受不了了

画图我会，题目里德epsilon_a公式如下：图中（5.5）
matlab里有bisection这个function吗？为什么我每次用bisect都说undefined function?
好像要自己倒入？
我根据这一页导入的

我的code是：

[size=12.800002098083496px]function [root,fx,ea,iter]=bisect(func,x1,xu,es,maxit,varargin)
[size=12.800002098083496px]if nargin<3,error('at least 3 input arguments required'),end
[size=12.800002098083496px]test = func(x1,varargin{:})*func(xu,varargin{:});
[size=12.800002098083496px]if test>0,error('no sign change'),end
[size=12.800002098083496px]if nargin<4|isempty(es),es=0.0001;end
[size=12.800002098083496px]if nargin<5|isempty(maxit),maxit=50;end
[size=12.800002098083496px]iter=0;xr=x1;ea=100;
[size=12.800002098083496px]while(1)
[size=12.800002098083496px]    xrold=xr;
[size=12.800002098083496px]    xr=(x1+xu)/2;
[size=12.800002098083496px]    iter=iter+1;
[size=12.800002098083496px]    if xr ~=0,ea=abs((xr-xrold)/xr)*100;end
[size=12.800002098083496px]    test=func(x1,varargin{:})*func(xr,varargin{:});
[size=12.800002098083496px]    if test<0
[size=12.800002098083496px]        xu=xr;
[size=12.800002098083496px]    elseif test>0
[size=12.800002098083496px]        x1=xr;
[size=12.800002098083496px]    else
[size=12.800002098083496px]        ea=0;
[size=12.800002098083496px]    end
[size=12.800002098083496px]    if ea<=es | iter >=maxit,break,end
[size=12.800002098083496px]end
[size=12.800002098083496px]root = xr; fx=func(xr,varargin{:});

[size=12.800002098083496px]fm=sin(m)-m.^3;
[size=12.800002098083496px][~, fx ea iter]=bisect(fm,0.5,1);[size=12.727272033691406px]

[size=12.727272033691406px]然后怎么操作？把这个M-file保存后再编辑另一个即可？
我试着模仿下面这个例题，未遂。。。


求高手解答！
我从网上看的很多地方用abs(b-a)/2<2%作为停止的标准，a,b是初始区间断点。虽然code中间已经重置了a,b,但我还是看不出为什么abs(b-a)/2和abs((x_new-x_old)/x_new)的联系。


谢谢！

hrbatanu 发表于 2013-7-2 00:16
I am with you, but I need to fit the code into my question: sin(x)-x^3,  thx anyway~

针对你的问题，这样就可以啦
fm=@(m)sin(m)-m^3;
[mass, fx ea iter]=bisect(fm,0.5,1)
结果是：
mass = 0.9286
fx = -4.5446e-007
ea = 5.1349e-005
iter =20

新建一个bisect.m就可以啦

fm=@(m)sin(m)-m^3;[mass, fx ea iter]=bisect(fm,0.5,1)
mass = 0.9286
fx = -4.5446e-007
ea = 5.1349e-005
iter =20

fm=@(m)sqrt(9.81*m/0.25)*tanh(sqrt(9.81*0.25/m)*4)-36;
[mass, fx ea iter]=bisect(fm,40,200)
mass =142.7377
fx =4.6089e-007
ea =5.3450e-005
iter =21

24578901 发表于 2013-7-1 14:41
fm=@(m)sin(m)-m^3;[mass, fx ea iter]=bisect(fm,0.5,1)
mass = 0.9286
fx = -4.5446e-007

Undefined function 'bisect' for input arguments of type 'function_handle'.

24578901 发表于 2013-7-1 14:41
fm=@(m)sin(m)-m^3;[mass, fx ea iter]=bisect(fm,0.5,1)
mass = 0.9286
fx = -4.5446e-007

is bisection built in matlab? or we should implement by ourself?

hrbatanu 发表于 2013-7-1 23:19
is bisection built in matlab? or we should implement by ourself?

1、要想知道Matlab有没有bisection，可以在help（软件界面上的?号）查询

2、当你在命令窗口运行你的代码，发现提示Undefined function 'bisect'，这时候你应该建立一个bisect.m之后在运行就可以了

yes, but how to build a bisect.m? I've tried that as I said in the question, I copy from my textbook, but I do not fully understand that and thus fail to build it.

hrbatanu 发表于 2013-7-1 23:43
yes, but how to build a bisect.m? I've tried that as I said in the question, I copy from my textbook ...

步骤：
1、点击软件左上角的新建

2、把以下内容粘贴下去
function [root,fx,ea,iter]=bisect(func,x1,xu,es,maxit,varargin)
if nargin<3,error('at least 3 input arguments required'),end
test = func(x1,varargin{:})*func(xu,varargin{:});
if test>0,error('no sign change'),end
if nargin<4|isempty(es),es=0.0001;end
if nargin<5|isempty(maxit),maxit=50;end
iter=0;xr=x1;ea=100;
while(1)
    xrold=xr;
    xr=(x1+xu)/2;
    iter=iter+1;
    if xr ~=0,ea=abs((xr-xrold)/xr)*100;end
    test=func(x1,varargin{:})*func(xr,varargin{:});
    if test<0
        xu=xr;
    elseif test>0
        x1=xr;
    else
        ea=0;
    end
    if ea<=es | iter >=maxit,break,end
end
root = xr; fx=func(xr,varargin{:});

3、按保存，名字会自动保存为bisect.m

4、在命令窗口输入
fm=@(m)sqrt(9.81*m/0.25)*tanh(sqrt(9.81*0.25/m)*4)-36;
[mass, fx ea iter]=bisect(fm,40,200)

5、done！！

24578901 发表于 2013-7-1 23:47
步骤：
1、点击软件左上角的新建

I am with you, but I need to fit the code into my question: sin(x)-x^3,  thx anyway~