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.



Taylor expansion of the determinant of a Riemannian metric
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.


