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## -*- texinfo -*-
## @deftypefn {} {} invhilb (@var{n})
## Return the inverse of the 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
##
## @noindent
## 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 @nospell{J. W. Demmel},
## @cite{Applied Numerical Linear Algebra}, p.@: 92.
##
## Compare this with the numerical calculation of @code{inv (hilb (n))},
## which suffers from the ill-conditioning of the Hilbert matrix, and the
## finite precision of your computer's floating point arithmetic.
## @seealso{hilb}
## @end deftypefn

function retval = invhilb (n)

  if (nargin < 1)
    print_usage ();
  elseif (! isscalar (n))
    error ("invhilb: N must be a scalar integer");
  endif

  ## 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

endfunction


%!assert (invhilb (1), 1)
%!assert (invhilb (2), [4, -6; -6, 12])
%!test
%! result4 = [16  , -120 , 240  , -140;
%!            -120, 1200 , -2700, 1680;
%!            240 , -2700, 6480 , -4200;
%!            -140, 1680 , -4200, 2800];
%! assert (invhilb (4), result4);
%!assert (invhilb (7) * hilb (7), eye (7), sqrt (eps))

%!error <Invalid call> invhilb ()
%!error <N must be a scalar integer> invhilb ([1, 2])
