# subexpressions (OT: math)

Wildemar Wildenburger wildemar at freakmail.de
Mon Jun 4 01:03:50 CEST 2007

```Stebanoid at gmail.com wrote:
> if you are discordant read more :P :
> sine is a dimensionless value.
> if we expand sine in taylor series sin(x) = x - (x^3)/6 + (x^5)/120
> etc.
> you can see that sin can be dimensionless only if x is dimensionless
> too.
>
> I am a professional physicist and a know about what I talk
>
>
No you don't. I'm a student of physics, and I know better:

First of all, what you have presented here is called the MacLaurin
series. It is however a special case of the Taylor series, so you are
correct. I just thought I'd let you know. (Sorry to sound like a bitch
here, i love smartassing ;))

Let me start by saying that *if* x had a dimension, none of the terms in
your expansion would have the same dimension. A well well-versed
physicist's head should, upon seeing such a thing, explode so as to warn
the other physicists that something is terribly off there. How (ye
gods!) do you add one metre to one square-metre? You don't, that's how!

OK, the *actual* form of the MacLaurin series for some function f(x) is

f(x) = f(0) + x/1! f'(0) + x^2/2! f''(0) + ...

So  in each term of the sum you have a derivative of f, which in the
case of the sine function translates to sine and cosine functions at the
point 0. It's not like you're rid of the function just by doing a
polynomial expansion. The only way to *solve* this is to forbid x from
having a dimension. At least *I* see no other way. Do you?

/W
(Don't take this as a personal attack, please. I'm a good guy, I just
like mathematical nitpicking.)

```