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This is solution to one of problems in Numerical Analysis. This is matlab code. Its helpful to students of Computer Science, Electrical and Mechanical Engineering. This code also help to understand algorithm and logic behind the problem. This code includes: Newton, Rophson, Method, Algorithm, Initial, Approximation, Tolerance, Maximum, Iterations, Solution, Function, Derivative
Typology: Exercises
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% To find a solution to f(x) = 0 given an % initial approximation p0: % % INPUT: initial approximation p0; tolerance TOL; % maximum number of iterations NO. % % OUTPUT: approximate solution p or a message of failure syms('OK', 'P0', 'TOL', 'NO', 'FLAG', 'NAME', 'OUP', 'F0'); syms('I', 'FP0', 'D','x','s'); TRUE = 1; FALSE = 0; fprintf(1,'This is Newtons Method\n'); fprintf(1,'Input the function F(x) in terms of x\n'); fprintf(1,'For example: cos(x)\n'); s = input(' ','s'); F = inline(s,'x'); fprintf(1,'Input the derivative of F(x) in terms of x\n'); s = input(' '); FP = inline(s,'x'); OK = FALSE; fprintf(1,'Input initial approximation\n'); P0 = input(' '); while OK == FALSE fprintf(1,'Input tolerance\n'); TOL = input(' '); if TOL <= 0 fprintf(1,'Tolerance must be positive\n'); else OK = TRUE; end end OK = FALSE; while OK == FALSE fprintf(1,'Input maximum number of iterations - no decimal point\n'); NO = input(' '); if NO <= 0 fprintf(1,'Must be positive integer\n'); else OK = TRUE; end end if OK == TRUE fprintf(1,'Select output destination\n'); fprintf(1,'1. Screen\n'); fprintf(1,'2. Text file\n'); fprintf(1,'Enter 1 or 2\n'); FLAG = input(' '); if FLAG == 2 fprintf(1,'Input the file name in the form - drive:\name.ext\n'); fprintf(1,'For example: A:\OUTPUT.DTA\n'); NAME = input(' ','s'); OUP = fopen(NAME,'wt');
else OUP = 1; end fprintf(1,'Select amount of output\n'); fprintf(1,'1. Answer only\n'); fprintf(1,'2. All intermediate approximations\n'); fprintf(1,'Enter 1 or 2\n'); FLAG = input(' '); fprintf(OUP, 'Newtons Method\n'); if FLAG == 2 fprintf(OUP, ' I P F(P)\n'); end F0 = F(P0); % STEP 1 I = 1; OK = TRUE; % STEP 2 while I <= NO & OK == TRUE % STEP 3 % compute P(I) FP0 = FP(P0); D = F0/FP0; % STEP 6 P0 = P0 - D; F0 = F(P0); if FLAG == 2 fprintf(OUP,'%3d %14.8e %14.7e\n',I,P0,F0); end % STEP 4 if abs(D) < TOL % procedure completed successfully fprintf(OUP,'\nApproximate solution = %.10e\n',P0); fprintf(OUP,'with F(P) = %.10e\n',F0); fprintf(OUP,'Number of iterations = %d\n',I); fprintf(OUP,'Tolerance = %.10e\n',TOL); OK = FALSE; % STEP 5 else I = I+1; end end if OK == TRUE % STEP 7 % procedure completed unsuccessfully fprintf(OUP,'\nIteration number %d',NO); fprintf(OUP,' gave approximation %.10e\n',P0); fprintf(OUP,'with F(P) = %.10e not within tolerance %.10e\n',F0,TOL); end if OUP ~= 1 fclose(OUP); fprintf(1,'Output file %s created successfully \n',NAME); end end