Fix $x>0$ and $c\in\mathbb{N}$. Let $f(x):=\frac{c}{4c-2+2x^2}$ and $$m_N(x):=\frac{1}{N} \sum_{i=0}^{f(x)N} \log(\frac{c N}{2}-i(2c-1))$$

I'm pretty sure $m_N(x)\to\infty$ as $N\to\infty$.

I would like to know the asymptotic behaviour of $m_N(x)$, and I expect to find something like

$$m_N(x)=f(x) \log{N} + g(x) + o(1)\ \ \text{ as }N\to\infty$$

Can you confirm this result? If this is the case, can you help me to compute the constant $g(x)$?

  • $\begingroup$ Unless you can give some background and context to this question, you are not likely to get any useful answers, and your question is likely to be closed. $\endgroup$
    – Lee Mosher
    Jun 11, 2012 at 23:28
  • $\begingroup$ I'm trying to compute a limit in a Statistical Mechanics model. After some computations, I need to compute that $g(x)$. Just knowing it is true that $m_N(x)=f(x) logN + g(x) + o(1)$ for some $g(x)$ would be a good result. $\endgroup$
    – user22980
    Jun 11, 2012 at 23:39
  • 1
    $\begingroup$ Rather than do the computation, I'll just tell you how I would do it: the sum looks a lot like a Riemann sum. So write the Riemann sum and the integral to which it is an approximation side by side and carefully estimate the differences. $\endgroup$ Jun 12, 2012 at 4:26
  • $\begingroup$ I have to guess that $i$ is an integer so, why is the upper limit of the sum the real number $f(x)N$? Is it a mistake or is it so? $\endgroup$
    – Jon
    Jun 12, 2012 at 7:36
  • $\begingroup$ Jon you're right: the sum is over integer $i$. The correct upper bound of the sum would be $$[f(x) N + k(x)]+1$$ where $k(x)=\frac{-x^2}{2c-1+x^2}$ and $[]$ denotes the integer part. But I think it's not important for the asymptotic behaviour of the mean. $\endgroup$
    – user22980
    Jun 12, 2012 at 9:52

1 Answer 1


Let us consider the sum $$ m_N(x)=\frac{1}{N}\sum_{i=0}^{[f(x)]N}\log\left(\frac{cN}{2}-i(2c-1)\right). $$ The first step to get an asymptotic approximation is to extract the leading term in $N$ to obtain $$ m_N(x)=[f(x)]\log N+\frac{1}{N}\sum_{i=0}^{[f(x)]N}\log\left(\frac{c}{2}-\frac{i}{N}(2c-1)\right). $$ When $N$ is finite, we recognize a Riemann series and apply the average theorem. So, there exists a value of argument of the logarithm such that $$ m_N(x)=[f(x)]\log N+[f(x)]\log[z(x)]. $$ We can take $z(x)=\frac{c}{2}-t\[f(x)\](2c-1)$ being $t\in (0,1)$.

Indeed, we can define a partition with $x_i=x_{i-1}+\frac{1}{\[f(x)\]N}$ and so $$ \frac{1}{N}\sum_{i=0}^{[f(x)]N}\log\left(\frac{c}{2}-\frac{i}{N}(2c-1)\right)=[f(x)]\Delta x\sum_{i=0}^{[f(x)]N}\log\left(\frac{c}{2}-i\[f(x)\](2c-1)\Delta x\right) $$ being $\Delta x=\frac{1}{\[f(x)\]N}$. But this, in the given limit, is nothing else than $$ \int_{\frac{c}{2}}^{\frac{c}{2}-\[f(x)\](2c-1)}\log(z)dz<\infty $$ as it should.

  • $\begingroup$ Yes in the end it's just Riemann sums $\endgroup$
    – user22980
    Jun 14, 2012 at 15:20

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