Consider the following functions:

$$f(n)=2^{\log^*n} \text{ and } g(n)=\sqrt{2}^{\log{n}}$$

Using $\log{}$ properties I think that $g(n) < f(n)$, since:

  1. $f(n)\sim n$,
  2. $g(n)\sim n^{\frac{1}{2}}$, and
  3. $n^{\frac{1}{2}}<n$.

However the book that I'm reading says otherwise.

What have I gotten wrong?


1 Answer 1


You have gotten wrong the 1st property, that $2^{\log^*(n)} \sim ~n$. This is not the case.

The notation $\log^*(n)$ is the iterated logarithm, and the iterated logarithm grows much much slower than the logarithm, so $f$ grows much slower than linearly.

  • $\begingroup$ I'm aware of the definition of log star, so I guess the question is don't log* and log have the same properties? because if they do then I can do this: $2^{log^*n}=n^{log^*2}=n^1=n$. $\endgroup$
    – Titan
    Jan 25, 2021 at 10:54
  • 1
    $\begingroup$ No, $2^{\log^* 100} \sim 8$, whereas $100^{\log^* 2} \sim 100$. $\endgroup$
    – Pål GD
    Jan 25, 2021 at 11:53

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