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Diffstat (limited to 'doc/LeastSquares.dox')
| -rw-r--r-- | doc/LeastSquares.dox | 17 |
1 files changed, 11 insertions, 6 deletions
diff --git a/doc/LeastSquares.dox b/doc/LeastSquares.dox index e2191a22f..ddbf38dec 100644 --- a/doc/LeastSquares.dox +++ b/doc/LeastSquares.dox @@ -16,7 +16,7 @@ equations is the fastest but least accurate, and the QR decomposition is in betw \section LeastSquaresSVD Using the SVD decomposition -The \link JacobiSVD::solve() solve() \endlink method in the JacobiSVD class can be directly used to +The \link BDCSVD::solve() solve() \endlink method in the BDCSVD class can be directly used to solve linear squares systems. It is not enough to compute only the singular values (the default for this class); you also need the singular vectors but the thin SVD decomposition suffices for computing least squares solutions: @@ -30,14 +30,17 @@ computing least squares solutions: </table> This is example from the page \link TutorialLinearAlgebra Linear algebra and decompositions \endlink. +If you just need to solve the least squares problem, but are not interested in the SVD per se, a +faster alternative method is CompleteOrthogonalDecomposition. \section LeastSquaresQR Using the QR decomposition The solve() method in QR decomposition classes also computes the least squares solution. There are -three QR decomposition classes: HouseholderQR (no pivoting, so fast but unstable), -ColPivHouseholderQR (column pivoting, thus a bit slower but more accurate) and FullPivHouseholderQR -(full pivoting, so slowest and most stable). Here is an example with column pivoting: +three QR decomposition classes: HouseholderQR (no pivoting, fast but unstable if your matrix is +not rull rank), ColPivHouseholderQR (column pivoting, thus a bit slower but more stable) and +FullPivHouseholderQR (full pivoting, so slowest and slightly more stable than ColPivHouseholderQR). +Here is an example with column pivoting: <table class="example"> <tr><th>Example:</th><th>Output:</th></tr> @@ -61,9 +64,11 @@ Finding the least squares solution of \a Ax = \a b is equivalent to solving the </tr> </table> -If the matrix \a A is ill-conditioned, then this is not a good method, because the condition number +This method is usually the fastest, especially when \a A is "tall and skinny". However, if the +matrix \a A is even mildly ill-conditioned, this is not a good method, because the condition number of <i>A</i><sup>T</sup><i>A</i> is the square of the condition number of \a A. This means that you -lose twice as many digits using normal equation than if you use the other methods. +lose roughly twice as many digits of accuracy using the normal equation, compared to the more stable +methods mentioned above. */ |