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author | Rik <rdrider0-list@yahoo.com> |
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date | Fri, 27 Mar 2009 22:31:03 -0700 |
parents | eb63fbe60fab |
children | 3140cb7a05a1 |
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## Copyright (C) 1993, 1994, 1995, 1996, 1997, 1998, 1999, 2000, 2002, ## 2004, 2005, 2006, 2007, 2008, 2009 Dirk Laurie ## ## 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 ## <http://www.gnu.org/licenses/>. ## -*- texinfo -*- ## @deftypefn {Function File} {} invhilb (@var{n}) ## Return the inverse of a Hilbert matrix of order @var{n}. This can be ## computed exactly using ## @tex ## $$\eqalign{ ## A_{ij} &= -1^{i+j} (i+j-1) ## \left( \matrix{n+i-1 \cr n-j } \right) ## \left( \matrix{n+j-1 \cr n-i } \right) ## \left( \matrix{i+j-2 \cr i-2 } \right)^2 \cr ## &= { p(i)p(j) \over (i+j-1) } ## }$$ ## where ## $$ ## p(k) = -1^k \left( \matrix{ k+n-1 \cr k-1 } \right) ## \left( \matrix{ n \cr k } \right) ##$$ ## @end tex ## @ifnottex ## @example ## @group ## ## (i+j) /n+i-1\ /n+j-1\ /i+j-2\ 2 ## A(i,j) = -1 (i+j-1)( )( ) ( ) ## \ n-j / \ n-i / \ i-2 / ## ## = p(i) p(j) / (i+j-1) ## ## @end group ## @end example ## where ## @example ## @group ## k /k+n-1\ /n\ ## p(k) = -1 ( ) ( ) ## \ k-1 / \k/ ## @end group ## @end example ## @end ifnottex ## ## The validity of this formula can easily be checked by expanding ## the binomial coefficients in both formulas as factorials. It can ## be derived more directly via the theory of Cauchy matrices: ## see J. W. Demmel, Applied Numerical Linear Algebra, page 92. ## ## Compare this with the numerical calculation of @code{inverse (hilb (n))}, ## which suffers from the ill-conditioning of the Hilbert matrix, and the ## finite precision of your computer's floating point arithmetic. ## @seealso{hankel, vander, sylvester_matrix, hilb, toeplitz} ## @end deftypefn ## Author: Dirk Laurie <dlaurie@na-net.ornl.gov> function retval = invhilb (n) if (nargin != 1) print_usage (); endif nmax = length (n); if (nmax == 1) ## The point about the second formula above is that when vectorized, ## p(k) is evaluated for k=1:n which involves O(n) calls to bincoeff ## instead of O(n^2). ## ## We evaluate the expression as (-1)^(i+j)*(p(i)*p(j))/(i+j-1) except ## when p(i)*p(j) would overflow. In cases where p(i)*p(j) is an exact ## machine number, the result is also exact. Otherwise we calculate ## (-1)^(i+j)*p(i)*(p(j)/(i+j-1)). ## ## The Octave bincoeff routine uses transcendental functions (gammaln ## and exp) rather than multiplications, for the sake of speed. ## However, it rounds the answer to the nearest integer, which ## justifies the claim about exactness made above. retval = zeros (n); k = [1:n]; p = k .* bincoeff (k+n-1, k-1) .* bincoeff (n, k); p(2:2:n) = -p(2:2:n); if (n < 203) for l = 1:n retval(l,:) = (p(l) * p) ./ [l:l+n-1]; endfor else for l = 1:n retval(l,:) = p(l) * (p ./ [l:l+n-1]); endfor endif else error ("invhilb: expecting scalar argument, found something else"); endif endfunction %!test %! result4 = [16, -120, 240, -140; %! -120, 1200, -2700, 1680; %! 240, -2700, 6480, -4200; %! -140, 1680, -4200, 2800]; %! %! assert((invhilb (1) == 1 && invhilb (2) == [4, -6; -6, 12] %! && invhilb (4) == result4 %! && abs (invhilb (7) * hilb (7) - eye (7)) < sqrt (eps))); %!error invhilb ([1, 2]); %!error invhilb (); %!error invhilb (1, 2);