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MATLAB Root-Finding Function

  1. Sep 20, 2011 #1
    1. The problem statement, all variables and given/known data
    Create and test a MATLAB function called bisection that implements the bisection root finding method. You will have to use a function handle and exit flag coding.

    2. Relevant equations

    3. The attempt at a solution

    Code (Text):
    % bisection.m

    function [root, err, iter, exitFlag] = bisection (fun, lb, ub, errMax, iterMax)

    % This function performs the bisection algorithm on a function to
    % locate a root of the function (location where it crosses zero). The
    % function must be of a single variable and must return a scalar result
    % (i.e., a scalar function of a scalar argument).
    % The approximation loop is controlled by three termination criteria:
    %  - estimate of the approximation error is less than errTol
    %  - number of iterations is greater than iterMax
    %  - function value magnitude is less than funTol
    % The initial bracket controls the search. The initial bracket must be
    % defined such that the sign of the function is opposite at each end of the
    % bracket (to ensure a zero crossing exists between the two. THERE IS NO
    % ERROR CHECKING for this in the current code.
    % Inputs:  @fun = function that is the subject of the root-finding
    %          lb = lower bound of the bisection search bracket
    %          ub = upper bound of the bisection search bracket
    %          errMax = maximum acceptable relative approximation error
    %          iterMax = maximum number of iterations
    % Outputs: root = approximation of root location
    %          err = relative approximation error of solution
    %          iter = number of iterations taken to find the solution
    %          exitFlag = indicates termination status of function
    % Exit Flag Encoding: 1: Alrogithm terminated normally (due to error being
    %                        sufficiently small)
    %                     0: Alrogithm terminated due to maximum number of
    %                        iterations being reached
    %                    -1: Algorithm terminated due to invalid bracket
    %                        specification (no root in bracket)
    %                    -2: Algorithm terminated due to invalid return value
    %                        from function fun (e.g., NaN, Inf, empty brakets)

    iter = 0;       % iteration counter variable
    done = false;   % Boolean variable to control looping
    root = (ub+lb)/2; % initial approximation of root location
    err = 100*abs([ub-lb]/[ub+lb]); % estimate of error

    if fun(lb)*fun(ub)>0
        exitFlag = -1;
        error('invalid bracket: exitFlag = %g\n',exitFlag);

    elseif fun(root)+errMax>0 && fun(root)-errMax<0;
        exitFlag = 1;
        err = 100*abs([ub-lb]/[ub+lb]); % estimate of error
    elseif fun(lb)+errMax>0 && fun(lb)-errMax<0;
        exitFlag = 1;
        err = 100*abs([ub-lb]/[ub+lb]); % estimate of error
    elseif fun(ub)+errMax>0 && fun(ub)-errMax<0;
        exitFlag = 1;
        err = 100*abs([ub-lb]/[ub+lb]); % estimate of error

    elseif fun(lb)==NaN || Inf || -Inf || isempty(lb);
        exitFlag = -2;
        error('invalid lower bound: exitFlag = %g\n',exitFlag);
        done = true;

    elseif fun(ub)==NaN || Inf || -Inf || isempty(ub);
        exitFlag = -2;
        error('invalid upper bound: exitFlag = %g\n',exitFlag);
        done = true;


    while ~done
        iter = iter+1;      % increment iteration counter
            % determine which way to go with next bracket
        if fun(lb)*fun(root)<0
            ub = root;
            lb = root;
        % update approximation of root location based on new bracket
        root = (ub+lb)/2;    
        err = 100*abs([ub-lb]/[ub+lb]); % estimate of error
        % loop control -- check for termination criteria
        if err<errMax
            exitFlag = 1;
            done = true;
        elseif iter >= iterMax
            exitFlag = 0;
            % if the next line executes, it means at least one of the
            % termination criteria was true
            done = true;    

    Example of function call:
    Code (Text):
    [root, err, iter, exitFlag] = bisection (@(x)x^3-3*x,-4,2,.0001, 1000)

    My problem with the code is that the error flag -2 keeps popping up when I am trying to find a root
  2. jcsd
  3. Sep 20, 2011 #2


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Two things: first, Doing fun(lb)==NaN doesn't work. Try it out, NaN==NaN returns 0, not 1. There's a function isnan that you should use instead,

    Second, you have a serious parenthesis problem. fun(lb)==NaN || Inf doesn't do (fun(lb)==NaN)||(fun(lb)==Inf), it does (fun(lb)==NaN) || Inf and Inf counts as true so your program always takes that part of the if statement and runs it
  4. Sep 20, 2011 #3
    Ok, thanks that helps. Between you and a TA I got it to work so thank you!
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