[EMAIL PROTECTED] 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.) -- http://mail.python.org/mailman/listinfo/python-list
