Algebra 101 essentials

1. 2.2. ALGEBRA 101 ✞ ans = 1 ✌ ✆ Again, changing the function Q2 = 2 − 3s, we have f(s) = 1 s3 ∗ (2 − 3s) Expanding the relation f(s) = 1 2s3 + 3 4s2 + 9 8s + 27 16 + 81 32 s + . . . The coefficient b1, i.e. the numeric coefficient of 1/s is called the residue of the function. Here, it is 1.125. ✞ --> s=poly (0,’s’); 2 --> P=1; --> Q1=s^3; 4 --> Q2 =(2 -3*s); --> residu(p, Q1 , Q2) ✌ ✆ ✞ ans = 1.125 ✌ ✆ 2.2.17 Roots of Polynomial (roots) Roots of a polynomial are those values which on substituted in place of variable gives polynomial value zero. A polynomial having degree of ‘n’ has n roots. Root values may be real numbers (either integers or fractions) or complex numbers. Roots of Quadratic Equation An equation of variable ‘x’ having degree ‘2’ is known as quadratic equation. Roots of the quadratic equation (Eg ax2 + bx + c = 0) can be obtained by using Shridharacharys Formula x = −b ± √ b2 − 4ac 2a Scilab example is given below ✞ --> x=poly (0,’x’) 2 --> roots(x^2 -3 * x + 2) ✌ ✆ ✞ ans = 2. 1. ✌ ✆ All the roots are real in integers. If equation is modified then roots becomes

2. 102 Linear Algebra ✞ 1 --> x=poly (0,’x’) --> roots(x^2 -3 * x + 3) ✌ ✆ ✞ ans = 1.5 + 0.8660254 i 1.5 - 0.8660254 i ✌ ✆ Roots of Polynomials Roots of higher degree polynomials can be obtained as ✞ 1 --> x=poly (0,’x’) --> roots(x^3 - 6 * x^2 + 11 * x -6) ✌ ✆ ✞ ans = 3. 2. 1. ✌ ✆ All are the real roots. roots function also calculates the complex roots. ✞ --> x=poly (0,’x’) 2 --> roots(x^3 - 6 * x^2 + 11 * x + 1) ✌ ✆ ✞ ans = 3.04 + 1.50 i 3.04 - 1.50 i - 0.08 ✌ ✆ roots returns all the roots of a polynomial function. Roots of a polynomial function are those possible values which satisfy to the given polynomial to zero. For example, x2 − 2x + 1 = 0 is a polynomial of degree ‘2’ and is known as quadratic equation. The roots of this polynomial are x = +1 and x = +1. When in place of ‘x’, 1 is put then polynomial value is zero. p(1) = 12 − 2 × 1 + 1 = 0 Roots of a polynomial is obtained either by fraction method or by substitution and re- duction method. ✞ --> p=poly ([1,2,3], ’s’) 2 --> roots(p) ✌ ✆ ✞ ans = 3. 2. 1. ✌ ✆

3. 2.2. ALGEBRA 103 2.2.18 Simplification (simp) simp simplifies rational function. It returns numerator & denominator as result. This functions takes two inputs as its arguments. First argument is numerator of algebraic function and second argument is denominator of the algebraic function. For example, assume a polynomial rational fraction f = (s + 1) ∗ (s + 2) (s + 1) ∗ (s − 2) On simplification of it, the result is f = s + 2 s − 2 Scilab code for this function is given below: ✞ --> s=poly (0,’s’); 2 --> [n,d]= simp ((s+1) *(s+2) ,(s+1)*(s-2)) ✌ ✆ ✞ d = - 2 + s n = 2 + s ✌ ✆ 2.2.19 Flip Matrix Dimension (flipdim) flipdim function flips the matrix with respect to the dimension as specified in the function. It is similar to the row or column interchange in the matrix during solving of a problem. Consider a matrix A as given below: A =

7. 1 2 3 4

11. When this matrix is flipped about first row, then flipped matrix becomes B as: B =

15. 3 4 1 2

19. See the example as given below: ✞ --> x=[1 2 3 4; 5 6 7 8] 2 --> dim =1; --> y=flipdim (x,dim) ✌ ✆ ✞ x = 1. 2. 3. 4. 5. 6. 7. 8. y = 5. 6. 7. 8. 1. 2. 3. 4. ✌ ✆

20. 104 Linear Algebra 2.2.20 Permutation (permute) In mathematics, the notion of permutation relates to the act of arranging all the mem- bers of a set into some sequence or order, or if the set is already ordered, rearranging (reordering) its elements, a process called permuting. permute permutes matrix elements or array elements into new dimensions of an array. ✞ --> x=[1 2 3;4 5 6]; 2 --> y=permute (x,[2 1]) ✌ ✆ ✞ y = 1. 4. 2. 5. 3. 6. ✌ ✆ 2.2.21 Matrix Replication (repmat) repmat replicates the given matrix and makes new matrix. This function has three arguments. First is the elements, second and third are the rows and columns of the matrix to be arranged respectively. First argument should be a list, array or vector. ✞ --> repmat (1:3 ,2 ,2) ✌ ✆ ✞ ans = 1. 2. 3. 1. 2. 3. 1. 2. 3. 1. 2. 3. ✌ ✆ 2.2.22 Cumulative Product (cumprod) It returns the cumulative product of array elements of a matrix. It takes two arguments, first, a matrix reference and second, cumulative sum orientation. Cumulative product orientation has three values. ‘*’ For the direction of cumulative product along-the-column-then-row by default. ‘1’ For the direction of cumulative product along-the-column. ‘2’ For the direction of cumulative product along-the-row. Assume a matrix which is given below and orientation argument in the cumprod function is omitted, then the cumulative product is performed along-the-column-then-row. A =

24. a11 a12 a21 a22

28. gives cumulative product B as B =

32. a11 a11 ∗ a21 ∗ a12 a11 ∗ a21 a11 ∗ a21 ∗ a12 ∗ a22

36. For a given matrix of size m × n, the cumulative product for an element aij is given by

37. 2.2. ALGEBRA 105 ✞ 1 /* Perform cumulative product columnwise for an element */ for (l = 0; l < n; l++) {// columns 3 if (l < j) { /* All row elements of previous columns */ 5 for (k = 0; k < m; k++) { r *= a[k][l]; 7 } } else if (l == j) { 9 /* Row element of current col at and before this row*/ for (k = 0; k <= i; k++) { 11 r *= a[k][l]; } 13 } } ✌ ✆ The example for cumulative product is given below: ✞ --> A=[1 ,2;3 ,4]; 2 --> a=cumprod (A) ✌ ✆ ✞ a = 1. 6. 4. 10. ✌ ✆ The same matrix which is given above is supplied to the cumprod function along with second argument ‘1’, then the cumulative product is performed along-the-column. A =

41. a11 a12 a21 a22

45. gives cumulative product B as B =

49. a11 a12 a11 ∗ a21 a12 ∗ a22

53. The example for cumulative product is given below: ✞ 1 --> A=[1 ,2;3 ,4]; --> b=cumprod (A,1) ✌ ✆ ✞ b = 1. 2. 4. 6. ✌ ✆ Second argument in the function cumprod is orientation of cumulative product. The possible values are ‘1’ for right orientation and ‘2’ for center orientation. Similarly, the same matrix is supplied to the cumprod function along with second argument ‘2’, then the cumulative product is performed along-the-rows. A =

57. a11 a12 a21 a22

62. 106 Linear Algebra gives cumulative product B as B =

66. a11 a11 ∗ a12 a21 a21 ∗ a22

70. 2.2.23 Cumulative Summation (cumsum) It returns the cumulative sum of array elements of a matrix. It takes two arguments, first, a matrix reference and second, cumulative summation orientation. Cumulative summation orientation has three values. ‘*’ For the direction of cumulative sum along-the-column-then-row by default. ‘1’ For the direction of cumulative sum along-the-column. ‘2’ For the direction of cumulative sum along-the-row. Assume a matrix which is given below, and orientation argument in the cumsum function is omitted, then the cumulative sum is performed along-the-column-then-row. A =

74. a11 a12 a21 a22

78. gives cumulative sum B as B =

82. a11 a11 + a21 + a12 a11 + a21 a11 + a21 + a12 + a22

86. For a given matrix of size m × n, the cumulative sum for an element aij is given by ✞ 1 /* Perform cumulative sum columnwise for an element */ for (l = 0; l < n; l++) {// columns 3 if (l < j) { /* All row elements of previous columns */ 5 for (k = 0; k < m; k++) { r += a[k][l]; 7 } } else if (l == j) { 9 /* Row elementd of current col at and before this row*/ for (k = 0; k <= i; k++) { 11 r += a[k][l]; } 13 } } ✌ ✆ The example for cumulative product is given below: ✞ --> A=[1 ,2;3 ,4]; 2 --> a=cumsum(A) ✌ ✆ ✞ a = 1. 6. 4. 10. ✌ ✆

87. 2.2. ALGEBRA 107 The same matrix which is given above is supplied to the cumsum function along with second argument ‘1’, then the cumulative sum is performed along-the-column. A =

91. a11 a12 a21 a22

99. a11 a12 a11 + a21 a12 + a22

103. The example for cumulative product is given below: ✞ 1 --> A=[1 ,2;3 ,4]; --> b=cumsum(A,1) ✌ ✆ ✞ b = 1. 2. 4. 6. ✌ ✆ Second argument in the function is orientation of cumulative sum. The possible values are ‘1’ for right orientation and ‘2’ for center orientation. Similarly, the same matrix is supplied to the cumsum function along with second argument ‘2’, then the cumulative sum is performed along-the-rows. A =

107. a11 a12 a21 a22

115. a11 a11 + a12 a21 a21 + a22

119. 2.2.24 Kronekar Product (kron) kron is a Kronekar product of a matrix. A matrix A of order 2 × 2 is A =

123. a11 a12 a21 a22

127. kron of this matrix is given by δ =

131. a11A a12A a21A a22A

135. ✞ 1 --> A=[1 ,2;3 ,4]; --> kron (A,A) ✌ ✆ ✞ ans = 1. 2. 2. 4. 3. 4. 6. 8. 3. 6. 4. 8. 9. 12. 12. 16. ✌ ✆

136. 108 Linear Algebra 2.2.25 Product (prod) prod returns the product of an array of elements of a matrix. Second argument to this function is the orientation of the product. If orientation argument is not provided then prod returns the product of all elements mutually. If product orientation argument is ‘1’ then product is performed along the column and if product orientation argument is ‘2’ then product is performed along the row. Product of a matrix, without orientation argument, is performed is shown below: A =

140. a11 a12 a21 a22

144. gives cumulative product as a = a11 ∗ a21 ∗ a12 ∗ a22 Product of a matrix, with orientation argument ‘1’, is performed is shown below: A =

148. a11 a12 a21 a22

152. gives cumulative product as a =

154. a11 ∗ a21 a12 ∗ a22

156. Product of a matrix, with orientation argument ‘2’, is performed is shown below: A =

160. a11 a12 a21 a22

164. gives cumulative product as a =

168. a11 ∗ a12 a21 ∗ a22

172. A Scilab example is given below: ✞ 1 --> A=[1 ,2;3 ,4]; --> a=prod (A) 3 --> b=prod (A,1) --> c=prod (A,2) ✌ ✆ ✞ a = 24. b = 3. 8. c = 2. 12. ✌ ✆

173. 2.2. ALGEBRA 109 2.2.26 Summation (sum) sum returns the sum of elements of a given matrix. Second argument to this function is the orientation of the summation. If summation orientation argument is ‘1’ then summation is performed along the column and if summation orientation argument is ‘2’ then summation is performed along the row. Summation of a matrix, without orientation argument, is performed is shown below: A =

177. a11 a12 a21 a22

181. gives cumulative summation as a = a11 + a21 + a12 + a22 Summation of a matrix, with orientation argument ‘1’, is performed is shown below: A =

185. a11 a12 a21 a22

189. gives cumulative summation as a =

191. a11 + a21 a12 + a22

193. Summation of a matrix, with orientation argument ‘2’, is performed is shown below: A =

197. a11 a12 a21 a22

201. gives cumulative summation as a =

205. a11 + a12 a21 + a22

209. A Scilab example is given below: ✞ 1 --> A=[1 ,2;3 ,4]; --> a=sum(A) 3 --> b=sum(A,1) --> c=sum(A,2) ✌ ✆ ✞ a = 10. b = 4. 6. c = 3. 7. ✌ ✆

210. 110 Linear Algebra 2.3 Matrices A matrix is arrangement of coefficients of a set of algebraic equations in rows and columns. Each row represents to a distinct algebraic equation while each column represents to the coefficients of same variable of each equation. For example, take a set of algebraic equations ax + by = p; cx + dy = q The coefficients of this set of algebraic equation shall be arranged in matrix form as a b c d x y = q q The word “matrix” refers only to the arrangement of coefficients of x and y variables only. So, matrix is a b c d First row has coefficients of equation ax + by = p and second row has coefficients of equation cx + dy = q. First column has coefficients of variable x and second column has coefficients of variable y. The number of rows is equal to the number of algebraic equation in the given set and number of columns is equal to the number of distinct variables. Numerical computational software have deep rooted application of matrices or matrix structure. Thus we must have familiarity with matrices, their operations and reading/writing matrix elements. Before, explaining the matrix operations, we will discuss here about the reading/writing of matrix elements. For this purpose we take experimental matrix A =   1 2 3 4 5 6 7 8 9   ✞ 1 -- A=[1 2 3; 4 5 6; 7 8 9] ✌ ✆ ✞ A = 1. 2. 3. 4. 5. 6. 7. 8. 9. ✌ ✆ Elements of matrix can be accessed by supplying indices in two ways. (i) have only one argument, i.e. element-wise and (ii) have two arguments for row and column indices each separated by comma. First argument is for rows and second argument is for columns. In Scilab, indices starts from ‘1’ not from ‘0’. For example, if only one argument is supplied as ✞ -- A(1:4) ✌ ✆ Then it will return elements column-wise, i.e. from top to bottom in first column and then second column and so on.

211. 2.3. MATRICES 111 ✞ ans = 1. 4. 7. 2. 5. ✌ ✆ If two arguments are supplied as ✞ -- A(1:1 ,2:2) ✌ ✆ Then it will return elements from row one to row one and from column two to column two. Thus it will only return element ‘2’. ✞ ans = 2. ✌ ✆ If column range is changed from column two to column three, then output will be changed. ✞ -- A(1:1 ,2:3) // First row and 2nd and 3rd columns ✌ ✆ ✞ ans = 2. 3. ✌ ✆ For row or column access, respective row or column index is provided and other argument is supplied as range operator ‘:’. ✞ -- A(1,:) // First row and all columns ✌ ✆ ✞ ans = 1. 2. 3. ✌ ✆ ✞ -- A(:,2) // All rows and second columns ✌ ✆ ✞ ans = 2. 5. 8. ✌ ✆ There are several types of matrices. The most common type of matrix is square matrix. In square matrix, rows and columns are of equal size. ✞ -- x=1:1:9; 2 -- matrix(x,3,3) ✌ ✆

212. 112 Linear Algebra ✞ ans = 1. 4. 7. 2. 5. 8. 3. 6. 9. ✌ ✆ Two or more matrices can be added, subtracted, multiplied and divided if matrices follow respective rules. If A, B and C are three matrices of order n × n then sum of all the matrices is Sij = Aij + Bij + Cij ✞ -- A=[1 ,2 ,3;4 ,5 ,6;7 ,8 ,9] 2 -- B=[5 ,6 ,7;8 ,9 ,10;11 ,12 ,13] -- C=[10 ,11 ,12;13 ,14 ,15;16 ,17 ,18] 4 -- S=A+B+C ✌ ✆ ✞ A = 1. 2. 3. 4. 5. 6. 7. 8. 9. B = 5. 6. 7. 8. 9. 10. 11. 12. 13. C = 10. 11. 12. 13. 14. 15. 16. 17. 18. S = 16. 19. 22. 25. 28. 31. 34. 37. 40. ✌ ✆ Similarly subtraction of the two matrices can be obtained. Dij = Aij − Bij ✞ -- A=[1 ,2 ,3;4 ,5 ,6;7 ,8 ,9] 2 -- B=[5 ,6 ,7;8 ,9 ,10;11 ,12 ,13] -- D=A-B ✌ ✆ ✞ A = 1. 2. 3. 4. 5. 6. 7. 8. 9. B = 5. 6. 7. 8. 9. 10. 11. 12. 13.

213. 2.3. MATRICES 113 D = - 4. - 4. - 4. - 4. - 4. - 4. - 4. - 4. - 4. ✌ ✆ Sum and difference of the matrices are elementwise operations, hence matrices undergoing addition and subtraction must be of same order. Product of the matrices is of two type. (i) Dot (scalar) product and (ii) Cross (vector) product. Dot product or scalar product is elementwise operation. It gives resultant matrix by multiplying elements by elements. Due to elementwise operation, matrix addition, dot product are commutative. Subtraction of matrices is not commutative as 2−3 6= 3−2. The dot product of matrices symbol is group of dot and asterisk symbol (.*). Pij = Aij. ∗ Bij Example of dot product is given below: ✞ -- A=[1 ,2 ,3;4 ,5 ,6;7 ,8 ,9] 2 -- B=[5 ,6 ,7;8 ,9 ,10;11 ,12 ,13] -- P=A.*B ✌ ✆ ✞ A = 1. 2. 3. 4. 5. 6. 7. 8. 9. B = 5. 6. 7. 8. 9. 10. 11. 12. 13. P = 5. 12. 21. 32. 45. 60. 77. 96. 117. ✌ ✆ Cross product is special operation of the product of matrix. For cross product, number of rows of one matrix must be equal to the number of columns of the other matrix. A = [1, 2]; B = 3 4 The product A × B is possible. Using cross product rules, the result is C = [1, 2] × 3 4 = [11] ✞ -- A=[1 ,2] 2 -- B=[3;4] -- C=A*B ✌ ✆

214. 114 Linear Algebra ✞ A = 1. 2. B = 3. 4. C = 11. ✌ ✆ If this rule is not followed then Scilab shows error of inconsistent multiplication. Let we have two matrices A and B as A = [1, 2]; B = [3, 4] ✞ 1 -- A=[1 ,2] -- B=[3 ,4] 3 -- C=A*B ✌ ✆ ✞ A = 1. 2. B = 3. 4. C=AB !--error 10 Inconsistent multiplication . ✌ ✆ Cross product may or may not commutative. Most of matrices gives A× B 6= B × A. Let two matrices are A and B of order n × n then A B = A B × B−1 B−1 = A × B−1 I = A × B−1 Here, cross product of matrix and its inverse matrix is identity matrix. Note that dot product of a matrix and its inverse matrix is not equal to identity matrix. See the following Scilab codes for the division of the matrices. ✞ 1 -- A=[1 ,2;3 ,4] //a 2x2 matrix -- B=[5 ,6;7 ,8] //a 2x2 matrix 3 -- r=A/B // division of matrix p by matrix q -- s=inv(B) // inverse of matrix q 5 -- t=A*s // cross product of p and inverse of q ✌ ✆ ✞ A = 1. 2. 3. 4. B = 5. 6. 7. 8. r =

215. 2.3. MATRICES 115 3. - 2. 2. - 1. s = - 4. 3. 3.5 - 2.5 t = 3. - 2. 2. - 1. ✌ ✆ 2.3.1 Determinant The determinant det(A) or |A| of a square matrix A is a number encoding certain prop- erties of the matrix. A matrix is invertible if and only if its determinant is nonzero. Determinant of the matrix A =

221. a b c d e f g h i

227. is given by |A| = a(e × i − h × f) − b(d × i − g × f) + c(d × h − g × e) It is a numerical value and crucial for the solution of the given algebraic equations or matrix. Scilab example is ✞ 1 -- det ([2 ,3 ,5;8 ,1 ,5;9 ,7 ,2]) ✌ ✆ ✞ ans = 256. ✌ ✆ 2.3.2 Transpose Matrix Transpose matrix is the change of rows into columns and vice versa of a square matrix. ✞ -- x=1:1:9; 2 -- y=matrix(x,3,3),y’ ✌ ✆ ✞ y = 1. 4. 7. 2. 5. 8. 3. 6. 9. ans = 1. 2. 3. 4. 5. 6. 7. 8. 9. ✌ ✆

228. 116 Linear Algebra 2.3.3 Diagonal Matrix A matrix which has only non-zero diagonal elements and other elements are zero, is called diagonal matrix. Mathematically A = aij when i = j 0 when i 6= j In Scilab diagonal matrix is ✞ -- diag ([1 ,2 ,3]) // elements in [] are diagonal elements . ✌ ✆ ✞ ans = 1. 0. 0. 0. 2. 0. 0. 0. 3. ✌ ✆ 2.3.4 Identity Matrix Identity matrix has only unity diagonal elements and other elements are zeros. Mathe- matically I = aij = 1 when i = j = 0 when i 6= j In Scilab identity matrix is ✞ -- eye (2,3)// eye(row , columns) ✌ ✆ ✞ ans = 1. 0. 0. 0. 1. 0. ✌ ✆ The product of a matrix and identity matrix is that matrix, i.e. A × I = I × A = A 2.3.5 Inverse of Matrix From the definition of inverse, if x is a number and y is its inverse then x × y = 1. Similarly, if P is a invertible square matrix and P−1 is its inverse then PP−1 = I Where I is unit identity matrix having same order as the matrix P has. A matrix should be square matrix for being invertible. All square matrices are not invertible. The square matrix which has an inverse is called invertible or non-singular matrix. A square matrix is invertible when its determinant is not zero, i.e. det(A) 6= 0. Inverse of a square matrix is given by A−1 = Adj(A) Det(A)

229. 2.3. MATRICES 117 Adj(A) of the matrix A is transpose matrix of the co-factor matrix of matrix A. Co-factor of a matrix of m × n order in respect of ith row and jth column is given by Aij that is equal to the product of (−1)i×j and determinant of remaining matrix after eliminating ith row and jth column. Let a matrix is given like A = 1 2 3 4 The co-factors of the matrix are a11 = (−1)1×1 × 4 a11 = (−1)1×2 × 3 a21 = (−1)2×1 × 2 a22 = (−1)2×2 × 1 Now co-factors matrix of matrix A is Acf = 4 −3 −2 1 Now Adj(A) of matrix A is Adj(A) = 4 −2 −3 1 Now the determinant of the matrix A is Det(A) = −2 Finally, inverse of matrix A is A−1 = Adj(A) Det(A) = −2 1 1.5 −0.5 Inverse of a 3 × 3 square matrix P =    1 3 5 4 8 9 2 1 6    is find by using Scilab as ✞ 1 -- P=[1 ,3 ,5;4 ,8 ,9;2 ,1 ,6] -- inv(P) ✌ ✆ ✞ ans = - 1. 0.3333333 0.3333333 0.1538462 0.1025641 - 0.2820513 0.3076923 - 0.1282051 0.1025641 ✌ ✆

230. 118 Linear Algebra 2.3.6 Normalization of Matrix A matrix T is said to be normalized if T 2 = I where, I is identity matrix. Take a matrix A as A =

234. a b c d

238. Let k is factor that must be multiply to matrix A such that, T = kA and kA × kA = I. Now, kA × kA shall be transform in such manner that A2 = KI. Now the new form of the matrix is k2 A2 = I. It give k2 KI = I Or k = 1 √ K Now, the normalized matrix is T = kA, which gives T 2 = I. Take a matrix A =

242. 3 2 −2 −3

246. Its A2 is A × A =

250. 5 0 0 5

254. This matrix can be transformed in form of KI as shown below: A × A = 5 ×

258. 1 0 0 1

262. Now, the value of k is 0.4472136. Now the transform matrix is T + kA T = kA =

266. 1.3416408 0.8944272 −0.8944272 −1.3416408

270. See the Scilab example as given below: ✞ -- A = [3,2;-2,-3] 2 -- k = 1/5^0.5 -- N = k*A 4 -- N*N ✌ ✆ ✞ A = 3. 2. - 2. - 3. k = 0.4472136 N = 1.3416408 0.8944272 - 0.8944272 - 1.3416408

271. 2.3. MATRICES 119 ans = 1. 0. 0. 1. ✌ ✆ A matrix may or may not be normalized. 2.3.7 Normalzation Factor (norm) norm returns the normalized form of a matrix. A matrix T is said to be normalized if T 2 = I where, I is identity matrix. Take a matrix A as A =

275. a b c d

279. Let k is factor that must be multiply to matrix A such that, T = kA and kA × kA = I. Now, kA × kA shall be transform in such manner that A2 = KI. Now the new form of the matrix is k2 A2 = I. It give k2 KI = I The norm function returns the value of K associated with a matrix, i.e. ✞ -- K=norm (A) ✌ ✆ Further proceeding successive steps for finding of the normalized matrix. k = 1 √ K Now, the normalized matrix is T = kA which givens T 2 = I. Take a matrix A =

283. 3 2 −2 −3

287. Its A2 is A × A =

291. 5 0 0 5

295. This matrix can be transformed in form of KI as shown below: A × A = 5 ×

299. 1 0 0 1

303. Now, the value of k is 0.4472136. Now the transform matrix is T + kA T = kA =

307. 1.3416408 0.8944272 −0.8944272 −1.3416408

311. See the Scilab example as given below: ✞ 1 -- A=[3,2;-2,-3]; -- k=1/ norm (A)^0.5 3 -- N=k*A -- N*N ✌ ✆

312. 120 Linear Algebra ✞ A = 3. 2. - 2. - 3. k = 0.4472136 N = 1.3416408 0.8944272 - 0.8944272 - 1.3416408 ans = 1. 0. 0. 1. ✌ ✆ There are following types of normalization of a matrix. Norm Type Description norm(x,2) Largest singular value of x. It is computed like (max(svd(x))) norm(x,1) The l 1 norm x. It is sum of largest column. Its value is max(sum(abs(x), ‘r′ )). norm(x,‘inf’) The infinity norm of x. It is computed by max(sum(abs(x), ‘c′ )). norm(x,‘fro’) Frobenius norm. Computed by sqrt(sum(diag(x′ ∗ x))) for a vector. norm(v,p) The l p norm. It is computed from relation (sum(v(i)p ))(1/p) . 2.3.8 Permutation Transposition (pertrans) pertrans is the permutation and transposition of the elements of a matrix simultaneously. Take a matrix A =

316. 1 2 3 3 4 4

320. Its permutation and transposition, rows are transform into columns and then columns are arranged from right to left in matrix form. For example, the permutation and transposition of matrix, A is B (say) B =

326. 4 3 4 2 3 1

332. ✞ -- A = [1 ,2 ,3;3 ,4 ,4]; 2 -- pertrans (A) ✌ ✆

333. 2.3. MATRICES 121 ✞ ans = 4. 3. 4. 2. 3. 1. ✌ ✆ Similarly, the permutation and transposition of matrix A =

341. 1 2 3 3 4 4 7 6 8 0 9 8

349. is B (say) B =

355. 8 8 4 3 9 6 4 2 0 7 3 1

361. ✞ -- A = [1 ,2 ,3;3 ,4 ,4;7 ,6 ,8;0 ,9 ,8]; 2 -- pertrans (A) ✌ ✆ ✞ ans = 8. 8. 4. 3. 9. 6. 4. 2. 0. 7. 3. 1. ✌ ✆ 2.3.9 Orthogonal Matrix An orthogonal matrix is a square matrix with real entries whose columns and rows are orthogonal unit vectors. Equivalently, a matrix A is orthogonal if its transpose is equal to its inverse: AT = A−1 We can find the orthogonal matrix of a given matrix by decomposing it as explained in function svd. It is equal to the Matrix U of the svd decomposition. ✞ -- orth ([1 ,2;3 ,4]) ✌ ✆ ✞ ans = - 0.4045536 - 0.9145143 - 0.9145143 0.4045536 ✌ ✆ 2.3.10 Complex Matrix Complex matrix is a matrix whose at least one element is a complex number. A complex numbers are written in form of z = a + ib. A real number is also a complex number with zero imaginary part, i.e. z = 2 + i0. The function complex accepts two matrices of equal shape and size as its argument. First matrix represents real part of the complex number, while second matrix represents to the imaginary part of the complex numbers.

362. 122 Linear Algebra ✞ 1 -- // First part is real , second part is imaginary -- complex ([1 2;3 4],[3 4;5 8]) ✌ ✆ ✞ ans = 1. + 3.i 2. + 4.i 3. + 5.i 4. + 8.i ✌ ✆ 2.3.11 Matrix Product Vector product of two matrices (different from elementwise product) is performed by special mathematical operations. Matrix product is possible, if number of rows in left matrix is equal to the number of columns in right matrix. Let we have two matrices of equal size 2 × 2 as given below: A =

366. a b c d

370. ; B =

374. e f g h

378. The vector product of the matrices is given by C = A × B as Cij = mn X ij AijBji Here corresponding elements of columns of right matrix are multiplied by respective elements of rows of left matrix and their respective products are added with each other. The vector product of matrices A and B is A × B =

382. ae + bg af + bh ce + dg cf + dh

386. ✞ 1 -- a=[1 ,3 ,5;4 ,8 ,9;2 ,1 ,6]; -- b=[1 ,3 ,5;4 ,8 ,9;2 ,1 ,6]; 3 -- a*b ✌ ✆ ✞ ans = 23. 32. 62. 54. 85. 146. 18. 20. 55. ✌ ✆ Scalar matrix product (element-wise matrix product) is performed by multiplying corresponding elements of the two matrices. Scalar product is possible if both matrices are of equal size. The scalar matrix product of two matrices is given by C = A · B as Cij = Aij · Bij Assume two matrices A =

390. a b c d

394. ; B =

398. e f g h

403. 2.3. MATRICES 123 The element-wise matrix product of these two matrices is A · B =

407. ae bf cg dh

411. ✞ -- a=[1 ,3 ,5;4 ,8 ,9;2 ,1 ,6]; 2 -- b=[1 ,3 ,5;4 ,8 ,9;2 ,1 ,6]; -- a.*b ✌ ✆ ✞ ans = 1. 9. 25. 16. 64. 81. 4. 1. 36. ✌ ✆ 2.3.12 Eigenvalues of Matrix An eigenvector of a square matrix A is a non-zero vector v that, when the matrix is multiplied by v, yields a constant multiple of v, the multiplier being commonly denoted by λ. That is: Av = λv The number λ is called the eigenvalue of A corresponding to Av. To find the eigenvalues, we solve above relations as Av − λv = 0 ⇒ (A − λI)v = 0 To get the values of λ we will solve the relation A − λI. Note that, v can not be equal to zero. So, A − λI must be zero. Matrix equation will be found if determinant of A − λI is equal to zero. |A − λI| = 0 Expand the determinants and find all the values of λ. Values of λ are eigenvalues of the given matrix. Assume a matrix A as A = 1 2 3 4 For its eigenvalues, we have

415. 1 2 3 4 − λ 1 0 0 1

419. = 0 Or

423. 1 − λ 2 3 4 − λ

427. = 0 It gives λ2 − 5λ − 2 = 0 On solving it, we have λ = −0.37 and λ = 5.37. The Scilab codes are

428. 124 Linear Algebra ✞ -- spec ([1 ,2;3 ,4]) // eigenvectors ✌ ✆ ✞ ans = - 0.3722813 5.3722813 ✌ ✆ 2.3.13 Triangular Lower Matrix (tril) tril returns the lower triangular matrix. If a matrix is given like A =   a f g f b h g h c   Now the lower triangular matrix of matrix A is A =   a 0 0 f b 0 g h c   ✞ 1 -- s=poly (0,’s’); -- tril ([s,s;s,1]) ✌ ✆ ✞ ans = s 0 s 1 ✌ ✆ 2.3.14 Triangular Upper Matrix (triu) tril returns the upper triangular matrix. If a matrix is given like A =   a f g f b h g h c   Now the lower triangular matrix of matrix A is A =   a f g 0 b h 0 0 c   ✞ -- s=poly (0,’s’); 2 -- triu ([s,s;s,1]) ✌ ✆

429. 2.3. MATRICES 125 ✞ ans = s s 0 1 ✌ ✆ 2.3.15 Lower Upper Matrix (lu) This function factorize a given matrix in lower triangular and upper triangular matrices. It is used as ✞ -- [L,U,E]=lu(matrix name ); ✌ ✆ Where input matrix is m×n real or complex matrix. L is lower triangular real or complex matrix of size m × min(m, n) and U is upper triangular real or complex matrix of size min(m, n) × n. E is a n × n permutation matrix. ✞ 1 -- A=[1 ,3;2 ,1] -- [L,U,E]=lu(A) ✌ ✆ ✞ E = 0. 1. 1. 0. U = 2. 1. 0. 2.5 L = 1. 0. 0.5 1 ✌ ✆ 2.3.16 Diagonal Matrix (diag) For a matrix A =

435. a b c d e f g h i

441. The diagonal elements are given as d = [a, e, i]. This function returns the diagonals of a matrix or diagonal matrix. Diagonal elements are taken along the main diagonal of the matrix. Its syntax is ✞ 1 -- diag ( matrix name ) ✌ ✆ Take a matrix A of size 3 × 3 A =

447. 1. 2. 3. 4. 1. 5. 8. 8. 9.

454. 126 Linear Algebra The diagonals of the matrix ‘A’ is d =

460. 1. 1. 9.

466. If matrix name supplied to this function is of m × 1 size then it convert it into the matrix of m×m size with all the input elements at the diagonals of the output matrix. Elements other than diagonal are sets to zero. Take input matrix of type d =

472. 1. 1. 9.

478. Then diagonal matrix shall be D =

484. 1. 0. 0. 0. 1. 0. 0. 0. 9.

490. Scilab example is ✞ 1 -- A=[1, 2, 3; 4, 1, 5; 8, 8, 9] -- d=diag (A) ✌ ✆ ✞ A = 1. 2. 3. 4. 1. 5. 8. 8. 9. d = 1. 1. 9. ✌ ✆ The diagonal matrix is constructed from the diagonals of the given matrix ‘A’ by using the diag() function. ✞ 1 -- D=diag (d) ✌ ✆ ✞ D = 1. 0. 0. 0. 1. 0. 0. 0. 9. ✌ ✆ If a matrix is not a square matrix, then rows and columns from the top-left position are chosen, which ultimately form a perfect square matrix. For example, if a non-square matrix is A =

498. 1. 2. 3. 4. 1. 5. 8. 8. 9. 4. 8. 5.

507. 2.3. MATRICES 127 then diag() function assumes the given matrix as A =

516. 1. 2. 3. 4. 1. 5. 8. 8. 9. . . . ... . . ..

525. for extraction of diagonal from the given matrix. The diagonal of the matrix ‘A’ is obtained by using diag() function as d =

531. 1. 1. 9.

537. Here elements are taken along the main diagonal of the matrix. 2.3.17 Jordan Canonical Form (bdiag) It is also called diagonalization of a matrix. A matrix is said to be diagonalizable if it can be transform into Jordan canonical (normal) form as given below: A = P · D · P−1 Where D is diagonal matrix. Matrix function can be used with diagonalized matrix as f(A) = P · f(D) · P−1 Here, matrix function is operated only with diagonal matrix not with P or P−1 . For example, suppose a matrix A as A = 1 3 2 1 It can be transformed into P · D · P−1 form as A = P 0.344 0 0 −1.449 # P−1 Where P = 0.774 −0.790 0.632 0.645 Now, the matrix function operation, for example inverse of matrix A is given as A−1 = P · D−1 · P−1 And solving above relation by substituting the values of P, D we get A−1 and it is A−1 = −0.2 0.6 0.4 −0.2

538. 128 Linear Algebra ✞ -- A=[1 ,3;2 ,1]; 2 -- [D,P,d]= bdiag(A) ✌ ✆ ✞ d = 1. 1. P = 0.7745967 - 0.7905694 0.6324555 0.6454972 D = 3.4494897 0. 0. - 1.4494897 ✌ ✆ Here, d is diagonals of the matrix A. On computation of P · D · P−1 ), we get the matrix A. ✞ 1 -- P*D*inv(P) ✌ ✆ ✞ ans = 1. 3. 2. 1. ✌ ✆ Applying the inverse matrix operation, we get the same result by P ·D−1 ·P−1 ) and A−1 . ✞ 1 -- I=P*inv(D)*inv(P) -- i=inv(A) ✌ ✆ ✞ I = - 0.2 0.6 0.4 - 0.2 i = - 0.2 0.6 0.4 - 0.2 ✌ ✆ 2.3.18 Cholensky Factorization (chol) The Cholesky decomposition of a Hermitian positive-definite matrix A, is a decomposition of the form A = LL ′ where L is a lower triangular matrix with real and positive diagonal entries, and L∗ denotes the conjugate transpose of L. A positive-definite matrix has positive determinant.

539. 2.3. MATRICES 129 Scilab function chol is used for the cholensky factorization of a matrix. Let we have a matrix 3 2 3 4 Its determinant is |A| = 12 − 6 = 6 0. So, it is positive-definite matrix. Now, The A = LL ′ decomposition is 3 2 3 4 = 1.7320508 0. 1.1547005 1.6329932 1.7320508 1.1547005 0. 1.6329932 ✞ 1 --A=[3 ,2;3 ,4] --chol (A) ✌ ✆ ✞ A = 3. 2. 3. 4. ans = 1.7320508 1.1547005 0. 1.6329932 ✌ ✆ Every Hermitian positive-definite matrix has a unique Cholesky decomposition. Let we have a Hermitian matrix A = 4 2 + 2i 2 − 2i 3 Its determinant |A| 0, hence it has Cholesky decomposition. Scilab codes for this matrix are given below: ✞ 1 -- A=[4 ,2+2* %i ;2-2*%i ,3] -- R=chol (x) ✌ ✆ ✞ A = 4. 2. + 2.i 2. - 2.i 3. R = 2. 1. + i 0. 1. ✌ ✆ 2.3.19 Determinant of Matrix (det) det returns the determinant of a square vector matrix. If a matrix is given like A =   1 2 3 4 5 6 7 8 9  

540. 130 Linear Algebra Then, determinant of this matrix A is |A| = 1(45 − 48) − 2(36 − 42) + 3(32 − 35) = 0 Hence determinant of the matrix A is zero. Scilab call of determinant of the matrix is ✞ 1 -- det ([1 ,2 ,3;4 ,5 ,6;7 ,8 ,9]) ✌ ✆ ✞ ans = 6.661D-16 ✌ ✆ 2.3.20 Inverse Matrix (inv) inv returns the inverse of a square matrix. Inverse of a matrix can be obtained if it is convertible. All square matrices are not convertible. The square matrix which has an inverse is called convertible or non-singular matrix. A square matrix is convertible when its determinant is not zero, i.e. det(A) 6= 0. If A is a matrix then A × A−1 = I i.e. vector product of matrix and its inverse matrix is unitary matrix. Dot product of matrix and its inverse matrix is not unitary matrix. ✞ -- A=[1 ,2;3 ,4] // Matrix of 2x2 order 2 -- B=A*inv(A) // Cross product of matrix and its inverse -- C=A.* inv(A) // Dot product of matrix and its inverse ✌ ✆ ✞ A = 1. 2. 3. 4. B = 1. 0. 8.882D-16 1. C = - 2. 2. 4.5 - 2. ✌ ✆ Inverse of a square matrix is given by A−1 = Adj(A) Det(A) Adj(A) of the matrix A is transpose matrix of the co-factor matrix of matrix A. Co-factor of a matrix of m × n order in respect of ith row and jth column is given by Aij that is

541. 2.3. MATRICES 131 equal to the product of (−1)i×j and determinant of remaining matrix after eliminating ith row and jth column. Let a matrix is given like A = 1 2 3 4 The co-factors of the matrix are a11 = (−1)1×1 × 4; a11 = (−1)1×2 × 3; a21 = (−1)2×1 × 2; a22 = (−1)2×2 × 1; Now co-factors matrix of matrix A is Acf = 4 −3 −2 1 Now Adj(A) of matrix A is Adj(A) = 4 −2 −3 1 Now the determinant of the matrix A is Det(A) = −2 Finally, inverse of matrix A is A−1 = Adj(A) Det(A) = −2 1 1.5 −0.5 This result can be obtained by calling Scilab function inv like ✞ -- inv ([1 ,2;3 ,4]) ✌ ✆ ✞ ans = - 2. 1. 1.5 - 0.5 ✌ ✆ 2.3.21 Orthogonal Matrix (orth) In linear algebra, an orthogonal matrix is a square matrix whose columns and rows are orthogonal unit vectors (orthonormal vectors). Orthogotal matrix is represented as AT A = AAT = I Where, AT is transpose of the matrix A. I is identity matrix. From the relation A−1 A = AA−1 = I we can say that a matrix A is orthogonal if its transpose is equal to its inverse, i.e. AT = A−1 . For example, 1 0 0 1 0.96 −0.28 0.28 0.96

542. 132 Linear Algebra An orthogonal basis for an inner product space V is a basis for V whose vectors are mutually orthogonal. If the vectors of an orthogonal basis are normalized, the resulting basis is an orthonormal basis. In Scilab, orth returns the orthogonal basis of a given matrix. To illustrate method of orth function, consider a matrix A = a b c d where, u = a c v = b d The orthogonal basis for first column shall be found by dividing all column elements by column normalisation value as shown in below matrix. c1 =     a √ a2 + c2 c √ a2 + c2     Second column is calculated by using relation c2 = v − u · v u · u × u Or c2 = b d − ab + cd a2 + b2 × a c Now, divide all elements of second columns by column normalisation value. The orthogonal basis shall be matrix constructed with columns c1 and c2. ✞ 1 B=orth (A) ✌ ✆ returns an orthonormal basis for the range of A. The columns of B span the same space as the columns of A, and the columns of B are orthogonal, so that BT ∗ B = eye(rank(A)). The number of columns of B is the rank of A. Take matrix A = 3 2 1 2 where, u = 3 1 v = 2 2 The orthogonal basis for first column shall be found by dividing all column elements by column normalisation value as shown in below matrix. c1 =      3 √ 32 + 12 = 0.948683298 1 √ 32 + 12 = 0.316227766      Second column is calculated by using relation c2 = 2 2 − 3 × 2 + 1 × 2 32 + 12 × 3 1 This gives c2 = −0.4 1.2

543. 2.3. MATRICES 133 Dividing it with normalised column value, we get c2 =      −0.4 p (−0.4)2 + 1.22 = −0.316227766 1.2 p (−0.4)2 + 1.22 = 0.948683298      The orthogonal basis shall be matrix constructed with columns c1 and c2. c2 = 0.948683298 −0.316227766 0.316227766 0.948683298 ✞ 1 -- A= [3 2;1 2] -- orth (A) ✌ ✆ ✞ ans = - 0.8649101 - 0.5019268 - 0.5019268 0.8649101 ✌ ✆ This result is different from what we have calculated. This is due to different way of Gram-Schmidt Orthonormal Method implementation. Orthogonal basis implementation of above illustrated method in scilab script is shown below: ✞ 1 -- function A = myOrth(A) -- k=size (A,2); 3 -- function w = proj (u,v) -- w = (sum(v.*u)/sum(u.*u)) * u; 5 -- endfunction -- for r = 1:1: k 7 -- A(:,r) = A(:,r) / norm (A(:,r)) -- for c = r+1:1:k 9 -- ; u v -- A(:,c) = A(:,c) - proj (A(:,r),A(:,c)) 11 -- end -- end 13 -- endfunction -- A = [3 ,2;1 ,2] 15 -- myOrth(A) ✌ ✆ 2.3.22 Rank of Matrix (rank) rank determines the minimum non-zero rows of a matrix when reduces to echelon form. Suppose a matrix A is in echelon form as A = 1 0 0 1

544. 134 Linear Algebra and it has two non zero rows. Now the rank of matrix is 2. Similarly, rank of following matrix B =   1 2 0 0 0 1 0 0 0   is also two. Rank of a matrix can be obtained by Scilab by using function rank function like ✞ 1 -- rank ([1, 0; 0, 1]) ✌ ✆ ✞ ans = 2. ✌ ✆ Example with other matrix of 3 × 3 order: ✞ -- rank ([1 ,3 ,5;4 ,8 ,9;2 ,1 ,6]) ✌ ✆ ✞ ans = 3. ✌ ✆ 2.3.23 Eigenvalues Eigenvectors (spec) spec evaluates eigenvalues and eigenvectors of a square matrix. The eigenvalues of a matrix is given by |A − λI| = 0 Assume a matrix of order 2 × 2 A = 1 2 3 4 Now its eigenvalues are given by |A − λI| = 0 ie

548. 1 2 3 4 − λ 1 0 0 1

552. = 0 Or

556. 1 − λ 2 3 4 − λ

560. = 0 Or (1 − λ) × (4 − λ) − 6 = 0 On solving it λ = −0.372281; 5.372281 Or eigenvalues in matrix form, when they are arranged in descending order is d = 5.372281 0.000000 0.000000 −0.372281

561. 2.3. MATRICES 135 For λ = 5.372281, eigenvector (v1) is (A − λI)v1 = 0. So, 1 − 5.372281 2 3 4 − 5.372281 x y = 0 −4.372281 2 3 −1.372281 x y = 0 Or −4.372281x + 2y = 0; 3x − 1.372281y = 0 To get solutions, put x = 1 in −4.372281x + 2y = 0, we get the value of y = 2.186140. To get eigenvectors, we shall normalize these two values as x = 1 √ 12 + 2.1861402 ; y = 2.186140 √ 12 + 2.1861402 It gives, x = 0.415973 and y = 0.909376. As coefficients of above two eigenvector equations are of opposite signs, hence values of x and y shall be either both negative or both positive. Now, we shall submit x and y values in equation g = 3x − 1.372281y to get minimum positive value. g = 3 × 0.415973 − 1.372281 × 0.909376 = 0.000000407 Taking sign convention, the other possible set of solution be x = −0.415973 and y = −0.909376. g = 3 × −0.415973 − 1.372281 × −0.909376 = 0.000000407 When x = −0.415973 and y = −0.909376, we have positive f value. This gives first eigenvector corresponding to λ = 5.372281. v1 = −0.415973 −0.909376 We can also find normalized x and y solution from 3x − 1.372281y = 0 and minimum positive value of f = −4.372281x+2y can be found by substituting x and y values taking sign conventions. For λ = −0.372281, eigenvector (v2) is (A − λI)v2 = 0. So, 1 − (−0.372281) 2 3 4 − (−0.372281) x y = 0 1.372281 2 3 4.372281 x y = 0 Or 1.372281x + 2y = 0; 3x + 4.372281y = 0 As coefficients of above two eigenvector equations are of same signs, hence values of x and y shall be in opposite signs. On solving these two algebraic equations, as explained for λ = 5.372281, we have x = −0.824564 and y = 0.565767 or x = 0.824564 and y =

562. 136 Linear Algebra −0.565767. From the numberline, acceptable values are x = −0.824564 and y = 0.565767. This gives second eigenvector corresponding to λ = −0.372281. v2 = −0.824564 0.565767 The corresponding eigenvector matrix from above two eigenvectors (v1 and v2) is v = −0.415973 −0.824564 −0.909376 0.565767 From Scilab, eigenvalues and eigenvectors are obtained by using function spec like ✞ -- [vec , val]= spec ([1 ,2; 3,4]) ✌ ✆ ✞ val = - 0.3722813 0 0 5.3722813 vec = - 0.8245648 - 0.4159736 0.5657675 - 0.9093767 ✌ ✆ 2.3.24 Square Root of Matrix Assume a matrix of order 2 × 2 A = 1 2 3 4 Now its eigenvalues are given by |A − λI| = 0 i.e.

566. 1 2 3 4 − λ 1 0 0 1

570. = 0 Or

574. 1 − λ 2 3 4 − λ

578. = 0 Or (1 − λ) × (4 − λ) − 6 = 0 On solving it λ = −0.372281; 5.372281 Or eigenvalues in matrix form, when they are arranged in descending order is d = 5.372281 0.000000 0.000000 −0.372281 For λ = 5.372281, eigenvector (v1) is (A − λI)v1 = 0. So, 1 − 5.372281 2 3 4 − 5.372281 x y = 0

579. 2.3. MATRICES 137 −4.372281 2 3 −1.372281 x y = 0 Or −4.372281x + 2y = 0; 3x − 1.372281y = 0 To get solutions, put x = 1 in −4.372281x + 2y = 0, we get the value of y = 2.186140. To get eigenvectors, we shall normalize these two values as x = 1 √ 12 + 2.1861402 ; y = 2.186140 √ 12 + 2.1861402 It gives, x = 0.415973 and y = 0.909376. As coefficients of above two eigenvector equations are of opposite signs, hence values of x and y shall be either both negative or both positive. Now, we shall submit x and y values in equation g = 3x − 1.372281y to get minimum positive value. g = 3 × 0.415973 − 1.372281 × 0.909376 = 0.000000407 Taking sign convention, the other possible set of solution be x = −0.415973 and y = −0.909376. g = 3 × −0.415973 − 1.372281 × −0.909376 = 0.000000407 When x = −0.415973 and y = −0.909376, we have positive f value. This gives first eigenvector corresponding to λ = 5.372281. v1 = −0.415973 −0.909376 For λ = −0.372281, eigenvector (v2) is (A − λI)v2 = 0. So, 1 − (−0.372281) 2 3 4 − (−0.372281) x y = 0 1.372281 2 3 4.372281 x y = 0 Or 1.372281x + 2y = 0; 3x + 4.372281y = 0 As coefficients of above two eigenvector equations are of same signs, hence values of x and y shall be in opposite signs. On solving these two algebraic equations, as explained for λ = 5.372281, we have x = −0.824564 and y = 0.565767 or x = 0.824564 and y = −0.565767. This gives second eigenvector corresponding to λ = −0.372281. v2 = −0.824564 0.565767 The corresponding eigenvector matrix from above two eigenvectors (v1 and v2) is v = −0.415973 −0.824564 −0.909376 0.565767

580. 138 Linear Algebra Note that, each column of eignevectors is arranged to the corresponding eigenvalues. Now, the matrix A can be written as A = v × d × v−1 . The square root of the matrix is given by A 1 2 = v × d 1 2 × v−1 ✞ -- A=[1 ,2; 3,4] 2 -- [vec , val]= spec (A) -- B=vec*sqrt (val)*inv(vec) 4 -- M=B*B ✌ ✆ ✞ A = 1. 2. 3. 4. val = - 0.372281 0 0 5.372281 vec = - 0.824564 - 0.415973 0.565767 - 0.909376 B = 0.553688 + 0.464394 i 0.806960 - 0.212426 i 1.210441 - 0.318639 i 1.764129 + 0.145754 i M = 1. + 1.110D-16i 2. + 4.163D-17i 3. + 8.327D-17i 4. ✌ ✆ Here, matrix M is same as matrix A. 2.3.25 Hermitian Factorisation (sqroot) A Hermitian matrix (A) (or self-adjoint matrix) is a complex square matrix that is equal to its own conjugate transpose (AT ). In other words, the ijth element of Hermitian matrix is equal to the complex conjugate of the jith element of Hermitian matrix for all indices i and j. Elements of a Hermitian matrix are satisfy aij = aji. If A is a Hermitian matrix then sqroot of matrix A returns a matrix B such that A = BBT .In Scilab, sqroot function is used to find the square root of the Hermitian matrix. ✞ -- A=[1 ,2;2 ,1] 2 -- B=sqroot(A) -- B*B’ ✌ ✆ ✞ A = 1. 2. 2. 1. B = - 1.2247449 0.7071068 - 1.2247449 - 0.7071068 ans =

581. 2.3. MATRICES 139 1. 2. 2. 1. ✌ ✆ 2.3.26 Signaular Matrix A matrix is said to be singular whose determinant is zero. A two dimensional singular matrix is A =

585. 1 1 1 1

589. Its determinant is |A| = 1 − 1 = 0. Inverse of a singular matrix is not possible as its determinant is zero. 2.3.27 Singular Value Approximation (sva) sva is acronym of Singular Value Approximation of a matrix. ✞ 1 -- sva ([1 ,2;3 ,4]) ✌ ✆ ✞ ans = -0.4045536 -0.9145143 -0.9145143 0.4045536 ✌ ✆ This returns left singular vectors of the given matrix. Mathematically, sva is matrix U of the Singular Value Decomposition (see svd). 2.3.28 Singular Value Decomposition (svd) svd acronym of Singular Value Decomposition of a matrix. Singular Value Decomposition (SVD) is a factorization of a real or complex matrix. Singular value decomposition of a rectangular matrix Am×n where m is rows represents genes and n is column represents the experimental conditions. The SVD theorem states than Am×n = Um×m Sm×n V T n×n Where U and V are orthogonal, i.e. UT U = Im×m and V T V = In×n. Here, the columns of U are the left singular vectors; S has singular values and is diagonal; and V T has rows that are the right singular vectors. SVD calculation consists 1. Eigenvalues and Eigenvectors of AAT and AT A. 2. Eigenvectors of AAT make up the columns of U. 3. The Eigenvectors of AT A make up the columns of V . 4. The singular values in S are square roots of eigenvalues from AAT or AT A.

590. 140 Linear Algebra The singular values are the diagonal entries of the S matrix and are arranged in descending order. The singular values are always real numbers. If the matrix A is a real matrix, then U and V are also real. Let a matrix A and its transpose matrix AT are respectively A = 1 2 3 4 AT = 1 3 2 4 Now, AAT is 1 2 3 4 1 3 2 4 = 5 11 11 25 Eigenvalues of this matrix is found when |AAT − λI| = 0. So,

594. 5 − λ 11 11 25 − λ

598. = (5 − λ)(25 − λ) − 121 = 0 This gives eigenvalues of the matrix AAT and eigenvalues are arranged in descending order. val = 29.866069 0.000000 0.000000 0.133931 and find the corresponding eigenvectors as

602. 5 − 29.866069 11 11 25 − 29.866069

606. x y = 0 It gives, −24.866069x + 11y = 0; 11x − 4.866069y = 0 The eigenvector (v1) corresponding to λ = 29.866069 is found by solving these two equations. Recall the solutions of algebraic equations, these equations have only one solution, that is x = 0 and y = 0. This is true if we take x, y ∈ I. If x, y ∈ R, then we can approximate the values of x and y so that above algebraic equations are satisfied approximately to zero, i.e. left hand side solution tends to zero. As x and y have “opposite sign” coefficients, therefore, values of x and y may be either both positive or both negative. To solve the relation, put x = 1 in −24.866069x + 11y = 0, we have −24.866069 × 1 + 11y = 0 This gives y = 2.260551. Normalizing these values we have, x = 1 √ 12 + 2.2605512 ; y = 2.260551 √ 12 + 2.2605512 Or x = 0.40455369; y = 0.914514249 Substituting these values in equation g = 11x − 4.866069y, we have g+ 1 = 11 × 0.40455369 − 4.866069 × 0.914514249 = −0.000001153 g− 1 = 11 × −0.404552907 − 4.866069 × −0.914514595 = 0.000001153

607. 2.3. MATRICES 141 Note that there are two equations for λ = 29.866069 therefore approximate values of x and y may be found either of the two equations. But here, we shall find x and y from each equation separately and their values will be put in other equation to check which solution set is more close to zero. Take x = 1, in 11x − 4.866069y = 0, we get the value of y 11 × 1 − 4.866069y = 0 This gives y = 2.260551587. Normalizing these values we have, x = 1 √ 2.2605515872 + 12 ; y = 2.260551587 √ 2.2605515872 + 12 Or x = 0.404553602; y = 0.914514288 Substituting values of x and y in equation f = −24.866069x + 11y, we have f+ 1 = −24.866069 × 0.404553602 + 11 × 0.914514288 = −0.000000614 f− 2 = −24.866069 × −0.404553602 + 11 × −0.914514288 = 0.000000614 Here, x = −0.40455369, y = −0.914514249, give minimum positive value. Why we consider minimum positive value? To understand it, consider modulo function f(x) = −x when x 0 x when x ≥ 0 It means 0 is positive side. That’s why, 0.000000614 is considered more closure to zero than −0.000000614. Hence acceptable eignevector is v1 = −0.404553 −0.914514 Now, we shall find the corresponding eigenvectors for eigenvalue λ = 0.133931.

611. 5 − 0.133931 11 11 25 − 0.133931

615. x y = 0 It gives, 4.866068x + 11y = 0; 11x + 24.866068y = 0 The eigenvector (v2) corresponding to λ = 0.133931 is found by solving these two equations. Recall the solutions of algebraic equations, these equations have only one solution, that is x = 0 and y = 0. This is true if we take x, y ∈ I. If x, y ∈ R, then we can approximate the values of x and y so that above algebraic equations are satisfied approximately to zero, i.e. left hand side solution tends to zero. As x and y have “same sign” coefficients, therefore, values of x and y shall have opposite signs. We know that any number when multiplied by a constant |k| 1, i.e. −1 k +1, result approaches to zero as k → 0. So, that the solutions of these two equations shall be within (−1, +1). To solve the relation, take x = 1 in 4.866068x + 11y = 0, we have 4.866068 × 1 + 11y = 0

616. 142 Linear Algebra This gives y = −0.442369818. Normalizing these values we have, x = 1 p (−0.442369818)2 + 12 ; y = −0.442369818 p (−0.442369818)2 + 12 Or x = 0.914514319; y = −0.404553533 Substituting these values in equation g = 11x + 24.866068y, we have g+ 1 = 11 × 0.914514319 + 24.866068 × −0.404553533 = 0.000001848 g− 1 = 11 × −0.914514319 + 24.866068 × 0.404553533 = −0.000001848 To solve the second relation 11x + 24.866068y = 0, take x = 1, we have 11 × 1 + 24.866068y = 0 This gives y = −0.442369899. Normalizing these values we have, x = 1 p (−0.442369899)2 + 12 ; y = −0.442369899 p (−0.442369899)2 + 12 Or x = 0.914514291; y = −0.404553595 Substituting these values in equation f = 4.866068x + 11y, we have f+ 1 = 4.866068 × 0.914514291 + 11 × −0.404553595 = −0.000000818 f− 1 = 4.866068 × −0.914514291 + 11 × 0.404553595 = 0.000000818 The minimum positive value is obtained when x = −0.914514291 and y = 0.404553595. So, the eignevector is v2 = −0.914514 0.404553 Eigenvector of the matrix AAT is column arrangement of v1 and v2. vec = −0.404553 −0.914514 −0.914514 0.404553 Here, U = −0.404553 −0.914514 −0.914514 0.404553 Similarly, eigenvalues and eigenvectors of AT A shall be computed as computed for AAT above. The matrix relation AT A is 1 3 2 4 1 2 3 4 = 10 14 14 20 Eigenvalues of this matrix is found when |AT A − λI| = 0. So,

620. 10 − λ 14 14 20 − λ

624. = (10 − λ)(20 − λ) − 196 = 0

625. 2.3. MATRICES 143 This gives eigenvalues and corresponding eigenvectors as val = 29.866069 0.000000 0.000000 0.133931 ; vec = −0.576048 0.817415 −0.817415 −0.576048 Here, V = −0.576048 0.817415 −0.817415 −0.576048 S is the square root of the eigenvalues from AAT or AT A. So, S = √ 29.866069 0.000000 0.000000 √ 0.133931 = 5.464985 0.000000 0.000000 0.365966 Here, each column of val matrix represents the eignevalues of AAT or AT A matrices and corresponding columns of U or V are eigenvectors corresponding to the eigenvalues respectively. Relation between U, S and V with A as A = USV T Now, from the above explained example, the modified values of U, S and V are U = −0.404553 −0.914514 −0.914514 0.404553 V = −0.576048 0.817415 −0.817415 −0.576048 S = 5.464985 0.000000 0.000000 0.365966 In Scilab, svd of given matrix is given by ✞ 1 -- [U,S,V]= svd ([1 ,2;3 ,4]) ✌ ✆ This function has three outputs as given below: ✞ V = - 0.5760484 0.8174156 - 0.8174156 - 0.5760484 S = 5.4649857 0. 0. 0.3659662 U = - 0.4045536 - 0.9145143 - 0.9145143 0.4045536 ✌ ✆ 2.3.29 Trace of Matrix (trace) Tracing of a square matrix is sum of its diagonal elements. If A is a square matrix of order n × n then T r(A) = n X i=1 aii

626. 144 Calculus If λi are the eigenvalues of a matrix A, then tracing of the matrix is T r(A) = X i λi In Scilab, trace returns the tracing of a matrix. ✞ 1 -- A=rand (3,3) -- trace(A) ✌ ✆ ✞ A = 0.3312931 0.7221236 0.3707945 0.0518477 0.0774625 0.2116117 0.4149242 0.5855878 0.1903269 ans = 0.5990825 ✌ ✆

627. 3.1. DERIVATIVE CALCULUS 145 3Calculus 3.1 Derivative Calculus 3.1.1 Derivative Derivative of a function f(x) is given by f′ (x) = lim h→0 f(x + h) − f(x) h where, h is approximately zero but not zero. Symbolic derivative of a function in Scilab is calculated by derivat function. ✞ -- s=poly (0,’s’); 2 -- derivat (1/s) ✌ ✆ ans = −1 s2 derivative function calculates the numerical derivative of a function. ✞ -- function y=myF(x) 2 -- y = x*x; -- endfunction 4 -- x = 1; -- fp = derivative (myF ,x) ✌ ✆ ✞ fp = 2. ✌ ✆ 3.1.2 Numeric Derivative Second method of finding the derivative of a function is numdiff that is more accurate than the derivative function. Mathematically, it is given by numdiff (f)|x=c = df dx

631. x=c It always returns numeric value rather than symbolic output.

632. 146 Calculus ✞ -- function y=myF(x) 2 -- y = x*x; -- endfunction 4 -- x = 1; -- fp = numdiff(myF ,x) ✌ ✆ ✞ fp = 2. ✌ ✆ 3.1.3 Numeric Difference diff function is used to find the difference between two consecutive elements continuously. This function is used with a vector having only numeric values. In mathematics, first order discrete difference is given by ∆f = fn+1 − fn A function table and its first order difference is given in the following table. f = 1 8 27 64 125 216 ∆f = 7 19 37 61 91 The diff function does the same work. Numerically, it is reqpresented like diff(f) = fn+1 − fn ✞ -- v=(1:8) ^3; 2 -- diff (v) ✌ ✆ ✞ ans = 7. 19. 37. 61. 91. 127. 169. ✌ ✆ 3.1.4 Function Evaluation (feval) Multiple evaluation of a function for one or two arguments of vector type. The syntax is ✞ -- feval(var 1, var 2, ...., var n, n var func ) ✌ ✆ Example is ✞ 1 -- function [z]=f(x,y) -- z=x+y; 3 -- endfunction -- a=1:3; 5 -- b=1:3; -- feval(a,b,f) ✌ ✆

633. 3.2. INTEGRAL CALCULUS 147 ✞ ans = 2. 3. 4. 3. 4. 5. 4. 5. 6. ✌ ✆ 3.2 Integral Calculus 3.2.1 Integration Integration is also called anti-derivative. It is a method of finding reverse of differentiation. x y f(x) dx b x = a b x = b The symbolic representation of integral of a function is given by I = Z f(x)dx Where symbol Z represent to the integration. f(x) is function on which integration is to be performed and dx is the base in which d represents the small element of function and x the base variable against which integration is to be performed. Variables other than x are considered as constants. Indefinite integral has global scope while definite integral is computed within the given limits of independent variable. I = Z b a f(x)dx Scilab does not support symbolic integration but it has one of the most precisive calcu- lations in numerical integration i.e. definite integration. Integration of a simple function f(x) = x is ✞ --I=integrate (’x’,’x’ ,0,1) ✌ ✆ ✞ I = 0.5 ✌ ✆ In this integration command, first term is integrand function, second term is base variable, third is lower limit and fourth term is upper limit. Applying these observations, the integration of t2 is

634. 148 Calculus ✞ --I=integrate (’t*t’,’t’ ,0,1) ✌ ✆ ✞ I = 0.3333333 ✌ ✆ Similarly integration of a sine function is ✞ -- I=integrate (’sin(t)’,’t’ ,0,1) ✌ ✆ ✞ I = 0.4596977 ✌ ✆ Remember that lower limit and upper limit in trigonometric integration are taken by trigonometric functions in radian unit of angle. Definite integration of trigonometric function is computed by using intg function. ✞ -- function y=f(x) 2 -- y=sin(x)/x -- endfunction 4 -- I=intg (0,2*%pi ,f) ✌ ✆ ✞ I = 1.4181516 ✌ ✆ There are several method of finding the integration by interpolation method. One of them is Trapezoidal method. ✞ -- t=0:0.1: %pi; 2 -- inttrap (t,sin(t)) ✌ ✆ ✞ ans = 1.9974689 ✌ ✆ The quadrature method of integration is given by 3.2.2 Double Integral (int2d) Double integral is used to integrate a function along two dimensions. For example, double integral is given by A = Z Z f(x, y)dx dy where dx and dy are length and width of element in x-axis and y-axis. Double integral gives volume of region bounded between two axes and f(x, y).

635. 3.2. INTEGRAL CALCULUS 149 x y z f(x, y) dx dy int2d is used to integrate two dimensional functions. The syntax is ✞ -- [I,e]= int2d( .. 2 -- a three dimensional N array for abscissa , .. -- a three dimensional N array for ordinate , .. 4 -- external function .. -- ) ✌ ✆ Here I is integrated value and e is estimated error. Example is ✞ 1 -- function [z]=f(x,y) -- z=x+y; 3 -- endfunction -- X=[0 ,0;1 ,1;1 ,0]; 5 -- Y=[0 ,0;0 ,1;1 ,1]; -- int2d(X,Y,f) ✌ ✆ ✞ ans = 1. ✌ ✆ 3.2.3 Tripple Integration (int3d) int3d is used to integrate three dimensional functions. The syntax is ✞ -- [I,e]= int3d( .. 2 -- a four dimensional array for abscissae , .. -- a four dimensional array for ordinate , .. 4 -- a four dimensional array for z-axis , .. -- external function .. 6 -- ) ✌ ✆ Here I is integrated value and e is estimated error. ‘external function’ is a function or list or string which defines the integrand f(xyz, nf), where xyz is the vector of a point coordinates and nf is the number function. By default nf is 1. Example is ✞ -- function v=f(xyz , numfun) 2 -- v=xyz ’* xyz; -- endfunction 4 -- // Tetrahedron coordinates

636. 150 Calculus -- //(0,0,0) ,(1,0,0) ,(0,1,0) ,(0,0,1) 6 -- X=[0;1;0;0]; -- Y=[0;0;1;0]; 8 -- Z=[0;0;0;1]; -- [I,e]= int3d(X,Y,Z,f ,1) ✌ ✆ ✞ e = 5.551D-16 I = 0.05 ✌ ✆ Another example is ✞ -- function v=f(F, numfun) 2 -- //x+y+z=1 -- v=1; 4 -- endfunction -- // Tetrahedron coordinates 6 -- //(0,0,0) ,(1,0,0) ,(0,1,0) ,(0,0,1) -- X=[0;1;0;0]; // Vector of abscissa values 8 -- Y=[0;0;1;0]; // Vector of ordinate values -- Z=[0;0;0;1]; // Vector of z values 10 -- [I,e]= int3d(X,Y,Z,f ,1) ✌ ✆ ✞ e = 1.850D-15 I = 0.1666667 ✌ ✆ Another example is ✞ -- function v=f(F, numfun) 2 -- //x+y+z=6 -- v=6; 4 -- endfunction -- // tetrahedron coordinates 6 -- //(0,0,0) ,(2,0,0) ,(0,3,0) ,(0,0,1) -- X=[0;2;0;0]; // Vector of abscissa values 8 -- Y=[0;0;3;0]; // Vector of ordinate values -- Z=[0;0;0;1]; // Vector of z values 10 -- [I,e]= int3d(X,Y,Z,f ,1) ✌ ✆ ✞ e = 6.661D-14 I = 6. ✌ ✆

637. 3.2. INTEGRAL CALCULUS 151 3.2.4 Integrate (integrate) integrate function calculates the sum of values of a function for all the point range within the limits by using quadrature method. Plane integration is simple area bounded by function and the axis of limits. It is given by Z f(x) dx = n X i=0 f(xi) × h Where h is width of the two consecutive lower upper bound limits. ✞ -- x0 =0; 2 -- x1 =2; -- X= integrate (’sin(x)’, ’x’, x0 , x1) ✌ ✆ ✞ ans = 1.4161468 ✌ ✆ 3.2.5 Definite Integration (intg) It return the definite integration of an external function. If f(x) is a function bounded between close interval [a b] then definite integral of the function is given by I = Z b a f(x) dx Definite integration of given relation I = Z 2π 0 x sin(30x) q 1 − x 2π 2 dx is −2.5432. Now the Scilab call of definite integral of the above relation is ✞ -- function y=f(x) 2 -- y=x*sin (30* x)/sqrt (1-((x/(2* %pi))^2)) -- endfunction 4 -- I=intg (0,2*%pi ,f) ✌ ✆ ✞ ans = - 2.5432596 ✌ ✆ 3.2.6 Cauchy’s Integration (intc) If f is a complex-valued function, then intc computes the integral over the curve in the complex plane from lower limit z1 to upper limit z2 of the complex function f(z) along the line z1z2. Since line integrals of analytic functions are independent of the path, function intc can be used to evaluate integrals of any analytic function. Syntax of Cauchy’s Line integral is given below:

638. 152 Calculus ✞ -- [I,e]= intc (.. 2 -- var z1 ,.. // complex number -- var z2 ,.. // complex number 4 -- external function of two vars (f) .. -- ) ✌ ✆ Here I is integrated value and e is estimated error. Let z = x(t) + iy(t) is a parametric curve of t in complex plane C, where z1 ≤ t ≤ z2. Let f(z) is continuous function on complex plane C. 1 −1 1 2 −1 −2 x y b b b b b b b b b b b z0 z1 z2 zn−1 zn 1 −1 1 2 −1 −2 x y b b b b b b b b b b b z1 z2 ∆zi bc Si The t is sub divided into n subdivisions as a = t0, t1, . . ., tn = b and corresponding subdivisions of curve on C is z0, z1, . . ., zn. The width of two consecutive subdivisions of curve is ∆zi = zi − zi−1, where 0 i n. If function is continuous and limit exists over each subdivision of curve, then bounded area of curve within given bounded limits is sum of areas of all subdivisions. So, Sn = n X i=1 f(zi)∆zi Here, Sn is line integral of function f(z) and it is represented by I = Z C f(z) dz = Z z2 z1 f(z) dz Where dz = |∆zi| → 0. Scilab uses following relation for computation of complex integral in intc function. Z C f(z) dz = Z z2 z1 f(C(z))C′ (z) dz To parameterize this relation, put z = z1 + (z2 − z1)t. When z = z1, t = 0 and when z = z2, t = 1. This gives Z C f(z) dz = Z 1 0 f(z1 + (z2 − z1)t) (z2 − z1) dt Real and imaginary part of integrand are computed separately and their sum is desired result.

639. 3.2. INTEGRAL CALCULUS 153 Illustrated Example Take a function f(z) = i. We have to find Cauchy’s integral along the line za = 1 − i to zb = 1 + i. The Cauchy’s line integral for the curve shall be I = Z C f(z) dz = Z 1+i 1−i i dz I = [iz] 1+i 1−i = i [(1 + i) − (1 − i)] This gives I = i × 2i = 2i2 = −2 This is desired result. ✞ 1 -- function [I]=f(z) -- I=%i; 3 -- endfunction -- intc (1-%i ,1+%i ,f) ✌ ✆ ✞ ans = - 2 ✌ ✆ Illustrated Example Take a function f(z) = 1 + i. We have to find Cauchy’s integral along the line za = 1 − i to zb = 1 + i. The Cauchy’s line integral for the curve shall be I = Z C f(z) dz = Z 1+i 1−i (1 + i) dz I = [(1 + i)z] 1+i 1−i = (1 + i) [(1 + i) − (1 − i)] This gives I = (1 + i) × 2i = 2(−1 + i) This is desired result. ✞ -- function [I]=f(z) 2 -- I=1+%i; -- endfunction 4 -- intc (1-%i ,1+%i ,f) ✌ ✆ ✞ ans = - 2 + 2 i ✌ ✆ Illustrated Example Let the given function is f(z) = z2 . We have to integrate it along the line 0 to 1 + i in complex plane. The Cauchy’s line integral intc along the curve is given by Z C f(z) dz = Z 1+i 0 f(z) dz

640. 154 Calculus Parameterization of this integral, we will substitute z = z1 + (z2 − z1)t. When z = z1, t = 0 and when z = z2, t = 1. Again dz = (z2 − z1)dt Thus Z C f(z) dz = Z 1 0 (z1 + (z2 − z1)t)2 × (z2 − z1) dt Substituting the value of z1 and z2. Z C f(z) dz = Z 1 0 (1 + i)2 t2 × (1 + i) dt = Z 1 0 (1 + i)3 t2 dt Or Z C f(z) dz = (1 + i)3 t3 3

644. 1 0 = (1 + i)3 3 = 2i(1 + i) 3 This is desired result. The Scilab code for this example are ✞ -- function y=f(z) 2 -- y=z^2 -- endfunction 4 -- intc (0,1+%i ,f) ✌ ✆ ✞ ans = - 0.6666667 + 0.6666667 i ✌ ✆ 3.2.7 Cauchy Integration (intl) If f is a complex-valued function, then intl computes the integral of f(z) over the line in the complex plane, along the curve defined by z = z0 + reiθ for θ1 ≤ θ ≤ θ2 and |z − z0| ≤ r. The syntax of this function is ✞ -- [y]= intl ( .. 2 -- var a, .. // lower limit , real number -- var b, .. // upper limit , real number 4 -- origin point as complex number (z0), .. -- positive real number (r), .. 6 -- external function (f) .. -- ) ✌ ✆ Here, z0 is center of circle in complex plane and r is radius of the circle within exponential phase between θ1 and θ2. The Cauchy’s path integral of complex function f(z) dz along the curve C(θ) = z0 + reiθ is given by

645. 3.2. INTEGRAL CALCULUS 155 1 1 2 3 x f(x) z1 z2 θ1 θ2 1 1 2 3 x f(x) z1 z2 b b θ θ1 θ2 Z C f(z) dz = Z θ2 θ1 f(C(θ))C′ (θ) dθ Where, θ1 ≤ θ ≤ θ2. To convert it into parametric form, put θ = θ1 + (θ2 − θ1)t Now, when θ = θ1, t = 0 and when θ = θ2, t = 1. Thus the limit in parametric form is 0 ≤ t ≤ 1. This gives C(t) = z0 + rei[θ1+(θ2−θ1)t] and Z C f(z) dz = Z 1 0 f(C(t))C′ (t) dt Now, real and imaginary parts of the above relation are integrated separately and final result is equal to the sum of real and imaginary integrals. Scilab uses following intl codes. ✞ 1 -- function r=intl (a,b,z0 ,r,f) -- function y=real1(t) 3 -- z=r*exp(%i *((1-t)*a+t*b)) -- y=real (f(z+z0)*%i*(b-a)*z) 5 -- endfunction -- function y=imag1(t) 7 -- z=r*exp(%i *((1-t)*a+t*b)) -- y=imag (f(z+z0)*%i*(b-a)*z) 9 -- endfunction -- r=intg (0,1, real1)+%i*intg (0,1, imag1) 11 -- endfunction ✌ ✆ Illustrated Example Let the given function is f(z) = z2 . We have to integrate it along the unit circle, i.e. |z| = r ≤ 1 and 0 ≤ θ ≤ 2π. The Cauchy’s line integral intl along the curve is given by Z C f(z) dz = Z b a f(z) dz The curve of integral about z0 origin is z = z0 + reiθ ; θ1 ≤ θ ≤ θ2 Parametrization of this curve, we will substitute θ = θ1 + (θ2 − θ1)t

646. 156 Calculus in z, we get z = z0 + rei(θ1+(θ2−θ1)t) From θ = θ1 + (θ2 − θ1)t When θ = θ1, t = 0 and when θ = θ2, t = 1. Again dz = (θ2 − θ1)rei(θ1+(θ2−θ1)t) dt Thus Z C f(z) dz = Z 1 0 z0 + rei(θ1+(θ2−θ1)t) 2 × (θ2 − θ1)rei(θ1+(θ2−θ1)t) dt Substituting the value of θ1 = 0 and θ2 = 2π, and taking origin z0 = 0 + i0 and radius of unit circle r ≤ 1. Z C f(z) dz = Z 1 0 e2πti 2 × 2πe2πti dt = Z 1 0 2πe6πti dt Or Z C f(z) dz = 2π × 6πi × e6πti

648. 1 0 = 12π2 i × (1 − 1) = 0 + 0i This is desired result. The Scilab code for this example are ✞ 1 -- function y=f(z) -- y=z^2 3 -- endfunction -- intl (0,2 %pi ,0,1,f) ✌ ✆ ✞ ans - 3.384D-17 - 5.272D-16i ✌ ✆ 3.3 Differential Equations An equation said to be differential equation, if its at least one term has differential part. For example, dy dx + 2y = 0 is a differential equation. 3.3.1 Ordinary Differential Equation (ode) A first order ordinary differential equation is given by df dt = f(x, t) and second order ordinary differential equation is given by d2 f dt2 = f(x, t) ode function is used to solve the ordinary differential equation. The syntax used is

649. 3.3. DIFFERENTIAL EQUATIONS 157 ✞ -- y=ode( .. 2 -- initial dependent value , .. -- initial variable value , .. 4 -- point where function is calculated , .. -- function .. 6 -- ) ✌ ✆ For illustration purposes, we will solve two examples of each first order and second order ODE. Consider, first order ODE for practice and introduction dy dt = −ky (3.1) Where, at t = 0, y = 0.5, i.e. y(0) = 0.5. Firstly, the above relation is re-arranged like dy y = −k dt Integration of this equation is ln y = −kt + ln c Where ln c is an auxiliary constant of integration and can be obtained by putting initial boundary values, y(0) = 0.5. Hence y = c e−kt Or c = 0.5 It gives the result of the relation (3.1) y = 0.5 e−kt It is the required solution. An example is ✞ -- // Here we will use dy/dt=-2*y relation . 2 -- // Its solution is y = y0 e^{-k*t} -- function ydot = f (t, y) 4 -- ydot = -2*y; //k=2 -- endfunction 6 -- t = 0:0.1:3; -- y0 = 0.5; // c=0.5 value 8 -- t0 = 0; -- y = ode(y0 ,t0 ,t,f) 10 -- plot (t,y) ✌ ✆ 0 0.5 1.0 1.5 2.0 2.5 0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 Second order differential equations are solved by using dassl function.

650. 158 Calculus 3.3.2 Differential Algebraic System Solver (dassl) dassl is used to solve the set of differential-algebraic equation. The function must be in form of fnc(t, y, ydot) = 0. This returns a matrix of y and ydot, with each row of y corresponding to one of the elements in the time vector t. The first element of t should be t0 and correspond to the initial state of the system y(0) and its derivative ẏ(0), so that the first row of the output y is y(0) and the first row of the output ẏ is ẏ(0). Now, take a second order ODE for problem solving exercise d2 y dt2 + y = 0 (3.2) where, at t = 0, y = 0.5, i.e. y(0) = 0.5 and at t = 0, ẏ = 0.25, i.e. ẏ(0) = 0.25. Auxiliary equation of second order ODE, (3.2) is m2 + 1 = 0 Root of this second order algebraic equation for variable m are m = ±i Now the solution of the second order differential equation, (3.2) is y = c1ei t + c2e−i t (3.3) On derivating and substituting the boundary values for ẏ in above equation ẏ = c1iei t − c2ie−i t and c1i − c2i = 0.25 Similarly substituting the boundary values for y in the equation 3.3 c1 + c2 = 0.5 On solving relations for c1 and c2 c1 = 0.5 i + 0.25 2 i c2 = 0.5 i − 0.25 2i Replacing these constants, c1 and c2 in equation of solution (3.3) of the second order ODE, (3.2), the result is y = 0.5 i + 0.25 2 i ei t + 0.5 i − 0.25 2i e−i t Using, rule of complex mathematics along-with Demoire Theorem, solution of second order ODE, (3.2) can be written in terms of trigonometric functions like y = 0.25 sin(t) + 0.5 cos(t) This is the solution of second order ODE, (3.2). In Scilab, dassl function is

651. 3.3. DIFFERENTIAL EQUATIONS 159 ✞ -- [y] = dassl (y0 , t0 , tt , fcn) ✌ ✆ The ‘fcn’ is described in form of f(t, y, ydot) = 0 i.e. function body should have first degree ODE equations. The function starts from initial conditions , i.e. y(t0) and ydot(t0) and iterates for each time value t0. As function body has multiple first order differential equations, thus y0 and ydot are matrices, i.e. one value for each first order differential equation. t is scalar and for each t value there is corresponding result row as y and ydot. It returns solution matrices y and ydot with each row in the result matrices corresponding to one of the elements in the vector t. The first element of t should be t0 and correspond to the initial state of the system y0 and its derivative ydot0, so that the first row of the output y is y0 and the first row of the output ydot is ydot0. To understand it, take above example explained in the beginning of this section. d2 y dt2 + y = 0 To solve this second order differential algebraic equation, first we will resolve in first order differential equations an(t)y(n) + an−1(t)y(n−1) + . . . . . . + a0(t)y + r(t) = 0 By using algorithm y1 = y Here, y(1) of Scilab dassl function is y1. y2 = dy1 dt = dy dt Here, x(2) of Scilab dassl function is y2 and ydot(1) of Scilab dassl function is dy1 dt or dy dt . y3 = dy2 dt = d2 y dt2 Here, x(3) of Scilab dassl function is y3 and ydot(2) of Scilab dassl function is dy2 dt or d2 y dt2 . . . . . . . yn = dyn−1 dt = dn−1 y dtn−1 y ′ n = − an−1(t) an(t) yn − an−2(t) an(t) yn−1 − . . . − a0(t) an(t) y1 + r(t) an(t) So, from the given second order differential equation d2 y dt2 + y = 0

652. 160 Calculus Let y1 = y = y(1) then on derivating it about t, we have dy1 dt = dy dt = ydot(1) Now, put y2 = dy1 dt = y(2), then ydot(1) = y(2) and on derivating it about t, we have dy2 dt = d2 y1 dt2 = d2 y dt2 = ydot(2) As we knew that d2 y dt2 = −y Then ydot(2) = −y(1). This equation must be in form of f(t, y, ydot) = 0 so, set of first order differential algebraic equation of reduced second order differential equation shall be ydot(1) = y(2); ydot(2) = −y(1) From the initial conditions, when t = 0, y = 0.5, it means y(1) = 0.5 and from above set of equations, ydot(2) = −0.5. When t = 0, dy dt = 0.25, i.e. ydot(1) = 0.25 and from above set of equations, we have y(2) = 0.25. So, initial conditions of y0 and ydot0 are [0.5; 0.25] and [0.25; −0.5] respectively. Note that there are two parameters of y and ydot, hence y0 and ydot0 are either row matrix or column matrix as the equation set are arranged either in row or column respectively. In Scilab, ydot is computed internally. The time iteration is in ten steps between t = 0 to t = 1. The complete set of example is given below: ✞ 1 -- y0 =[0.5;0.25]; -- t0 =0; 3 -- tt =0:0.1:1; -- function [r,ires ]=f(t,y,ydot ) 5 -- r=[ ydot (1) - y(2) , ydot (2) + y(1)] -- ires =0 7 -- endfunction -- y=dassl(y0 ,t0 ,tt ’,f); 9 -- y’ ✌ ✆ ✞ ans = ; --------- y -------- ------- ydot ------- ; t y ydot ydot yddot ;----------------------------------------------------------- 0. 0.5 0.25 0. 0. 0.1 0.5224602 0.1988344 0.1988336 - 0.5224595 0.2 0.5397004 0.1456821 0.145682 - 0.5397003 0.3 0.5515481 0.0910742 0.0910742 - 0.5515482

653. 3.3. DIFFERENTIAL EQUATIONS 161 0.4 0.5578849 0.0355562 0.0355562 - 0.5578849 0.5 0.5586475 - 0.0203170 - 0.0203170 - 0.5586472 0.6 0.5538283 - 0.0759871 - 0.0759871 - 0.5538279 0.7 0.5434754 - 0.1308981 - 0.1308978 - 0.5434759 0.8 0.5276923 - 0.1845012 - 0.1845012 - 0.5276925 0.9 0.5066366 - 0.2362608 - 0.2362610 - 0.5066364 1. 0.4805188 - 0.2856597 - 0.2856599 - 0.4805187 ✌ ✆ Verify it by substituting the time value t = 0.5 in equation y = 0.25 sin(t) + 0.5 cos(t) We have y = 0.25 sin(0.5) + 0.5 cos(0.5) = 0.55865 and dy/dt at t = 0.5 is dy dt

657. t=0.5 = −0.5 sin(t) + 0.25 cos(t)|t=0.5 = −0.2032 Thus dassl method is proved for solving of differential algebraic equations. Solved Problem 3.1 Solve second order differential equation y ′′ − y = 0 where, at t = 0, y = 0.5, i.e. y(0) = 0.5 and at t = 0, ẏ = 0.25, i.e. ẏ(0) = 0.25. Solution From the given second order differential equation d2 y dt2 − y = 0 Let y1 = y = y(1) then on derivating it about t, we have dy1 dt = dy dt = ydot(1) Now, put y2 = dy1 dt = y(2), then ydot(1) = y(2) and on derivating it about t, we have dy2 dt = d2 y1 dt2 = d2 y dt2 = ydot(2) As we knew that d2 y dt2 = y Then ydot(2) = y(1). This equation must be in form of f(t, y, ydot) = 0 So, set of first order differential algebraic equation of reduced second order differential equation shall be ydot(1) = y(2); ydot(2) = y(1)

658. 162 Calculus From the initial conditions, when t = 0, y = 0.5, it means y(1) = 0.5 and from above set of equations, ydot(2) = 0.5. When t = 0, dy dt = 0.25, i.e. ydot(1) = 0.25 and from above set of equations, we have y(2) = 0.25. So, initial conditions of y0 and ydot0 are [0.5; 0.25] and [0.25; 0.5] respectively. Note that there are two parameters of y and ydot, hence y0 and ydot0 are either row matrix or column matrix as the equation set are arranged either in row or column respectively. The time iteration is in ten steps between t = 0 to t = 1. The complete set of example is given below: ✞ 1 -- y0 =[0.5;0.25]; -- t0 =0; 3 -- tt =0:0.1:1; -- function [r,ires ]=f(t,y,ydot ) 5 -- r=[ ydot (1) - y(2) , ydot (2) - y(1)] -- ires =0 7 -- endfunction -- y=dassl(y0 ,t0 ,tt ’,f); 9 -- y’ ✌ ✆ ✞ ans = ; --------- y -------- ------- ydot ------- ; t y ydot ydot yddot ;----------------------------------------------------------- 0. 0.5 0.25 0. 0. 0.1 0.5275440 0.3013347 0.3013351 0.5275436 0.2 0.5603678 0.3556850 0.3556868 0.5603689 0.3 0.5988000 0.4135953 0.4135954 0.5988002 0.4 0.6432250 0.4756448 0.4756448 0.6432246 0.5 0.6940875 0.5424548 0.5424543 0.694087 0.6 0.7518968 0.6146937 0.6146937 0.7518967 0.7 0.8172313 0.6930848 0.693085 0.8172314 0.8 0.8907449 0.7784126 0.7784126 0.8907450 0.9 0.9731734 0.8715309 0.8715311 0.9731736 1. 1.0653418 0.9733718 0.9733719 1.065342 ✌ ✆ The solution of the given differential algebraic equation is y(t) = 0.375 et + 0.125 e−t At t = 0.2, solution is y(0.2) = 0.375 e0.2 + 0.125 e−0.2 = 0.5603678 Hence result is verified. Solved Problem 3.2 Solve second order differential equation y ′′ + 2y ′ + y = 0, where, at t = 0, y = 0.5, i.e. y(0) = 0.5 and at t = 0, ẏ = 0.25, i.e. ẏ(0) = 0.25. Solution From the given second order differential equation d2 y dt2 + 2 dy dt + y = 0

659. 3.3. DIFFERENTIAL EQUATIONS 163 Let y1 = y = y(1) then on derivating it about t, we have dy1 dt = dy dt = ydot(1) Now, put y2 = dy1 dt = y(2), then ydot(1) = y(2) and on derivating it about t, we have dy2 dt = d2 y1 dt2 = d2 y dt2 = ydot(2) As we knew that d2 y dt2 = −2 dy dt − y Then ydot(2) = −2 ydot(1) − y(1). This equation must be in form of f(t, y, ydot) = 0 So, set of first order differential algebraic equation of reduced second order differential equation shall be ydot(1) = y(2); ydot(2) = −2 ydot(1) − y(1) From the initial conditions, when t = 0, y = 0.5, it means y(1) = 0.5 and from above set of equations, ydot(2) = −1. When t = 0, dy dt = 0.25, i.e. ydot(1) = 0.25 and from above set of equations, we have y(2) = 0.25. So, initial conditions of y0 and ydot0 are [0.5; 0.25] and [0.25; −1] respectively. Note that there are two parameters of y and ydot, hence y0 and ydot0 are either row matrix or column matrix as the equation set are arranged either in row or column respectively. The time iteration is in ten steps between t = 0 to t = 1. The complete set of example is given below: ✞ 1 -- y0 =[0.5;0.25]; -- t0 =0; 3 -- tt =0:0.1:1; -- function [r,ires ]=f(t,y,ydot ) 5 -- r=[ ydot (1) - y(2) , ydot (2) + 2* ydot (1) + y(1)] -- ires =0 7 -- endfunction -- y=dassl(y0 ,t0 ,tt ’,f); 9 -- y’ ✌ ✆ ✞ ans = ; ------ y ------ ---- ydot ----- ; t y ydot ydot yddot ;--------------------------------------------------- 0.00000 0.50000 0.25000 0.25000 -1.00000 0.10000 0.52028 0.15835 0.15835 -0.83697 0.20000 0.53218 0.08187 0.08187 -0.69592

660. 164 Elementary Functions 0.30000 0.53709 0.01852 0.01852 -0.57413 0.40000 0.53626 -0.03352 -0.03352 -0.46922 0.50000 0.53071 -0.07582 -0.07582 -0.37908 0.60000 0.52137 -0.10976 -0.10976 -0.30185 0.70000 0.50900 -0.13656 -0.13656 -0.23588 0.80000 0.49426 -0.15727 -0.15727 -0.17973 0.90000 0.47772 -0.17279 -0.17279 -0.13214 1.00000 0.45985 -0.18394 -0.18394 -0.09197 istate = 3 msg = integration to tout completed by stepping past TOUT ✌ ✆ The solution of the given differential algebraic equation d2 y dt2 + 2 dy dt + y = 0 is y(t) = (k1 + k2t) e−t Substitute t = 0 and y(0) = 0.5, we get k1 = 0.5. Now, derivate this relation and substitute t = 0 and dy/dt = 0.25, we get k2 = 0.75. Now, the solution of the given differential equation is y(t) = (0.5 + 0.75t) e−t At t = 0.2, solution is y(0.2) = (0.5 + 0.75 × 0.2) e−0.2 = 0.53217 Similarly, at t = 0.9, solution is y(0.2) = (0.5 + 0.75 × 0.9) e−0.9 = 0.47772 Hence result is verified.

661. 4.1. DATA STRUCTURE 165 4Elementary Functions This section includes the elementary functions. 4.1 Data Structure Data means elements arranged in structured or non structured form. Scilab accepts each input as vector of integer, float, string etc type data. For example, a matrix data in Scilab is represented as: ✞ 1 -- t = [1,2,3] ✌ ✆ While non matrix data is supplied as: ✞ 1 -- t = 1,2,3 ✌ ✆ Both of these data are different in meaning. First is matrix data while second is assign- ment of values to variables. The output of first data is ✞ t = 1. 2. 3. ✌ ✆ while output of second data is ✞ t = 1. ans = 2. ans = 3. ✌ ✆ 4.1.1 Comparison of Data Data comparison in Scilab is performed elementwise. The type of data structure is automatically assessed by the Scilab itself. For example, if Scilab sense the vector parenthesis then input is considered as vector automatically. The element wise matrix comparison of data is performed by using double equal operator as shown in the following code snippet. ✞ -- [1 ,2]==[1 ,3] ✌ ✆ The output is ✞ ans = T F ✌ ✆

662. 166 Elementary Functions In above example first element of matrices being compared are same hence first answer is true but comparison of second elements (2 3) of both matrices is false. Comparison between vector and scalar is performed like ✞ -- [1 ,2]==1 ✌ ✆ The output is ✞ ans = T F ✌ ✆ In this case both elements of left hand side matrix elements are compared with ‘1’. First element comparison (1 1) is true while second comparison (2 1) is false. 4.1.2 Data Generation In Scilab, range operator is used to create data values automatically. The range operator, i.e. colon symbol is used for this purpose. See the following code snippet: ✞ -- a=1:1:10; 2 -- a(a5) ✌ ✆ The output is ✞ ans = 1. 2. 3. 4. ✌ ✆ 4.1.3 Creation of Matrix Cell Command cell(m) creates a blank cell matrix of m × m order. But the same command cell(m, n) creates a blank cell matrix of m × n order. ✞ -- a=cell (2) ✌ ✆ The output is ✞ ans = !{} {} ! ! ! !{} {} ! ✌ ✆ cell() is equal to cell(0) and produce null matrix. iscell() checks whether a variable is a cell or not. ✞ -- a=cell (2) 2 -- iscell(a) ✌ ✆ ✞ ans = T ✌ ✆

663. 4.1. DATA STRUCTURE 167 4.1.4 Creation of List A list can be produced by using list() command ✞ -- k = list () ✌ ✆ The output is ✞ k = () ✌ ✆ It produces empty list. The index ‘0’ of a list is also called empty list and list() is equivalent to list(0). Each list element is separated by comma symbol. The elements of list may be constant, variable, vector or matrix etc. See the example of a list with mixed type elements. Index for non null list elements started with index 1. ✞ -- k = list (1,[a b]) ✌ ✆ The output is ✞ k = l(1) 1. l(2) !a b ! ✌ ✆ The first element of list ‘k’ is ‘1’ and second element is matrix ‘!a b !’. We can add another element by selecting the index and defining the list element for that index. ✞ 1 -- k = list (1,[a b]); -- k(3)=Arun ✌ ✆ ✞ k = k(1) 1. k(2) !a b ! k(3) Arun ✌ ✆ iscell() checks whether a element is a cell or not. ✞ 1 -- k = list (1,[a b]); -- iscell(k) ✌ ✆ ✞ ans = F ✌ ✆ Here ‘k’ is an element list not a cell matrix.

664. 168 Elementary Functions 4.2 Bitwise operations Bitwise operations are used to perform logic operations. Followings are the main com- mands used most prominently. 4.2.1 Bitwise AND (bitand) bitand is used for logical AND operation. The binary output is true if and only if all the binary inputs are true. The AND table of two binary input is given below. X Y Z 0 0 0 0 1 0 1 0 0 1 1 1 ✞ -- x=86; // 1010110 in binary 2 -- y=91; // 1011011 in binary -- z=bitand(x,y) // 1010010 in binary ✌ ✆ ✞ z = 82. ✌ ✆ The matrix method of bitwise and operation is ✞ --x = uint8 ([0 1 0 1]); 2 --y = uint8 ([0 0 1 1]); --z = bitand(x, y) ✌ ✆ ✞ z = 0 0 0 1 ✌ ✆ 4.2.2 Bitwise Complements (bitcmp) bitcmp returns the complements of a binary number. This function has two arguments, first is decimal number and second is bit places for complements. This functions intakes two decimal numbers as arguments. First argument is converted into binary number. Second argument controls the number of binary bits to be selected from LSB side. This selected binary number is complemented and converted into decimal number as result. The numerical example is ✞ -- bitcmp (10, 5) 2 -- // 10 5 are decimal numbers. -- // 1010 (binary)=10 (decimal) 4 -- //

665. 4.2. BITWISE OPERATIONS 169 -- // 01010 (five binary bits from LSB) 6 -- // -------------------------------- -- // 10101 (complement binary number) 8 -- // = 21 (decimal equivalent ) ✌ ✆ ✞ ans = 21 ✌ ✆ Here ‘10’ is an decimal number and ‘5’ is number of bits to be extracted from the binary equivalent to the decimal number ‘10’. The calculation performed as: 1. First read the decimal number and converts it into binary number. 2. Extracts ‘n’ bits from LSB side. If there are deficiency of bits, use bit ‘0’. 3. Now make complements of the bits extracted. 4. Converts this binary number into decimal number. This is the final result. If ‘n’ is not supplied in this function, then it is calculated by using the relation k = log 2(bitmax) + 1. The maximum value of ‘n’ is ‘54’ because ‘54’ bit binary number and its equivalent decimal number is the largest floating point value. 4.2.3 Get a Position Bit (bitget) bitget return the bit at specific position. First argument in this function is decimal number and second argument is the index of bit to be obtained. ✞ -- dec2bin (13) 2 -- bitget(uint8(13) ,4:-1:1) ✌ ✆ ✞ ans = 1 1 0 1 ✌ ✆ 4.2.4 Bitwise OR (bitor) bitor is used for logical OR operation. The output is ‘true’ if any of the input is ‘true’. The logical OR table is X Y Z 0 0 0 0 1 1 1 0 1 1 1 1 Table 4.1: Logical OR Table.

666. 170 Elementary Functions ✞ -- x=86; // 1010110 in binary 2 -- y=91; // 1011011 in binary -- z=bitor(x,y) // 1011111 in binary ✌ ✆ ✞ z = 95. ✌ ✆ 4.2.5 Set Bit In Position (bitset) bitset is used to set a bit at any position. The syntax of this function is ✞ -- bitset( .. 2 -- decimal number , .. -- bit index , .. 4 -- bit value .. -- ) ✌ ✆ ✞ 1 -- // Decimal 25 is equal to binary 11001 -- bitset (25, 5, 0) // Binary 01001 ✌ ✆ ✞ ans = 9. ✌ ✆ 4.2.6 Bitwise Exclusive OR (bitxor) bitxor is used for logical exclusive OR operations. The truth table of Exclusive OR operation is X Y Z 0 0 0 0 1 1 1 0 1 1 1 0 Table 4.2: Exclusive OR Table. ✞ -- // decimal 25 - binary 11001 2 -- // decimal 5 - binary 00101 -- // -------------------------- 4 -- // Exclusive OR is - 11100 -- bitxor (25, 5) // 11100 - 28 ✌ ✆

667. 4.3. COMPLEX 171 ✞ ans = 28. ✌ ✆ 4.3 Complex Complex numbers are those numbers that contains real and imaginary parts in addition or subtraction. 4.3.1 Complex Number (complex) complex command is used the create a complex number from two or more inputs. First argument to this function is real part of the complex number and second argument is imaginary part of the complex number. ✞ -- c = complex ([1 2 3], 4) ✌ ✆ ✞ c = 1. + 4.i 2. + 4.i 3. + 4.i ✌ ✆ By changing the order of matrices. ✞ -- c = complex (1, [2 3 4]) ✌ ✆ ✞ c = 1. + 2.i 1. + 3.i 1. + 4.i ✌ ✆ Remember that, some time complex numbers are also represented as a + ib or a + jb. Where i = j = √ −1. Symbol j is mostly used in electric engineering where complex numbers are used to solve an electrical circuit. Conjugate of a complex number ✞ -- c = complex (1, [2 3 4]); 2 -- conj (c) ✌ ✆ ✞ c = 1. - 2.i 1. - 3.i 1. - 4.i ✌ ✆ 4.3.2 Complex Imaginary (imag) imag return the imaginary part of a complex number. If A = x + iy is a complex number then imag(A) will return the value y. In Scilab ✞ -- c = complex (1, 2); 2 -- [y]= imag (c) ✌ ✆

668. 172 Elementary Functions ✞ y = 2. ✌ ✆ 4.3.3 Iota Multiplication (imult) imult returns the ‘i’ multiplication to the complex number. If A = x + iy then imult returns the value of A as Ai. ✞ -- c = complex (1, 2); 2 -- [y]= imult(c) ✌ ✆ ✞ y = - 2. + i ✌ ✆ 4.3.4 Is Real (isreal) isreal checks if a variable is real or complex entries. If entry is real then it return ‘true’ otherwise returns ‘false’. ✞ -- isreal (1+%s) ✌ ✆ ✞ ans = T ✌ ✆ 4.3.5 Complex Real (real) real returns the real part of a complex number. If A = x + iy is a complex number then real(A) will return the value x. In Scilab ✞ -- x=86; 2 -- [y]= real (x) ✌ ✆ ✞ y = 86. ✌ ✆ 4.3.6 Complex Conjugate (conj) conj returns conjugate value of a complex number. If a + bi is a complex number then its complex conjugate value is a − bi and vice-verse. ✞ -- A=[1, %i;%i ,4]; 2 -- conj (A) ✌ ✆

669. 4.4. DISCRETE MATHEMATICS 173 ✞ ans = 1. - i - i 4. ✌ ✆ Remember that, some time complex numbers are also represented as a + ib or a + jb. Where i = j = √ −1. Symbol j is mostly used in electric engineering where complex numbers are used to solve an electrical circuit. 4.4 Discrete mathematics 4.4.1 Binomials (binomial) Binomial series expansion is used to get the exponent of base in series form. For example, 2.410 can be computed by using exponent relation as (2 + 0.4)10 . A binomial expansion of a number k = a + b with power n is given by kn or (a + b)n = n C0an b0 + n C1an−1 b1 + . . . + n Cna0 bn Number of terms in binomial expansion are n+ 1. In Scilab, function binomial() expands series (a + b)n such that (a + b)n = 1. Hence, either a or b should be between 0 and 1. First argument of this function is either a or b and they are between 0 and 1. See the example given below: ✞ 1 -- binomial (0.1, 3) ✌ ✆ Output of this function is coefficients of each term. As (a + b)n = 1, hence sum of all coefficients is always equal to one. ✞ ans = 0.729 0.243 0.027 0.001 ✌ ✆ S = 0.729 + 0.243 + 0.027 + 0.001 = 1 Second argument of binomial() function is exponent n. Number of terms in binomial expansion are n + 1 always. 4.4.2 Factors (factor) factor returns all the prime co-factors of supplied number. Co-factors of a given number are those factors which are not mutually divisible. For example, the co-factor of 12 are 2, 2, 3 and not 2 and 6, as 2 and 6 are mutually divisible. ✞ -- y=factor (620) ✌ ✆ ✞ y = 2. 2. 5. 31. ✌ ✆

670. 174 Elementary Functions 4.4.3 Factorial (factorial) factorial function returns factorial of a number. Factorial of a number n is given by following relation. n! = n × (n − 1) × (n − 2) × . . . × 3 × 2 × 1 ✞ -- y=factorial (10) ✌ ✆ ✞ y = 3628800. ✌ ✆ 4.4.4 Permutations (perms) perms function returns all possible permutations of a vector component. The possible maximum number of permutations on taking all n items of a set are n!. Note that, a single element is used more than once in a permutation set. Permutation may be learn as per-mutation. It means if digits are interchanged then outcome is mutated. For example, 12 and 21 are two different numbers even if same digits are sued. ✞ -- x=[4, 7, 10]; 2 -- y=perms(x) ✌ ✆ ✞ y = 10. 7. 4. 10. 4. 7. 7. 10. 4. 7. 4. 10. 4. 10. 7. 4. 7. 10. ✌ ✆ 4.4.5 Primes (primes) primes returns all the prime numbers less than the number given by user. ✞ 1 -- y=primes (10) ✌ ✆ ✞ y = 2. 3. 5. 7. ✌ ✆ 4.5 Floating point This section includes the function used in numeric systems.

671. 4.5. FLOATING POINT 175 4.5.1 Ceiling To (ceil) ceil round up a number to the next larger integer. Symbolic representation of ceil of a number x is ⌈x⌉. If supplied value is a complex number z then computation used in this function is ceil(z) = ceil(real(x)) + ceil(imag(x)) ∗ i An example of this function for positive number is ✞ -- ceil ([1.3 1.5 1.7 2.5 3.7]) ✌ ✆ ✞ ans = 2. 2. 2. 3. 4. ✌ ✆ For negative numbers ✞ -- ceil ([ -1.3 -1.5 -1.7 -2.5 -3.7]) ✌ ✆ ✞ ans = - 1. - 1. - 1. - 2. - 3. ✌ ✆ ceil function returns the right side integer to the given real number based on number-line. In the following figure, ceiling of −2.25 is circled by cyan colour circle. Here, ceiling value of −2.25 is −2. 0 1 2 3 4 0 −1 −2 −3 −4 b ceil(-2.25) 4.5.2 Double (double) It converts the integer value into double integer value. Actually, double makes an integer a floating point value. ✞ -- x=int8 ([0 12 140]); 2 -- double(x) ✌ ✆ ✞ ans = 0 12 -116 ans = 0. 12. -116. ✌ ✆ 4.5.3 Number as Integer (int8) It converts a number into an 8 bits long integer value. See the following example: ✞ -- x=double ([0 12 140]),int8 (x) ✌ ✆

672. 176 Elementary Functions ✞ x = 0. 12. -116. ans = 0 12 - 116 ✌ ✆ 4.5.4 Fix to Zero (fix) fix rounds a number towards zero. In other words rounded up to the next smaller integer. It is equivalent to rounding towards zero. If supplied value is a complex then computation used is ✞ fix (real (x)) + fix (imag (x)) * i ✌ ✆ Here an example is ✞ 1 -- fix ([1.3 1.5 1.7 2.5 3.7]) ✌ ✆ ✞ ans = 1. 1. 1. 2. 3. ✌ ✆ 4.5.5 Round to Lower Integer (floor) It rounds downward to a number i.e. rounded to lower integer. Symbolic representation of floor of a number x is ⌊x⌋. It is equivalent to rounding towards negative infinity. If supplied value is a complex number then computation in result is ✞ floor (real (x)) + floor (imag (x)) * i ✌ ✆ Example is ✞ 1 -- floor ([1.3 1.5 1.7 2.5 3.7]) ✌ ✆ ✞ ans = 1. 1. 1. 2. 3. ✌ ✆ This function is similar to the Greatest Integer Function (gif). floor function returns the left side integer of the given real number based on numberline. In the following figure, floor of −2.25 is circled by cyan colour circle. Here, floor value of −2.25 is −3. Floor value of 2.5 is 2. 0 1 2 3 4 0 −1 −2 −3 −4 b gi b floor(2.5)

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