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So I am given the following:

$n = \sum_{i=1}^{k}m_{i}$

I am also given

$x = \sum_{i=1}^{k}log(m_{i}) = log\prod_{i=1}^{k}m_{i}$

I was only given the first part, however I believe that is a valid substitution and feel it may be of some value. I was hoping that there was a way to solve for $x$ in terms of $n$ and/or $k$. This is for a larger problem, however I am not sure how to tackle this part of it. I am not very familiar with $\prod$ identities or $\sum$ identities.

If anyone had any thoughts it would be much appreciated. Wolfram mentioned something about Log[Pochhammer[1, k]], however, I am not sure how to make sense of this. Thank you!

EDIT: Would it be safe to say that $m\log m = \log (m!) = \log\prod_{i=1}^{k}m_{i}$ ? If so, then I believe my problem is solvable from there.

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What do you want to show in your exercise? – Jimmy R. Mar 16 '14 at 11:23
I have to find an algorithm that runs in $O(k*\sum_{i=1}^{k}log(m_{i}))$. I am trying to make sense of what the summation actually means. – BoZa Mar 16 '14 at 11:42
For your edit, no; note that e.g. if $m=3$ then these expressions are $\log 3 + \log 3 + \log 3$ and $\log 1 + \log 2 + \log 3$ respectively, which are clearly not equal. – MPW Mar 16 '14 at 12:04
My problem is that $m_{i}$ is a value at the index i. So I can't represent it as $\log m^{k}$ either – BoZa Mar 16 '14 at 12:45

Firstly you have that $$\log (m_1\cdot m_2)= \log m_1 + \log m_2$$ Generalizing for $k$ elements you have that $$\log \prod_{i=1}^{k} m_i=\sum_{i=1}^{k}\log m_i$$ which reads "the logarithm of the product is the sum of the logarithms". Secondly, you have that $$\log m^a=a \log m$$ So in your case $$m \log m=\log m^m \neq \log m! \neq \prod_{i=1}^{k} m_i$$ which means that the implication that you have in your EDIT does not hold.

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So this tells me that the last part was incorrect, however, I am looking to make sense of the relation, which this answer does not do. Can you elaborate? – BoZa Mar 16 '14 at 11:44
@BoZa Which relation do you mean? I explained the relation between logarith of product and sum of logarithms – Jimmy R. Mar 16 '14 at 11:50

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