most recent 30 from http://mathoverflow.net 2013-05-23T09:25:53Z http://mathoverflow.net/feeds/question/99319 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/99319/asymptotic-behaviour-of-a-mean xn--qwertyuiop-86a 2012-06-11T23:15:20Z 2012-06-12T17:28:24Z

<p>Fix $x>0$ and $c\in\mathbb{N}$. Let $f(x):=\frac{c}{4c-2+2x^2}$ and $$m_N(x):=\frac{1}{N} \sum_{i=0}^{f(x)N} \log(\frac{c N}{2}-i(2c-1))$$</p> <p>I'm pretty sure $m_N(x)\to\infty$ as $N\to\infty$.</p> <p>I would like to know the asymptotic behaviour of $m_N(x)$, and I expect to find something like</p> <p>$$m_N(x)=f(x) \log{N} + g(x) + o(1)\ \ \text{ as }N\to\infty$$ </p> <p>Can you confirm this result? If this is the case, can you help me to compute the constant $g(x)$?</p>

http://mathoverflow.net/questions/99319/asymptotic-behaviour-of-a-mean/99360#99360 Jon 2012-06-12T12:31:03Z 2012-06-12T17:28:24Z

<p>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 $t\in (0,1)$. </p> <p>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 $$ \int_{\frac{c}{2}}^{\frac{c}{2}-[f(x)](2c-1)}\log(z)dz&lt;\infty $$ as it should.</p>