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用matlab程序写用二分法求方程根
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二分法在很多地方应该都会见到,这里是通过二分法迭代逼近的方法求出一个方程的根。
function xc = bisection(f,a,b,tol)
% use the bisection method to find the root of the function
% Page 30,computer problem 7(Bisection method)
% input:
% f:the function that transform from the equation
% a,b:the left and right value of the interval which the root is in% tol:the accuracy
% output:
% xc:the solution of the equationif sign(f(a)) * sign(f(b)) >=0
error('f(a)f(b)<0 not satisfied!')endif nargin < 3
disp('The function should at least include 3 parameters');endif nargin == 3
tol = 10^-6;endwhile (b-a)/2 > tol
c = (a + b)/2;
if f(c) == 0 % when f(c) == 0,c is a root of the function
break
end
if f(a) * f(c) < 0
% a and c form a new interval
b = c; else % c and b form a new interval
a = c; endendxc = (a+b)/2; % the mid_rang is the root that we find
编程问题,已经给出完整代码,没有相关配图可以配,希望谅解。
参考资料
CSDN.CSDN[引用时间2018-1-9]
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matlab源程序如下:
function erfenfa(a,b)%a,b为区间
s=(a+b)/2;
while b-a>1e-5
if fun(a)*fun(s)>0
a=s;
elseif fun(a)*fun(s)<0
b=s;
elseif fun(s)==0
disp(s);
end
s=(a+b)/2;
end
disp(s);
function y=fun(x)
y=2*x^2+x-4;%这里是函数表达式,例如:y=2*x^2+x-4
end
end
function erfenfa(a,b)%a,b为区间
s=(a+b)/2;
while b-a>1e-5
if fun(a)*fun(s)>0
a=s;
elseif fun(a)*fun(s)<0
b=s;
elseif fun(s)==0
disp(s);
end
s=(a+b)/2;
end
disp(s);
function y=fun(x)
y=2*x^2+x-4;%这里是函数表达式,例如:y=2*x^2+x-4
end
end
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