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given: $$m_1+m_2+...+m_k=m$$ How can I find $\Theta(log(m_1)+...+log(m_k))$ as related to $m$?

I know that i can doing that: $O(log(m_1)+...log(m_k))=O(log(m)+...+log(m))=O(k \cdot log(m))$ , but can I find something without $k$ ?

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    $\begingroup$ Is $k$ a fixed constant? If not, you can't give a big-Theta bound, since you could have $k=m$ and $m_1 = \dots = m_k = 1$, giving $\sum_i\log m_i = 0$. $\endgroup$ – David Richerby Jun 13 '18 at 18:55
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The $\log$ function is concave so, by Jensen's inequality, $$\frac{1}{k}\sum_{i=1}^k\log m_i \leq \log\Big(\frac1k\sum_{i=1}^k m_i\Big)$$ i.e., $$\sum_{i=1}^k \log m_i \leq k\log\frac{m}{k} = k\,(\log m - \log k)\,,$$ which is basically what you got by bounding $\log m_i\leq \log m$. We can't do any better than this because we could have $m_1 = \dots = k_m = \tfrac{m}{k}$, in which case $\sum_i\log m_i = k\log\tfrac{m}{k}$.

However, if $k$ is a fixed constant, then $k\log\tfrac{m}{k}=O(\log m)$.

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