Engineering Mathematics Mohamed Altemaly

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(2) Elminia University Faculty of Engineering. Engineering Mathematics Part 1. Dr. Ali Mohamed Eltamaly.

(3) Preface Most complex scientific and engineering models of the real world are Differential equations. Here are some that I know of: heat flow, electrostatic potential, waves (radio, light, sound, water), metal beam bending, quantum mechanics, hydrogen bombs, electrons in telegraph wires, optics, classical mechanics, general relativity, distributions of organisms, ice sheets, tsunamis, air flow, ocean currents, weather, auroras, blood flow, plate tectonics, supernovas. For this reason we introduce this notes for students in faculty of engineer. By the end of the course the student should: • be familiar with the concept of a complex number and be able perform algebraic operations on complex numbers, both with numeric and symbolic entries, solve simple equations with complex roots, and in particular describe geometrically the roots of unity; • be familiar with the concept of a matrix and be able to perform algebraic operations on matrices, both with numeric and symbolic entries, be able to define a determinant and calculate one both directly and by using row and column operations, understand the definition and use of the inverse of a non-.

(4) singular matrix, and be able to solve simple systems both using inverses and reduction to triangular form, and be able to compute inverses using Gaussian reduction and explain the method in terms of elementary matrices; • be familiar with many topics in calculus like limits, differentiations, and all methods of integrations; • be familiar with first order differential equations (linear and nonlinear) and their solution by many techniques; • be familiar with many engineering applications of first order differential equations like falling bodies, the time rate of change in temperature of an object varies as the difference in temperature between the object and surroundings, Chemical Applications, time required for liquid tanks to get empty, Half Life Of Nuclear Materials, and Electrical Circuits; • be familiar with solution of higher order linear differential equations with constant coefficients and Cuchy differential equation and their solution by many techniques; • be familiar with many engineering applications of higher order differential equations like, Free Oscillation of suspended bodies, Bending of Beams, and Electrical Circuits; • be familiar with the benefits of Laplace and inverse Laplace transforms, using Laplace transform for solving differential equations, finding Laplace transform of any periodical and non.

(5) periodical waveforms, using Laplace transform to solve many application problems like Free Oscillation of suspended bodies, Bending of Beams, and Electrical Circuits; • be familiar with finding Fourier transform of any waveform by advanced techniques like jump technique; • be familiar with curve fitting by using least square technique and using this technique to fiend Fourier transform for any waveform numerically; • be familiar with using power series for solving linear differential equations of second order, and Bessel function; • be familiar with partial differentiation and solving partial differential equations by many techniques as separation of variables, Laplace transform, and Fourier transform; • be familiar with solving differential equations which governs the conduction of heat in solids; • be familiar with eigen values and eigen vectors and using them for solving simultaneous linear differential equations; • be familiar with special functions like Gamma and Beta functions; • be familiar with many topics in numerical analysis like Numerical solution of equations by many techniques like Simple Iteration, Bisection, false position, Newton Raphson and secant method; and.

(6) • be familiar with polynomial interpolation and numerical solution of differential equations by many techniques like Euler’s and RungeKutta’s method..

(7) Contents. Part 1 Chapter. Title. No.. Page No.. 1. Mathematical Numbers. 1. 2. Matrices. 42. 3. Calculus. 74. 4. Ordinary Differential. 101. Equations 5. Linear Differential Equations. 151. Of Higher Order 6. Laplace Transforms. 190. 7. Fourier Series. 238. 8. Least Square Technique. 259.

(8) References. Contents. Part 2 Chapter. Title. No. 9. Page No.. Power Series Solution Of. 285. Differential Equations 10. Partial Differential Equations. 346. 11. Simultaneous Linear. 389. Differential Equations 12. Special Functions. 459. 13. Numerical Analysis. 478. Appendix. 543. References. 574.

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(10) Chapter 1 Mathematical Numbers 1.1 Natural Numbers Natural numbers known as counting numbers are the numbers beginning with 1, with each successive number greater than its predecessor by 1. If the set of natural numbers is denoted by N, then N = { 1, 2, 3, ......} 1.2 Whole Numbers Whole numbers are the numbers beginning with 0, with each successive number greater than its predecessor by 1. It combines the set of natural numbers and the number 0. If the set of whole numbers is denoted by W, then W = { 0, 1, 2, 3, .......} 1.3 Integer Numbers Integers are the numbers that are in either (1) the set of whole numbers, or (2) the set of numbers that contain the negatives of the natural numbers. If the set of integers is denoted by I, then I = {......, -3, -2, -1, 0, 1, 2, 3, ......} Positive integers are the numbers in I greater than 0. Negative numbers are the numbers in I less than 0. The number zero is neither positive nor negative, i.e., it is both non-positive and non -negative..

(11) 2 Mathematical Numbers Given the above definitions, the following statements about integers can be made: (1) N is the set of positive integers. (2) W is the union of N and the number 0. (3) The set of numbers that contain the negatives of the numbers in N is the set of negative integers. (4) I is the union of W and the set of negative integers. 1.4 Real Number Line The set of real numbers can be pictorially represented by the real number line. It is a straight line, whose "origin" is designated by the number 0, and continues in both directions. All the positive integers are ordered, in ascending order from left to right, to the right side of 0; all the negative integers are ordered, in descending order from right to left, to the left side of 0. Notches are marked to denote the position of these integers in the following figure (Fig.1).. Fig.1 Every point on the line corresponds to a real number, and every real number can be paired with a point on this number line. If the real number is an integer, its point on the number line coincides with one of the notches for an integer; otherwise, its point lies between two successive notches. All real numbers represented by points to.

(12) Chapter One. 3 the right of the number 0 are positive, while all real numbers represented by points to the left of the number 0 are negative. 1.5 Absolute Values The absolute value of a real number is the distance between its corresponding point on the number line and the number 0. The absolute value of the real number a is denoted by |a|. From the diagram shown in Fig.2, it is clear that the absolute value of non-negative numbers is the number itself, while the absolute value of negative integers is the negative of the number. Thus, the absolute value of a real number can be defined as follows: For all real numbers a, (1) If a > 0, then a = a . (2) If a < 0, then a = − a .. Fig.2 Example 1: |2|=2 | -4.5 | = 4.5 |0|=0.

(13) 4 Mathematical Numbers 1.6 Complex Numbers 1.6.1 Introduction The solution of a second order equation ax 2 + bx + c = 0 can be. − b ± b 2 − 4ac obtained by the famous formula, x = 2a For example, if x 2 + x − 2 = 0 , then we have:. x=. −1±. (1 + 8) 2. ∴ x = 1 or. =. −1± 9 −1± 3 = 2 2. −2. As we see there is no problems with solving the above equation. But if we solve the equation 5 x 2 − 6 x + 5 = 0 in the same way, we get:. x=. 6±. (36 − 100) 10. =. 6 ± − 64 10. And the next stage is now to determine the square root of (-64). Is it (i) 8, (ii) -8, (iii) neither? It is, of course, neither, since + 8 and − 8 are the square roots of 64 and not of (-64). In fact,. (− 64). cannot be represented by an. ordinary number, for there is no real number whose square is a negative quantity. However, − 64 = −1 * 64 And therefore we can write:. (− 64) = (− 1 * 64) =. − 1 * 64 = 8 − 1.

(14) Chapter One. 5. (− 1) ,. Of course, we are still faced with. which cannot be. evaluated as a real number, for the same reason as before, but, if we. − 1 with the letter j, then. can replace. (− 64) = (− 1) * 8 =. j8. We now have a way of finishing off the quadratic equation we started before as following:. 5x 2 − 6x + 5 = 0 x=. 6±. (36 − 100) 10. ∴ x = 0.6 + j 0.8. =. 6 ± − 64 6 ± j8 = 10 10. or x = 0.6 − j 0.8. 1.6.2 Powers of j. j = −1 j 2 = −1. ( ) 2 j 4 = ( j 2 ) = (− 1)2 = 1. j 3 = j 2 * j = −1 * j = − j. Note especially the last result: j 4 = 1 . Every time a factor j 4 occurs, it can be replaced by the factor 1, so that the power of j is reduced to one of the four results above. In the same way we can replace j 2 = −1 with –1. The complex number x = 1 + j 4 , consists of two separate terms, 1, and j 4 These terms cannot be combined any further, since the second is an imaginary number (due to its having the factor j)..

(15) 6 Mathematical Numbers In such an expression as x = 1 + j 4. 1 is called the real part of x 4 is called the imaginary part of x The two together form what is called a complex number. So, a Complex number= (Real part) + j(Imaginary part) Complex numbers is very important especially in some engineering application like electrical and mechanical engineering. So we have to fully understand how to carry out the usual arithmetical operations. 1.6.3 Addition and Subtraction of Complex Numbers. Addition and Subtraction are quite easy as shown in the following. example: Example 2 Find the results of the following arithmetical operations.. (3 +. j 7 ) + (6 − j 2 ) .. Solution :. (3 +. j 7 ) + (6 − j 2 ) = 3 + j 7 + 6 − j 2 = (3 + 6 ) + j (7 − 2 ) = 9 + j 5 So, in general, (a + jb ) + (c + jd ) = (a + c ) + j (b + d ) 1.6.4 Multiplication of. Complex Numbers. The following example illustrate the multiplication process in complex numbers. Example 3 Find the results of the following arithmetical operations.. (2 + j3)(5 + J 7 ).

(16) Chapter One. 7 Solution: These are multiplied together in just the same way as. you would determine the product (2 + j 3)(5 + j 7 ) . Form the product terms of. (2 + j3)(5 + j 7 ) = 2 * 5 +. j3 * 5 + j 2 * 7 + j 2 3 * 7. = 10 + j15 + j14 + j 2 21 = 10 + j 29 − 21 = −11 + j 29 If the expression contains more than two factors, we multiply the factors together in stages:. (2 + j3)(5 +. (. ). j 7 )(1 − j 2) = 10 + j15 + j14 + j 2 21 (1 − j 2) = (10 + j 29 − 21)(1 − j 2) = (− 11 + j 29 )(1 − j 2) = −22 + j 22 + j 29 − j 2 58 = −22 + j 51 + 58 = 36 + j 51. Example 4 Find the results of the following arithmetical. operations. (5 + j8)(5 − j8) Solution:. (5 + j8)(5 − j8) = 25 +. j 40 − j 40 − j 2 64. = 25 + 64 = 89 In spite of what we said above, here we have a result containing no imaginary term. The result is therefore entirely real. This is rather an exceptional case. If we look at the two complex numbers we can.

(17) 8 Mathematical Numbers find that they are identical except for the middle sign in the brackets. are different. These two complex numbers called conjugate complex numbers and the product of two conjugate complex numbers is always entirely real. In general we can say:. (a + b )(a − b ) = a 2 − b 2. difference of two squares and there is no. any imaginary part. 1.6.5 Divison of. Complex Numbers. Division of a complex number by a real number is easy enough.. 5 − j4 5 4 = − j = 1.67 − j1.33 3 3 3 But how do we manage with dividing complex number with other complex one? If we could, somehow, convert the denominator into a real number, we could divide out as in the above example. So our problem is really, how can we convert (4 + j3) into a completely real denominator and this is explained in the previous item. We know that we can convert (4 + j3) into a completely real number by multiplying it by its conjugate (4 - j3). But if we multiply the denominator by (4 − j 3) , we must also multiply the numerator by the same factor.. 7 − j 4 (7 − j 4)(4 − j 3) 28 − j 37 − 12 16 − j 37 = = = 4 + j 3 (4 + j 3)(4 − j 3) 16 + 9 25. 16 37 −j = 0.64 − j1.48 25 25.

(18) Chapter One. 9 Then, to divide one complex number by another, therefore, we. multiply numerator and denominator by the conjugate of the denominator. This will convert the denominator into a real number and the final step can then be completed. Example 5 Simplify the following expression:. (2 + j3)(1 − j 2) 3 + j4. Solution:. (2 + j3)(1 − j 2) = 2 − 3 + j4. 8 − j 3 − j4 j +6 8− j = = * 3 + j4 3 + j4 3 + j4 3 − j4 24 − j 35 − 4 20 − j 35 = = = 0.8 − j1.4 9 + 16 25. Equal Complex Numbers. Now let us see what we can find out about two complex numbers which we are told are equal. Let the numbers a + jb, and c + jd are equal ∴ a + jb = c + jd Rearranging terms, we get ∴ a − c = j (d − b ) In this last statement the quantity on the left hand side is entirely real, while that on the right hand side is entirely imaginary, i.e. a real quantity equals an imaginary quantity. This seems contradictory and in general it just cannot be true. But there is one special case for which the statement can be true. That is when each side is zero.. ∴ a − c = j (d − b ) can be true only if a − c = 0, ie. a = c and if d − b = 0, ie. b = d.

(19) 10Mathematical Numbers So we get this important result, If two complex numbers are equal. then, (i) the two real parts are equal (ii) the two imaginary parts are equal For example, if x + jy = 5 + j 4 , then we know x = 5 and y = 4 . 1.6.6 Graphical Representation of a Complex Numbers. Although we cannot evaluate a complex number as a real number, we can represent it diagrammatically, as we shall now see. In the usual system of plotting numbers, the number 4 could be represented by a line from the origin to the point 4 on the scale. Likewise, a line to represent (-4) would be drawn from the origin to the point (-4). These two lines are equal in length but are drawn in opposite directions. Therefore, we put an arrow head on each to distinguish between them as shown in Fig.3. -4 -4. -3. -2. 4 -1. 0. 1. 2. 3. 4. Fig.3 A line which represents a magnitude (by its length) and direction (by the arrow head) is called a vector. We shall be using this word quite a lot. Any vector therefore must include both magnitude (or size) and direction. If we multiply (+4) by the factor (-1), we get (-4), i.e. the factor (-1) has the effect of turning the vector through 180o as shown in Fig.4..

(20) Chapter One. 11. 180 o 4. -4 -4. -3. -1. -2. 0. 1. 2. 3. 4. Fig.4 Multiplying by (-1) is equivalent to multiplying by j 2 , i.e. by the factor j twice. Therefore multiply in a single factor j will have half the effect and rotate the vector through only 90o . So, the factor j always turns a vector through 90o in the positive direction measuring angles, i.e. anticlockwise. If we now multiply j 4 by a further factor j, we get j 2 4 , i.e. (-4) and the following diagram (Fig.5) agrees with this result. If we multiply (-4) by a further factor j, sketch showing this new vector (− j 4 ) is shown in Fig.6. 4 3 j4. 4. 2. 3. 1 -4. 180. o. -4 -4. -3. -2. 0. 1. -3. -2. 4 0. -1. 1. -1. 2. 3. 4. -2 -j4. 4 -1. o. -4. j4 2 1. 180. 2. -3. 3. 4. Fig.5. -4. Fig.6. Let us denote the two reference lines by XX, and YY, as usual. You will see that: (i) The scale on the X-axis represents real numbers, XX1 is therefore called the real axis..

(21) 12Mathematical Numbers (ii) The scale on the Y-axis represents imaginary numbers, YY1 is. therefore called the imaginary axis. If we now wish to represent 2+ 3 as the sum of two vectors, we must draw them as a chain, the second vector starting where the first one finishes as shown in Fig.7. 5 2. 3. 1. 0. 2. 3. 4. 5. Fig.7 The two vectors, 2 and 3 are together equivalent to a single vector drawn from the origin to the end of the final vector (giving naturally that 2+3=5). If we wish to represent the complex number (3 + j2), then we add together the vectors which represent 3 and j2. Notice that the 3 is now multiplied by a factor j which turns that vector through 90o . The equivalent single vector to represent (2 + j3) is therefore the vector from the beginning of the first vector (origin) to the end of the last one. This graphical representation constitutes an Argand. diagram as shown in Fig.8. 3 j3. 2 1 2 0. 1. 2. Fig.8. 3.

(22) Chapter One. 13 Example 6 Draw an Argand diagram to represent the following. vectors: z1 = 3 + j 2 , z 2 = −3 + j1 , z3 = 2 − j 4 , and z 4 = −4 − j 4 Solution: The Argand diagram of the above vectors are shown in. the following Fig.9. 3. z1. 2 j1. z2. j2. 1 3. -4. -3. j4. -2. 0. 3 -1. 1. 2. -1. 3. 4 j4. -2 -j4 -3. z4. -4. z3. Fig.9.. 1.6.7 Graphical Addition of Complex Numbers. Let us find the sum of z1 = 3 + j 2 and z 2 = 2 − j 4 by Argand diagram. If we are adding vectors, they must be drawn as a chain. We therefore draw at the end of z1 , a vector representing z2 in magnitude and direction, and is parallel to it. In the same way, we therefore draw at the end of z2 , a vector representing z1 in magnitude and direction, and is parallel to it. Therefore we have a parallelogram. Thus the sum of z1 and z2 is given by the vector joining the starting point to the end of the last vector..

(23) 14Mathematical Numbers. The complex numbers z1 and z2 can thus be added together by drawing the diagonal of the parallelogram formed by z1 and. z 2 .Thus, z1 + z 2 = 3 + j 2 + 2 − j 4 = 5 − j 2 which is clear that this results is the same as obtained from Fig10. So the sum of two vectors on an Argand diagram is given by the diagonal of the parallelogram of vectors.. z1. 2. j2. 1 3 0. 1. -1. 2. 3. 4. z1 + z2. 5. -2 -j4 -3 -4. z2 Fig.10. Regarding to the subtraction it is quite similar to addition but the only trick is simply this: z1 − z 2 = z1 + (− z 2 ) That is, we draw the vector representing z1 and the negative vector of z2 and add them as before. The negative vector of z 2 is simply a vector with the same magnitude (or length) as z 2 but pointing in the opposite direction..

(24) Chapter One. 15. Example 7 If z1 = 3 + j 2 and z 2 = 2 − j 4 Find z1 − z 2 Solution:. It is clear from Argand diagram (Fig.11) that. z1 − z 2 = 1 + j 6 . We can now check for the above results:. Z1 − Z 2 = 3 + j 2 − (2 − j 4) = 3 + j 2 + (−2 + j 4) = 1 + j 6 6. z1 − z 2. 5 4. u. Fig.11. 3. − z2. z1. 2. j2. 1 3 3. 0. 3 -1. -2. 1. -1. 2. 3. 4. -2 -j4 -3 -4. z2. 1.6.8 Polar Form of a Complex Numbers. Complex numbers in the form a + jb is called rectangular form. Sometimes, it is convenient to express it in a different form. On an Argand diagram shown in Fig.12, let OA be a vector a + jb . Let r = length of the vector and θ the angle made with OX. Since. z = a + jb ,. this. can. be. written. z = r cos θ + jr sin θ. or. z = r (cos θ + j sin θ ) This is called the polar form of the complex number a + jb , where: r =. (a 2 + b 2 ),. b and , θ = tan −1   a.

(25) 16Mathematical Numbers. A. y r. o. b. θ. a. x. Fig.12 Example 8 Express z = 4 + j 3 in polar form. Solution: First draw a sketch diagram (that always helps). We can.  3 see that: r = 4 2 + 32 = 16 + 9 = 5 and θ = tan −1   = 36.87 o  4. (. z = a + jb = r (cos θ + j sin θ ) ∴ z = 5 cos 36.87 o + j sin 36.87 o. ). (i) r is called the modulus of the complex number z and is often abbreviated to mod( z ) or indicated by z. (. ). Thus if z = 3 + j 4 , ∴ z = 32 + 4 2 = 9 + 16 = 5 (ii) θ is called the argument of the complex number and can be abbreviated to arg ( z ) . So, if z = 5 + j 5 then arg( z ) = θ = 45o Warning: In finding θ , there are of course two angles between 0o. and 360 o , the tangent of which has the value θ . We must be careful to use the angle in the correct quadrant. Always draw a sketch of the vector to ensure you have the right one. The follwing table and Fig.13 show the correct angle range and quadrant..

(26) Chapter One. 17. Value of a. Value of b. Angle range. Quuadrant. +ve. +ve. 0o < θ < 90o. First. -ve. +ve. 90 o < θ < 180 o. Second. -ve. -ve. 180o < θ < 270o. Third. +ve. -ve. 270o < θ < 360o. Fourth. y. First. Second. o. 0 < θ < 90o. 90 o < θ < 180o o o. 180 < θ < 270. o. o. 270 < θ < 360. o. x. Fourth. Third. Fig.13 Example 9 Find arg( z ) when z = −3 − j 4 Solution: The vector has been drawn as shown in Fig. 14.. θ. is measured from OX to OP. We first find E the equivalent. acute angle from the trinngle shown.. tan E =. 4 = 1.3333 then E = 53.13o 3.

(27) 18Mathematical Numbers. But from Fig.14. This angle is 180 o < θ < 270 o then we have to add 180o to the angle E to get angle θ .. ∴θ = 180o + E = 233.13o. Fig.14 Example 10 Find arg( z ) when z = −5 + j 2 Solution: The vector has been drawn as shown in Fig.15.. Fig.15. ∴ tan E =. 2 = 0.4 ∴ E = 21.8o 5. In this particular case, θ = 180 − E o ∴θ = 158.2 o Complex numbers in polar form are always of the same shape and differ only in the actual values of r and θ . We often use the.

(28) Chapter One. 19. shorthand version r∠θ o to denote the polar form as shown in the following examples: Then, if z = −5 + j 2 r =. θ = 158.2 .. above. (. Then,. (25 + 4) = the. 29 = 5.385 and from. full. ). polar. form. is. z = 5.385 cos158.2 o + j sin 158.2 o and this can be shortened to z = 5.385∠158.2o . Example 11 express 4 − j 3 in shortened form. Solution: r =. (42 + 32 ) = 5. tan E = 0.75 , ∴ E = 36.87 o ∴θ = 360 − E = 323.13o. (. ). ∴ z = 5 cos 323.13o + j sin 323.13o = 5∠323.8o Of course, given a complex number in polar form, you can convert it into the basic rectangular form a + jb simply by evaluating the cosine and the sine and multiplying by the value of r. Example 12 Find the rectangular form of the following: z = 5∠35o. (. ). z = 5 cos 35o + j sin 35o = 5(0.8192 + j 0.5736) Solution: = 4.096 + j 3.868 1.6.9 Exponential Form of a complex numbers. There is still another way of expressing a complex number which we must deal with. We shall Expalin it this way: Many functions can be expressed as series. For example,.

(29) 20Mathematical Numbers ∞. xm x 2 x3 x 4 =1+ x + + + + ...... e = ∑ ! 2 ! 3 ! 4 ! m m=0 x. (1). ∞. (−1) m x 2 m x2 x4 cos x = ∑ =1− + − +......... ( 2 )! 2 ! 4 ! m m=0. (2). (−1) m x 2 m +1 x3 x5 sin x = ∑ = x− + − +......... ( 2 + 1 )! 3 ! 5 ! m m=0. (3). ∞. If we now take the series for e x and write jθ in place of x, we get the following series. e. jθ. =1+. ∴e. jθ. ∴e. jθ. ∴e. jθ. ( jθ )2 ( jθ )3 ( jθ )4 ( jθ )5 + + + jθ + 2!. 3!. 4!. 5!. (4). j 2 (θ )2 j 3 (θ )3 j 4 (θ )4 j 5 (θ )5 = 1 + jθ + + + + ++ 2! 3! 4! 5! = 1+. ( θ )2 j (θ )3 (θ )4 j (θ )5 jθ − − + + −− 2!. 3!.   (θ )2 (θ )4 = 1 − + − .... +   2! 4!  . 4!. 5!.   j (θ )3 j (θ )5  j θ− + − ....   3! 5!  . (5). It is clear from (2),(3) and (5) that the first bracket is in the form of cosine and the second bracket in the form of sine.. ∴ e jθ = cos θ + j sin θ. (6). Therefore, r (cos θ + j sin θ ) can now be written as re jθ . This is called the exponential form of the complex number. It can be.

(30) Chapter One. 21 obtained from the polar form quite easily since the r value is the. same and the angle θ is the same in both. The three ways of expressing a complex number are therefore (i) z = a + jb (ii) z = r (cos θ + j sin θ ). (Rectangular form) (Polar form). (iii) z = re jθ (Exponential form) And now a ward about negative angles. We know that:. e jθ = cos θ + j sin θ if we replace θ by − θ in this result, we get the following:. e − jθ = cos(− θ ) + j sin (− θ ) = cos θ − j sin θ So, e jθ = cos θ + j sin θ And e − jθ = cos θ − j sin θ There is one operation that we have been unable to carry out with complex numbers before this. That is to find the logarithm of a complex number. The exponential form now makes this possible, since the exponential form consists only of products and powers. For, if we have z = r.e jθ we can say: ln z = ln r + jθ Example 13 Express e1− jπ / 4 in the rectangular form: Solution: Well now, we can write  π  π  e1− jπ / 4 = e1e − jπ / 4 = e cos  − j sin     4   4. 1  e  1 (1 − j ) = e −j = 2 2  2.

(31) 22Mathematical Numbers Since every complex number in polar form is of the same shape,. i.e. r (cos θ + j sin θ ) = r∠θ o and differs from another complex number simply by the values of r and θ , we have a shorthand method of quoting the result in polar form. Example 14 Express z = 4 − j 3 in the polar form. Solution: The vector has been drawn as shown in Fig. 16.. r=. (42 + 32 ) = 5. Fig.16. 3 = −0.7, ∴ E = −36.87 o 4 ∴θ = 360 o − 36.87 o = 323.13o. From this r = 5 , tan E = −. (. ). Then, the polar form is z = 5 cos 323.13o + j sin 323.13o . And the polar form is z = 5.e form is. (. j 323.13o. And the the shortened. z = 5∠323.13o . In this last example, we have. z = 5 cos 323.13o + j sin 323.13o. ).

(32) Chapter One. 23. Fig.17. But the direction of the vector, measured from OX, could be given as − 36.87 o , the minus sign showing that we are measuring the angle in the opposite direction sense from the usual positive. ( (. ). (. )). direction. We could write z = 5 cos − 36.87 o + j sin − 36.87 o . But. you. already. sin (− θ ) = − sin (θ ) .. ( (. known. ). as. (. ∴ z = 5 cos 36.87 o − j sin 36.87 o. cos(− θ ) = cos(θ ). and. )). i.e. very much like the polar form but with a minus sign in the middle. This comes about whenever we use negative angles. In the same way we can say the following:. (. ) ( (. ). (. z = 5 cos 250o + j sin 250o = 5 cos − 110o + j sin − 110o. )). It is sometimes convenient to use this form when the value of θ is greater than 180 o , i.e. in the 3rd and 4th quadrants. In the same way we can write the following:.

(33) 24Mathematical Numbers. ( ) ( ) z = 4(cos 290o + j sin 290o ) = 4(cos 70 o − j sin 70 o ) = 4∠ − 70 o .. z = 3 cos 230o + j sin 230o = 3 cos130o − j sin130o = 3∠ − 130o. The polar form at first sight seems to be a complicated way of representing a complex number. However it is very useful as we shall see. Suppose we multiply together two complex numbers in this form: Let z1 = r1 (cos θ1 + j sin θ1 ) and z 2 = r2 (cos θ 2 + j sin θ 2 ). ∴ z1 * z 2 = r1 (cos θ1 + j sin θ1 ) * r2 (cos θ 2 + j sin θ 2 ). ∴ z1 * z 2 = r1.r2 (cos θ1 cos θ 2 + j sin θ1 cos θ 2 + j cos θ1 sin θ 2 + j 2 sin θ1 sin θ 2. ). Rearranging the terms and remmembering j 2 = −1 we get:. (cos θ1 cos θ 2 − sin θ1 sin θ 2 )  z1 * z 2 = r1.r2   + j (sin θ1 cos θ 2 + cos θ1 sin θ 2 ) Now. the. brackets. (cosθ1 cosθ 2 − sin θ1 sin θ 2 ) and. (sin θ1 cos θ 2 + cos θ1 sin θ 2 ) ought to ring the bell. What are they? (cos θ1 cos θ 2 − sin θ1 sin θ 2 ) = cos(θ1 + θ 2 ). (sin θ1 cos θ 2 + cos θ1 sin θ 2 ) = sin (θ1 + θ 2 ) ∴ z1 * z 2 = r1.r2 [cos(θ1 + θ 2 ) + j sin (θ1 + θ 2 )] Note: This is important result. Then we can say that to multiply. together two complex numbers in the polar form, (i) Multiply the r's together, (ii) Add the angles, θ , together it is just easy as that..

(34) Chapter One. 25 Example 15 Find the result of the following in the polar form:. 2∠30 * 3∠40 Solution: It is easy to do that as we get:. (. ). 2∠30 * 3∠40 = (2 * 3)∠ 30o + 40 o = 6∠70o Now let us see if we can discover a similar set of rules for division. We already know that to simplify 5 + j 4 we first obtain a denominator that is entirely real by multiplying top and bottom by the conjugate of the denominator i.e (5 − j 4 ) Right. Then let us do the same thing with z1 and z2 as following:. z1 r (cos θ1 + j sin θ1 ) = 1 z 2 r2 (cos θ 2 + j sin θ 2 ). ∴. z1 r (cos θ1 + j sin θ1 ) (cos θ 2 − j sin θ 2 ) = 1 * z 2 r2 (cos θ 2 + j sin θ 2 ) (cos θ 2 − j sin θ 2 ). ∴. z1 r1 (cosθ1 cosθ 2 + j sin θ1 cosθ 2 − j cosθ1 sin θ 2 + sin θ1 sin θ 2 ) = z 2 r2 cos 2 θ 2 + sin 2 θ 2. (. ). ∴. z1 r1 (cosθ1 cosθ 2 + sin θ1 sin θ 2 ) + j (sin θ1 cosθ 2 − cosθ1 sin θ 2 ) = 1 z 2 r2. ∴. z1 r1 r = (cos(θ1 − θ 2 ) + j sin (θ1 − θ 2 )) = 1 ∠(θ1 − θ 2 ) z 2 r2 r2. So, for division the rule is divide the r's and subtract the angles θ 's. Example 16 Simplify the following expressions: Solution:. 10∠95o 2∠44 o. (. ). = 5∠ 95o − 44 o = 5∠51o. 10∠95o 2∠44 o. ,.

(35) 26Mathematical Numbers 1.6.10 DeMoivre’s Theorem. There is very important rule is called DeMoivre’s Theorem. It says that to raise a complex number in polare form to any power n, we raise the r to the power n and multiply the angle by n.. ∴ [r (cos θ + j sin θ )]n = r n (cos nθ + j sin nθ ) Example 17 Use DeMoivre’s Theorem to find the results of the. following expression in polar form: Solution:. [3(cos110. o. + j sin 110o. )]3. [3(cos110 + j sin 110 )] = 3 (cos(3 *110 ) + j sin (3 *110 )) ∴ [3(cos 110 + j sin 110 )] = 27(cos(330 ) + j sin (330 )) = 27∠330 o 3. o. 3. o 3. o. o. o. o. o. This is where the polar form really comes into its own. For. DeMoivre's theorem also applies when we are raising the complex number to a fractional power, i.e. when we are finding the roots of a complex number as shown in the following example. Example 18 Find the square root of z = 9∠44 o Solution:.  44o   = 3∠22o We have z = 9∠44 = 9∠  2    Expansion of sin nθ and cos nθ o. By DeMoiver's theorem, we know that:. cos nθ + j sin nθ = (cos θ + j sin θ )n where n is a positive integer.. o.

(36) Chapter One. 27 The method is simply to expand the right hand side as a binomial. series, after which we can equate real and imaginary parts. An example will soon show you how it is done. Example 19 To find expansions for cos 3θ and sin 3θ Solution: We have cos 3θ + j sin3θ = (cosθ + j sinθ )3 = (c + js)3. Where c = cos θ and s = sin θ just for simplicity. Now expand this by the binomial series so that:. cos 3θ + j sin 3θ = (cosθ + j sin θ )3 = c 3 + 3c 2 ( js ) + 3c( js )2 + ( js )3. cos 3θ + j sin 3θ = c 3 + j 3c 2 s − 3cs 2 − js 3. (. ) (. cos 3θ + j sin 3θ = c 3 − 3cs 2 + j 3c 2 s − s 3. ). Now equating real parts and imaginary parts we get:. cos 3θ = cos3 θ − 3 cos θ sin 2 θ sin 3θ = 3 cos 2 θ sin θ − sin 3 θ If. we. wish,. (. we. ). can. replace. (. sin 2 θ = 1 − cos 2 θ. ). and. cos 2 θ = 1 − sin 2 θ . So that we could write the results above as following: cos 3θ = 4 cos3 θ − 3 cos θ , ∴ sin 3θ = 3 sin θ − 4 sin 3 θ . While these results are useful, it is really the method that counts. So now do this one in just the same way as done before, obtain an expansion for cos 4θ in terms of cos θ .. cos 4θ + j sin 4θ = (cos θ + j sin θ )4 = (c + js )4.

(37) 28Mathematical Numbers. = c 4 + 4c 3 ( js ) + 6c 2 ( js )2 + 4c( js )3 + ( js )4 = c 4 + j 4c 3 s − 6c 2 s 2 − j 4cs 3 + s 4. (. ) (. = c 4 − 6c 2 s 2 + s 4 + j 4c 3 s − 4cs 3 Equating real parts:. (. ∴ cos 4θ = c 4 − 6c 2 s 2 + s 4. (. ). ) (. = c 4 − 6c 2 1 − c 2 + 1 − c 2. ). )2. = c 4 − 6c 2 + 6c 4 + 1 − 2c 2 + c 4 = 8c 4 − 8c 2 + 1 = 8 cos 4 θ − 8 cos 2 θ + 1. (. ). (. Similarly, sin 4θ = 4c 3 s − 4cs 3 = 4cs c 2 − s 2. ). ( ) = 4cs (1 − s 2 ) = 4cs * c 2 = 4c 3 s = 4cs c 2 − s 2 − c 2 − s 2 + 1. ∴ sin 4θ = 4 cos3 θ * sin θ Expansions for cos n θ and sin n θ in terms of sines and cosines of muiltiples of θ .. 1 = z −1 = cos θ − j sin θ z 1 1 ∴ z + = 2 cos θ and , z − = j 2 sin θ z z Also by DeMoivre's theorem z n = cos nθ + j sin nθ. z = cos θ + j sin θ , ∴.

(38) Chapter One. 29. 1. = cos nθ − j sin nθ zn 1 1 ∴ z n + n = 2 cos nθ And, z n − = j 2 sin nθ z zn And,. Let us collect theses four results together: z = cos θ + j sin θ. 1 = 2 cos θ z 1 z n + n = 2 cos nθ z. z+. 1 = j 2 sin θ z 1 z n − n = j 2 sin nθ z z−. Example 20 Expand cos 3 θ Solution: From the previous results, 3. 1 1  Q z + = 2 cos θ ∴  z +  = (2 cos θ )3 z z   1  1 1 ∴ (2 cos θ )3 = z 3 + 3z 2   + 3z  2  + 3 z z  z 1 1 = z 3 + 3z + 3 + 3 z z Now here is the trick: we rewrite this, collecting the terms up in pairs from the two extreme ends, thus:. (2 cos θ )3 =  z 3 + . 1   1 + + 3 z    z z3  . But from the previous results. ∴z +. 1 1 = 2 cos θ , and z 3 + = 2 cos 3θ 3 z z.

(39) 30Mathematical Numbers. ∴ (2 cos θ )3 = 2 cos 3θ + 3 * 2 cos θ. ∴ 8 cos3 θ = 2 cos 3θ + 6 cos θ 1 ∴ cos3 θ = (cos 3θ + 3 cos θ ) 4 Example 21 Expand sin 4 θ Solution: 1 1 Q z − = j 2 sin θ , and, z n − n = j 2 sin nθ z z. ∴ ( j 2 sin θ ). 4. 1  = z −  z . 4.  1   1  1 1 = z 4 − 4 z 3   + 6 z 2  2  − 4 z 3  + 4 z z  z  z 1   1   =  z 4 + 4  − 4 z 2 + 2  + 6  z   z  Now, Q z n +. 1 zn. = 2 cos nθ. 1   1   ∴ ( j 2 sin θ )4 =  z 4 + 4  − 4 z 2 + 2  + 6  z   z  = 2 cos 4θ − 4 * 2 cos 2θ + 6 ∴ 16 sin 4 θ = 2 cos 4θ − 4 * 2 cos 2θ + 6. ∴ sin 4 θ =. 1 (cos 4θ − 4 cos 2θ + 3) 8.

(40) Chapter One. 31. 1.6.11 The Roots of Unity. The problem here is to solve the equation z n = 1 , where n is usually a positive whole number. Write both sides of the equation in polar form. Let z have polar form. z = r (cos θ + j sin θ ) ∴ z n = r n (cos nθ + j sin nθ ) We know that 1 = 1(cos 0 + j sin 0 ) . So our equation becomes:. ∴ z n = r n (cos nθ + j sin nθ ) = 1 * (cos 0 + j sin 0) Now two complex numbers in standard polar form are equal if and only if their modulus and arguments are equal. In the case of the argument this statement has to be handled with care. It means are equal if reduced to the proper range. So, for example 10o and 370 o count as equal from this point of view. So we can say that r n = 1 and that nθ and 0 are equal up to the addition of some multiple of. 2π radians. r n = 1. nθ = 0 + 2kπ Where k is some whole. number. Since r is real and positive, the only possibility for r is r= 1. The other equation gives us: θ = 0 + 2π. k n. This, in principle, gives us infinitely many answers one for each possible whole number k. But not all the answers are different. Remember that changing the angle by 2π does not change the number z. The distinct solutions, of which there are n, are given by r = 1and.

(41) 32Mathematical Numbers. θ = 2π. k , n. k = 0, 1, 2, 3,.........n − 1. and we can write these. solutions as following: z k = cos θ k + j sin θ k Where θ = 2π. k , n. k = 0, 1, 2, 3,.........n − 1. That looks rather complicated. It becomes a lot simpler if you think in terms of the Argand diagram. All the solutions have modulus 1 and so lie on the circle of radius 1 centered at the origin. The solution with k = 1 is just z = 1. The other solutions are just n − 1 other points equally spaced round this circle, with angle 2π / n between one and the next. This is shown in Fig.18 for n = 17 .. Fig.18 The n roots of 1. Let's look at some specific examples. The cube roots of unity are the solutions to z 3 = 1 . There are three of them and they are:. zo = 1,. z1 = cos 2π / 3 + j sin 2π / 3,. z 2 = cos 4π / 3 + j sin 4π / 3.

(42) Chapter One. 33. Fig.19: The three cube roots of 1. Note that z 2 = z1 , z 2 = z12 and 1 + z1 + z 2 = 0 . The roots are shown in Fig.19. Similarly the fourth roots of unity are the solutions of z 4 = 1 and these are: z = 1 , z = j , z = −1 , and z = − j A picture for n = 4 together with those for n = 5 and n= 6 is given in Fig.20.. Fig.20 The nth roots of 1 for n = 4;5;6. We can do other equations like this in much the same way..

(43) 34Mathematical Numbers Example 22 Find the solutions of the equation z 4 = j . Solution:. Put z = r (cos θ + j sin θ ) . Then z 4 = r 4 (cos 4θ + j sin 4θ ) . We know that: j = 1(cos π / 2 + j sin π / 2 ) . So our equation becomes:. z 4 = r 4 (cos 4θ + j sin 4θ ) = 1(cos π / 2 + j sin π / 2 ) Therefore; r = 1 and 4θ =. π 2. + 2kπ. or θ =. π 8. +. kπ 2. There are 4 distinct solutions, given by k = 0;1;2;3. They form a square on the unit circle. 1.7 Polynomials. We have learned how to manipulate complex numbers, and suggested that they will prove valuable in engineering calculations. The original motivation for introducing them was to give the equation x 2 = −1 two roots, namely j and − j , rather than it having no roots. It turns out that this is all we have to do to ensure that every polynomial has the right number of roots. We now discuss this, and a number of other basic results about polynomials that are quite useful to know. A polynomial in x is a function of the form:. p ( x ) = an x n + an −1 x n −1 + .... + a1 x + ao.

(44) Chapter One. 35. where the a's are (real or complex) numbers and an ≠ 0 . For example: p ( x ) = x 3 − 2 x + 4,. q (t ) = 5t 8 − t 4 + 6t 3 − 1. The highest power in the polynomial is called the degree of the polynomial. The above examples have degrees 3 and 8. A number a (real or complex) is said to be a root of the polynomial. p( x ) if p (a ) = 0 . Thus x = 1 is a root of x 2 − 2 x + 1 = 0 The first important result about polynomials is that a number a (real or complex) is a root of the polynomial p ( x ) if and only if ( x − a ) is a factor of p(x), in the sense that we can write p ( x ) as:. p ( x ) = ( x − a )q( x ) . Where q( x ) is another polynomial. This result is often called the remainder theorem . For example, x = 2 is a root of p ( x ) = x 3 + x 2 − 7 x + 2 and it turns out that. (. ). p(x ) = (x − 2) x 2 + 3x − 1. Note that necessarily the polynomial q has degree one less than the degree of p. It may be the case that you can pull more than one factor of x − a out of the polynomial. For example, 2 is a root of. p ( x ) = x 3 − x 2 − 8 x + 12 and it turns out that p ( x ) = ( x − 2 )( x − 2 )( x + 3) In such cases a is said to be a multiple root of p ( x ) . The multiplicity of the root is the number of factors ( x − a ) that you can take out. In the above example, 2 is a root of multiplicity 2, or a double root. A.

(45) 36Mathematical Numbers root is called a simple root if it produces only one factor. Multiple. roots are a considerable pain in the neck in many applications. There is a simple test for multiplicity. Suppose a is a root of. p( x ) , so that p (a ) = 0 . If, in addition, p′(a ) = 0 (derivative) then a is a multiple root. To take the above example:. Q p ( x ) = x 3 − x 2 − 8 x + 12 ∴ p′( x ) = 3 x 2 − 2 x − 8. and. p (2 ) = 0. and we have p′(2 ) = 0 , so we know that 2 is a multiple root. 2.9. Theorem (Fundamental Theorem of Algebra).. Let p be any polynomial of degree n. Then p can be factored into a product of a constant and n factors of the form ( x − a ) , where a may be real or complex. Also, the factorization is unique; you cannot find two essentially different factorizations for the same polynomial. The factors need not all be different because of multiple roots. The fact that there cannot be more than n such factors is fairly obvious, since we would have the wrong degree. What is not at all obvious is that we have all the factors that we want. Note that this result does not tell you how to find these factors; just that they must be there! The result is often stated loosely as: a polynomial of degree n must have exactly n roots. You have to allow complex roots or the theorem is not true. For example p ( x ) = x 2 + 1 has no real roots at.

(46) Chapter One. 37. all. Its roots are x = ± j and it factorizes as p ( x ) = ( x − j )( x + j ) . In fact, if ω ≠ 0 then p ( z ) = z n − ω. (n ≥ 1). always has exactly n. distinct roots because we know that it must have n roots in all and it cannot have any multiple roots because p′( z ) = nz n −1 has only 0 as a root and 0 is not a root of p ( z ) . There is one other result about roots of polynomials that is worth knowing. Suppose we have a polynomial with real , as opposed to complex, coefficients. Suppose that the complex number z is a root of the polynomial. Then the complex conjugate z is also a root. So you get two roots for the price of one. You can see this in the example of the previous paragraph. x 2 + 1 has j as a root, so it automatically must have − j as a root as well. Example 23 Let p ( z ) = z 4 − 4 z 3 + 9 z 2 − 16 z + 20 . Given that 2 +. j is a root, express p( z ) as a product of real quadratic factors and list all four roots, drawing attention to any conjugate pairs. Solution:. Since p has real coefficients, and complex roots occur in pairs consisting of a root and its complex conjugate. Given that 2 + j is a root, it follows that 2 − j must also be a root, and so the quadratic:.

(47) 38Mathematical Numbers. (z − (2 + j ))(z − (2 − j )) = z 2 − 4 z + 5. must be a factor. Dividing. the given polynomial by this factor gives. (. )(. p ( z ) = z 4 − 4 z 3 + 9 z 2 − 16 z + 20 = z 2 − 4 z + 5 z 2 + 4. ). The roots of z 2 + 4 are 2j and its complex conjugate, − 2 j . Thus the given polynomial, of degree four, has two pairs of complex conjugate roots. Example 24 Express z 5 − 1 as a product of real linear and quadratic. factors. Solution:. We rely on our knowledge of the nth roots of unity from the previous section. Let.  2π   2π   2π   = cos  + j sin    5   5   5 . α = exp j. Then the roots of z 5 − 1 = 0 are α , α 2 , α 3 , α 4 , and, 1.. (. ). (. )(. )(. z 5 − 1 = ( z − 1) z 4 + z 3 + z 2 + z + 1 = ( z − 1)( z − α ) z − α 2 z − α 3 z − α 4. ). For convenience, write β = α 2 , and note that β = α 3 while. (. ). α = α 4 . Our problem is to factorize z 4 + z 3 + z 2 + z + 1 as a product of real quadratic factors. We know the roots are. α ,α , β , and β . Now construct the quadratic with roots α and α . We have: ( z − α )( z − α ) = z 2 − (α + α ) + αα = z 2 − 2ℜ(α ) + 1.

(48) Chapter One. 39. where ℜ(α ) is the real part of α . Since ( z − β )(z − β ) behaves in the same way, we have:. (. )(. ). z 5 − 1 = ( z − 1) z 2 − 2ℜ(α ) + 1 z 2 − 2ℜ(β ) + 1.   2π    4π   ∴ z 5 − 1 = ( z − 1) z 2 − 2 cos  + 1 z 2 − 2 cos  + 1  5    5    and this is a product of real linear and quadratic factors. Problems. ١) Express the complex number z = 1 − 3 j exactly in modulus - argument form. Hence find the modulus and principal argument of z 4 . ٢) Find all solutions w to the equation ω 3 = −27 j and mark them on an Argand diagram. ٣) Let z = 1 − j 2 ω = 3 + j be complex numbers. Express each of the following complex numbers in the rectangular form zω ,. ω z+2+ j. ,. 1 + 3 j − zz. ٤) Express the complex number 2 + 2 j exactly in. modulus - argument form. Hence find all solutions. w to the equation ω 3 = −2 + 2 j and mark them on an Argand diagram..

(49) 40Mathematical Numbers. ٥) Let z = 3 + j and ω = 1 − 7 j . Express rectangular form. Find also z ,. ω ω+z. in a. ω. ω,. z. ٦) Express the complex number − 2 + 2 j in polar form. Hence solve the equation. z 3 = −2 + 2 j. expressing the solutions in polar form and marking them in the Argand Diagram. ٧) Let p ( z ) = z 5 − 5 z 4 + 8 z 3 − 2 z 2 − 8 z + 8 Show that p (2 ) = 0 . Show also that z 2 − 2 z + 2 is a factor of p ( z ) . Hence write p as a product of linear factors. ٨) Show that z − (1 + j ) is a factor of the real polynomial p ( z ) = z 3 + 2 z 2 − 6 z + 8 Hence write p as a product of linear factors. ٩) Let p ( z ) = z 4 − 3 z 3 + 5 z 2 − 27 z − 36 Show that. p (3 j ) = 0 . Hence write p as a product of linear factors. ١٠). Express in polar z = −5 − j 3. ١١). Express in rectangular 2∠156 o and 5∠ − 37 o. ١٢). If z1 = 12 cos 125o + j sin 125o and. (. ).

(50) Chapter One. (. 41. ). z 2 = 5 cos 72 o + j sin 72 o Then, find (i) z1 * z 2 and results in polar form. (. z1 giving the z2. ). If z1 = 12 cos 125o + j sin 125o , find z 3 and. ١٣) 1 z3. ١٤). If z = x + jy , find the equations of the two loci. defined by: (i) z − 4 = 3 and (ii) arg( z + 2 ) = ١٥). π 6. If z = x + jy , find the value of x and y when :. 3z 3z 4 + = 1− j j 3− j ١٦). Express 2 + j 3 and 1 − j 2 in polar form and. apply DeMoiver’s theorem to evaluate. (2 + j 3)4 . 1 − j2. Express the result in rectangular and exponential form. ١٧). Find the fifth roots of − 3 + j 3 in polar and. exponential form. ١٨). Express 5 + j12 in polar form and hence. evaluate the principle value of 3 (5 + j12 ) giving the results in rectangular form..

(51) 42Mathematical Numbers ١٩) Obtain the expansion of sin 7θ in terms of. sin θ ..

(52) Chapter 2 Matrices 2.1 Introduction A matrix is, by definition, a rectangular array of numeric or algebraic quantities, which are subject to mathematical operations. So a real matrix is an arrangement of real numbers into rows and columns. Matrices can be defined in terms of their dimensions (number of rows and columns). Let us take a look at a matrix with 4 rows and 3 columns (we denote it as a 4x3 matrix and call it A):. 7 5 A = 2  9. 6 8 12 5. 1 1  0  0. The dimensions of this matrix are 4 by 3. The dimensions of a matrix tell you the size of the matrix because they tell you the number of rows and columns in the matrix. By convention, we list the number of rows before the number of columns. Definition 1 The dimensions of a matrix are the number of rows and columns (listed in that order) of the matrix. Each element of the matrix is named according to its position. Typically, capital letters represent matrices and small letters with subscripts represent elements in the matrix. Since vectors can be.

(53) Chapter Two. 43 considered to be matrices with only one row or one column, they could be labeled with capital letters also. However, small letters usually represents vectors. The element 6 is in the position a12 (read a one two) because it is in row 1 and column 2. Also by convention, we list the row number of the element before the column number. An element in row i and column j would be denoted by aij . This gives us a compact way to refer to specific elements of a matrix. Can you represent the same information as before in a 3 by 4 matrix? Yes, you can. It would look like the matrix B which follows.. 7 B = 6  1. 5. 2. 8 1. 12 0. 9 5  0. Matrix B is the transpose of A, and A is the transpose of B. Transposing a matrix results in writing the columns as rows and the rows as columns, but what really happens is that element aij is placed in the position b ji of the new matrix. Therefore, a12 moves to the position b12 when we form the transpose of A. The transpose of A is denoted by AT (read A transpose). Therefore, matrix B is. AT . Definition 2 By the transpose of the m by n matrix A, denoted by. AT , we mean the n by m matrix, which has aij as its (i, j )th element..

(54) 44 Matrices Definition 3 We say that two m by n matrices, A and B are equal if their corresponding elements are equal. In other words, A = B if A and B have the same dimensions and. a11 = b11 , a12 = b12 , etc. Is A = AT ? Usually not, but we have a special word for a matrix which satisfies A = AT . Definition 4 A matrix is said to be symmetric if A = AT . Observe that the following matrix is symmetric:. 9 2 A= 5  1. 2 5 1 7 0 8  0 4 6  8 6 3. Notice that aij = a ji for all i and j; as is true for all symmetric matrices. Symmetric matrices are easy to spot because if you draw a line down the main diagonal (from 9 to 3 in this matrix), then the two halves are mirror images of each other. Symmetric matrices have many special qualities that will be used when you study matrices in more detail. The matrix A, given above, has another special property; it is a square matrix because A has the same number of rows as columns. Notice that A is a 4 by 4 square matrix. We said that the main diagonal for A runs from 9 to 3. For any square matrix, the main diagonal runs from the upper left corner to the lower right corner. Definition 5 We say that an m by n matrix is square if m = n ..

(55) Chapter Two. 45. 2.2 Addition and Subtractions of Matrices Definition 6 Matrices of the same dimensions are added by adding corresponding elements. For instance, aij corresponds to bij because they both lie in the ith row and jth column of their respective matrices. Therefore, we would add, aij + bij to obtain the (i, j ) th element of A + B : Example 1 Find the result of the following:. 7 5 A+ B = 2  9 Solution:. 6 8 12 5. 1 8 1 9 + 0 5   0 11. 6 6 9 4. 1 0  1  0. 6+6 1 + 1  15 12 2 7+8 5+9 8+6 1 + 0  14 14 1 =  A+ B =  2+5 12 + 9 0 + 1  7 21 1     5+4 0 + 0 20 9 0 9 + 11 Think about the similarities between addition and subtraction. How do you think matrices are subtracted? Definition 7 Matrices of the same dimensions are subtracted by subtracting corresponding elements. 2.3 Multiplication of Matrices Multiplying a matrix by a scalar value involves multiplying every element of the matrix by that value. Here we multiply our 4x3 matrix A by a scalar value k:.

(56) 46 Matrices. 7 5 k * A =k * 2  9 The multiplication. 1 k * 7 k *6 1  k * 5 k *8 = 12 0 k * 2 k *12   5 0  k * 9 k *5 operation on matrices differs 6 8. k *1 k *1  k * 0  k * 0 significantly from. its real counterpart. One major difference is that multiplication can be performed on matrices with different dimensions. The first restriction is that the first matrix has to have the same amount of columns as the second has rows. The reason for this will become clear shortly. Another thing to note is that matrix multiplication is not commutative i.e, (CD) does not equal (DC). The procedure for matrix multiplication is rather simple. First, we determine the dimensions of the resultant matrix. All we require is that there are as many columns in the first matrix as there are rows in the second. A simple way of determining is to look at the nearest and farthest dimensions of two matrix symbols written next to each other, for instance: C[2x3] D[3x2]. The nearest dimensions are both equal to 3, and so we know that the operation is possible. The farthest dimensions will give us the dimensions of the product matrix, so our result will be a 2x2 matrix. The general rule says that in order to perform the multiplication AB, where A is a mxn matrix and B a kxl matrix, we must have n=k. The result will be a mxl matrix..

(57) Chapter Two. 47 Performing the operation product involves multiplying the cells. of a particular rows in the first matrix by the cells of a particular column in the second matrix, adding the products, and storing the result in the cell of the resultant matrix whose coordinates correspond to the row of the first matrix and the column of the second matrix. For instance, in AB = C, if we want to find the value of c12, we must multiply the cells of row 1 in the first matrix by the cells of column 2 in the second matrix and sum the results. There are several interesting things to notice about matrix multiplication. We multiplied a 1 by 3 matrix by a 3 by 4 matrix and got a 1 by 4 matrix. The following picture expresses the requirements on the dimensions:. Let's also look closely at how we multiply the matrices because we will multiply matrices with larger dimensions later. This is a hands on activity. Take your left pointer finger and place it at the beginning of the first row of the first matrix (the only row we have in this case). Take your right pointer finger and place it on the first number of the first column of the second matrix. Multiply the two numbers to which you are pointing. Each time you move, your left.

(58) 48 Matrices hand will go across the row, and your right hand will go down the column. When you reach the end of the row and column, add the numbers you have obtained from the multiplications. This number goes in the first row and first column of your product matrix. This is the same as taking the inner product of the first row of first matrix and the first column of the second matrix. Now you can move to the first row, second column doing the same thing. This number will go in the first row, second column of your product matrix. In short, position ij of your product matrix consists of the inner product of the ith row of your first matrix and the jth column of the second matrix. This is a lot easier to do than it is to describe! Your left hand will move across and your right hand will move down. Do this for every row and column combination to get your product matrix.. This. picture depicts the motions necessary to find a product: Inner product of row i with column j equals position ij Definition 8 An identity matrix is a square matrix with ones along the main diagonal and zeros elsewhere. Example 2.   2  1 If S = [1 4 3] And R = 2  4  2 . 2 0 1. 1 3 4 1 1 3.  0   0  Find S * R  2  .

(59) Chapter Two. 49. Solution:.  2 1  2  1 3 S * R = [1 4 3] 2 0 4  4 1  2 1 1  3 1 Column 1 of S , R = 1 * 2 + 4 * 2 + 3 * 2 = 17 4 Column 2 of S , R. = 1 * 2 + 4 * 0 + 3 *1 = 5. Column 3 of S , R. 3 1 = 1 *1 + 4 * + 3 *1 = 8 3 4. Column 4 of S , R. = 1* 0 + 4 * 0 + 3 * 2 = 6.  0   0   2  . ∴ S * R = [17 5 8 6] Example 3 Multiply the following matrices.   2  1 2  4  2 . 2 0 1. 1 3 4 1 1 3.  0  17  8  0 *   13   2  4      2 * 17 + 2 * 8 + 1 * 13 + 0 * 4    63      1 3 = 2 * 17 + 0 * 8 + * 13 + 0 * 4 = 48    4 4  67 1   1 2 * 17 + 1 * 8 + 1 * 13 + 2 * 4   3  3  .

(60) 50 Matrices Example 4 Multiply the following matrices.   2  1 R * F = 2  4  2 . 2 0 1. 1 3 4 1 1 3.  0   17  8 0    13  2  4 .   510  63 1250  70    =  48 1215 90      1 120  67 1450   3. 2.4 Equations Solving equations is an important part of mathematics. If we are working with more than one unknown at a time, we need to solve systems of equations. You may already know how to solve a system of linear equations, but matrices provide a more compact way to arrive at the solution. Matrices are also easier to manipulate on a computer or calculator. Both of these facts will become more important when you work with larger systems. Example 5 Solve the following system of equations:. 5 x1 + 3 x2 = 93 − 4 x1 − 2 x2 = −66 Solution: Let's look at a system of linear equations:. 5 x1 + 3 x2 = 93 − 4 x1 − 2 x2 = −66 Can be written in matrix form as AX = B where.

(61) Chapter Two.  5 A= − 4. 51. 3  x1   93  ; X =   , and B =    −2 − 66  x2 . When you learned to solve systems of linear equations, you learned that (a) You arrive at the same solution no matter which equation you write first, (b) The solution doesn't change if you multiply an equation by a scalar other than zero, and, (c) You can replace an equation with the sum of that equation and another equation without changing the solution. These may not be exactly the words you used when you were solving a system of linear equations, but you did all these things. Experiment with the system above to convince yourself that these statements are true. We can also solve this system entirely in matrix form. We use the same rules, and we call them Elementary Row Operations (EROs). The EROs tell us that we can (a) Interchange any two rows; (b) Multiply any row by a non-zero scalar; and (c) Replace any row by the sum of that row and any other row. Proper use of EROs will leave us with a system that has the same solution as our original system, but is much easier to solve. If you were presented the system. x1 = a, x2 = b.

(62) 52 Matrices You would be able to solve it instantly because you only have to read the solution. If this system were written using matrix notation, it would look like this:. 1 0 . 0  x1  a  The matrix = 1   x2  b . 1 0 . 0 is the 2 by 2 identity 1 . matrix. Because you can just read of the solution when a system is in this form, our first goal is to transform our system into this form. Let's solve the system above using matrices. We can represent this entire system with a 2 by 3 matrix, which looks like this:.  5  − 4. 93   . This is called an augmented matrix because we −2 − 66 3. combined 2 matrices (a matrix and a vector for this system). In this case, we combined the 2 by 2 coefficient matrix which is made of the coefficients for our unknowns and the 2 by 1 matrix from the right-hand side of the equations into one 2 by 3 matrix. In other words, we put A to the left of the bar and put b to the right of the bar. The application of an ERO to the augmented matrix does not change the solution set of the linear system that the augmented matrix represents because whatever you do to the left side of an equation, you also do to the right side. Therefore, we will arrive at the same solution whether we use augmented matrices or not, and augmented matrices are more compact to write. Using matrix notation, our goal is to transform our system into one that looks like the following form:.

(63) Chapter Two.  1   0. 53. 0 a    1  b  . In other words, we want the identity matrix to the left of the bar and the solution to the right of the bar. Remark 1 The bar is not a formal part of the matrix, so it is not necessary. It is placed there so that we can refer to the different parts of the augmented matrix and easily move back and forth between the augmented matrix and the linear system that it represents. In this book, r1 represents row 1 and so on..  5  − 4. 93   Original augment matrix −2 − 66. 1  − 4. 0.6 18.6   r1 ÷ 5 −2 − 66. 1  0. 0.6 18.6  4 r1 + r2 0. 4 8. 4 . 1  0. 0.6 18.6  1 21 . 1  0. 0 6  1 21. 3. r2 ÷ 0.4 −0.6 * r2 + r1. When we convert this from augmented matrix notation back to the algebraic notation for a system of equations, it looks like this:.

(64) 54 Matrices. 1x1 + 0 x2 = 6 0 x1 + 1x2 = 21 This tells us that x1 = 6 and x2 = 21 . Substitute this solution into the system to assure yourself that we are correct. If we systematically use elementary row operations (ERO) to obtain the identity matrix to the left of the bar, we call this the Gauss Jordan Elimination Method. Example 6 Now, let's solve the system using Gauss Jordan elimination.. 5 x1 + 3 x2 = 70 − 4 x1 − 2 x2 = −56  5  − 4. 70   Original augmented matrix. −2 − 56. 1  − 4. 0.6 14   −2 − 56. 1  0 1  0. 0.6 14  0. 4 0  0.6 14  1 0. 3. r1 ÷ 5. 4 * r1 + r2 r2 ÷ 4. 0 14 1 − 0.6r2 + r1   0 1 0   Let's look at the scalar version of this equation, ax = b ; to help us find a general method for matrices. We know that x = a −1b if.

(65) Chapter Two. 55. A ≠ 0 because a −1 = 1 / a where a −1 is called the multiplicative inverse or the reciprocal. There is something analogous to this with matrices. It is also called the inverse. With scalars, a −1a = aa −1 = 1. Definition 9 The matrix A−1 (called A inverse) is the inverse of a square matrix A if A −1 A = AA−1 = I where I is the identity matrix. Once we find A1; Ax = b can be solved by matrix multiplication rather than Gauss Jordan elimination. We follow the algebraic steps below to find an expression for x:. Ax = b. ∴ A−1 Ax = A−1 b. ∴ I * x = A−1 b. This means that if we find A−1 ; we only need to multiply to solve systems with the same matrix A for different b vectors. Please remember that A−1b ≠ b A−1 , so you must multiply in the correct order. Remark 2 In computational mathematics, the inverse is very seldom found because other methods exist that serve the same purpose and require fewer steps. However, the inverse will serve our needs at this level and is important in the theory of matrices. Example 7 Using the Gauss Jordan elimination method, let's find. A−1. 0 2 4 where A 4 2 3   1 3 6. Solution:.

(66) 56 Matrices. 0 2 4 1 0 0   4 2 3 0 1 0   Original augmented matrix. 1 3 6 0 0 1. Switch r1 and r3 because we cannot have a zero on the main diagonal, and we would prefer 1 rather 4.. 1  0 0 1  0 0 1  0 0 1  0 0. 3 6 0 0 1   − 10 − 21 0 1 − 4 − 4r1 + r2 2 4 1 0 0  3 6 0 0 1   1 2.1 0 − 0.1 0.4 r2 / (− 10 ) 2 4 1 0 0  1   1 2.1 0 − 0.1 0.4  − 2r2 + r3 0 − 0.2 1 0.2 − 0.8 3. 6. 0. 0 1   − 0.1 0.4 r3 / (− 0.2 ) 1 − 5 −1 4 . 3 6 0 1 2.1 0 0. 0.

(67) Chapter 3 Calculus 3.1 Limits The concept of limits is essential to calculus. A good understanding of limits will help explain many theories in calculus. So, it is recommended to start studying calculus from limits. Consider a function f defined for values of x, as x gets close to a number a, not necessarily true for x = a . If the value of f ( x ) approaches a number b as x approaches a, then the limit of f ( x ) as x approaches a is equal to b, denoted as :. lim f ( x) = b. (1). x→a. Example 1 Find the limit of f ( x) = 5 x + 2 as x approaches 3. Solution: It is clear that as x approaches 3, 5x approaches 15, and. 5 x + 2 approaches 17. Thus; lim 5 x + 2 = 17 x →3. Example 2 Find the limits of f ( x) =. 1 as x approaches 5. 2 x − 10. Solution: It is clear as x approach 5, 2 x − 10 approaches zero the. 1 1 approaches which is undefined. Thus; 2 x − 10 0 1 = ∞ (undefiend ) x → 5 2 x − 10 lim.

(68) Chapter Three. 75. This limit lim f ( x) = f ( x) represents a horizontal line, which x→a. says that as x approaches a, and f ( x) = f ( x) or c = c where c is a constant. Then, lim f ( x) = c x→a. Then, as x approaches a, f ( x ) also approaches c. Limits can be approached from the negative ( or left ) or the positive ( or right ) side of a number denoted as:. lim f ( x) = b or lim f ( x) = b. x→a −. x→a +. If lim f ( x) ≠ lim f ( x ) Then, x→a −. x→a +. lim f ( x) = does not exist. x→a. lim f ( x) = lim f ( x) = b. x→a −. x→a +. lim f ( x) = b. x→a. If the value of f ( x ) gets larger and larger without bound as x approaches a, then: lim f ( x) = ∞ x→a. Similarly; If the value of f ( x ) gets smaller and smaller without bound as x approaches a, then: lim f ( x) = ∞ x→a. Consider a function f defined for large positive ( or negative ) values of x, as x increases indefinitely in the positive ( or negative ) direction. If the value of f ( x ) approaches a number b as x increases.

(69) 76 Calculus. (or decreases ) indefinitely, then the limit of f ( x ) as x increases (or decreases ) indefinitely is equal to b, denoted as :. lim. x → +∞. f ( x) = b. or. lim. x → −∞. f ( x) = b. A function f ( x ) is continuous at x = a if f is defined at x = a and either; f is not defined anywhere near a, or f is defined arbitrarily near x = a and, lim f ( x) = f ( x) x→a. Conversely, A function f ( x ) is discontinuous at x = a if f ( x ) is defined at x = a and f ( x ) is not continuous at x = a . 3.2 Derivatives Suppose y = f (x) is shown in Fig.1, the slope of the curve is the slope of the secant line between point A and another point P on the graph is shown in the following equation:. m AP =. ( f (x + h ) − f (x )) = ( f (x + h ) − f (x )) (x + h ) − x h. Notice that h can change and with it the location of point P, therefore h is the limiting factor of the slope of the curve. As h gets close to point A, the slope of the curve becomes the tangent of the graph at point A. The tangent line of f at point A is: lim. ( f (x + h ) − f (x )). h →0. So,. the Differentiation ( f (x + h ) − f (x )) lim h h →0. of. function. h f. at. x. is:.

(70) Chapter Three. 77 If this limit exists, then it is called the derivative of function f. at x, which is denoted by f ′( x) or So, f ′( x) or. dy = lim dx h →0. dy . dx. ( f (x + h ) − f (x )) h. Fig.1 The Approximate slope of the curve at point A.. Fig.2 The slope of the curve at point A. So, general rules of differentiation are shown in the appendix of this book before going in the following example you have to take a look to the rules of differentiation in the appendix. Example 3 Find from the first principles. (. d tan ( x ) e dx. ).

(71) 78 Calculus Solution: Put u = tan ( x ) ∴ y = eu ∴ But from chain rule,. ∴. (. dy du = eu and = sec 2 x du dx. dy dy dy = . , dx du dx. ). d tan ( x ) e = e tan ( x ) * sec 2 x dx. Example 4 Find. d  3 2 − x   dx . Solution: 2. .  1. d  3 2 2  3 −1 2  − 3  =− x − x  = − x  3 3 dx  Example 5 Find. d  − 3  dx  x 2 − 1 . Solution: 1 3 − −  3 d  − 3  d  2 2 = − 3 * x − 1 2 = * x − 1 2 * 2x     2 2 dx  x − 1  dx  . (. (. ). ). 3x 3x * x 2 − 1 d  − 3  ∴ = = 2 dx  x 2 − 1  x 2 − 1 x 2 − 1 x2 − 1. (. Example 6 Solution:. Find. ). (. ). d (5 x + 7 )4 dx. d (5 x + 7 )4 = 4(5 x + 7)3 * 5 = 20(5 x + 7 )3 dx.

(72) Chapter Three. 79. Example 7 Find Solution:. d (sin (5 x + 6)) = 5 cos(5 x + 6) dx. Example 8 Find Solution:. ( ( )). d cos x 2 dx. ( ( )). ( ). d (ln(3 − 4 cos x )) dx. d 1 4 sin x (ln(3 − 4 cos x )) = * (4 sin x ) = dx 3 − 4 cos x 3 − 4 cos x. Example 10 Find Solution:. ( ). d cos x 2 = − sin x 2 * 2 x = −2 x * sin x 2 dx. Example 9 Find Solution:. d (sin (5 x + 6)) dx. d (log10 (2 x − 1)) dx. d 1 2 (log10 (2 x − 1)) = *2 = (2 x − 1) ln(10) (2 x − 1) ln(10) dx. Example 11 Find. (. ). d 5x e * ln (2 x − 1) dx. Solution: Assume y = e5 x * ln (2 x − 1) , u = e5 x and v = ln (2 x − 1). ∴ y = uv , ∴. dy dv du =u +v dx dx dx. 1 dy = e5 x * 2 + ln (2 x − 1) * 5 e5 x (2 x − 1) dx. dy 2e 5 x ∴ = + 5 e5 x ln (2 x − 1) dx (2 x − 1).

(73) 80 Calculus Example 12 Find. (. ). d 3 x 5 ln(sin x ) dx. Solution: Assume y = 3 x 5 ln (sin x ) , u = 3 x 5 and v = ln (sin x ). ∴ y = uv ,. dy dv du =u +v dx dx dx. (. ). (. ). ∴. d 1 * cos x + ln (sin x ) * 3 * 5 x 4 3x 5 ln (sin x ) = 3x 5 dx sin x. ∴. d 3x 5 ln (sin x ) = 3 x 5 cot x + 15 x 4 * ln (sin x ) dx. d  e 2 x  Example 13 Find dx  ln (3x )  e2x , u = e 2 x and v = ln(3 x ) Solution: Assume y = ln (3x ) dv du −u v dy = dx 2 dx dx v 3 2x − e2x * 2 x  ln (3 x ) * 2e  d e 3x = ∴  2   dx  ln (3 x )  (ln(3x )) 1 e 2 x (2 ln ( x ) + 2 ln (3) − ) e 2 x (2 x ln ( x ) + 2 x ln (3) − 1) x = = 2 (ln(3x )) x(ln (3 x ))2 d  x 2 sinh( 2 x)  Example 14 Find dx  cosh (3 x ) .

(74) Chapter Three. 81. x 2 sinh( 2 x) Solution: Assume y = , u = x 2 , v = sinh (2 x ) , cosh (3 x ) and w = cosh (3x ) d  uv  uv  1 du 1 dv 1 dw  Where + −  =  . dx  w  w  u dx v dx w dx  d  x 2 sinh( 2 x)  x 2 sinh( 2 x)  1 1 ∴ = *  2 * 2x + * 2 cosh (2 x ) dx  cosh (3 x )  cosh (3 x )  x sinh (2 x ) − ∴.  1 * 3 sinh (3 x ) cosh (3 x ) . d  x 2 sinh(2 x)  x 2 sinh(2 x)  2 2 cosh(2 x ) 3 sinh (3x )  = − * +  dx  cosh(3 x )  cosh(3 x )  x sinh (2 x ) cosh(3 x ) . d  x 2 sinh(2 x)  x 2 sinh(2 x)  2 2 cosh (2 x ) 3 sinh (3x )  * + ∴ = −  dx  cosh (3 x )  cosh (3x )  x sinh (2 x ) cosh (3x )  d  x 2 sinh( 2 x)  2 x sinh(2 x) x 2 2 cosh (2 x ) = + ∴ dx  cosh (3x )  cosh (3x ) cosh (3x ) 3x 2 sinh(2 x) tanh (3x ) − cosh (3x ) Example 15 Find. (. d 5 x sin 2 x cos 4 x dx. ). Solution: Assume y = uvw = x 5 sin 2 x cos 4 x , where u = x 5 , v = sin 2 x , and. w = cos 4 x Take the logarithm for both sides we get:.

(75) 82 Calculus. (. ) ( ). ∴ ln ( y ) = ln x 5 sin 2 x cos 4 x = ln x 5 + ln ( sin 2 x ) + ln ( cos 4 x ) By differentiating both sides of the above equation we get:. 1 dy 1 1 1 (− 4 sin 4 x ) 2 cos(2 x ) + = 5 5x 4 + ( cos 4 x ) sin 2 x y dx x 1 dy 5 ∴ = + 2 cot 2 x − 4 tan 4 x y dx x dy  5 ∴ = x 5 sin 2 x cos 4 x *  + 2 cot 2 x − 4 tan 4 x  dx  x ∴. Example 16 Find. d (1 + tan 2 x )3 dx. d (1 + tan 2 x )3 = 3(1 + tan 2 x )2 * d (1 + tan 2 x ) dx dx d d   ∴ (1 + tan 2 x )3 = 3(1 + tan 2 x )2 *  0 + sec 2 (2 x ) (2 x ) dx dx   d ∴ (1 + tan 2 x )3 = 3(1 + tan 2 x )2 * 2 * sec 2 (2 x ) dx d ∴ (1 + tan 2 x )3 = 6 sec 2 (2 x ) * (1 + tan 2 x )2 dx d  2  x   1 − cot Example 17 Find   dx   3 Solution:. (. ). Solution: 1   −1 2  x   2 . d  2  x   1  − 1 cot    = 1 − cot    dx   3 2  3 . *. d   x  1 − cot 2    dx   3 .

(76) Chapter Three. ∴. ∴. ∴. d  2  x   − 1 cot   = dx  3. 83.   x  d   x   0 − 2 cot   *  cot      3  dx   3       x  2 1 − cot 2     3  . d  dx .  x   1   x  − 2 cot  *  − csc 2      x  3   3  3  1 − cot 2    = 3   x  2 1 − cot 2     3  . d  dx .  x  x cot  * csc 2    x 3 3 1 − cot 2    =  3   x  3 * 1 − cot 2     3  . 3.2.1 Implicit differentiation So far, all the functions being differentiated are explicit functions, meaning that one of the variables was specifically given in terms of the other variable.. f ( x) = 5 x + 2, then f ′( x) = 5 However, not all functions are given explicitly and are only implied by an equation. Example 18 xy = 1 is an equation given implicitly, explicitly it is. y = 1 / x . Now to find dy / dx for xy = 1, simply solve for y and differentiate..

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