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Math Help - Limits

  1. #1
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    Limits

    For all k>=0 show that  n^k \in o(n^a)
    where:
    <br />
a=log2(n)<br />

    I get to here but i am not sure were to go next:
    <br />
  \lim_{n \to \infty} \frac{n^k}{n^a}\<br />
    Last edited by ukrobo; February 24th 2010 at 05:00 AM.
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  2. #2
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    Quote Originally Posted by ukrobo View Post
    For all k>=0 show that  n^k \in o(n^a)
    where:
    <br />
a=log2(n)<br />

    I get to here but i am not sure were to go next:
    <br />
\lim_{n \to \infty} \frac{n^k}{n^a}\<br />

    What does n^k\in o(n^a) mean, anyway? I know n^k=o(n^a) , but belongs...?

    Tonio
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  3. #3
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    Quote Originally Posted by tonio View Post
    What does n^k\in o(n^a) mean, anyway? I know n^k=o(n^a) , but belongs...?

    Tonio
    http://en.wikipedia.org/wiki/Big_O_notation#Equals_sign

    Basically it means the same either way.
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  4. #4
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    Quote Originally Posted by drumist View Post
    Big O notation - Wikipedia, the free encyclopedia

    Basically it means the same either way.

    Ok then, thanx. But then as a=\log_2 n =\frac{1}{\log_n 2} , we get n^a=n^\frac{1}{\log_n2} ,and thus 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
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  5. #5
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    Reply

    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!
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  6. #6
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    Quote Originally Posted by ukrobo View Post
    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!

    I don't understand what you want: I thought you want to show that limit in your first post equals zero and I just wrote what the condition on k must be for that to be true.

    Tonio
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