If $(K,\leq)$ is a totally ordered field with $P\!=\!\{\alpha\!\in\!K;\, 0\!\leq\!\alpha\}$, how is the valuation associated to $P$ defined?
I was searching through Prestel & Delzell's Positive Polynomials and Engler & Prestel's Valued Fields, but didn't find anything. Perhaps I didn't search thoroughly enough. Google also didn't provide much. Any reference is welcome.
I must calculate the valuation $v$ on $\mathbb{R}(x)$ associated to $P\!=\!\{x^kf(x);\, k\!\in\!\mathbb{Z}, f\!\in\!\mathbb{R}(x), 0\!<\!f(0)\!<\!\infty\}$.
Furthermore, given an ordering $\leq$ and valuation $v$ on $K$, when is $v$ compatible with $\leq$ (definition)?
This doesn't really deserve to be an answer but I don't have enough points to make a comment, so here it is:
As navigetor23 says, a valuation is order compatible if $0<a\leq b$ implies $v(a)\geq v(b)$.
But, for me at least, a nicer equivalent formulation is that the valuation ring $\mathcal{O}_v$ corresponding to $v$ is convex.
There are three more equivalent conditions on page 3 of the following link (along with other useful results):
http://math.usask.ca/~fvk/bookch10.pdf