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Some exercises for myself,

(1) $ln((1+\frac{1}{n})^n) \sim 1$ so $ln((1+1/n)^n)=1+o(1)$ as $n \rightarrow \infty$?

(2) $(1+o(1)) ln(n+1) + o(n^2)=1+ln(n+1)+o(n^2)$

(3) $(f(n)+o(n^2))^k = \sum_{i=0}^{k} \binom{k}{i} f^{k-i}(n)o(n^{2i}) $ Can it be further simplified?

LLT
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1 Answers1

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(1) is correct, although $1+O(1) = O(1)$. For (2) note that $n+1$ and thus certainly $\ln(n+1)$ are $O(n^2)$. So this would simply be $O(n^2)$. (3) depends a bit on $f$, so it is a bit hard to say. For example $f(n)+O(n^2)$ could be $0$. But you can at least say that this is either $O(n^{2k})$ if $f(n)=O(n^2)$ or $O(f(n)^k)$ if $n^2 = O(f(n))$.

Lazy
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  • Thank you very much. For (3), if $f(n)=1+ln(n+1)$ (so it is $o(n^2)$, then is this correct, $(1+ln(n+1)+o(n^2))^k = \sum_{i=0}^{k} \binom{k}{i} (1+ln(n+1))^{k-i}(n)+o(n^{2k}) $ ? – LLT Nov 07 '21 at 21:45
  • If $f(n) = O(n^2)$ then $(f(n)+O(n^2))^k = O(n^{2k})$. – Lazy Nov 07 '21 at 22:33