subexpressions (OT: math)

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

Stebanoid at 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?

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

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