SOLUTIONS MANUAL
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, Contents
Preface page ix
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1 Preliminary algebra 1
2 Preliminary calculus 17
3 Complex numbers and hyperbolic functions 39
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4 Series and limits 55
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5 Partial differentiation 71
6 Multiple integrals 90
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7 Vector algebra 104
8 Matrices and vector spaces 119
C
9 Normal modes 145
10 Vector calculus 156
11 Line, surface and volume integrals 176
,12 Fourier series 193
13 Integral transforms 211
14 First-order ODEs 228
15 Higher-order ODEs 246
16 Series solutions of ODEs 269
PR
17 Eigenfunction methods for ODEs 283
18 Special functions 296
19 Quantum operators 313
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20 PDEs: general and particular solutions 319
FD
21 PDEs: separation of variables and other methods 335
22 Calculus of variations 353
23 Integral equations 374
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24 Complex variables 386
C
25 Applications of complex variables 400
26 Tensors 420
27 Numerical methods 440
28 Group theory 461
29 Representation theory 480
,30 Probability 494
31 Statistics 519
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vii
, C
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, 1
Preliminary algebra
PR
Polynomial equations
1.1 It can be shown that the polynomial
O
— —
1
and three real roots altogether. Continue
2
FD
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
O
values for the other two roots.
3x — 6x — k
3 2
(c) Show that 4x has only one real root unless
—5 ≤ k ≤ 4 .
7
C
(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
49 — 16
—7 ± 8 .
α1,2 =
(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 — 114 , respectively. Thus g(x) can have up to 4
subtracted from it or up to 114 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 ≤4 7 , f(x) [= g(x)+1 — k] has only one real root.
1.3 Investigate the properties of the polynomial equation
PR
by proceeding as follows.
(a) By writing the fifth-degree polynomial appearing in the expression for fr(x)
in the form 7x5 30x4 a(x — b)2 c, show that there is in fact only one
O
negative values of x, determine what you can about the positions of the real
FD
(a) We start by finding the derivative of f(x) and note that, because f contains no
linear term, fr can be written as the product of x and a fifth-degree polynomial:
f(x) = x7 + 5x6 + x4 — x3 + x2 — 2 = 0,
fr(x) = x(7x5 + 30x4 + 4x2 — 3x + 2)
O
= x[ 7x5 + 30x4 + 4(x — 38)2 — 4( 38 )2 + 2 ]
= x[ 7x5 + 30x4 + 4(x — 38 )2 + 23
16
].
C
Since, for positive x, every term in this last expression is necessarily positive, it
follows that fr(x) can have no zeros in the range 0 < x <œ . 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(+ œ ) = +œ and f(0) = —2 < 0 and so f(x) = 0
has at least one root in 0 < x < œ . 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.
PR
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 i 2 —1.
Starting in each case from the ‘product of factors’ form of the quadratic equation,
(x — α1)(x — α2) = 0, we obtain:
O
(a) (x + 6)(x + 3) = x2 + 9x + 18 = 0;
(b) (x — 0)(x — 4) = x2 — 4x = 0;
FD
(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.
O
Trigonometric identities
C
1.7 Prove that
12 2 2
by considering
the sum of the sines of π/3 and π/6,
the sine of the sum of π/3 and π/4.
(a) Using
A+B A—B
sin A + sin B = 2 sin cos ,
2 2
3
, PRELIMINARY ALGEBRA
we have
π π π π
sin + sin = 2 sin cos ,
3, 6 4 12
3 1 1 π
+ = 2 , cos ,
2 2 ,2 12
π 3+1
cos = , .
12 2 2
(b) Using, successively, the identities
sin(A + B) = sin A cos B + cos A sin B,
sin(π — θ) = sin θ
PR
and cos( 21 π — θ) = sin θ,
we obtain
π π π π π π
sin + = sin cos + cos sin ,
3 4 3 4 3 4
7π ,3 1 1 1
12 ,2 ,2 + 2 ,2 ,
O
sin =
5π 3+1
sin = 2, 2 ,
12 ,
FD
π 3+1
cos = , .
12 2 2
O
1.9 Find the real solutions of
(a) 3 sin θ — 4 cos θ 2,
C
12 sin θ — 5 cos θ = —6.
We use the result that if
a sin θ + b cos θ = k
then
k
θ = sin—1 — φ,
K
where
b
K2 = a2 + b2 and φ = tan—1 .
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 25— tan—1 —34 = 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 —136 — tan—1 —125 = —0.0849 or — 2.267.
PR
1.11 Find all the solutions of
O
that lie in the range — θ ≤ π. What is the multiplicity of the solution θ 0?
FD
Using
sin(A + B) = sin A cos B + cos A sin B,
A+B A—B
and cos A — cos B = —2 sin sin ,
2 2
O
and recalling that cos(—φ) = cos(φ), the equation can be written successively as
5θ 3θ 5θ θ
C
2 sin cos — = 2 sin cos — ,
2 2 2 2
5θ 3θ θ
sin cos — cos = 0,
2 2 2
5θ θ
—2 sin sin θ sin = 0.
2 2
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
PR
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C
, Contents
Preface page ix
PR
1 Preliminary algebra 1
2 Preliminary calculus 17
3 Complex numbers and hyperbolic functions 39
O
4 Series and limits 55
FD
5 Partial differentiation 71
6 Multiple integrals 90
O
7 Vector algebra 104
8 Matrices and vector spaces 119
C
9 Normal modes 145
10 Vector calculus 156
11 Line, surface and volume integrals 176
,12 Fourier series 193
13 Integral transforms 211
14 First-order ODEs 228
15 Higher-order ODEs 246
16 Series solutions of ODEs 269
PR
17 Eigenfunction methods for ODEs 283
18 Special functions 296
19 Quantum operators 313
O
20 PDEs: general and particular solutions 319
FD
21 PDEs: separation of variables and other methods 335
22 Calculus of variations 353
23 Integral equations 374
O
24 Complex variables 386
C
25 Applications of complex variables 400
26 Tensors 420
27 Numerical methods 440
28 Group theory 461
29 Representation theory 480
,30 Probability 494
31 Statistics 519
PR
O
FD
O
C
vii
, C
O
FD
O
PR
, 1
Preliminary algebra
PR
Polynomial equations
1.1 It can be shown that the polynomial
O
— —
1
and three real roots altogether. Continue
2
FD
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
O
values for the other two roots.
3x — 6x — k
3 2
(c) Show that 4x has only one real root unless
—5 ≤ k ≤ 4 .
7
C
(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
49 — 16
—7 ± 8 .
α1,2 =
(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 — 114 , respectively. Thus g(x) can have up to 4
subtracted from it or up to 114 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 ≤4 7 , f(x) [= g(x)+1 — k] has only one real root.
1.3 Investigate the properties of the polynomial equation
PR
by proceeding as follows.
(a) By writing the fifth-degree polynomial appearing in the expression for fr(x)
in the form 7x5 30x4 a(x — b)2 c, show that there is in fact only one
O
negative values of x, determine what you can about the positions of the real
FD
(a) We start by finding the derivative of f(x) and note that, because f contains no
linear term, fr can be written as the product of x and a fifth-degree polynomial:
f(x) = x7 + 5x6 + x4 — x3 + x2 — 2 = 0,
fr(x) = x(7x5 + 30x4 + 4x2 — 3x + 2)
O
= x[ 7x5 + 30x4 + 4(x — 38)2 — 4( 38 )2 + 2 ]
= x[ 7x5 + 30x4 + 4(x — 38 )2 + 23
16
].
C
Since, for positive x, every term in this last expression is necessarily positive, it
follows that fr(x) can have no zeros in the range 0 < x <œ . 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(+ œ ) = +œ and f(0) = —2 < 0 and so f(x) = 0
has at least one root in 0 < x < œ . 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.
PR
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 i 2 —1.
Starting in each case from the ‘product of factors’ form of the quadratic equation,
(x — α1)(x — α2) = 0, we obtain:
O
(a) (x + 6)(x + 3) = x2 + 9x + 18 = 0;
(b) (x — 0)(x — 4) = x2 — 4x = 0;
FD
(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.
O
Trigonometric identities
C
1.7 Prove that
12 2 2
by considering
the sum of the sines of π/3 and π/6,
the sine of the sum of π/3 and π/4.
(a) Using
A+B A—B
sin A + sin B = 2 sin cos ,
2 2
3
, PRELIMINARY ALGEBRA
we have
π π π π
sin + sin = 2 sin cos ,
3, 6 4 12
3 1 1 π
+ = 2 , cos ,
2 2 ,2 12
π 3+1
cos = , .
12 2 2
(b) Using, successively, the identities
sin(A + B) = sin A cos B + cos A sin B,
sin(π — θ) = sin θ
PR
and cos( 21 π — θ) = sin θ,
we obtain
π π π π π π
sin + = sin cos + cos sin ,
3 4 3 4 3 4
7π ,3 1 1 1
12 ,2 ,2 + 2 ,2 ,
O
sin =
5π 3+1
sin = 2, 2 ,
12 ,
FD
π 3+1
cos = , .
12 2 2
O
1.9 Find the real solutions of
(a) 3 sin θ — 4 cos θ 2,
C
12 sin θ — 5 cos θ = —6.
We use the result that if
a sin θ + b cos θ = k
then
k
θ = sin—1 — φ,
K
where
b
K2 = a2 + b2 and φ = tan—1 .
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 25— tan—1 —34 = 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 —136 — tan—1 —125 = —0.0849 or — 2.267.
PR
1.11 Find all the solutions of
O
that lie in the range — θ ≤ π. What is the multiplicity of the solution θ 0?
FD
Using
sin(A + B) = sin A cos B + cos A sin B,
A+B A—B
and cos A — cos B = —2 sin sin ,
2 2
O
and recalling that cos(—φ) = cos(φ), the equation can be written successively as
5θ 3θ 5θ θ
C
2 sin cos — = 2 sin cos — ,
2 2 2 2
5θ 3θ θ
sin cos — cos = 0,
2 2 2
5θ θ
—2 sin sin θ sin = 0.
2 2
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
