positive terms ) en > 2 .... (1) To prove that en+1 > en for all n 2 Z + , consider
³
1+
Hence, from (1), 2 6 en < en+1
1 1 1 1 2 1¡ + 1¡ 1¡ 2! n 3! n n ³ ´ ³ ´ 1 1 n¡1 +:::: + 1¡ :::: 1 ¡ n! n n
) en = 1 + 1 +
³
) en+1
³ ´ 1 n
³ ´ n(n ¡ 1) 1
³n ¡ 1´
1 +:::: + n!
f (x) is strictly increasing for all x > 0
)
³ ´ 1 n
n 2! n2 ³ ´ n(n ¡ 1)(n ¡ 2) 1 + + :::: 3! n3 ³ 1 ´ n(n ¡ 1)(n ¡ 2)::::(n ¡ [n ¡ 1]) + n! nn ³n ¡ 1´ 1 ³n ¡ 1´³n ¡ 2´ 1 ) en = 1 + 1 + + 2! n 3! n 2
c
¡1 x(x + 1)2
x!1
k.
´n
1
¡(x + 1) + x x(x + 1)2
) g 0 (x) < 0 for all x > 0 ) g(x) > lim g(x) = ln 1 ¡ 0
n!1
³
b
=
L3 = k p 3 L= k
a
g 0 (x) =
)
k L2
L=
)
6
k L2
2L = L +
1+
100
)
black
Y:\HAESE\IB_HL_OPT-Calculus\IB_HL_OPT-Calculus_an\170IB_HL_OPT-Calculus_an.cdr Wednesday, 13 March 2013 4:13:07 PM BRIAN
IB HL OPT 2ed Calculus
WORKED SOLUTIONS
³
´
1 n n!1 n ³ n ¡ 1 ´n = lim n!1 n lim
1¡
= lim
d
n=1 1
P
³ m ´m+1
n=1
freplacing n by m + 1g
m+1
m!1
)
³
lim
m!1
m+1 m
n
n! = nn
)
³n ¡ 1´³n ¡ 2´
n! 6 nn
) )
0
0,
´2 ³ 1 ´
n! lim =0 n!1 nn
)
´
1 converges fp-series testg n2
³
)
n
³
´n ³
P
an2 4n4 + 12n3 + 9n2 n3 = £ 2 7 5+n 4 bn
³3´³2´³1´
n ¡ 2 of these
n! 6 nn
1 P 1 1 ¡ where n n=1 n2
n=1 1
1 ¡ 2 n n
=
³ n ¡ 1 ´n¡2 ³ 1 ´ n
1 P
=
1 ³ P 1
)
|
´
1 diverges and n
n! n(n ¡ 1)(n ¡ 2) £ :::: £ 3 £ 2 £ 1 = nn n £ n £ n £ :::: £ n £ n £ n
ii
1 n2
2n2 + 3n 2 2 Let an = p and bn = p 5 + n7 n3
´m
1 = e
1
¡
n=1
m lim m!1 m+1
=
1 ³ P 1
171
black
Y:\HAESE\IB_HL_OPT-Calculus\IB_HL_OPT-Calculus_an\171IB_HL_OPT-Calculus_an.cdr Friday, 8 March 2013 2:49:57 PM BRIAN
IB HL OPT 2ed Calculus
172 4
WORKED SOLUTIONS a
1
p
n(n + 1)(n + 2)
1 P
1
n=1
n1:5
1 P
)
1 1 = 1:5 and < p n n3
From the graphs of y = x and y = ln x ln x < x for all x > 0
1
p
converges
n(n + 1)(n + 2)
n=1
fComparison testg
1 1 > for all x > 0 ln x x
)
1 1 > for all n 2 Z + , n > 2 ln n n
But
n(n + 1)(n ¡ 1) = n3 ¡ n < n3
b
)
converges fp-series testg
p 3
)
n(n + 1)(n ¡ 1) < n
1 > p 3 n n(n + 1)(n ¡ 1)
)
1 P
)
n=2
where
n=2
1 P
5
c As 0 6 sin2 µ 6 1 for all µ 2 R , 1 sin2 n p 6 1:5 n n n
P
)
n=1
sin2 n p n n
fComparison testg
also converges
) 2
1 ¡ ¢ P 2 n 3
n=1
Hence f
06x
0:5 n¡1 n 1 P 1 But diverges fp-series testg 0:5 n n=2 p 1 P n ) also diverges fComparison testg n=2 n ¡ 1
n=1
j2 sin xj < 1 ¡1 < 2 sin x < 1
or
d Since
But
) )
fp-series testg
converges n1:5
j2 sin xjn which is a geometric series,
n=0
or
1
n=1 1
1 P
2n jsinn xj =
fComparison testg
converging for jrj < 1
fComparison testg
Since
1 diverges also ln n
n=2
1 diverges n
diverges
n(n + 1)(n ¡ 1)
1 P
P
fp-series testg
diverges
n=0
1
p 3
1 P
n
n=2 1
)
1
1 P 1
1 + 3n
) ) )
j1 + cj > 1 1 + c > 1 or 1 + c < ¡1 c > 0 or c < ¡2 p 3¡1 Thus c = 2 x is continuous for all x > 1 .... (1) 7 a f (x) = 2 x +1
converges fComparison testg
y y=x y = ln x 1
1 0 ) x2 + 1 cannot be 0g
x
f 0 (x) =
1(x2 + 1) ¡ x(2x) 1 ¡ x2 = 2 (x2 + 1)2 (x + 1)2
cyan
magenta
yellow
95
100
50
75
25
0
5
95
100
50
75
25
0
5
95
100
50
75
25
0
5
95
100
50
75
25
0
5
But x > 1 ) x2 > 1 ) 1 ¡ x2 6 0
black
Y:\HAESE\IB_HL_OPT-Calculus\IB_HL_OPT-Calculus_an\172IB_HL_OPT-Calculus_an.cdr Friday, 8 March 2013 2:50:27 PM BRIAN
IB HL OPT 2ed Calculus
173
WORKED SOLUTIONS Thus f 0 (x) 6 0 for all x > 1 ) f (x) is decreasing for all x > 1 .... (2)
) f (x) is decreasing when ln x > 1 ff 0 (x) < 0g that is, when x>e
And as x > 1, f (x) > 0 for all x > 1 ) f (x) is positive for all x > 1 .... (3)
)
ln x is continuous for x > 1 .... (3) x From (1), (2), and (3), the Integral Test applies for x > 3. Also f (x) =
From (1), (2), and (3) we can apply the Integral Test.
Z
1
x dx x2 + 1
Now 1
µZ
b
b!1
1
µ Z 1 2
= lim
b!1
= lim
¶
1
Now 3
ln x dx = lim b!1 x
1 (ln(b2 b!1 2
b!1
which DNE, fas b ! 1, ln(b + 1) ! 1g )
x
2
b f (x) = xe¡x =
2
ex
2 fex
2
Thus
n=3
6= 0 for all x > 1g
2
2
f 0 (x) = 1e¡x + xe¡x (¡2x) = e¡x (1 ¡ 2x2 ) ) f 0 (x) =
1 ¡ 2x2 ex
2
Since
n=1
n=1
2
2x > 1
1 ¡ 2x2
) f 0 (x) = )
1 .... (2)
f 0 (x) =
And f (x) is positive for all x > 1 .... (3)
)
From (1), (2), and (3) we can apply the Integral Test.
Z
1
Now
µZ
2
b
xe¡x = lim
b!1
1
= lim
µZ µ
b
³
)
1 P
2
ne¡n
1
¡ 12
¶
2
³
1
2
e¡b ¡
1 e
1 dx = lim b!1 x ln x
)
´´
µZ
b
2
µZ
b!1
magenta
¶
1 du dx u dx
ln b
£
ln 2
¶
1 du u
¤ ln b
ln u
ln 2
b!1
)
yellow
1 P
95
which DNE fas b ! 1, ln(ln b) ! 1g 1 diverges. n ln n
100
50
75
25
0
n=2
5
95
100
50
25
0
5
¡(ln x + 1) [x ln x]2
= lim (ln(ln b) ¡ ln(ln 2))
1 ¡ ln x = x2
75
x ¡ ln x(1)
=
du 1 fletting u = ln x, = g dx x
= lim
x2
95
1
b!1
100
50
75
25
0
5
95
100
50
75
25
0
5
1 x
= lim
³ ´
cyan
³ ´
[x ln x]2
2
ln x is positive .... (1) x
and f (x) =
1 P ln n ln 1 ln 2 + + , 1 2 n=3 n
0(x ln x) ¡ 1(1 ln x + x
Now
n=1
0
=
1 is positive for x > 2 .... (1) x ln x fln x > 0 for x > 1g
Z
1 2e
1 x
diverges.
f 0 (x) < 0 when ln x + 1 < 0, which is when x > e¡1 0 f (x) is decreasing for x > 2 .... (2)
is convergent.
c For x > 2, f (x) =
[ln b]2 ! 1g 2
From (1), (2), and (3), the Integral Test applies for x > 2.
ib ¶ 2
h
¶
Also f (x) is continuous for all x > 2 .... (3)
¡ 12 e¡x (¡2x) dx
¡ 12 e¡x
b!1
=
)
1
= lim
b!1
2
xe¡x dx
= lim
b!1
¶
n
3
[ln b]2 [ln 3]2 ¡ 2 2
diverges also.
n
d f (x) =
) 1 ¡ 2x2 < 0
n
1 P ln n
1 P ln n
< 0 for all x > 1
Now if x > 1 then
1 P ln n
dx
¸b
[ln x]2 2
which DNE fas b ! 1,
is continuous for all x > 1 .... (1)
x
¶
" " f (x) f 0 (x)
µ
b!1
n diverges fIntegral Testg n2 + 1
n=1
[ln x]
³1´
3
= lim
2
1 P
b 1
= lim
+ 1) ¡ ln 2)
ln x dx x
3
·
1
¶
b
µZ
b!1
¤ b´
= lim
µZ
= lim
¶
2x dx x2 + 1
ln(x2 + 1)
2
b!1
b
1
³ £ 1
Z
x dx x2 + 1
= lim
f (x) is decreasing for x > 3 .... (2)
black
Y:\HAESE\IB_HL_OPT-Calculus\IB_HL_OPT-Calculus_an\173IB_HL_OPT-Calculus_an.cdr Friday, 8 March 2013 12:50:21 PM BRIAN
IB HL OPT 2ed Calculus
174
WORKED SOLUTIONS 1 is positive for all x .... (1) 1 + x2
)
¡2x (1 + x2 )2
1
1
Z
1
Now 1
1 . 2
.... (¤)
1 dx = lim b!1 1 + x2 = lim
µZ
b
1
£
Z
1
1
1 dx 1 + x2
= So, in ¤,
¼ 4
p = n fas n 6= 0g n p Since 0 6 n 6 Sn , p and lim n DNE,
¤b
1
n!1
by the Comparison test fSn g diverges.
+
) )
P 1
1 P
ln n > 1
)
n ln n > n
) 1 P
an is convergent
n=1
)
)
magenta
yellow
25
0
5
95
100
50
75
25
0
50
75
25
0
5
95
100 cyan
lim
and
3 ¡ 13 (7) 2
2 =
9¡7 2 = = 3£2 3£2
1 3
P1 is true.
12 22 32 42 k2 + + + + :::: + 31 32 33 34 3k 3 ¡ 3¡k (k2 + 3k + 3) = (*) 2
12 22 32 42 k2 (k + 1)2 + 2 + 3 + 4 + :::: + + 31 3 3 3 3k 3(k+1)
)
lim an = 0
1 3
3 ¡ 3¡1 (12 + 3(1) + 3)
(2) If Pk is true, then
1 DNE n!1 an 1 P 1 is divergent ) a n=1 n
75
50
=
=
3 ¡ 3¡k (k2 + 3k + 3) (k + 1)2 + 2 3(k+1)
=
3 ¡ 3¡(k+1) (3k2 + 9k + 9) + 3¡(k+1) 2(k2 + 2k + 1) 2
=
3 ¡ 3¡(k+1) [(3k2 + 9k + 9) ¡ (2k2 + 4k + 2)] 2
n!1
)
25
RHS =
1 P 1 1 1 = p + converges. ln n 2 ln 2 n=3 np ln n
5
1 P
(1) If n = 1, LHS = p
95
n=2
3 ¡ 3¡n (n2 + 3n + 3) , n 2 Z+ 2
Proof: (By the Principle of Mathematical Induction)
1 also converges fComparison testg np ln n np
12 22 32 42 n2 i2 = 1 + 2 + 3 + 4 + :::: + n 3i 3 3 3 3 3 =
1 converges for p > 1 fp-series testg np
P
n P i=1
1 1 < p np ln n n 1 1 P P 1 1 < p p n ln n n n=3 n=3
95
)
n=3 1
the series diverges (by definition).
a Pn : Sn =
12
50
P
)
1 p = lim Sn if this limit exists, n n!1
but from c the sequence of partial sums diverges )
75
n=3 1
p
100
But
n=1
1 diverges fComparison testg np ln n
If p > 1, and n > 3,
)
0
1 diverges. n ln n
np ln n < n ln n
n=2
5
1 P
1 1 > np ln n n ln n 1 1 P P 1 1 > p n ln n n ln n n=2 n=2
)
10
n!1
¼ . 4
n=2
)
1 lim p = 0 (from a), this has no bearing n on the convergence or otherwise of the series.
d Even though
9 If p = 1, we showed in 7 d that If p < 1, np < n
1 if k < ng n
> p
= p
b!1 ¼ ¡ ¼4 2 ¼ 4
1 < 1 + n2
1 n
1
= lim (arctan b ¡ arctan 1) =
1
fsince p
1 dx 1 + x2
arctan x
b!1
1
1 2 3 1 1 1 1 > p + p + p + :::: + p n n n n
.... (3)
1 P 1 1 dx < < a1 + 1 + x2 1 + n2 n=1
where a1 = f (1) =
p i
= p + p + p + :::: + p
From (1), (2), and (3), the Integral Test applies for x > 1. 1
n P 1 i=1
Also f (x) is continuous for all x 2 R
Z
n!1
b Sn =
f 0 (x) < 0 for all x > 1 .... (2)
Now
1 lim p = 0 n
)
) f 0 (x) = ¡(1 + x2 )¡2 (2x) =
p n!1
a As n ! 1,
11
100
8 f (x) =
black
Y:\HAESE\IB_HL_OPT-Calculus\IB_HL_OPT-Calculus_an\174IB_HL_OPT-Calculus_an.cdr Wednesday, 13 March 2013 4:13:50 PM BRIAN
fusing *g
IB HL OPT 2ed Calculus
WORKED SOLUTIONS To be correct to 6 decimal places, we require
=
3 ¡ 3¡(k+1) (k2 + 5k + 7) 2
=
3 ¡ 3¡(k+1) (k2 + 2k + 1 + 3k + 3 + 3) 2
=
3 ¡ 3¡(k+1) [(k + 1)2 + 3(k + 1) + 3] 2
1 < 0:000 000 5 3k3 ) 3k3 > 2 £ 106 ) k3 >
b
) k > 88 )
since 2 £ 3n grows much faster than n2 + 3n + 3,
2
1 P
c Yes,
3 n!1 2
n!1 n2
n=1
3n
Hence
n2
n=1
3n
=
n¡1 n¡2 ¡ n+1 n n2 ¡ n ¡ [n2 ¡ n ¡ 2] = n(n + 1) 2 = , n > 2, n 2 Z + n(n + 1)
) an =
3 2
converges, by definition, since from b the 3 . 2
3 . 2
=
a1 + a2 + a3 + a4 + ::::
b
=0+
f (x) dx < Rk
Pn is true fPOMIg
n!1
2 £ 106 3
r
Thus Pk+1 is true whenever Pk is true, and P1 is true. )
175
black
Y:\HAESE\IB_HL_OPT-Calculus\IB_HL_OPT-Calculus_an\175IB_HL_OPT-Calculus_an.cdr Wednesday, 13 March 2013 4:14:22 PM BRIAN
IB HL OPT 2ed Calculus
176
WORKED SOLUTIONS m ! 1 ) S2m ! 1 2 Thus Sn is divergent.
1 2 3 k k+1 + + + :::: + + 2! 3! 4! (k + 1)! (k + 2)!
)
(k + 1)! ¡ 1 k+1 + (k + 1)! (k + 2)!
= =
c As m ! 1, 1 +
³ k + 2 ´ µ (k + 1)! ¡ 1 ¶ k+2
(k + 1)!
+
EXERCISE K.3
k+1 (k + 2)!
1
a
(k + 2)! ¡ (k + 2) + k + 1 (k + 2)!
=
1 P n=1
)
n=1
) Pn is true fPOMIg 1 d Since Sn = 1 ¡ , lim Sn = 1 (n + 1)! n!1 1 P n ) =1 (n + 1)! n=1 a
S16 = 1 + +
) S16 > 1 + S21 = 1 + S22 = 1 + S23 = 1 + S24 = 1 +
9
+
1 2
+
¡1
) S16 > 1 + b S20 = 1 +
+
¡1
) S16 > 1 + +
1 2
¢ 1
3
+
1 10
+
1 11
4
+
1 4
¡1
1 P
4
+
5
1 12
+
¢
¡1
+
+
1 7
1 13
+
1 14
+
1 8
+
1 8
+
1 16
+
¡1 8
1 1 1 + 16 + 16 + 16 16 1 + 12 + 12 + 12 2 4 2
0 2 1 2 2 2 3 2 4 2
1 6
+
+
n=1 n
=
+
+ 1 16
1 15
1 8
+
n=1 n 1
+
¢
1 16
1 16
+
P 1 1
)
¢
n=1 1
n=1
=
2
1 P 1
=
Proof: (by induction on m) (1) If m = 0, S20 = S1 = 1 + 02 = 1 X ) P0 is true k (2) If Pk is true, S2k > 1 + 2 Now ³ ´ 1 1 1 S2k+1 = S2k + + :::: + k + k k¡1 2 +1 2 ¡1 2
k + 2
1 1 1 + + :::: + 2k 2k 2k
|
{z
n
) S2k+1 ) S2k+1
¡
diverges.
n2
P n¡1
¡
n2
1 P 1
n=1
n
¡
1 P 1 n=1
¸
n2
which converges fp-series testg
1 P (¡1)n
ln n
where bn =
1 ln n
1 P
(¡1)n bn
n=2
Consider y = ln x, x > 0, x 2 R y
¶
ln x is increasing for all x 2 R , x > 0. y = ln x
}
1
) x
1 is decreasing ln x for all x 2 R , x > 0
1 is decreasing for all n > 2, n 2 Z + ln n lim bn = 0
Thus bn = Also
n!1
)
magenta
yellow
1 P
95
(¡1)n converges ln n
100
50
75
25
0
5
95
100
50
75
25
0
5
95
n=1 1
2k
100
50
75
25
0
5
95
100
50
75
25
n2
1 P 1¡n
·
n2
n=2
0
n
which is an alternating series of the form
) Pn is true fPOMIg
5
n=1
P 1
n=1
1 P 1
n=2
Thus P1 is true and Pk+1 is true whenever Pk is true
cyan
¡
converges
2k¡1 of these
) S2k+1
n2
¸
| {z }
³1´ k¡1
k >1+ +2 2 k > 1 + + 12 2 k+1 >1+ 2
n=1 1
1 P 1
1 1 1 1 ¡ + ¡ + :::: ln 2 ln 3 ln 4 ln 5
a
m , m 2 Z , m > 0. 2
n
n2
1 P 1
n=1
¡
n
P 1
c
n=1
) S2k+1 > 1 +
¡
diverges
=
µ
¡
P 1
n=1
So, we conjecture: S2m = 1 +
·
n=1
| {z }
¢
1 16
1 P 1¡n
¡
n=1 n
8
+
1 P 1
=2
¢ 1
n=1
1 converges n2
1¡n diverges. n2
1 P 1
b
1 P
1 diverges fp-series testg n
n=1
Thus P1 is true and Pk+1 is true whenever Pk is true
4
1 P
and
(k + 2)! ¡ 1 = (k + 2)!
¡1
1 1 P P 1¡n 1 1 = ¡ where 2 2 n n n n=1 n=1
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Y:\HAESE\IB_HL_OPT-Calculus\IB_HL_OPT-Calculus_an\176IB_HL_OPT-Calculus_an.cdr Wednesday, 13 March 2013 4:14:54 PM BRIAN
IB HL OPT 2ed Calculus
WORKED SOLUTIONS 1 p 2 x n , consider f (x) = b For (¡1)n¡1 n+4 x+4 n=1
1 P
1 1 x¡ 2
where f 0 (x) = 2
(x + 4) ¡
1 x 2 (1)
p
f 0 (x) 6 0 for all x > 4 f (x) is decreasing for all x > 4 p n then bn is decreasing for all n > 4 Thus if bn = n+4
d
n!
n
³
1 = 1+ n
1 P
Hence,
e
an is divergent fRatio Testg
³¼ ´
n=1
b Consider f (x) = sin f 0 (x) = cos
x
³¼ ´ x
£
n!1
for all x > 1
¡¼ = x2
³¼´
(¡1)n sin
n
n=2
¡¼ cos
³¼´
Hence
x
cyan
f Let f (x) =
which is < 1
(¡1)n is convergent fRatio Testg 2n n!
x2 +1
x3
) f 0 (x) =
and for n > 2, 2 ¡ x3 6 0
yellow
50
75
25
0
5
95
100
50
75
95
magenta
is convergent.
50
n
75
(¡1)n sin
25
³¼´
25
0
5
100
95
1 P n=1
fAlternating Series Testg
n=1
50
¯ ¯ ¯ an+1 ¯ = 0 ¯ ¯ n!1 a lim
n
is convergent
0
)
75
is convergent fRatio Testg
(¡1)n 2n n! ¯ ¯ a 1 2n n! ¯ n+1 ¯ = ) ¯ £ ¯ an 2n+1 (n + 1)! 1 1 = 2(n + 1)
x2
5
P 1
100
1 P
Hence,
25
(2n ¡ 1)!
2x(x3 + 1) ¡ x2 (3x2 ) (x3 + 1)2
=
2x4 + 2x ¡ 3x4 (x3 + 1)2
=
x(2 ¡ x3 ) (x3 + 1)2
lim bn = 0
Also
0
(2n ¡ 1)!
(¡1)n¡1
Let an =
)
) f 0 (x) 6 0 for all x > 2, x 2 R ³¼ ´ ) bn = sin is decreasing for all n > 2, n 2 Z + n
5
1 P (¡1)n¡1
fAlternatively the Alternating Series Test can be usedg
which is > 1
n
1 1 1 1 1 ¡ + ¡ + ¡ :::: 1! 3! 5! 7! 9!
n!
n=1
95
)
n! nn
´n
¯ ¯ ¯ an+1 ¯ = e ¯ ¯ a n!1 lim
=
n!
)
f 0 (x) 6 0 for all x > 2
)
f (x) is decreasing for all x > 2
100
)
´
Let an =
(¡1)n nn
=
which is < 0 for all x > 2
¯ ¯ 1 (2n ¡ 1)! ¯ an+1 ¯ = ) ¯ £ ¯ an (2n + 1)! 1 ¯ ¯ an+1 ¯ 1 ¯ ) ¯ ¯= an (2n + 1)(2n) ¯ ¯ ¯ an+1 ¯ = 0 which is < 1 Thus lim ¯ ¯ n!1 an ³ n¼ ´ 1 sin P 2
p (¡1)n¡1 n is convergent n+4
£
n¼ 2
n=1
fAlternating series testg
)
x
(¡1)n¡1 is convergent. p 3 ln n
=
n!1
¯ ¯ ¯ an+1 ¯ = (n + 1)n+1 ¯ ¯ an (n + 1)! ³ n + 1 ´n
³
P sin n=1
lim bn = 0
Let an =
1 P
n=2
1
1 p n !0 Also, as n ! 1, bn = 4 1+ n
a
4 3x(ln x) 3
³1´
fAlternating Series Testg
) )
3
¡1
lim bn = 0,
4¡x 2 x(x + 4)2
n=1
0
n!1
= p
1 P
4
) f (x) is decreasing for all x > 2 1 Hence bn = p is decreasing for all n > 2 and since 3 ln n
x 2 = (x + 4)2
)
= [ln x]
) f (x) = ) f (x) =
p ¡ x
x
)
¡3
ln x
(x + 4)2 p2 ¡
1
1
¡ ¡ 13 [ln x] 3
0
(x + 4)2
1p p2 2 x+ x
=
c Consider f (x) = p 3
177
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Y:\HAESE\IB_HL_OPT-Calculus\IB_HL_OPT-Calculus_an\177IB_HL_OPT-Calculus_an.cdr Friday, 8 March 2013 1:02:28 PM BRIAN
IB HL OPT 2ed Calculus
178
WORKED SOLUTIONS 1 P
So, if
(¡1)n+1
n=1
n2 = n3 + 1
1 P
So, by the Estimation Theorem, jS ¡ S4 j 6 b5 ) S4 ¡ b5 6 S 6 S4 + b5 ) 0:841 465 6 S 6 0:841 471
(¡1)n+1 bn then bn
n=1
is decreasing for all n > 2 and also
1 n
lim bn = lim
n!1
n!1
1 P
Hence, )
(¡1)
n+1
(¡1)
n+1
n=1
n=1
n!
As 0
0, fbn g is
yellow
a S2 = (¡1)0 b1 + (¡1)1 b2 = b1 ¡ b2 But bn+1 6 bn for all n 2 Z + ) b2 6 b1 )
S2 > 0
50
b S4 = (¡1)0 b1 + (¡1)1 b2 + (¡1)2 b3 + (¡1)3 b4 = b1 ¡ b2 + b3 ¡ b4
25
0
lim bn = 0.
n!1
5
95
100
50
75
25
0
5
95
50
100
magenta
¡
n=1
is convergent
75
25
0
5
95
cyan
1 384
¼ 0:000 021 70 < 0:000 05
=1 = 0:875 ¼ 0:912 037 0 ¼ 0:896 412 0 ¼ 0:904 412 0
75
n=1
100
50
75
25
0
1 1
6 We are given S =
fAlternating Series Testg 1 1 1 1 1 Now S = ¡ + ¡ + ¡ :::: 1! 3! 5! 7! 9! 1 and b5 ¼ ¼ 0:000 002 76 < 0:000 05 9! and S4 ¼ 0:841 468 25
5
(¡1)n¡1 = n3
lim bn = 0
P
+
An estimate of the error in using S10 to approximate S is b11 = 1113 ¼ 0:000 751 3.
n!1
Thus
1 P
5
n=1
(¡1)n¡1
1 48
¡
and S6 ¼ 0:606 510 42
0 < bn+1 < bn for all n 2 Z + ) fbn g is a decreasing sequence, n 2 Z + .
1
1 8
+
But, by the Estimation Theorem jS ¡ S6 j 6 b7 ) S6 ¡ b7 6 S 6 S6 + b7 ) 0:606 489 6 S 6 0:606 532 ) S ¼ 0:6065
n=1
Also
1 2
1 46 084
where b7 =
So, by the Estimation theorem jS ¡ S5 j 6 b6 ) ¡0:001 39 6 S ¡ 0:633 33 6 0:001 39 ) 0:631 94 6 S 6 0:634 72 Thus S ¼ 0:63
Now as 0
1
Thus
1
n2 is convergent fas a1 = 12 g n3 + 1
0 < bn+1 6 bn for all n 2 Z + Also
1 (¡1)n is an alternating series with bn = n 2n n! 2 n!
¡ Now bn+1 ¡ bn = n+1 2 (n + 1)! 2n n!
95
a
(¡1)n+1
S ¼ 0:8415
1 P
100
4
1 P
1 n3 n2 is convergent n3 + 1
1+
n=2 1 P
) =0
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Y:\HAESE\IB_HL_OPT-Calculus\IB_HL_OPT-Calculus_an\178IB_HL_OPT-Calculus_an.cdr Friday, 8 March 2013 1:02:33 PM BRIAN
IB HL OPT 2ed Calculus
WORKED SOLUTIONS Thus S4 ¡ S2 = b1 ¡ b2 + b3 ¡ b4 ¡ (b1 ¡ b2 ) = b3 ¡ b4 which is > 0 fas fbn g is decreasingg
¼
1 P
But
n=1
Thus S2n > S2n¡2 > S2n¡4 > :::: > S4 > S2 So 0 6 S2 6 S4 6 :::: 6 S2n¡4 6 S2n¡2 6 S2n
)
c S2n = b1 ¡ b2 + b3 ¡ b4 + :::: ¡ b2n¡2 + b2n¡1 ¡ b2n ) S2n = b1 ¡ (b2 ¡ b3 ) ¡ (b4 ¡ b5 ) ¡ :::: ¡ (b2n¡2 ¡ b2n¡1 ) ¡ b2n
1 P n=1
)
1 P
)
1 P n=1
S2n 6 b1
d From b and c, fS2n g is increasing and has an upper bound of b1 lim S2n = S.
n!1
= b1 ¡ b2 + b3 ¡ b4 + :::: + b2n+1 ¡ b2n + b2n+1
}
)
S2n S2n+1 = S2n + b2n+1
)
n!1
n!1
n!1
=S
a If an =
)
n!
,
¯ ¯ ¯ ¯ 3n+1 an+1 ¯ ¯ lim ¯ = lim ¯¯ ¯ n!1 n!1
¯ n! ¯¯ (n + 1)! 3n ¯
an
= lim
n!1
¯ ¯ ¯ an+1 ¯ ¯ ¯ n!1
(¡1)n 2n , n2 + 1
b If an =
lim
2n+1 = lim n!1 (n + 1)2 + 1 = lim 2 n!1
0
n!1
=2£
cyan
1 + n12 1+
magenta
2 n
+
2 n2
4
fa convergent GSg fComparison testg
is absolutely convergent.
ln x x
1 x x
¡ ln x
=
x2
1 ¡ ln x x2
f 0 (x) < 0 for all x such that ln x > 1, that is, x > e
n ln n o n
is decreasing for all n > 3, n 2 Z + .... (1) 1
lim
1 P n=1
n2 + 1 2n
¶
f Let bn =
1 , then n ln n
Now consider f (x) =
¶
(¡1)n+1 ln n n
is absolutely
lim bn = 0
n!1
1 = [x ln x]¡1 , x > 2 x ln x
³
) f 0 (x) = ¡[x ln x]¡2 1 ln x + x
1
¡(ln x + 1) = (x ln x)2
A
³ 1 ´´ x -
2
So, f 0 (x) < 0 for all x > 2 1 is decreasing for all n > 2, n 2 Z + ) bn = n ln n Thus
1 P
yellow
95
50
75
n=2
95
50
75
25
0
4n + 3
¡ 3 ¢n
ln x = lim x = 0 x!1 x n!1 1 ln n ) lim = 0, n 2 Z + .... (2) n!1 n
1 1
5
95
100
50
75
25
0
95
100
50
75
25
0
5
n=1
3 + 4n
By l’Hˆopital’s Rule,
=2 (¡1)n 2n is divergent fRatio testg n2 + 1
5
)
1 P
µ
n2 + 1 n2 + 2n + 2
= lim 2@
jan j is convergent
³ 3n ¡ 1 ´n
convergent. fAlternating Series Testg
an
µ
jan j
0, b4 ¡b5 > 0, ...., b2n¡2 ¡b2n¡1 > 0, and b2n > 0
e
jarctan nj 2 6 3 n3 n
jan j =
and as S2n ¡ S2n¡2 = b2n¡1 ¡ b2n > 0 S2n ¡ S2n¡2 > 0 for all n 2 Z + .
)
(¡1)n arctan n then n3
c If an =
Likewise S6 ¡ S4 > 0, S8 ¡ S6 > 0, etc,
)
179
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Y:\HAESE\IB_HL_OPT-Calculus\IB_HL_OPT-Calculus_an\179IB_HL_OPT-Calculus_an.cdr Friday, 8 March 2013 2:53:42 PM BRIAN
IB HL OPT 2ed Calculus
180
WORKED SOLUTIONS g Let br = ) br =
p p r+1¡ r
¡p
r+1¡
p
r
¢
p ¶ r+1+ r p p r+1+ r
µp
r+1¡r p r+1+ r 1 1 ) br = p p > p r+1+ r 2 r+1 1 P r=1 1
P
)
br >
1 2
br >
1 2
r=1
P 1
where
n=2 1 P
)
1 P
Hence
1 P
c
lim
n!1
1 P
)
fn = r + 1g
2n 2 = lim 8n ¡ 5 n!1 8 ¡
n=1
1 p is divergent fp-series testg n
d Let an =
cos
p p (¡1)r¡1 ( r + 1 ¡ r) is divergent.
xn for x 2 R . n!
a Let an =
¯ ¯ ¯ ¯ an+1 ¯ = ¯¯ xn+1 ¯ a ¯ ¯ (n + 1)! n ¯ ¯ ¯ an+1 ¯ = 0 lim ¯ ¯ n!1 a
¯
)
jan j 6
)
06
)
n=1
)
)
n=0
)
xn is absolutely convergent and ) convergent. n!
an is convergent )
n=0
e
lim an = 0
n!1
n3 n4
1 P
9
a
1 P n=0
)
10n is n!
1 P n=0 1 P
by 8 a,
n=0
1
(n + 1)(n + 1)
1,
n=1
1 = n+1
n=1
µ
P 1
n=1
1 n
2
is absolutely convergent and
n3 +1 > 4 for all n 2 Z + , n > 2 ¡1 n 1 P n3 + 1 1 > n4 ¡ 1 n n=2
1 P 1 P
1 diverges n
fp-series testg
n3 + 1 diverges n4 ¡ 1
fComparison testg
n! 2 £ 5 £ 8 £ :::: £ (3n + 2) ¯a ¯ (n + 1)! ¯ n+1 ¯ = ) ¯ ¯ an 2 £ 5 £ 8 £ :::: £ (3n + 5) 2 £ 5 £ 8 £ :::: £ (3n + 2) £ n! n+1 = 3n + 5 Let an =
)
¶
fComparison testg
convergent.
n=2
xn with x = 10 n!
1 n2
¡n¢
n2 + 4n
n=2
)
1 P n=1
cos
n=2
But
and n2 + 4n > n2 , n 2 Z +
jan j is convergent
1 P
)
xn is convergent n=0 n! xn =0 ) lim n!1 n!
)
jan j 6
1 P
)
b By the Converse of the Test for divergence 1 P
2
1 is convergent fp-series testg n2
n=1
n
1 P
1 P
6= 0
1 n2
1 P
1 P
where
n! ¯¯ jxj = xn ¯ n+1
1 4
n2 + 4n
n=1
Now
=
¡n¢
¯ ³ ´¯ ¯cos n ¯ 6 1 ¯ ¯ 2
Now
br is divergent
5 n
2n is divergent. 8n ¡ 5
r=1
8
diverges
n(n + 1)
fComparison test for seriesg
r=1
)
1
p
n=1
1 p r+1 r=1 p n n=2
diverges fp-series testg, it follows that
1 diverges. n+1
n=1
1 P
1 P 1
1 n
n=1
1 P
) br = p
)
1 P
Since
)
¯ ¯ ¯ an+1 ¯ = lim ¯ ¯ n!1 n!1 a lim
n
by the Ratio test
¡ 1, which is the
1 P
1+ 3+
1 n 5 n
=
1 3
6= 0
an converges.
n=0
cyan
magenta
yellow
95
100
50
75
25
0
5
95
100
50
75
25
0
5
95
100
50
75
25
0
5
95
100
50
75
25
0
5
Harmonic Series with the first term removed.
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Y:\HAESE\IB_HL_OPT-Calculus\IB_HL_OPT-Calculus_an\180IB_HL_OPT-Calculus_an.cdr Friday, 8 March 2013 1:08:05 PM BRIAN
IB HL OPT 2ed Calculus
181
WORKED SOLUTIONS 1 , n2
10 If an =
¯ ¯ ¯ an+1 ¯ = lim ¯ ¯ n!1 n!1 lim
1 n2 (n + 1)2 1
an
n2 n2 + 2n + 1 1
= lim
n!1
= lim
by the Ratio test
1 1 1 1 = 1 + 2 + 4 + 6 + :::: is geometric with a1 = 1 2n x x x x n=0 1 and r = 2 . x )
1 n2
1 P
n=0
=1
1 P n=1
divergent.
1 P
2 1+ n + n12
n!1
)
2
But, from the p-series test it is convergent. Likewise, if an =
1 , n
¯ ¯ ¯ an+1 ¯ = lim ¯ ¯ n!1 n!1 lim
an
= lim
n!1
)
¯ 2¯ ¯x ¯ > 1
)
jxj > 1
)
1 n n+1 1
)
1
3
x2 x2
¶
1 x2 ¯ ¯ ¯ 1 ¯ 1 or x < ¡1
the interval of convergence is ]¡1, ¡1[ [ ]1, 1[ .
a If an = n5n xn ,
¯ ¯ ¯a ¯ ¯ (n + 1)5n+1 xn+1 ¯ ¯ n+1 ¯ ¯ ¯ lim ¯ lim ¯ ¯ = n!1 ¯ n!1 an n5n xn ¯³ ¯ ´ 1 ¯ ¯ = lim ¯ 1 + 5x¯ n!1
1 1+ n
=1 So, once again the ratio test is inconclusive.
n
= j5xj
EXERCISE K.4 1
a
1 P
= 5 jxj
x3n = 1+x3 +x6 +x9 +x12 +:::: which is geometric
with u1 = 1 and r = 1 P
)
x3n =
n=0
x3 .
b
(2 ¡
x)n
1 u1 = 1¡r 1 ¡ x3
provided
and diverges for jxj >
jrj < 1
)
¯ 3¯ ¯x ¯ < 1
)
jxj < 1
)
= 1 + (2 ¡ x) + (2 ¡
x)2
+ (2 ¡
x)3
each case (x = If x =
1 , 5
+ ::::
P
1 P
j2 ¡ xj < 1 jx ¡ 2j < 1 ¡1 < x ¡ 2 < 1 1 2, n 2 Z + ) bn = n ln n and
magenta
P 1
)
n=2
¯ n ln n ¯¯ £ (2x + 3)n ¯
lim bn = 0
So,
1 P n=2
(¡1)n is convergent fAlternating Series testg n ln n (¡1)n (2x + 3)n is convergent for n ln n
¡2 < x 6 ¡1. Hence, the radius of convergence is convergence is ]¡2, ¡1].
1 2
and the interval of
yellow
95
50
75
25
0
5
95
100
50
75
25
0
ln n ! 1g ln(n + 1)
5
95
100
50
75
25
0
1 = [x ln x]¡1 x ln x
)
n ln n £ n+1 ln(n + 1)
n ! 1 and n+1
5
95
100
50
75
n=2
(¡1)n 1 where bn = , n ln n n ln n
) f 0 (x) < 0 for all x > 2 fln x + 1 > 0, [x ln x]2 > 0g
= j2x + 3j
25
1 P
1 is divergent. n ln n
n!1
n!1
0
1 P
(ln x + 1) =¡ [x ln x]2
(¡1)n (2x + 3)n d If an = , ¯n ln n ¯
5
ln 2
) f 0 (x) = ¡[x ln x]¡2 1 ln x + x
It has an infinite radius of convergence and its interval of convergence is R .
cyan
when x = 2, u = ln 2 when x = b, u = ln b
¤ ln b ´
by the Integral test,
n > 2.
an is convergent for all x 2 R .
fas
ln juj
For x = ¡1, we have
n=0
n!1
ln 2
1 du u
n=2
, then
= lim j2x + 3j £
¶ ½
ln b
du 1 = g dx x
b!1
=0
1 P
³£
fu = ln x,
= lim (ln(ln b) ¡ ln(ln 2))
¯ ¯ ¯ ¯ ¯ an+1 ¯ = lim ¯¯ x2n+1 (2n ¡ 1)! ¯¯ lim ¯ ¯ n!1 ¯ n!1 an (2n + 1)! x2n¡1 ¯ ¯ ¯ ¯ ¯ x2 ¯ ¯ = lim ¯ n!1 2n(2n + 1) ¯ )
¶
which DNE
is convergent for ¡ 13 6 x 6
It has a radius of convergence of
1 du dx u dx
100
)
3n xn
b
2
µZ
b!1
(¡1)n (n + 1)2
which converges by the Alternating series test. 1 P
1 dx x ln x
2
µZ
¶
b
= lim
1 converges. (n + 1)2
3n ¡ 13
µZ
= lim
1 converges fp-series testg n2
1 P
n=2
1 n ln n
1 dx x ln x
= lim
1 1 < where (n + 1)2 n2 n=0
1 P
1
Now
1 P
n=0
If x = ¡ 13 ,
1 , 3
1 P
=
1)2
(n +
by the Comparison test
1 P
If x = ¡2, we have
fRatio testg
As the Ratio test is inconclusive for jxj =
)
¡1 < 2x + 3 < 1 ¡4 < 2x < ¡2
j3xj < 1 ) jxj
1 also
)
Thus
P
and
£ (¡1)n
n=1
R
n=0 1
lim an 6= 0
P
p
p R.
n=0
n!1 1
)
1 P
7
1 P
where
jan j > 1 fas each fraction is > 1g
)
)
¡8¢
1 P
(cn + dn )xn =
n=0
2 £ 4 £ 6 £ 8 £ :::: £ (2n)(¡1)n 1 £ 3 £ 5 £ 7 £ :::: £ (2n ¡ 1)
¡4¢
the radius of convergence is
1 P
7
As the Ratio test is inconclusive for x = §1, we examine each of these cases separately.
=
cn x2n converges when jxj
1 or x < ¡1 ) x 2 ]¡1, ¡1[ [ ]1, 1[
=0 xn¡1
1 P
£ x2n ¯ =
x2(n+1)
) interval of convergence
x
¶
or
(¡1)n x2n
jxj = lim n!1 n 1 P
0
(0:1)n ¡0 n
1 n n(n ¡ 1)! 10 n=1
n
¶
0:1
1 P
¯ ¯ ¯ ¯ an+1 ¯ = ¯ ¯ a ¯ ¯
n=1 (n ¡ 1)!
¯ ¯ ¯ ¯ ¯ an+1 ¯ = lim ¯¯ xn (n ¡ 1)! ¯¯ ¯ ¯ n¡1 ¯ n!1 n!1 ¯ lim
where
1 P
dx
(n ¡ 1)!
1 = (n ¡ 1)! n=1
dx n!
n=1
n=1
1 P
´ 1 ³ P d xn
=
¶
1 P xn¡1
1 = (n ¡ 1)! n=1
nxn¡1 converges for all x 2 ]¡3, 3], and has radius n2 3n
1 P xn
xn¡1 dx is defined (n ¡ 1)!
1 P
of convergence 3.
µ
µ
0
converges conditionally.
n=0
0:1
)
(¡1)n¡1 When x = ¡3, an = and 3n 1 1 P P (¡1)n¡1 an = which 3n n=1 n=1 1 P
n=1
Z
divergent. fp-series testg
)
P
0:1 1
)
When x = ¡1,
1 P
(¡1)2n =
n=0
by Example 38 a this has an
since
1 P
n=0
1n =
1 P
1 diverges,
n=0
lim 1 6= 0.
n!1
cyan
magenta
yellow
95
100
50
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25
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5
95
100
50
75
25
0
5
95
100
50
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25
0
5
95
100
50
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25
0
5
infinite radius of convergence.
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Y:\HAESE\IB_HL_OPT-Calculus\IB_HL_OPT-Calculus_an\184IB_HL_OPT-Calculus_an.cdr Friday, 8 March 2013 1:10:43 PM BRIAN
IB HL OPT 2ed Calculus
WORKED SOLUTIONS 1 P
When x = 1,
1 P
12n =
n=0
1 diverges,
n=0
lim 1 6= 0.
since
n!1
)
the interval of convergence is ]¡1, 1[
Z
1 2
µ
0
¶
1 P
x2n
1 P
dx =
n=0
ÃZ
1 2
2n + 1
n=0 1 P
1 = n=0 2n + 1
Z 2µ
) 0
1 P
0
³¡ ¢2n+1 1
)
0:5805 < sin
¡0
2
1 (2n + 1) £ 22n+1 n=0
¡ sin
5
¡¼¢
¡ ¼ ¢5 5
¡¼¢
¡ ¼5¢ 5
< 0:0065
< 0:0065
< ¡0:5805
< 0:5935
the approximation is accurate to 1 decimal place only. x5 x3 + + R5 (x : 0) 3! 5!
¯ ¯ ¯ f (6) (c)x6 ¯ ¯ ¯ jR5 (x : 0)j = ¯ ¯ 6! ¯ ¯ ¯ ¡ sin c £ x6 ¯ ¯ ¯ =¯ ¯ 6! =
x2n has an interval of convergence
jsin cj jxj6 6!
and c lies between 0 and x. But jxj 6 0:3 and max jsin cj = sin(0:3)
dx is not defined as the interval [0, 2]
x
¡0:5935 < ¡ sin
where
´
¶ 2n
)
2 sin x ¼ x ¡
n=0
of ]¡1, 1[ .
¡0:0065 < 0:5870 ¡ sin
1 P
1 P
11 From question 10 part c,
)
)
x2n dx
0 1 · 2n+1 ¸ 12 1 P x @ A
=
¡0:0065 < T3
0
n=0
=
!
¡¼¢
)
185
n=0
sin(0:3)(0:3)6 6!
)
jR5 (x : 0)j
0 for all k, 1 + uk 6 euk (1 + u1 )(1 + u2 )(1 + u3 ) :::: (1 + un ) 6 eu1 eu2 eu3 ::::eun
yellow
95
´³
´
x x x2 1+ =1¡ 2 ¼ ¼ ¼ ³ ´³ ´ x x x2 1¡ 1+ =1¡ 2¼ 2¼ 4¼ 2 ³ ´³ ´ x x x2 1¡ 1+ =1¡ , and so on. 3¼ 3¼ 9¼ 2 1¡
100
50
75
25
0
5
95
100
50
75
25
0
5
95
100
50
75
25
0
5
95
100
50
75
25
0
= eu1 +u2 +u3 +::::+un
5
x2 (2n + 3)(2n + 2)
c The zeros of ³ ´³ ´³ ´³ ´³ ´ x x x x x 1¡ 1+ 1¡ 1+ 1¡ :::: ¼ ¼ 2¼ 2¼ 3¼ are §¼, §2¼, §3¼, §4¼, ....
) ex ¡ (1 + x) > 0 for all x > 0
magenta
¯ ¯ ¯ an+1 ¯ ¯ ¯ n!1 an ¯ ¯ ¯ x2n+2 (2n + 1)! ¯ ¯ ¯ = lim ¯ n!1 (2n + 3)! x2n ¯ lim
n!1
x3 x4 x2 + + + :::: 2! 3! 4! x2 x3 x4 Now for x > 0, + + + :::: is > 0 2! 3! 4!
cyan
(¡1)n x2n , (2n + 1)!
¸
µ3 µ5 µ7 + ¡ + :::: 3! 5! 7!
a ex = 1 + x +
)
(2n + 1)!
= lim
b eiµ = cos µ + i sin µ 6
1 P (¡1)n x2n
¸
·
(1, 0)
)
iµ i2 µ2 i3 µ3 i4 µ 4 i5 µ 5 + + + + + :::: 1! 2! 3! 4! 5!
+i µ¡
(-1, 0)
sin x = 0 , x = k¼, k 2 Z , x 6= 0 x sin x = 0 , x = §¼, § 2¼, § 3¼, .... x x3 x5 x7 sin x = x ¡ + ¡ + :::: 3! 5! 7!
b
k!
= 1¡
y
x
1 (iµ)k P k=0
(1 + uk ) has an upper bound of eL
(1 + uk ) is increasing and has an upper bound, it
zk for all z k!
k=0
(1 + uk ) 6 eL
k=1
¶
a0 = 0, a1 = 1, a2 = 0, a3 = tan x ¼ x +
un converges
n=1
k=1
By equating coefficients we get
)
Q
As
= a0 + a1 x + (a2 ¡ 12 a0 )x2 + (a3 ¡ 12 a1 )x3 + (a4 ¡ 12 a2 +
n Q
1 P
k=1 n
sin x , sin x = cos x tan x cos x
as tan x =
n Q
Thus
¡ :::: and
5!
(1 + uk ) is increasing and as
k=1
x2 x4 cos x = 1 ¡ + ¡ :::: 2! 4! Let tan x = a0 + a1 x + a2 x2 + a3 x3 + a4 x4 + :::: )
n Q
)
¼ That is, for x 2 R , x 6= (2k + 1) , k 2 Z . 2 x5
(1 + uk ) 6 eu1 +u2 +u3 +::::+un
k=1
b cos x has R = 1 and is defined for all x 2 R . x equals its full Maclaurin series expansion for all ) cos x x is defined. x 2 R such that cos x
x3
n Q
)
black
Y:\HAESE\IB_HL_OPT-Calculus\IB_HL_OPT-Calculus_an\192IB_HL_OPT-Calculus_an.cdr Friday, 8 March 2013 3:02:46 PM BRIAN
IB HL OPT 2ed Calculus
193
WORKED SOLUTIONS
³
)
x ¼
1¡
µ
´³
x ¼
1+
x2 ¼2
1¡
=
´³
¶µ
1¡
1¡
x2 4¼ 2
x 2¼
´³
¶µ
1¡
x2 9¼ 2
´³
x 2¼
1+
1¡
¶
x 3¼
´
y = 2x ¡ 2 + ce¡x is a solution of
)
::::
b We need graphs of y y y y y
::::
sin x has zeros of §¼, §2¼, §3¼, .... x ³ ´³ ´³ ´³ ´ x x x x From c, 1 ¡ 1+ 1¡ 1+ :::: has ¼ ¼ 2¼ 2¼ From a,
= 2x ¡ 2 = 2x ¡ 2 + e¡x = 2x ¡ 2 ¡ e¡x = 2x ¡ 2 + 2e¡x = 2x ¡ 2 ¡ 2e¡x
c=2
the same zeros.
dy = 2x ¡ y dx (c = 0) (c = 1) (c = ¡1) (c = 2) (c = ¡2)
y 4
This evidence supports Euler’s claim that
sin x x2 x4 x6 =1¡ + ¡ + :::: is equal to x 3! 5! 7!
µ
x2 1¡ 2 ¼
e Now,
µ
x2 1¡ 2 ¼
µ
x2 1¡ 4¼ 2
¶µ
x2 1¡ 4¼2
¶µ 2
x ¼2
1¡
and
¶µ
h
1¡
³1 2
1 + ¼2 4¼ 2
= 1¡x =1 ¡ x2
x2 4¼2
³1
¼2
+
¶µ
2
=1¡x
¶µ 1¡
´
¶
x2 1¡ 9¼ 2
¶
:::: (1)
³1
´
c=0
¶
x2 1¡ 9¼ 2
¶
1 9
1 16
+
+ ::::
That is, y = ¡x + 1
+ :::: 2
1 P 1 1 = = 2 2 (2n) 4n n=1 n=1 1 P
1 ) = 2 n=1 (2n)
n=1
n2
=
µ
1 4
¼2 6
¶
p
x2 + c then y 2 = x2 + c
1 P 1 n=1
)
n2
2r
+
p x dy = . x2 + c is a general solution of dx y p b As y(3) = 4, 4 = 9 + c ) 16 = 9 + c ) c=7 p Thus y = x2 + 7
1 P r=1
1 , (2r ¡ 1)2
1 P
1 ¼2 ¼2 ¼2 = ¡ = 2 6 24 8 n=1 (2n ¡ 1)
3
a
y 2 1
EXERCISE M 1
a If y = 2x ¡ 2 + ce¡x then -2
dy = 2 + ce¡x (¡1) dx
magenta
yellow
2 x
95
100
50
75
25
0
5
95
100
50
75
25
0
5
95
1
-2
100
50
75
25
0
5
95
100
50
75
25
0
cyan
-1 -1
= 2 ¡ ce¡x = 2 ¡ [y ¡ 2x + 2] = 2 ¡ y + 2x ¡ 2 = 2x ¡ y
5
dy = 2x dx dy x = dx y
Thus y =
¼2 = 24
´ 1 ³ P 1 2 r=1
1 4
1 P ¼2 ¼2 1 = + 6 24 n=1 (2n ¡ 1)2
)
a If y =
) 2y
1 P
1 P 1
y¡1 = ¡1 x¡0
) the equation is
¼2 ) = 2 6 n=1 n
and as
dy = 2(0) ¡ (1) = ¡1 dx
d At (0, 1),
1 P 1
f
c = -2
c If (0, 1) lies on y = 2x ¡ 2 + ce¡x then 1 = ¡2 + c ) c = 3 ) y = 2x ¡ 2 + 3e¡x
´
+
x
-4
1 1 1 =¡ + + + :::: 3! ¼2 4¼2 9¼ 2 1 4
4
c = -1
³1
¼2 =1+ 6
)
2 -2
By equating the coefficients of x2 in (1) we get ¡
-2
´
iµ
1 1 + 4¼ 2 9¼2
-4
1 + + :::: ¼2 4¼ 2
x2 9¼ 2
+ ::::
2
c=1
black
Y:\HAESE\IB_HL_OPT-Calculus\IB_HL_OPT-Calculus_AN\193IB_HL_OPT-Calculus_an.cdr Wednesday, 13 March 2013 2:43:29 PM BRIAN
IB HL OPT 2ed Calculus
194
WORKED SOLUTIONS b
6
y
a
y
2
2
1
1 -3
x -1
-2
1
-2
-1 1
2
2
x
3
-1 -1 -2 -2
y
¡2 ¡1 0 1 2
¡2 0:70 0:35 0 ¡0:35 ¡0:70
dy is undefined when y = 5x ¡ 10 dx dy = 0 when x2 + 4y 2 = 1 ii dx
b
dy 4 = 10y tan x, x in degrees dx
x 0 0 0 0 0 0
¡1 0:35 0:17 0 ¡0:17 ¡0:35
1 ¡0:35 ¡0:17 0 0:17 0:35
i
2 ¡0:70 ¡0:35 0 0:35 0:70
y
y = 5x - 10
2 1 -3
-2
-1
x2 + 4y2 = 1
y 2
1
2
x
3
-1 -2
1
-1
2 x
1
nx
n+1 = xn + h yn+1 = yn + hf (xn , yn ) where h = 0:2
and f (xn , yn ) = 1 + 2xn ¡ 3yn
-1
) yn+1 = yn + 0:2(1 + 2xn ¡ 3yn ) = 0:2 + 0:4xn + 0:4yn
-2
2
x0 x1 x2 x3 x4 x5
2
5 If k = 0,
x + y ¡ 1 = 0 ) y = ¡x + 1
If k = 2, If k = 4,
x2 + y ¡ 1 = 2 ) y = ¡x2 + 3 x2 + y ¡ 1 = 4 ) y = ¡x2 + 5 y
8 -4
n+1 = xn + 0:1 yn+1 = yn + 0:1(sin(xn + yn ))
x0 x1 x2 x3 x4 x5
x y = -x2 + 5 -4 y = -x2 + 3
y = -x2 + 1
-2
=0 = 0:1 = 0:2 = 0:3 = 0:4 = 0:5
dy = 1, y(4) = 3 dx dy 1 ) = dx 2¡x Z 1 ) y=¡ dx x¡2
a (2 ¡ x)
1
1 1
2
4 x
3
magenta
yellow
95
50
75
25
0
5
95
100
50
75
25
0
5
95
100
50
) y = ¡ ln jx ¡ 2j + c
75
25
0
¼ 0:5 ¼ 0:547 94 ¼ 0:608 30 ¼ 0:680 61 ¼ 0:763 69 ¼ 0:855 52
EXERCISE N
-1-1
5
95
100
50
75
25
0
5
y0 y1 y2 y3 y4 y5
So, y(0:5) ¼ y5 ¼ 0:856
y
-2 -3 -4 -5
cyan
=1 = 0:6 = 0:52 = 0:568 = 0:6672 = 0:786 88
nx
4
-4 -3
y0 y1 y2 y3 y4 y5
So, y(1) ¼ y5 ¼ 0:787
4
5 4 3 2
=0 = 0:2 = 0:4 = 0:6 = 0:8 =1
100
-2
7
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Y:\HAESE\IB_HL_OPT-Calculus\IB_HL_OPT-Calculus_AN\194IB_HL_OPT-Calculus_an.cdr Friday, 15 March 2013 10:11:13 AM BRIAN
IB HL OPT 2ed Calculus
WORKED SOLUTIONS
Z
But when x = 4, y = 3 ) 3 = ¡ ln 2 + c ) c = 3 + ln 2 Hence, y = 3 + ln 2 ¡ ln jx ¡ 2j
But, when x = e, y =
x¡2
dy ¡ 3x sec y = 0, y(1) = 0 dx dy ) = 3x sec y dx dy cos y = 3x dx Z Z dy ) cos y dx = 3x dx dx
Z
)
2
a
3 2
) c= Hence, sin y =
2
3 2
¡
£3
) y = arcsin
2
=
3 (x2 2
)
+c
Z
¡ 32
)
¡ 1)
¤
(x2 ¡ 1)
Z
1 4
ey dy = y e +3
1 4
4x + 4 dx 2x2 + 4x + 1
) ln jey + 3j =
1 4
ln ¯2x2 + 4x + 1¯ + c
Z
Thus,
4x + 4 dx + 4x + 1
ey dy dx = y e + 3 dx
)
¯
)
¯
¯ ¯1 + 3) = ln ¯2x2 + 4x + 1¯ 4
) ey =
+c
d
Hence y =
µ ¯1 ¯ 2 ¯2x + 4x + 1¯ 4
Z
¶
3
(e2 + 3)
j2x2 + 4x + 1j (e2 + 3) ¡ 3
hp 4
a
(x ¡ 1) then y = ¡1 (x + 1)2 ¡(x ¡ 1) then y = 1 (x + 1)2
1¡x (x + 1)2
dT / T ¡ R, t > 0 dt
dT = k(T ¡ R) dt 1 dT ) =k T ¡ R dt Z Z dT 1 ) dt = k dt T ¡ R dt
i
j2x2 + 4x + 1j (e2 + 3) ¡ 3
dy = cos2 y, y(e) = dx dy 1 ) sec2 y = dx x sec2 y
(x ¡ 1) (x + 1)2
From a,
x
)
´
2 dx x+1
jx ¡ 1j (x + 1)2
which is true.
p 4
) y = ln
´
¯ ¯ ¯ x¡1 ¯ ¯ (x + 1)2 ¯
When x = 0 and y =
c = ln(e2 + 3)
Thus, ln(ey + 3) = ln
x¡1
¡
which is false.
ln(e2 + 3) = ln 1 + c
)
Z ³ 1
Check: When x = 0 and y =
But when x = 0, y = 2 )
³
ln jyj = ln ¯¯
) y=§
But ey + 3 > 0 for all y 2 R ln(ey
3¡x x2 ¡ 1
1 dy dx = y dx
) jyj =
2x2
Z
=
But when x = 0, y = 1 ) 0 = 0 ¡ 2(0) + c ) c=0
dy c e (2x + 4x + 1) = (x + 1)(ey + 3), y(0) = 2 dx ³ ey ´ dy x+1 ) = ey + 3 dx 2x2 + 4x + 1
Z
x + 1 ¡ 2x + 2 (x ¡ 1)(x + 1)
) ln jyj = ln jx ¡ 1j ¡ 2 ln jx + 1j + c
2
)
=
dy 3y ¡ xy 3¡x = 2 =y dx x ¡1 x2 ¡ 1 1 dy 1 2 = ¡ ffrom ag y dx x¡1 x+1
b
But when x = 1, y = 0 ) 0 =
y
1 2 (x + 1) ¡ 2(x ¡ 1) ¡ = x¡1 x+1 (x ¡ 1)(x + 1)
3x dx
3x2 ) sin y = +c 2
3x2
) 1=1+c ) c=0
¼ 4
) tan y = ln jxj ) y = arctan(ln jxj)
Z
cos y dy =
1 dx x
) tan y = ln jxj + c
¯ ¯ ¯ 2 ¯+3 y = ln ¯ ¯
b
Z sec2 y dy =
)
195
dy dx = dx
Z
¼ 4
Z
)
1 dT = T ¡R
Z
k dt
) ln jT ¡ Rj = kt + c
1 dx x
) T ¡ R = §ekt+c ) T = §ec ekt + R
cyan
magenta
yellow
95
100
50
75
25
0
5
95
100
50
75
25
0
5
95
100
50
75
25
0
5
95
100
50
75
25
0
5
) T = Aekt + R .... (1)
black
Y:\HAESE\IB_HL_OPT-Calculus\IB_HL_OPT-Calculus_an\195IB_HL_OPT-Calculus_an.cdr Wednesday, 13 March 2013 12:22:08 PM BRIAN
IB HL OPT 2ed Calculus
196
WORKED SOLUTIONS b But when t = 0, R = 18, and T = 82
5
AP : PB = 2 : 1 ) A is (3x, 0) and B is (0, 32 y)
y
) 82 ¡ 18 = Ae0 ) A = 64 Thus (1) becomes T = 64ekt + 18 .... (2)
B
But when t = 6, T = 50 ) 50 = 64e6k + 18
P(x, y)
1
2
) 64e6k = 32 )
e6k =
)
ek =
¡ 1 ¢ 16 2
¡ 1 ¢ 6t
Thus (2) becomes, T = 64
2
+ 18
t ¡6
or T = 64 £ 2
+ 18
Z
t
¡6
Now when T = 26,
26 = 64 £ 2
+ 18
=8
) 64 £ 2
t
¡6
)
2
)
=
1 8
=2
+ 18
) 64 £ 2
=2
t
)
2
¡6
)
=
1 32
1 x
) y = §p
= 2¡5
So, from t = 18 to t = 30, T decreases from
26±
to
20± .
This takes place over a 12 minute time interval.
y x gradient of tangent at x P(x, y) is ¡ y x dy =¡ dx y
Gradient of [OP] = ) P(x, y)
)
6
a
Z (0, 0)
Hence, y
Z )
)
x
Q
) ln jmj = kt + c ) jmj = ekt+c ) m = §ec ekt
y dy = ¡
x dx
) m = Aekt .... (1) But at t = 0, m = m0 ) m0 = A
Z
y2 2
=¡
x2 2
)
b At t = 30, m(t) = ) (2) becomes
But when x = 1, y = 2 ) 1 + 4 = 2c ) 2c = 5
magenta
yellow
=
95
4 5
1 ¡ 4 ¢ 30
5
100
50
75
25
0
5
95
100
50
75
25
0
5
95
100
50
75
25
0
= m0 e30k
) ek =
5
95
100 cyan
.... (2)
4 m 5 0
4 m 5 0 30k
) e
x2 + y2 = 5
75
50
(1) becomes m = m0 ekt
That is, m(t) = m0 ekt
+c
) x2 + y 2 = 2c
25
k dt
x dx
)
0
Z
dy dx = ¡ dx
)
5
1 dm dt = m dt
dy = ¡x dx
Z
)
fas x must be > 0g
dm /m dt dm ) = km dt 1 dm ) =k m dt
Z
y
1
¡2
1 Check: When x = 1 and y = p , y = 1 X x 1 When x = 1 and y = ¡ p , y = ¡1 x 1 ) y= p x
t = 30
y
1 dx x
Thus, ln jyj = ¡ 12 ln jxj = ln jxj
t
¡6
Z
But when x = 1, y = 1 ) 0=0+c ) c=0
t
¡6
20 = 64 £ 2
1 dy dx = ¡ 12 y dx
) ln jyj = ¡ 12 ln jxj + c
¡3
t = 18
and when T = 20,
¡ 32 y dy OB ¡y ) =¡ = = dx OA 3x 2x 1 dy ¡1 ) = y dx 2x
)
t
¡6
4
x
A
1 2
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Y:\HAESE\IB_HL_OPT-Calculus\IB_HL_OPT-Calculus_AN\196IB_HL_OPT-Calculus_an.cdr Friday, 15 March 2013 10:15:50 AM BRIAN
IB HL OPT 2ed Calculus
WORKED SOLUTIONS Hence in (2), m(t) = m0
¡ 4 ¢ 30t
Since 1 + v2 > 0 for all v,
5
m(t) =
Now when
1
arctan v = ln(1 + v2 ) 2 + ln jxj + c
1 m , 2 0
)
(0:8)
) arctan
= 0:5
t ln(0:5) = 30 ln(0:8)
) )
c
x¡y dy = a dx x Let y = vx,
dv Hence x +v dx dv ) x dx 1 dv ) 1 ¡ 2v dx Z 1 ) dv 1 ¡ 2v
dy y2 ¡ x2 = dx 2xy Let y = vx
dv dy = x+v dx dx
)
) x
1 dx x
)
Z
1 ln j1 ¡ 2vj = ln jxj + c ¡2 ¯ ¯ 2y ¯ ¯ ) ln ¯1 ¡ ¯ = ¡2 ln jxj ¡ 2c x ¯ ¯ ¯ x ¡ 2y ¯ + ln ¯¯x2 ¯¯ = ¡2c ) ln ¯ ¯ x ¯ ¯ ¯ x ¡ 2y £ x2 ¯ = ¡2c ) ln ¯ ¯ x
Z
2v dv dx = ¡ 1 + v 2 dx
)
)
¯
¯µ ¶ ¯ ¯ ¯ y2 ¯ ln ¯ 1 + 2 x¯¯ = c x
)
) x+
8
) x2 ¡ 2xy = A, a constant
) x
dy dv = x+v dx dx dv x + vx 1+v +v = = ) x dx x ¡ vx 1¡v dv 1+v ) x = ¡v dx 1¡v dv 1 + v ¡ v(1 ¡ v) 1 + v2 = = ) x dx 1¡v 1¡v )
)
cyan
y dy = y + e x then dx y³ ´ dy y 1 = + ex dx x x 1 dv 1 So, by a, = 2 ev dx x
¯
¯
Z
)
1 dx x
magenta
yellow
95
1 dv = ev 1 ¡v e ¡1
=
Z
x¡2 dx
x¡1 +c ¡1
100
50
75
25
0
)
5
95
100
50
ln ¯1 + v 2 ¯ = ln jxj + c
75
1 2
25
95
50
75
25
0
5
95
100
50
75
25
0
5
) arctan v ¡
2v dv = 1 + v2
0
1 2
5
1 dv ¡ 1 + v2
100
)
Z
dv = f (v)g(x) dx 1 dv g(x) = and so is separable. f (v) dx x
b If x
1 ¡ v dv 1 = 1 + v2 dx x Z ³ Z ´ dv 1 v 1 ) ¡ dx = dx 1 + v2 1 + v 2 dx x
Z
dv + v = v + f (v)g(x) dx
) x
Thus,
Z
dv dy = x+v dx dx ³y´ dy y = +f g(x) becomes dx x x
a If y = vx,
x+y dy = dx x¡y Let y = vx
y2 = §ec = A, say x
) x2 + y2 = Ax, where A is a constant
¯ 2 ¯ ¯x ¡ 2xy¯ = e¡2c
) x2 ¡ 2xy = §e¡2c
b
¯
1 dx x
) ln ¯1 + v2 ¯ = ¡ ln jxj + c
) ln jx(x ¡ 2y)j = ¡2c )
dv v2 ¡ 1 = ¡v dx 2v
dv v 2 ¡ 1 ¡ 2v 2 ¡v 2 ¡ 1 = = dx 2v 2v 2v dv ¡1 = 1 + v 2 dx x
1 x
Z
dy dv = x+v dx dx
dv v 2 x2 ¡ x2 v2 ¡ 1 +v = = dx 2x(vx) 2v
) x
= 1 ¡ 2v
=
)
) x
y =1¡ =1¡v x
=
x2 + y 2 + c
= ln
t ¼ 93:2
It would take 93:2 days (approximately). 7
¯ r ¯ ¯ ¯ y2 ¯ ¯ = ln ¯x 1 + 2 ¯ + c x x ¯ ¯ ¯p ¯ ¯ ¯ = ln ¯ x2 + y 2 ¯ + c p
³y´
t
m0 (0:8) 30 = 0:5m0 t 30
197
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Y:\HAESE\IB_HL_OPT-Calculus\IB_HL_OPT-Calculus_an\197IB_HL_OPT-Calculus_an.cdr Wednesday, 13 March 2013 12:34:59 PM BRIAN
IB HL OPT 2ed Calculus
198
WORKED SOLUTIONS y
¡x
) ¡e
=¡
y
¡x
) e
=
h
Z
1 +c x
1 ¡c x
) y = x ¡ 1 + 12 ex + ce¡x
i
But when x = 1, y = 1 ) 1 = 1 ¡ 1 + 12 e + ) 1¡
h 1 ¡ cx i¡1
) y = x ¡ 1 + 12 ex + e1¡x ¡ 12 e2¡x d x dy + P (x)y = Q(x) dx
dy + 4y = 12 is of the form dx with P (x) = 4
R
)
4 dx
the integrating factor is e
Multiplying the DE by
e4x
= e4x .
= 12
)
¡1¢ 4
) y = 3 + ce
4x
R
Z
=
) xy = x sin x ¡
) y=
) y = sin x +
¡ 12
+c ) c=
1 5 Thus y = ¡ ex + e3x . 2 2 c
dy + y = x + ex dx
R
has I(x) = e
)
5 2
= ex
We integrate by parts with:
Z e2x dx
) xex y = ex (x ¡ x2 ) ¡
magenta
= ex+ln x = ex £ x
yellow
u0 = ex u = ex
95
v = x ¡ x2 v 0 = 1 ¡ 2x
ex (1 ¡ 2x) dx
n
u0 = ex u = ex
100
50
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5
95
100
50
75
25
0
R
ex (x ¡ x2 ) dx
) xex y = ex (x ¡ x2 ) ¡ [ex (1 ¡ 2x) ¡
5
95
v=x v0 = 1
100
50
We integrate by parts with:
75
xex dx +
75
25
0
5
95
100
50
75
25
0
5
cyan
1+ x dx
n
(xex + e2x ) dx
u0 = ex u = ex
y = 1 ¡ x where
Z
xex dx we integrate by parts with:
n
R ¡ x 1¢
) xex y =
Z For
³x + 1´
dy + xex y = ex (x ¡ x2 ) dx x dy ) xex + ex (x + 1)y = ex (x ¡ x2 ) dx d ) (xex y) = ex (x ¡ x2 ) dx
Z
R
dy = x ¡ x2 dx
³x + 1´
) xex 1 dx
dy + dx
I(x) = e
dy + ex y = xex + e2x dx d x ) (e y) = xex + e2x dx
=
sin x dx
cos x c + x x
(x + 1)y + x
2
) ex
) ex y =
v=x v0 = 1
¢
1 e¡2x + c ¡2 ¡ 12 e¡2x+3x + ce3x ¡ 12 ex + ce3x
But when x = 0, y = 2 ) 2=
u0 = cos x u = sin x
= x sin x ¡ (¡ cos x) + c
dy ¡ 3ye¡3x = ex e¡3x dx d ) (ye¡3x ) = e¡2x dx ) y=
x cos x dx
We integrate by parts with: e¡3x
dx
fsame as (1)g
Z
¡3 dx
1 x
= eln x =x
dy + y = x cos x dx d (xy) = x cos x dx
) e¡3x
¡
x
R
y = cos x which has I(x) = e
n
+c
e
has I(x) = e
) ye¡3x =
³1´
) xy =
¡4x
dy ¡ 3y = ex b dx
dy + dx
) x
gives
dy + 4ye4x = 12e4x dx d ) (ye4x ) = 12e4x dx ) ye
dy + y = x cos x .... (1) dx
)
e4x
4x
e2 2
) y = x ¡ 1 + 12 ex + e¡x (e ¡ 12 e2 )
EXERCISE O a
c e
e c = 2 e
) c=e¡
x h x i ) y = x ln 1 ¡ cx
1
e2x dx
ex dx +
= xex ¡ ex + 12 e2x + c
y 1 ¡c ¡ = ln x x h 1 ¡ cx i ) y = ¡x ln x ) y = x ln
Z
) ex y = xex ¡
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Y:\HAESE\IB_HL_OPT-Calculus\IB_HL_OPT-Calculus_an\198IB_HL_OPT-Calculus_an.cdr Wednesday, 13 March 2013 12:35:13 PM BRIAN
v = 1 ¡ 2x v 0 = ¡2
R
(¡2ex ) dx]
IB HL OPT 2ed Calculus
WORKED SOLUTIONS ) xex y = ex (x ¡ x2 ) ¡ ex (1 ¡ 2x) ¡ 2ex + c
But y 0 (x) = 2x +
1 ¡ 2x 2 c ¡ + x x xex 1 2 c = 1¡x¡ +2¡ + x x xex 3 c = 3¡x¡ + x xex
) y = 1¡x¡
a )
) Tn (x) = ¡1 + 1(x ¡ 1) +
) Tn (x) = ¡1 + x ¡ 1 + 2x2 ¡ 4x + 2
dy e¡y = ¡1 dx 3 ³ ´ d2 y 1 ¡y dy = ¡e 3 dx2 dx dy = ¡ 13 e¡y dx
µ
=
1 ¡y e 3
b
+ e¡y
¶
d2 y dx2
¡
d2 y dx2
1 3
¡ 1 = ¡ 23
¶
)
) T3 (x) = ¡ 23 x + 19 x2 +
1 + 2e¡x 3
¶
, ey =
1 3
3 + 23 e¡x
.... (1) ) 1
) x
1
When x = 1, y(1) = 0, y 0 (1) = 3, y 00 (1) =
)
T3 (x) = y(1) + y0 (1)(x ¡ 1) +
)
d3 y = 0, etc dx3
+
=0
6(x ¡ 1)2 24(x ¡ 1)3 + 2 6
yellow
95
100
50
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25
0
= 3(x ¡ 1) ¡ 3(x ¡ 1)2 + 4(x ¡ 1)3
5
95
y00 (1)(x ¡ 1)2 2!
y000 (1)(x ¡ 1)3 3!
= 0 + 3(x ¡ 1) ¡
100
50
75
25
0
5
95
50
75
100
magenta
3¡3(3) 1
= ¡6
d2 y =4 dx2
25
0
d3 y d2 y = ¡4 dx3 dx2
¡4(¡6) and y000 (1) = = 24 1
5
95
100
50
75
d2 y d3 y d2 y +x = ¡3 2 3 dx dx dx2 ) x
y00 (1)(x ¡ 1)2 Tn (x) = y(1) + ¡ 1) + 2! fas all other terms are zerog
25
d2 y dy = 3¡3 dx2 dx
and differentiating again
y 0 (1)(x
0
dy d2 y dy +x = 3¡2 dx dx2 dx
ffrom (1)g
dy dy +x = 4x + dx dx2 dx
cyan
X
dy 3x ¡ 2y = dx x dy ) x = 3x ¡ 2y dx
d2 y
dx(n)
¡1 = 2 + c c = ¡3
y dy = 4x ¡ 3 = 2x + (2x ¡ 3) = 2x + dx x
a
dy e¡y = ¡1 X ) dx 3 dy y = 2x + dx x dy ) x = 2x2 + y dx
d(n) y
1 x
1 dy 1 ¡ y=2 x dx x2 ³ ´ d y ) =2 dx x Z y ) = 2 dx = 2x + c x
Check:
1 3 x 81
dy = ¡ 23 e¡x dx dy ) ey = 13 ¡ ey dx dy 1 = ¡1 ) dx 3ey
) 1
= e¡ ln x
But when x = 1, y = ¡1 ) ) ) y = 2x2 ¡ 3x
2 9
) ey
a
1
¡ x dx
y = 2x2 + cx
¡ ¢ 2 d3 y = 13 49 ¡ 29 = 27 3 dx x2 x3 Now T3 (x) = f (0) + f 0 (0)x + f 00 (0) + f 000 (0) 2! 3! µ
x
R
y = 2x has I(x) = e
=
d2 y dy = ¡ 13 = dx2 dx
b If y = ln
³1´
¡1
dx
dx dy = dx
dy ¡ dx
= eln x
³ dy ´2
µ³ ´2 dy
At x = 0, y = 0,
5
4(x2 ¡ 2x + 1) 2
) Tn (x) = 2x2 ¡ 3x
d3 y = ¡ 13 ¡e¡y dx3
2
y ) y0 (1) = 2 + y x ) 1=2+y ) y = ¡1
EXERCISE P 1
199
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Y:\HAESE\IB_HL_OPT-Calculus\IB_HL_OPT-Calculus_an\199IB_HL_OPT-Calculus_an.cdr Wednesday, 13 March 2013 12:36:38 PM BRIAN
IB HL OPT 2ed Calculus
200
WORKED SOLUTIONS 3x ¡ 2y 2 dy = =3¡ y dx x x
b As
R
where I(x) = e
2 x
dy 2 + y=3 dx x
then
1 dy sin x 1 + y= cos x dx cos2 x cos2 x ³ y ´ d ) = sec2 x dx cos x
)
dx
= e2 ln x 2
= eln x = x2
y = cos x
)
dy + 2xy = 3x2 dx d 2 ) (x y) = 3x2 dx
c x2
Note:
But when x = 1, y = 0 ) 0 = 1 + c ) c = ¡1 1 x2
y =x¡
sin x = x ¡
x3 x5 + ¡ :::: 3! 5!
cos x = 1 ¡
x2 x4 + ¡ :::: 2! 4!
)
sin x + 2 cos x ¼x¡
dy + y sin x = 1, y(0) = 2 dx Differentiating with respect to x gives dy d2 y dy ¡ sin x + cos x 2 + sin x + y cos x = 0 dx dx dx
a cos x
µ
) cos x
d2 y +y dx2
¶
p(p ¡ 1)x2 p(p ¡ 1)(p ¡ 2)x3 + + :::: 2! 3! dy 2p(p ¡ 1)x 3p(p ¡ 1)(p ¡ 2)x2 =0+p+ + + :::: ) dx 2! 3! p(p ¡ 1)(p ¡ 2)x2 + :::: = p + p(p ¡ 1)x + 2! dy p(p ¡ 1)(p ¡ 2)x3 ) x = px + p(p ¡ 1)x2 + + :::: dx 2! dy dy dy = +x , ) (1 + x) dx dx dx
= 0 for all x
y = 1 + px +
dy dy + 2(0) = 1 ) = 1, dx dx d2 y = ¡2 dx2
fa sum of two convergent seriesg = p + [p(p ¡ 1) + p]x
d3 y = ¡1 dx3 x2 x3 Now T3 (x) = y(0) + y0 (0)x + y 00 (0) + y 000 (0) 2! 3!
·
and
+
·
dy + y sin x = 1 dx ³ sin x ´ dy 1 + y= dx cos x cos x
cos x
R
where I(x) = e
R
¡
=e
sin x cos x
=
¡ sin x cos x
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95
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5
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50
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5
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100
magenta
¸
1 p(p ¡ 1) :::: (p ¡ n + 1)xn P dy = p+p dx n! n=1
= py
75
50
1 cos x
25
0
5
95
100
50
75
25
0
5
p(p ¡ 1) :::: (p ¡ n + 1) n x n!
(1 + x)
=e
cyan
·
Hence
dx
¡ ln(cos x)
=
p(p ¡ 1) :::: (p ¡ n + 1) [(p ¡ n) + n]xn n!
=p
dx
¸
p(p ¡ 1) :::: (p ¡ n) p(p ¡ 1) :::: (p ¡ n + 1) n + x n! (n ¡ 1)!
as required
100
)
¸
p(p ¡ 1)(p ¡ 2) + p(p ¡ 1) x2 + :::: 2!
where the general term is:
) T3 (x) ¼ 2 + x ¡ x2 ¡ 16 x3
b
+ ::::
a For ¡1 < x < 1,
5
But when x = 0, y = 2 ) cos 0
1 4 x 12
which checks with the answer in a.
d3 y dy =¡ dx3 dx
and
x3 x5 2x2 2x4 + +2¡ + + :::: 3! 5! 2! 4!
¼ 2 + x ¡ x2 ¡ 16 x3 +
d2 y = ¡y dx2
)
cos x
´
+c
But when x = 0, y = 2 ) 2 = 0 + c ) c=2 ) y = sin x + 2 cos x
3x2 dx = x3 + c
) y =x+
4
³ sin x
) y = sin x + c cos x
Z
)
sec2 x dx = tan x + c
) y = cos x
) x2
) x2 y =
Z
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Y:\HAESE\IB_HL_OPT-Calculus\IB_HL_OPT-Calculus_an\200IB_HL_OPT-Calculus_an.cdr Thursday, 14 March 2013 4:54:16 PM BRIAN
IB HL OPT 2ed Calculus
WORKED SOLUTIONS b
dy = py dx 1 dy p = y dx 1+x
(1 + x)
Z
1 dy = y
)
Z
n!1
= lim
n!1
= =
for some constant A.
d
p(p ¡ 1) :::: (p ¡ n + 1)xn = (1 + x)p n!
n=0
But )
REVIEW SET A 1 ln x and x are continuous for all x > 0 The limit has type
5
so we can use l’Hˆopital’s Rule.
ln x = lim x x!1 x x
lim
x!1
1 P 1 n=1 1
P
fl’Hˆopital’s Ruleg
2 As x ! 0, ex sin x ! 0 also The limit has type )
lim
5 6+ n ¡ n72
so we can use l’Hˆopital’s Rule.
n2
3x + x¡2
x!0
j3xj < jx ¡ 2j j3xj2 < jx ¡ 2j2 9x2 ¡ (x ¡ 2)2 < 0 (3x + x ¡ 2)(3x ¡ x + 2) < 0 (4x ¡ 2)(2x + 2) < 0
¡ +
=
0¡0¡2 0+0+7
=
¡ 27
2 n 6 n
cyan
b
´ 1 ³ P 3x n
n=0
+7
Qw
x
1 2
=
u1 1¡r 1 1¡
3x x¡2
£
x¡2 x¡2
x¡2 x ¡ 2 ¡ 3x x¡2 = ¡2x ¡ 2 2¡x = f¡1 < x < 12 g 2x + 2 =
1 + 2n which diverges as n ! 1 n
magenta
yellow
95
100
50
75
25
0
5
95
100
50
75
25
0
5
95
100
50
75
25
0
x¡2
=
1 + n[1 + (¡1)n ] diverges n
5
95
100
50
75
25
0
3+
+
Note: The ratio test could have also been used.
¡2
b If n is even, un = 3 +
5
-1
4 + 6n + 7n2
8 n2 = lim n!1 4 n2
´
+
)
) ¡1 < x < lim
´2 ³
+
8 ¡ 2n ¡ 2n2
n!1
)
3x x¡2
fl’Hˆopital’s Ruleg
=1 a
³
fComparison testg
)
= 1(0) + 1(1)
3
fp-series testg
converges
n converges n3 + 1
) ) )
ex sin x
x ex sin x + ex cos x = lim x!0 1
¼ 2
3 3x + :::: is a geometric x¡2 3x series with u1 = 1 and r = x¡2 ¯ ¯ ¯ 3x ¯ < 1 ) the series converges when jrj = ¯ ¯ x¡2
=0
0 , 0
q
2 + 13 n
lim arctan n =
a 1+
1
)
5 6+ n ¡ n72
n
2 p 6 p 6 3
n=1
1 , 1
2n + 13
1 2 3 4 + + + + :::: 13 + 1 23 + 1 33 + 1 43 + 1 1 1 1 P P P n n 1 = which is 6 = 3 3 2 n + 1 n n n=1 n=1 n=1
4
for all jxj < 1, that is x 2 ]¡1, 1[ .
´
q
n!1
Now y(0) = 1 ) 1 = A(1 + 0)p ) A=1 1 P
³
5 n2 6 + n ¡ n72
= lim
p(p ¡ 1) :::: (p ¡ n + 1)xn = A(1 + x)p n!
Thus y(x) =
r
2n + 13
6n2 + 5n ¡ 7 2n + 13
p dx 1+x
solution to the differential equation,
n=0
= lim
n!1
Since jxj < 1, we have y = A(1 + x)p , where A = ec is a constant. 1 P p(p ¡ 1) :::: (p ¡ n + 1)xn is a c Since y(x) = n! n=0 1 P
p
n!1
) ln jyj = p ln j1 + xj + c, where c is a constant
y(x) =
lim
c
201
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Y:\HAESE\IB_HL_OPT-Calculus\IB_HL_OPT-Calculus_an\201IB_HL_OPT-Calculus_an.cdr Friday, 15 March 2013 3:15:01 PM BRIAN
IB HL OPT 2ed Calculus
202
WORKED SOLUTIONS c Differentiating the result in b with respect to x:
´ 1 ³ d P 3x n = dx n=0 x ¡ 2 µ
1 P
)
³
d 3x dx x ¡ 2
n=0
´n ¶
n
x¡2
n=0
³
´
1 2 1 2
d 2¡x dx x + 1
· =
³ 3x ´n¡1 1 P n
x¡2
n=0
)
6 x+
´
1 2
=
n+1 n=0 (x ¡ 2)
·
1 2
1 P
an converges for
) ) ) )
¡3 (x + 1)2
dy x = = k where k = 0, §1, §4 dx y ) we have isoclines x = 0, y = §x, y = § 14 x 10
x , but we will 1¡x
)
Z
2
sin(x ) dx = 0
¼
1 dy dx = y dx
)
1 3
1 42
+
1 1320
¯ ¯
) ln ¯¯
x=0
x!
= e¡¸ =e
¸1
11
dy ¡ dx
x=0
x!
³1´ x
R
p x has I(x) = e
1
¡ x dx
= e¡ ln x ¡1
= eln x =
1 x
1 1 dy 1 ¡ ¡ 2 y=x 2 x dx x ³ ´ 1 d y ¡ ) =x 2 dx x Z 1 y ¡ ) = x 2 dx x
)
1
is the Maclaurin series for e¸ g )
y x2 = 1 +c x 2
magenta
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100
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5
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100
50
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5
95
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50
p ) y = 2x x + cx
75
25
0
5
95
100
50
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25
0
5
y=
y = 2(x ¡ 1)ex¡2
1 75 600
= e0 =1
cyan
)
0
x!
1 P ¸x
y = §ex¡2 2(x ¡ 1)
But x = 2, y = 2 does not satisfy the negative solution
e
fas
¯
¯ y ¯=x¡2 2(x ¡ 1) ¯
)
1 P ¸x
x=0 ¡¸ ¸
ln 2 = 2 + c c = ln 2 ¡ 2
¯ ¯ ¯ y ¯ x¡2 ¯ ¯ ¯ 2(x ¡ 1) ¯ = e
)
e¡¸ ¸x 8 P(X = x) = where x = 0, 1, 2, 3, 4, .... x! 1 P e¡¸ ¸x
´
1 dx x¡1
) ln jyj ¡ ln j2(x ¡ 1)j = x ¡ 2
1 2
¡
1+
) ln jyj = x + ln jx ¡ 1j + ln 2 ¡ 2
x3 x7 x11 x15 ¡ + ¡ + :::: 3 42 1320 75 600 ¡
Z ³
) ln jyj = x + ln jx ¡ 1j + c
¼ 0:310 Check: GDC gives ¼ 0:310 268
)
dy xy = dx x¡1 x dy = dx x¡1 x¡1+1 1 dy = =1+ dx x¡1 x¡1
But, when x = 2, y = 2 ) )
x6 x10 x14 x18 + ¡ + ¡ :::: 3! 5! 7! 9!
·
1
)
1 y 1 y
)
x3 x5 x7 x9 + ¡ + ¡ :::: 3! 5! 7! 9!
) sin(x2 ) = x2 ¡
y = -x
x=0
jxj2 < j1 ¡ xj2 2 x ¡ (1 ¡ x)2 < 0 (x + 1 ¡ x)(x ¡ 1 + x) < 0 2x ¡ 1 < 0
sin(x) = x ¡
y = -Vr
¸
x 1¡x
x
1 + r2
r=1 1
r
2
1 2k
´ i ¼
=1
2k
diverges.
P 1+r 1 + r2
a Sn =
1 X 1
r r=1
diverges
n P k=3
diverges
fp-series testg fComparison testg
(¡1)k+1 ln(k ¡ 1)
Consider bk =
1 where k > 3 ln(k ¡ 1)
yellow
95
50
75
25
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5
95
100
50
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25
0
Now y = ln x is an increasing function for all x > 0.
5
95
100
50
75
25
0
5
95
³
¡2
¡¼¢
¡
1+r 1 > 1 + r2 r
P1
r=1
¡2 = ¡ 19 lim un = n!1 3£6
100
50
75
25
0
5
)
1 n
1 2
r + r2 > 1 for all r > 1 1 + r2
4
100
p
h³
¶
fTest for divergenceg
(n + 5) ¡ (n ¡ 1) p n+5+ n¡1
where
(k ¡ 1)¼ 2k
³ (k ¡ 1)¼ ´
= p
)
1 n and bn ! 0
alternating series with bn =
a If n is even, un =
c un =
provided jxj < 1
an
¡1 < x < 1 and is divergent for jxj > 1.
-1
1
n+1
n!1
= jxj
4
-2
203
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Y:\HAESE\IB_HL_OPT-Calculus\IB_HL_OPT-Calculus_an\203IB_HL_OPT-Calculus_an.cdr Friday, 15 March 2013 3:15:32 PM BRIAN
IB HL OPT 2ed Calculus
204
WORKED SOLUTIONS ) )
1 is decreasing for all x > 0 ln x bk is decreasing for all k > 3
The equation of the tangent is y = mx + c dy ) y= x + 3x2 y3 dx dy = y ¡ 3x2 y3 fsee Exercise N question 8g ) x dx dy dv Let y = vx, so = x+v dx dx
And, as bk > 0 for all k > 3 n P
Sn =
k=3
(¡1)k+1 converges fAlternating series testg ln(k ¡ 1)
dv + vx = vx ¡ 3x2 v 3 x3 dx dv ) x2 = ¡3x2 v 3 x3 dx 1 dv ) = ¡3x3 v3 dx Z Z dv ) v ¡3 dx = ¡3 x3 dx dx
b Now jS ¡ S10 j 6 b11 when S = lim Sn
Hence, x2
n!1
fAlternating Series Estimation Theoremg ) jS ¡ S10 j 6
1 ln 10
) jS ¡ S10 j 6 0:4343 f4 s.f.g 6 The Taylor series expansion of ex is: ex = 1 + x +
x2 x3 x4 + + + :::: 2! 3! 4!
) ex¡1 = 1 + (x ¡ 1) +
(x ¡ 1)4
+
4!
(x ¡ 1)2 2!
+
µ ¶2
3!
)
+ ::::
)
) y¡2
)
Z
dy dx = dx
)
=
c = 10
3 4 x 2
¡ 20
x2 3x4 ¡ 40 = y2 2
)
) y2 =
2x2 ¡ 40
3x4
r
) y=§
Z
2x2 (3x4 ¡ 40)
p8
When x = 2, y = §
r
(2x ¡ 1) dx
y=
Z
y ¡2 dy =
)
x y
)
dy = 2xy2 ¡ y 2 = (2x ¡ 1)y2 dx 1 dy = 2x ¡ 1 y 2 dx
Z
= ¡ 34 x4 + c
¡ 12 (4) = ¡12 + c
µ ¶2
dy = 4x ¡ 2y dx
then a = 4x ¡ 2(ax + b) for all x ) a = (4 ¡ 2a)x ¡ 2b for all x ) 4 ¡ 2a = 0 and a = ¡2b ) a = 2 and b = ¡1 8
x y
¡ 12
But when x = 2, y = 1
) (x ¡ 1)ex¡1 ¼ (x ¡ 1) + (x ¡ 1)2 + 12 (x ¡ 1)3 7 If y = ax + b is a solution of
v¡2 ¡3x4 = +c ¡2 4
)
(x ¡ 1)3
8
= §1 indicates that the solution is
2x2 . 3x4 ¡ 40
(2x ¡ 1) dx
y ¡1 2x2 = ¡x+c ¡1 2 ¡1 ) = x2 ¡ x + c y ¡1 ) y= 2 x ¡x+c
10 From equation a,
)
)
dy = 1 at (0, 0) dx
a has slope field B.
From equation b, )
dy = 0 at (2, 2) dx
b has slope field C.
Consequently c has slope field A.
9
y
11
(x, y)
a
1 1 400 ¡ P + P + = P 400 ¡ P P (400 ¡ P ) =
3x2y3
400 P (400 ¡ P )
(2, 1)
b Notice that as P is growing
³
x
cyan
magenta
yellow
95
400
dP >0 dt
>0
100
50
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5
95
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25
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5
) 0:2P 1 ¡
´ P
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Y:\HAESE\IB_HL_OPT-Calculus\IB_HL_OPT-Calculus_an\204IB_HL_OPT-Calculus_an.cdr Friday, 15 March 2013 10:01:32 AM BRIAN
IB HL OPT 2ed Calculus
WORKED SOLUTIONS )
P (400 ¡ P ) >0 400
12
) P (400 ¡ P ) > 0
) f (0) =
Sign diagram of P (400 ¡ P ) is: -
+
=
-
0
)
µ ³
´
b
dP P = 0:2P 1 ¡ dt 400 ³ 400 ¡ P ´ dP ) = 15 P dt 400 400 dP = 15 P (400 ¡ P ) dt Now as
) ) )
Z ³ 1
P
+
1 400 ¡ P
1 + P 400 ¡ P 1 ¡1
) ln jP j +
´ dP
´ dP
=
dt
1 5
ln j400 ¡ P j =
µ
=
³
P 400 ¡ P
´
=
)
t 400 ¡ P ¡ ¡c =e 5 P
1 5
=
1 t 5
dt
+c
=
¡
1 + Ae
)
c When t = 20, P = t ¡5
d As t ! 1, e )
1 P
where r =
ri
t 5
x2 1 + x2
1 P
¯ ¯ ¯1 + x2 ¯
¯
¯
)
1 < ¯1 + x2 ¯
) ) )
1 < 1 + x2 0 < x2 x 6= 0
d For x = 0, f (0) = 0 from a. For x = 6 0, 1 x2 P ri 1 + x2 i=0
x2 1 + x2
x2 = 1 + x2
¼ 389 people
=
yellow
³ 1 ´
fGeometric seriesg
1¡r
0
1
x2 1 + x2
95
f (x) =
µ µ
1 + x2 1 + x2 ¡ 1 1 + x2 x2
n 1, 0,
¶
¶
x 6= 0 x=0
100
50
75
25
0
5
95
100
50
75
25
0
5
95
0 and 400 ¡ P > 0 ffrom ¤g ) ln
x2 (1 + x2 )i i=1
1 + x2 x2
1 + x2 = x2
Z
dt =
dt
1 P
i=1
P
400
0 < P < 400 .... (¤)
³1
f (x) =
a
205
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Y:\HAESE\IB_HL_OPT-Calculus\IB_HL_OPT-Calculus_an\205IB_HL_OPT-Calculus_an.cdr Thursday, 14 March 2013 5:45:33 PM BRIAN
IB HL OPT 2ed Calculus
206
WORKED SOLUTIONS e
y
3 y = f(x)
x
1 1 = ln(n2 ) 2 ln n 1 1 P P 1 1 ) = 12 2) ln(n ln n n=2 n=2
.... (¤)
But n > ln n for all n 2 Z +
1 1 < for all n 2 Z + n ln n 1 1 P P 1 1 ) < n ln n n=2 n=2 )
REVIEW SET C 1
p
n2
a n¡
+n=
=
³
p
n2
n¡
³
n¡
n!1
+n
n+ n+
p
n2 + n
!
p
n2 + n
¡n
´
p
)
n2 + n
q
h
n2 + n
1 n
1+
Thus 4
¡n
q
h
1 1+ n
n 1+
n!1
1+
q
1 1+ n
fsince n 6= 0g =
¡1 1+1
P 1
fsince
lim
n!1
1 = 0g n
= ¡ 12
P
1 diverges 2) ln(n n=2
) both
b
n
n
n
1
(3n ) n
)
6 (3 + n
n
1
1
2n ) n
3n ) n
) 31 6 (3n + 2n )
1 n
6 (2 £
1 P
6 3£2
1 P
)
1 ³ P
an ¡
1
converges .... (¤)
n=1
1 n
´2
lim
Thus
n!1
= 3 fSqueeze theoremg
an = 0 for all a > 0 fTheoremg, n!1 n! en lim =0 n!1 n!
d Let un =
(¡1)n ne¡n
If n is even, If n is odd,
=
(¡1)n
(2) is convergent as
3
1 r
Thus
an ¡
1 P
1 r
5
magenta
a
n2
| {z } n=1
(3)
1 P
an 6 n
1 P
an
fComparison testg
n=1
1 n
´2
converges.
an2 is divergent.
95
1 x(x + 1)
100
50
75
25
0
5
95
yellow
1 P
(¡1) p . n
1 1 x+1¡x ¡ = x x+1 x(x + 1) =
100
50
75
25
0
5
95
100
50
75
25
0
5
95
(2)
n=1
1 1 P
n=1
1
100
50
75
25
0
(1)
n=1
r=1
5
| {z }
an is convergent, but
3 r is not convergent fTest of divergenceg
cyan
n
+
| {z }
For a counter example, consider an =
lim ar = 30 = 1 6= 0
1 P
1 a P n
b No, we can only apply the Limit Comparison test if an > 0 for all n. n
r!1
)
1 ³ P
n=1
r=1
Now
an2 ¡ 2
(3) is convergent fp-series testg
n un = n ! 0+ as n ! 1 e ¡n un = n ! 0¡ as n ! 1 e
we let ar = 3
P
n=1
n en
n!1
1 P
n=1 1
´
(1) is convergent ffrom ¤g
lim (¡1)n ne¡n = 0
)
2an 1 + 2 n n
an2 ¡
n=1
lim
c As
2 For
+
1 ³ P
= =
1
2n ) n
bn converge
n=1
fLimit Comparison testg an2
n=1
1 n
where 2 n ! 1 as n ! 1 (3n
ffrom ¤ g
1 P
an and
n=1
Now 3 6 3 + 2 6 3 + 3 n
fComparison testg
diverges
ln n
a We use the Limit Comparison test which states: “If an > 0 and bn > 0 for all n 2 Z + and if bn lim = c where c is a real constant, then an and bn n!1 an either converge or diverge together.” Suppose lim an = L and consider bn = an2 n!1 bn As an > 0 and an2 > 0 and = an an bn then lim = lim an = L n!1 an n!1
i
¡1
= lim
fp-series testg
diverges
n
n=2 1 n=2 1
i
= lim
n!1
1 P 1
But
p
n+
n 1+ lim
Ã
n2 ¡ (n2 + n)
=
)
´
black
Y:\HAESE\IB_HL_OPT-Calculus\IB_HL_OPT-Calculus_an\206IB_HL_OPT-Calculus_an.cdr Friday, 15 March 2013 3:16:10 PM BRIAN
IB HL OPT 2ed Calculus
WORKED SOLUTIONS b Consider f (x) =
1 . x(x + 1)
Thus f (n+1) (c) =
If x > 1, f (x) is > 0 and is continuous and
Z
1
1
1 dx = lim b!1 x(x + 1)
Z b³ 1 1
£
= lim = lim = lim
b!1
dx
ln
x+1
³ ³ b ´ ln
b+1
à Ã
1
= lim
ln
b!1
1+
1
¡ ln
where
2
Thus jRn (x : 0)j =
!
=
+ ln 2
1 b
Hence,
n=1
c
i
= ln 2 1 is convergent fIntegral testg n(n + 1)
Sn
P n
=
i=1
=
i=1
³
¡
i
1 1
1 i+1
ffrom ag
´
+
³
´
³
´
Thus
Z ) )
1 P
1 ii = lim Sn n(n + 1) n!1 n=1
1 P
)
lim
n!1
1 =0 n+1
x0 x1 x2 x3 x4 x5 x6
yellow
95
50
75
25
0
5
95
100
50
75
25
0
)
5
95
100
50
75
(¡1)n n! (1 + x)n+1
25
0
5
95
100
50
75
0
¸ n+1 x
t n+1
0
= ln(1 + x) ¡ ln 1
0
1 P (¡1)n xn+1
n+1 x2 2
+
x3 x4 x5 ¡ + ¡ :::: 3 4 5
Now xn+1 = xn + 0:1 and yn+1 = yn + 0:1f (xn , yn ) = yn + 0:1(xn ¡ 2yn ) = 0:1xn + 0:8yn
) f (n+1) (x) = (¡1)(¡2)(¡3) :::: (¡n)(1 + x)¡n¡1
25
¤x
dy = x ¡ 2y and when x = 1, y = 2 dx
8
f (4) (x) = (¡1)(¡2)(¡3)(1 + x)¡4
0
0
1 dt 1+t
tn dt = ln(1 + t)
n+1
f 000 (x) = (¡1)(¡2)(1 + x)¡3
5
£
x
=x¡
1 = (1 + x)¡1 1+x
magenta
(¡1)n
(¡1)n
x
n=0
f 00 (x) = ¡1(1 + x)¡2
cyan
Z
(¡t) dt =
Z
·
1 for ¡1 < x < 1 1+x
n
n=0
n=0
f (x) = ln(1 + x), 0 6 x < 1
) f (n+1) (x) =
P
´ 1
1¡
= 1, since
(¡x)n =
) ln(1 + x) =
³
n!1
) f 0 (x) =
1 n+1
=1 = 1:1 = 1:2 = 1:3 = 1:4 = 1:5 = 1:6
y0 y1 y2 y3 y4 y5 y6
=2 = 1:7 = 1:47 = 1:296 = 1:1668 = 1:073 44 = 1:008 752
y6 ¼ 1:009 fto 4 s.f.g
100
= lim
0
1 P
n=0 x 1
n=0
1 P
6
1+c
u1 for jrj < 1 1¡r 1 = for jxj < 1 1+x
So, its sum is
1 1 1 + 12 ¡ + 13 ¡ 1+1 2+1 3+1 ³ 1 ´ ³1 ´ 1 1 + :::: + ¡ + ¡ n¡1 n¡1+1 n n+1 1 1 1 1 1 1 1 1 = 1 ¡ 2 + 2 ¡ 3 + 3 ¡ 4 + :::: + ¡ + n¡1 n n 1 ¡ n+1 1 =1¡ n+1 =
³ x ´n+1
7 1 ¡ x + x2 ¡ x3 + x4 ¡ x5 + :::: is a geometric series with u1 = 1 and r = ¡x
1 i(i + 1)
n P 1
n! jxjn+1 n+1 (1 + c) (n + 1)!
But 0 6 x < 1 and 1 < 1 + c < 2 x ) 06 61 1+c ³ ´n+1 x 1 fas ! 0g ) jRn j 6 n+1 1+c
= ln 1 + ln 2 1 P
¯ ¯ ¯ f (n+1) (c)xn+1 ¯ ¯ ¯ jRn (x : 0)j = ¯ ¯ (n + 1)!
and c lies between 0 and 1 ) 1 < 1 + c < 2
¡ 1 ¢´
!
1
first n terms of Taylor series
¤b
h ³ x ´ib
(¡1)n n! (1 + c)n+1
Now ln(1 + x) = Tn (x) + Rn (x : 0)
´
ln x ¡ ln(x + 1)
b!1
b!1
1 ¡ x x+1
207
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Y:\HAESE\IB_HL_OPT-Calculus\IB_HL_OPT-Calculus_an\207IB_HL_OPT-Calculus_an.cdr Wednesday, 13 March 2013 3:09:07 PM BRIAN
IB HL OPT 2ed Calculus
208
WORKED SOLUTIONS a If V is the volume of water remaining then 4 ¡ V escaped.
9
Z
has
)
dv dx = dx
v
v2 = ln jxj + c 2
)
y2 = 2 ln jxj + 2c x2
)
V = volume remaining
11
p d (4 ¡ V ) / h dt 1 dV ) ¡ = kh 2 dt
) sin x
4 +c k 1
But when t = 0, h = 1 ) 0 = ¡ ) c=
¼ , 2
8 k
a un =
1
8 (0:1) k
Consider an = ) an+1 ¡ an =
p h) fin (2)g
sin x
2n + 1 3n ¡ 1
2(n + 1) + 1 2n + 1 ¡ 3(n + 1) ¡ 1 3n ¡ 1
=
6n2 + 7n ¡ 3 ¡ 6n2 ¡ 7n ¡ 2 (3n + 2)(3n ¡ 1)
=
¡5 (3n + 2)(3n ¡ 1)
magenta
yellow
fan g is strictly decreasing 7 9 11 , , , 8 11 14
95
) )
.... decrease
100
50
75
25
0
5
95
100
50
75
25
0
5
95
100
) an+1 ¡ an < 0 for all n ) an+1 < an for all n
75
50
¡ 14 (2 cos2 x)
=
dv dy = x+v dx dx
25
0
5
95
100
50
cyan
= ¡ 14
2n + 3 2n + 1 ¡ 3n + 2 3n ¡ 1 (2n + 3)(3n ¡ 1) ¡ (2n + 1)(3n + 2) = (3n + 2)(3n ¡ 1)
when h = 0, t = 20 min
dv x vx x+v = + dx vx x dv 1 x+v = +v dx v dv 1 ) v = dx x
75
25
)
1 (¡1) 4
3 £ 5 £ 7 £ :::: £ (2n + 1) 2 £ 5 £ 8 £ :::: £ (3n ¡ 1)
x y dy = + dx y x
)
cos 2x + c
REVIEW SET D
8 ) = 20 k
Let y = vx, so
(¡ cos 2x) + c
cos2 x ) y=¡ 2 sin x
8 +c k
and when t = 2, h = 0:81 ) 2 =
10
2
y=0
) y=
p 8 Thus t = (1 ¡ h) .... (2) fin (1)g k
0
) y sin x =
¡ 14
2
) y sin x = ¡ 14 (cos 2x + 1)
2
5
) y sin x =
) c=
8p h + c .... (1) ) t=¡ k
)
sin 2x dx
) 0 = ¡ 14 cos ¼ + c
1 h2
t = 20(1 ¡
1 2
¡ ¢ 1 1
But when x =
dt 4 ¡1 =¡ h 2 dh k
)
dx
Z
) y sin x =
1 dV dh =4 = ¡kh 2 dt dt dh k 1 ) = ¡ h2 dt 4
) t=¡
cos x sin x
dy + cos x £ y = sin x cos x dx d ) (y sin x) = 12 sin 2x dx
) V = 4h
c From b,
cot x dx
=e = sin x
But V = 2 £ 2 £ h m3
)
R
dy + (cot x)y = cos x has I(x) = e R dx =e
ln(sin x)
1 dV = ¡kh 2 m3 /min dt
) b
) y 2 = 2x2 (ln jxj + c)
2m 2m
1 dx x
) hm 1m
Z
black
Y:\HAESE\IB_HL_OPT-Calculus\IB_HL_OPT-Calculus_an\208IB_HL_OPT-Calculus_an.cdr Thursday, 14 March 2013 5:51:48 PM BRIAN
IB HL OPT 2ed Calculus
WORKED SOLUTIONS
¡ 3 ¢¡ 7 ¢n
Hence, un
4
(0:3)3 = 0:0045 so we use n = 4 But 3!
¼ 1:350
)
c Suppose e is rational, that is, e =
dy = 1 + x + y2 where y(1) = 0 dx dy dv Let y = vx ) = x+v dx dx
positive integers.
6 xy
x+v
´
v2 2
)
8
2
) y = ¡1 ¡ 2x + 2cx But when x = 1, y = 0 ) 0 = ¡1 ¡ 2 + 2c ) 2c = 3
dy + dx
cyan
f (n+1) (c)xn+1 (n + 1)!
=
magenta
3 x
dx
3 ln x
=e = x3
8x8 +c 8 c ) y = x5 + 3 x
But when x = 1, y = 0 ) 0 = 1 + c ) c = ¡1 Thus, y = x5 ¡
yellow
95
50
75
25
0
5
95
100
50
75
8x7
) x3 y =
ec xn+1
25
0
5
95
100
50
75
0
95
100
50
75
25
0
5
k=0
x
R y = 8x4 , y(1) = 0 has I(x) = e
) x3 y =
ec 1 + , 0 0g
R
2
e y = f (x) is half of a parabola since x =
Now (1, 5) lies on the curve
10
2r
) r =x+c
) ln jyj = ¡ 12 ln jxj + c
)
211
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Y:\HAESE\IB_HL_OPT-Calculus\IB_HL_OPT-Calculus_an\211IB_HL_OPT-Calculus_an.cdr Thursday, 14 March 2013 5:55:07 PM BRIAN
IB HL OPT 2ed Calculus
INDEX
INDEX
yellow
95
100
50
75
25
0
5
95
100
50
25
0
5
95
50
75
25
0
5
95
100
50
75
25
0
5
100
magenta
75
9 79 61 76 78 76 36 11 100 24 60 9 51 69 61 68 79 20 50 66 41 25, 34 105 131 20 50 66 136 75 20 109 45, 46 18 106 68 73 114 50 36 18, 29 66 110 9 41 70 36 117 23 85 108 91 26 29
absolute value function absolutely convergent Algebra of Limits Theorems alternating series Alternating Series Estimation Theorem Alternating Series Test antiderivative Archimedean Property binomial series bounded function bounded sequence closed interval Comparison Test for integrals Comparison Test for limits Comparison Test for sequences Comparison Test for series conditionally convergent continuous function convergent integral convergent series definite integral differentiable function differential equation direct proof discontinuous function divergent integral divergent series equivalence error essential discontinuity Euler's Method Fundamental Theorem of Calculus Fundamental Trigonometric Limit general solution geometric series harmonic series homogeneous differential equation improper integral indefinite integral indeterminate form infinite series initial value problem integer integrable Integral Test integrand integrating factor Intermediate Value Theorem interval of convergence isocline Lagrange form left-hand derivative l'Hopital's Rule ˆ
cyan
limit limit laws limit of a sequence Maclaurin polynomial Maclaurin series Mean Value Theorem Monotone Convergence Theorem monotone sequence natural number open interval nth partial sum particular solution partition power series product of series proof by contradiction p-series radius of convergence Ratio Test rational function rational number real number removable discontinuity Riemann sum right-hand derivative Rolle's theorem separable differential equation sequence slope field Squeeze Theorem Squeeze Theorem for sequences Taylor polynomial Taylor series Taylor's formula with remainder Taylor's Theorem term Test for Divergence Theorem of Absolute Convergence Triangle Inequality truncation error
black
Y:\HAESE\IB_HL_OPT-Calculus\IB_HL_OPT-Calculus_AN\213IB_HL_OPT-Calculus_an.cdr Monday, 18 March 2013 3:51:40 PM BRIAN
213
12, 141 15 57 90, 120 89 35 64 64 9 9 66 106 38 84 102 132 73 85 81 16 9 9 20 38 26 34 112 57 107 17 60 90, 120 89 91 94 57 67 80 10 78
IB HL OPT 2ed Calculus
ERRATA MATHEMATICS FOR THE INTERNATIONAL STUDENT MATHEMATICS HL (Option): Calculus
First edition - 2015 third reprint The following erratum was made on 27/Feb/2017 page 55 SECTION I DECREASING AND INCREASING FUNCTIONS, Last line before Example 20 should specify:
Similarly, for any negative continuous function which is increasing on [a, 1[ , Z 1 1 1 P P f(i) 6 f (x) dx 6 f (i + 1) = U . L= i=a
a
i=a
ERRATA MATHEMATICS FOR THE INTERNATIONAL STUDENT MATHEMATICS HL (Option): Calculus
First edition - 2015 second reprint The following errata were made on 14/Nov/2016 page 106 SECTION M Example 47, solutions to parts c and d should read:
c y = ce3x ¡ 1 is a general solution to the differential equation. The particular solution passes through (0, 2), so 2 = ce3£0 ¡ 1 ) c=3 ) the particular solution is y = 3e3x ¡ 1 dy = 3 + 3y dx
d
y
) at the point (0, 2),
dy =3+3£2=9 dx
) the gradient of the tangent to the particular solution y = 3e3x ¡ 1 at (0, 2), is 9. ) the equation of the tangent is
-2
2 y = 3e 3x - 1 -1
x
y¡2 =9 x¡0
)
1 -2
y = 9x + 2
y = 9x + 2
page 114 SECTION N Example 53, first line of solution should read:
Since OP = PQ, triangle OPQ is isosceles. Hence [PA] is the perpendicular bisector of [OQ].
The following errata were made on or before 09/Dec/2015 page 171 Worked Solutions EXERCISE K.1, Question 1 d had an invalid first step, replace whole solution with: 1
d Suppose
1 ³ P 1
n=1
)
1 ³ P 1 n=1
n
¡
This is false )
n 1 n2
¡
´
1 n2
+
´
1 P n=1
1 ³ P 1
n=1
converges
n
1 P 1 1 = 2 n n n=1
¡
1 n2
converges
´
diverges.
ERRATA MATHEMATICS FOR THE INTERNATIONAL STUDENT MATHEMATICS HL (Option): Calculus
First edition - 2014 initial print The following errata were made on or before 18/Aug/2015 page 19 EXERCISE B.3, Question 3 c was bad. Replace with 3 d and add Question 4:
3 Use the Squeeze Theorem to prove that a g(x) = x2 cos 4
¡
1 x2
¢
lim g(x) = 0 for:
x!0
b g(x) = x sin
¡1¢
c g(x) =
x
jxj 1 + x4
¡ 1¢ ¡ lim e x sin x = 0.
a Use the Squeeze Theorem to prove that x!0+ ¡ 1¢ ¡ b Explain why lim e x sin x does not exist. x!0
page 34 SECTION F ROLLE'S THEOREM, Rolle's Theorem doesn't require f(a) = f(b) = 0:
ROLLE’S THEOREM Suppose function f : D ! R is continuous on the closed interval [a, b], and differentiable on the open interval ] a, b [.
y y = f(x)
If f (a) = f(b), then there exists a value c 2 ] a, b [ such that f 0 (c) = 0.
a
c
b
x
Proof of Rolle’s theorem: Since f is continuous on [a, b], it attains both a maximum and minimum value on [a, b]. If f takes values greater than f(a) on [a, b], let f (c) be the maximum of these. Now f (b) = f(a) and f(c) > f(a), so c 2 ] a, b [ . Since f is differentiable at c, f must have a local maximum at x = c. ) f 0 (c) = 0. Similarly, if f takes values less than f (a) on [a, b], let the minimum of these be f(c). It follows that f has a local minimum at x = c, and therefore f 0 (c) = 0. Finally, if f (x) = f (a) for all x 2 [a, b] then clearly f 0 (c) = 0 for all c 2 ] a, b [ . By taking f (a) = f (b) = 0, Rolle’s theorem guarantees that between any two zeros of a differentiable function f there is at least one point at which the tangent line to the graph y = f (x) is horizontal.
A lemma is a proven proposition which leads on to a larger result.
Rolle’s theorem is a lemma used to prove the Mean Value Theorem. page 49 EXERCISE H question 8, should ask for definite intervals within the domain of F(x):
Z
8 Let F (x) = a F 0 (x)
1
x
2
cos(et ) dt, x > 1. Find exactly: ¡p ¢ b F 0 ( 2) c F 0 ln ¼
d F 00 (x)
e F 00 (2)
page 61 SECTION J SEQUENCES, Remove Comparison Test for Sequences:
COMPARISON TEST FOR SEQUENCES Consider sequences fan g, fbn g such that 0 6 an 6 bn . If fan g diverges, then fbn g diverges. If fbn g converges, then fan g converges. note: Counter example {an} = 1,2,1,2,1,2, ... and {bn} = 3, 3, 3, 3, ...
page 147 Worked Solutions EXERCISE B.1 question 7 c, the limits do not exist: 7
c
y f(x) = sin(_Qv_), x 6 = 0
x
f (x) does not approach any value as x ! 0 from above or below. i lim f (x) DNE ii lim f (x) DNE x!0¡
x!0+
iii lim f (x) DNE x!0
page 149 Worked Solutions EXERCISE B.3 question 3 d becomes 3 c, and question 4 is added: 1
4
a As ¡1 6 sin x 6 1 and e x > 0 for all x 2 R , ¡
1
sin x
6
1 ex
e
à lim
Ã
= lim
1
1
¡x
lim e
!
1
!
=0
1
x!0+
ex
¡ )
1
1 ! 1 and e x ! 1 x
1
¡
x!0+
1
ex
But as x ! 0+ ,
)
1
6
1 x
ex
¢
sin x = 0 fSqueeze Theoremg
x!0+
1
b As ¡1 6 sin x 6 1 and e x > 0 for all x 2 R , ¡
1
)
e
1
6
1 x
1
ex
1
¡x
lim e
Now )
sin x
6
1 ex
x!0+
1 ¡x
lim e
1
= 0 but
¡x
lim e
x!0¡
is undefined.
is undefined.
x!0
1 ¡x
lim e
x!0
sin x is undefined.
page 161 Worked Solutions EXERCISE H question 8, answers need to change according to question alteration:
Z
8 F (x) =
x
2
cos(et ) dt, x > 1 1
2
a F 0 (x) = cos(ex ), x > 1 b F 0 (2) = cos(e4 ) ¡p ¢ ¡ ¢ c F0 ln ¼ = cos eln ¼ = cos ¼ = ¡1 2
2
d F 00 (x) = ¡ sin(ex ) ex 2x 2
2
= ¡2x ex sin(ex ) e F 00 (2) = ¡4e4 sin(e4 )
fFTOCg
page 179 Worked Solutions EXERCISE K.3 questions 7 e and f, the series are conditionally convergent: 7
e Consider f (x) = Now f 0 (x) =
ln x x
1 x x
¡ ln x x2
=
1 ¡ ln x x2
Now consider f (x) =
f 0 (x) < 0 for all x such that ln x > 1, that is, x > e
)
n
Thus
ln n n
o
1 P
)
n=1
(¡1)n+1 ln n n
1 P
But
n=1
)
1 P n=1
ln n n
is convergent
(¡1)n+1 ln n n
is conditionally convergent.
³ 1 ´´ x
¡(ln x + 1) = (x ln x)2 f 0 (x)
-
< 0 for all x > 2 1 is decreasing for all n > 2, n 2 Z + ) bn = n ln n So,
Thus
2
1 P n=2
fAlternating Series Testg
is divergent fIntegral testg
1 = [x ln x]¡1 , x > 2 x ln x
³
1
ln x = lim x = 0 x!1 x n!1 1 ln n = 0, n 2 Z + .... (2) ) lim n!1 n lim
lim bn = 0
n!1
) f 0 (x) = ¡[x ln x]¡2 1 ln x + x
is decreasing for all n > 3, n 2 Z + .... (1)
By l’Hˆopital’s Rule,
1 , then n ln n
f Let bn =
But
1 P n=2
)
1 P n=2
(¡1)n n ln n
1 n ln n
(¡1)n n ln n
is convergent. is divergent fIntegral testg
is conditionally convergent.
page 184 Worked Solutions EXERCISE K.4 question 8, the last line of the solution should have correct interval: 8
.... 1 P
)
n=0
nxn¡1 converges for all x 2 [¡3, 3[ , and has radius n2 3n
of convergence 3.
page 187 Worked Solutions EXERCISE L.2 question 2 a, should read: 2
a
f (x) = e¡x , f 0 (x) = ¡e¡x , f 00 (x) = e¡x , f 000 (x) = ¡e¡x ,
) f (0) = 1 ) f 0 (0) = ¡1 ) f 00 (0) = 1 ) f 000 (0) = ¡1
By Maclaurin’s theorem, f (x) = e¡x = f (0) + f 0 (0)x + :::: +
....
f 00 (0)x2 f 000 (0)x3 + + 2! 3!
f (n) (0)xn + Rn (x : 0) n!