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楼主 |
发表于 2007-11-20 15:38
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%%% example01 %%%%
z=linspace(-0.1,0.1);
a=0.1;
kesi=(z+sqrt(z.^2-a^2))./a;
for j=1:length(z)
f=@(b) log(1-((b+sqrt(b.^2-a^2))./a)./kesi(j)); % 被积函数
I(j)=quadl(f,-0.1+eps,0.1+eps,1e-8);
end
%%% example02 %%%%
z=linspace(-0.1,0.1);
a=0.1;
kesi=(z+sqrt(z.^2-a^2))./a;
for i=1:length(z)
k=kesi(i);
fun=strcat('log(1-((b+sqrt(b.^2-0.1^2))./0.1)./',num2str(k),')');
II(i)=quadl(inline(fun),-0.1+eps,0.1+eps);
end
%%% example03 %%%%
z=linspace(-0.1,0.1);
a=0.1;
kesi=(z+sqrt(z.^2-a^2))./a;
for i=1:length(z)
k=kesi(i);
fun=inline(subs('log(1-((b+sqrt(b^2-0.1^2))/0.1)/kesi)','kesi',k));
III(i)=quadl(fun,-0.1+eps,0.1+eps);
end
%%% example04 %%%%
z=linspace(-0.1,0.1);
a=0.1;
kesi=(z+sqrt(z.^2-a^2))./a;
ff=@(k) ['log(1-((b+sqrt(b.^2-0.1^2))./0.1)./',num2str(k),')']; % 被积函数
f=@(k) quadl(ff(k),-0.1+eps,0.1+eps);
for i=1:length(z)
IIII(i)=f(kesi(i));
end
%%% example05 %%%%
clear all;clc;
a=0.1;
f=@(kesi) quadl(@(b) log(1-((b+sqrt(b.^2-a^2))./a)./kesi),-0.1+eps,0.1+eps)
z=linspace(-0.1,0.1);
kesi=(z+sqrt(z.^2-a^2))./a;
y=zeros(size(z));
for ii=1:length(z)
IIIII(ii)=f(kesi(ii));
end
%%example06 %%%%
a=0.1;
z=linspace(-0.1,0.1);
kesi=(z+sqrt(z.^2-a^2))./a;
IIIIII=arrayfun(@(kesi) quad(@(b) log(1-((b+sqrt(b.^2-a^2))./a)./kesi),-0.1+eps,0.1+eps),kesi);
以上程序都可以积出来,其中1,3,5,6结果相同,2,4相同。
也就是说凡是用到NUM2STR的结果与其他不一样。
这也正是LS的那位的迷惑,我也同样迷惑
再次感谢LS。
希望高人能答疑解惑
[ 本帖最后由 sigma665 于 2007-11-20 17:39 编辑 ] |
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