# How to simplify the summation of log

I have a summation that involve log. I don't know how to solve this summation. I want to find an expression (even a good approximation is enough) for this summation.

$\sum_{k=0}^{n}{log(a_k)}$ or $log(\prod_{k=0}^{n}{a_k})$

I only know that $\sum_{k=0}^{n}{a_k}= N$

Edit

Let $a_k$ is a random variable which can take its value according to binomial (or normal) distribution. Then how to solve the above log summation problem.

• Assuming $a_k > 0$ the AM-GM inequality gives $a_1a_2\ldots a_n \leq (\sum a_k / n)^n = (N/n)^n$ so $\sum \log a_k \leq n\log (N/n)$. Without any more information about the problem this is as close to a bound you can get. – Winther Apr 29 '15 at 0:43
• Aside from an upper bound, there is not even a good approximation based on this information. The answer could change by orders of magnitude depending on exactly how the total "weight" $N$ is distributed among the $a_k$; for example, consider $a_0 = a_1 = 1$ vs. $a_0 = 2-10^{-20}$ and $a_1 = 10^{-20}$. Perhaps a much better answer would be possible if you explain why you are interested in this sum. – David K Apr 29 '15 at 0:45
• I designed an algorithm in which the memory required by the algorithm depends on this sum. The AM-GM inequality is the worst case in which the weight N is distributed equally i.e each $a_i$ is equal to $\frac{N}{n}$. What if the weight N is distributed according to some well know distributions. May be like binomial, normal etc – user3212493 Apr 29 '15 at 1:02
• @user3212493 If $a_k$ is a random variable distributed according to some given distribution then one can calculate the expectation value (and variance) for the sum (but if one can do it analytically depends on the distribution). However if this is your question you should edit the question with this new info. – Winther Apr 29 '15 at 1:34
• Is this okay now? – user3212493 Apr 29 '15 at 1:51

When you characterize this as a probability question rather than a general series question, if you avoid certain pitfalls you can use the Law of Large Numbers to characterize the sum. That is, having defined a probability distribution for $a_k$, this implies a probability distribution of $\log(a_k)$; then from the expectation and variance of $\log(a_k)$ you can estimate the expectation and variance of the sum for large $n$.
There is a discussion of the case where $a_k$ is (approximately) normally distributed on stats.SE (see this question) and a discussion of the case where $a_k$ has a binomial distribution on mathoverflow (see this question). One technique that is recommended is to use the Taylor series of $\log(x)$ centered at $\mathbb E(a_k)$.
You cannot actually let $a_k$ have a normal distribution, because any normal distribution has negative values. In fact, you really want to prevent $a_k$ from getting close to zero, because the resulting large negative logarithms will cause problems. (This is one of the above-mentioned "pitfalls" to avoid.) A binomial distribution with only positive integer values would produce a much better-behaved distribution of the logarithm.