Ain Shams University General Mathematics (1) 2022-2023 PDF

Ain Shams University College for Women Art, Sciences and Education Mathematics department General mathematics (1) First level Biochemistry; Chem_Zoo and Chem_Bot Dr. Hanaa M. S. Abdelgawad...

Ain Shams University College for Women Art, Sciences and Education Mathematics department General mathematics (1) First level Biochemistry; Chem_Zoo and Chem_Bot Dr. Hanaa M. S. Abdelgawad 2022-2023  CONTENTS : CHAPTER ONE: COMPLEX NUMBERS 1.1. The general form of a complex number 1.2.Arithmetic Operations 1.2.1 Basic Algebraic Properties 1.3. Complex Plane 1.4. Polar Form of Complex Numbers 1.4. 1. Products and Quotients in Polar Form 1.5. Integer Power of z and Roots Exercise 1 :CHAPTER TWO: DETERMINANTS AND MATRIES Part I : DETERMINANTS 2.1 Definition of matrix 2.2 Determinant of matrix 2.2.1 Determinants Of Second Order 2.2.2 Determinants OF Third Order 2.2.3 Triangular matrices 2.3 Conditions That Yield a Zero Determinant Exercise2.1 Part II: MATRIES 2.4 Operations With Matrices 2.4.1 Equality of matrices 2.4.2 Matrix Addition 2.4.3 Scalar Multiplication 2.4.4 Matrix Multiplication 2.5 Elementary Row Operations 2.6. Inverse OF A Matrix 2.7 The System OF m Linear Equations in n unknowns 2.8 Solving A System OF Linear Equations Exercise2 2  CHAPTER THREE :: MATHEMATICAL INDUCTION 3.1. Principle Of Mathematical Induction 3.2. Proof By Mathematical Induction Solved examples Exercise 3 CHAPTER FOURTH: PARTIAL FRACTIONS 4.1 Introduction 4.2 Polynomial 4.3 Rational fraction 4.3 Process of Finding Partial Fraction 4.3.1 Type I Solved examples Exercise 4.1 4.3.2 Type II Solved examples Exercise 4.2 4.3.3 Type III Solved examples Exercise 4.1 CHAPTER FIFTH : DERIVATIVES OF POLYNOMIALS AND EXPONENTIAL FUNTION 5.1 Basic Differentiation Rules 5.2 Derivatives of Natural Exponential Function 5.3 The Product and Quotient Rules 5.3.1 The Product Rule 5.3.2 The Quotient Rule Exercises 5.1 5.4 Derivatives of Trigonometric Functions Exercises 5.2 CHAPTER SIXTH : INDEINITE INDEFINTE INTEGER 6.1 Table of Indeinite Integrals 3  Solved examples Exercise 6.1 6.2 Techniques of Integration 6.2.1 Integration by Parts Exercise 6.2 6.2.2 Integration of Rational Functions by Partial Fraction 6.2.2.1 Integration Proper Rational Function 6.2.2.2 Integration imProper Rational Function Exercises 6.3 4  CHAPTER ONE COMPLEX NUMBERS In this chapter, we survey the algebraic and geometric structure of the complex number system. We assume various corresponding properties of real numbers to be known 1.1 The general form of a complex number Definition 1.1 A complex number can be represented by an expression of the form z = a + ib where a and b are real numbers and i is the imaginary unit with the property that 𝑖 2 = −1. The real number a in z = a+ ib is called the real part of z; the real number b is called the imaginary part of z.The real and imaginary parts of a complex number z are abbreviated Re(z) and Im(z), respectively For example, if z = 4 − 9i, then Re(z) = 4 and Im(z) = −9 Definition 1.2 Equality Complex numbers 𝑧1 = 𝑎1 + i𝑏1 and 𝑧2 = 𝑎2 + i𝑏2 are equal, 𝑧1 = 𝑧2 , if 𝑎1 =𝑎2 and 𝑏1 =𝑏2. In terms of the symbols Re(z) and Im(z), Definition 1.2 states that , 𝑧1 = 𝑧2 if Re(𝑧1 ) = Re(𝑧2 ) and Im(𝑧1 ) = Im(𝑧2 ). The totality of complex numbers or the set of complex numbers is usually denoted by the symbol C 1.2.Arithmetic Operations etic Operations Complex numbers can be added, subtracted, multiplied, and divided. If 𝑧1 = 𝑎1 + i𝑏1 and 𝑧2 = 𝑎2 + i𝑏2 , these operations are defined as follows. Addition : 𝑧1 + 𝑧 2 = (𝑎1 + i𝑏1 ) + (𝑎2 + i𝑏2 ) = (𝑎1 + 𝑎2 ) + i(𝑏1 + 𝑏2 ) Subtraction : 𝑧1 - 𝑧 2 = (𝑎1 + i𝑏1 ) - (𝑎2 + i𝑏2 ) = (𝑎1 - 𝑎2 ) + i(𝑏1 - 𝑏2 ) Multiplication: 𝑧1 ∙ 𝑧 2 = (𝑎1 + i𝑏1 ) (𝑎2 + i𝑏2 ) = 𝑎1 𝑎2 - 𝑏1 𝑏2 + i(𝑏1 𝑎2 + 𝑎1 𝑏2 ) 𝑧1 𝑎 +𝑏 Division: = 1 1 𝑎2 ≠ 0, or 𝑏2 ≠ 0, 𝑧2 𝑎2 + 𝑖𝑏2 𝑎1 𝑎2 + 𝑏1 𝑏2 𝑏1 𝑎2 − 𝑎1 𝑏2 = 2 + 𝑏2 +𝑖 𝑎2 2 𝑎2 2 + 𝑏2 2 Addition, Subtraction, and Multiplication (i ) To add (subtract ) two complex numbers, simply add (subtract ) the corresponding real and imaginary parts. (ii) To multiply two complex numbers, use the distributive law and the fact that 𝑖 2 = −1. 5 1.2.1 Basic Algebraic Properties Various properties of addition and multiplication of complex numbers are the same as for real numbers. The familiar commutative, associative, and distributive laws hold for complex numbers Commutative laws: Associative laws: Distributive law: Example 1 If 𝑧1 = 2 + 4i and 𝑧2 = −3 + 8i, find (a) 𝑧1 + 𝑧2 and (b) 𝑧1 𝑧2 Solution (a) By adding real and imaginary parts, the sum of the two complex numbers 𝑧1 and 𝑧2 is 𝑧1 + 𝑧2 = (2+4i) + (−3 + 8i) = (2 − 3) + (4 + 8)i = −1 + 12i.. (b) By the distributive law and 𝑖 2 = −1, the product of 𝑧1 and 𝑧2 is 𝑧1 𝑧2 = (2 + 4i) (−3 + 8i) = (2 + 4i) (−3) + (2 + 4i) (8i) = −6 − 12i + 16i + 32𝑖 2 = (−6 − 32) + (16 − 12)i = −38 + 4i Zero and Unity The zero in the complex number system is the number 0 + 0i and the unity is 1 + 0i.The zero and unity are denoted by 0 and 1, respectively z + 0 = (a + ib) + (0 + 0i) = a+0+i(b + 0) = a + ib = z. (additive identity) z ・1 = z ・ (1 + 0i) = z. (multiplicative identity) Conjugate If z is a complex number, the number obtained by changing the sign of its imaginary part is called the complex conjugate, or simply conjugate, of z and is denoted by the symbol 𝑧. ̅ = a − ib. ̅ In other words, if z = a + ib, then its conjugate is 𝑧. Example 2 If z = 6 + 3i, then 𝑧. ̅ = 6− 3i; If z = −5 − i, then 𝑧. ̅ = −5 + i. If z is a real number, say, z = 7, then 𝑧. ̅= 7. We have the following additional properties: 1- the conjugate of the sum is the sum of the conjugates: 2- The conjugate of any finite product of complex numbers is the product of the conjugates. 6  3- The definitions of addition and multiplication show that the sum and product of a complex number z with its conjugate 𝑧. ̅ is a real number: 4- The difference of a complex number z with its conjugate 𝑧. ̅is a pure imaginary number Since a = Re(z) and b = Im(z), (3) and (5) yield two useful formulas: Division : To divide z1 by z2 , multiply the numerator and denominator of z1 / z2 by the conjugate of z2. That is, Example 3: If 𝑧1 = 2− 3i and 𝑧2 = 4+6i, find 𝑧1 / 𝑧2 Solution We multiply numerator and denominator by the conjugate 𝑧. ̅ 2 = 4− 6i of the denominator 𝑧2 = 4 + 6i and then use (4): −1+3𝑖 Example 4: Express the number 2−𝑖 in the form a + bi Inverses In the complex number system, every number z has a unique additive inverse. As in the real number system, the additive inverse of z = a + ib is its negative, −z, where −z = −a − ib. For any complex number z, we have z + (−z) = 0.Similarly , every nonzero complex number z has a multiplicative inverse.In symbols, for z ≠ 0 there exists one and only one nonzero complex number 𝑧 −1 such that z 𝑧 −1 = 1.The multiplicative inverse 𝑧 −1 is the same as the reciprocal 1/z Example 5 Find the reciprocal of z = 2− 3i. Solution By the definition of division we obtain 7 You should take a few seconds to verify the multiplication Remark Usual derivation and for the roots of the quadratic equation 𝑎𝑥 2 + 𝑏𝑥 + 𝑐 = 0 are valid even when 𝑏 2 − 4𝑎𝑐 < 0 : Example 6 Find the root of the equation 𝑥 2 + 𝑥 + 1 = 0 Solution Using the quadratic formula ,we have 1.3 Complex Plane A complex number z = x + iy is uniquely determined by an ordered pair of real numbers (x, y).The first and second entries of the ordered pairs correspond, in turn, with the real and imaginary parts of the complex number. For example, the ordered pair (2, −3) corresponds to the complex number z = 2 – 3i.Conversely, z = 2 – 3i determines the ordered pair (2, −3).The numbers 7, i, and −5i are equivalent to (7, 0), (0, 1), (0,−5), respectively. In this manner we are able to associate a complex number z = x + iy with a point (x, y) in a coordinate plane The horizontal or x-axis is called the real axis because each point on that axis represents a real number. The vertical or y-axis is called the imaginary axis because a point on that axis represents a pure imaginary number. It is customary to denote a complex number (x, y) by z, so that z = (x, y). (1) Re z = x, Im z = Y. (2) 8 Vectors A complex number z = x + iy can be viewed as a two dimensional position vector, that is, a vector whose initial point is the origin and whose terminal point is the point (x, y).This vector interpretation prompts us to define the length of the vector z as the distance from the origin to the point (x, y). Modulus Definition 1.3 The modulus of a complex number z = x + iy, is the real number |𝑧| = √𝑥 2 + 𝑦 2 (1) The number Iz1 is the distance between the point (x, y) and the origin, or the length of the vector representing z. It also follows from definition (1) that the real numbers I z 1, Re z = x, and Im z = y are related by the equation Example 7 : Find the modulus of the number z a) Z = 2 – 3i b) Z = -9i Solution a) b) Example 8 Evaluate and Solution Remark The modulus of a complex number z has the additional properties. Note that when 𝑧1 = 𝑧2 = z, then Distance Again The addition of complex numbers 𝑧1 = 𝑥1 + i𝑦1 and 𝑧2 = 𝑥2 + i𝑦2 in terms of ordered pairs: 9  The difference 𝑧2 –𝑧1 can be drawn either starting from the terminal point of 𝑧1 and ending at the terminal point of 𝑧2 that the distance between two points 𝑧1 = 𝑥1 + i𝑦1 and 𝑧2 = 𝑥2 + i𝑦2 in the complex plane is the same as the distance between the origin and the point (𝑥2 – 𝑥1 , 𝑦2 – 𝑦1 ); that is |z| = |𝑧2 − 𝑧1 | = |(𝑥2 – 𝑥1 ) + i(𝑦2 – 𝑦1 )| or When𝑧1 = 0, we see again that the modulus |𝑧2 | represents the distance between the origin and the point 𝑧2. Example 9 Describe the set of points z in the complex plane that satisfy |z| = |z – i|. Solution write |z| = |z – i| as 1 𝑦= Is an equation of the horizontal line 2 1 Complex numbers satisfying |z| = |z – i| can then be written as z = x + 𝑖 2 Inequalities We turn now to the triangle inequality, which provides an upper bound for the modulus of the sum of two complex numbers 𝑧1 and 𝑧2 : (I) This important inequality is geometrically since Triangle with vector sides "It is statement that the length of one side of a triangle is less than or equal to the sum of the lengths of the other two sides" An immediate consequence of the triangle inequality is the fact that 01 Proof : Then (1) (2) From (1), (2) then (II) i.e the length of one side of a triangle is greater than or equal to the difference of the lengths of the other two sides. Because , one can replace 𝑧2 by – 𝑧2 that |𝑧1 + (−𝑧2 )| ≤ |𝑧1 | + |(−𝑧2 )| = |𝑧1 | + |𝑧2 | To summarize these results in a particularly useful form Example 10 Find an upper bound for Solution Since |−1| = 𝟏 Example 11 Find the equation of circle whose center is 𝑧0 = (1, -3) and whose radius is R = 2. Solution the circle with center 𝑧0 and radius R thus satisfy the equation 00  |𝑧 − 𝑧0 | = 𝑅 |𝑧 − (1, −3)| = 2 → |𝑧 − (1 − 3𝑖)| = 2 → |𝑧 − 1 + 3𝑖| = 2 1,4. Polar Form in complex number Let r and 𝜃 be polar coordinates of the point (x, y) that corresponds to a nonzero complex number z = x + iy. Since x = r cos 𝜃 and y = r sin 𝜃, the number z can be written in polar form as z = r(cos 𝜃 + i sin 𝜃) (1). 𝑦 Where 𝑟 = |𝑧| = √𝑥 2 + 𝑦 2 and tan 𝜃 = 𝑥 The angle 𝜃 is called the argument of z and we write 𝜃 = arg(𝑧) Not that arg(z) is not unique ; any two argument f z differ by an inteer multiple of 2 𝜋 Example 12 02 1.4.1 Products and Quotients in polar Form Example 13 Solution From example 12 03 1.5. Integer Power of z and Roots Another important result that can be obtained formally by applying rules for real numbers to z = 𝑟𝑒 𝑖𝜃 is Example 14 Solution 12 04 Example 15 using de Moivre's formula. Remark Example 16 Roots of complex numbers An n the root of the complex number z is a complex number w such that 𝑤0 = 𝑧 05 Example 17: Find the three cube roots of z = i. Solution Keep in mind that we are basically solving the equation 𝑤 3 = i. Now with r = 1, θ = arg(i) = π/2, a polar form of the given number is given by z = cos(π/2) + i sin(π/2). From (4), with n = 3, we then obtain Hence the three roots are , 06 EXAMPLE 18 Find the four fourth roots of z = 1+i. Solution In this case we obtain With the aid of a calculator we find The four roots lie on a circle centered at the origin of radius and are spaced at equal angular intervals of 2π/4 = π/2 radians, beginning with the root whose argument is π/16. EXAMPLE 19 07 EXERCISES 1 08 13. Find the solution of the following equations 09  Chapter two Determinants And Matrices Part 1: Determinants 2.1. Definition of a Matrix 2.2. Determinant of matrix Every square matrix can be associated with a real number called its determinant. Determinants have many uses, several of which will be explored in this chapter. The first two sections of this chapter concentrate on procedures for evaluating the determinant of a matrix. 2.2.1 Determinants Of Second Order The symbol consisting of the four numbers 𝒂𝟏𝟏 , 𝒂𝟐𝟏 , 𝒂𝟏𝟐 , 𝒂𝟐𝟐 arranged in two rows and two columns, is called a determinant of second order or determinant of order two. The four numbers are called elements of the determinant. By definition, Thus 2 3  2(2)  3(1)  1 1 2 2.2.2. Determinants OF Third Order 𝒂𝟏𝟏 𝒂𝟏𝟐 𝒂𝟏𝟑 The symbol |𝒂𝟐𝟏 𝒂𝟐𝟐 𝒂𝟐𝟑 | 𝒂𝟑𝟏 𝒂𝟑𝟐 𝒂𝟑𝟑 consisting of nine numbers arranged in three rows and three columns is called a determinant of third order. Definitions of Minors and Cofactors of a Matrix 21 Example 1 Find all the minors and cofactors of Solution To find the minor 𝑴𝟏𝟏 delete the first row and first column of A and evaluate the determinant of the resulting matrix. 20 Example 2 Find the determinant of a) Solution b) c) There is an alternative method commonly used for evaluating the determinant of a 3× 3 matrix A To apply this method, copy the first and second columns of A to form fourth and fifth columns. The determinant of A is then obtained by adding (or subtracting) the products of the six diagonals, as shown in the following diagram Example 3 Find the determinant of Solution 22 2.2.3 Triangular Matrices If A is a triangular matrix of order n then its determinant is the product of the entries on the main diagonal. That is, To find the determinant of a triangular matrix, simply form the product of the entries on the main diagonal. Example 4 Find the determinate of The determinant of this upper triangular matrix is given by │A│ = 2 (-1) (3) = -6 2.3 Conditions That Yield a Zero Determinant If A s a square matrix and any one of the following conditions is true, then det (A) = 0 1. An entire row (or an entire column) consists of zeros. 2. Two rows (or columns) are equal. 3. One row (or column) is a multiple of another row (or column). Example 5 23  Exercise 2.1 1- Evaluate the following determinants. 2- Find the values of x for which 3- Evaluate each of the following determinants. 4. Show that 24  Part II Matrices 2.4 Operations with Matrices The next section use matrices to solve systems of linear equations. Matrices, however, can be used to do much more than that. There is a rich mathematical theory of matrices, and its applications are numerous. This section introduce some fundamentals of matrix theory. 2.4.1. Equality of Matrices Example 1: If Then y = 9 and x = 9 2.4.2 Matrix Addition Example 2 25 2.4.3 Scalar Multiplication Example 3; 2.4.4 Matrix Multiplication 26  This definition means that the entry in the th row and the th column of the product is obtained by multiplying the entries in the th row of by the corresponding entries in the th column of and then adding the results. The next example illustrates this process. Example 4 : Find the product where Solution First note that the product AB is defined because A has size 3 × 2 and B has size 2 × 2Moreover, the product AB has size 3 × 2 and will take the form Continuing this pattern produces the results shown below. 27 Example 5 2.5 Elementary Row Operations Two matrices are said to be row equivalent if one can be obtained from the other by a sequence of elementary row operations. Elementary Row Operations 1-Interchange two rows. 2-Multiplying a row by a nonzero constant. 3-Add a multiple of a row to another row. A matrix in row-echelon form has the following properties. 1. All rows consisting entirely of zeros occur at the bottom of the matrix. 2. For each row that does not consist entirely of zeros, the first nonzero entry is 1 (called a leading 1). 3. For two successive (nonzero) rows, the leading 1 in the higher row is farther to the left than the leading 1 in the lower row R E M A R K : A matrix in row-echelon form is in reduced row-echelon form if every column that has a leading 1 has zeros in every position above and below its leading 1. Example 6 The matrices below are in row-echelon form 28  The matrices shown in parts (b) and (d) are in reduced row-echelon form The matrices listed below are not in row-echelon form Example 7. Use elementary row operations to put the matrix A in reduced row-echelon form when ∼ is the symbol used between two matrices to indicate that the two matrices are row equivalent. R2 in front of a matrix means that the row following it was row 2 in the previous matrix. R3 − 3R1 in front of a matrix means that the row following it was obtained from the previous matrix by subtracting 3 times row 1 from row 3. The reduced row-echelon form of matrix 2.6 The Inverse of a Matrix A square matrix A has an inverse if there is a matrix A–1 such that AA–1 = A–1A = I To find the inverse, if it exists, of a square matrix A I. Inverse of 2 × 2 matrix If the matrix is invertible , then 29 Note Example 8 : The matrix has inverse ? Solution The matrix has no inverse because (4)(1) –(2)(2) = 0 Example 13 Find the inverse of the matrix Solution II. The inverse matrix using the adjoint If A is an invertible matrix (|𝑨 | ≠ 𝟎) then With where 𝑪𝒊𝒋 is the cofactor of the element 𝒄𝒊𝒋 For 3×3 matrix we have Example 9 Find the inverse of the matrix Solution 31  III. Finding the Inverse of a Matrix by Gauss-Jordan Elimination Example 10 : Solution Example 11: Find the inverse of the matrix Solution 30 Example 12: Show that the matrix has no inverse. Solution Because the “ portion” of the matrix has a row of zeros, you can conclude that it is not possible to rewrite the matrix in the form This means that A has no inverse, or is noninvertible (or singular) Theorem : 32 Another solution of example 17 𝟏 𝟐 𝟎 Since |𝑨| = | 𝟑 −𝟏 𝟐 | = −𝟒 + 𝟒 = 𝟎 then the matrix A is noninvertible (or −𝟐 𝟑 −𝟐 singular) i.e the matrix A has no inverse 2.7 The system of m linear equations in n unknowns that is AX = B, where A = [aij ] is the coefficient matrix of the system 𝑥1 𝑏1 𝑥2 𝑏 𝑋 = [ ⋮ ] is the vector of unknowns and 𝐵 = [ 2 ] is the vector of constants. ⋮ 𝑥𝑛 𝑏𝑛 Number of Solutions of a System of Linear Equations For a system of linear equations in variables, precisely one of the following is true. 1. The system has exactly one solution (consistent system). 2. The system has an infinite number of solutions (consistent system). 3. The system has no solution (inconsistent system). 2.8 Solving a System of Linear Equations A system of linear equations can have exactly one solution, an infinite number of solutions, or no solution. For square systems (those having the same number of equations as variables), you can use the theorem below to determine whether the system has a unique solution I. Solving a System of Equations Using an Inverse Matrix If A is an invertible matrix, then the system of linear equations AX = B has a unique solution given by 𝑿 = 𝑨−𝟏 𝑩 33 Example 1 Solve the system Solution The system can be written as 𝟐 𝟏 𝒙 𝟏 𝑨𝑿 = 𝑩 , 𝑨 = [ ] , 𝑿 = [𝒚] , 𝒂𝒏𝒅 𝑩 = [ ] −𝟏 𝟑 𝟏 The system has a unique solution since A is invertible and the unique solution is given by 𝟑 −𝟏 𝟐 𝟏 𝟐 𝟑 𝑿 = 𝑨−𝟏 𝑩 = [𝟕𝟏 𝟕 𝟐 ] [ ] = [𝟕𝟑] 𝒕𝒉𝒆𝒏 𝒙 = , 𝒚= 𝟏 𝟕 𝟕 𝟕 𝟕 𝟕 Example 2 Use an inverse matrix to solve each system Solution First note that the coefficient matrix for each system is Using Gauss-Jordan elimination, you can find e 𝑨−𝟏 to be 𝑿 = 𝑨−𝟏 𝑩 II. CRAMER'S RULE Cramer’s Rule generalizes easily to systems of n linear equations in n variables. The value of each variable is the quotient of two determinants. The denominator is the determinant of the coefficient matrix, and the numerator is the determinant of the matrix formed by replacing the column corresponding to the variable being solved for with the column representing the constants. For example, the solution for𝑥3 in the system 34  Example 3 Use Cramer’s Rule to solve the system of linear equations for x Solution The determinant of the coefficient matrix is Because |𝑨 | ≠ 𝟎) you know the solution is unique, and Cramer’s Rule can be applied to solve for x as follows The Gauss-Jordan algorithms We now describe the GAUSS-JORDAN ALGORITHMS. This is a process which starts with a given matrix A and produces a matrix B in reduced row-echelon form, which is row-equivalent to A. If A is the augmented matrix of a system of linear equations, then B will be a much simpler matrix than A from 35 III. Gaussian Elimination with Back-Substitution 1. Write the augmented matrix of the system of linear equations. 2. Use elementary row operations to rewrite the augmented matrix in row-echelon form. 3. Write the system of linear equations corresponding to the matrix in row-echelon form, and use back-substitution to find the solution. Example 4 : Solution 36  Example 5: solution Example 6 Solve the system solution Example 7: Solution 37  Exercise2.2 1- Find (a) A + B, (b) A – B, (c) 3A, and (d) 5A – 2B when 2- Find, if they exist, (a) AB, (b) BA, and (c) 𝑨𝟐 when 3- Find AB, if possible. 4- Write each matrix in reduced row-echelon form 5- Find the inverse, if it exists, for each matrix. 38 6- 7- Write the matrix equation AX = B and use it to solve the system 8- Solve each system of equations using matrices 9- Solve for the unknowns in each of the following systems.. 10- Find nontrivial solutions for the system if they exist 11- For what values of k will the system have nontrivial solutions? 12- Write the minors and cofactors of the elements in the 4th row of the determinant 39  Chapter Three Mathematical Induction Some statements are defined on the set of positive integers. To establish the truth of such a statement, we could prove it for each positive integer of interest separately. However, since there are infinitely many positive integers, this case-by-case procedure can never prove that the statement is always true. A procedure called mathematical induction can be used to establish the truth of the state for all positive integers. 3.1. Principle Of Mathematical Induction Let P(n) be a statement that is either true or false for each positive integer n. If the following two conditions are satisfied: 1. P(1) is true, and 2. Whenever for n = k, P(k) is true implies P(k + 1) is true. Then P(n) is true for all positive integers n. 3.2. Proof By Mathematical Induction To prove a theorem or formula by mathematical induction there are two distinct steps in the proof. 1. Show by actual substitution that the proposed theorem or formula is true for some one positive integer n, as n = 1, or n = 2, etc. 2. Assume that the theorem or formula is true for n = k. Then prove that it is true for n = k + 1. Once steps (1) and (2) have been completed, then you can conclude the theorem or formula is true for all positive integers greater than or equal to a, the positive integer from step (1). Example 1 Show that if n is a positive integer , then Solution 1. P(1) is true because 2. Whenever for n = k, P(k) is true.That is, we assume that Under this assumption ,it must be shown that P(k + 1) is true namely, that 41 Is also true. When add k + 1 to both sides of the equation P(k) , we obtain This last equation show that P(k +1) is true under the assumption that P(k) is true. Example 2 Conjecture a formula for the sum of the first n positive odd integer. Then prove your conjecture using mathematical induction. Solution The sums of the first n positive odd integer for n = 1,2,3,4,5 are n=1 1=1 , n=2  1 + 3 = 4 = 22 n = 3  1 + 3 + 5 = 9 = 32 n = 4  1 + 3 + 5 + 7 = 16 = 42 n = 5  1 + 3 + 5 + 7 + 9 = 25 = 52 Not that sum = square of the number of elements 1. P(1) states that " the sum of the first one positive is 12. This is true because the sume of the first odd positive integer is 1. 2. We asume that P(k) is true , that is 1 + 3 + 5 + …+(2k -1) = 𝑘 2 Not that P(k + 1) is the statement that 1 + 3 + 5 + …+(2k -1) + (2k + 1) = (𝑘 + 1)2 So assuming that P(k) is true , it follows that 1 + 3 + 5 + …+(2k -1) + (2k + 1) = ( 1 + 3 + 5 + …+(2k -1)) + (2k + 1) = 𝑘 2 + (2k + 1) = 𝑘 2 + 2𝑘 + 1 = (𝑘 + 1)2 Example 3 Prove the following assertions using the Principle of Mathematical Induction 1. For a complex number z, {𝒛̅}𝒏 = ̅̅̅̅ 𝒛𝒏 for n1. 2. Let A be an nxn matrix and let A’ be the matrix obtained by replacing a row R of A with cR for some real number c. Use the definition of determinant to show det(A’) = c det(A). Solution. 𝒛𝒏. The base case P(1) is {𝒛̅}𝟏 = ̅̅̅̅ 1. We let P(n) be the formula {𝒛̅}𝒏 = ̅̅̅̅ 𝒛𝟏 , which reduces to z = z which is true. We now assume P(k) is true, that is, we assume {𝒛 ̅ }𝒌 = ̅̅̅̅ 𝒛𝒌 and attempt 40 to show that P(k+1) is true. Since{𝒛̅}𝒌+𝟏 = {𝒛̅}𝒌 𝒛̅ , we can use the induction hypothesis and write{𝒛̅}𝒌 = ̅̅̅̅ 𝒛𝒌. Hence, {𝒛̅}𝒌+𝟏 = {𝒛̅}𝒌 𝒛̅ =̅̅̅̅ 𝒛𝒌 𝒛̅= ̅̅̅̅̅̅̅ 𝒛𝒌+𝟏.. This establishes {𝒛̅}𝒌+𝟏 = ̅̅̅̅̅̅̅ 𝒛𝒌+𝟏 so that P(k + 1) is true. Hence, by the Principle of Mathematical Induction, ( z )n = z n for all n  1. 2. To prove this determinant property, we use induction on n, where we take P(n) to be that the property we wish to prove is true for all nxn matrices. For the base case, we note that if A is a 1x1 matrix, then A = [a] so A’ = [ca]. By definition, det(A) = a and det(A’) = ca so we have det(A’) = c det(A) as required. Now suppose that the property we wish to prove is true for all kxk matrices. Let A be a (k+1)x(k+1) matrix. We have two cases, depending on whether or not the row R being replaced is the first row of A. n Case 1: The row R being replaced is the first row of A. By definition, det(A’) =  a '1 p C '1 p p 1 where the 1p cofactor of A’ is C’1p = (-1)(1+p) det(A’1p) and A’1p is the kxk matrix obtained by deleting the 1st row and pth column of A’. Since the first row of A’ is c times the first row of A, we have a’1p = c a1p. In addition, since the remaining rows of A’ are identical to those of A, A’1p = A1p. (To obtain these matrices, the first row of A’ is removed.) Hence det(A’1p)= det (A1p), so that C’1p = C1p. As a result, we get n n n det(A’) =  a '1 p C '1 p =  c a1 p C1 p = c  a1 p C1 p = c det(A); p 1 p 1 p 1 as required. Hence, P(k + 1) is true in this case, which means the result is true in this case for all natural numbers n  1. (You'll note that we did not use the induction hypothesis at all in this case. It is possible to restructure the proof so that induction is only used where it is needed. While mathematically more elegant, it is less intuitive, and we stand by our approach because of its pedagogical value.) Case 2: The row R being replaced is the not the first row of A. By definition, n det(A’) =  a '1 p C '1 p ; where in this case, a’1p = a1p, since the first rows of A and A’ are the same. p 1 The matrices A’1p and A1p, on the other hand, are different but in a very predictable way - the row in A’1p which corresponds to the row cR in A’ is exactly c times the row in A1p which corresponds to the row R in A. In other words, A’1p and A1p are kxk matrices which satisfy the induction hypothesis. Hence, we know det A’1p_= c det (A1p) and C’1p = cC1p. We get n n n det(A’) =  a '1 p C '1 p =  a1 p c C1 p = c  a1 p C1 p = c det(A); p 1 p 1 p 1 which establishes P(k + 1) to be true. Hence by induction, we have shown that the result holds in this case for n  1 and we are done. 42  Exercise 3 1. Prove by mathematical induction that, for all positive integers n, 1 + 2 + 22 + ⋯ + 2𝑛 = 2𝑛+1 − 1 , 𝑛 ≥ 0 2. Prove by mathematical induction that the sum of n terms of an arithmetic sequence a, a + d, That is 3. Prove by mathematical induction that, for all positive integers n, 3n > 100 n for n > 5. 4. Prove by mathematical induction that, for all positive integers n, 5. Prove by mathematical induction that, for all positive integers n, 5 - Prove by mathematical induction that is divisible by a + b when n is any positive integer. 43  Chapter fourth Partial Fractions 4.2 Introduction To express a single rational fraction into the sum of two or more single rational fractions is called Partial fraction resolution. For example, 4.2 Polynomial: The degree of a polynomial is the power of the highest term in x. 4.3 Rational fraction: 𝑝(𝑥) We know that 𝑞 ≠ 0 is called a rational number. Similarly the quotient of two 𝑞(𝑥) 𝑁(𝑥) polynomials 𝐷 ≠ 0 with no common factors, is called a rational fraction. 𝐷(𝑥) A rational fraction is of two types 1. If the degree of the numerator N(x) is less than the degree of the denominator D(x) the fraction is said to be a proper fraction 2. If the degree of the numerator N(x) is greater than or equal to the degree of the denominator D(x) the fraction is said to be an improper fraction Note: An improper fraction can be expressed, by division, as the sum of a polynomial and a proper fraction. For example 6𝑥 3 +5 𝑥 2 −7 8𝑥−4 : = (2𝑥 + 3) + 3𝑥 2 −2𝑥−1 𝑥 2 −2𝑥−1 44 4.3 Process of Finding Partial Fraction: A proper fraction N(x) ,D(x) can be resolved into partial fractions as: (I) If in the denominator D(x) a linear factor (ax + b) occurs and is non-repeating, its partial 𝐴 fraction will be of the form where A is a constant whose value is to be determined. 𝑎𝑥+𝑏 (II) If in the denominator D(x) a linear factor (ax + b) occurs n times, i.e., (𝑎𝑥 + 𝑏)𝑛 , then there will be n partial fractions of the form (III) If in the denominator D(x) a quadratic factor a𝑥 2 + bx + c occurs and is non-repeating, its 𝐴𝑥+𝐵 partial fraction will be of the form , where A and B are constants whose values 𝑎𝑥 2 +𝑏𝑥+𝑐 are to be determined. (IV) If in the denominator a quadratic factor a𝑥 2 + bx + c occurs n times, i.e., (𝑎𝑥 2 + 𝑏𝑥 + 𝑐)𝑛 ,then there will be n partial fractions of the form Note: The evaluation of the coefficients of the partial fractions is based on the following theorem: "If two polynomials are equal for all values of the variables, then the coefficients having same degree on both sides are equal," for example , if 4.3.1 Type I When the factors of the denominator are all linear and distinct i.e., non repeating. Example 1: 7𝑥 −25 Resolve into partial fractions x2 −7x+12 Solution: 45 Comparing the co-efficient of like powers of x on both sides, we have 7 = A + B and –25 = – 4A – 3B Solving these equation we get A = 4 and B = 3 Hence the required partial fractions are: Alternative Method: Since 7x – 25 = A(x – 4) + B(x – 3) Put x -4 = 0,  x = 4 in equation (2) 7(4) – 25 = A(4 – 4) + B(4 – 3) 28 – 25 = 0 + B(1) ,  B = 3 Put x – 3 = 0 ,  x = 3 in equation (2) 7(3) – 25 = A(3 – 4) + B(3 – 3) 21 – 25 = A(–1) + 0 –4=–A, A=4 Hence the required partial fractions are Example 2 6𝑥 3 +5𝑥 2 −7 Resolve into partial fractions 3𝑥 2 −2 𝑥−1 Solution: This is an improper fraction first we convert it into a polynomial and a proper fraction by 6𝑥 3 +5 𝑥 2 −7 8𝑥−4 division = (2𝑥 + 3) + 3𝑥 2 −2𝑥−1 𝑥 2 −2𝑥−1 8𝑥 − 4 8𝑥 − 4 𝑨 𝑩 = = + 𝑥 2 − 2𝑥 − 1 (3𝑥 + 1)(𝑥 − 1) 𝑥 − 1 3𝑥 + 1 Multiplying both sides by (x – 1)(3x + 1) we get 8x – 4 = A(3x + 1) + B(x – 1) (I) Put x – 1 = 0,  x = 1 in (I), we get The value of A 8(1) – 4 = A(3(1) + 1) + B(1 – 1) 8 – 4 = A(3 + 1) + 0  2 = 4A  A =1 1 Put 3x + 1 = 0  x = − in (I) 3 46  1 1 8(- ) − 4 = 𝐵 ( − − 1) 3 3 −20 −4 = B( ) 𝐵 = 5 3 3 Hence the required partial fractions are 8x − 4 1 5 = (2x + 3) + + 𝑥 2 − 2𝑥 − 1 x − 1 3x + 1 Example 3 : Resolve into partial fraction Solution: Multiplying both sides by L.C.M. i.e., x(x – 4)(x + 2) 8x – 8 = A(x – 4)(x + 2) + Bx(x + 2) + Cx(x – 4) (I) Put x = 0 in equation (I), we have 8 (0) – 8 = A(0 – 4)(0 + 2) + B(0)(0 + 2)+C(0)(0 – 4)  –8 = –8A + 0 + 0  A = 1 Put x – 4 = 0  x = 4 in Equation (I), we have 8 (4) – 8 = B (4) (4 + 2) 32 – 8 = 24B  24 = 24B  B = 1 Put x + 2 = 0  x = – 2 in Eq. (I), we have 8(–2) –8 = C(–2)( –2 –4) –16 –8 = C(–2)( –6)  –24 = 12C  C = –2 Hence the required partial fractions Exercise 4.1 Resolve into partial fraction: 47 4.3.2 Type II: When the factors of the denominator are all linear but some are repeated. Example 1: Resolve into partial fractions: Solution: 48 Example 2 Resolve into partial fractions Solution Put 1 + x = 0  x = –1 in eq. (I), we get4 + 7 (–1) = C ( 2 – 3) 4 – 7 = C(–1)  – 3 = – C  C = 3 Comparing the co-efficient of 𝑥 2 on both sides A + 3B = 0  – 6 + 3B = 0  3B = 6  B=2 Hence the required partial fraction will be 49 4.3.3 Type III: When the denominator contains ir-reducible quadratic factors which are non-repeated. Example 1: Resolve into partial fractions Solution: Comparing the co-efficient of like powers of x on both sides A+B=0 3B + C = 9 51  Exercise 4.3 Resolve into partial fraction: , , , 50  Chapter Fifth Derivatives of Polynomials and Exponential Functions in this chapter we develop rules for finding derivatives without having to use the definition directly. These differentiation rules enable us to calculate with relative ease the derivatives of polynomials, rational functions, algebraic functions, exponential and logarithmic functions, and trigonometric and inverse trigonometric functions. 5.1 Basic Differentiation Rules In this section we learn how to differentiate constant functions, power functions, polynomials, and exponential functions Definition.1: [Basic Differentiation Rules] Let f, g be two differentiable functions and n, c are real numbers Example 1. Find the derivative of the following Solution 52 Example 2. Solution 2𝑥 3 −3𝑥 2 −12𝑥 Example 3. For What values of x does the curve 𝑓(𝑥) = have any horizontal 6 tangents? Now Hence f has horizontal tangent line at x= -1 ,x = 2 53 Example 4 Let find Solution: First note that Example 5. Let find Solution: 5.2 Derivatives of Natural Exponential Function Example 6. For What value of x does the curve f (x) = 2x − 𝑒 𝑥 , have any horizontal tangents? Also for what value of x does the tangent line to the curve parallel to y = −3x. Solution First note that the horizontal tangents occur where Hence f has horizontal tangent line at x = ln2 Note that the tangent line is parallel to y = 3x occur where equal to the slope of the line (−3.) Hence f has a tangent line parallel to y = −3x at x= ln 5 54  5.3 The Product and Quotient Rules 5.3.1 The Product Rule Theorem 1: [The Product Rule] Let f(x) and g(x) be differentiable and n be a real number. Then Example 1. Find the derivative of Solution: Example 2. Find the derivative of 𝑓(𝑥) = (𝑥 2 + 3)(𝑥 3 − 3𝑥 + 1) Solution: Example 3. Solution 55 Example 4: Let where g is a differentiable function with g(2) =-1 and Solution Example 5: Find an equation for the tangent line to the curve Solution: To find an equation for a line you need a point and slope. The point is (1, f(1)) and the slope is m = Hence Thus we have the point (1, e) and m = 2e. Therefore Hence the equation of the tangent line is y = 2ex – e 5.3.2 The Quotient Rule Let f(x) and g(x) be differentiable and n be a real number. Then Example 6 Find the derivative of Solution 56 Example 7: Find the derivative of Solution 2 𝑒𝑥 Example 8: Find an equation for the tangent line to the curve 𝑓(𝑥) = 𝑎𝑡 𝑥 = 1 𝑥+1 Solution: To find an equation for a line you need a point and slope. The point is (1, f (1)) and the slope is Exercises 5.1 In Exercises 1-6 Find 57  (7) Find the first and second derivatives (8) Suppose f and g are function of x are differentiable at x = -2 and that Find the value of the following derivatives (9) (10) 58  5.4 Derivatives of Trigonometric Functions We collect all the differentiation formulas for trigonometric functions in the following table. Remember that they are valid only when is measured in radians. 59 61 Example 5 60  Exercises 5.2 1. Differentiate function 2. Using the Quotient Rule to prove that 3. Find an equation of the tangent line to the curve at the given point 4. 62  Chapter sixth Indefinite Integer 6.1 Table of Indeinite Integrals Remark: To prove these laws, we can find the first derivative of the right hand side and show that it equal to the integrated function. Now we shall prove the integration of some trigonometric functions: 63 1  tan x dx   sin x dx     sin x dx    d cos x   ln cos x  ln cos x 1  c cos x cos x cos x  ln sec x  c  cot xdx Similarly, we can find sec x  tan x sec2 x  sec x tan x 2  sec xdx   sec x dx   dx , sec x  tan x  0 sec x  tan x sec x  tan x d sec x  tan x    ln sec x  tan x   c sec x  tan x  csc x dx Similarly, we can find The following examples illustrate how to apply the laws of integration Example 1 Solution Example 2: Solution Example 3 Solution Example 4 Solution: 64   1  tan x  sec x  2 Example 5: Find dx , dx x Solution: dx  2 sec x 2.. x 2 dx  2 sec x 2 d x 2  sec x 1 1 1 1 1 1 1  x dx   sec x 2 x 2 2    2 ln sec x  tan x  c  1  tan x dx   1  2 tan x  tan xdx   1  2 tan x  sec x  1dx 2 2 2  2 tan x dx   sec x dx  2 ln sec x  tan x  c 2 Special case: If the desired integrated function is of the form  dx , x its numerator is the differential of the function under the square root of the denominator ,then the value of the integration equal twice square root in the denominator. i.e.  dx  2 x  c x Example 6 Evaluate the following integrals: 2x  1 sec x tan x  5x  5x  2 2 dx ,  sec x  1 dx Solution: 2 x 1 1 10 x  5 2  5 x2  5 x  2 dx   5 5x 5x  2 2 dx  5 5 x2  5 x  2  c sec x tan x  sec x  1 dx  2 sec x  1  c 65  dx Example 7: Find the following integral:  1  4x 2 , Solution: dx 1  1  4x  sin 1 2 x  c 2 2 3dx 3 dx 3 dx 3 2 5x  5x 2   2  4 4 5x  ( ) 4 5x 2  ( )( ) tan 1 4 5 2 c 1 ( ) 1 4 2 Example 8 Calculate Solution Example 9 66 Example 10 Exercise 6.1 67  6.2 Techniques of Integration In this chapter we develop techniques for using these basic integration formulas to obtain indefinite integrals of more complicated functions. We learned the most important method of integration, integration by parts,. Then we learn methods that are special to particular classes of functions, such as trigonometric functions and rational functions 6.2.3. Integration by Parts 68 Example 1 Solution Example 2 Solution 69 Example 3 Solution Let Example 4 Solution 71 Example 5 Solution Exercise 6.2 70  6.2.2 Integration of Rational Functions by Partial Fraction. In this section we show how to integrate any rational function (a ratio of polynomials) by expressing it as a sum of simpler fractions, called partial fractions, that we already know how to integrate. Case 3 Q(x) contains irreducible quadratic factors, none of which is repeated 72 Case4 where S and R are also polynomials As the following example illustrates, sometimes this preliminary step is all that is required. 6.2.2.1 Integration Proper Rational Function 𝑥 2 +2𝑥−1 Example 1 Express the function 2𝑥 3 +3 𝑥 2 −2 𝑥 as a sum of partial fractions and 𝑥 2 +2𝑥−1 find∫ 3 𝑑𝑥 2𝑥 +3 𝑥 2 −2 𝑥 Solution 73 Example 2 Solution Example 3 Solution 74 Example 4 Solution 75  6 6.2.2.2 Integration Proper Rational Function Example 5 Solution E Example 6 Solution 76  Exercises 6.3 1. 77 2. References 1. James Ward Brown and Ruel V. Churchill "COMPLEX VARIABLES AND APPLICATIONS "SEVENTH EDITION 1996 2. Dennis G.Zill and Patrick D. Shanahan:" A First Course in Complex Analysis with Applications "Copyright © 2003 by Jones and Bartlett Publishers, Inc 3. James stewart "Calculus Early Transcendentals" SIXTH EDITION COPYRIGHT © 2008, 2003 78 4. RON LARSON and DAVID C. FALVO; "Elementary Linear Algebra" SIXTH EDITION Copyright ⓒ 2009 by HOUGHTON MIFFLIN HARCOURT PUBLISHING COMPANY Boston New York 79