二分法matlab编程代码及注释(MATLAB二分法求方程的近似解)

matlab二分法求解实例

关注:95  答案:2  mip版

解决时间 2021-01-28 01:44

提问者人潮拥挤你不在

2021-01-27 18:57

求函数 f=x^3+2*x^2+x-5 (-2，2)为区间起点和终点，10^-4为精度

>> a=-2;

b=2;

x=a:b;

f = @(x)x^3+2*x^2+x-5;

c=(a+b)./2;

while abs(b-a)>1e-6

if f(c)*f(b)<0

a=c;

else

b=c;

end

c=(a+b)./2;

x=c;

end

fprintf(‘\n x = %.5f, f(x) = %.5f \n’, x, f(x));

如果区间有两个，分别有一个零点，即x1,x2;需要怎么改动输出两个零点值啊

最佳答案

二级知识专家青春敷年華

2021-01-27 19:25

那调用两次不就ok了。区间你是用a、b定义的嘛。

全部回答

1楼雨落轻尘

2021-01-27 19:40

>> f=inline(‘x^2-x-2’);

>> [c,err,yc]=bisect(f,0,3,0.01)

c =

2.0010

err =

0.0059

yc =

0.0029

———–

%使用二分法 求解上面超越方程

%下面是二分法的函数文件,你直接设置输入参数就可以了

function [c,err,yc]=bisect(f,a,b,delta)

%input – f is the function

% – a and b are the left and right endpoints

% – delta is the tolerance

%output – c is the zero

% – yc= f(c)

% – err is the error estimate for c

%if f is defined as an m-file function use the @ notation

% call [c,err,yc]=bisect(@f,a,b,delta).

%if f is defined as an anonymous function use the

% call [c,err,yc]=bisect(f,a,b,delta).

% numerical methods: matlab programs

% (c) 2004 by john h. mathews and kurtis d. fink

% complementary software to accompany the textbook:

% numerical methods: using matlab, fourth edition

% isbn: 0-13-065248-2

% prentice-hall pub. inc.

% one lake street

% upper saddle river, nj 07458

ya=f(a);

yb=f(b);

if ya*yb > 0,return,end

max1=1+round((log(b-a)-log(delta))/log(2));

for k=1:max1

c=(a+b)/2;

yc=f(c);

if yc==0

a=c;

b=c;

elseif yb*yc>0

b=c;

yb=yc;

else

a=c;

ya=yc;

end

if b-a < delta, break,end

end

c=(a+b)/2;

err=abs(b-a);

yc=f(c);

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