86 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES x 6 = – 1.93375. Thus, the approximation value to the root is –1.93375, correct up to five decimals. 2.9 METHODS FOR COMPLEX ROOTS We now consider methods for determining complex roots of non-linear equations. Even if all coefficients of a non-linear equation are real, the equation can have complex roots. The iterative methods like the Secant method or the Newton-Raphson method are applicable to complex roots also, provided complex arithmetic is used. Starting with the complex initial approximation, if the iteration converges to a complex root, then the asymptotic convergence rate is the same as that for a real root. The problem of finding a complex root of f(z) = 0, where z is a complex variable, is equivalent to finding real values x and y, such that f(z)= f (x + iy) = u (x, y) + iv (x, y) = 0 Where u and v are real functions. This problem is equivalent to solving a system of two non-linear equations in two real unknowns x and y, u (x, y) = 0, v (x, y) = 0 Which can be solved using the methods discussed in previous section. Example 6. Find all roots of the equation f(x) = x 3 + 2x 2 – x + 5 using Newton-Raphson method. Use initial approximations x 0 = – 3 for real root and x 0 = 1 + i for complex root. Sol. Given f(x)= x 3 + 2x 2 – x + 5 f´(x)= 3x 2 + 4x – 1 Newton-Raphson formula is given by x n+1 = x n – () () ′ n n fx fx For real root: Taken initial approximation as x 0 = – 3. First approximation: x 1 = x 0 – () () ′ 0 0 fx fx x 1 = – 3 – () − 1 14 = – 2.928571429 Second approximation: x 2 = x 1 – () () ′ 1 1 fx fx x 2 = – 2.928571429 + () 0.035349848 13.01530612 x 2 = – 2.925855408 Third approximation: x 3 = x 2 – () () ′ 2 2 fx fx ALGEBRAIC AND TRANSCENDENTAL EQUATION 87 x 3 = – 2.925855408 + () 0.000050045 12.97846797 x 3 = – 2.925851552 Since the second and third approximations are same for five decimals hence the real root is –2.92585 correct up to five decimals. For complex root: Initial approximation is x 0 = 1 + i First approximation: x 1 = x 0 – () () 0 0 fx fx ′ x 1 = 1 + i – () ()() ()() 32 121 15 31 41 1 iii ii ++ +−++ ++ +− x 1 = () 53 114 109 i + = 0.486238 + i (1.045871) Thus 0.486238 + i (1.045871) is the first approximation value of the root. Proceeding similarly, we get next iterations as x 2 = 0.448139 + i (1.23665) x 3 = 0.462720 + i (1.22242) x 4 = 0.462925 + i (1.22253) x 5 = 0.462925 + i (1.22253) Since the last two iterations are similar, we take 0.462925 + i(1.22253) as the value of the complex root. 2.10 MULLER’S METHOD This method can also used to determine the both real and complex root of equation f(x) = 0. Let x i–2 , x i–1 , x i be three distinct approximations to a root of f(x) = 0 and let y i–2 , y i–1 , y i be the corresponding value of y = f(x). Let =−+−+ 2 1 () ( ) ( ) ii px Ax x Bx x y is a parabola passing through the points 211 2 ( , ), ( , ), ( , ), ii ii ii xy xy xy −−− − we have 2 11 1 ()() iiiiii yAxxBxxy −− − =−+−+ (1) −− − =−+−+ 2 22 2 ()() iiiiii y AxxBxxy (2) Equations (1) and (2) can be written as 2 111 ()() ii iiii Ax x Bx x y y −−− −+ −=− −−− −+ −=− 2 222 ()() ii iiii AxxBxxyy Therefore 21 12 121 2 ()()()() ()()() iiiiiii i iiiiii xxyyxxyy A xxxxxx −− −− −−− − −−−−− = −−− −− −− −−− − −−−−− = −−− 22 21 12 211 2 ()()()() ()()() iiiiiiii iiiiii xxyyxxyy B xxxxxx 88 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES From A and B, the quadratic equation 2 ()()() 0 iii p xAxx Bxx y =−+−+= gives the next approximation 2 1 4 2 i ii BB Ay xx A + −± − −= (3) But direct solution from (3) lead to loss of accuracy therefore for maximum accuracy, equation (3) can be written as 1 2 2 4 i ii i y x x BB A y + −=− ±− Note: If B > 0, we use +ve sign with square root of the equation and if B < 0, we use –ve sign with square root of the equation. Example 7. Find the root of the equation =−−−= 32 y(x) x x x 1 0  Muller’s method, taking initial approximations as x 0 = 0, x 1 = 1, x 2 = 2 Sol. Let x i – 2 = 0, x i – 1 = 1, x i = 2 Then 21 1, 2, 1 iii y yy −− =− =− = 21 12 121 2 ()()()( ) ()()) iiiiiiii iiiiii xxyyxxyy A xxxxxx −− −− −−− − −−−−− = −−− (0 2)( 2 1) (1 2)( 1 1) (1 0)(1 2)(0 2) A −−−−−−− = −−− A = 2 −− −− −− − − −−−−− = −−− 22 21 12 11 2 2 ()()()() ()()() iiiiiiii iiiiii xxyyxxyy B xxxxxx 22 (0 2) ( 2 1) (1 2) ( 1 1) (0 1)(1 2)(0 2) B −−−−−−− = −− − B = 5 The next approximation to the desired root is 1 2 2 4 i ii y xx BB Ayi + =− ±− 1 21 2 525421 i x + × =− ±−×× (taking +ve sign) 1 2 2 1.7807 76 5 4.123106 i x + =− = + The procedure can now be repeated with three approximations as 1, 2, 1.780776. Let 21 1, 2, 1.780776 iii xxx −− === Then 21 2, 1, 0.3048 08 iii y yy −− =− = =− 21 12 121 2 ()()()() ()()() iiiiiiii iiiii i xxyyxxyy A xxxxxx −− −− −−− − −−−−− = −−− ALGEBRAIC AND TRANSCENDENTAL EQUATION 89 A = −+−−−+ −− − (1 1.780776)(1 0.304808) (2 1.780776)( 2 0.304808) (2 1)(2 1.780776)(1 1.780776) A=3.780773 B = −−− − −−− − −−−−− −−− 22 112 2 211 2 ()()()() ()()() iiiiiiii iiiiii xxyyxxyy xxxxxx B = −+−−−+ −− − 22 (1 1.780776) (1 0.304808) (2 1.780776) ( 2 0.304808) (1 2)(2 1.780776)(1 1.780776) B = 5.123098 The next approximation to the desired root is 1 2 2 4 i ii i y xx BB Ay + =− ±− 1 2 2 ( 0.304808) 1.780776 5.123098 (5.123098) 4 3.780773 ( 0.304808) i x + ×− =− +−××− 1 1.837867 i x + = The procedure can repeated with three approximations as 2, 1.780776, 1.837867. Let 21 2, 1.780776, 1.837867 ii i xx x −− == = Then 21 1, 0.304808, 0.007757 ii i yy y −− ==− =− A = −− −− −−− − −−−−− −−− 21 12 121 2 ()()()() ()()() iiiiiiii iiiiii xxyyxxyy xxxxxx A = −−+−− + −−− (2 1.837867)( 0.304808 0.007757) (1.780776 1.837867)(1 0.007757) (1.780776 2)(1.780776 1.837867)(2 1.837867) A = 4.619024 B = −− −− −−− − −−−−− −−− 22 21 12 22 11 ()()()() ()()() iiiiiiii iiiiii xxyyxxyy xxxxxx B = −−+−− + −−− 22 (2 1.837867) ( 0.304808 0.007757) (1.780776 1.837867) (1 0.007757) (2 1.780776)(1.780776 1.837867)(2 1.837867) B = 5.467225 The next approximation to the desired root is 1 2 2 4 i ii i y xx BB Ay + =− ±− + ×− =− +−×− 1 2 2 ( 0.007757) 1.837867 5.467225 (5.467225) 4 4.619024 (0.007757) i x x i+1 = 1.839284 The procedure can now be repeated with three approximations as 1.780776, 1.837867, and 1.839284. 90 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES Let 21 1.780776, 1.837867, 1.839284 iii xxx −− === Then y i-2 = – 0.304808, y i-1 = – 0.007757, y i = – 0.000015 A = 21 12 121 2 ()()()() ()()() iiiiiiii iiiiii xxyyxxyy xxxxxx −− −− −−− − −−−−− −−− (1.780776 1.839284)( 0.007757 0.000015) (1.837867 1.839284)( 0.304808 0.000015) (1.837867 1.78.776)(1.837867 1.839284)(1.780776 1.839284) A −−+−−−+ = −− − A = 4.20000 22 21 12 211 2 ()()()() ()()() iiiiiiii iiiiii xxyyxxyy B xxxxxx −− −− −−− − −−−−− = −−− 22 (1.780776 1.839284) ( 0.007757 0.000015) (1.837867 1.839284) ( 0.304808 0.000015) (1.780776 1.837867)(1.837867 1.839284)(1.780776 – 1.839284) B −−+−−−+ = −− B = 5.20000 The next approximation to the desired root is x i+1 = x i – 2 2 4 i i y BB Ay ±− x i+1 = 1.839284 – () () ( ) 2 2 0.000015 5.2 5.2 4 4.2 0.000015 ×− +−××− x i+1 = 1.839287 Hence the required root is 1.839287 Example 8. Using Muller’s method, find the root of the equation y(x) = x 3 – 2x – 5 = 0, which lies between 2 and 3. Sol. Let x i–2 = 1.9, x i–1 = 2, x i = 2.1 Then y i–2 = – 1.941, y i–1 = – 1, y i = 0.061 A = () () ()() ()()() 212 1 121 2 iii i i ii i iiiiii xxyyxxyy xxxxxx −−− − −−− − −−−− − −−− A = ()( )()( ) ()( )( ) 0.2 1.061 0.1 2.002 0.1 0.1 0.2 −− −−− −− A = 0.2122 0.2002 6 0.002 − = B = ()( )( )( ) ()()() 22 212 211 2 1 iii i i ii i iiiiii xxy y x xy y xxxxxx −−− −−− − −−−−− − −−− B = ()( )()( ) ()()() 22 0.2 1.061 0.1 2.002 0.1 0.1 0.2 −−−−− −−− B = 0.04244 0.02002 0.002 −+ − = 11.21 ALGEBRAIC AND TRANSCENDENTAL EQUATION 91 The next approximation to the desired root is x i+1 = x i – 2 2 4 i i y BB Ay ±− x i+1 = 2.1 – () () () 2 2 0.061 11.21 11.21 24 0.061 × ±−× x i+1 = 2.1 – 0.122 11.21 11.1445 + = 2.094542 The procedure can now be repeated with three approximations as 2, 2.1 and 2.094542. Let x i–2 = 2, x i–1 = 2.1, x i = 2.094542 Then y i–2 = – 1, y i–1 = 0.061, y i = – 0.0001058 A = ()()()() ()()() 21 1 2 121 2 iiiiiiii iiiiii xxyyxxyy xxxxxx −− −− −−− − −−−−− −−− A = () ()()() ()( ) () 2 2.094542 0.061 0.0001058 2.1 2.094542 1 0.0001058 2.1 2 2.1 2.094542 2 2.094542 −+−−−+ −− − A = 6.194492 B = ()() () ()()() 2 2 2112 21 1 2 () iiiiiiii ii i ii i xxyyxxyy xx x xx x −−−− −− − − −−−−− −−− B = ()()() ()( )( ) 22 0.094542 0.0611058 0.005458 ( 0.9998942) 0.1 0.005458 0.094542 −−− −− B = 11.161799 The next approximation to the desired root is x i+ 1 = x i – 2 2 4 i i y BB Ay ±− x i+1 = 2.094542 – () ()()( ) 2 2 0.0001058 11.161799 11.161799 4 6.194492 0.0001058 ×− ±−− x i+1 = 2.094542 + 0.0002116 11.161799 11.161916 + = 2.094551 Hence the required root is 2.0945 correct up to 4 decimal places. 2.11 LIN BAIRSTOW METHOD Let the polynomial equation be p n (x)= a 0 x n + a 1 x n–1 + a 2 x n–2 + a 3 x n–3 + + a n–1 x + a n = 0 (1) where a 0 ≠ 0 and all a i ′ s are real. For polynomials, if the coefficients are all real valued then the complex roots occurs in conjugate pair. Therefore we extract the quadratic factors that are the products of the pairs of 92 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES complex roots, and then complex arithmetic can be avoided because such quadratic factors have real coefficients. This method extracts a quadratic factor from polynomial given by equation (1), which gives a pair of complex roots or a pair of real roots. Let us divide the given polynomial p n (x) by a quadratic factor x 2 + px + q, we obtain a quotient polynomial Q n–2 (x) of degree (n–2) and a linear remainder of the form Rx + S. Therefore p n (x)= (x 2 + px + q)Q n – 2 (x) + (Rx + S) (2) where Q n–2 (x)= b 0 x n–2 + b 1 x n–3 + b 2 x n–4 + +b n–3 x + b n–2 If (x 2 + px + q) is a factor of equation (1) then the remainder terms must vanish, therefore the problem is then to find p and q such that R(p, q) = 0 and S (p, q) = 0 (3) If we regularly change the values of p and q, we can make the remainder zero or at least make its coefficient smaller, however this equation (3) will normally not be so, for the approximated values of p and q. Since R and S are both functions of the two parameters p and q then the improved values are given by R(p + ∆p, q + ∆q) = 0 (4) S(p + ∆p, q + ∆q) = 0 Expand equation (4) by Taylor’s series for a function of two variables, were the second and higher order terms are neglected. We get R(p, q) + RR p pq ∂∂ ∆+ ∂∂ ∆q = 0 S(p, q) + SS p pp ∂∂ ∆+ ∂∂ ∆q = 0 (5) On solving equation (5), we get ∆p = – SR RS qq RS RS pq qp   ∂∂ −   ∂∂    ∂∂ ∂∂ −  ∂∂ ∂∂  ∆q = – RS SS pp RS RS pq qp   ∂∂ −   ∂∂    ∂∂ ∂∂ −  ∂∂ ∂∂  (6) Now the coefficients of b i ’s, R and S can be obtained by comparing the like powers of x in (2). i.e., a 0 x n + a 1 x n–1 + a 2 x n–2 + a 3 x n–3 + a n–2 x 2 + a n–1 x + a n = (x 2 + px + q) × (b 0 x n–2 + b 1 x n–3 + b 2 x n–4 + b n–4 x 2 + b n–3 x + b n–2 ) + (Rx + S) ALGEBRAIC AND TRANSCENDENTAL EQUATION 93 ⇒ a 0 = b 0 a 1 = b 1 + pb 0 a 2 = b 2 + pb 1 + qb 0 a 3 = b 3 + pb 2 + qb 1 a r = b r + pb r–1 + qb r–2 a n–1 = pb n–2 + qb n–3 + R a n = S + qb n–2 Hence, b 0 = a 0 b 1 = a 1 – pb 0 b 2 = a 2 – pb 1 – qb 0 b 3 = a 3 – pb 2 – qb 1 . b r = a r – pb r-1 – qb r–2 R = a n–1 – pb n–2 – qb n–3 S = a n – qb n-2 (7) Using b r = a r – pb r–1 – qb r–2 , r = 1, 2, 3, n (8) where b 0 = a 0 , b – 1 = 0, Also from (8) b n–1 = a n–1 – pb n–2 – qb n–3 Therefore a n–1 = b n–1 + pb n–2 + qb n–3 and b n = a n – pb n–1 – qb n–2 Therefore a n = b n – pb n–1 – qb n–2 So equation (7) becomes R = b n–1 and S = b n + pb n–1 (9) For partial derivaties of R and S equation (8) can be used i.e., differentiate equation (8) with respect to ‘p’ and ‘q’. – r b p ∂ ∂ = b r–1 + p 12rr bb q pp −− ∂∂ + ∂∂ or 0 1 bb pp − ∂∂ = ∂∂ = 0 (10) – r b q ∂ ∂ = b r–2 + p 12rr bb q qq −− ∂∂ + ∂∂ or 0 1 bb qq − ∂∂ = ∂∂ = 0; where r = 1, 2, 3, 4, (11) 94 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES Now, an equation (10) and (11) shows that 1 + ∂∂ = ∂∂ rr bb qp = 0 (12) Now set, – r b p ∂ ∂ = C r–1 so that – r b p ∂ ∂ = C r–2 ; r = 1, 2, 3, (13) Thus from (10) r b p ∂ − ∂ = 12 1 −− − ∂∂ ++ ∂∂ rr r bb bp q pp C r–1 = b r–1 – pC r–2 – qC r–3 or r C = 1 2 rr r b pC qC −− −− (14) where 1 C − = 0 and ∂ ∂∂ =− =− − =− = ∂∂ ∂ 0 1 010 00 () () or C pb b C a pb b pp p (15) From equations (9) and (15), R p ∂ ∂ = 1 2 () nn R bC pp −− ∂∂ ⇒=− ∂∂ R q ∂ ∂ = 13 () −− ∂∂ ⇒=− ∂∂ nn R bC qq S p ∂ ∂ = 1 ( ) n n b pb pp − ∂ ∂ + ∂∂ ⇒ 1 11 () − −− ∂ ∂∂ =− + + ∂∂∂ n nn b S Cp b p ppp = 121 0 −−− −− + nnn CpCb S q ∂ ∂ = 1 ( ) n n b pb qq − ∂ ∂ + ∂∂ = – C n–2 – pC n–3 On substituting these values in equation (6), we get p ∆ = −− − −− − − − − −  −++ −  +−−−−+  23 3 22 3 3 1 2 1 () ()[][ ] nn n nn n n n n n RC pC SC C C pC C C pC b Since R = b n–1 and S = b n + pb n–1 , therefore p ∆ = −− −− − −− −− − − −− −− −−  −− ++ −  ++ − − +  12 13 3 13 22 2 3 31 23 13 )] nn nn nn nn nn n n nn nn nn bC pbC bC pbC C C pC pC C C pC C b C p ∆ = () −−− −−−− − − −− 312 2 2311 nn n n nnnn bC b C CCCb (16) ALGEBRAIC AND TRANSCENDENTAL EQUATION 95 Similarly q ∆ = () () −− − − −−−− −− − −− 11 1 2 2 2311 nn n nn nnnn bC b bC CCCb Then the improved values of p and q are given by p 1 = p + p ∆ and q 1 = q + q ∆ The method for computation of b r and C r can be given by 01 2 3 2 1 012 3 2 1 01 4 3 2 01 2 3 2 1 012 3 2 01 . −− −−− −−− −− −− −−− − − − − − −− −− − − − − −−− − − − − − −− −− − nn n nnn nnn nn n nn aa a a a a a ppbpbpb pbpbpb qqbqb qbqbqb bb b b b b b ppCpCpC pCpC qqCqC 43 01 2 3 2 1 −− −− −− − nn nn qC qC CC C C C C The quotient polynomial 2 ( ) n Qx − = 2 ()/( ) n p xx px q ++ = 23 01 3 nn n b xbx b −− − ++ + can be obtained when p and q have been determined to the desired accuracy. This polynomial is called the defaulted polynomial. Another quadratic factor is of obtained using this default polynomial. If the initial approximation of p and q are not known then the last three terms of given polynomial a n – 2 x 2 + a n–1 x + a n = 0 can be used to get approximations as p 0 = 1 2 n n a a − − , q 0 = 2 n n a a − Example 9. Find the quadratic factor of the equation x 4 – 6x 3 + 18x 2 – 24x + 16 = 0 using Bairstow’s method where p 0 = – 1.5 and q 0 = 1. Also, find all the roots of the equation. Sol. Let the quadratic factor of the equation be x 2 + px + q. Using Bairstow’s method we find the values of p and q. First approximation: Let p 0 and q 0 be the initial approximations, then the first approximation can be obtained by p 1 = p 0 + p ∆ and q 1 = q 0 + q ∆ . Because given equation is of the degree four then. p ∆ = – () − −− 41 32 2 2133 bC bC CCCb (1) q ∆ = – () () −− −− 33 3 42 2 2133 bC b bC CCCb (2) . in conjugate pair. Therefore we extract the quadratic factors that are the products of the pairs of 92 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES complex roots, and then complex arithmetic can. be avoided because such quadratic factors have real coefficients. This method extracts a quadratic factor from polynomial given by equation (1), which gives a pair of complex roots or a pair of. of p and q, we can make the remainder zero or at least make its coefficient smaller, however this equation (3) will normally not be so, for the approximated values of p and q. Since R and S are