Gerhard Wesp wrote:
On Fri, Jun 09, 2006 at 06:19:58PM -0700, Geoffrey Irving wrote:
polynomial regression). All the nice Taylor series example seem unitless.
They have to be, don't they? Because you add up different powers of the argument. I took this once as a "heuristic" explanation to myself why the transcendental functions only work for dimensionless arguments.
Yes, that's right.
On the other hand, the square root can be approximated by a series as well, and this function does make sense with dimensional arguments.
sqrt(x) has no Taylor series expansion at x = 0, but sqrt(1+x) does. You'll find that if you use the latter to compute the square root of a dimensioned quantity, the dimension can factored out of the series, so that the series itself is completely dimensionless.
Regards -Gerhard