How to find first negative solution with the bisection method

f(x) = x + 1 −2 sin(πx) = 0

How to find first negative solution with the given bisection method

function [c, n, err] = Bisection_method(g, a, b, tol, N)
    c = [];
    n = 0;
    err = inf;
    FA = g(a);
    FB = g(b);
    if(a > b)
        err = inf;
        c = [];
    elseif (FA*FB>= 0)
    else
        while ((abs(err) > abs(tol)) && (n <= N))
            n = n+1;
            c = (a + b) / 2;
            fmid = g(c);
            err = abs(fmid);
            if(fmid * g(a) > 0)
                a = c;       
            
            else
                b = c;
             
            end
        end
    end
end

[KSSV]

You should play with the intervel [a,b].

Answer 1

[Chunru]

g = @(x) x + 1 -2 * sin(pi*x) ;

% Use interval [-1.5, 0] for example
[c, n, err] = Bisection_method(g, -1.5, 0, 1e-6, 1000)

c = -1.0000

n = 22

err = 8.6822e-07

function [c, n, err] = Bisection_method(g, a, b, tol, N)
    c = [];
    n = 0;
    err = inf;
    FA = g(a);
    FB = g(b);
    if(a > b)
        err = inf;
        c = [];
    elseif (FA*FB>= 0)
    else
        while ((abs(err) > abs(tol)) && (n <= N))
            n = n+1;
            c = (a + b) / 2;
            fmid = g(c);
            err = abs(fmid);
            if(fmid * g(a) > 0)
                a = c;       
            
            else
                b = c;
             
            end
        end
    end
end