Numerical Methods 1

1,131 views

Published on

Introduction of Numerical method

Published in: Education, Technology, Business
0 Comments
3 Likes
Statistics
Notes
  • Be the first to comment

No Downloads
Views
Total views
1,131
On SlideShare
0
From Embeds
0
Number of Embeds
2
Actions
Shares
0
Downloads
138
Comments
0
Likes
3
Embeds 0
No embeds

No notes for slide

Numerical Methods 1

  1. 1. Numerical Methods N. B. Vyas Department of Mathematics, Atmiya Institute of Tech. and Science, Rajkot (Guj.)N.B.V yas − Department of M athematics, AIT S − Rajkot
  2. 2. Introduction There are two types of functions: (i) Algebraic function and (ii) Transcendental function N.B.V yas − Department of M athematics, AIT S − Rajkot
  3. 3. Introduction There are two types of functions: (i) Algebraic function and (ii) Transcendental function An algebraic function is informally a function that satisfies a polynomial equation N.B.V yas − Department of M athematics, AIT S − Rajkot
  4. 4. Introduction There are two types of functions: (i) Algebraic function and (ii) Transcendental function An algebraic function is informally a function that satisfies a polynomial equation A function which is not algebraic is called a transcendental function. N.B.V yas − Department of M athematics, AIT S − Rajkot
  5. 5. Introduction There are two types of functions: (i) Algebraic function and (ii) Transcendental function An algebraic function is informally a function that satisfies a polynomial equation A function which is not algebraic is called a transcendental function. The values of x which satisfy the equation f (x) = 0 are called roots of f (x). N.B.V yas − Department of M athematics, AIT S − Rajkot
  6. 6. Introduction There are two types of functions: (i) Algebraic function and (ii) Transcendental function An algebraic function is informally a function that satisfies a polynomial equation A function which is not algebraic is called a transcendental function. The values of x which satisfy the equation f (x) = 0 are called roots of f (x). If f (x) is quadratic, cubic or bi-quadratic expression, then algebraic formulae are available for getting the solution. N.B.V yas − Department of M athematics, AIT S − Rajkot
  7. 7. Introduction There are two types of functions: (i) Algebraic function and (ii) Transcendental function An algebraic function is informally a function that satisfies a polynomial equation A function which is not algebraic is called a transcendental function. The values of x which satisfy the equation f (x) = 0 are called roots of f (x). If f (x) is quadratic, cubic or bi-quadratic expression, then algebraic formulae are available for getting the solution. If f (x) is a higher degree polynomial or transcendental function then algebraic methods are not available. N.B.V yas − Department of M athematics, AIT S − Rajkot
  8. 8. Errors It is never possible to measure anything exactly. N.B.V yas − Department of M athematics, AIT S − Rajkot
  9. 9. Errors It is never possible to measure anything exactly. So in order to make valid conclusions, it is good to make the error as small as possible. N.B.V yas − Department of M athematics, AIT S − Rajkot
  10. 10. Errors It is never possible to measure anything exactly. So in order to make valid conclusions, it is good to make the error as small as possible. The result of any physical measurement has two essential components : N.B.V yas − Department of M athematics, AIT S − Rajkot
  11. 11. Errors It is never possible to measure anything exactly. So in order to make valid conclusions, it is good to make the error as small as possible. The result of any physical measurement has two essential components : i) A numerical value N.B.V yas − Department of M athematics, AIT S − Rajkot
  12. 12. Errors It is never possible to measure anything exactly. So in order to make valid conclusions, it is good to make the error as small as possible. The result of any physical measurement has two essential components : i) A numerical value ii) A degree of uncertainty Or Errors. N.B.V yas − Department of M athematics, AIT S − Rajkot
  13. 13. Errors Exact Numbers: There are the numbers in which there is no uncertainty and no approximation. N.B.V yas − Department of M athematics, AIT S − Rajkot
  14. 14. Errors Exact Numbers: There are the numbers in which there is no uncertainty and no approximation. Approximate Numbers: These are the numbers which represent a certain degree of accuracy but not the exact value. N.B.V yas − Department of M athematics, AIT S − Rajkot
  15. 15. Errors Exact Numbers: There are the numbers in which there is no uncertainty and no approximation. Approximate Numbers: These are the numbers which represent a certain degree of accuracy but not the exact value. These numbers cannot be represented in terms of finite number of digits. N.B.V yas − Department of M athematics, AIT S − Rajkot
  16. 16. Errors Exact Numbers: There are the numbers in which there is no uncertainty and no approximation. Approximate Numbers: These are the numbers which represent a certain degree of accuracy but not the exact value. These numbers cannot be represented in terms of finite number of digits. Significant Digits: It refers to the number of digits in a number excluding leading zeros. N.B.V yas − Department of M athematics, AIT S − Rajkot
  17. 17. Bisection Method Consider a continuous function f (x). N.B.V yas − Department of M athematics, AIT S − Rajkot
  18. 18. Bisection Method Consider a continuous function f (x). Numbers a < b such that f (a) and f (b) have opposite signs. N.B.V yas − Department of M athematics, AIT S − Rajkot
  19. 19. Bisection Method Consider a continuous function f (x). Numbers a < b such that f (a) and f (b) have opposite signs. Let f (a) be negative and f (b) be positive for [a, b]. N.B.V yas − Department of M athematics, AIT S − Rajkot
  20. 20. Bisection Method Consider a continuous function f (x). Numbers a < b such that f (a) and f (b) have opposite signs. Let f (a) be negative and f (b) be positive for [a, b]. Then there exists at least one point(root), say x, a < x < b such that f (x) = 0. N.B.V yas − Department of M athematics, AIT S − Rajkot
  21. 21. Bisection Method Consider a continuous function f (x). Numbers a < b such that f (a) and f (b) have opposite signs. Let f (a) be negative and f (b) be positive for [a, b]. Then there exists at least one point(root), say x, a < x < b such that f (x) = 0. Now according to Bisection method, bisect the interval [a, b], a+b x1 = (a < x1 < b). 2 N.B.V yas − Department of M athematics, AIT S − Rajkot
  22. 22. Bisection Method Consider a continuous function f (x). Numbers a < b such that f (a) and f (b) have opposite signs. Let f (a) be negative and f (b) be positive for [a, b]. Then there exists at least one point(root), say x, a < x < b such that f (x) = 0. Now according to Bisection method, bisect the interval [a, b], a+b x1 = (a < x1 < b). 2 If f (x1 ) = 0 then x1 be the root of the given equation. N.B.V yas − Department of M athematics, AIT S − Rajkot
  23. 23. Bisection Method Consider a continuous function f (x). Numbers a < b such that f (a) and f (b) have opposite signs. Let f (a) be negative and f (b) be positive for [a, b]. Then there exists at least one point(root), say x, a < x < b such that f (x) = 0. Now according to Bisection method, bisect the interval [a, b], a+b x1 = (a < x1 < b). 2 If f (x1 ) = 0 then x1 be the root of the given equation. Otherwise the root lies between x1 and b if f (x1 ) < 0. N.B.V yas − Department of M athematics, AIT S − Rajkot
  24. 24. Bisection Method Consider a continuous function f (x). Numbers a < b such that f (a) and f (b) have opposite signs. Let f (a) be negative and f (b) be positive for [a, b]. Then there exists at least one point(root), say x, a < x < b such that f (x) = 0. Now according to Bisection method, bisect the interval [a, b], a+b x1 = (a < x1 < b). 2 If f (x1 ) = 0 then x1 be the root of the given equation. Otherwise the root lies between x1 and b if f (x1 ) < 0. OR the root lies between a and x1 if f (x1 ) > 0. Then again bisect this interval to get next point x2 . N.B.V yas − Department of M athematics, AIT S − Rajkot
  25. 25. Bisection Method Consider a continuous function f (x). Numbers a < b such that f (a) and f (b) have opposite signs. Let f (a) be negative and f (b) be positive for [a, b]. Then there exists at least one point(root), say x, a < x < b such that f (x) = 0. Now according to Bisection method, bisect the interval [a, b], a+b x1 = (a < x1 < b). 2 If f (x1 ) = 0 then x1 be the root of the given equation. Otherwise the root lies between x1 and b if f (x1 ) < 0. OR the root lies between a and x1 if f (x1 ) > 0. Then again bisect this interval to get next point x2 . Repeat the above procedure to generate x1 , x2 , . . . till the root upto desired accuracy is obtained. N.B.V yas − Department of M athematics, AIT S − Rajkot
  26. 26. Bisection MethodCharacteristics: 1 This method always slowly converge to a root. N.B.V yas − Department of M athematics, AIT S − Rajkot
  27. 27. Bisection MethodCharacteristics: 1 This method always slowly converge to a root. 2 It gives only one root at a time on the the selection of small interval near the root. N.B.V yas − Department of M athematics, AIT S − Rajkot
  28. 28. Bisection MethodCharacteristics: 1 This method always slowly converge to a root. 2 It gives only one root at a time on the the selection of small interval near the root. 3 In case of the multiple roots of an equation, other initial interval can be chosen. N.B.V yas − Department of M athematics, AIT S − Rajkot
  29. 29. Bisection MethodCharacteristics: 1 This method always slowly converge to a root. 2 It gives only one root at a time on the the selection of small interval near the root. 3 In case of the multiple roots of an equation, other initial interval can be chosen. 4 Smallest interval must be selected to obtain immediate convergence to the root, . N.B.V yas − Department of M athematics, AIT S − Rajkot
  30. 30. Bisection Method- ExampleEx. Find real root of x3 − x − 1 = 0 correct upto three decimal places using Bisection method N.B.V yas − Department of M athematics, AIT S − Rajkot
  31. 31. Bisection Method- ExampleSol. Let f (x) = x3 − x − 1 = 0 N.B.V yas − Department of M athematics, AIT S − Rajkot
  32. 32. Bisection Method- ExampleSol. Let f (x) = x3 − x − 1 = 0 f (0) = N.B.V yas − Department of M athematics, AIT S − Rajkot
  33. 33. Bisection Method- ExampleSol. Let f (x) = x3 − x − 1 = 0 f (0) = −1 < 0 N.B.V yas − Department of M athematics, AIT S − Rajkot
  34. 34. Bisection Method- ExampleSol. Let f (x) = x3 − x − 1 = 0 f (0) = −1 < 0 f (1) = N.B.V yas − Department of M athematics, AIT S − Rajkot
  35. 35. Bisection Method- ExampleSol. Let f (x) = x3 − x − 1 = 0 f (0) = −1 < 0 f (1) = −1 < 0 N.B.V yas − Department of M athematics, AIT S − Rajkot
  36. 36. Bisection Method- ExampleSol. Let f (x) = x3 − x − 1 = 0 f (0) = −1 < 0 f (1) = −1 < 0 f (2) = N.B.V yas − Department of M athematics, AIT S − Rajkot
  37. 37. Bisection Method- ExampleSol. Let f (x) = x3 − x − 1 = 0 f (0) = −1 < 0 f (1) = −1 < 0 f (2) = 5 > 0 N.B.V yas − Department of M athematics, AIT S − Rajkot
  38. 38. Bisection Method- ExampleSol. Let f (x) = x3 − x − 1 = 0 f (0) = −1 < 0 f (1) = −1 < 0 f (2) = 5 > 0 ∴ since f (x) is continuous function there must be a root in the lying in the interval (1, 2) N.B.V yas − Department of M athematics, AIT S − Rajkot
  39. 39. Bisection Method- ExampleSol. Let f (x) = x3 − x − 1 = 0 f (0) = −1 < 0 f (1) = −1 < 0 f (2) = 5 > 0 ∴ since f (x) is continuous function there must be a root in the lying in the interval (1, 2) Now according to Bisection method, the next approximation is obtained by taking the midpoint of (1, 2) N.B.V yas − Department of M athematics, AIT S − Rajkot
  40. 40. Bisection Method- ExampleSol. Let f (x) = x3 − x − 1 = 0 f (0) = −1 < 0 f (1) = −1 < 0 f (2) = 5 > 0 ∴ since f (x) is continuous function there must be a root in the lying in the interval (1, 2) Now according to Bisection method, the next approximation is obtained by taking the midpoint of (1, 2) 1+2 c= = 1.5, f (1.5) = 2 N.B.V yas − Department of M athematics, AIT S − Rajkot
  41. 41. Bisection Method- ExampleSol. Let f (x) = x3 − x − 1 = 0 f (0) = −1 < 0 f (1) = −1 < 0 f (2) = 5 > 0 ∴ since f (x) is continuous function there must be a root in the lying in the interval (1, 2) Now according to Bisection method, the next approximation is obtained by taking the midpoint of (1, 2) 1+2 c= = 1.5, f (1.5) = 2 a+b No. of a b c= f(c) 2 iterations (f (a) < 0) (f (b) > 0) (< 0, > 0) 1 1 2 1.5 - N.B.V yas − Department of M athematics, AIT S − Rajkot
  42. 42. Bisection Method- ExampleEx. Find real root of x3 − 4x + 1 = 0 correct upto four decimal places using Bisection method N.B.V yas − Department of M athematics, AIT S − Rajkot
  43. 43. Bisection Method- ExampleEx. Find real root of x2 − lnx − 12 = 0 correct upto three decimal places using Bisection method N.B.V yas − Department of M athematics, AIT S − Rajkot
  44. 44. Newton-Raphson Method(N-R Method)Graphical derivation of the Method: Consider the portion of the graph y = f (x) which crosses X − axis at R corresponding to the equation f (x) = 0. N.B.V yas − Department of M athematics, AIT S − Rajkot
  45. 45. Newton-Raphson Method(N-R Method)Graphical derivation of the Method: Consider the portion of the graph y = f (x) which crosses X − axis at R corresponding to the equation f (x) = 0. Let B be the point on the curve corresponding to the initial guess x0 at A. N.B.V yas − Department of M athematics, AIT S − Rajkot
  46. 46. Newton-Raphson Method(N-R Method)Graphical derivation of the Method: Consider the portion of the graph y = f (x) which crosses X − axis at R corresponding to the equation f (x) = 0. Let B be the point on the curve corresponding to the initial guess x0 at A. The tangent at B cuts the X − axis at C which gives first approximation x1 . N.B.V yas − Department of M athematics, AIT S − Rajkot
  47. 47. Newton-Raphson Method(N-R Method)Graphical derivation of the Method: Consider the portion of the graph y = f (x) which crosses X − axis at R corresponding to the equation f (x) = 0. Let B be the point on the curve corresponding to the initial guess x0 at A. The tangent at B cuts the X − axis at C which gives first approximation x1 . Thus AB = f (x0 ) and AC = x0 − x1 N.B.V yas − Department of M athematics, AIT S − Rajkot
  48. 48. Newton-Raphson Method(N-R Method)Graphical derivation of the Method: Consider the portion of the graph y = f (x) which crosses X − axis at R corresponding to the equation f (x) = 0. Let B be the point on the curve corresponding to the initial guess x0 at A. The tangent at B cuts the X − axis at C which gives first approximation x1 . Thus AB = f (x0 ) and AC = x0 − x1 AB Now ∠ACB = α, tanα = AC N.B.V yas − Department of M athematics, AIT S − Rajkot
  49. 49. Newton-Raphson Method(N-R Method)Graphical derivation of the Method: Consider the portion of the graph y = f (x) which crosses X − axis at R corresponding to the equation f (x) = 0. Let B be the point on the curve corresponding to the initial guess x0 at A. The tangent at B cuts the X − axis at C which gives first approximation x1 . Thus AB = f (x0 ) and AC = x0 − x1 AB Now ∠ACB = α, tanα = AC f (x0 ) ∴ f (x0 ) = x0 − x1 N.B.V yas − Department of M athematics, AIT S − Rajkot
  50. 50. Newton-Raphson Method(N-R Method)Graphical derivation of the Method: Consider the portion of the graph y = f (x) which crosses X − axis at R corresponding to the equation f (x) = 0. Let B be the point on the curve corresponding to the initial guess x0 at A. The tangent at B cuts the X − axis at C which gives first approximation x1 . Thus AB = f (x0 ) and AC = x0 − x1 AB Now ∠ACB = α, tanα = AC f (x0 ) ∴ f (x0 ) = x0 − x1 f (x0 f (x0 ∴ x0 − x1 = ⇒ x0 − = x1 f (x0 ) f (x0 ) N.B.V yas − Department of M athematics, AIT S − Rajkot
  51. 51. Newton-Raphson Method(N-R Method)Graphical derivation of the Method: Consider the portion of the graph y = f (x) which crosses X − axis at R corresponding to the equation f (x) = 0. Let B be the point on the curve corresponding to the initial guess x0 at A. The tangent at B cuts the X − axis at C which gives first approximation x1 . Thus AB = f (x0 ) and AC = x0 − x1 AB Now ∠ACB = α, tanα = AC f (x0 ) ∴ f (x0 ) = x0 − x1 f (x0 f (x0 ∴ x0 − x1 = ⇒ x0 − = x1 f (x0 ) f (x0 ) f (x0 ) ∴ x1 = x0 − f (x0 ) N.B.V yas − Department of M athematics, AIT S − Rajkot
  52. 52. Newton-Raphson Method(N-R Method)Graphical derivation of the Method: Consider the portion of the graph y = f (x) which crosses X − axis at R corresponding to the equation f (x) = 0. Let B be the point on the curve corresponding to the initial guess x0 at A. The tangent at B cuts the X − axis at C which gives first approximation x1 . Thus AB = f (x0 ) and AC = x0 − x1 AB Now ∠ACB = α, tanα = AC f (x0 ) ∴ f (x0 ) = x0 − x1 f (x0 f (x0 ∴ x0 − x1 = ⇒ x0 − = x1 f (x0 ) f (x0 ) f (x0 ) ∴ x1 = x0 − f (x0 ) f (xn ) In general xn+1 = xn − ; where n = 0, 1, 2, 3, . . . f (xn ) N.B.V yas − Department of M athematics, AIT S − Rajkot
  53. 53. Derivation of the Newton- Raphson method. f(x) y= f(x) AB tan(α ) = f(x0 ) B AC f ( x0 ) f ( x 0) = x0 − x1 f ( x0 ) R C α A x1 = x0 − X f ′( x0 ) x1 x04 N.B.V yas − Department of M athematics, AIT S − Rajkot
  54. 54. Newton-Raphson method f(x) f(x0 ) f(xn )  x0, f ( x0 ) xn +1 = xn -   f ′ (xn ) f(x 1) α x2 x1 x0 X3 N.B.V yas − Department of M athematics, AIT S − Rajkot
  55. 55. N-R Method:Advantages: Converges Fast (if it converges). Requires only one guess. N.B.V yas − Department of M athematics, AIT S − Rajkot
  56. 56. Drawbacks: Divergence at inflection point. Selection of the initial guess or an iteration value of the root that is close to the inflection point of the function f (x) may start diverging away from the root in the Newton-Raphson method. N.B.V yas − Department of M athematics, AIT S − Rajkot
  57. 57. N-R Method:(Drawbacks) Division by zero N.B.V yas − Department of M athematics, AIT S − Rajkot
  58. 58. N-R Method:(Drawbacks) Results obtained from the N-R method may oscillate about the local maximum or minimum without converging on a root but converging on the local maximum or minimum. For example for f (x) = x2 + 2 = 0 the equation has no real roots. N.B.V yas − Department of M athematics, AIT S − Rajkot
  59. 59. N-R Method:(Drawbacks) Root Jumping N.B.V yas − Department of M athematics, AIT S − Rajkot
  60. 60. N-R MethodIterative formula for finding q th root: 1 xq − N = 0 i.e. x = N q N.B.V yas − Department of M athematics, AIT S − Rajkot
  61. 61. N-R MethodIterative formula for finding q th root: 1 xq − N = 0 i.e. x = N q Let f (x) = xq − N N.B.V yas − Department of M athematics, AIT S − Rajkot
  62. 62. N-R MethodIterative formula for finding q th root: 1 xq − N = 0 i.e. x = N q Let f (x) = xq − N ∴ f (x) = qxq−1 N.B.V yas − Department of M athematics, AIT S − Rajkot
  63. 63. N-R MethodIterative formula for finding q th root: 1 xq − N = 0 i.e. x = N q Let f (x) = xq − N ∴ f (x) = qxq−1 f (xn ) (xn )q − N Now by N-R method, xn+1 = xn − = xn − f (xn ) q(xn )q−1 N.B.V yas − Department of M athematics, AIT S − Rajkot
  64. 64. N-R MethodIterative formula for finding q th root: 1 xq − N = 0 i.e. x = N q Let f (x) = xq − N ∴ f (x) = qxq−1 f (xn ) (xn )q − N Now by N-R method, xn+1 = xn − = xn − f (xn ) q(xn )q−1 1 N = xn − xn + q q(xn )q−1 N.B.V yas − Department of M athematics, AIT S − Rajkot
  65. 65. N-R MethodIterative formula for finding q th root: 1 xq − N = 0 i.e. x = N q Let f (x) = xq − N ∴ f (x) = qxq−1 f (xn ) (xn )q − N Now by N-R method, xn+1 = xn − = xn − f (xn ) q(xn )q−1 1 N = xn − xn + q q(xn )q−1 1 N xn+1 = (q − 1)xn + ; n = 0, 1, 2, . . . q (xn )q−1 N.B.V yas − Department of M athematics, AIT S − Rajkot
  66. 66. N-R MethodIterative formula for finding q th root: 1 xq − N = 0 i.e. x = N q Let f (x) = xq − N ∴ f (x) = qxq−1 f (xn ) (xn )q − N Now by N-R method, xn+1 = xn − = xn − f (xn ) q(xn )q−1 1 N = xn − xn + q q(xn )q−1 1 N xn+1 = (q − 1)xn + ; n = 0, 1, 2, . . . q (xn )q−1 N.B.V yas − Department of M athematics, AIT S − Rajkot
  67. 67. N-R MethodIterative formula for finding reciprocal of a positive numberN: 1 1 x= i.e. − N = 0 N x 1 Let f (x) = − N x N.B.V yas − Department of M athematics, AIT S − Rajkot
  68. 68. Secant Method In N-R method two functions f and f are required to be evaluate per step. N.B.V yas − Department of M athematics, AIT S − Rajkot
  69. 69. Secant Method In N-R method two functions f and f are required to be evaluate per step. Also it requires to evaluate derivative of f and sometimes it is very complicated to evaluate f . N.B.V yas − Department of M athematics, AIT S − Rajkot
  70. 70. Secant Method In N-R method two functions f and f are required to be evaluate per step. Also it requires to evaluate derivative of f and sometimes it is very complicated to evaluate f . Often it requires a very good initial guess. N.B.V yas − Department of M athematics, AIT S − Rajkot
  71. 71. Secant Method In N-R method two functions f and f are required to be evaluate per step. Also it requires to evaluate derivative of f and sometimes it is very complicated to evaluate f . Often it requires a very good initial guess. To overcome these drawbacks, the derivative of f of the function f (xn−1 ) − f (xn ) f is approximated as f (xn ) = xn−1 − xn N.B.V yas − Department of M athematics, AIT S − Rajkot
  72. 72. Secant Method In N-R method two functions f and f are required to be evaluate per step. Also it requires to evaluate derivative of f and sometimes it is very complicated to evaluate f . Often it requires a very good initial guess. To overcome these drawbacks, the derivative of f of the function f (xn−1 ) − f (xn ) f is approximated as f (xn ) = xn−1 − xn Therefore formula of N-R method becomes N.B.V yas − Department of M athematics, AIT S − Rajkot
  73. 73. Secant Method In N-R method two functions f and f are required to be evaluate per step. Also it requires to evaluate derivative of f and sometimes it is very complicated to evaluate f . Often it requires a very good initial guess. To overcome these drawbacks, the derivative of f of the function f (xn−1 ) − f (xn ) f is approximated as f (xn ) = xn−1 − xn Therefore formula of N-R method becomes f (xn ) xn+1 = xn − f (xn−1 )−f (xn ) xn−1 −xn N.B.V yas − Department of M athematics, AIT S − Rajkot
  74. 74. Secant Method In N-R method two functions f and f are required to be evaluate per step. Also it requires to evaluate derivative of f and sometimes it is very complicated to evaluate f . Often it requires a very good initial guess. To overcome these drawbacks, the derivative of f of the function f (xn−1 ) − f (xn ) f is approximated as f (xn ) = xn−1 − xn Therefore formula of N-R method becomes f (xn ) xn+1 = xn − f (xn−1 )−f (xn ) xn−1 −xn xn−1 − xn ∴ xn+1 = xn − f (xn ) f (xn−1 ) − f (xn ) where n = 1, 2, 3, . . ., f (xn−1 ) = f (xn ) N.B.V yas − Department of M athematics, AIT S − Rajkot
  75. 75. Secant Method This method requires two initial guesses N.B.V yas − Department of M athematics, AIT S − Rajkot
  76. 76. Secant Method This method requires two initial guesses The two initial guesses do not need to bracket the root of the equation, so it is not classified as a bracketing method. N.B.V yas − Department of M athematics, AIT S − Rajkot
  77. 77. Secant MethodGeometrical interpretation of Secant Method: Consider a continuous function y = f (x) N.B.V yas − Department of M athematics, AIT S − Rajkot
  78. 78. Secant MethodGeometrical interpretation of Secant Method: Consider a continuous function y = f (x) Draw the straight line through the points (xn , f (xn )) and (xn−1 , f (xn−1 )) N.B.V yas − Department of M athematics, AIT S − Rajkot
  79. 79. Secant MethodGeometrical interpretation of Secant Method: Consider a continuous function y = f (x) Draw the straight line through the points (xn , f (xn )) and (xn−1 , f (xn−1 )) Take the x−coordinate of intersection with X−axis as xn+1 N.B.V yas − Department of M athematics, AIT S − Rajkot
  80. 80. Secant MethodGeometrical interpretation of Secant Method: Consider a continuous function y = f (x) Draw the straight line through the points (xn , f (xn )) and (xn−1 , f (xn−1 )) Take the x−coordinate of intersection with X−axis as xn+1 From the figure ABE and DCE are similar triangles. N.B.V yas − Department of M athematics, AIT S − Rajkot
  81. 81. Secant MethodGeometrical interpretation of Secant Method: Consider a continuous function y = f (x) Draw the straight line through the points (xn , f (xn )) and (xn−1 , f (xn−1 )) Take the x−coordinate of intersection with X−axis as xn+1 From the figure ABE and DCE are similar triangles. AB AE f (xn ) xn − xn+1 Hence = ⇒ = DC DE f (xn−1 ) xn−1 − xn+1 N.B.V yas − Department of M athematics, AIT S − Rajkot
  82. 82. Secant MethodGeometrical interpretation of Secant Method: Consider a continuous function y = f (x) Draw the straight line through the points (xn , f (xn )) and (xn−1 , f (xn−1 )) Take the x−coordinate of intersection with X−axis as xn+1 From the figure ABE and DCE are similar triangles. AB AE f (xn ) xn − xn+1 Hence = ⇒ = DC DE f (xn−1 ) xn−1 − xn+1 xn−1 − xn ∴ xn+1 = xn − f (xn ) f (xn−1 ) − f (xn ) N.B.V yas − Department of M athematics, AIT S − Rajkot
  83. 83. Secant MethodNOTE: Do not combine the secant formula and write it in the form as follows because it has enormous loss of significance errors xn f (xn−1 ) − xn−1 f (xn ) xn+1 = f (xn−1 ) − f (xn ) N.B.V yas − Department of M athematics, AIT S − Rajkot
  84. 84. Secant MethodGeneral Features: The secant method is an open method and may not converge. N.B.V yas − Department of M athematics, AIT S − Rajkot
  85. 85. Secant MethodGeneral Features: The secant method is an open method and may not converge. It requires fewer function evaluations. N.B.V yas − Department of M athematics, AIT S − Rajkot
  86. 86. Secant MethodGeneral Features: The secant method is an open method and may not converge. It requires fewer function evaluations. In some problems the secant method will work when Newton’s method does not and vice-versa. N.B.V yas − Department of M athematics, AIT S − Rajkot
  87. 87. Secant MethodGeneral Features: The secant method is an open method and may not converge. It requires fewer function evaluations. In some problems the secant method will work when Newton’s method does not and vice-versa. The method is usually a bit slower than Newton’s method. It is more rapidly convergent than the bisection method. N.B.V yas − Department of M athematics, AIT S − Rajkot
  88. 88. Secant MethodGeneral Features: The secant method is an open method and may not converge. It requires fewer function evaluations. In some problems the secant method will work when Newton’s method does not and vice-versa. The method is usually a bit slower than Newton’s method. It is more rapidly convergent than the bisection method. This method does not require use of the derivative of the function. N.B.V yas − Department of M athematics, AIT S − Rajkot
  89. 89. Secant MethodGeneral Features: The secant method is an open method and may not converge. It requires fewer function evaluations. In some problems the secant method will work when Newton’s method does not and vice-versa. The method is usually a bit slower than Newton’s method. It is more rapidly convergent than the bisection method. This method does not require use of the derivative of the function. This method requires only one function evaluation per iteration. N.B.V yas − Department of M athematics, AIT S − Rajkot
  90. 90. Secant MethodDisadvantages: There is no guaranteed error bound for the computed value. N.B.V yas − Department of M athematics, AIT S − Rajkot
  91. 91. Secant MethodDisadvantages: There is no guaranteed error bound for the computed value. It is likely to difficulty of f (x) = 0. This means X−axis is tangent to the graph of y = f (x) N.B.V yas − Department of M athematics, AIT S − Rajkot
  92. 92. Secant MethodDisadvantages: There is no guaranteed error bound for the computed value. It is likely to difficulty of f (x) = 0. This means X−axis is tangent to the graph of y = f (x) Method may converge very slowly or not at all. N.B.V yas − Department of M athematics, AIT S − Rajkot

×