0

$$L_{n}\left[\alpha,c\right]:=e^{\left(c + o(1)\right)(\ln n)^\alpha(\ln \ln n)^{1-\alpha}}$$

$f$ is in o(1) if and only if for every $c\in\left[0,\infty\right)$, there exists some $N\in\mathbb{N}$ such that for every $n\in\mathbb{N}$ with $n\geq N$, we have $f(n)< c \cdotp 1$.

So, can't we just replace the $\left(c + o(1)\right)$ with just $k$ a new constant? o(1) `vanishes' as $n\rightarrow\infty$ What extra information does$\left(c + o(1)\right)$ add?

Edit to clarify:

$e^{\left(c + o(1)\right)(\ln n)^\alpha(\ln \ln n)^{1-\alpha}}=e^{c(\ln n)^\alpha(\ln \ln n)^{1-\alpha}}e^{o(1)(\ln n)^\alpha(\ln \ln n)^{1-\alpha}}\rightarrow e^{c(\ln n)^\alpha(\ln \ln n)^{1-\alpha}}$ as $n\rightarrow\infty$

So, $e^{(c+o(1))(\ln n)^\alpha(\ln \ln n)^{1-\alpha}}\leq e^{(c+\epsilon)(\ln n)^\alpha(\ln \ln n)^{1-\alpha}}$ eventually, so let $k=c+\epsilon$, $$L_{n}\left[\alpha,c\right]:=O\left(e^{k(\ln n)^\alpha(\ln \ln n)^{1-\alpha}}\right)$$

djones
  • 27
  • What is exactly your suggestion for replacing the RHS? – md5 Apr 21 '20 at 12:00
  • $L_{n}\left[\alpha,k\right]:=\mathcal{o}\left(e^{(k)(\ln n)^\alpha(\ln \ln n)^{1-\alpha}}\right)$? or something like it? – djones Apr 21 '20 at 12:08
  • How would that make sense as a definition? What is $k$? – md5 Apr 21 '20 at 12:10
  • sorry, was editing as I went, k a non-negative real – djones Apr 21 '20 at 12:12
  • Your clarification still does not really make sense. How do you choose $\epsilon$? – md5 Apr 21 '20 at 12:29
  • Well it would work for any $\epsilon\in(0,\infty)$, as o(1) means eventually smaller than any strictly positive constant. – djones Apr 21 '20 at 12:37
  • Well if you want to give an alternate definition of $L_n[\alpha,c]$, you need to specifiy which $\epsilon$ you choose. If you define $L_n[\alpha,c]=O(\exp((c+0.1)(\ln n)^\alpha(\ln\ln n)^{1-\alpha}))$, you are going to get more functions than in the original definition of $L_n[\alpha,c]$ (for example the function $n\mapsto \exp((c+0.1)(\ln n)^\alpha(\ln\ln n)^{1-\alpha})$). – md5 Apr 21 '20 at 12:41

0 Answers0