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Topic: Error : Inner matrix dimensions must agree When computing the three dimensional integral.
Replies: 7   Last Post: Jun 24, 2013 11:14 AM

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 yolanda Posts: 5 Registered: 6/19/13
Re: Error : Inner matrix dimensions must agree When computing the three dimensional integral.
Posted: Jun 20, 2013 12:47 PM

Hi,

You are right. I also checked the triplequad function detailedly yesterday. And as I defined, z should be a vector and then I can use the function triplequad.

So as you said, the two matrixs don't work. Then how should I define the f function?
The exponentional part of the f function is just the part of the three dimension
joint Gaussian distributation as following:

exp( ([(x-m1),(y-m2),(z-m3)]*(cov1)*[(x-m1);(y-m2);(z-m3)])./(-2) )./((sqrt(2*pi)).^3.*(sqrt(det(cov))))

It has three matrixs operation. So how should I define?
Thanks for your nice help!!!

"Steven_Lord" <slord@mathworks.com> wrote in message <kpv4sv\$otv\$1@newscl01ah.mathworks.com>...
>
>
> "The first input, fun, is a function handle. fun(x,y,z) must accept a vector
> x and scalars y and z, and return a vector of values of the integrand."
>
> When your function is called with a vector z and scalar x and y, the
> expressions (x-m1) and (y-m2) will be scalars but the expression (z-m3) will
> be a vector. One of the concatenation steps will not work:
>
> [(x-m1),(y-m2),(z-m3)] % Doesn't work if z is a column vector
> [(x-m1);(y-m2);(z-m3)] % Doesn't work if z is a row vector
>
> I believe z will be a row vector, so the latter step will not work.
>
> --
> Steve Lord
> slord@mathworks.com
> To contact Technical Support use the Contact Us link on
> http://www.mathworks.com

Date Subject Author
6/19/13 yolanda
6/19/13 Curious
6/19/13 yolanda
6/19/13 yolanda
6/19/13 yolanda
6/20/13 Steven Lord
6/20/13 yolanda
6/24/13 Steven Lord