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MCS 471 --- Computer Problem 2 --- Fall 2004

Zeros of Nonlinear Equations:
Computational Comparisons

Programs with Computer Outputs are Due Monday 18 October 2004 in Class.

  • General Description:This computer problem involves finding zeros of nonlinear equationsusing the TWO of the solution systems:

    1. MATLAB* interactive numerical computational (matrix) system;
    2. Maple interactive symbolic/numeric computational system;

    Test your code using three test function examples given below, but your programs should be fairly general and NOT tailored to these examples.

    *A programming language system (like C or Fortran or their variants) may be substituted for the MATLAB system ifpreferred.

  • Output Requirements: For the TWO Above Systems, submit

    1. If MATLAB system, a MATLAB M-File program, and Command Windowoutput for the three test examples, well-documented
      (if substitute C/Fortran system, a single program with six procedures,two each for the three sets of test functions and derivatives, withoutput well-documented source code and output);

    2. If Maple system, a Maple worksheet of code for the three test examples and output, well-documented;

  • Documentation:Document your own code with a brief description of the problem,an important variables dictionary and headings for important programmingsteps (use programming language comment lines).

  • Newton Method Comparison:Implement the Newton's algorithm given in class with the stopping criteriagiven later.However, if iterations run out of the current COSI interval [a,b],then do three (3) bisection iterations to reduce the current COSI interval in size, then continue with Newton's method with the reduced COSI. Count all function evaluations used:

    1. If MATLAB programming, set "format long", implement Newton's methodand compare the answer to MATLAB's "[x_zero,f_zero] := fzero(f,x0);" if"f" is an inline defined function else use the function handle version "@f" if a m-file defined function "f" or else use the expression for thefunction enclosed in single quotes in place of the "f" argument of "fzero",where "x0" is the x-value argument of the "min(f(a),f(b))",while plotting each of the three functions on the initial interval "[a,b]".
      (If C/F programming substitute, use double precision (64 bits = 8 bytes, double in C or real*8 in Fortran) for Newton's and the comparison method. You may use C (CC or C++), Fortran (F77 or F90+) for this problem (else, see Prof. Hanson). Compare your Newton results to the Method of Bisection algorithm (also presented in class) using the number of iterations and the difference infinal zero position answers and the approximate value of the functionat that point.)

    2. If Maple programming, set "Digits = 16;", implement Newton's methodand compare the answer to Maple's "x_zero := fsolve(f(x)=0,x=a..b);" andthe value "f_zero := f(x_zero);", while plotting each of the three functionson the initial interval "[a,b]".

  • Newton Method Hybrid Absolute-Relative Stopping Criterion:

      abs(xnew-xold) + abs(fnew) < tolabs + tolrel*(abs(xold)+abs(fold))

      with absolute tolerance tolabs = 0.5e - 7 for all subroutines and tolrel = 10*deps where deps is the double precision machine epsilon (found forComputer Problem#1) for the relative tolerance scale.Iteration limit is kmax = 80 (i.e., abort current function case with message if k > kmax) and the Tolerance Lower Limit for Derivative, |f'(x)| > told, Divide Check is told = 0.5E - 10.

  • Three Good Test Examples:

    1. f1(x) = exp(-7.627*x) - x*exp(-4.759); on [0.2, 1.2];

    2. f2(x) = exp(4.947) - x*exp(4.854) + exp(2.252*x); on [0.0,2.0];

    3. f3(x) = 80.18*exp(-75.88*(x-0.5714)^2) - 77.09*exp(-75.97*(x-0.4903)^2);on [0.1,1.1].
    In each case use for the starting value the end point of the initialinterval that has the minimum function absolute value.

  • Additional and Special Output Requirements:

    1. Name of Your Computer System, Processor, MATLAB version (or substitute Compiler, and Compile Command Line) and Maple version;

    2. For each test example and Newton's method print a nicely labeled table of

      kfeval,     xold,     xnew,     abs(xnew-xold),     f(xnew),     f'(xold),    

      for all iterates displaying 5 significant digits. The item (f'(xold))is only used for the Newton's method.Give the starting value for both xnew andf(xnew) for the appropriate comparison method. Here, kfeval is total number of functionevaluations for current test function. The abs(xnew-xold)and f(xnew) items of the table should be printed ine-notation, because errors should almost always be in e-notation.

  • Helpful Information:

    1. Help on MATLAB:

      • Prof. Hanson's Help for MATLAB,

    2. Help on Maple Worksheets:

      • Beginner's Introduction to MapleV Release 5, or for more information see

      • Prof. Hanson's Help for Maple Symbolic Tools Archive

    3. Substitute C Caution: Do NOT forget to include math.h headers, andadd the compiler option -lm to the gcc (or any othercc variant) option list if needed (i.e., to remove errors because the compiler can not find some math functions). Also, C needs the powmath function, since it lacks an exponentiation symbol like "^".

    4. Substitute Fortran Caution: Do NOT forget that EXTERNAL statements are needed in Fortran, in all subprograms in which other subprograms are called with FUNCTIONarguments; the syntax is like EXTERNAL f1, f2, ..., fN and must comebefore any executable statements. Also, Fortran uses "**" for exponentiation instead of "^".

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