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Mathematics

NNPC FSTP – ENGINEERS Phase 1. Mathematics. Nexus Alliance Ltd. 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. LEARNING OUTCOMES.

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Mathematics

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  1. NNPC FSTP – ENGINEERS Phase 1 Mathematics Nexus Alliance Ltd

  2. 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.

  3. 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 single-variable continuous systems.

  4. 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.

  5. 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.

  6. 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”)

  7. 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.

  8. 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; = uxy; (ux)y = uxy. u ≠ 0; (uv)x = uxvx;

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  24. ALGEBRAIC PROCESS - 19 QUADRATIC EQUATIONS (continued) Solution Methods (continued) whereD = b2 4ac (3) Use of the quadratic formula The result of the method of completing the squares is invoked directly to give x = 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

  25. ALGEBRAIC PROCESS - 20

  26. 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 = or

  27. ALGEBRAIC PROCESS - 22

  28. 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: • What is the minimum value of 3x2– 2x + 1 and for what value of x does it occur? • y = 3x2– 2x + 1 = • = • =

  29. ALGEBRAIC PROCESS - 24 QUADRATIC EQUATIONS (continued) Maximum and Minimum Values of a quadratic Function y = We see that the least value of (x - ) is 0 when x = .  The minimum or least value of y = when x = . 2. Find the maximum value of y = 5 – x – 2x2 . Y = -2x2 - x + 5 = -2(x2 + ½ x - ) = -2{x2 + ½x +(¼)² – (¼)² - } = -2 {(x² +¼)² - } = - (x + ¼)² + Hence when x = - ¼ , y has a max value of

  30. 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 = and β = , where D = b2 – 4ac .  + β = = = and β = = =

  31. 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

  32. ALGEBRAIC PROCESS - 27

  33. 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

  34. 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 ˂

  35. 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 ˂

  36. 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 • there are (n+1) terms • 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

  37. 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.

  38. ALGEBRAIC PROCESS - 32 • BINOMIAL EXPANSION (Continued) • 9. Each line is generated from the preceding line as follows: • 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.

  39. ALGEBRAIC PROCESS - 33 • BINOMIAL EXPANSION (Continued) • Note that • 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

  40. 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

  41. 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)

  42. ALGEBRAIC PROCESS - 36 • BINOMIAL EXPANSION (Continued)

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