How to Move the Big O Notation from the Denominator to the Numerator
Theorem
For $a \ne 0$, $p>0$, and $n \in \mathbb{N}$, the following holds. $$ {{1} \over { \sqrt[p]{a + O ( h^n ) } }} = {{1} \over { \sqrt[p]{a } }}+ O(h^n) $$
Explanation
This can come in handy as a lemma that turns a messy-looking denominator into a clean form.
If there is no constant term $a$, it rises cleanly as $\displaystyle {{1} \over { \sqrt[p]{ O ( h^n ) } }} = O \left( h^{ - {{n} \over {p}} } \right) $ even without the lemma, but this is usually of no use.
Proof
For $$ {{1} \over { \sqrt[p]{a + O ( h^n ) } }} = {{1} \over { \sqrt[p]{a + b h^n } }} = (a + b h^n )^{-{{1} \over {p}}} $$ let $\displaystyle q := {{1} \over {p}}$ and define the function $\displaystyle f(h) := (a + b h^n )^{-q}$.
Applying the Maclaurin expansion to $f$ gives $$ f(h) = f(0) + h f '(0) + {{h^2} \over {2}} f ''(0) + {{h^3} \over {6}} f^{(3)}(0) + \cdots $$ Since $\displaystyle f(h) = (a + b h^n )^{-q}$, $$ f(0) = (a + b \cdot 0^n )^{-q} = {{1} \over { \sqrt[p]{a } }} $$ Meanwhile, since $$ f ' (h) = -qbnh^{n-1} (a+ bh^{n})^{-q-1} $$ we have $f ' (0) = 0$, and since $$ f '' (h) = -qbn(n-1)h^{n-2} (a+ bh^{n})^{-q-1} + bnh^{n-1} (-q-1) bnh^{n-1} (a+ bh^{n})^{-q-2} $$ we have $f '' (0) = 0$. Computing the $k$th derivatives $f^{(k)}$ in this manner, we find that $f^{(k)}(0) = 0$ for $k=2,3, \cdots, (n-1)$. For some $g(h)$, $$ f^{(n)}(h) = -qbn!h^{n-n} (a+ bh^{n})^{-q-1} + h g(h) $$ so $$ f^{(n)} (h) = -qbn! {{1} \over { \sqrt[q+1]{a } }} $$ Therefore, $$ f(h) ={{1} \over { \sqrt[p]{a } }} + 0 + \cdots + 0 - {{h^n} \over {n!}} qbn! {{1} \over { \sqrt[q+1]{a } }} + {{h^{ (n+1) } } \over {(n+1)! }} f^{(n+1)}(0) + \cdots $$ Factoring $h^n$ out of all the terms appearing after the $0$s, we obtain the following. $$ {{1} \over { \sqrt[p]{a + O ( h^n ) } }} = {{1} \over { \sqrt[p]{a } }}+ O(h^n) $$
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