further mathematics uunniitt ffpp11 -...
TRANSCRIPT
Thomas Whitham Sixth Form
Further
Mathematics
UUnniitt FFPP11
Polynomial Equations
Mathematical Induction
Finite Series
Complex Numbers
Derivative by formula & by
using Logarithms
Matrices
1
Polynomial Equations
1. An nth
degree polynomial equation will have n roots
Example 072 2 xx is a quadratic equation. It is 2nd
degree
and it has two roots.
Example 01623 xx is a cubic equation. It is 3rd
degree and it
has three roots.
Example 0323 234 xxxx is a quartic equation. It is 4th
degree and has four roots.
2. Irrational roots will occur in pairs qp where 0q and is not a
perfect square.
Example The equation 0112 23 baxxx has one root
32 . Find the values of a and b, and hence solve the
equation completely.
If 32x is one root then 32x is a second root.
It follows that 3232 xx is a quadratic factor; this
simplifies to 142 xx
Hence cxxxbaxxx 214112 223
2x c 811 3c
x ca 42 14a
const cb 3b
The equation reduces to 0314112 23 xxx
Or 032142 xxx
By inspection
2
Full solution set 2
3 ,32 x
3. Imaginary roots will occur in conjugate pairs iba .
4. It follows from 1 and 3 that any polynomial equation of odd degree
will have at least one real root.
5. Recall the Factor Theorem (P2), restated her.
“If 0abP then bax is a factor of xP ; it follows that
a
bx is a solution of 0xP ”
6. If is a repeated root of 0xP , then is also a root of 0 xP .
NB the converse is not generally true.
Example Solve the equation 025412 23 xxx given that it
has two equal roots.
Let 25412 23 xxxxP
518125836 2 xxxxxp
0 xp for 2
1x and
18
5x
but 021 p
2
1x is a double root of the equation.
(you can readily check that 0185 p )
It follows that 231225412223 xxxxx
HHeennccee tthhee rroooottss ooff tthhee eeqquuaattiioonn aarree 21 ,, 2
1 ,, -- 23
By inspection
3
7. Graphical interpretations Real and distinct roots to 0xp
correspond to intersections of the graph of xpy with the x axis.
Repeated roots occur where the graph touches the x axis. Some
illustrative examples follow
Quadratics
Cubics If a cubic equation is to have more than one real root, the
corresponding cubic function must have two stationary points. This is a
necessary condition but is not in itself sufficient.
0 x
y
0 x
y
0 x
y
No real roots Real distinct roots Double roots
0 x
y
0 x
y
0 x
y
Three real and
distinct roots
Only one real root Adouble + one more
root
0 x
y
0 x
y
(a, 0)
inflexion ifpoint
stationary a
0
0
0
ap
ap
ap
Only one real root
(no stationary points)
4
Quartics Quartic graphs may have one, two or three stationary points. In
each of the following illustrations, the coefficient of 4x has been taken
to be positive so that, as x y
One SP
Two SPs
Three SPs
0 x
y
0 x
y
0 x
y
0 x
y
0 x
y
0 x
y
No real roots Two real distinct roots Four equal roots
Two real distinct roots
Triple +one root
Root (poi)
Double root
Two roots
Two distinct roots + one double root
Four roots
No real roots
5
8. Relationship between roots and coefficients of a polynomial
equation.
Quadratic 02 cbxax
Cubic 023 dcxbxax
Quartic
0234 edxcxbxax This time, using notation
a
e
a
d
a
c
a
b
Example Solve the equation 0367 23 xx given that one root
is double the other.
Let the roots be , 2 and
73
022 2
032
a
ca
b
Product
roots of sum
a
da
ca
b
product
pairs ofproduct
sum
6
0 032
Solve simultaneously 3 , 2
Check “” a
d 36263.2.
Therefore roots are 2,6,3 x
Example (i) Show that 22222
(ii) The roots of the equation 0133 xx are , , ; Find
the values of (a) 222 (b)
111
(i) 22222
2
222
222
222
(ii) 0
1
3
(a) Using the result (1) 320 2222
6222
(b) 31
3111
9. Equations with related roots Given the roots of a polynomial
equation the equation can be formed as follows.
Quadratic 02 xx
Cubic 023 xxx
7
Quartic 0234 xxxx
Etc..
Example The roots of the equation 05423 xxx are ,
and . Find the cubic equations whose roots are
(i) 2 , 2 , 2
(ii) 1 , 1 , 1
1
3
1
(i) For new equation 22222
164222222
408222
Equation 040162 23 xxx
(ii) For new equation
43111
1324
32
111
111111
71145
1
1
111
Equation 074 23 xxx
8
Mathematical Induction
Example Prove by induction that, where n is a positive integer
(i) 152 n is divisible by 24
(ii) 1216
1
1
2
nnnrn
r
(i) Let 152 nnf
nf
nf
n
nn
nnn
2
22
2212
5.24
155.24
1512415.25151
Hence if nf is divisible by 24, so is 1nf
When n = 1, 24125151 2 f , so the result is true for 1n
Hence it will be true successively for ,...4,3,2n
i.e. it is universally true.
(ii) Let the result be true for some value of kn say
i.e. kfkkkrk
r
1216
1
1
2
then
67216
1
161216
1
11216
1
2
2
21
1
2
kkk
kkkk
kkkkrk
r
132216
1 kfkkk
Hence if the result is true for kn , it is also true for 1 kn
9
With 1n LHS = 1
RHS = 13216
1
Hence result is true for 1n , and so will be successively true for
,...4,3,2n
Summation of finite series
(i) 12
1.....321
1
nnnrn
r
summing the AP
(ii) 1216
1.....321 2222
1
2
nnnnrn
r
Proof by induction, or as follows, using the identity
1331 233 nnnn
1131312
1232323
........
........
........
1232321
113131
13 31
233
233
233
233
233
nnnn
nnnn
nnnn
10
121
121
132
326
2136662
1333
3311
612
2
232
223
23223
22
nnnr
nnn
nnn
nnnr
nnnrnnn
nnnrnnn
nrrn
(iii) 2213 1 nnr
proof by induction or by the identity 14641 2344 nnnnn
Example Find the sum of n terms of the series ... 53 32 11
The rth
term 12 rrU r so we require
rrrrrrn
r
22
1
2212
141
31221
1121212
61
61
21
61
1
nnn
nnn
nnnnnrrn
r
(iv) nn
r
r nxxxxrx
....32.1 32
1
Here the coefficients are in AP and the terms in x are in GP. Let the sum
be nS
11
x
nx
x
xxS
nxxxxxSx
nxxnxxxS
nxxxxS
nn
n
nn
n
nn
n
n
n
11
1
........1
1.......2
.......32
1
2
132
132
32
Method to be learnt, not as a formula!
Note that if 1x then as n 0nx
21 x
xSn
i.e.
21 x
xS
In other words the series is convergent with a finite sum of 21 x
x
so
long as 1x
Method of differences This is based upon the following result or similar.
If 1 rfrfU r 01
fnfUn
r
r
It isn‟t really necessary to learn this as a formula
Example Find
n
r rr1 1
1
111
1
1
rrrr Cover up rule.
1
11
1
1
1 rrrr
n
r
12
11
11
1
11
......11
1141
31
31
21
21
n
n
n
n
n
nn
[As was stated, similar, 1 rfrfU r ]
COMPLEX NUMBERS
Real numbers consist of integers, rational numbers and irrational numbers; they
can be represented by points on a (real) number line.
A complex number is one which can be written in the form ibaz
where a and b are real and 1i
(i) 0,0 ba ; the number ibaz is said to be imaginary
(ii) 0,0 ba ; the number ibz is said to be purely imaginary
(iii) 0,0 ba ; the number az is said to be real
Given ibaz , a is known as the real part of z, and b is known as the
imaginary part of z. –both real!
Powers of i 12 i , ii 3 , 14 i , ii 5 , 16 i , etc…
Given ibaz , the imaginary number ibaz is said to be the
conjugate of z. Imaginary roots of polynomial equations occur in
conjugate pairs.
Example 0422 zz has roots 312
1642iz
i.e the roots are 31 i and 31 i
The zero complex number If ibaz and 0z then both 0a and
0b . i.e. both real parts and imaginary parts are zero.
13
Equal complex numbers If iqpiba then pa and qb . i.e.
real parts are equal and imaginary parts are equal.
Some algebraic operations on complex numbers
Addition qbipaiqpiba
Subtraction qbipaiqpiba
Multiplication by real number ikbkaibak
Product of two complex nos. aqbpibqapiqpiba
Example Show that for the complex number z and its conjugate z
(i) the sum is real
(ii) the difference is purely imaginary
(iii) the product is a real positive number.
Let ibaz ibaz
(i) azz 2 real
(ii) bizz 2 purely imaginary
(iii) 22 bazz real and positive
Example Find real numbers x and y such that 052143 iyix
yxiyxiyix 245352143
RP = 0, 053 yx
IP = 0, 024 yx
Division
idcidc
idciba
idc
ibaetc..
10,5
yx
Makes real denominator
14
Example 13
7
13
3
13
73
49
276
2323
232
23
22
2
ii
i
ii
ii
ii
i
i
Graphical representation of a complex number is by a point in the
Argand diagram, otherwise called the z plane.
P represents ibaz
Note the vector like use of the line
segment OP to denote the position
of P.
NB Not an i to be seen!
Addition
Parallelogram law
Subtraction 2121 zzzz
So use above law.
The modulus and argument of a complex number
Let P(x, y) represent iyxz
Modulus 22 yxrz
Argument zarg where is a directed
angle, in radians, and in the angle made by
OP with Ox measured from Ox where
P(a, b)
z
x
y
0
Axis for the
imaginary
part
(imaginary
axis)
Axis for the real part (real axis)
2z
x
y
0
1z
21 zz
P(x, y)
r
x
y
0
15
Take care with argument!
x
y1tan can lead to the wrong angle.
Always plot a point in the argand diagram to find the argument.
Example Find the modulus and argument of iz 3
213 z
3
1tan
6
6
5arg
z
NB If you are going to use your calculator when exact values are
required, work in degrees and convert to radians.
Note that
3
1tan 1 has a principle value of –30, not the angle which
we want… now refer to the diagram.
Modulus/ Argument form of a complex number
Using simple right angled triangle trig
cosrx and sinry
sincos irriyxz
sincos irz
P r
x
y
0
1
3
P(x, y)
r
x
y
0
16
Geometrically the difference of two complex numbers is very important
2121 PPzz
21arg zz angle between Ox and 12 PP
measured from Ox
Product and Quotient of two complex numbers
Given 1111 sincos irz and 2222 sincos irz it is quite
straight forward to show that
21212121 sincos irrzz
So that 2121 zzzz and 2121 argargarg zzzz
2121
2
1
2
1 sincos ir
r
z
z
So that 2
1
2
1
z
z
z
z and 21
2
1 argargarg zzz
z
Geometrically the quotient of two complex numbers is important
21
2
1 argargarg zzz
z
and is the angle
between 1OP and 2OP measured from
2OP
2z
r
1z
r
1P
x
y
0
2P
2z
r
1z
r
1P
x
y
0
2P
17
Example (a) Express iz 11 and 312 iz each in the form
sincos ir by plotting points 1P and 2P
representing 1z and 2z in an argand diagram. Show that
12arg zz is the acute angle where
2
13tan 1
(b) Write 2
1
z
z in the form iba and hence show that
13
13
12
5tan
2111 z
4
3arg 1
z
4
3sin
4
3cos21
iz
2312 z 3
arg 2
z
3sin
3cos22
iz
(a) 2
13
11
13gradtan 21
PP
2
13tan 1
(b)
ii
i
i
i
i
z
z
4
13
4
13
4
1313
31
31
31
1
2
1
2
1argz
z
3,12P
x
y
0
1,11 P
18
Since 12
5
34
3argargarg 21
2
1
zz
z
z then if r
z
z
2
1
4
13
12
5cos
r and
4
13
12
5sin
r
13
13
4
13
4
13
12
5tan
Extensions of products
Let kkkk rz sincos nk ,...3,2,1
Then
nnnn rrrrzzzz ..sin..cos...... 221221321321
Now if zzzz n ....21 say where sincos irz
ninrz nn sincos
Example If 31 iz express 5z in modulus argument form
231 z 325 z
32arg z
3
10arg 5
z ; but it is usual to express
argument in the range so we would take
32arg z
32
325 sincos32 z
32
3
(-1, 3)
19
Loci in the argand diagram Let P(x, y) represent iyxz in the agrand
diagram.
(i) Associated with modulus
Type azz 1 where 1z is a
constant complex number
represented by 1P ; and a is a real
constant
Locus of P is circle centre 1P , rad a.
Type 21 zzzz
Locus of P is perpendicular bisector
of 21PP
The above two are readily recognised, and their equations follow
Example 32 iz rewritten 32 iz gives a circle
with equation 91222 yx
However the equations of these, and other perhaps unfamiliar loci, can be
determined algebraically.
Example 32 ziz
Let iyxz
32 iyxiiyx
iyxyix 321
1P z
1z
P
a
1P
2P
P
20
35224330
3624412
341
321
22
2222
2222
2222
yxyx
yxxyyx
yxyx
yxyx
08335
3222 yxyx
This is the equation of a circle –the centre and radius can be found by
completing the square.
Example 64 zz
Let iyxz
64 iyxiyx 64 2222 yxyx
ellipsean 1
59
2
45925
459205
4202599
253
82012
12364
64
22
22
22
222
22
22
222222
2222
yx
yx
yxx
xxyx
xyx
xyx
yxyxyx
yxyx
(ii) Associated with argument
Type 1arg zz . The locus of P(x, y) will be a part line through P,
(representing 1z ) making an angle with 0x.
21
Example 4
1arg
iz
Write 4
1arg
iz and we have the locus as a part line through
(1, –1) making 4
with 0x.
Gradient = 14
tan
Equation. 111 xy
2 xy for 1x
Algebraic approach gives the equation of the line but will not reveal the
restriction. As above
4
1arg
iz 4
11arg
yix
4
tan1
1
x
y 1
1
1
x
y Hence 2 xy
So the diagram is essential!
Transformations of the Argand diagram
If iyxz is represented by P(x, y) in the z plane and zfw then
the function f represents a mapping of the point P to another point Q
where Q(u, v) represents ivuw . Consequently the locus of P
transforms to the locus of Q.
The procedure in finding the locus of Q is illustrated in the following:
4
1z
r
1P
x
y
0
P
22
Example The locus of P representing iyxz in the line 1x .
Show that the locus of Q representing w where z
w1
is a
circle which passes through the origin.
zw
1
iyxivu
1
iyx
iyx
iyxivu
1
2222 yx
yi
yx
xivu
22 yx
xu
,
22 yx
yv
Locus P is 1x 21
1
yu
,
21 y
yv
yuv i.e. u
vy
2
2
1
1
u
vu
22
2
vu
uu
uvu 22 022 vuu 4122
21 vu
Circle centre 0,21 radius =
21 and passing through 0.
Example The locus of P representing iyxz is the rectangular
hyperbola 422 yx . If 2zw find the locus of the
point Q representing w.
2zw xyiyxiyxivu 2222
,22 yxu xyv 2
but 422 yx 4u . Hence the locus of Q is the line 4u
23
CALCULUS: FURTHER DIFFERENTIATION
General definition of a derivative
x
xfxxf
xxf
0
lim
Proof –in essence a first principles procedure.
Let xfy
xf
x
xfxxf
x
y
x
xfxxf
x
y
xfxxfy
xxfyy
limlim
The formula might be otherwise expressed as
h
xfhxf
hxf
0
lim
Example Find using the formula the derivative of 12
1
x
1212
12212
0
lim
1212
2
0
lim
1212
12212
0
lim
12
1
12
1
0
lim
xhxh
hxx
h
xhxh
h
h
xhxh
hxx
h
h
xhx
hxf
24
212
2
1212
2
x
xx
Logarithmic differentiation
Example Obtain the derivative of xxsin
x
xxxx
x
xxxy
dx
dy
xxxx
dx
dy
y
xxy
xy
xy
x
x
x
sinlncos
sinlncos
1.sinlncos.
1
lnsinln
lnln
sin
sin
sin
MATRICES
3 x 3 matrices
(1) The determinant of
333
222
111
cba
cba
cba
A written Adet or A is given
by 33
22
1
33
22
1
33
22
1detba
bac
ca
cab
cb
cbaA
The 2x2 determinants 33
22
33
22
33
22,,
ba
ba
ca
ca
cb
cb are called the minors of
111 ,, cba respectively.
Implicit Product rule
25
In expanded form
231312123132213321det cbacbacbacbacbacbaA
And can be obtained schematically and certainly not by formula.
If 0det A , the matrix is said to be singular.
(2) Minors and cofactors of detA. The minor of an element of Adet is
found by deleting the row and column containing the element, and
forming a determinant of the remaining four elements. The cofactor
is then found by multiplying by +1 or –1 according to
111
111
111
e.g. From (2), the minor of 2a is 33
11
cb
cb and cofactor is
33
11
cb
cb
(3) The Adjugate matrix, written Aadj
Either (a) replace each element in A by its cofactor and transpose the
resulting matrix,
Or (b) transpose first and then replace elements by cofactors.
Example
150
132
011
A
1510
112
112
111
511
1022
adj
1
A
You could try it the other way round to check out the result!
(4) The inverse matrix, written 1
A is given by
AA
A adj.det
11
26
Example To find the inverse of
821
482
248
A
392)646416()168512(det A
adjA =
56124
286212
02856
56280
126228
412561
71
983
981
141
19631
983
141
71
1
0
56124
286212
02856
392
1A
You could check to see if IAA 1
(5) Solutions to simultaneous equations by matrices
Example Solve the equations
132
5
42
zyx
zyx
zyx
Write
1
5
4
132
111
121
z
y
x
1
5
4
132
111
1211
z
y
x
1
5
4
0
93
91
95
93
93
93
95
92
z
y
x
27
You will find
920x ,
31y ,
922z
Geometrically these equations represent three planes and this unique
solution corresponds to intersection of the planes in one point.
However other possibilities arise when solving
c
b
a
z
y
x
A
Here are some
(i) All three planes could be parallel
(ii) Two could be parallel and one not
(iii) The planes could intersect in three parallel lies to form an
enclosed triangular prism
In each of these cases there will be no values of x, y, z which satisfy the
equations simultaneously. The equations are said to be inconsistent and
0det A
(iv) The planes intersect in a line. In this case there is a line
solution and again 0det A . The equations are said to be not
independent.
Example Show that the equations 0352 zyx , 24 zyx
and 0457 zy are not independent.
First write as
0
2
0
570
411
352
z
y
x
and obtain 02556021010
570
415
352
det
A
Next
457
24
0352
zy
zyx
zyx
eliminate x 457 zy (this being the 3rd equation)
28
Hence the equations are not independent. The three planes intersect in a line, so
we can find a „line‟ solution.
In 457 zy let kz
7
45ky
Sub into 2nd
equation 247
45
k
kx
7
2310 kx
Line solution is
7
2310 kx ,
7
45ky , kz
You could check with the 1st equation if you wish!
Example Show that the equations 1 zyx , 4232 zyx
and 223 zyx are inconsistent .
First 0249463
123
232
111
det
A
Next
3rd and1st from eliminate
2first from eliminate
545
645
223
4232
1
x
x
zy
zy
zyx
zyx
zyx
These two equations are clearly inconsistent and consequently so are the
original three. Hence there are no solutions to these equations.
(6) Reduction to Echelon form
Example Solve the equations 7 zyx , 92 zyx ,
12 zyx using the reduction to echelon form.
First we write in augmented matrix form as follows:
29
1112
9211
7111
31 rr
1112
9211
8023
32 2rr
1112
11015
8023
21 2rr
1112
11015
14007
147 x 2x
115 yx 1y
12 zyx 4z
The matrix is said to be in echelon form when there are zeros above or
below the leading diagonal.
In the previous example, the matrix can be reduced to
28700
2120
7111
Try it and see!
30
Example Reduce the following equations to echelon form.
,132 zyx 32 zyx , zyx 554
Consider the following
(a) 1 (b) 7,1 (c) 7,1
554
321
1312
you will find reduces to
72200
53220
1312
(a) 1 There will be a unique solution to the equations.
(b) 7,1 Gives
7000
5120
1312
and the equations are inconsistent
(c) ) 7,1 Gives
0000
5120
1312
and the equations are consistent but not independent.
In this case there will be a line solution. You could find it in the form
kx 58 , ky , 53 kz .
(7) Translations in the x/y plane
We let the point (x, y) be represented by
1
y
x
31
Transformation matrices will be of the form
100
rqp
nml
M
1
'
'
1100
y
x
rqypx
nmylx
rqp
nml
z
y
x
M
so that rqypxnmylxyx ,,
In particular consider
(i)
100
10
01
b
a
M
11100
10
01
by
ax
y
x
b
a
byaxyx ,, which is a translation
b
a
The following 2x2 matrices can be found in the formula booklet
(ii) A rotation through anticlockwise as represented by
cossin
sincos becomes
100
0cossin
0sincos
(iii) A reflection in tanxy as represented by
2cos2sin
2sin2cos becomes
100
02cos2sin
02sin2cos
32
Invariance
For an invariant point, if any, put xx ' and yy ' so that
11
y
x
rqypx
nmylx
Example Find a 33 matrix corresponding to
(i) a rotation of anticlockwise about 0 (where is acute and 43tan )
followed by a translation
1
1
(ii) a translation
1
1 followed by a rotation of anticlockwise about 0
(where is acute and 43tan ).
For each transformation find the image of (x, y) and identify invariant
points, if any.
Can you describe either or both of these transformations in simpler
form?
(i) The rotation 2x2 matrix is
cossin
sincos where
53sin ,
54cos
The 3x3 matrix is
100
0
0
54
53
53
54
33
The translation matrix is
100
110
101
The matrix for the combined transformations will be
100
1
1
100
0
0
100
110
101
54
53
53
54
54
53
53
54
1
1
1
1100
1
1
54
53
53
54
54
53
53
54
yx
yx
y
x
1,1,54
53
53
54 yxyxyx
For invariance
1
2
05354351
05353451
54
53
53
54
y
x
yxyxyyxy
yxyxxyxx
(-2, -1) is an invariant point. Hence the transformation is an
anticlockwise rotation about (-2, -1)
(ii) The matrix is
100100
110
101
100
0
0
51
54
53
57
53
54
54
53
53
54
11100
51
54
53
57
53
54
51
54
53
57
53
54
yx
yx
y
x
51
54
53
57
53
54 ,, yxyxyx
34
For invariance
2
1
0131435
0737345
51
54
53
57
53
54
y
x
yxyxyyxy
yxyxxyxx
(-1, -2) is an invariant point. Hence the transformation is an
anticlockwise rotation about (-1, -2)
Example To find a single 3x3 matrix for reflection in the line
643 xy and the image of point (x, y)
This line is 234 xy
First translate
2
0 with matrix
100
210
001
Next reflect in xy34 with matrix
100
02cos2sin
02sin2cos
where 34tan and, therefore
2524
916
34
1
2sin
,
257
916
916
1
1cos
Finally translate
2
0 with matrix
100
210
001
The single matrix will be given by
y
x
(0,2)
0
35
100
100100
210
001
100
210
001
100
0
0
100
210
001
2536
257
2524
2548
2524
257
2514
257
2524
2548
2524
257
257
2524
2524
257
11100
2536
257
2524
2548
2524
257
2536
257
2524
2548
2524
257
yx
yx
y
x
2536
257
2524
2548
2524
257 ,, yxyxyx
Linear Transformations All a foregoing transformations are called linear
because straight lines map to straight lines.
An invariant line is one which all points map to some other point on the
same line.
A line of invariant points is one in which all points on the line map onto
themselves.
Example The transformation T of the x/y plane is equivalent to a
reflection in xy followed by a translation in which (1, 2) is mapped
onto (0, 1).
a) find the 3x3 matrix representing T
b) determine the fixed points of T
c) find the image of the line cxy
36
d) find the image of parabola xy 42
The reflection matrix is
100
001
010
The translation is given by
1
1
2
1
1
0
a) the matrix for T is
100
101
110
100
001
010
100
110
101
b)
1
1
1
1
x
y
y
x
c) Let the lien be described by parametric equations tx , cty
1
1
1
1
t
ct
y
x
d) Let the parabola be described by tytx 2,2
1
1
12
1
2t
t
y
x
For invariant points 11
1
yx
xy
yx
1 yx is the line of invariant points
cxyty
ctx
1
1
cxy is an invariant line for all c.
2
41
211
1
12
xy
ty
tx
another parabola
37
Notes
38
Notes
39
Notes