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I am told to solve the recurrence $$T(n)=10T(n/9)+n\lg(n)$$ using the Master theorem. I then try to use case 3. However, I am unable to show that for $f(n)=n\lg(n)$ then $10f(n/9) \leq cn\lg(n)$ for $c < 1$ and all sufficiently large $n$. Is it wrong to use case 3? Or does the Master theorem even apply?

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  • $\begingroup$ You should state the exat version of the master theorem you are using, there is a raft of them. $\endgroup$
    – vonbrand
    Commented Feb 19, 2021 at 21:19

1 Answer 1

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Case 3 does not apply. Indeed: $$ f(n) = n \log n \not\in \Omega(n^{\log_9 10}) = \Omega(n^{\log_b a}). $$

However case $1$ applies since, for $0<c \le 0.01$ $$ f(n) = n \log n \in O(n^{1.04 - 0.01}) \subset O(n^{\log_9 10 - 0.01} ) \subseteq O(n^{\log_b a - c} ). $$

This shows that $T(n) \in \Theta(n^{\log_9 10})$.

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