46 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES PROBLEM SET 2.1 1. Find the smallest root lying in the interval (1, 2) up to four decimal places for the equation x 6 – x 4 – x 3 – 1 = 0 by Bisection Method [Ans. 1.4036] 2. Find the smallest root of x 3 – 9x + 1 = 0, using Bisection Method correct to three decimal places. [Ans. 0.111] 3. Find the real root of e x = 3x by Bisection Method. [Ans. 1.5121375] 4. Find the positive real root of x – cos x = 0 by Bisection Method, correct to four decimal places between 0 and 1. [Ans. 0.7393] 5. Find a root of x 3 – x – 11 = 0 using Bisection Method correct to three decimal places which lies between 2 and 3. [Ans. 2.374] 6. Find the positive root of the equation xe x = 1 which lies between 0 and 1. [Ans. 0.5671433] 7. Solve x 3 – 9x + 1 = 0 for the root between x = 2 and x = 4 by the method of Bisection. (U.P.T.U. 2005) [Ans. 2.94282] 8. Compute the root of log x = cos x correct to 2 decimal places using Bisection Method. [Ans. 1.5121375] 9. Find the root of tan x + x = 0 up to two decimal places which lies between 2 and 2.1 using Bisection Method. [Ans. 2.02875625] 10. Use the Bisection Method to find out the positive square root of 30 correct to 4 decimal places. [Ans. 5.4771] 2.5 FALSE POSITION METHOD (OR REGULA FALSI METHOD) This method is essentially same as the bisection method except that instead of bisecting the interval. In this method, we choose two points x 0 and x 1 such that f(x 0 ) and f(x 1 ) are of opposite signs. Since the graph of y = f(x) crosses the X-axis between these two points, a root must lie in between these points. Consequently, f(x 0 ) f(x 1 ) < 0. Equation of the chord joining points {x 0 , f(x 0 )} and {x 1 , f (x 1 )} is () () () () 10 00 10 fx fx xxx xx − −= − − yf The method consists in replacing the curve AB by means of the chord AB and taking the point of intersection of the chord with X-axis as an approximation to the root. So the abscissa of the point where chord cuts y = 0 is given by () () () 10 20 0 10 xx xx fx fx fx − =− − Y X A{ , ( ) )} xfx 00 x 0 x 3 x 2 x 3 P( ) x B FIG. 2.2 ALGEBRAIC AND TRANSCENDENTAL EQUATION 47 The value of x 2 can also be put in the following form: () () () () 01 10 2 10 xf x xf x x fx fx − = − In general, the (i + 1)th approximation to the root is given by () ( ) () ( ) 11 1 1 iiii i ii xfx xfx x fx fx −− + − − = − 2.5.1 Procedure for the False Position Method to Find the Root of the Equation f ( x ) = 0 Step 1: Choose two initial guess values (approximations) x 0 and x 1 (where x 1 > x 0 ) such that f(x 0 ).f(x 1 ) < 0. Step 2: Find the next approximation x 2 using the formula () () () () 01 10 2 10 xf x xf x x fx fx − = − and also evaluate f(x 2 ). Step 3: If f(x 2 ) f(x 1 ) < 0, then go to the next step. If not, rename x 0 as x 1 and then go to the next step. Step 4: Evaluate successive approximations using the formula () ( ) () ( ) 11 1 1 , where = 2, 3, 4, iiii i ii xfx xfx xi fx fx −− + − − = − But before applying the formula for x i + 1 , ensure whether f(x i–1 ). f(x i ) < 0; if not, rename x i–2 as x i–1 and proceed. Step 5: Stop the evaluation when 1 , − −<ε ii xx where ε is the prescribed accuracy. 2.5.2 Order (or Rate) of Convergence of False Position Method The general iterative formula for False Position Method is given by () ( ) () ( ) 11 1 1 iiii i ii xfxxfx x fx f x −− + − − = − (1) where x i–1 , x i and x i+1 are successive approximations to the required root of f(x) = 0. The formula given in (1), can also be written as: () () () ( ) 1 1 1 ii i ii ii xx fx xx fx fx − + − − =− − (2) Let α be the actual (true) root of f(x) = 0, i.e., f( α ) = 0. If e i–1 , e i and e i+1 are the successive errors in (i – 1)th, ith and (i + 1)th iterations respectively, then e i–1 = x i–1 – α , e i = x i – α , e i + 1 = x i + 1 – α or x i–1 = α + e i–1 , x i = α + e i , x i+1 = α + e i+1 48 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES Using these in (2), we obtain ()() ()( ) 1 1 1 ii i ii ii ee f e ee fefe − + − −α+ α+ =α+ − α+ − α+ or () () () () 1 1 1 ii i ii ii ee f e ee fefe − + − −α+ =− α+ − α+ (3) Expanding f( α + e i ) and f(α + e i –1 ) in Taylor’s series around α , we have () () () () () () () () () () 1 2 1 1 22 1 2 22 i ii ii i ii ii e ee f ef f ee ee fef f fef f − − + −  ′′′ −α+α+α+    =−   ′′′ ′′ α+ α+ α+ − α+ ′α+ α+       i.e., () () () () () () () − + − −  ′′′ −α+α+α    =−   − ′′′ −α+ α      2 1 1 22 1 1 2 2 i ii i ii ii ii e ee f ef f ee ee ee f f , [on ignoring the higher order terms] i.e. () () () () () + −  ′′′ α+ α+ α   == +   ′′ α+ α     2 1 1 2 ' 2 i i ii ii e fef f ee ee ff i.e. () () () () −  ′′′ α+ α   − +   ′′′ α+ α     2 + 1 1 2 = 2 i i ii ii e ef f ee ee ff [since f( α ) = 0] i.e. () () () () 2 1 1 2 1 2 i i ii ii f e e f ee fee f + − ′′ α +  ′ α  =− ′′  α+  +   ′ α   , [on dividing numerator and denominator by f ′( α ) i.e. () () () () 1 2 1 1 1 22 iii iii ff eee eee ff − − +  ′′ ′′  αα +  =− + +    ′′ αα     i.e. () () () () 2 1 1 1 22 ii i iii ff eee eee ff − + ′′ ′′  αα +  =− + −    ′′ αα    ALGEBRAIC AND TRANSCENDENTAL EQUATION 49 i.e., () () () () 2 22 11 1 () () () 22()4 ii i i i i i ii ff f ee e e e e e ee ff f −− +  ′′ ′′ ′′   αα α ++   =− + −  ′′ ′ αα α     i.e., () () () + ′′ α =− + ′ α 1 2 1 0 2 iii i f eee e f If e i–1 and e i are very small, then ignoring 0(e 2 i ), we get () () 11 2 iii f eee f +− ′′ α = ′ α (4) which can be written as e i + 1 = e i e i–1 M, where M = () () 2 f f ′′ α ′ α and would be a constant (5) In order to find the order of convergence, it is necessary to find a formula of the type e i + 1 = Ae k i with an appropriate value of k. (6) With the help of (6), we can write e i = Ae 1 k i− or e i–1 = (e i /A) 1/k Now, substituting the value of e i+1 and e i–1 in (5), we get Ae k i = e i . 1/ . k i e M A    or e k i = MA –(1 + 1/k) .e i (1+1/k) (7) Comparing the powers of e i on both sides of (7), we get k = 1 + (1/k) or k 2 – k – 1 = 0 (8) From (8), taking only the positive root, we get k = 1.618 By putting this value of k in (6), we have i+1 1.618 1 1.618 or ii i e eAe A e + == Comparing this with 1 lim i k i i e A e + →∞  ≤    , we see that order (or rate) of convergence of false position method is 1.618. Example 1. Find a real root of the equation f(x) = x 3 – 2x – 5 = 0 by the method of false position up to three places of decimal. Sol. Given that f(x)= x 3 – 2x – 5 = 0 So that f(2) = (2) 3 – 2(2) – 5 = – 1 50 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES and f(3) = (3) 3 – 2(3) – 5 = 16 Therefore, a root lies between 2 and 3. First approximation: Therefore taking x 0 = 2, x 1 = 3, f(x 0 ) = – 1, f(x 1 )= 16, then by Regula- Falsi method, we get () () () 10 20 0 10 xx xx fx fx fx − =− − () 32 1 2 1 2 2.0588 16 1 17 − =− −=+ = + Now, f(x 2 )= f(2.0588) = (2.0588) 3 – 2 (2.0588) – 5 = – 0.3911 Therefore, root lies between 2.0588 and 3. Second approximation: Now, taking x 0 = 2.0588, x 1 = 3, f(x 0 ) = – 0.3911, f(x 1 ) = 16, then by Regula-Falsi method, we get () () () 10 30 0 10 xx xx fx fx fx − =− − = 2.0588 – 3 2.0588 16 0.3911 − + (– 0.3911) = 2.0588 + 0.0225 = 2.0813 Now, f(x 3 )= f (2.0813) = (2.0813) 3 – 2(2.0813) – 5 = – 0.1468 Therefore, root lies between 2.0813 and 3. Third approximation: Taking x 0 = 2.0813 and x 1 = 3, f(x 0 ) = – 0.1468, f(x 1 ) = 16. Then by Regula-Falsi method, we get () () () 10 40 0 10 xx xx fx fx fx − =− − = 2.0813 – 3 2.0813 16 0.1468 − + (– 0.1468) = 2.0813 + 0.0084 = 2.0897 Now, f(x 4 )= f(2.0897) = (2.0897) 3 – 2 (2.0897) – 5 = 9.1254 – 9.1794 = – 0.054 Therefore, root lies between 2.0897 and 3. Fourth approximation: Now, taking x 0 = 2.0897, x 1 = 3, f (x 0 ) = – 0.054, f (x 1 ) = 16, then by Regula-Falsi method, we get () () () 10 50 0 10 xx xx fx fx fx − =− − = 2.0897 – () 3 2.0897 0.054 16 0.054 − − + = 2.0897 + 0.0031 = 2.0928 ALGEBRAIC AND TRANSCENDENTAL EQUATION 51 Now, f(x 5 )= f(2.0928) = (2.0928) 3 – 2(2.0928) – 5 = 9.1661 – 9.1856 = – 0.0195 Therefore, root lies between 2.0928 and 3. Fifth approximation: Now, taking x 0 = 2.0928, x 1 = 3, f(x 0 ) = – 0.0195, f(x 1 ) = 16, then we get () () () 10 60 0 10 xx xx fx fx fx − =− − = 2.0928 – () 3 2.0928 0.0195 16 0.0195 − − + = 2.0928 + 0.0011 = 2.0939 Now, f(x 6 )= f(2.0939) = (2.0939) 3 – 2 (2.0939) – 5 = 9.1805 – 9.1879 = – 0.0074 Thus the root lies between 2.0939 and 3. Sixth approximation: Now, taking x 0 = 2.0939, x 1 = 3, f(x 0 ) = – 0.0074, f(x 1 ) = 16, then we get () () () 10 70 0 10 xx xx fx fx fx − =− − = 2.0939 – () 32.0939 0.0074 16 0.0074 − − + = 2.0939 + 0.00042 = 2.0943 Now, f(x 7 )= f(2.0943) = (2.0943) 3 – 2(2.0943) – 5 = 9.1858 – 9.1886 = – 0.0028 Therefore, root lies between 2.0943 and 3. Seventh approximation: Taking x 0 = 2.0943, x 1 = 3, f(x 0 ) = – 0.0028, f(x 1 ) = 16, then by Falsi position method, we get () () () 10 80 0 10 xx xx fx fx fx − =− − = 2.0943 – () 32.0943 0.0028 16 0.0028 − − + = 2.0943 + 0.00016 = 2.0945 Hence, the root is 2.094 correct to three decimal places. Example 2. Find the real root of the equation f(x) = x 3 – 9x + 1 = 0 by Regula-Falsi method. Sol. Let f(x)= x 3 – 9x + 1 = 0 (1) So that f(2) = (2) 3 – 9(2) + 1 = – 9 f(3) = (3) 3 – 9(3) + 1 = 1 52 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES Since f(2) and f(3) are of opposite signs, therefore the root lies between 2 and 3, so taking x 0 = 2, x 1 = 3, f(x 0 ) = – 9, f(x 1 ) = 1, then by Regula-Falsi method, we get First approximation: () () () 10 20 0 10 xx xx fx fx fx − =− − () 32 9 2922.9 19 10 − =− ×−=+ = + Now, f(x 2 )= f(2.9) = (2.9) 3 – 9 (2.9) + 1 = 24.389 – 25.1 = – 0.711 Second approximation: The root lies between 2.9 and 3. Therefore, taking x 0 = 2.9, x 1 = 3, f(x 0 ) = – 0.711, f(x 1 ) = 1. Then () () () 10 30 0 10 xx xx fx fx fx − =− − = 2.9 – () 32.9 0.711 1 0.711 − − + = 2.9 + 0.0416 = 2.9416 Now, f(x 3 )= f(2.9416) = (2.9416) 3 – 9(2.9416) + 1 = 25.4537 – 25.4744 = – 0.0207 Third approximation: The root lies between 2.9416 and 3. Therefore, taking x 0 = 2.9416, x 1 = 3, f(x 0 ) = – 0.0207, f(x 1 ) = 1. Then we get () () () 10 40 0 10 xx xx fx fx fx − =− − = 2.9416 – () 32.9416 0.0207 10.0207 − − + = 2.9416 + 0.0012 = 2.9428 Now, f(x 4 )= f (2.9428) = (2.9428) 3 – 9(2.9428) + 1 = 25.4849 – 25. 4852 = – 0.0003 Fourth approximation: The root lies between 2.9428 and 3. Therefore, taking x 0 = 2.9428, x 1 = 3, f (x 0 ) = – 0.0003, f (x 1 ) = 1. Then by False Position method, we have () () () 10 50 0 10 xx xx fx fx fx − =− − = 2.9428 – () 3 2.9428 0.0003 1 0.0003 − − + = 2.9428 + 0.000017 = 2.942817 Hence, the root is 2.9428 correct to four places of decimal. ALGEBRAIC AND TRANSCENDENTAL EQUATION 53 Example 3. Using the method of False Position, find the root of equation x 6 – x 4 – x 3 –1 = 0 up to four decimal places. Sol. Let f(x) = x 6 – x 4 – x 3 – 1 f(1.4) = (1.4) 6 – (1.4) 4 – (1.4) 3 – 1 = – 0.056 f(1.41) = (1.41) 6 – (1.41) 4 – (1.41) 3 – 1 = 0.102 Hence the root lies between 1.4 and 1.41. Using the method of False Position, () () () 10 20 0 10 xx xx fx fx fx − =− − = 1.4 – () 1.41 1.4 0.056 0.102 0.056 − − + = 1.4 + () 0.01 0.056 1.4035 0.158  =   Now, f (1.4035) = (1.4035) 6 – (1.4035) 4 – (1.4035) 3 – 1 f (x 2 ) = – 0.0016 (–ve) Hence the root lies between 1.4035 and 1.41. Using the method of False Position, () () () 12 32 2 12 xx xx fx fx fx − =− − = 1.4035 – − − + 1.41 1.4035 ( 0.0016) 0.102 0.0016 = 1.4035 + () 0.0065 0.0016 1.4036 0.1036  =   Now, f(1.4036) = (1.4036) 6 – (1.4036) 4 – (1.4036) 3 – 1 f(x 3 ) = – 0.00003 (–ve) Hence the root lies between 1.4036 and 1.41. Using the method of False Position, () () () 13 43 3 13 xx xx fx fx fx − =− − = 1.4036 – () 1.41 1.4036 0.00003 0.102 0.00003 − − + = 1.4036 + () 0.0064 0.00003 1.4036 0.10203  =   Since, x 3 and x 4 are approximately the same upto four places of decimal, hence the required root of the given equation is 1.4036. 54 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES Example 4. Find a real root of the equation f(x) = x 3 – x 2 – 2 = 0 by Regula-Falsi method. Sol. Let f(x)= x 3 – x 2 – 2 = 0 Then, f(0) = – 2, f(1) = – 2 and f(2) = 2 Thus, the root lies between 1 and 2. First approximation: Taking 1 00 1, 2, ( ) 2 x xfx== =− and 1 () fx = 2. Then by Regula-Falsi method, an approximation to the root is given by 10 20 0 1) 0 () (() xx xx fx fx fx − =− − 21 1 1(2)11.5 22 2 − =− − =+ = + Now, 2 () (1.5) fx f= = (1.5) 3 – (1.5) 2 – 2 = 3.375 – 4.25 = – 0.875 Thus, the root lies between 1.5 and 2. Second approximation: Taking 01 0 1.5, 2, ( ) 0.875 xxfx== =− and 1 ()2. fx = Then the next approximation to the root is given by 10 30 0 1) 0 () (() xx xx fx fx fx − =− − = 1.5 – () 21.5 0.875 20.875 − − + = 1.5 + 0.1522 = 1.6522 Now, f(x 3 )= f(1.6522) = (1.6522) 3 – (1.6522) 2 – 2 = 4.5101 – 4.7298 = – 0.2197 Thus, the root lies between 1.6522 and 2. Third approximation: Taking x 0 = 1.6522, x 1 = 2, f(x 0 ) = – 0.2197 and f(x 1 ) = 2. Then the next appoximation to the root is given by 10 40 0 1) 0 () (() xx xx fx fx fx − =− − = 1.6522 – () 2 1.6522 0.2197 2 0.2197 − − + = 1.6522 + 0.0344 = 1.6866 Now, f(x 4 )= f(1.6866) = (1.6866) 3 – (1.6866) 2 – 2 = 4.7977 – 4.8446 = – 0.0469 Thus, the root lies between 1.6866 and 2. ALGEBRAIC AND TRANSCENDENTAL EQUATION 55 Fourth approximation: Taking x 0 = 1.6866, x 1 = 2, f (x 0 ) = – 0.046 and f (x 1 ) = 2. Then the root is given by 10 50 0 1) 0 () (() xx xx fx fx fx − =− − = 1.6866 – () 1.6866 0.0469 2 0.0469 − − + = 1.6866 + 0.0072 = 1.6938 Now, f(x 5 )= f(1.6938) = (1.6938) 3 – (1.6938) 2 – 2 = 4.8594 – 4.8690 = – 0.0096 Thus, the root lies between 1.6938 and 2. Fifth approximation: Taking x 0 = 1.6938, x 1 = 2, f(x 0 ) = – 0.0096 and f(x 1 ) = 2. Then the next approximation to the root is given by 10 60 0 1) 0 () (() xx xx fx fx fx − =− − = 1.6938 – () 2 1.6938 0.0096 2 0.0096 − − + = 1.6938 + 0.0015 = 1.6953 Now, f(x 6 )= f(1.6953) = (1.6953) 3 – (1.6953) 2 – 2 = 4.8724 – 4.8740 = – 0.0016 Therefore, the root lies between 1.6953 and 2. Sixth approximation: Taking x 0 = 1.6953, x 1 = 2, f(x 0 ) = – 0.0016 and f(x 1 ) = 2. Then the next approximation to the root is 10 70 0 1) 0 () (() xx xx fx fx fx − =− − = 1.6953 – () 2 1.6953 0.0016 2 0.0016 − − + = 1.6953 + 0.0002 = 1.6955 Hence, the root is 1.695 correct to three places of decimal. Example 5. Find a real root of the equation f(x) = xe x – 3 = 0, using Regula-Falsi method correct to three decimal places. Sol. We have f(x)= xe x – 3 = 0 Then f(1) = 1e 1 – 3 = – 0.2817 and f(1.5) = (1.5)e (1.5) – 3 = 3.7225. ∴ The root lies between 1 and 1.5. Therefore, taking x 0 = 1, x 1 = 1.5, f(x 0 ) = – 0.2817 and f(x 1 ) = 3.7225. The first approximation to the root is . 1.4036 0.10203  =   Since, x 3 and x 4 are approximately the same upto four places of decimal, hence the required root of the given equation is 1.4036. 54 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES Example. method of false position up to three places of decimal. Sol. Given that f(x)= x 3 – 2x – 5 = 0 So that f(2) = (2) 3 – 2(2) – 5 = – 1 50 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES and. 0.0000 17 = 2.9428 17 Hence, the root is 2.9428 correct to four places of decimal. ALGEBRAIC AND TRANSCENDENTAL EQUATION 53 Example 3. Using the method of False Position, find the root of equation