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From: Deane Yang (deane_yang_at_[hidden])
Date: 2006-06-10 23:32:07

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

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