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Introduction of Numerical method

Introduction of Numerical method

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- 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. 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. 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. 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. 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. 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. 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. Errors It is never possible to measure anything exactly. N.B.V yas − Department of M athematics, AIT S − Rajkot
- 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. 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. 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. 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. 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. 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. 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. 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. Bisection Method Consider a continuous function f (x). N.B.V yas − Department of M athematics, AIT S − Rajkot
- 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. 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. 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. 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. 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. 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. 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. 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. Bisection MethodCharacteristics: 1 This method always slowly converge to a root. N.B.V yas − Department of M athematics, AIT S − Rajkot
- 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. 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. 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. 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. Bisection Method- ExampleSol. Let f (x) = x3 − x − 1 = 0 N.B.V yas − Department of M athematics, AIT S − Rajkot
- 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. N-R Method:(Drawbacks) Division by zero N.B.V yas − Department of M athematics, AIT S − Rajkot
- 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. N-R Method:(Drawbacks) Root Jumping N.B.V yas − Department of M athematics, AIT S − Rajkot
- 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. Secant Method This method requires two initial guesses N.B.V yas − Department of M athematics, AIT S − Rajkot
- 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. Secant MethodDisadvantages: There is no guaranteed error bound for the computed value. N.B.V yas − Department of M athematics, AIT S − Rajkot
- 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. 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

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