From An Introduction to the Theory of Numbers, Hardy, 6th ed, on Page 9, it says,

We may also observe that $f$ ~ $\phi$ is equivalent to $f = \phi + o(\phi)$ or to

$f= \phi{\{1 + o(1)\}}$

In these circumstances we say that $f$ and $\phi$ are asymptotically equivalent...

So what does $f= \phi{\{1 + o(1)\}}$ mean? I am especially confused about the brackets.


  • 1
    $\begingroup$ $f=\phi{\{1 + o(1)\}}$ means $f=\phi + o(\phi)$ as a form of factorisation. It says $f$ is $\phi$ times something which is asymptotically equivalent to $1$, i.e. $\frac f \phi \sim 1$ $\endgroup$
    – Henry
    Jan 15, 2021 at 10:45

1 Answer 1


It is a weird and non-standard choice of brackets, but clearly means nothing more than $φ(1+o(1)) = φ+φ·o(1) = φ+o(φ)$.


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