Is to 2nd order always related to Taylor series?

[Rasalhague]

What does "to 2nd order" mean?

In the limit as the number n of sides of the polygon increases and the angle a approaches zero, the value of \cos \alpha approaches 1 (to the second order), and the value of \sin \alpha approaches \alpha.

http://www.mathpages.com/rr/s2-11/2-11.htm

I thought "to second order" meant an approximation consisting of the initial terms of the Taylor series up to and including the second derivative term. For example, it seems to be used that way in the Wikipedia article Taylor series, in the section "Taylor series in several variables" [ http://en.wikipedia.org/wiki/Taylor_series ]. Is the quote above using an unusual definition of second order, or have I misunderstood something here?

 

[yossell]

edit: ok, sorry, bad suggestion

 

[Mute]

I would consider it a typo. Second order includes the second order term whenever I have heard it used.

 

[Rasalhague]

Thanks.

 

[paulfr]

In other contexts, 2nd order means "it can oscillate and therefore go unstable".

You need 2 storage elements to cause an oscillation and the differential equation that describes it
reduces to a 2nd order algebraic equation [quadratic] via Laplace Transformation.

JFYI