Hence a root lies between 1 and 2. Since f a is negative and f(x 0 ) is positive The root lies between a and x 0 i.e. 1 and 1.

Transcription

1 The Bisection method or BOLZANO s method or Interval halving method: Find the positive root of x 3 x = 1 correct to four decimal places by bisection method Let f x = x 3 x 1 Here f 0 = 1 = ve, f 1 = ve, f 2 = 5 = +ve Hence a root lies between 1 and 2 Here a = 1 and b = 2 a + b Here x 0 = = = 15, f x = f(15) = 0875 = +ve Since f a is negative and f(x 0 ) is positive The root lies between a and x 0 ie 1 and 15, put b = x 0 x 1 = = 125, f 125 = = ve 2 The root lies between x 1 and b ie 125 and 15, put a = x x 2 = = 1375, f 1375 = = +ve 2 Therefore the root lies between a and x 2 ie 125 and 1375 Proceeding like this and these values are put in a tabular form as follows i a b x n f(x n ) Since the values in 13 th and 14 th iteration are same, the root is 13248

2 Find the positive root of x cos x = 0 by bisection method Let f x = x cos x Here f 0 = 1 = ve, f 1 = 1 cos 1 = = +ve Hence a root lies between 0 and 1 Here a = 0 and b = 1 a + b Here x 0 = = = 05, f x = f 05 = = ve Since f x 0 is negative and f(b) is positive The root lies between x 0 and b ie 05 and 1, put a = x 0 x 1 = = 075, f 075 = = +ve 2 The root lies between a and x 1 ie 05 and 075, put b = x x 2 = = 0625, f 0625 = 0186 = ve 2 Therefore the root lies between x 2 and b ie 0625 and 075 Proceeding like this and these values are put in a tabular form as follows i a b x n f(x n ) Since the value of the last column is zero in last column of 11 th iteration, the root is Regula Falsi method or False position method: Solve for a positive root of x 3 4x + 1 = 0 by Regula Falsi method Let f x = x 3 4x + 1 Here f 0 = 1 = +ve, f 1 = 2 = ve A root lies between 0 and 1 Here a = 0 and b = 1

3 af b bf a x 1 = f b f a = 0 f 1 1 f 0 f 1 f 0 = = f x 1 = f = = ve Hence the root lies between 0 and ie a = 0 and b = (Since f 0 = +ve and f = ve) x 2 = af b bf a f b f a = 0 f f 0 f f 0 = f = = ve Hence the root lies between 0 and ie a = 0 and b = (Since f 0 = +ve and f = ve) Continuing in the same way and the values are put in a table below i a b f(a) f(b) x n f(x n ) Since the value of the last column of 4 th iteration is zero, the root is Find a positive root of xe x = 2 by the method of false position Let f x = xe x 2 Here f 0 = 2 = ve, f 1 = = +ve A root lies between 0 and 1 Here a = 0 and b = 1 x 1 = af b bf a f b f a = 0 f 1 1 f 0 f 1 f 0 = = f x 1 = f = = ve Hence the root lies between 0 and ie a = 0 and b = (Since f 0 = +ve and f = ve) x 2 = af b bf a f b f a = 0 f f 0 f f 0 = 02571

4 f = = ve Hence the root lies between 0 and ie a = 0 and b = (Since f 0 = +ve and f = ve) Continuing in the same way and the values are put in a table below i a b f(a) f(b) x n f(x n ) Since the value of the x n column of 5 th and 6 th iteration are same, the root is up to four decimal places Gauss Jacobi method: Solve by Gauss Jacobi method, the system 12x + y + z = 31,2 x + 8y z = 24, 3x + 4y + 10z = < 12, < 8, < 10 Therefore the given equations are diagonally dominant, so we can use Gauss Jacobi method Proceeding in the same manner and writing the solutions in the tabular form we get i x y z

5 Since the values in 7 th and 8 th iteration are same, the solution is x = 2, y = 3, z = 4 up to four decimal places Solve by Gauss Jacobi method, the system 9x y + 2z = 9, x + 10y 2z = 15, 2x 2y 13z = = 3 < 9, < 10, < 13 Therefore the given equations are diagonally dominant, so we can use Gauss Jacobi method 1 st iteration x = 1 (9 + y 2z) 9 y = 1 (15 x + 2z) 10 z = 1 (17 + 2x 2y) 13 x (1) = = 1 y (1) = = 15 z (1) = (0) 2(0) = nd iteration x (2) = = 08761

6 y (2) = = z (2) = (1) 2(15) = Proceeding in the same manner and writing the solutions in the tabular form we get i x y z Since the values in 8 th and 9 th iteration are same, the solution is x = 09174, y = 16473, z = up to four decimal places Gauss Seidel method Solve by Gauss Seidel method, the system 12x + y + z = 31,2 x + 8y z = 24, 3x + 4y + 10z = < 12, < 8, < 10 Therefore the given equations are diagonally dominant, so we can use Gauss Seidel method x = y z y = x + z

7 z = x 4y 1 st iteration x = = y = (0) = z = (25833) 4(23542) = nd iteration x = = y = 1 8 z = (20469) = (20469) 4(29987) = Proceeding in the same manner and writing the solutions in the tabular form we get i x y z Since the values in 5 th and 6 th iteration are same, the solution is x = 2, y = 3, z = 4 up to four decimal places Solve by Gauss Seidel method, the system 9x y + 2z = 9, x + 10y 2z = 15, 2x 2y 13z = 17

8 1 + 2 = 3 < 9, < 10, < 13 Therefore the given equations are diagonally dominant, so we can use Gauss Seidel method 1 st iteration x = 1 (9 + y 2z) 9 y = 1 (15 x + 2z) 10 z = 1 (17 + 2x 2y) 13 x (1) = = 1 y (1) = = 14 z (1) = (1) 2(14) = nd iteration x (2) = = y (2) = = z (2) = (08786) 2(16614) = Proceeding in the same manner and writing the solutions in the tabular form we get i x y z

9 Since the values in 5 th and 6 th iteration are same, the solution is x = 09174, y = 16473, z = up to four decimal places Gauss Elimination method Solve by Gauss Elimination method, the system 12x + y + z = 31,2 x + 8y z = 24, 3x + 4y + 10z = 58 ~ ~ A: B = R 2 R R 1 R 3 R R 1 R 3 R R 2 12x + y + z = y 117z = z = 4124 z = = 4 783y = 1883 y = = 3 Therefore the solution is x = 2, y = 3, z = 4 Solve by Gauss Elimination method, the system 12x = 31 x = = 2 9x y + 2z = 9, x + 10y 2z = 15, 2x 2y 13z = 17 A: B =

10 ~ R 2 R R 1 R 3 R R 1 ~ R 3 R R 2 9x y + 2z = y 22222z = z = 4124 z = = y = 14 y = 9x = 9 x = = = Therefore the solution is x = 09174, y = 16473, z = Gauss Jordon method Solve by Gauss Jordon method, the system 12x + y + z = 31,2 x + 8y z = 24, 3x + 4y + 10z = 58 ~ ~ ~ A: B = R 2 R R 1 R 3 R R 1 R 1 R R 2 R 3 R R 2 R 1 R R 3 R 2 R R 3

11 Therefore the solution is x = 2, y = 3, z = 4 Solve by Gauss Jordon method, the system 12x = 24 x = = 2 783y = 2351 y = = z = 4124 z = = 4 9x y + 2z = 9, x + 10y 2z = 15, 2x 2y 13z = 17 A: B = ~ R 2 R R 1 R 3 R R 1 ~ ~ R 1 R R 2 R 3 R R 2 R 1 R R 3 R 2 R R 3 9x = x = = y = y = = z = z = = Therefore the solution is x = 09174, y = 16473, z = 11954

12 NEWTON S METHOD OR NEWTON-RAPHSON METHOD Find the positive root of x 4 x = 10 correct to three decimal places using Newton- Raphson method Solution : Let f x = x 4 x 10 = 0 Now, f 0 = = 10 f 1 = = 10 f 2 = = +4 Therefore the root lies between 1 & 2 Let us take x 0 = 2 {Near to zero} The Newton- Raphson formula is ve ve +ve x n+1 = x n f x n f x n 1 Let f x = x 4 x 10 and f 1 x = 4 x 3 1 x 1 = x 0 f x 0 f x 0 x 1 = 2 f 2 f 2 x 1 = x 1 = x 1 = x 2 = x 1 f x 1 f x 1 x 2 = 2 f f x 2 = x 2 = x 2 = 1856 x 3 = x 2 f x 2 f x 2

13 x 3 = 1856 f 1856 f 1856 x 3 = x 3 = x 3 = 1856 The root of the equation x 4 x = 10 is 1856 Using Newton s iterative method to find the root between 0 and 1 of x 3 = 6x 4 correct to three decimal places Solution : Let f x = x 3 6x + 4 = 0 Now, f 0 = = +4 f 1 = = 1 Therefore the root lies between 0 & 1 Let us take x 0 = 1 {Near to zero} The Newton- Raphson formula is +ve ve x n+1 = x n f x n f x n 1 Let f x = x 3 6x + 4 and f 1 x = 3 x 2 6 x 1 = x 0 f x 0 f x 0 x 1 = 1 f 1 f 1 x 1 = x 1 = x 1 = 0666 x 2 = x 1 f x 1 f x 1 x 1 = 0666 f 0666 f 0666

14 x 2 = x 2 = x 1 = 073 x 3 = x 2 f x 2 f x 2 x 3 = 073 f 073 f 073 x 3 = x 3 = x 3 = The root of the equation x 3 6x + 4 = 0 is 0732 Find the real positive root of 3x cos x 1 = 0 correct to six decimal places using Newton method Solution : Let f x = 3x cos x 1 = 0 Now, f 0 = 3 0 cos 0 1 = 2 f 1 = 3 1 cos 1 1 = Therefore the root lies between 0 & 1 Let us take x 0 = 1 {Near to zero} The Newton- Raphson formula is ve ve Let x n+1 = x n f x n f x n 1 f x = 3x cos x 1 and f 1 x = 3 + sin x x 1 = x 0 f x 0 f x 0 x 1 = 1 f 1 f 1 x 1 = cos sin 1

15 x 1 = x 2 = x 2 = x 1 f x 1 f x 1 x 2 = f f cos sin x 2 = x 3 = x 3 = x 2 f x 2 f x 2 x 3 = f f cos sin x 3 = The root of the equation x 4 x = 10 is FIXED POINT ITERATION OR ITERATION METHOD The condition for convergence of a method Let f x = 0 be the given equation whose actual root is r The equation f x = 0 be written as x = g x Let I be the interval containing the root x = r If g x < 1 for all x in I, then the sequence of approximations x 0, x 1, x 2, x n will converge to r, if the initial starting value x 0 is chosen in I Note 1 Since x n r K x n 1 r where K is a constant the convergence is linear and the convergence is of order 1 Note 2 The sufficient condition for the convergence is g x < 1 for all x in I Find the positive root of x 2 2x 3 = 0 by Iteration method Solution : Let f x = x 2 2x 3 = 0 Now, f 0 = = 10 ve

16 f 1 = = 10 ve f 2 = = +4 +ve Therefore the root lies between 1 & 2 Let us take x 0 = 2 {Near to zero} x 2 2x 3 = 0 x 2 = 2x + 3 x = 2x + 3 Let x 0 = 2 x = g(x) = 2x + 3 x 1 = g x 0 = 2x = = x 2 = g x 1 = 2x = = x 3 = g x 2 = 2x = = x 4 = g x 3 = 2x = = x 5 = g x 4 = 2x = = x 6 = g x 5 = 2x = = x 7 = g x 6 = 2x = = x 8 = g x 7 = 2x = = x 9 = g x 8 = 2x = = x 10 = g x 9 = 2x = = Hence the root of the equation is x 2 2x 3 = 0 is Find the Real root of the equation x 3 + x by Fixed point iteration method Let f x = x 3 + x = 0 f 0 = = 100 ve f 1 = = 98 ve f 2 = = 88 ve f 3 = = 64 ve f 4 = = 20 ve f 5 = = +50 +ve The root lies between 4 & 5

17 Now, g x = Since x 3 + x = 0 x 2 x + 1 = 100 x 2 = 100 x + 1 x = g x = 10 x + 1 = 10 x x = 5 x + 1 ( 3 2 ) g 4 = ( 3 2 ) = < 1 g 5 = ( 3 2 ) = < 1 So that we can use the iteration method Let x 0 = 4 x 1 = g x 0 = 10 x = = = x 2 = g x 1 = 10 x = = = x 3 = g x 2 = 10 x = = x 4 = g x 3 = 10 x = = x 5 = g x 4 = 10 x = = x 6 = g x 5 = 10 x = = x 7 = g x 6 = 10 x = = x 8 = g x 7 = 10 x = = x 9 = g x 8 = 10 x = = x 10 = g x 9 = 10 x = = x 11 = g x 10 = 10 x = = Hence the root of the equation is x 3 + x = 0 is 43310

18 Find the real root of the equation cos x = 3x 1 correct to five decimal places using fixed point iteration method Let f x = cos x 3x + 1 = 0 f 0 = cos = 2 +ve f 1 = cos = ve The root lies between 0 & 1 Since cos x 3x + 1 = 0 3x = cos x + 1 x = cos x x = g x = cos x Now, g x = 1 3 sin x = 1 sin x 3 g 0 = 1 3 sin 0 = 0 < 1 So that we can use the iteration method Let x 0 = 4 g 1 = 1 sin 1 = 0284 < 1 3 x 1 = g x 0 = cos x 0 = 1 3 x 2 = g x 1 = cos x 1 = 1 3 x 3 = g x 2 = cos x 2 = = cos = cos = x 4 = g x 3 = cos x 3 = x 5 = g x 4 = cos x 4 = x 6 = g x 5 = cos x 5 = x 7 = g x 6 = cos x 6 = x 8 = g x 7 = cos x 7 =

19 x 9 = g x 8 = cos x 8 = Hence the root of the equation is cos x = 3x 1 is Triangularisation method or LU decomposition method 0r LU factorization method Solve the following system by triangularisation method x + y + z = 1, 4x + 3y z = 6, 3x + 5y + 3z = 4 The given system is equivalent to AX = B x y z = Let L = l l 31 l 32 1, U = u 11 u 12 u 13 0 u 22 u 23, Y = 0 0 u 33 y 1 y 2 y 3 Let LU = A l l 31 l 32 1 u 11 u 12 u 13 0 u 22 u 23 = 0 0 u u 11 u 12 u 13 l 21 u 11 l 21 u 12 + u 22 l 21 u 13 + u 23 = l 31 u 11 l 31 u 12 + l 32 u 22 l 31 u 13 + l 32 u 23 + u Equating the coefficients on both sides we get, u 11 = 1, u 12 = 1, u 13 = 1 l 21 u 11 = 4 l 21 1 = 4 l 21 = 4 l 21 u 12 + u 22 = u 22 = 3 u 22 = 1 l 21 u 13 + u 23 = u 23 = 1 u 23 = 1 4 = 5 u 23 = 5 l 31 u 11 = 3 l 31 1 = 3 l 31 = 3 l 31 u 12 + l 32 u 22 = l 32 1 = 5 l 32 = 3 5 l 32 = 2 l 31 u 13 + l 32 u 23 + u 33 = u 33 = 3 u 33 = u 33 = 10

20 AX = B LUX = B LY = B were UX = Y y 1 y 2 y 3 = By forward substitution method we get y 1 = 1 4y 1 + y 2 = y 2 = 6 y 2 = 2 3y 1 2y 2 + y 3 = y 3 = 4 y 3 = 5 UX = Y x y z = By backward substitution method we get 10z = 5 z = 1 2 y 5z = 2 y = 2 y = y = 1 2 x + y + z = 1 x = 1 x = 1 Hence x = 1, y = 1 2, z = 1 2 Solve the following system by triangularisation method x + 5y + z = 14, 2x + y + 3z = 13, 3x + y + 4z = 17 The given system is equivalent to AX = B x y z = Let L = l l 31 l 32 1, U = u 11 u 12 u 13 0 u 22 u 23, Y = 0 0 u 33 y 1 y 2 y 3 Let LU = A

21 1 0 0 l l 31 l 32 1 u 11 u 12 u 13 0 u 22 u 23 = 0 0 u u 11 u 12 u 13 l 21 u 11 l 21 u 12 + u 22 l 21 u 13 + u 23 = l 31 u 11 l 31 u 12 + l 32 u 22 l 31 u 13 + l 32 u 23 + u Equating the coefficients on both sides we get, u 11 = 1, u 12 = 5, u 13 = 1 l 21 u 11 = 2 l 21 1 = 2 l 21 = 2 l 21 u 12 + u 22 = u 22 = 1 u 22 = 9 l 21 u 13 + u 23 = u 23 = 3 u 23 = 1 l 31 u 11 = 3 l 31 1 = 3 l 31 = 3 l 31 u 12 + l 32 u 22 = l 32 9 = 1 l 32 = 14 9 l 31 u 13 + l 32 u 23 + u 33 = u 33 = 4 u 33 = u 33 = 5 9 AX = B LUX = B LY = B were UX = Y y 1 y 2 y 3 = By forward substitution method we get y 1 = 14 2y 1 + y 2 = y 2 = 13 y 2 = 15 3y y 2 + y 3 = y 3 = 17 y 3 = 5 3 UX = Y x y z = By backward substitution method we get

22 5 9 z = 5 3 z = 3 9y + z = 15 9y + 3 = 15 9y = 18 y = 2 x + 5y + z = 14 x = 14 x = 1 Hence x = 1, y = 2, z = 3