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Introduction of Numerical method

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### 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