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}\

$

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- Feb 24th 2010, 04:35 AMukroboLimits
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}\

$ - Feb 24th 2010, 05:24 AMtonio
- Feb 24th 2010, 05:52 AMdrumist
http://en.wikipedia.org/wiki/Big_O_notation#Equals_sign

Basically it means the same either way. - Feb 24th 2010, 06:05 AMtonio

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 - Feb 24th 2010, 11:59 AMukroboReply
I need to evaluate the expression as either 0 or a real number. Your equation seems to imply that the expression = 0 to evaulate properties of k. I need to show that the expression = 0. cheers!

- Feb 24th 2010, 07:05 PMtonio