matrix form numerical integration

2012 4 25
1 个回答
回答已采纳
3 次查看（30 天）

0 个投票

This code does not work
clear;
x = [1.0024 0.264 1.4334 1.46 0.0219 0.1;
1.0024 0.264 1.435 1.46 0.0220 0.1;
1.0024 0.264 1.4352 1.46 0.0221 0.1;
1.0024 0.264 1.4353 1.46 0.0222 0.1;
1.0024 0.264 1.4354 1.46 0.0222 0.1;
1.0024 0.264 1.4357 1.46 0.0223 0.1];
r = x(:,1); Y = x(:,2); q = x(:,3); K=x(:,4); L=x(:,5); c=x(:,6);
t = 0.1:0.1:0.6;
z = trapz(t,normcdf(((log(q./K)+t.*c.^2./2)./(c.*sqrt(t))),0,1));
If I just use one line of records, i.e
z = trapz(t,normcdf(((log(q(1)./K(1))+t.*c(1).^2./2)./(c(1).*sqrt(t))),0,1));
Then it works!!! What I dont understand is how to do this 'trapz' for each line of records in the vector 'x'.

请先登录，再回答此问题。

 采纳的回答

Thomas
Thomas 2012-4-25

0 个投票

you are trying to perform trapezoidal numerical integration over a number of different values but the output z is not a vector.. MAybe you need something like this.. where you get a z for each line of input data..
clear;
x = [1.0024 0.264 1.4334 1.46 0.0219 0.1;
1.0024 0.264 1.435 1.46 0.0220 0.1;
1.0024 0.264 1.4352 1.46 0.0221 0.1;
1.0024 0.264 1.4353 1.46 0.0222 0.1;
1.0024 0.264 1.4354 1.46 0.0222 0.1;
1.0024 0.264 1.4357 1.46 0.0223 0.1];
r = x(:,1); Y = x(:,2); q = x(:,3); K=x(:,4); L=x(:,5); c=x(:,6);
t = 0.1:0.1:0.6;
for i=1:length(x)
z(i) = trapz(t,normcdf(((log(q(i)./K(i))+t.*c(i).^2./2)./(c(i).*sqrt(t))),0,1));
end

更多回答（0 个）

类别

在 帮助中心 和 File Exchange 中查找有关 Numerical Integration and Differentiation 的更多信息

产品

标签

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by