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The following equation should be true: $O(n\log(n))=O(n^{1.1} )$

Based on the following article, one should see the equation as the left part being an element of the right part. What are the rules for equals signs with big-O and little-o?

Given that the equation is true, $n\log(n)$ should grow at most as fast as $n^{1.1}$. However, $\lim_{n \to \infty} \frac{n\log(n)}{n^{1.1}}=\lim_{n \to \infty} \frac{\log(n)}{n^{0.1}} \rightarrow \infty$. This does not correspond with the statement.

What mistake am I making?

Bernard
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Stef
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  • It goes to $0$ instead of to $\infty$. We have that $n^\epsilon$ grows much faster then $\log(n) $ for $\epsilon>0$. – kingW3 Sep 06 '19 at 14:50
  • Be careful with this kind of notation. In fact the point of the article is that you can use a consistent convention with big-O and little-o notation in which, if there's a function on the left and a growth class on the right, then = is interpreted as $\in$. What you've written does not adhere to that convention, and defies the expectation that any expression $A=B$ should be equivalent to $B=A$ when both expressions make sense. – mjqxxxx Sep 06 '19 at 15:07
  • Thanks @mjqxxxx! The expression is coming from an academic resource. I suppose the interpretation is the same as having a function on the left and a growth class on the right. – Stef Sep 08 '19 at 13:40
  • @kingW3 That's a good point, thank you! – Stef Sep 08 '19 at 13:41

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What is true, really, is this: $$O(n\log n)=o\bigl(n^{1.1}\bigr).$$

Bernard
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