For all $\displaystyle k>=0$ show that $\displaystyle n^k \in o(n^a) $
where:
$\displaystyle
a=log2(n)
$
I get to here but i am not sure were to go next:
$\displaystyle
\lim_{n \to \infty} \frac{n^k}{n^a}\
$
For all $\displaystyle k>=0$ show that $\displaystyle n^k \in o(n^a) $
where:
$\displaystyle
a=log2(n)
$
I get to here but i am not sure were to go next:
$\displaystyle
\lim_{n \to \infty} \frac{n^k}{n^a}\
$
http://en.wikipedia.org/wiki/Big_O_notation#Equals_sign
Basically it means the same either way.
Ok then, thanx. But then as $\displaystyle a=\log_2 n =\frac{1}{\log_n 2}$ , we get $\displaystyle n^a=n^\frac{1}{\log_n2}$ ,and thus $\displaystyle n^k=o\left(n^\frac{1}{\log_n2}\right)\Longleftrigh tarrow \frac{n^k}{n^\frac{1}{\log_n2}}=n^{k-\frac{1}{\log_n2}}\xrightarrow[n\to\infty]{}0\Longleftrightarrow k-\frac{1}{\log_n2}<0\Longleftrightarrow k<\frac{1}{\log_n2}$ .
Tonio