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#### find_the_fun

##### Active member

- Feb 1, 2012

- 166

- Thread starter find_the_fun
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- Thread starter
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- Feb 1, 2012

- 166

- Jan 26, 2012

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It is because for \(n>n_0\) where \(n_0>0\) is greater than the base of the logarithm (so \( \log(n)>1\) )

:

\[|T(n)|=|n \; b+c\; n \log(n)|<|b| \; n+|c|\; n \log(n)<|b|\; n \log(n)+ |c|\; n \log(n) = A\; n \log(n)\]

CB

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