Taylor expansion of the determinant of a Riemannian metric Let $(M,g)$ be a compact Riemannian manifold without boundary. Fix a point $x\in M$ and $N\ge 2$ large. Then there exists a metric $\tilde g$, conformal to $g$ such that $$ \det \tilde g=1+O(r^N)$$ for $r\le inj(x)$ the injectivity radius at $x$, where $r=|x|$ in the $\tilde g$-normal coordinates. References: Lee, Parker - the Yamabe Problem, THM 5.1 (existence of normal conformal coordinates)
Question: Does (and how) this imply
$$ \partial_r^{(i)}\log(\sqrt{\det \tilde g})=O(r^{N-i}).$$
It seems to me that this could come from the Taylor expansion of the determinant of the metric but i cannot find a proper proof. Gursky and Malchiodi in this paper http://arxiv.org/pdf/1401.3216.pdf (Lemma 2.6) use the notation $$ \det \tilde g=1+O^{(3)}(r^N)$$ and they argue $$ \partial_r \log(\sqrt{\det \tilde g})=O^{''}(r^{N-1})$$
but it is not clear to me if i can push it to higher order derivatives and yet why they use that notation since in Lee and Parker nothing like this seems to be proved.
 A: The answer to your question is yes. To see why, write $\det \tilde g = 1 + f$, where $f$ is a smooth function satisfying $f=O(r^N)$. Let $(u^i)$ be a smooth coordinate chart centered at $x$. Because $f$ is smooth, one version of Taylor's theorem says that near $x$, we can write
$$
f(u) = \sum_{i_1,\dots,i_N} h_{i_1,\dots,i_N}(u) u^{i_1}\dots u^{i_N},
$$
where the functions $h_{i_1,\dots,i_N}(u)$ are all smooth. It follows that every $j$th partial derivative of $f$ is $O(r^{N-j})$ for $0\le j\le N$.  
The problem with applying this directly to the question at hand is that $\partial_r$ is not a smooth vector field on a neighborhood of $x$.  However, $r\partial_r = \sum_j x^j \partial_{x^j}$ is smooth, and a simple inductive argument shows that $(r\partial_r)^k f=O(r^N)$ for all $k$.  Using the fact that 
$$(r\partial_r)^k = r^k(\partial_r)^k + a_{k-1} r^{k-1}(\partial_r)^{k-1} + \dots + a_1 r \partial_r$$ 
for some constants $a_1,\dots,a_{k-1}$,
another simple induction implies that $r^k(\partial_r)^k f = O(r^N)$, which is equivalent to the result you want.
