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NNPC FSTP – ENGINEERS Phase 1
Mathematics
Nexus Alliance Ltd
1
AIMS OF THE COURSE
To equip the trainee with mathematics knowledge
and skills required of any sound scientist or
engineer, and lay a solid foundation for the
Engineering Mathematics course.
2
LEARNING OUTCOMES
A trainee who has completed the course will be able to:
Apply algebraic techniques with confidence, handle,
with dexterity, polynomial, trigonometric and
exponential functions.
Fit mathematical models to experimental data.
Take full advantage of vectors and matrices in
scientific analysis, and efficiently manage sequences,
series and complex numbers, confidently apply
calculus in the modelling and simulation of singlevariable continuous systems.
3
COURSE CONTENT
Algebraic Processes;
Trigonometry;
Model Fitting;
Series; Vectors;
Matrices;
Complex Numbers;
Differentiation;
Numerical Solution of Algebraic Equations;
Integration;
Applications of Integration Theory;
Numerical Integration;
First-Order Ordinary Differential Equations (ODEs);
Laplace Transforms;
Ordinary Differential Equations:
Beyond the First Order.
4
EXAMINATION PLAN
There will be two one-hour examinations to assess the
trainees’ knowledge and skill levels. The two will be equally
weighted, and will have the following coverage:
Examination 1
Algebraic Processes; Trigonometry; Model Fitting; Series;
Vectors; Matrices; Complex Numbers.
Examination 2
Differentiation; Numerical Solution of Algebraic Equations;
Integration; Applications of Integration Theory; Numerical
Integration; Laplace Transforms; Ordinary Differential
Equations.
5
ALGEBRAIC PROCESS - 1
COVERAGE
Basic arithmetic rules and operations; indices and logarithms;
fractional expressions; linear algebraic equations; quadratic
equations; simultaneous linear and quadratic equations;
polynomial functions; factorization; binomial expansion;
polynomial equations; inequations; rational polynomial
functions.
ARITHMETIC OPERATIONS
Addition, Subtraction, Multiplication, Division, Exponentiation
(Treated under “Indices”)
6
ALGEBRAIC PROCESS - 2
RULES OF PRECEDENCE
Parentheses (brackets)  exponentiation  Multiplication &
Division  Addition & Subtraction - BODMAS.
Notes
•If adjacent operators are of the same level, operate from left to
right.
•In case of nested parentheses, start from the innermost.
Example
{2 + [(5  6  81  3  22)2  7  3  19]3}2  4  25 = {2  [(30  27  4)2
 7  3  19]3}2  100 = {2  [72  7  3  19]3}2  100 = {2  [21  19]3}2
 100 = {2  23}2  100 = 102  100 = 100  100 = 0.
7
ALGEBRAIC PROCESS - 3
INDICES
ux  u raised to power x.
In ux, u = the base; x = the power or index or exponent.
Rules of Indices
u0 = 1; ux uy = uxy;
1
x

u
u ≠ 0;
ux
u
x
y
 
 u
x
1
y
= uxy; (ux)y = uxy.
1
x
u x u

 u


x
(uv)x =
ux
uy
1
y
x

 


x
u
u


u
x
x

;


uv; v
 
vx  v 
 
x
 u
x
y

y
ux
x
vx
v
   x
u
u
8
ALGEBRAIC PROCESS - 4
INDICES (continued)
Example 1
3
 4  
2
 2   x
 x  




3
 1 

3 
2
x


2
 3x x
2
4
x3 x7

x4




3 2 ( 3)
2 ( 3)
1( 2 ) 3( 2 )
( 2 4)
( 3 7  4 )
4
x
x

2
x

3
x

x
=
=

64 x 6 x 6  4 x 6  3x 6  x 6
64 x (66) 1  4  3  1  64


=
Example 2
8
2
 
3
 125
3
 3
4

2
3 23 
 
 5
3 
 12 
3
4
 13 
45

1
9
9
ALGEBRAIC PROCESS - 5
LOGARITHMS
loguy  the power to which u is raised to give
y.
Thus, loguy = x  y = ux
(y > 0)
 logarithm is inverse of index function
 log u
u  
x
x
u logu
and
x
 x
Rules of logarithms
logu1 = 0, log 1 to any base is 0, u ≠1
loguu = 1. The log of the base is always 1, u≠1
y
logu(yz) = loguy + loguz ; logu  z  = loguy  loguz
1
logu(yx) = xloguy ; logu  y  = loguy

logvy =

log u y
log u v
10
ALGEBRAIC PROCESS - 6
LOGARITHMS Contd :
Notation
log y  log10y, common log;
log10( y x 10n) = n + log10y; n is called the characteristic
and log10y is called the mantissa part. The mantissa part
is read from the table and is always less than 1.
n may be +ve or –ve but the mantissa part is always
+ve
When n is –ve, it is written as n .
ln y  logey (e = 2.718281828)
11
ALGEBRAIC PROCESS - 7
LOGARITHMS (continued)
Illustrative Problems
Problem 1
Given that 7x+1  7x1 = 127, determine x to a precision
of 3 significant figures.
Problem 2
Simplify:
log 15 9  log 15 11  log 15 99  e
2 n ( x )
 
 n e
x2
Problem 3
Find s and t if
1
log 16 x  log 8  3   log 1 32 x   s log 2 x  t
4
x 
8
12
ALGEBRAIC PROCESS - 8
FRACTIONAL EXPRESSIONS
Simplification of fractional algebraic expressions is
similar to that of fractional numbers. Thus,
 For addition and subtraction, multiply both numerator
and denominator of each algebraic fraction by the
LCM of the denominators, then simplify.
 For multiplication and division, cancel out factors
common to both numerator and denominator.
Illustrative Problems
Problem 1
Simplify:
3 v2 v3


v v 1 v 1
Problem 2
2
Simplify:
2u  1
u4

u  1 u  12
u 6 u 3

u 1 u 1
13
ALGEBRAIC PROCESS - 9
POLYNOMIAL FUNCTIONS
Polynomial of degree n
pn(x)  anxn  an1xn1  ...  a1x  a0 (an  0)
Operations between polynomials
Addition or subtration:
operation is between terms of same order
Multiplication of pn(x) by qm(x):
each term of qm(x) multiplies every term of pn(x).
Division of pn(x) by qm(x):
apply algebraic (long) division if n  m;
otherwise, division is not possible
Example
Given that p(x)  32x5  243, q(x)  16x4  24x3  36x2 
54x  81, r(x)  2x  3 and s(x)  2x2  3x  4, find
(a) q(x) r(x)  p(x) and (b) q(x)  s(x).
14
ALGEBRAIC PROCESS - 10
POLYNOMIAL FUNCTIONS (continued)
Solution
(a)
16x4  24x3  36x2  54x

2x
48x4  72x3  108x2 162x
32x5  48x4  72x3  108x2 162x
32x5
32x5
= 0
(b)
 81
 3
243
243
243
8x2
2
2x2  3x  4 16x4  24x3  36x2  54x  81
16x4  24x3  32x2
4x2  54x  81
4x2  6x  8
48x  73
Thus, quotient = 8x2  2 and remainder = 48x  73
15
ALGEBRAIC PROCESS - 11
POLYNOMIAL FUNCTIONS (continued)
The factor theorem
If pn(x = ) = 0, then (x  ) is a factor of pn(x).
The remainder theorem
If pn(x) is divided by (x  ), the remainder is pn().
Example
The function y(x) = ax4  bx3  x2  x  c has (x  1) and (x  2)
as two of its factors. If the remainder when y(x) is divided by
( x 1) is 4, find a, b and c.
Solution
y(1) = 0  a  b  1  1  c = 0 
a  b  c = 0 (1)
y(2) = 0 16a  8b  4  2  c = 0 16a  8b  c = 6 (2)
y(1) = 4  a  b  1  1  c = 4 
a  b  c = 2 (3)
(3)  (1)  2b = 2  b = 1
(2)  (1)  15a  9b = 6  a = 1
From (1): c = b  a = 2
16
ALGEBRAIC PROCESS - 12
FACTORIZATION
Some useful identities
u2  v2 = (u  v)(u  v)
(difference-of-two-squares identity)
(u  v)2 = u2  2uv  v2
(square-of-sum identity)
(u  v)2 = u2  2uv  v2
(square-of-difference identity)
un  vn = (u  v)(un1  un2v  un3v2  …  uvn2  vn1) for all n
un  1 = (u  1)(un1  un2  un3  …  u  1) for all n
un  vn = (u  v)(un1  un2v  un3v2  …  uvn2  vn1) for odd n
un  1 = (u  1)(un1  un2  un3  …  u  1) for odd n
Examples
(1)
(x2 + 16)2  64x2 = (x2 + 16)2  (8x)2
= (x2 + 8x + 16)(x2  8x + 16) = (x + 4)2(x  4)2
(2) y8  256 = (y4)2  162 = (y4  16)(y4 + 16)
= [(y2)2  42](y4 + 16) = (y2  4)(y2 + 4)(y4 + 16)
= (y  2)(y + 2)(y2 + 4)(y4 + 16)
(3) f6  1 = (f3  1)(f3 + 1) = (f  1)(f2 + f + 1)(f + 1)(f2  f + 1)
 f6  1 = (f  1)(f + 1)(f2  f + 1)(f2 + f + 1)
17
ALGEBRAIC PROCESS - 13
FACTORIZATION (Continued)
Further tips on factorization of polynomials:
(1) xm is common to all terms  xm is a factor
(2) If four terms, pairing could reveal a common
factor
(3) Where possible, use difference of two
squares identity: a2 - b2  (a + b)(a - b)
(4) For a quadratic function or factor with b2 = 4ac:
2
ax + bx + c =
a  x  b 
2a 

2
(5)Use f(a) = 0 to identify (x - a) (a = -3, -2, …, 3)
as a factor
(6)After identifying a factor or some factors, apply
algebraic division to obtain the complementary
factor
19
ALGEBRAIC PROCESS - 14
FACTORIZATION (Continued)
Further tips on factorization of polynomials:
(1) xm is common to all terms  xm is a factor
(2) If four terms, pairing could reveal a common
factor
(3) Where possible, use difference of two
squares identity: a2 - b2  (a + b)(a - b)
(4) For a quadratic function or factor with b2 = 4ac:
ax2 + bx + c = a  x  2ba 
2


(5) Use f(a) = 0 to identify (x - a) (a = -3, -2, …, 3)
as a factor
(6) After identifying a factor or some factors, apply
algebraic division to obtain the complementary
factor
19
ALGEBRAIC PROCESS - 15
LINEAR ALGEBRAIC EQUATIONS
 In an algebraic equation, two algebraic expressions
containing at least one unknown are connected by an
equality sign.
 In a linear algebraic equation, each additive term is
either constant or contains only one unknown raised to
a power of one.
Valid Operations for Solving Equations
(1) Add the same quantity to both sides.
(2) Subtract the same quantity from both sides.
(3) Multiply both sides (term by term) by the same
quantity.
(4) Divide both sides (term by term) by the same
quantity.
(5) Take the same power of both sides (entire side)
(6) Apply the same function (log, exponential, etc.) to
both sides (entire side).
20
ALGEBRAIC PROCESS - 16
LINEAR ALGEBRAIC EQUATIONS (continued)
The Single-Variable Linear Equation
Illustrative Problem
2
5
z  4  3z  1  2 z  6
Solve for z: 3
7
Systems of Two Linear Equations
Considered: 3x  7y  1
(1)
x  2y  9
(2)
Methods Employed
(1) Elimination Method
(2) Substitution Method
(3) Graphical Method
Other Methods (for future treatment)
(1) Determinant method (Cramer’s rule)
(2) Inverse-matrix method
(3) Gaussian-reduction method
21
ALGEBRAIC PROCESS - 17
LINEAR ALGEBRAIC EQUATIONS
(continued)
Systems of Three Linear Equations
Considered:
4u  v  3w  13
(1)
3u  v  2w  19
(2)
u  v  4w  7
(3)
Method Employed
(1) Apply elimination method to reduce to two
linear equations.
(2) Solve the resultant two equations using any
valid method.
22
ALGEBRAIC PROCESS – 18
QUADRATIC EQUATIONS
In a single-variable quadratic equation, the only powers of
the variable contained in the constituent terms are zero (a
constant term), one and two.
General Form
ax2  bx  c  0
Solution Methods
(1) Factorization
 Write ax2  bx  c as ax2  px  qx  c such that
p  q  b and pq  ac
 Pair the terms to give (ax2  px)  (qx  c) or (ax2 
qx)  (px  c)
 Factorize by first factorizing each pair
(2) Method of completing the squares
b
c
2
x

x

0
ax  bx  c  0 
a
a
2
2
2
 2 b
 b   c  b 
  x  a x   2a    a   2a   0


23
ALGEBRAIC PROCESS - 19
QUADRATIC EQUATIONS (continued)
Solution Methods (continued)
b   b  c b 2  4ac
D

x






  
2a   2a  a
4a 2
4a 2

2
x
2
b  D

x
2a
2a
2  4ac
where
D
=
b
b D
2a
(3) Use of the quadratic formula
The result of the method of completing the squares
is invoked directly to give
b
D
x =
2a
Number of Real Roots
Define discriminant D  b2  4ac. If
D  0  2 real roots, a ˃ 0
D  0  1 real root (repeating root)
D  0  no real root, a = 0
24
ALGEBRAIC PROCESS - 20
POLYNOMIAL FUNCTIONS (continued)
The quadratic polynomial function
y(x)  ax2  bx  c (a  0)
2
b 
D

a
x



 y(x)  
2a 
4a
(D = b2 4ac)
Properties
(1) The quadratic function has one turning value: a
maximum for a < 0 and a minimum for a > 0.
(2) The function is symmetrical about the line x = b/(2a).
(3) The turning value of y is D/(4a) and occurs at the
line of symmetry.
(4) The curve of y(x) cuts the x-axis at 2 points if D > 0,
touches the x-axis at x = b/(2a) if D = 0, and makes
no contact with the x-axis if D < 0.
25
ALGEBRAIC PROCESS – 21
QUADRATIC EQUATIONS (continued)
Illustrative Problem
6x2  x  2  0
Solution by factorization
6x2  x  2 = 6x2  4x  3x  2 = (6x2  4x)  (3x  2)
= 2x(3x  2)  1(3x  2) = (2x  1)(3x  2) = 0
 2x  1 = 0 or 3x  2 = 0  x =

1
2
or
2
3
26
ALGEBRAIC PROCESS - 22
QUADRATIC EQUATIONS (continued)
Solution by completing the squares
1
1
2
x

x

0
6x  x  2  0 
6
3
2
1  1 1
 2 1
1
49

 x  x 



0
x


  12  144
6
144  3 144

2
1
7
1
7
1
2
 x  12   12  x  12  12   2 or 3
Solution by use of formula
D = (1)2  4(6)(2) = 49 (> 0  2 real roots)
x
 (1)  49 1  7
1
2

  or
2(6)
12
2
3
27
ALGEBRAIC PROCESS - 23
QUADRATIC EQUATIONS (continued)
Maximum and Minimum Values of a quadratic Function
The turning points (max and/or min points/values) of a quadratic function can be
determined by using ‘completing the square’ method. The calculus method will
be discussed later
Examples:
1. What is the minimum value of 3x2 – 2x + 1 and for what value of x does it
occur?
1
 2 2
y = 3x2 – 2x + 1 = 3 x  3 x  3 
=
=
2
2 2 2 2 1

3
(
x

x

(
) ( )  

3
6
6
3

1 2
1
1

 
3( x  ) 
3
9
3

28
ALGEBRAIC PROCESS - 24
QUADRATIC EQUATIONS (continued)
Maximum and Minimum Values of a quadratic Function
1 2
2

3
(
x

)

y = 


3
9
1
We see that the least value of (x -3 )
is 0 when x =
2
 The minimum or least value of y =3
1
.
3
when x = 1 .
3
2. Find the maximum value of y = 5 – x – 2x2 .
Y = -2x2 - x + 5 = -2(x2 + ½ x - 5 )
2
= -2{x2 + ½x +(¼)² – (¼)² -
5
2
}
41
= -2 {(x² +¼)² - 16
}
41
= - (x + ¼)² + 8
Hence when x = - ¼ , y has a max value of
41
8
29
ALGEBRAIC PROCESS - 25
QUADRATIC EQUATIONS (continued)
Sum and Product of the roots of quadratic equations
The sum and product of the roots can be found directly from the
equation.
If α and β are the roots of the eqn. ax2 + bx + c = 0 , then
α=
b  D
2a

and
and β =
b  D
2a
, where D = b2 – 4ac .
b
 2b
(b  D )  (b  D )
+β =
=
=
a
2a
2a
β =
b  D b  D
(
)(
)
2a
2a
b 2  b 2  4a 2c
=
4a 2
c
=
a
30
ALGEBRAIC PROCESS - 26
QUADRATIC EQUATIONS (continued)
Sum and Product of the roots of quadratic equations
So any quadratic eqn. with roots α and β can be
written as
x2 – (α + β)x + αβ = 0
ie
x2 – (sum of the roots)x + (product of the roots) = 0
31
ALGEBRAIC PROCESS - 27
SIMULTANEOUS LINEAR AND QUADRATIC
EQUATIONS
Example
x  3y  4  0
(1)
2x2  y2  6xy  8x  4y  3  0
(2)
Solution Procedure
(1) Solve Eq. (1) for x in terms of y.
(2) Substitute from Step 1 into Eq. (2) to eliminate x.
(3) Solve the quadratic in y from Step 2 for y.
(4) Compute x corresponding to y from Step 3.
Solution
From (1): x = 3y  4
(3)
(3) in (2): 2(3y  4)2  y2 –6y(3y  4)  8(3y  4)
 4y  3 = 0  y2  4y  3 = 0
 (y  1)(y  3) = 0  y = 1 or 3
From (3): x(y = 1) = 1 and x(y = 3) = 5
Thus,
(x, y) = (1, 1) or (5, 3).
32
ALGEBRAIC PROCESS - 27
INEQULITIES
x˃y, x greater than y; x – y is +ve
x≥y, x greater than or equal to y; x – y is +ve or zero
x˂y, x less than y; x – y is –ve
x≤y, x less than or equal to y; x – y is –ve or zero
Rules:
Add or subtract the same number from both sides, preserve the
inequality.
Multiply or divide both sides by the same +ve number preserve the
inequality.
Multiply or divide both sides by the same -ve number preserve the
inequality.
Never Find the values of x for which
33
ALGEBRAIC PROCESS - 28
INEQUALITIES (Continued)
Examples
1. Find the values of x for which
˃
is satisfied.
˃ 0
-
-
˃
0
˃ 0
The denominator (5x-1)(2x+1) ˃ 0
Hence 5x – 1 ˃ 0; x ˃
and
2x + 1 ˃ 0; x ˂
34
ALGEBRAIC PROCESS - 29
INQUALITIES (Continued)
2. Find the range of values of x for which
2x2 + 9x – 35 ˂ 0
ie (2x – 5)(x + 7) ˂ 0
Either 2x – 5 ˃ 0  x ˃
and x + 7 ˃ 0  x ˃ - 7
Or 2x - 5 ˂ 0  x ˂
and
x+7˃0 x˃-7
Consider the two sets of solutions
a) x ˃ and x ˂ -7
For any of these values, (2x - 5)(x + 7) ˃ 0
The given condition is not satisfied. Hence this solution is not correct.
b) x ˂
and x ˃ - 7
For any of these values, (2x - 5)(x + 7) ˂ 0
Hence this is the correct solution.
Thus the range is
-7 ˂ x ˂
35
ALGEBRAIC PROCESS - 30
BINOMIAL EXPANSION
For all values of x and a provided n is a +ve integer
= xn + nC1xn-1a + nC2xn-2a2 +……….+ n Crxn-rar + …………….+ nCn xn-nan (xn-n=x0=1)
Note that
1.there are (n+1) terms
2.the coeffs. are symmetrical; the 1st coeff is 1 (nC0) = the last coeff (nCn)
3. nCr =
; n! = n(n-1)(n-2)………1
4. 0! = 1
5. The expansion is homogeneous in x and a; ie the sum of the powers of x
and a in each term is equal to the power of the Binomial n
6 . The powers of x are in descending order while those of a are in ascending
order
ALGEBRAIC PROCESS - 31
BINOMIAL EXPANSION (Continued)
7. In this arrangement, a pattern is formed as follows:
Binomia l
No. of Terms
Coefficients
(x + a)1
2
1 1
(x + a)2
3
1 2 1
(x + a)3
4
1 3 3 1
(x + a)4
5
1 4 6 4 1
(x + a)5
6
1 5 10 10 5 1
(x + a)6
7
1 6 15 20 15 6 1
.
.
.
.
8. The second and the second last terms equal the degree n of
the Binomial.
ALGEBRAIC PROCESS - 32
BINOMIAL EXPANSION (Continued)
9. Each line is generated from the preceding line as follows:
(x + a)1 
(x + a)2 
(x + a)3 
(x + a)4 
11
121
1331
1 464 1
(x + a)5 
1 510105 1
This pattern of coeffs is called Pascal’s triangle. For odd
n, the triangle is flat at the
top because the two middle coeffs are equal.
The general Binomial theorem is given as
(1 + x)n = 1 + nx +
+
+……………
Where n is a rational number (+ve or –ve) and x is numerically less than 1.
ALGEBRAIC PROCESS - 33
BINOMIAL EXPANSION (Continued)
Note that
1.If n is not a +ve integer, the series will be infinite in ascending power of x as none
of the nos (n-1), (n-2), (n-3), ………… in the coeffs. will never be zero.
2. The coeffs can no longer be written in the form nCr.
3. The expansion is only valid if the numerical value of x is less than 1; ie -1 ˂ x ˂ 1.
4. The Binomial must be written in the form (1 + x)n .
Examples
1. Expand (a – b)6  [a + (-b)]6
Using the Pascal’s triangle, the 7 coeffs are 1 6 15 20 15 6 1
 (a – b)6 = a6 + 6a5(-b)1 + 15a4(-b)2 + 20a3(-b)3 + 5a2(-b)4 + 6a1(-b)5 + (-b)6
= a6 - 6a5b + 15a4b2 – 20a3b3 + 15a2b4 – 6ab5 + b6
ALGEBRAIC PROCESS - 34
BINOMIAL EXPANSION (Continued)
2. Find the exact value of (1.02)5 using Binomial theorem.
(1.02)5 = (1 + 0.02)5
= 1 + 5(0.02) +
+
+
+
= 1 + 5(0.02) + 10(0.0004) + 10(0.000008) + (0.00000016+(0.0000000032)
= 1.1040808032
ALGEBRAIC PROCESS - 35
BINOMIAL EXPANSION (Continued)
3. Find the value of (0.98)10 correct to five places of decimals.
(0.98)10 = (1 – 0.02)10  x = -0.02 and n =10
 (0.98)10 = 1 -10(2 x 10-2) + 45(2 x 10-2)2 – 120(2 x 10-2)3 + 210(2 x 10-2)4
= 1 – 0.2 + 180x10-4 – 960x10-6 + 3360x10-8
= 1 – 0.2 + 0.018 – 0.00096 + 0.0000336
= 0.81707 (five places of decimals)
ALGEBRAIC PROCESS - 36
BINOMIAL EXPANSION (Continued)
4
1. Give as far as the x term, the expansion of
1
(1+𝑥)3
1
(1+𝑥)3
= (1 + 𝑥)−3 ; n = -3
 (1 + 𝑥)
−3
=1+
+
(−3)
1!
𝑥+
−3 −4 (−5)
3!
−3 (−4)
2!
3
𝑥 +
𝑥2
−3 −4 −5 (−6)
4!
𝑥4
= 1 – 3x + 6x2 – 10x3 + 15x4
Recall that this expansion is valid only for -1˂ x ˂1