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view scripts/special-matrix/hilb.m @ 29358:0a5b15007766 stable
update Octave Project Developers copyright for the new year
In files that have the "Octave Project Developers" copyright notice,
update for 2021.
author | John W. Eaton <jwe@octave.org> |
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date | Wed, 10 Feb 2021 09:52:15 -0500 |
parents | 627da618dcc4 |
children | 7854d5752dd2 |
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######################################################################## ## ## Copyright (C) 1993-2021 The Octave Project Developers ## ## See the file COPYRIGHT.md in the top-level directory of this ## distribution or <https://octave.org/copyright/>. ## ## This file is part of Octave. ## ## Octave is free software: you can redistribute it and/or modify it ## under the terms of the GNU General Public License as published by ## the Free Software Foundation, either version 3 of the License, or ## (at your option) any later version. ## ## Octave is distributed in the hope that it will be useful, but ## WITHOUT ANY WARRANTY; without even the implied warranty of ## MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the ## GNU General Public License for more details. ## ## You should have received a copy of the GNU General Public License ## along with Octave; see the file COPYING. If not, see ## <https://www.gnu.org/licenses/>. ## ######################################################################## ## -*- texinfo -*- ## @deftypefn {} {} hilb (@var{n}) ## Return the Hilbert matrix of order @var{n}. ## ## The @math{i,j} element of a Hilbert matrix is defined as ## @tex ## $$ ## H(i, j) = {1 \over (i + j - 1)} ## $$ ## @end tex ## @ifnottex ## ## @example ## H(i, j) = 1 / (i + j - 1) ## @end example ## ## @end ifnottex ## ## Hilbert matrices are close to being singular which make them difficult to ## invert with numerical routines. Comparing the condition number of a random ## matrix 5x5 matrix with that of a Hilbert matrix of order 5 reveals just how ## difficult the problem is. ## ## @example ## @group ## cond (rand (5)) ## @result{} 14.392 ## cond (hilb (5)) ## @result{} 4.7661e+05 ## @end group ## @end example ## ## @seealso{invhilb} ## @end deftypefn function retval = hilb (n) if (nargin != 1) print_usage (); elseif (! isscalar (n)) error ("hilb: N must be a scalar integer"); endif ## Very elegant solution by N. Higham ## https://nhigham.com/2020/06/30/what-is-the-hilbert-matrix/ j = 1:n; retval = 1 ./ (j' + j - 1); endfunction %!assert (hilb (2), [1, 1/2; 1/2, 1/3]) %!assert (hilb (3), [1, 1/2, 1/3; 1/2, 1/3, 1/4; 1/3, 1/4, 1/5]) %!error hilb () %!error hilb (1, 2) %!error <N must be a scalar integer> hilb (ones (2))