# How to find first negative solution with the bisection method

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ken on 9 May 2022
Answered: Chunru on 9 May 2022
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 on 9 May 2022
You should play with the intervel [a,b].

Chunru on 9 May 2022
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