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<!-- Additional documentation for the optim package for Octave.

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<a name="wsolve"></a>
<div class="header">
<p>
Previous: <a href="LinearRegression.html#LinearRegression" accesskey="p" rel="prev">LinearRegression</a>, Up: <a href="Residual-optimization.html#Residual-optimization" accesskey="u" rel="up">Residual optimization</a> &nbsp; [<a href="Function-index.html#Function-index" title="Index" rel="index">Index</a>]</p>
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<hr>
<a name="Function-wsolve_002c-another-linear-solver"></a>
<h3 class="section">2.14 Function wsolve, another linear solver</h3>
<a name="index-wsolve"></a>

<a name="Helptext_003a-10"></a>
<h4 class="subheading">Helptext:</h4>

<a name="XREFwsolve"></a><pre class="verbatim">[x,s] = wsolve(A,y,dy)

Solve a potentially over-determined system with uncertainty in
the values. 

    A x = y +/- dy

Use QR decomposition for increased accuracy.  Estimate the 
uncertainty for the solution from the scatter in the data.

The returned structure s contains

   normr = sqrt( A x - y ), weighted by dy
   R such that R'R = A'A
   df = n-p, n = rows of A, p = columns of A

See polyconf for details on how to use s to compute dy.
The covariance matrix is inv(R'*R).  If you know that the
parameters are independent, then uncertainty is given by
the diagonal of the covariance matrix, or 

   dx = sqrt(N*sumsq(inv(s.R'))')

where N = normr^2/df, or N = 1 if df = 0.

Example 1: weighted system

   A=[1,2,3;2,1,3;1,1,1]; xin=[1;2;3]; 
   dy=[0.2;0.01;0.1]; y=A*xin+randn(size(dy)).*dy;
   [x,s] = wsolve(A,y,dy);
   dx = sqrt(sumsq(inv(s.R'))');
   res = [xin, x, dx]

Example 2: weighted overdetermined system  y = x1 + 2*x2 + 3*x3 + e

   A = fullfact([3,3,3]); xin=[1;2;3];
   y = A*xin; dy = rand(size(y))/50; y+=dy.*randn(size(y));
   [x,s] = wsolve(A,y,dy);
   dx = s.normr*sqrt(sumsq(inv(s.R'))'/s.df);
   res = [xin, x, dx]

Note there is a counter-intuitive result that scaling the
uncertainty in the data does not affect the uncertainty in
the fit.  Indeed, if you perform a monte carlo simulation
with x,y datasets selected from a normal distribution centered
on y with width 10*dy instead of dy you will see that the
variance in the parameters indeed increases by a factor of 100.
However, if the error bars really do increase by a factor of 10
you should expect a corresponding increase in the scatter of 
the data, which will increase the variance computed by the fit.

</pre>




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