📂Lemmas

How to Move the Big O Notation from Denominator to Numerator

Theorem

The following holds for $a \ne 0$, $p>0$, and $n \in \mathbb{N}$. $${{1} \over { \sqrt[p]{a + O ( h^n ) } }} = {{1} \over { \sqrt[p]{a } }}+ O(h^n)$$

Explanation

It serves as a handy lemma for converting complex denominators into cleaner forms.

If there were no constant term $a$, it could neatly rise to $\displaystyle {{1} \over { \sqrt[p]{ O ( h^n ) } }} = O \left( h^{ - {{n} \over {p}} } \right)$ even without the lemma, but that is usually not very useful.

Proof

Let’s set $\displaystyle q := {{1} \over {p}}$ for $${{1} \over { \sqrt[p]{a + O ( h^n ) } }} = {{1} \over { \sqrt[p]{a + b h^n } }} = (a + b h^n )^{-{{1} \over {p}}}$$ and define the function $\displaystyle f(h) := (a + b h^n )^{-q}$.

When Maclaurin expanding for $f$, $$f(h) = f(0) + h f '(0) + {{h^2} \over {2}} f ''(0) + {{h^3} \over {6}} f^{(3)}(0) + \cdots$$ Since $\displaystyle f(h) = (a + b h^n )^{-q}$, $$f(0) = (a + b \cdot 0^n )^{-q} = {{1} \over { \sqrt[p]{a } }}$$ Meanwhile, $$f ' (h) = -qbnh^{n-1} (a+ bh^{n})^{-q-1}$$ Therefore, it’s $f ' (0) = 0$, and $$f '' (h) = -qbn(n-1)h^{n-2} (a+ bh^{n})^{-q-1} + bnh^{n-1} (-q-1) bnh^{n-1} (a+ bh^{n})^{-q-2}$$ Hence, it’s $f '' (0) = 0$. Proceeding this way, when calculating the $k$th derivative $f^{(k)}$, we know that for $k=2,3, \cdots, (n-1)$ it’s $f^{(k)}(0) = 0$. For some $g(h)$, $$f^{(n)}(h) = -qbn!h^{n-n} (a+ bh^{n})^{-q-1} + h g(h)$$ Therefore, $$f^{(n)} (h) = -qbn! {{1} \over { \sqrt[q+1]{a } }}$$ Thus, $$f(h) ={{1} \over { \sqrt[p]{a } }} + 0 + \cdots + 0 - {{h^n} \over {n!}} qbn! {{1} \over { \sqrt[q+1]{a } }} + {{h^{ (n+1) } } \over {(n+1)! }} f^{(n+1)}(0) + \cdots$$ After $0$, grouping all terms that appear later under $h^n$, we obtain the following. $${{1} \over { \sqrt[p]{a + O ( h^n ) } }} = {{1} \over { \sqrt[p]{a } }}+ O(h^n)$$

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