36 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES fa () fb () a x 2 y f x =() x 3 x 1 b X Y FIG. 2.1 Then, we bisect the interval and continue the process till the root is found to be desired accuracy. In the above figure, f(x 1 ) is positive; therefore, the root lies in between a and x 1 . The second approximation to the root now is x 2 = 1 2 (a + x 1 ). If f (x 2 ) is negative as shown in the figure then the root lies in between x 2 and x 1 , and the third approximation to the root is x 3 =(x 2 + x 1 )/2 and so on. This method is simple but slowly convergent. It is also called as Bolzano method or Interval halving method. 2.4.1 Procedure for the Bisection Method to Find the Root of the Equation f ( x ) = 0 Step 1: Choose two initial guess values (approximation) a and b (where (a > b)) such that f(a) . f(b) < 0. Step 2: Evaluate the mid point x 1 of a and b given by x 1 = 1 2 (a + b) and also evaluate f(x 1 ). Step 3: If f(a) . f(x 1 ) < 0, then set b = x 1 else set a = x 1 . Then apply the formula of step 2. Step 4: Stop evaluation when the difference of two successive values of x 1 obtained from step 2, is numerically less than the prescribed accuracy. 2.4.2 Order of Convergence of Bisection Method In Bisection Method, the original interval is divided into half interval in each iteration. If we take mid points of successive intervals to be the approximations of the root, one half of the current interval is the upper bound to the error. In Bisection Method, e i + 1 = 0.5e i or 1 0.5 i i e e + = ALGEBRAIC AND TRANSCENDENTAL EQUATION 37 Here e i and e i + 1 are the errors in i th and (i + 1) th iterations respectively. Comparing the above equation with 1 lim i k i i e A e + →∞ ≤ We get k = 1 and A = 0.5. Thus the Bisection Method is first order convergent or linearly convergent. Example 1. Find the root of the equation x 3 – x – 1 = 0 lying between 1 and 2 by bisection method. Sol. Let f(x)= x 3 – x – 1 = 0 Since f(1) = 1 3 – 1 – 1 = – 1, which is negative and f (2) = 2 3 – 2 – 1 = 5, which is positive Therefore, f(1) is negative and f(2) is positive, so at least one real root will lie between 1 and 2. First iteration: Now using Bisection Method, we can take first approximation 1 12 3 1.5 22 x + === Then, f(1.5) = (1.5) 3 – 1.5 – 1 = 3. 375 – 1.5 – 1 = 0.875 ∴ f (1.5) > 0 that is, positive So root will now lie between 1 and 1.5. Second iteration: The Second approximation is given by + === 2 11.5 2.5 1.25 22 x . Then, f(1.25) = (1.25) 3 – 1.25 – 1 = 1.953 – 2.25 = – 0.297 < 0 ∴ f (1.25) is negative. Therefore, f(1.5) is positive and f(1.25) is negative, so that root will lie between 1.25 and 1.5. Third iteration: The third approximation is given by 3 1.25 1.5 1.375 2 x + == x 3 = 1.375 Now f(1.375) = (1.375) 3 – 1.375 – 1 f(1.375) = 0.2246 ∴ f(1.375) is positive. ∴ The required root lies between 1.25 and 1.375. Fourth iteration: The fourth approximation is given by 4 1.25 1.375 1.313 2 x + == Now f(1.313) = (1.313) 3 – 1.313 – 1 f(1.313) = – 0.0494 38 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES Therefore, f(1.313) is negative and f(1.375) is positive. Thus root lies between 1.313 and 1.375. Fifth iteration: The fifth approximation is given by 5 1.313 1.375 1.344 2 x + == ∴ f(1.344) = (1.344) 3 – 1.344 – 1 = 0.0837 ∴ f(1.313) > 0 ∴ f(1.313) is negative and f(1.344) is positive, so root lies between 1.313 and 1.344. Sixth iteration: The sixth approximation is given by 6 1.313 1.344 1.329 2 x + == ∴ f(1.329) = (1.329) 3 – 1.329 – 1 = 0.0183 ∴ f(1.329) > 0 ∴ f(1.313) is negative and f(1.329) is positive, so that the required root lies between 1.313 and 1.329. Seventh iteration: The seventh approximation is given by 7 1.313 1.329 1.321 2 x + == ∴ f(1.321) = (1.321) 3 – 1.321 – 1 = – 0.0158 ∴ f(1.321) < 0 ∴ f(1.321) is negative and f(1.329) is positive, so that the required root lies between 1.321 and 1.329. Eighth iteration: The eighth approximation is given by 8 1.321 1.329 1.325 2 x + == From above iterations, the root of 3 () 1 0 fx x x=−−= up to three places of decimals is 1.325, which is of desired accuracy. Example 2. Find the root of the equation x 3 – x – 4 = 0 between 1 and 2 to three places of decimal by Bisection method. Sol. Given 3 () 4 fx x x=−− We want to find the root lie between 1 and 2. At 0 1 x = ⇒ 3 0 ()(1) 14 4 =−−=−fx negative At 1 2 x = ⇒ 3 1 ()(2) 24 2 fx =−−= positive This implies that root lies between 1 and 2. First iteration: Here, + == === 01 2 12 3 1, 2, 1.5 22 xx x Now, =− = 01 () 4,()2 fx fx . Then, =−−=− 3 2 ( ) (1.5) 1.5 4 2.125 fx . ALGEBRAIC AND TRANSCENDENTAL EQUATION 39 Since f(1.5) is negative and f(2) is positive. So root will now lie between 1.5 and 2. Second iteration: Here, 012 1.5 2 1.5, 2, 1. 75 2 x xx + ==== Also, =− = 01 ( ) 2.125, ( ) 2 fx fx then, =−−=− 3 2 ( ) (1.75) 1.75 4 0.39062 fx Since f(1.75) is negative and f(2) is positive, therefore the root lies between 1.75 and 2. Third iteration: Here, + ==== 012 1.75 2 1.75, 2, 1.875 2 xxx Also, =− = 01 ( ) 0.39062, ( ) 2 fx fx then, =−−= 3 2 ( ) (1.875) 1.875 4 0.71679 fx Since f(1.75) is negative and f(1.875) is positive, therefore the root lies between 1.75 and 1.875. Fourth iteration: Here, 01 2 1.75 1.875 1.75, 1.875, 1.8125 2 xx x + == = = Also, =− = 01 ( ) 0.39062, ( ) 0.71679 fx fx then, =−−= 3 2 ( ) (1.8125) 1.8125 4 0.14184 fx Since f(1.75) is negative and f(1.8125) is positive, therefore the root lies between 1.75 and 1.8125. Fifth iteration: Here, 01 2 1.75 1.8125 1.75, 1.8125, 1.78125 2 xx x + == = = Also, =− = 01 ( ) 0.39062, ( ) 0.14184 fx fx then, =−−=− 3 2 ( ) (1.78125) 1.78125 4 0.12960 fx Since f(1.78125) is negative and f(1.8125) is positive, therefore the root lies between 1.78125 and 1.8125. Repeating the process, the successive approximations are x 6 = 1.79687, x 7 = 1.78906, x 8 = 1.79296, x 9 = 1.79491, x 10 = 1.79589, x 11 = 1.79638, x 12 = 1.79613 From the above discussion, the value of the root to three decimal places is 1.796. Example 3. Using Bisection Method determine a real root of the equation 3 () 8 2 1 0. fx x x=−−= Sol. It is given that 3 () 8 2 1 0 fx x x=−−= . Then 3 (0) 8(0) 2(0) 1 1 f =−−=− and 3 (1) 8(1) 2(1) 1 5 f =−−= Therefore, f(0) is negative and f(1) is positive so that the root lies between 0 and 1. First approximation: First approximation to the root is given by 1 01 0.5 2 x + == ∴ 3 (0.5) 8(0.5) 2(0.5) 1 1 f =−−=− , which is negative. Thus f(0.5) is negative and f(1) is positive. Then the root lies between 0.5 and 1. 40 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES Second approximation: The second approximation to the root is given by 2 0.5 1 0.75 2 x + == ∴ 3 (0.75) 8(0.75) 2(0.75) 1 f =−− = 2.265 – 2.5 = 0.875, which is positive. Since f(0.5) is negative, while f(0.75) is positive. Therefore, the root lies between 0.5 and 0.75. Third approximation: The third approximation to the root is given by 3 0.5 0.75 0.625 2 x + == ∴ 3 (0.625) 8(0.625) 2(0.625) 1 f =−− =1.935 – 2.25 = – 0.297, which is negative. Therefore f(0.75) is positive, while f(0.625) is obtained negative. Therefore, the root lies between 0.625 and 0.75. Fourth approximation: The fourth approximation to the root is given by + == 4 0.625 0.75 0.688 2 x ∴ 3 (0.688) 8(0.688) 2(0.688) 1 f =−− = 2.605 – 2.376 = 0.229, which is positive Therefore f(0.688) is obtained positive, while f(0.625) is negative. Therefore, the root lies between 0.625 and 0.688. Fifth approximation: The fifth approximation to the root is given by x 5 = 0.625 0.688 2 + = 0.657 ∴ f(0.673) = 8(0.657) 3 – 2(0.673) – 1 = 2.269 – 2.314 = – 0.045, which is negative. Therefore f(0.657) is negative and f(0.688) is positive so the root lies between 0.657 and 0.688. Sixth approximation: The sixth approximation to the root is given by 6 0.657 0.688 0.673 2 x + == ∴ 3 (0.673) 8(0.673) 2(0.673) 1 f =−− = 2.439 2.346 1.093,−= which is positive. Therefore f (0.673) is positive and f(0.657) is negative so the root lies between 0.657 and 0.673. ALGEBRAIC AND TRANSCENDENTAL EQUATION 41 Seventh approximation: The seventh approximation to the root is given by 7 0.657 0.673 0.665 2 x + == ∴ =−− 3 (0.665) 8(0.665) 2(0.665) 1 f = 2.353 – 2.33 = 0.023, which is positive. Therefore f(0.665) is positive and f(0.657) is negative so that the root lies between 0.657 and 0.665. Eighth approximation: The eighth approximation to the root is given by 8 0.657 0.665 0.661 2 x + == From last two approximations, i.e., 7 0.6 65 x = and 8 0.6 61 x = it is observed that the approximate value of the root of f(x) = 0 up to two decimal places is 0.66. Example 4. Perform five interactions of Bisection method to obtain the smallest positive root of equation 3 () 5 1 0 fx x x=−+= . Sol. Let f(2.1) = –ve, f(2.15) = +ve. Therefore the root lies between 2.1 and 2.15. First approximation to the root is 1 2.1 2.15 2.1 25 2 x + == Now, (2.125) vef =− Therefore the root lies between 2.215 and 2.15. Second approximation to the root is 2 2.125 2.15 2.1375 2 x + == Now, (2.1375) vef =+ Therefore the root lies between 2.125 and 2.1375. Third approximation to the root is 3 2.125 2.1375 2.13125 2 x + == Now, (2.13125) vef =+ Therefore the root lies between 2.125 and 2.13125. Fourth approximation to the root is 4 2.125 2.13125 2.1281 2 x + == Now, (2.1281) vef =− 42 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES Therefore the root lies between 2.1281 and 2.13125. Fifth approximation to the root is 5 2.1281 2.13125 2.1 29 2 x + == Hence the required root is 2.129. Example 5. Find the real root of equation = 10 xlog x 1.2 by Bisection Method. Sol. Let =−= 10 () log 1.2 0 fx x x So that =−=−< 10 (1) 1 log 1 1.2 1.2 0 f and =−=− 10 (2) 2log 2 1.2 0.602 1.2 f =− <0.598 0 and =− 10 (3) 3log 3 1.2 f =−=>3(0.4771) 1.2 0.2313 0 Thus f (2) is negative and f (3) is positive, therefore, the root will lie between 2 and 3. First approximation: The first approximation to the root is 1 23 2.5 2 x + == Again, 10 (2.5) 2.5log 2.5 1.2 f =− = 2.5 (0.3979) – 1.2 = 0.9948 – 1.2 = – 0.2052 < 0 Thus, f (2.5) is negative and f (3) is positive, therefore, the root lies between 2.5 and 3. Second Approximation: The second approximation to the root is 2 2.5 3 2.75 2 x + == Now, 10 (2.75) 2.75 log 2.75 1.2 f =− =−=−=>2.75(0.4393) 1.2 1.2081 1.2 0.0081 0 Thus, f (2.75) is positive and f (2.5) is negative, therefore, the root lies between 2.5 and 2.75. Third approximation: The third approximation to the root is 3 2.5 2.75 2.625 2 x + == Again, 10 ( 2.625) 2.625log 2.625 1 .2 f =− =−=−=−<2.625(0.4191) 1.2 1.1001 1.2 0.0999 0 Thus, f(2.625) is found to be negative and f(2.75) is positive, therefore, the root lies between 2.625 and 2.75. Fourth approximation: The fourth approximation to the root is 4 2.625 2.75 2.6875 2 x + == Again, f(2.6875) = 2.6875 log 10 2.6875 – 1.2 = 2.6875(0.4293) – 1.2 = 1.1537 – 1.2 = – 0.0463 < 0 ALGEBRAIC AND TRANSCENDENTAL EQUATION 43 Thus, f(2.6875) is negative and f(2.75) is positive, therefore, the root lies between 2.6875 and 2.75. Fifth approximation: The fifth approximation to the root is 5 2.6875 2.75 2.7188 2 x + == Now, f(2.7188) = 2.7188 log 10 2.7188 – 1.2 = 2.7188 (0.4344) – 1.2 = 0.1810 – 1.2 = – 0.019 < 0 Thus, f(2.7188) is negative and f(2.75) is positive, therefore, the root lies between 2.7188 and 2.75 Sixth approximation: The sixth approximation to the root is 6 2.7188 2.75 2.7344 2 x + == Again, f(2.734) = 2.734 log 10 2.734 – 1.2 = 2.734 (0.4368) – 1.2 = 1.1942 – 1.2 = – 0.0058 < 0 Thus, f(2.734) is negative and f(2.75) is positive, therefore, the root lies between 2.734 and 2.75 Seventh approximation: The seventh approximation to the root is 7 2.734 2.75 2.742 2 x + == Again, f(2.742) = 2.742 log 10 2.742 – 1.2 = 2.742 (0.4381) – 1.2 = 1.2012 – 1.2 = 0.0012 > 0 Thus, f(2.742) is positive and f(2.734) is negative, therefore, the root lies between 2.734 and 2.742. Eighth approximation: The eighth approximation to the root is 8 2.734 2.742 2.738 2 x + == Hence, from the approximate value of the roots x 7 and x 8 , we observed that, up to two places of decimal, the root is 2.74 approximately. Example 6. Using Bisection Method, find the real root of the equation f(x) = 3x – 1sinx 0+= Sol. The given equation f(x)= 3x – 1sin 0x+= is a transcendental equation. Given f(x)= 3x – 1sin 0x+= (1) Then f(0) = 0 – 1sin0 1 +=− and f(1) = 3 – 1sin13 1.84 14 +=− = 3 – 1.3570 = 1.643 > 0 Thus f(0) is negative and f(1) is positive, therefore, a root lies between 0 and 1. 44 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES First approximation: The first approximation to the root is given by 1 01 0.5 2 x + == Now, f(0.5) = 3(0.5) – () 1sin0.5 + = 1.5 – 1.4794 = 1.5 – 1.2163 = 0.2837 > 0 Thus, f(0.5) is positive, while f(0) is negative, therefore, a root lies between 0 and 0.5. Second approximation: The second approximation to the root is given by 2 00.5 0.25 2 x + == Again, f(0.25) = 3(0.25) – () 1 sin 0.25 + = 0.75 – 1.2474 = 0.75 – 1.1169 = – 0.3669 < 0 Thus, f(0.25) is obtained to be negative and f (0.5) is positive; therefore, a root lies between 0.25 and 0.5. Third approximation: The third approximation to the root is given by 3 0.25 0.5 0.375 2 x + == Now, f(0.375) = 3(0.375) – () 1sin0.375 + = 1.125 – 1.3663 = 1.125 – 1.1689 = – 0.0439 < 0 Thus, f(0.375) is negative and f(0.5) is positive, therefore, a root lies between 0.375 and 0.5. Fourth approximation: The fourth approximation to the root is given by 4 0.375 0.5 0.4375 2 x + == Now, f (0.4375) = 3(0.4375) – () 1 sin 0.4375 + = 1.3125 – 1.4237 = 1.3125 – 1.1932 = 0.1193 > 0 Thus, f(0.4375) is positive, while f(0.375) is negative, therefore, a root lies between 0.375 and 0.4375. Fifth approximation: The fifth approximation to the root is given by 5 0.375 0.4375 0.4063 2 x + == Again, f(0.4063) = 3(0.4063) – () 1 sin 0.4063 + = 1.2189 – 1.3952 = 1.2189 – 1.1812 – 0.0377 > 0 Thus, f(0.4063) is positive, while f(0.375) is negative, therefore, a root lies between 0.375 and 0.4063. ALGEBRAIC AND TRANSCENDENTAL EQUATION 45 Sixth approximation: The sixth approximation to the root is given by 6 0.375 0.4063 0.39 07 2 x + == Again, f(0.3907) = 3(0.3907) – () 1 sin 0.390 7 + = 1.1721 – 1.380 8 = 1.1721 – 1.1751 = – 0.003 < 0 Thus, f (0.3907) is negative, while f (0.4063) is positive, therefore, a root lies between 0.3907 and 0.4063. Seventh approximation: The seventh approximation to the root is given by 7 0.3907 0.4063 0.3985 2 x + == From the last two observations, that is, x 6 = 0.3907 and x 7 = 0.3985, the approximate value of the root up to two places of decimal is given by 0.39. Hence the root is 0.39 approximately. Example 7. Find a root of the equation f(x) = x 3 – 4x – 9 = 0, using the Bisection method in four stages. Sol. Given f(x) = x 3 – 4x – 9 = 0 Then f(2) = 2 3 – 4(2) – 9 = – 9 and f(3) = (3) 3 – 4 (3) – 9 = 6 Therefore, the root lies between 2 and 3. First approximation: First approximation to the root is given by 1 23 2.5 2 x + == Thus f(2.5) = (2.5) 3 – 4(2.5) – 9 = 15.625 – 19 = – 3.375 Therefore, the root lies between 2.5 and 3. Second approximation: Second approximation to the root is given by 2 2.5 3 2.75 2 x + == Thus f(2.75) = (2.75) 3 – 4 (2.75) – 9 = 20.797 – 20 = 0.797 Therefore, f(2.75) is positive and f(2.5) is negative. Thus the root lies between 2.5 and 2.75. Third approximation: Third approximation to the root is given by 3 2.5 2.75 2.625 2 x + == Now, f(2.625) = (2.625) 3 – 4 (2.625) – 9 = 18.088 – 19.5 = – 1.412 Therefore, f(2.625) is negative while f(2.75) is positive. Thus the root lies between 2.625 and 2.75. Fourth approximation: Fourth approximation to the root is given by 4 2.625 2.75 2.6875 2 x + == Hence, after the four steps the root is 2.6875 approximately. . between 0 .65 7 and 0 .66 5. Eighth approximation: The eighth approximation to the root is given by 8 0 .65 7 0 .66 5 0 .66 1 2 x + == From last two approximations, i.e., 7 0 .6 65 x = and 8 0 .6 61 x =. negative. Thus f(0.5) is negative and f(1) is positive. Then the root lies between 0.5 and 1. 40 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES Second approximation: The second approximation. positive. Therefore f (0 .67 3) is positive and f(0 .65 7) is negative so the root lies between 0 .65 7 and 0 .67 3. ALGEBRAIC AND TRANSCENDENTAL EQUATION 41 Seventh approximation: The seventh approximation to the