TAOCP 1.2.11.2 Exercise 10
Let $u(z)=O(z^m)$ as $z\to 0$, with $m>0$.
Section 1.2.11.2: Euler's Summation Formula
Exercise 10. [HM22] Make a statement similar to that in exercise 9 about $\ln(1+O(z^m))$.
Verified: yes
Solve time: 45s
Solution
Let $u(z)=O(z^m)$ as $z\to 0$, with $m>0$. Then there exist positive constants $L$ and $z_0$ such that $|u(z)| \le L|z|^m$ for all $|z|\le z_0$, and hence $u(z)\to 0$ as $z\to 0$.
For $|u|<1$, the expansion
$$ \ln(1+u)=u+O(u^2) $$
holds, since $\ln(1+u)$ has a convergent Taylor expansion at $u=0$ with first neglected term of order $u^2$.
Substituting $u=u(z)$ gives
$$ \ln(1+u(z)) = u(z) + O(u(z)^2). $$
From $u(z)=O(z^m)$ it follows that $u(z)^2=O(z^{2m})$, since
$$ |u(z)^2| \le L^2 |z|^{2m} $$
for all $|z|\le z_0$. Hence $O(z^{2m}) \subseteq O(z^m)$ as $z\to 0$ because $|z|^{2m} \le |z|^m$ for sufficiently small $z$.
Therefore
$$ \ln(1+u(z)) = O(z^m) + O(z^m), $$
which yields
$$ \ln(1+u(z)) = O(z^m). $$
This completes the proof. ∎