(Solution Manual) Mathematical methods for physics and engineering

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P1: JZP 0521842077pre CB1005/Chen 0 521 84207 7 January 29, 2006 14:8 This page intentionally left blankStudent Solutions Manual for Mathematical Methods for Physics and Engineering , third edition Mathematical Methods for Physics and Engineering , third edition, is a highly ac- claimed undergraduate textbook that teaches all the mathematics needed for an undergraduate course in any of the physical sciences. As well as luciddescriptions of the topics and many worked examples, it contains over 800exercises. New stand-alone chapters give a systematic account of the ‘specialfunctions’ of physical science, cover an extended range of practical applica-tions of complex variables, and give an introduction to quantum operators. This solutions manual accompanies the third edition of Mathematical Meth- ods for Physics and Engineering . It contains complete worked solutions to over 400 exercises in the main textbook, the odd-numbered exercises that are pro-vided with hints and answers. The even-numbered exercises have no hints,answers or worked solutions and are intended for unaided homework prob-lems; full solutions are available to instructors on a password-protected website,www.cambridge.org/9780521679718. Ken Riley read mathematics at the University of Cambridge and proceeded to a Ph.D. there in theoretical and experimental nuclear physics. He became a research associate in elementary particle physics at Brookhaven, and then, having taken up a lectureship at the Cavendish Laboratory, Cambridge, continued thisresearch at the Rutherford Laboratory and Stanford; in particular he was involvedin the experimental discovery of a number of the early baryonic resonances. Aswell as having been Senior Tutor at Clare College, where he has taught physicsand mathematics for over 40 years, he has served on many committees concernedwith the teaching and examining of these subjects at all levels of tertiary and undergraduate education. He is also one of the authors of 200 Puzzling Physics Problems . Michael Hobson read natural sciences at the University of Cambridge, spe- cialising in theoretical physics, and remained at the Cavendish Laboratory to complete a Ph.D. in the physics of star-formation. As a research fellow at Trinity Hall, Cambridge and subsequently an advanced fellow of the Particle Physicsand Astronomy Research Council, he developed an interest in cosmology, andin particular in the study of fluctuations in the cosmic microwave background.He was involved in the first detection of these fluctuations using a ground-basedinterferometer. He is currently a University Reader at the Cavendish Laboratory,his research interests include both theoretical and observational aspects of cos- mology, and he is the principal author of General Relativity: An Introduction for Physicists . He is also a Director of Studies in Natural Sciences at Trinity Hall and enjoys an active role in the teaching of undergraduate physics and mathematics. Student Solutions Manual for Mathematical Methods for Physics and Engineering Third Edition K .F .R I L E Ya n dM .P .H O B S O N cambridge university press Cambridge, New York, Melbourne, Madrid, Cape Town, Singapore, São Paulo Cambridge University Press The Edinburgh Building, Cambridge cb22ru,U K First published inprint format isbn-13 978-0-521-67973-2isbn-13 978-0-511-16804-8©K.F.Riley and M. P.Hobson 2006 2006Information on this title: www.cambrid ge.or g/9780521679732 This publication isincopyright. Subject tostatutory exception and tothe provision of relevant collective licensing agreements, no reproduction ofany part may take place without the written permission ofCambridge University Press. isbn-10 0-511-16804-7 isbn-10 0-521-67973-7 Cambridge University Press has no responsibility for the persistence oraccuracy ofurl s for external orthird-party internet websites referred tointhis publication, and does not guarantee that any content on such websites is,orwill remain, accurate orappropriate.Published inthe United States ofAmerica by Cambridge University Press, New York www.cambridge.org paperbackeBook (EBL) eBook (EBL) paperbackContents Preface page ix 1 Preliminary algebra 1 2 Preliminary calculus 173 Complex numbers and hyperbolic functions 39 4 Series and limits 55 5 Partial differentiation 716 Multiple integrals 907 Vector algebra 104 8 Matrices and vector spaces 119 9 Normal modes 14510 Vector calculus 15611 Line, surface and volume integrals 176 v CONTENTS 12 Fourier series 193 13 Integral transforms 21114 First-order ODEs 22815 Higher-order ODEs 24616 Series solutions of ODEs 26917 Eigenfunction methods for ODEs 28318 Special functions 29619 Quantum operators 31320 PDEs: general and particular solutions 31921 PDEs: separation of variables and other methods 33522 Calculus of variations 35323 Integral equations 37424 Complex variables 38625 Applications of complex variables 40026 Tensors 42027 Numerical methods 44028 Group theory 46129 Representation theory 480 vi CONTENTS 30 Probability 494 31 Statistics 519 vii Preface The second edition of Mathematical Methods for Physics and Engineering carried more than twice as many exercises, based on its various chapters, as did the first.In the Preface we discussed the general question of how such exercises shouldbe treated but, in the end, decided to provide hints and outline answers to allproblems, as in the first edition. This decision was an uneasy one as, on the onehand, it did not allow the exercises to be set as totally unaided homework thatcould be used for assessment purposes, but, on the other, it did not give a full explanation of how to tackle a problem when a student needed explicit guidance or a model answer. In order to allow both of these educationally desirable goals to be achieved, we have, in the third edition, completely changed the way this matter is handled.All of the exercises from the second edition, plus a number of additional ones testing the newly added material, have been included in penultimate subsections of the appropriate, sometimes reorganised, chapters. Hints and outline answersare given, as previously, in the final subsections, but only to the odd-numbered exercises . This leaves all even-numbered exercises free to be set as unaided homework, as described below. For the four hundred plus odd-numbered exercises, complete solutions are avail- able, to both students and their teachers, in the form of thismanual; these are in addition to the hints and outline answers given in the main text. For each exercise,the original question is reproduced and then followed by a fully worked solution.For those original exercises that make internal reference to the text or to other(even-numbered) exercises not included in this solutions manual, the questionshave been reworded, usually by including additional information, so that the questions can stand alone. Some further minor rewording has been included to improve the page layout. In many cases the solution given is even fuller than one that might be expected ix PREFACE of a good student who has understood the material. This is because we have aimed to make the solutions instructional as well as utilitarian. To this end, we have included comments that are intended to show how the plan for the solution is formulated and have provided the justifications for particular intermediatesteps (something not always done, even by the best of students). We have alsotried to write each individual substituted formula in the form that best indicateshow it was obtained, before simplifying it at the next or a subsequent stage.Where several lines of algebraic manipulation or calculus are needed to obtain afinal result, they are normally included in full; this should enable the student to determine whether an incorrect answer is due to a misunderstanding of principles or to a technical error. The remaining four hundred or so even-numbered exercises have no hints or answers (outlined or detailed) available for general access. They can therefore be used by instructors as a basis for setting unaided homework. Full solutions to these exercises, in the same general format as those appearing in this manual(though they may contain references to the main text or to other exercises), areavailable without charge to accredited teachers as downloadable pdf files on thepassword-protected website http://www.cambridge.org/9780521679718. Teacherswishing to have access to the website should contact solutions@cambridge.orgfor registration details. As noted above, the original questions are reproduced in full, or in a suitably modified stand-alone form, at the start of each exercise. Reference to the maintext is not needed provided that standard formulae are known (and a set of tables is available for a few of the statistical and numerical exercises). This means that, although it is not its prime purpose, this manual could be used as a test or quizbook by a student who has learned, or thinks that he or she has learned, thematerial covered in the main text. In all new publications, errors and typographical mistakes are virtually unavoid- able, and we would be grateful to any reader who brings instances to ourattention. Finally, we are extremely grateful to Dave Green for his considerableand continuing advice concerning typesetting in L ATEX. Ken Riley, Michael Hobson, Cambridge, 2006 x 1 Preliminary algebra Polynomial equations 1.1It can be shown that the polynomial g(x)=4 x3+3x2−6x−1 has turning points at x=−1andx=1 2and three real roots altogether. Continue an investigation of its properties as follows. (a)Make a table of values of g(x)for integer values of xbetween −2and2. Use it and the information given above to draw a graph and so determinethe roots of g(x)=0 as accurately as possible. (b)Find one accurate root of g(x)=0by inspection and hence determine precise values for the other two roots. (c)Show that f(x)=4 x 3+3x2−6x−k=0has only one real root unless −5≤k≤7 4. (a) Straightforward evaluation of g(x)a ti n t e g e rv a l u e so f xgives the following table: x −2−10 1 2 g(x)−94 −103 1 (b) It is apparent from the table alone that x= 1 is an exact root of g(x)=0a n d sog(x) can be factorised as g(x)=(x−1)h(x)=(x−1)(b2x2+b1x+b0). Equating the coefficients of x3,x2,xand the constant term gives 4 = b2,b1−b2=3 , b0−b1=−6a n d −b0=−1, respectively, which are consistent if b1= 7. To find the two remaining roots we set h(x)=0 : 4x2+7x+1=0 . 1 PRELIMINARY ALGEBRA The roots of this quadratic equation are given by the standard formula as α1,2=−7±√ 49−16 8. (c) When k= 1 (i.e. the original equation) the values of g(x) at its turning points, x=−1a n d x=1 2,a r e4a n d −11 4, respectively. Thus g(x) can have up to 4 subtracted from it or up to11 4added to it and still satisfy the condition for three (or, at the limit, two) distinct roots of g(x) = 0. It follows that for koutside the range −5≤k≤7 4,f(x)[ =g(x)+1 −k] has only one real root. 1.3Investigate the properties of the polynomial equation f(x)=x7+5x6+x4−x3+x2−2=0 , by proceeding as follows. (a)By writing the fifth-degree polynomial appearing in the expression for f/prime(x) in the form 7x5+3 0x4+a(x−b)2+c, show that there is in fact only one positive root of f(x)=0. (b)By evaluating f(1),f(0)andf(−1), and by inspecting the form of f(x)for negative values of x, determine what you can about the positions of the real roots of f(x)=0. (a) We start by finding the derivative of f(x) and note that, because fcontains no linear term, f/primecan be written as the product of xand a fifth-degree polynomial: f(x)=x7+5x6+x4−x3+x2−2=0 , f/prime(x)=x(7x5+3 0x4+4x2−3x+2 ) =x[7x5+3 0x4+4 (x−3 8)2−4(3 8)2+2] =x[7x5+3 0x4+4 (x−3 8)2+23 16]. Since, for positive x, every term in this last expression is necessarily positive, it follows that f/prime(x) can have no zeros in the range 0 <x< ∞.C o n s e q u e n t l y , f(x) can have no turning points in that range and f(x) = 0 can have at most one root i nt h es a m er a n g e .H o w e v e r , f(+∞)=+ ∞andf(0) = −2<0a n ds o f(x)=0 has at least one root in 0 <x< ∞. Consequently it has exactly one root in the range. (b)f(1) = 5 ,f(0) = −2a n d f(−1) = 5, and so there is at least one root in each of the ranges 0 <x< 1a n d −1<x< 0. There is no simple systematic way to examine the form of a general polynomial function for the purpose of determining where its zeros lie, but it is sometimes 2 PRELIMINARY ALGEBRA helpful to group terms in the polynomial and determine how the sign of each group depends upon the range in which xlies. Here grouping successive pairs of terms yields some information as follows: x7+5x6is positive for x> −5, x4−x3is positive for x>1a n d x<0, x2−2 is positive for x>√ 2a n d x< −√ 2. Thus, all three terms are positive in the range(s) common to these, namely −5<x< −√ 2a n d x>1. It follows that f(x) is positive definite in these ranges a n dt h e r ec a nb en or o o t so f f(x) = 0 within them. However, since f(x) is negative for large negative x, there must be at least one root αwith α<−5. 1.5Construct the quadratic equations that have the following pairs of roots: (a)−6,−3; (b) 0 ,4; (c) 2 ,2; (d) 3 + 2 i,3−2i,w h e r e i2=−1. Starting in each case from the ‘product of factors’ form of the quadratic equation, (x−α1)(x−α2) = 0, we obtain: (a) ( x+6 ) ( x+3 )= x2+9x+1 8=0 ; (b) ( x−0)(x−4) = x2−4x=0 ; (c) ( x−2)(x−2) = x2−4x+4=0 ; (d) ( x−3−2i)(x−3+2 i)=x2+x(−3−2i−3+2 i) +( 9 −6i+6i−4i2) =x2−6x+1 3=0 . Trigonometric identities 1.7Prove that cosπ 12=√ 3+1 2√ 2 by considering (a)the sum of the sines of π/3andπ/6, (b)t h es i n eo ft h es u mo f π/3andπ/4. (a) Using sinA+s i n B=2s i n/parenleftbiggA+B 2/parenrightbigg cos/parenleftbiggA−B 2/parenrightbigg , 3 PRELIMINARY ALGEBRA we have sinπ 3+s i nπ 6=2s i nπ 4cosπ 12, √ 3 2+1 2=21 √ 2cosπ 12, cosπ 12=√ 3+1 2√ 2. (b) Using, successively, the identities sin(A+B)=s i n AcosB+c o s AsinB, sin(π−θ)=s i n θ and cos(1 2π−θ)=s i n θ, we obtain sin/parenleftBigπ 3+π 4/parenrightBig =s i nπ 3cosπ 4+c o sπ 3sinπ 4, sin7π 12=√ 3 21 √ 2+1 21 √ 2, sin5π 12=√ 3+1 2√ 2, cosπ 12=√ 3+1 2√ 2. 1.9Find the real solutions of (a) 3 sin θ−4c os θ=2, (b) 4 sin θ+3c o s θ=6, (c) 12 sin θ−5c os θ=−6. We use the result that if asinθ+bcosθ=k then θ=s i n−1/parenleftbiggk K/parenrightbigg −φ, where K2=a2+b2and φ=t a n−1b a. 4 PRELIMINARY ALGEBRA Recalling that the inverse sine yields two values and that the individual signs of aandbhave to be taken into account, we have (a)k=2 , K=√ 32+42=5 , φ=t a n−1(−4/3) and so θ=s i n−12 5−tan−1−4 3=1.339 or −2.626. (b)k=6 , K=√ 42+32=5 .S i n c e k>K there is no solution for a real angle θ. (c)k=−6,K=√ 122+52= 13, φ=t a n−1(−5/12) and so θ=s i n−1−6 13−tan−1−5 12=−0.0849 or −2.267. 1.11Find all the solutions of sinθ+ sin4 θ= sin 2 θ+ sin3 θ that lie in the range −π<θ ≤π. What is the multiplicity of the solution θ=0 ? Using sin(A+B)=s i n AcosB+c o s AsinB, and cos A−cosB=−2s i n/parenleftbiggA+B 2/parenrightbigg sin/parenleftbiggA−B 2/parenrightbigg , and recalling that cos( −φ)=c o s ( φ), the equation can be written successively as 2s i n5θ 2cos/parenleftbigg −3θ 2/parenrightbigg =2s i n5θ 2cos/parenleftbigg −θ 2/parenrightbigg , sin5θ 2/parenleftbigg cos3θ 2−cosθ 2/parenrightbigg =0, −2s i n5θ 2sinθsinθ 2=0. The first factor gives solutions for θof−4π/5,−2π/5, 0, 2 π/5a n d4 π/5. The second factor gives rise to solutions 0 and π, whilst the only value making the third factor zero is θ=0 .T h es o l u t i o n θ= 0 appears in each of the above sets and so has multiplicity 3. 5 PRELIMINARY ALGEBRA Coordinate geometry 1.13Determine the forms of the conic sections described by the following equa- tions: (a)x2+y2+6x+8y=0 ; (b) 9 x2−4y2−54x−16y+2 9=0 ; (c) 2 x2+2y2+5xy−4x+y−6=0 ; (d)x2+y2+2xy−8x+8y=0. (a)x2+y2+6x+8y=0.The coefficients of x2andy2are equal and there is noxyterm; it follows that this must represent a circle. Rewriting the equation in standard circle form by ‘completing the squares’ in the terms that involve xand y, each variable treated separately, we obtain (x+3 )2+(y+4 )2−(32+42)=0 . The equation is therefore that of a circle of radius√ 32+42=5c e n t r e do n (−3,−4). (b) 9 x2−4y2−54x−16y+ 29 = 0. This equation contains no xyterm and so the centre of the curve will be at (54 /(2×9),16/[2×(−4)] ) = (3 ,−2), and in standardised form the equation is 9(x−3)2−4(y+2 )2+2 9 −81 + 16 = 0 , or (x−3)2 4−(y+2 )2 9=1. The minus sign between the terms on the LHS implies that this conic section is a hyperbola with asymptotes (the form for large xandyand obtained by ignoring the constant on the RHS) given by 3( x−3) = ±2(y+ 2), i.e. lines of slope ±3 2 passing through its ‘centre’ at (3 ,−2). (c) 2x2+2y2+5xy−4x+y−6=0 .As an xyterm is present the equation cannot represent an ellipse or hyperbola in standard form. Whether it representstwo straight lines can be most easily investigated by taking the lines in the forma ix+biy+1 = 0, ( i=1,2) and comparing the product ( a1x+b1y+1)(a2x+b2y+1) with −1 6(2x2+2y2+5xy−4x+y−6). The comparison produces five equations which the four constants ai,bi,(i=1,2) must satisfy: a1a2=2 −6,b 1b2=2 −6,a 1+a2=−4 −6,b 1+b2=1 −6 and a1b2+b1a2=5 −6. 6 PRELIMINARY ALGEBRA Combining the first and third equations gives 3 a2 1−2a1−1 = 0 leading to a1and a2having the values 1 and −1 3, in either order. Similarly, combining the second and fourth equations gives 6 b2 1+b1−2 = 0 leading to b1andb2having the values 1 2and −2 3, again in either order. Either of the two combinations ( a1=−1 3,b1=−2 3,a2=1 , b2=1 2)a n d( a1=1 , b1=1 2,a2=−1 3,b2=−2 3) also satisfies the fifth equation [note that the two alternative pairings do not do so]. That a consistent set can be found shows that the equation does indeed represent a pair of straight lines, x+2y−3=0a n d 2x+y+2=0 . (d)x2+y2+2xy−8x+8y=0.We note that the first three terms can be written as a perfect square and so the equation can be rewritten as (x+y)2=8 (x−y). The two lines given by x+y=0a n d x−y= 0 are orthogonal and so the equation is of the form u2=4av,w h i c h ,f o rC a r t e s i a nc o o r d i n a t e s u, v,r e p r e s e n t s a parabola passing through the origin, symmetric about the v-axis ( u=0 )a n d defined for v≥0. Thus the original equation is that of a parabola, symmetric about the line x+y= 0, passing through the origin and defined in the region x≥y. Partial fractions 1.15Resolve (a)2x+1 x2+3x−10, (b)4 x2−3x into partial fractions using each of the following three methods: (i)Expressing the supposed expansion in a form in which all terms have the same denominator and then equating coefficients of the various powers of x. (ii)Substituting specific numerical values for xand solving the resulting simul- taneous equations. (iii)Evaluation of the fraction at each of the roots of its denominator, imagining a factored denominator with the factor corresponding to the root omitted – often known as the ‘cover-up’ method. Verify that the decomposition obtained is independent of the method used. (a) As the denominator factorises as ( x+5)(x−2), the partial fraction expansion must have the form 2x+1 x2+3x−10=A x+5+B x−2. 7 PRELIMINARY ALGEBRA (i) A x+5+B x−2=x(A+B)+( 5 B−2A) (x+5 ) ( x−2). Solving A+B=2a n d −2A+5B= 1 gives A=9 7andB=5 7. (ii) Setting xequal to 0 and 1, say, gives the pair of equations 1 −10=A 5+B −2;3 −6=A 6+B −1, −1=2 A−5B;−3=A−6B, with solution A=9 7andB=5 7. (iii) A=2(−5) + 1 −5−2=9 7;B=2(2) + 1 2+5=5 7. All three methods give the same decomposition. (b) Here the factorisation of the denominator is simply x(x−3) or, more formally, (x−0)(x−3), and the expansion takes the form 4 x2−3x=A x+B x−3. (i) A x+B x−3=x(A+B)−3A (x−0)(x−3). Solving A+B=0a n d −3A= 4 gives A=−4 3andB=4 3. (ii) Setting xequal to 1 and 2, say, gives the pair of equations 4 −2=A 1+B −2;4 −2=A 2+B −1, −4=2 A−B;−4=A−2B, with solution A=−4 3andB=4 3. (iii) A=4 0−3=−4 3;B=4 3−0=4 3. Again, all three methods give the same decomposition. 8 PRELIMINARY ALGEBRA 1.17Rearrange the following functions in partial fraction form: (a)x−6 x3−x2+4x−4, (b)x3+3x2+x+1 9 x4+1 0x2+9. (a) For the function f(x)=x−6 x3−x2+4x−4=g(x) h(x) the first task is to factorise the denominator. By inspection, h(1) = 0 and so x−1 is a factor of the denominator. Write x3−x2+4x−4=( x−1)(x2+b1x+b0). Equating coefficients: −1=b1−1, 4 = −b1+b0and −4= −b0, giving b1=0 andb0= 4. Thus, f(x)=x−6 (x−1)(x2+4 ). The factor x2+ 4 cannot be factorised further without using complex numbers and so we include a term with this factor as the denominator, but ‘at the price of’ having a linear term, and not just a number, in the numerator. f(x)=A x−1+Bx+C x2+4 =Ax2+4A+Bx2+Cx−Bx−C (x−1)(x2+4 ). Comparing the coefficients of the various powers of xin this numerator with those in the numerator of the original expression gives A+B=0 ,C−B=1a n d 4A−C=−6 ,w h i c hi nt u r ny i e l d A=−1,B=1a n d C= 2. Thus, f(x)=−1 x−1+x+2 x2+4. (b) By inspection, the denominator of x3+3x2+x+1 9 x4+1 0x2+9 factorises simply into ( x2+9 ) ( x2+ 1), but neither factor can be broken down further. Thus, as in (a), we write f(x)=Ax+B x2+9+Cx+D x2+1 =(A+C)x3+(B+D)x2+(A+9C)x+(B+9D) (x2+9 ) ( x2+1 ). 9 PRELIMINARY ALGEBRA Equating coefficients gives A+C=1, B+D=3, A+9C=1, B+9D=1 9. From the first and third equations, A=1a n d C= 0. The second and fourth yield B=1a n d D= 2. Thus f(x)=x+1 x2+9+2 x2+1. Binomial expansion 1.19Evaluate those of the following that are defined: (a)5C3,(b)3C5,(c)−5C3, (d)−3C5. (a)5C3=5! 3! 2!= 10. (b)3C5. This is not defined as 5 >3>0. For (c) and (d) we will need to use the identity −mCk=(−1)km(m+1 )···(m+k−1) k!=(−1)km+k−1Ck. (c)−5C3=(−1)35 + 3 −1C3=−7! 3! 4!=−35. (d)−3C5=(−1)55 + 3 −1C5=−7! 5! 2!=−21. Proof by induction and contradiction 1.21Prove by induction that n/summationdisplay r=1r=1 2n(n+1 ) a n dn/summationdisplay r=1r3=1 4n2(n+1 )2. To prove that n/summationdisplay r=1r=1 2n(n+1 ), 10 PRELIMINARY ALGEBRA assume that the result is valid for n=Nand consider N+1/summationdisplay r=1r=N/summationdisplay r=1r+(N+1 ) =1 2N(N+1 )+( N+1 ),using the assumption, =(N+1 ) (1 2N+1 ) =1 2(N+1 ) ( N+2 ). This is the same form as in the assumption except that Nhas been replaced by N+ 1; this shows that the result is valid for n=N+ 1 if it is valid for n=N. But the assumed result is trivially valid for n= 1 and is therefore valid for all n. To prove that n/summationdisplay r=1r3=1 4n2(n+1 )2, assume that the result is valid for n=Nand consider N+1/summationdisplay r=1r3=N/summationdisplay r=1r3+(N+1 )3 =1 4N2(N+1 )2+(N+1 )3,using the assumption, =1 4(N+1 )2[N2+4 (N+1 )] =1 4(N+1 )2(N+2 )2. This is the same form as in the assumption except that Nhas been replaced by N+ 1 and shows that the result is valid for n=N+ 1 if it is valid for n=N. But the assumed result is trivially valid for n= 1 and is therefore valid for all n. 1.23Prove that 32n+7,w h e r e nis a non-negative integer, is divisible by 8. As usual, we assume that the result is valid for n=Nand consider the expression with Nreplaced by N+1 : 32(N+1)+7=32N+2+7+32N−32N =( 32N+7 )+32N(9−1). By the assumption, the first term on the RHS is divisible by 8; the second is clearly so. Thus 32(N+1)+ 7 is divisible by 8. This shows that the result is valid forn=N+ 1 if it is valid for n=N. But the assumed result is trivially valid for n= 0 and is therefore valid for all n. 11 PRELIMINARY ALGEBRA 1.25Prove by induction that n/summationdisplay r=11 2rtan/parenleftbiggθ 2r/parenrightbigg =1 2ncot/parenleftbiggθ 2n/parenrightbigg −cotθ. (∗) Assume that the result is valid for n=Nand consider N+1/summationdisplay r=11 2rtan/parenleftbiggθ 2r/parenrightbigg =1 2Ncot/parenleftbiggθ 2N/parenrightbigg −cotθ+1 2N+1tan/parenleftbiggθ 2N+1/parenrightbigg . Using the half-angle formula tanφ=2r 1−r2,where r=t a n1 2φ, to write cot( θ/2N) in terms of t=t a n ( θ/2N+1), we have that the RHS is 1 2N/parenleftbigg1−t2 2t/parenrightbigg −cotθ+1 2N+1t=1 2N+1/parenleftbigg1−t2+t2 t/parenrightbigg −cotθ =1 2N+1cot/parenleftbiggθ 2N+1/parenrightbigg −cotθ. This is the same form as in the assumption except that Nhas been replaced by N+ 1 and shows that the result is valid for n=N+ 1 if it is valid for n=N. But, for n=1 ,t h eL H So f( ∗)i s1 2tan(θ/2). The RHS can be written in terms of s=t a n ( θ/2): 1 2cot/parenleftbiggθ 2/parenrightbigg −cotθ=1 2s−1−s2 2s=s 2, i.e. the same as the LHS. Thus the result is valid for n=1a n dh e n c ef o ra l l n. 1.27Establish the values of kfor which the binomial coefficientpCkis divisible by pwhen pis a prime number. Use your result and the method of induction to prove thatnp−nis divisible by pfor all integers nand all prime numbers p. Deduce thatn5−nis divisible by 30 for any integer n. Since pCk=p! k!(p−k)!, its numerator will always contain a factor p. Therefore, the fraction will be divisible by punless the denominator happens to contain a (cancelling) factor of p.S i n c e pis prime, this latter factor cannot arise from the product of two or more terms in the denominator; nor can phave any factor that cancels with a 12 PRELIMINARY ALGEBRA term in the denominator. Thus, for cancellation to occur, either k!o r( p−k)! must contain a term p; this can only happen for k=pork=0 ;f o ra l lo t h e r values of k,pCkwill be divisible by p. Assume that np−nis divisible by prime number pforn=N. Clearly this is true forN=1a n da n y p. Now, using the binomial expansion of ( N+1 )p,c o n s i d e r (N+1 )p−(N+1 )=p/summationdisplay k=0pCkNk−(N+1 ) =1+p−1/summationdisplay k=1pCkNk+Np−N−1. But, as shown above,pCkis divisible by pfor all kin the range 1 ≤k≤p−1, andNp−Nis divisible by p, by assumption. Thus ( N+1 )p−(N+ 1) is divisible bypif it is true that Np−Nis divisible by p. Taking N= 1, for which, as noted above, the assumption is valid by inspection for any p, the result follows for all positive integers nand all primes p. Now consider f(n)=n5−n. By the result just proved f(n) is divisible by (prime number) 5. Further, f(n)=n(n4−1) = n(n2−1)(n2+1 )= n(n−1)(n+1 ) ( n2+1 ) . Thus the factorisation of f(n) contains three consecutive integers; one of them must be divisible by 3 and at least one must be even and hence divisible by 2.Thus, f(n) has the prime numbers 2, 3 and 5 as its divisors and must therefore be divisible by 30. 1.29Prove, by the method of contradiction, that the equation xn+an−1xn−1+···+a1x+a0=0, in which all the coefficients aiare integers, cannot have a rational root, unless that root is an integer. Deduce that any integral root must be a divisor of a0and hence find all rational roots of (a)x4+6x3+4x2+5x+4=0 , (b)x4+5x3+2x2−10x+6=0 . Suppose that the equation has a rational root x=p/q, where integers pandq have no common factor and qis neither 0 nor 1. Then substituting the root and multiplying the resulting equation by qn−1gives pn q+an−1pn−1+···+a1pqn−2+a0qn−1=0. But the first term of this equation is not an integer (since pandqhave no factor 13