Great info, thanks Roger. If it was up to me, I'd DEFINITELY include that in the Vocabulary, is it even documented anywhere else?
On 23 April 2014 17:33, Roger Hui <rogerhui.can...@gmail.com> wrote: > %. x for a vector x is the same as ($x)$%.,.x, and the key expression is > %.,.x, the "matrix inverse" of a 1-column matrix. b=.y%.x on a tall matrix > x is solving a least-squares problem, the coefficients b that minimizes the > sum of squares of y - x +/ .* b . > > In addition, for a non-zero vector x, (%.x) +/ .* x is 1, a special case of > that (%.x)+/ .* x is an identity matrix, whence one can deduce that for > vector x, %.x is x%+/x^2. > > ] x=: 7 ?.@$ 100 > 94 56 8 6 85 48 66 > %. x > 0.00362137 0.00215741 0.000308202 0.000231152 0.00327465 0.00184921 > 0.00254267 > (%.x) +/ .* x > 1 > x % +/x^2 > 0.00362137 0.00215741 0.000308202 0.000231152 0.00327465 0.00184921 > 0.00254267 > > M=: 7 3 ?.@$ 100 > (%.M) +/ .* M > 1 5.55112e_17 _2.77556e_17 > _1.21431e_16 1 1.11022e_16 > _4.85723e_17 1.94289e_16 1 > > > > On Wed, Apr 23, 2014 at 9:13 AM, alexgian <alexg...@blueyonder.co.uk> > wrote: > > > Just wondering: > > %. 2 3 4 > > 0.0689655 0.103448 0.137931 > > > > Which is fair enough enough at one level, I suppose, since the dot > product > > of the two arrays IS 1, but what system/equation is being solved here? > > Obviously, there are infinite solutions. Why that one? > > IOW, which "matrix" is being inverted here? > > > > Thanks > > ---------------------------------------------------------------------- > > For information about J forums see http://www.jsoftware.com/forums.htm > > > ---------------------------------------------------------------------- > For information about J forums see http://www.jsoftware.com/forums.htm > ---------------------------------------------------------------------- For information about J forums see http://www.jsoftware.com/forums.htm
