Exam (elaborations) TEST BANK FOR Mathematical Methods for Physics and Engineering 3rd Edition By Riley

Exam (elaborations) TEST BANK FOR Mathematical methods for physics and engineering 3rd Edition By Riley, Kenneth Franklin_ Hobson, Michael Paul (Solution Manual) Student Solutions Manual for Mathematical Methods for Physics and Engineering Third Edition K. F. RILEY andM. P. HOBSON Contents Preface page ix 1 Preliminary algebra 1 2 Preliminary calculus 17 3 Complex numbers and hyperbolic functions 39 4 Series and limits 55 5 Partial differentiation 71 6 Multiple integrals 90 7 Vector algebra 104 8 Matrices and vector spaces 119 9 Normal modes 145 10 Vector calculus 156 11 Line, surface and volume integrals 176 v CONTENTS 12 Fourier series 193 13 Integral transforms 211 14 First-order ODEs 228 15 Higher-order ODEs 246 16 Series solutions of ODEs 269 17 Eigenfunction methods for ODEs 283 18 Special functions 296 19 Quantum operators 313 20 PDEs: general and particular solutions 319 21 PDEs: separation of variables and other methods 335 22 Calculus of variations 353 23 Integral equations 374 24 Complex variables 386 25 Applications of complex variables 400 26 Tensors 420 27 Numerical methods 440 28 Group theory 461 29 Representation theory 480 vi 1 Preliminary algebra Polynomial equations 1.1 It can be shown that the polynomial g(x) = 4x3 + 3x2 − 6x − 1 has turning points at x = −1 and x = 1 2 and 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 x between −2 and 2. Use it and the information given above to draw a graph and so determine the roots of g(x) = 0 as accurately as possible. (b) Find one accurate root of g(x) = 0 by inspection and hence determine precise values for the other two roots. (c) Show that f(x) = 4x3 + 3x2 − 6x − k = 0 has only one real root unless −5 ≤ k ≤ 7 4 . (a) Straightforward evaluation of g(x) at integer values of x gives the following table: x −2 −1 0 1 2 g(x) −9 4 −1 0 31 (b) It is apparent from the table alone that x = 1 is an exact root of g(x) = 0 and so g(x) can be factorised as g(x) = (x−1)h(x) = (x−1)(b2x2+b1x+b0). Equating the coefficients of x3, x2, x and the constant term gives 4 = b2, b1 − b2 = 3, b0 − b1 = −6 and −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 = −1 and x = 1 2, are 4 and −11 4 , respectively. Thus g(x) can have up to 4 subtracted from it or up to 11 4 added to it and still satisfy the condition for three (or, at the limit, two) distinct roots of g(x) = 0. It follows that for k outside the range −5 ≤ k ≤ 7 4 , f(x) [= g(x) + 1 − k] has only one real root. 1.3 Investigate 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(x) in the form 7x5 + 30x4 + 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) and f(−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 f contains no linear term, f can be written as the product of x and a fifth-degree polynomial: f(x) = x7 + 5x6 + x4 − x3 + x2 − 2 = 0, f  (x) = x(7x5 + 30x4 + 4x2 − 3x + 2) = x[ 7x5 + 30x4 + 4(x − 3 8 )2 − 4( 3 8 )2 + 2] = x[ 7x5 + 30x4 + 4(x − 3 8 )2 + 23 16 ]. Since, for positive x, every term in this last expression is necessarily positive, it follows that f(x) can have no zeros in the range 0 x ∞. Consequently, f(x) can have no turning points in that range and f(x) = 0 can have at most one root in the same range. However, f(+∞) = +∞ and f(0) = −2 0 and so 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) = −2 and f(−1) = 5, and so there is at least one root in each of the ranges 0 x 1 and −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 x lies. Here grouping successive pairs of terms yields some information as follows: x7 + 5x6 is positive for x −5, x4 − x3 is positive for x 1 and x 0, x2 − 2 is positive for x √ 2 and x − √ 2. Thus, all three terms are positive in the range(s) common to these, namely −5 x − √ 2 and x 1. It follows that f(x) is positive definite in these ranges and there can be no roots of f(x) = 0 within them. However, since f(x) is negative for large negative x, there must be at least one root α with α −5. 1.5 Construct the quadratic equations that have the following pairs of roots: (a) −6,−3; (b) 0, 4; (c) 2, 2; (d) 3 + 2i, 3 − 2i, where 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 + 18 = 0; (b) (x − 0)(x − 4) = x2 − 4x = 0; (c) (x − 2)(x − 2) = x2 − 4x + 4 = 0; (d) (x − 3 − 2i)(x − 3 + 2i) = x2 + x(−3 − 2i − 3 + 2i) + (9 − 6i + 6i − 4i2) = x2 − 6x + 13 = 0. Trigonometric identities 1.7 Prove that cos π 12 = √ 3 + 1 2 √ 2 by considering (a) the sum of the sines of π/3 and π/6, (b) the sine of the sum of π/3 and π/4. (a) Using sinA + sinB = 2sin
A + B 2  cos
A − B 2  , 3 PRELIMINARY ALGEBRA we have sin π 3 + sin π 6 = 2sin π 4 cos π 12 , √ 3 2 + 1 2 = 2 √1 2 cos π 12 , cos π 12 = √ 3 + 1 2 √ 2 . (b) Using, successively, the identities sin(A + B) = sinAcosB + cosAsin B, sin(π − θ) = sinθ and cos( 1 2π − θ) = sinθ, we obtain sin π 3 + π 4  = sin π 3 cos π 4 + cos π 3 sin π 4 , sin 7π 12 = √ 3 2 √1 2 + 1 2 √1 2 , sin 5π 12 = √ 3 + 1 2 √ 2 , cos π 12 = √ 3 + 1 2 √ 2 . 1.9 Find the real solutions of (a) 3 sin θ − 4 cos θ = 2, (b) 4 sin θ + 3cos θ = 6, (c) 12 sin θ − 5 cos θ = −6. We use the result that if a sin θ + b cos θ = k then θ = sin −1
k K  − φ, where K2 = a2 + b2 and φ = tan −1 b a . 4 PRELIMINARY ALGEBRA Recalling that the inverse sine yields two values and that the individual signs of a and b have to be taken into account, we have (a) k = 2, K = √ 32 + 42 = 5, φ = tan−1(−4/3) and so θ = sin −1 2 5 − tan −1 −4 3 = 1.339 or − 2.626. (b) k = 6, K = √ 42 + 32 = 5. Since k K there is no solution for a real angle θ. (c) k = −6, K = √ 122 + 52 = 13, φ = tan−1(−5/12) and so θ = sin −1 −6 13 − tan −1 −5 12 = −0.0849 or − 2.267. 1.11 Find all the solutions of sin θ + sin 4θ = sin 2θ + sin 3θ that lie in the range −π θ ≤ π. What is the multiplicity of the solution θ = 0? Using sin(A + B) = sinAcosB + cosAsin B, and cosA − cosB = −2 sin
A + B 2  sin
A − B 2  , and recalling that cos(−φ) = cos(φ), the equation can be written successively as 2 sin 5θ 2 cos
−3θ 2  = 2sin 5θ 2 cos
−θ 2  , sin 5θ 2
cos 3θ 2 − cos θ 2  = 0, −2 sin 5θ 2 sin θ sin θ 2 = 0. The first factor gives solutions for θ of −4π/5, −2π/5, 0, 2π/5 and 4π/5. The second factor gives rise to solutions 0 and π, whilst the only value making the third factor zero is θ = 0. The solution θ = 0 appears in each of the above sets and so has multiplicity 3. 5 PRELIMINARY ALGEBRA