On Tue, 2008-08-19 at 09:49 +0200, Grégory Lielens wrote: > Using __call__ as matmul: > b = a.I - ( (a.I)(u) / (c.I + (v/a)(u)) )(v) / a
oups, of course you do not have right-divide in this case, it would thus read b = a.I - (a.I) (u) ( ( c.I + (v)(a.I)(u) ).I ) (v) (a.I) hum, given the amount of re-read I had to do, even with parenthesis-highlighting, I think that visual parsing for complex formula was the biggest drawback ;-) _______________________________________________ Numpy-discussion mailing list [email protected] http://projects.scipy.org/mailman/listinfo/numpy-discussion
