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4068 Fortran Programming Coursework 6.2 Newton’s method for root-finding Name: Zora Wing Fong Law Student ID: 00310913, SSTP 1. Flow chart (see also the Numerical Implementation section) Given a function f(x), want to find it root(s) Assume the initial guess is given, x = a Proceed to find a better guess b by considering the intersection of the tangent to the f (a) function f(a) with x-axis. b a f ' (a) f (a) 2. Derivation of b a f ' (a) Let the given function be y = f(x), the equation of the tangent is thus y = f’(x) + c. At the initial guess a, f(a) = f’(a)a + c, Since b is the intersection of the tangent to the function f(a) with x-axis, f’(a)b + c = 0, i.e. c = - f’(a)b. f (a) Hence f(a) = f’(a)a - f’(a)b, i.e. b a . f ' (a) 3. Numerical Implementation This program initially requires 3 input parameters: tolerance <xacc>, maximum number of iterations <jmax> and the initial guess <xint>. It calls two functions f, fprime which calculate the value of f(x) and f’(x) respectively. It generates an error warning if f’(x) = 0 For each iteration, the number of iterations j, guess x, and the value of f(x) are printed on screen. The final solution is stated with the number of iterations taken. 4. Results & Analysis Function, tolerance FORTRAN Mathematica Initial guess No. of Solution Solution Iterations f = x2+5x+2, 1e-4 -0.43844718 -0.438447 -0.5 2 -1.0, 0.0, 0.5, 1.0 3 -1.5, 1.5, 2.0, 2.5 4 -2.0 5 f = x3 +x +5, 1e-4 -1.51598024 -1.51598 -1.5 2 -2.0, -1.8, -1.3 3 -1.0 4 -0.5, 0.5 5 1.0 6 0.0, 1.5 7 Referring to the above data, we see the number of iterations varies with the initial choice of a. The cubic equation seems to be more sensitive to the initial guess than the quadratic case, especially it take 7 iterations for a = 0.0 to generate a tolerable solution for f = x3 +x +5, while it takes 5 iterations for a = -2.0 for f = x2+5x+2. 5. Results using method of Bracketing & Bisection Function, tolerance FORTRAN Bracketing & Bisecting Solution Initial guess Solution Initial guess f = x2+5x+2, 1e-4 -0.43844718 -0.5 -0.250000 -1.0 & 0.0 f = x3 +x +5, 1e-4 -1.51598024 -1.5 -1.500000 -2.0 & -1.0 6. Exact Solution of the Cubic equation f(x) = f = x3 +x +5 Exact: 1 3 1 3 1 45 2037 2 2 x , 3 45 2037 32 3 1 1 3 1 3 45 2037 2 x 2 32 3 1 3 , 1 3 22 3 3 45 2037 1 1 3 1 3 45 2037 2 x 2 32 3 1 3 1 3 22 3 3 45 2037 Numerical: {{x-1.51598},{x0.75799 -1.65035 },{x0.75799 +1.65035 }} Comparing with the exact result, the solution obtained by using the FORTRAN program is fairly accurate.

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posted: | 5/1/2010 |

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