User ogn - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T10:01:15Z http://mathoverflow.net/feeds/user/12981 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/118515/derivative-indicator-function Derivative indicator function ogn 2013-01-10T10:07:52Z 2013-04-07T08:22:00Z <p>I am wondering what is the derivative of the following function with respect to $x(t)$ in sense of distributions. $$I\left(\int_0^t x(\tau)d\tau \leq c\right)$$ where $I$ is the indicator function and $c$ is a constant.</p> http://mathoverflow.net/questions/102627/strengthening-an-inequality Strengthening an inequality ogn 2012-07-19T08:38:28Z 2012-07-19T09:38:40Z <p>Let $k$ be an integer. The following inequality is standard. $$(a+b)^{k+1} - b^{k+1} \leq (k+1)a(a+b)^k$$ for $a,b > 0$.</p> <p>However, does the following inequality still hold $$(a+b)^{k+1} - b^{k+1} \leq (k+1)a\left(a+ \frac{b}{(k+1)^{1/(k+1)}} \right)^k$$ for $a,b > 0$? While $k \rightarrow \infty$, the term $(k+1)^{1/(k+1)} \rightarrow 1$ so that becomes the first inequality. What about if $k$ is large enough? </p> http://mathoverflow.net/questions/92599/distribution-of-primes-in-small-intervals Distribution of primes in small intervals ogn 2012-03-29T19:05:59Z 2012-03-30T03:41:47Z <p>Let $\pi(x)$ be the number of primes smaller than $x$. Do there exist unconditionally universal constants $c > d$ such that $$\lim_{x \rightarrow \infty} \frac{\pi(x + \log^c x) - \pi(x)}{\log^{c-d} x} \geq 1$$</p> <p>We know that by Maier Theorem, it is not possible that $c = d+1$. </p> <p>By Selberg theorem, for any function $y(x)$ grows faster than $\log^2 x$, it holds that $$\lim_{x \rightarrow \infty} \frac{\pi(x + y) - \pi(x)}{y/\log x} = 1$$ for \emph{almost} $x$ (assuming the Riemann hypothesis). Does it hold for \emph{all} $x$ if $y(x) = \log^c x$ for some constant $c$ (with Riemann hypothesis)? </p> http://mathoverflow.net/questions/91850/generator-density-in-mathbbz-p Generator density in $\mathbb{Z}^*_p$ ogn 2012-03-21T19:26:28Z 2012-03-21T22:00:11Z <p>Hello,</p> <p>Consider the multiplicative group $(\mathbb{Z}/p)^*$ for a given prime $p$. We know that the number of generators in this group is $\phi(p-1)$ --- the Euleur totient function. The question is, for $0 \leq a &lt; a + \log^{c} p &lt; b \leq p-1$ where $c$ is a constant (say $c=10$), how many generators of the group belongs to $[a,b]$? In other words, what is the density of generators in a given interval $[a,b]$ (compared to the density $\phi(p-1)/p-1$)? Is it easier if $b=p-1$? </p> <p>For a given prime $p$, what is the densest interval in term of generators?</p> http://mathoverflow.net/questions/58980/coefficient-bounds-of-an-inequality Coefficient bounds of an inequality ogn 2011-03-20T15:34:34Z 2011-04-11T16:10:44Z <p>Hello,</p> <p>Given positive integers $k$ and $n$. Are there upper bounds on coefficients $A$ and $B$ such that they depends only on $k$ (eg., $2 k^k$) and for all non-negative integer sequences $(a_i)_{1}^n, (b_i)_{1}^n$ and non-negative increasing real sequence $(p_i)_{1}^n$, the following inequality holds?</p> <p>$$\sum_{i=1}^n b_i \left(\sum_{j=1}^i a_j p_j \right)^k \leq A \sum_{i=1}^n a_i \left(\sum_{j=1}^i a_j p_j \right)^k + B \sum_{i=1}^n b_i \left(\sum_{j=1}^i b_j p_j \right)^k$$</p> <p>Do you know any result or reference related to the question?</p> <p>Edit: 11/4 Due to the asymmetry of the left-hand side, we can prove the inequality for $A = k/(k+1)$ and $B = \Theta(k^k)$. Is it possible for the same kind of $A,B$ (up to a constant) such that</p> <p>\begin{eqnarray*} \sum_{i=1}^n &amp;b_i&amp; (a_1p_1 + \ldots + a_{j-1}p_{j-1} + (a_j + \ldots + a_n) p_j)^k \ &amp;\leq&amp; A \sum_{i=1}^n a_i (a_1p_1 + \ldots + a_{j-1}p_{j-1} + (a_j + \ldots + a_n) p_j)^k \ &amp;+&amp; B \sum_{i=1}^n b_i(b_1p_1 + \ldots + b_{j-1}p_{j-1} + (b_j + \ldots + b_n) p_j)^k \end{eqnarray*}</p> <p>The difficulty is due to the tail $(a_{j+1} + \ldots a_n)p_n$ (idem for $b$).</p> http://mathoverflow.net/questions/60943/root-estimation Root estimation ogn 2011-04-07T14:20:20Z 2011-04-11T09:07:42Z <p>What is the estimation for the positive root of the following equation $$ax^k = (x+1)^{k-1}$$ where $a > 0$ (specifically $0 &lt; a \leq 1$). </p> <p>Could you point out some reference related to the question?</p> http://mathoverflow.net/questions/60907/inner-product-space-on-sphere Inner product space on sphere ? ogn 2011-04-07T09:08:29Z 2011-04-07T13:07:52Z <p>I do not know if the following question makes sense.</p> <p>Is it possible to define an inner product (that gives real values) for vectors on a sphere $S^n$ (let say $S^1$)? The set of these vectors is not a real vector space but probably we can handle somehow to create an inner product.</p> <p>Do you know any example? </p> <p>Edit: </p> <p>I define a map on a real vector space (say $\mathbb{R}^n$) and I would like to prove that it is indeed an inner product to use the Cauchy-Schwarz inequality. But it fails to satisfy the definite-positiveness over the whole vector space. However, restricted on a sphere of this vector space, the definite-positiveness holds. But in this case, let $v$ be a vector in the sphere, a vector $\alpha v$ for $\alpha \in \mathbb{R}^n$ is not in the sphere. So the inner-product is not well-defined. How I get rid of that? </p> <p>Besides, the only things I want is the use of Cauchy-Schwarz inequality - the property "like an inner product". Do you know something related?</p> http://mathoverflow.net/questions/55455/summation-of-an-expression Summation of an expression ogn 2011-02-14T22:06:02Z 2011-02-20T15:34:39Z <p>Hi,</p> <p>Does anyone have an idea about an exact or approximate formulae for the following summation? $$\sum_{j=1}^n \frac{j^k}{(j-1)!}$$ where k is a positive integer (the denominator of the j^th term is of course $\Gamma(j)$).</p> http://mathoverflow.net/questions/91850/generator-density-in-mathbbz-p Comment by ogn ogn 2012-03-21T22:08:57Z 2012-03-21T22:08:57Z I have just relaxed the length of $[a,b]$, the length can be lower-bounded by $(\log p)^c$ for some small constant $c$. Is there any nontrivial answer? http://mathoverflow.net/questions/58980/coefficient-bounds-of-an-inequality/59028#59028 Comment by ogn ogn 2011-04-11T16:12:16Z 2011-04-11T16:12:16Z @fedja: see edit of question http://mathoverflow.net/questions/60943/root-estimation/61022#61022 Comment by ogn ogn 2011-04-08T12:05:57Z 2011-04-08T12:05:57Z Put $y = ax$, we have $(ax)^n = (ax + a)^{n-1} \leq (ax + 1)^{n-1}$. Thus, $x \leq c \cdot \frac{n}{a \log n}$ for some constant $c$. % However, I think the bound could be strengthened. http://mathoverflow.net/questions/60943/root-estimation/61022#61022 Comment by ogn ogn 2011-04-08T11:43:45Z 2011-04-08T11:43:45Z Your estimation is correct. We can take an estimation $W(ak) = \Theta(\log(ak))$ and deduce $x = \Theta(k/\log(ak))$. However, I am wondering what happens if $a$ is a function of $k$, let say $a = 2^{-k}$. In this case $\log(ak)$ is negative, but there exits a positive root. http://mathoverflow.net/questions/60943/root-estimation/60963#60963 Comment by ogn ogn 2011-04-07T16:37:48Z 2011-04-07T16:37:48Z Your estimation is correct, but I think we can do better. For example, if $a = 1$ then a good estimation of $x$ is $\Theta(k/\log k)$. Let $a$ is a fix constant, what is the best estimation for $x$ as a function of $k$ (of course, for $k$ large) http://mathoverflow.net/questions/60907/inner-product-space-on-sphere Comment by ogn ogn 2011-04-07T16:18:46Z 2011-04-07T16:18:46Z I just want to make sure that I understand you. Is the following what you meant If a bilinear map on $\mathbb{R}^n$ satisfies the positive-definiteness on $S^{n-1}$ then we can apply C-S for vector $x,y \in S^{n-1}$. If it is true, could you please point out some references? http://mathoverflow.net/questions/60907/inner-product-space-on-sphere Comment by ogn ogn 2011-04-07T13:08:21Z 2011-04-07T13:08:21Z See edit of the question. http://mathoverflow.net/questions/60907/inner-product-space-on-sphere Comment by ogn ogn 2011-04-07T13:04:56Z 2011-04-07T13:04:56Z Besides, the only things I want is the use of Cauchy-Schwarz inequality - the property &quot;like an inner product&quot;. Do you know something related? http://mathoverflow.net/questions/60907/inner-product-space-on-sphere Comment by ogn ogn 2011-04-07T13:03:50Z 2011-04-07T13:03:50Z I define a map on a real vector space (say $\mathbb{R}^n$) and I would like to prove that it is indeed an inner product to use the Cauchy-Schwarz inequality. But it fails to satisfy the definite-positiveness over the whole vector space. However, restricted on a sphere of this vector space, the definite-positiveness holds. But in this case, let $v$ be a vector in the sphere, a vector $\alpha v$ for $\alpha \in \mathbb{R}$ is not in the sphere. So the inner-product is not well-defined. How I get rid of that? http://mathoverflow.net/questions/58980/coefficient-bounds-of-an-inequality/59028#59028 Comment by ogn ogn 2011-03-31T14:41:42Z 2011-03-31T14:41:42Z Yes, you are right! http://mathoverflow.net/questions/58980/coefficient-bounds-of-an-inequality/59028#59028 Comment by ogn ogn 2011-03-30T17:50:45Z 2011-03-30T17:50:45Z I play with this kind of inequality. I am wondering if the following inequality holds for some C depending only on k (always with the same assumptions on the sequences). \begin{align*} \sum_{i=1}^n &amp; b_i(a_1p_1+ \ldots + a_ip_i+ (a_{i+1}+ \ldots + a_n)p_i)^k \\ ≤&amp; C\sum_{i=1}^n a_i(a_1p_1+ \ldots + a_ip_i+(a_{i+1} +\ldots +a_n)p_i)^k \\ &amp;+ C\sum_{i=1}^n b_i(b_1+…+b_ip_i+(b_{i+1}+…+b_n)p_i)^k \end{align*} I think that is true but the technique above does not give a proof. http://mathoverflow.net/questions/58980/coefficient-bounds-of-an-inequality/59028#59028 Comment by ogn ogn 2011-03-22T10:45:08Z 2011-03-22T10:45:08Z Thank you very much! I understand now; I learn a new trick, that's nice. http://mathoverflow.net/questions/58980/coefficient-bounds-of-an-inequality/59028#59028 Comment by ogn ogn 2011-03-21T17:59:27Z 2011-03-21T17:59:27Z Hello, I am slow so I don't understand why the first inequality is linear in $q_i$ so we just need to check the second inequality? Please explain.