* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Download nnpc – fstp- maths_eng 1
Linear algebra wikipedia , lookup
Eisenstein's criterion wikipedia , lookup
Factorization of polynomials over finite fields wikipedia , lookup
Cubic function wikipedia , lookup
Field (mathematics) wikipedia , lookup
Signal-flow graph wikipedia , lookup
Motive (algebraic geometry) wikipedia , lookup
Quartic function wikipedia , lookup
Quadratic form wikipedia , lookup
Elementary algebra wikipedia , lookup
Homomorphism wikipedia , lookup
Fundamental theorem of algebra wikipedia , lookup
Quadratic equation wikipedia , lookup
System of linear equations wikipedia , lookup
System of polynomial equations wikipedia , lookup
Factorization wikipedia , lookup
History of algebra wikipedia , lookup
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 = uxy; 1 x u u ≠ 0; ux u x y u x 1 y = uxy; (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 (66) 1 4 3 1 64 = Example 2 8 2 3 125 3 3 4 2 3 23 5 3 12 3 4 13 45 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 7x1 = 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 v2 v3 v v 1 v 1 Problem 2 2 Simplify: 2u 1 u4 u 1 u 12 u 6 u 3 u 1 u 1 13 ALGEBRAIC PROCESS - 9 POLYNOMIAL FUNCTIONS Polynomial of degree n pn(x) anxn an1xn1 ... 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)(un1 un2v un3v2 … uvn2 vn1) for all n un 1 = (u 1)(un1 un2 un3 … u 1) for all n un vn = (u v)(un1 un2v un3v2 … uvn2 vn1) for odd n un 1 = (u 1)(un1 un2 un3 … 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