TEST BANK FOR Mathematical methods for physics and engineering 3rd Ed By Riley, Kenneth Hobson

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.