Program 3-2: User-defined function. Newton's method. function Xs = NewtonRoot (Fun, FunDer, Xest, Err, imax) * NewtonRoot finds the root of Fun = 0 near the point Xest using Newton's method. * Input variables: % Fun Name of a user-defined function that calculates Fun for a given x. * FunDer Name of a user-defined function that calculates the derivative of Fun for a given x. 8 % Xest Initial estimate of the solution. % Err Maximum error. * imax Maximum number of iterations Output variable: % Xs Solution for i=1: imax end Xi Xest - Fun (Xest)/FunDer (Xest); Eq. (3.14). if abs ((Xi - Xs = Xi; Xest)/Xest) < Err Eq. (3.18). break end Xest = Xi; if i = = imax fprintf('Solution was not obtained in %i iterations.\n', imax) Xs ('No answer'); end x+1 =x- 8–4.5(x – sinx) -4.5(1-cos.x;) x2 = 2- 8-4.5(2-sin(2))- 2.48517 -4.5(1 cos(2)) X3 = 2.48517- 8-4.5(2.48517- sin (2.48517)) 2.43099 -4.5(1 cos(2.48517)) function y = FunExample2 (x) - y 8 4.5* (x - sin(x)); function y FunDerExample2 (x) y = -4.5 +4.5*cos(x); The user-defined functions are entered as function handles. >> format long >> xSolution = NewtonRoot (@Fun Example2, @FunDerExample2,2,0.0001, 10) xSolution = 2.430465741723630 ('x), f x = 1+x f(x)) f(x) 10 5 0 -5 -10 -155 0 1 2x 3 clear all F(x) 8-4.5* (x-sin(x)); Define Ax) as an anonymous function. a = 2; b = 3; imax = 20; tol = 0.001; Fa = F(a); Fb = F(b); if Fa*Fb > 0 Assign initial values to a and b, define max number of iterations and tolerance. disp (Error: The function has the same sign at points a and b.') else Stop the program if the function has the same sign at points a and b. disp('iteration a b (xNS) Solution f(xNS) Tolerance') for i = 1; imax xNS toli (a+b)/2; Calculate the numerical solution of the iteration, xNS. Calculate the current tolerance. Calculate the value of f(xNS) of the iteration. FxNS (ba)/2; F(xNS); fprintf('%3i %11.6f $11.6f $11.6f $11.6f $11.6f\n', i, a, b, xNS, FXNS, toli) if FxNS == 0 fprintf('An exact solution x =11.6f was found',>NS) end break if toli < tol Stop the program if the true solution, f(x) 0, is found. break end if i = = imax end H Stop the iterations if the tolerance of the iter- ation is smaller than the desired tolerance. fprintf('Solution was not obtained in &i iterations',imax) break if F(a) *FxNS < 0 b = xNS; else a = xNS; end end end Stop the iterations if the solution was not obtained and the num- ber of the iteration reaches imax. Determine whether the true solution is between a and xNS, or between xNS and b. and select a and b for the next iteration. iteration a b (xNS) Solution 1 2 3 2.000000 2.000000 2.250000 3.000000 2.500000 f(xNS) -0.556875 Tolerance 0.500000 2.500000 2.250000 1.376329 0.250000 2.500000 2.375000 0.434083 0.125000 4 2.375000 2.500000 2.437500 -0.055709 0.062500 5 2.375000 2.437500 2.406250 0.190661 0.031250 6 2.406250 2.437500 2.421875 0.067838 0.015625 10 7899 2.421875 2.437500 2.429688 0.006154 0.007813 2.429688 2.437500 2.433594 -0.024755 0.003906 2.429688 2.433594 2.431641 -0.009295 0.001953 2.429688 2.431641 2.430664 -0.001569 0.000977 The numerical solution. The value of the function at the numerical solution. The last tolerance (satisfies the prescribed tolerance). SNX 9+0 Tolerance = b-a 2 x x rsina α 3r x 4 r α C α x