subexpressions (OT: math)
wildemar at freakmail.de
Mon Jun 4 11:15:26 CEST 2007
Erik Max Francis wrote:
> Wildemar Wildenburger wrote:
>> 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?
> That was precisely his point. The Maclaurin series (not MacLaurin) only
> makes any sense if the independent variable is dimensionless. And thus,
> by implication, so it is also the case for the original function.
No that was not his point. Maybe he meant it, but he said something
profoundly different. To quote:
Stebanoid at gmail.com wrote:
> if we expand sine in taylor series sin(x) = x - (x3)/6 + (x5)/120 etc.
> you can see that sin can be dimensionless only if x is dimensionless too.
This argumentation gives x (and sin(x)) the option of carrying a unit.
That however is *not* the case. This option does not exist. Until
someone proves the opposite, of course.
Geez, I love stuff like that. Way better than doing actual work. :D
(PS: THX for the MacLaurin<>Maclaurin note. I can't help that
appearantly; I also write 'LaGrange' all the time.)
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