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Re: leasqr problem

 From: Jose Rodriguez Subject: Re: leasqr problem Date: Mon, 26 Jan 2009 16:14:14 +0000 User-agent: Thunderbird 2.0.0.19 (X11/20081209)

```On Sat, Jan 24, 2009 at 1:12 PM, Jose Rodriguez
```
```
I'm trying to solve a problem with leasqr, where the observed
value, instead of being a set of points y=f(x), is the result of a
definite integral. I found that I can use leasqr if I adjust the
dimensions of the function to fit and its expected values to the
dimension of the x range:

x=(0:0.1:pi);

P=@(x)  1/2 * ( 3*cos(x).^2-1 );

F=inline( " ones ( size(x,1), 1 ) * trapz( x, P(x).*
exp(-a*P(x)).*sin(x) ) ", "x", "a" );

y=0.5 * ones ( size(x,1), 1 );
pin=1;

[f,p,kvg,iter,corp,covp,covr,stdresid,Z,r2]=leasqr(x, y, pin, F);

If I run this I obtain the value 'a' for which the integral:

trapz( x, P(x) .* exp(-a*P(x)) .* sin(x) )

is 0.5.

My problem is that I also want to include a normalisation
condition, like:

1 = trapz( x, exp(-a*P(x)).*sin(x) )

Could somebody give me some pointers as to how to proceed? Or am I
completely out of track?

```
```

```
```Hi, I think you're a little muddled about how you're approaching the
problem here.
```
```
Most definitely so :)

```
```You are trying to solve the overdetermined system

F(a) = 0.5
G(a) = 1

where F(a) and G(a) are your two integrals, you only have one
observation point, the vector [0.5;1]. So the best you can do is try
to find the a s.t. the norm of [F(a); G(a)] - [0.5; 1] is minimal.

In this case it's not so hard to just search.
```
```
See below.

Jaroslav Hajek wrote:

```
```I'm not sure I understand what you're trying to do.
The condition 1 == trapz( x, exp(-a*P(x)).*sin(x) ) is generally
enough to determine a,
because you have a single equation with a single variable.
So is the other equation,
trapz( x, P(x) .* exp(-a*P(x)) .* sin(x) ) == 0.5, but generally, you
can't satisfy a system of two equations
simultaneously if you only have one variable. Such a system may be
solved in a least-squares sense, but
is that what you need?
```
```
```
My original explanation wasn't very clear. I actually have two variables, only that I called them a=[a(1) a(2)] to comply with what leasqr expects; i.e. a function f(x,p) where p can be written like above to fit multiple variables. The equations look better like this:
```
P(x)=some function of x;

known_value=trapz(x, a*exp(b*P(x)).*P(x).*sin(x));

1=trapz(x, a*exp(b*P(x)).*sin(x));

Being 'a' and 'b' what I want to find.

```
It's true that for this example I could do with some plotting, but this is the simplest situation I can have. The general case involves n Pn(x) that go in the exponential, each of them with its coefficient to fit and its 'known_value' arising from the integral, plus the normalisation condition. The number of variables to fit always equals the number of equations I have.
```
```
Lest squares is what other people have used to solve this very same problem, but I don't know how to put it in Octave.
```

Regards

Jose

```