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I'm learning a math post How limit $\lim_{n \to \infty} \frac{ \log^bn }{n^a} = 0$ and being stuck at $\log^bn$.

I've skimmed the wikipedia page but didn't find a clue.

Could some help me on this?

JJJohn
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2 Answers2

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It's $(\log n)^b$, which we don't write as $\log n^b$ as that's $b\log n$.

J.G.
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In general, $f^n(x)$ means $\left( f(x) \right)^n.$

So, $\sin^n(x)$ means $\left( \sin(x) \right)^n,\ $ which can also be written as $\left( \sin x \right)^n,\quad $ $\log^n(x)$ means $\left( \log(x) \right)^n\ $ which can also be written as $\left( \log x \right)^n\ $ etc etc

Adam Rubinson
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  • Although $f^n$ may also denote function concatenation, with $f^{-1}$ an inverse. Hopefully, however, concatenation uses the notation $f^{\circ n}$. – J.G. Feb 21 '22 at 14:23