Find a function which grows slower (but not by a polynomial factor slower) than $n^{\log_27}$

Question

So if I divide $f(n) = n^{\log_27}$ by $log_27$, it should be growing slower than $f(n)$ (and still not by a polynomial factor).

$g(n) = f(n)/\log_27$

Is my assumption true? If not what should I do instead?

Answer 1

If the function $f(n)=n^{\log_2 7}$ grows faster than $n^2$ but slower than $x^3$. Let $g(n)=f(n)~p(n)$ it will grow slower than $f(n)$ for non-polynomial functions(factors) such as $p(n)=\frac{1}{1+n^k}. k>0.$ Let $k$ be non-integral: $k=0.1, 0.2, 0.3, 0.5, 5/7, 7/5.$

Answer 2

1. It depends on what is meant by "grows more slowly".

2. Since $n^{-k}\ln n\to 0$ as $n\to \infty$ for any fixed $k>0,$ you can let $g(n)=f(n)/\ln n$ for $n>1.$ Then $g(n)\to \infty$ and $g(n)/f(n)\to 0,$ but $[g(n)/f(n)]\cdot n^k\to \infty$ for any constant $k>0.$

3. Instead, for non-negative integer $m$ and for $10^m\le n<10^{m+1}$ let $g(n)=f(n)/(m+1) .$