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Probably a very easy question, but I can't find a simplification for the following:

$$\sum_{i=1}^n\log(a_i)$$

Can this summation be simplified?

tony1kenobi
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    Do you consider $\log(a_1 \cdots a_n)$ (or $\log(a_1\cdots a_n)+2\pi i k_n$ for some integer $k_n$, if $a_k$'s are allowed to be complex-valued) as simplified? – Sangchul Lee Apr 12 '17 at 17:49
  • not really, I am doing an max likelihood estimator problem where the next step is to differentiate the above multiplied by x and wrt x. I am then left with the above. – tony1kenobi Apr 12 '17 at 17:53
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    Honestly, with no extra information supplied, I can hardly believe that your expression can be simplified further. – Sangchul Lee Apr 12 '17 at 17:56
  • Thanks I will use as is in the solution. – tony1kenobi Apr 12 '17 at 18:04

2 Answers2

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Hint.

$$\log(a) + \log(b) = \log(ab)$$

Now, extend it to $a_n$ elements.

Enrico M.
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  • Remember that it holds as long as all the $a_n$ are greater than zero. – Enrico M. Apr 12 '17 at 17:49
  • do you mean $\log(a_1 \cdots a_n)$ – tony1kenobi Apr 12 '17 at 17:59
  • @tony1kenobi Suppose your sum ends at $n = 3$ then you have

    $$\log(a_1) + \log(a_2) + \log(a_3)$$

    Which you may manipulate as

    $$\log(a_1\cdot a_2) + \log(a_3)$$

    hence

    $$\log(a_1\cdot a_2\cdot a_3)$$

    Eventually you'll end up in obtaining the logarithm of a productory.

     – Enrico M. Apr 12 '17 at 18:01
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\begin{align} & \phantom{\Big(}\log(a_1) + \log(a_2) + \log(a_3) + \log(a_4) + \log(a_5) +\log(a_6) \\[10pt] = {} & \Big(\log(a_1)+\log(a_2)\Big) + \log(a_3) + \log(a_4) + \log(a_5) +\log(a_6) \\ = {} & \Big(\log(a_1a_2)\Big) + \log(a_3) + \log(a_4) + \log(a_5) +\log(a_6) \\[10pt] = {} & \Big(\log(a_1a_2) + \log(a_3)\Big) + \log(a_4) + \log(a_5) +\log(a_6) \\ = {} & \Big( \log(a_1a_2a_3) \Big) + \log(a_4) + \log(a_5) +\log(a_6) \\[10pt] = {} & \Big( \log(a_1a_2a_3) + \log(a_4)\Big) + \log(a_5) +\log(a_6) \\ = {} & \Big( \log(a_1a_2a_3a_4)\Big) + \log(a_5) +\log(a_6) \\[10pt] = {} & \Big( \log(a_1a_2a_3a_4) + \log(a_5)\Big) +\log(a_6) \\ = {} & \Big( \log(a_1a_2a_3a_4a_5)\Big) +\log(a_6) \\[10pt] = {} & \cdots \cdots \end{align}

You can make the above argument into a proof by mathematical induction if you want to.

Michael Hardy
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