User nikita sidorov - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T08:48:16Z http://mathoverflow.net/feeds/user/8131 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/121915/linear-numeration-systems/121918#121918 Answer by Nikita Sidorov for Linear numeration systems Nikita Sidorov 2013-02-15T16:47:25Z 2013-02-15T16:47:25Z <p>Some results on the quantity in question can be found in </p> <p><em>J. M. Dumont, N. Sidorov and A. Thomas, Number of representations related to a linear recurrent basis, Acta Arithmetica 88 (1999), 371-394.</em></p> <p>We are mainly interested in the summatory function but there are also some upper bounds for the quantity itself. Our main assumption is that the corresponding root (of $x^4=x^3+x^2+x-1$ in your case) is a Perron number (in your example it's even Salem, so our results apply). </p> http://mathoverflow.net/questions/118811/hutchinsons-formula-for-asymptotically-homogeneous-cantor-sets Hutchinson's formula for asymptotically homogeneous Cantor sets Nikita Sidorov 2013-01-13T14:10:21Z 2013-01-13T17:00:25Z <p>As everyone knows, the standard middle-thirds Cantor set is constructed by dividing the interval into three equal parts, removing the middle one, then applying the same procedure to the remaining two intervals, etc. </p> <p>The resulting set has Hausdorff dimension $s=\log 2/\log 3$, in view of Hutchinson's formula: $$\sum_{i=1}^m r_i^s=1,$$ where in our case $m=2, r_1=r_2=1/3$. </p> <p>Now, assume that $m=2$ but that on level $n$ we remove the middle-thirds interval whose relative measure is not exactly $1/3$ but $1/3+\delta_n$, where $\delta_n\to0$ sufficiently fast. (It may also depend on a position of the interval but there is always a uniform upper bound which tends to 0 sufficiently fast.) </p> <p>Is it still true that the Hausdorff dimension of such a set is $\log2/\log3$? </p> <p>This is actually a toy question', since in my set-up the corresponding iterated function system is infinite countable. However, Hutchinson's formula works for such IFS just as well (with $m=\infty$), so I'm sure the conclusion should be the same. If it helps, the uniform upper bound for the $\delta_n$ in my case is a double exponent, i.e., $\frac1n \log (-\log \delta_n)\to \text{const}$ as $n\to+\infty$. </p> http://mathoverflow.net/questions/115509/when-we-use-bernstein-polynomials-in-application/115510#115510 Answer by Nikita Sidorov for When we use Bernstein polynomials in application Nikita Sidorov 2012-12-05T16:17:48Z 2012-12-05T16:17:48Z <p>The only practical advantage of Bernstein polynomials is their universality. They really work for any continuous function $f$. However, it is well known that if $$\|f-B_n(f)\|_\infty=o(1/n),\quad n\to\infty,$$ then $f(x)=ax+b$. In other words, one cannot hope to approximate a non-linear function by a Bernstein polynomial with an error term better than $1/n$ - which is impractical. </p> http://mathoverflow.net/questions/114745/monic-polynomial-with-integer-coefficients-with-roots-on-unit-circle-not-roots-r/114759#114759 Answer by Nikita Sidorov for Monic polynomial with integer coefficients with roots on unit circle, not roots root of unity? Nikita Sidorov 2012-11-28T11:32:35Z 2012-11-28T11:32:35Z <p>Just a couple of minor top-ups to Dmitri's nice answer. </p> <ol> <li>For each even $n\ge2$ the polynomial $p_n(x)=x^n-x^{n-1}-\dots-x+1$ is a Salem polynomial.</li> <li>It is not known whether for any $\delta>0$ there exists a Salem polynomial such that $\theta&lt;1+\delta$ for its largest root $\theta$ (which is a Salem number). <a href="http://en.wikipedia.org/wiki/Lehmer%27s_conjecture" rel="nofollow">Lehmer's conjecture</a> suggests that the answer is no. </li> </ol> http://mathoverflow.net/questions/112419/an-identity-which-involves-eulers-totient-function An identity which involves Euler's totient function Nikita Sidorov 2012-11-14T21:46:24Z 2012-11-14T21:46:24Z <p>It is well known that $$\sum_{n=1}^\infty \frac{\phi(n)x^n}{1-x^n}=\frac x{(1-x)^2},$$ where $\phi$ is Euler's totient function and $|x|&lt;1$ - see [Hardy and Wright, Theorem 309]. For $x=\frac12$ this immediately yields $$\sum_{n=1}^\infty \frac{\phi(n)}{2^n-1}=2.$$ What I need for my research is the analytic value for $$\sum_{n=1}^\infty \frac{\phi(n)}{(2^n-1)^2}.$$ Numerically it is $1.1659457\dots$, which doesn't look like something familiar to me (or to Google, for that matter). </p> <p>Any ideas? </p> http://mathoverflow.net/questions/112050/logistic-map-periodic-point/112054#112054 Answer by Nikita Sidorov for Logistic map periodic point Nikita Sidorov 2012-11-11T02:46:17Z 2012-11-11T02:46:17Z <p>Essentially you have proved that the logistic map is conjugate to the doubling map $Tx=2x\bmod 1$. Now, $T$ is in turn conjugate to the shift map $\sigma:\Sigma\to\Sigma$, where $\Sigma$ is the space of infinite 0-1 words. </p> <p>More precisely, if $$\pi(w_1,w_2,\dots)=\sum_{n=1}^\infty w_n2^{-n},$$ then you have $$\pi \sigma = T\pi.$$ Since $\pi$ is 1-1, except for a countable set of finite words, you can just take any word $u$ of length $m$ and the corresponding infinite word $w=uuuu\dots$. Clearly, $\sigma^m(w)=w$, i.e., $w$ is $\sigma$-periodic of period $m$. (With a little effort you can make this the <em>smallest</em> period.) </p> <p>For instance, $w=001001001\dots$ is of period 3. </p> <p>Now take $x:=\pi(w)$; it is $T$-periodic of period $m$. Finally, use you conjugate map to turn $x$ into an $m$-periodic point for the logistic map. </p> http://mathoverflow.net/questions/107653/prove-log-of-eigenvalues-are-dense-in-r/107691#107691 Answer by Nikita Sidorov for Prove log of eigenvalues are dense in R? Nikita Sidorov 2012-09-20T15:15:59Z 2012-09-20T15:15:59Z <p>In addition to Doug's nice answer above: it is probably even easier to show that the set of simple Parry numbers is dense in $(1,\infty)$. More precisely, let $\beta>1$ and let $(d_n)_{n=1}^\infty$ be the <em>greedy $\beta$-expansion</em> of 1, i.e., $$1=\sum_{n=1}^\infty d_n\beta^{-n},$$ where $d_1=\lfloor \beta\rfloor, d_2=\lfloor\beta\ \text{frac}(\beta) \rfloor, d_3=\lfloor \beta\ \text{frac}(\text{frac}(\beta))\rfloor$, etc. (Here $\lfloor\cdot\rfloor$ stands for the integer part and frac$(\cdot)$ for the fractional part.)</p> <p>A number $\beta$ is called a <em>simple Parry number</em> (also known as a simple $\beta$-number) if $(d_n(\beta))_1^\infty$ has only a finite number of nonzero terms (i.e., ends with $0^\infty$). It is known that any Parry number is a Perron number; also, it is obvious that the Parry numbers are dense, since for any $\beta$ with an infinite $(d_n(\beta))_1^\infty$ we can truncate this sequence at any term and get a $d_n(\beta')$ for some simple Parry number $\beta'$. Since $(d_n(\beta))_1^\infty$ and $d_n(\beta')_1^\infty$ are close (in the topology of coordinate-wise convergence), so are $\beta$ and $\beta'$.</p> <p>For more details and some references you may read the first couple of pages of <a href="http://www-fourier.ujf-grenoble.fr/PUBLIS/publications/REF_709.pdf" rel="nofollow">this paper</a>, for instance. </p> http://mathoverflow.net/questions/100799/liouvilles-theorem-in-diophantine-approximation/100813#100813 Answer by Nikita Sidorov for Liouville's Theorem in Diophantine Approximation Nikita Sidorov 2012-06-27T21:43:14Z 2012-06-27T21:43:14Z <p>The constant $c=1/\sqrt5$ (with $n=2$) works for any $\alpha$. If $\alpha$ is not, roughly speaking, the golden ratio, then $c$ can be improved to $1/2\sqrt2$, etc. If one removes a certain infinite sequence of quadratic irrationals, one can take $c=1/3$, but this is the best you can do in a general setting. </p> <p>A nice exposition can be found in [Cassels, An Introduction to Diophantine Approximation]. You may also start with a <a href="http://en.wikipedia.org/wiki/Markov_spectrum" rel="nofollow">Wikipedia article</a>. </p> http://mathoverflow.net/questions/91544/eigenvalues-for-toral-anosov-automorphisms/91554#91554 Answer by Nikita Sidorov for Eigenvalues for toral Anosov automorphisms Nikita Sidorov 2012-03-18T17:25:06Z 2012-03-18T17:25:06Z <p>Given $k &lt; d$, one can always construct a monic polynomial irreducible over $\mathbb Q$ with exactly $k$ roots less than 1 in modulus and $d-k$ roots greater than 1 in modulus. This follows from the general construction of algebraic units, namely, each group of units of an algebraic field contains a unit with a given $k$ -- see, e.g., [Borevich and Shafarevich]. </p> <p>Then you can simply take the companion matrix of such a polynomial. </p> <p>Or do you need an explicit construction?</p> http://mathoverflow.net/questions/88621/what-is-the-adic-realization-of-a-bernoulli-shift/88626#88626 Answer by Nikita Sidorov for What is the adic realization of a Bernoulli shift ? Nikita Sidorov 2012-02-16T13:18:59Z 2012-02-16T13:18:59Z <p>The short answer is that nobody knows. The reason is that Vershik's proof uses Rokhlin's towers and is thus virtually non-constructive. </p> <p>As far as I know, the only known examples of explicit adic realizations are substitutional dynamical systems and the irrational rotations of the circle. Even for a simple ergodic rotation of the 2-torus this is an open question, let alone Bernoulli shifts. </p> http://mathoverflow.net/questions/85844/large-geodesically-convex-subsets-of-tori Large geodesically convex subsets of tori Nikita Sidorov 2012-01-16T21:00:23Z 2012-01-17T15:20:27Z <p>Let $X=\mathbb T^d=\mathbb R^d/\mathbb Z^d$ and let $E$ be a proper open subset of $X$. We say $E$ is <em>geodesically convex</em> if for any $x,y\in E$ the <strong>shortest</strong> geodesic connecting $x$ and $y$ lies in $E$. </p> <p><strong>Question.</strong> How large can the Haar/Lebesgue measure of $E$ can be?</p> <p>For example, is $d=2$, then it seems that this cannot exceed $1/2$. Say, $[0,1)\times [0,s)$ is geodesically convex if and only if $s\leq1/2$. (If $s>1/2$, then $[x,x+\delta]$ is not the shortest geodesic for any $\delta\in(1/2,s)$ and any $x\in(0,1)$.) </p> <p>Is it true for any $d\ge2$ that the measure of such an $E$ cannot exceed $1/2$? </p> http://mathoverflow.net/questions/85324/bound-for-zeros-of-a-polynomial-with-bounded-integer-coefficients/85326#85326 Answer by Nikita Sidorov for bound for zeros of a polynomial with bounded integer coefficients Nikita Sidorov 2012-01-10T11:14:50Z 2012-01-10T12:00:22Z <p>Just a brief remark that if $M=2$ and the constant term is $\pm2$, then these are called <em>Garsia numbers</em>. It is known that $z=1$ is a limit point for this set (and some computational results as well). Perhaps, you'll find the <a href="http://www.math.uwaterloo.ca/~kghare/Preprints/PDF/P36_Garsia.pdf" rel="nofollow">following recent paper</a> useful as far as the techniques are concerned. </p> http://mathoverflow.net/questions/76546/how-to-prove-the-hahn-banach-constructively/76562#76562 Answer by Nikita Sidorov for How to prove the Hahn-Banach constructively Nikita Sidorov 2011-09-27T21:34:22Z 2011-09-27T21:34:22Z <p>The idea is to show that one can extend a linear functional from an $n$-dimensional space to a space of dimension $n+1$ without increasing its norm. See, for instance, <a href="http://www.maths.manchester.ac.uk/~nikita/31002/hahn-banach.pdf" rel="nofollow">my notes</a> (Lemma E.2)</p> <p>In fact, by doing so, you can prove THBT constructively for any separable space. </p> http://mathoverflow.net/questions/63504/other-realms-for-studying-symbolic-dynamics/63522#63522 Answer by Nikita Sidorov for Other realms for studying symbolic dynamics Nikita Sidorov 2011-04-30T13:19:33Z 2011-04-30T13:19:33Z <p>I believe what you're referring to here are called <b>$\mathbb Z^d$-actions</b> (with $d=2$ in your setting). This is a pretty large area of (algebraic) dynamics with people like Klaus Schmidt, Doug Lind, Thomas Ward and Manfried Einsiedler (and many others) actively working in it.</p> <p>Perhaps, the following short survey paper by Klaus Schmidt could help you get going: <a href="http://www.mathematik.uni-bielefeld.de/~rehmann/ECM/cdrom/3ecm/pdfs/pant3/schdtk.pdf" rel="nofollow">http://www.mathematik.uni-bielefeld.de/~rehmann/ECM/cdrom/3ecm/pdfs/pant3/schdtk.pdf</a> </p> http://mathoverflow.net/questions/58113/kronecker-approximation-theorem-and-fibonacci-numbers/58119#58119 Answer by Nikita Sidorov for Kronecker Approximation theorem and Fibonacci numbers Nikita Sidorov 2011-03-11T00:35:57Z 2011-03-11T01:13:02Z <p>Well, as Gerry has pointed out, this is certainly not true for all $\alpha$. On the other hand, this is true for a.e. $\alpha$. More precisely, the sequence $2^n\alpha$ is equidistributed mod 1 for a.e. $\alpha$.</p> <p>I believe this result is due to H. Weyl and can be found in Cornfeld, Fomin and Sinai Ergodic Theory'. (I don't have it with me.) </p> <p>The same must be true for the Fibonacci sequence, I'm sure. </p> <p>So, what you probably need is for this to be true for all $\alpha$, except some small (countable?) set. After all, $\|2^n\alpha\|&lt;\varepsilon$ is indeed much weaker than equidistibution. </p> <p><strong>Update.</strong> Come to think about it, the answer is as follows: let $$\alpha = \sum_{k=1}^\infty a_k2^{-k}$$ be the binary expansion of $\alpha$. Then the sequence $2^n\alpha\bmod 1$ gets arbitrarily close to 0 if and only if the sequence $(a_k)$ has unbounded strings of 0s. In particular, any rational $\alpha$ is out of the picture, apart from the binary rationals, of course.</p> <p>All in all, your set of $\alpha$'s is indeed of full measure, but the exceptional set is of Hausdorff dimension 1, i.e., pretty big.</p> <p>For the Fibonacci sequence you'll need to replace binary expansion with the $\beta$-expansion, where $\beta=(\sqrt5-1/)2$, with the same conclusion. </p> http://mathoverflow.net/questions/55620/commuting-matrices-in-gln-z Commuting matrices in GL(n,Z) Nikita Sidorov 2011-02-16T14:02:14Z 2011-02-16T17:37:21Z <p>Suppose $M$ is a "hyperbolic" matrix in $GL(n,\mathbb Z)$, i.e., that its characteristic polynomial $p$ is irreducible over $\mathbb Z$ and has no roots of modulus 1. </p> <p>Is there a closed description of the set of elements of $GL(n,\mathbb Z)$ which commute with $M$? </p> <p>I have a vague recollection that it is somewhat similar to the Dirichlet theorem on the units of an algebraic field, but it is really vague so a reference would be appreciated. </p> <p>The case I'm most interested in is when $p$ has only one root of modulus greater than 1. Can $M$ commute with another matrix $M'$ with the same property (and $M, M'$ not being powers of the same matrix in $GL(n,\mathbb Z)$)? </p> http://mathoverflow.net/questions/53122/mathematical-urban-legends/53582#53582 Answer by Nikita Sidorov for Mathematical "urban legends" Nikita Sidorov 2011-01-28T09:03:16Z 2011-01-28T09:03:16Z <p>In a fabulous French comedy <a href="http://www.imdb.com/title/tt0071863/" rel="nofollow">La moutarde me monte au nez</a> Pierre Richard's character is a maths teacher at a local girls' college. One day he accidentally ends up in a mansion that belongs to an American movie star (Jane Birkin). Frighened of a potentially dangerous intruder, she sets her pet cheetah on him, and the academic promptly jumps on a huge chandelier. </p> <p>To prove to her that he was indeed a maths lecturer, he had to answer a few questions. First - to expand $(ax+b)^2$ (which he did) and then - to integrate $\sin(ax+b)$ (again, success). All that - hanging from a chandelier, with a cheetah pacing below. </p> <p>Only after that she allows him to climb down to the floor... and then they have a romantic dinner, needless to say. </p> http://mathoverflow.net/questions/50798/the-sum-of-integers-being-a-bijection/50819#50819 Answer by Nikita Sidorov for The sum of integers being a bijection Nikita Sidorov 2010-12-31T18:24:07Z 2010-12-31T18:54:45Z <p>If you accept that 0 is not a natural number, then there is a very simple answer to your question: take $P$ to be all numbers whose expansions base 4 contain only digits 0 and 1 and $Q$ to contain only digits 0 and 2. Then $P\cap Q={0}$, which we have boldly excluded. </p> <p>Also, both sets have the lowest possible asymptotic density of order $1/\sqrt n$, which is kinda nice. </p> http://mathoverflow.net/questions/49979/statistics-of-a-simple-markov-chain/50007#50007 Answer by Nikita Sidorov for Statistics of a simple Markov chain Nikita Sidorov 2010-12-20T22:34:21Z 2010-12-20T22:34:21Z <p>The measure $\mu_\lambda$ on an interval whose distribution is given by the random variable $$\sum_{n=1}^\infty \epsilon_n\lambda^n,$$ where the $\epsilon_n$ assume the values 0 and 1 (or $\pm1$) independently with probabilities $(p,1-p)$ is called a <em>biased Bernoulli convolution</em>. </p> <p>If one assumes $\lambda\in(0,1/2)$, then $\mu_\lambda$ is supported by a Cantor set and is consequently singular. If $\lambda=1/2$, then it is a well known singular measure on $[0,1]$. (It is invariant and ergodic under the doubling map $\tau x=2x\ \bmod 1$ and so is the Lebesgue measure.)</p> <p>The most interesting case is $\lambda\in(1/2,1)$. Here if $p\in[1/3,2/3]$, then for a.e. $\lambda$ the measure $\mu_\lambda$ is known to be equivalent to the Lebesgue measure. If $\lambda^{-1}$ is a Pisot number, then it is singular. (Which was essentially proved by Erdős in 1939.)This is almost all that is known about these measures. </p> <p>For more detail see, e.g., </p> <p><strong>B. Solomyak</strong>, <a href="http://www.math.washington.edu/~solomyak/PREPRINTS/mandel2.pdf" rel="nofollow">Notes on Bernoulli convolutions</a></p> http://mathoverflow.net/questions/48477/never-appeared-forthcoming-papers/48482#48482 Answer by Nikita Sidorov for Never appeared forthcoming papers Nikita Sidorov 2010-12-06T20:16:28Z 2010-12-06T20:16:28Z <p>A. Bertrand-Mathis, Le $\theta$-shift sans peine</p> http://mathoverflow.net/questions/47140/characteristic-polynomials-for-k-bonacci-numbers-whats-their-name/47154#47154 Answer by Nikita Sidorov for Characteristic polynomials for $K$-Bonacci numbers: what's their name? Nikita Sidorov 2010-11-23T22:13:20Z 2010-11-23T22:29:57Z <p>The dominant root of such a polynomial is often referred to as a <strong>multinacci number</strong>. These numbers are known to be <a href="http://mathworld.wolfram.com/PisotNumber.html" rel="nofollow">Pisot numbers</a> and, indeed, tend to 2. </p> http://mathoverflow.net/questions/46856/a-free-subgroup-of-gl2-z A free subgroup of GL(2,Z)? Nikita Sidorov 2010-11-21T20:21:46Z 2010-11-22T10:40:14Z <p>Is the subgroup of $GL(2,\mathbb Z)$ generated by the matrices $$\left( \begin{array}{cc} 1 &amp; 1 \\ 1 &amp; 0 \end{array} \right) \ \ \text{and} \ \ \left( \begin{array}{cc} 2 &amp; 1 \\ 1 &amp; 0 \end{array} \right)$$ <strike>free</strike> of exponential growth? More generally, how does one find all the relations between two matrices? </p> <p>I am sure this is well known, so any relevant references will be appreciated. </p> <p>My motivation comes from dynamical systems where these matrices specify two automorphisms of the 2-torus; I am interested in studying the orbits of their joint action. </p> http://mathoverflow.net/questions/45020/non-holder-continuous-devils-staircases Non-Hölder continuous devil's staircases Nikita Sidorov 2010-11-06T01:51:31Z 2010-11-06T03:20:24Z <p>Let $f:[0,1]\to[0,1]$ be a devil's staircase in the <a href="http://en.wikipedia.org/wiki/Singular_function" rel="nofollow">usual sense</a>. (That is, $f$ is continuous, non-decreasing, $f'=0$ on a set of full Lebesgue measure.) We also require the complement to the set where $f'$ vanishes to have <strong>Hausdorff dimension zero</strong>.</p> <p><em>Question</em>. Is it true that $f$ is <strong>not</strong> Hölder continuous? </p> <p>(This looks plausible, since $f$ has very little room' where it can grow so it has to grow very fast - at least, at some points.)</p> http://mathoverflow.net/questions/44192/fourier-dimension-of-the-sum-of-sets/44198#44198 Answer by Nikita Sidorov for Fourier dimension of the sum of sets Nikita Sidorov 2010-10-30T00:28:05Z 2010-10-30T00:28:05Z <p>Regarding the Hausdorff dimension of sumsets, you might want to have a look at an important paper by Peres and Shmerkin:</p> <p><a href="http://arxiv.org/abs/0705.2628" rel="nofollow">http://arxiv.org/abs/0705.2628</a></p> <p>There are plenty of useful references there, too. </p> http://mathoverflow.net/questions/42201/what-are-the-zero-entropy-invariant-measures-for-an-anosov-geodesic-flow/42216#42216 Answer by Nikita Sidorov for What are the zero entropy invariant measures for an Anosov geodesic flow? Nikita Sidorov 2010-10-14T21:33:45Z 2010-10-14T21:33:45Z <p>OK, I am sure that there must be something easier for the shifts but... In <a href="http://www.ma.umist.ac.uk/nikita/unique.ps" rel="nofollow">my paper with Glendinning</a> we briefly mention a construction of a subshift on two symbols of zero topological entropy. Namely, let $\sigma$ denote the shifted <a href="http://en.wikipedia.org/wiki/Thue%25E2%2580%2593Morse_sequence" rel="nofollow">Thue-Morse sequence</a>, i.e., $$\sigma = 1101001100101101001011001101001 \dots$$ Define $X$ to be the set of all two-sided 0-1 sequences $(x_n)$ such that $$(x_n, x_{n+1},\dots) \prec \sigma \ \ \ \forall n\in\mathbb Z$$ and $$(1-x_n, 1-x_{n+1},\dots) \prec \sigma \ \ \ \forall n\in\mathbb Z,$$ where $\prec$ stands for lexicographically smaller''. The set $X$ has zero Hausdorff dimension in the standard metric on ${0,1}^{\mathbb Z}$. In fact, the number of admissible words of length $n$ in $X$ grows approximately as $n^{\log n}$. </p> <p>To endow $X$ with a shift-invariant measure, one can take the sequence of subshifts of finite type converging to $X$ (by truncating $\sigma$) and $\mu$ to be the weak limit of the corresponding measures of maximal entropy, say. </p> http://mathoverflow.net/questions/41680/subtracting-greatest-possible-prime/41703#41703 Answer by Nikita Sidorov for subtracting greatest possible prime Nikita Sidorov 2010-10-10T18:08:32Z 2010-10-10T18:08:32Z <p>If I understand you correctly, this is effectively the question about (greedy) systems of numeration for the natural numbers. These are well understood in the case when $A$ satisfies some recurrence relation -- like the Fibonacci sequence (Zeckendorf) or the denominators $q_n$ of the CF convergents for some irrational $\alpha$ (Ostrowski). </p> <p>If $A$ grows subexponentially, this is usually not good news for the ergodic'' questions like this. </p> http://mathoverflow.net/questions/41577/how-do-i-explain-the-number-e-to-a-ten-year-old/41595#41595 Answer by Nikita Sidorov for How do I explain the number e to a ten year old? Nikita Sidorov 2010-10-09T14:06:19Z 2010-10-09T14:06:19Z <p>When I was this age, I loved the factorials. So why not try to explain it via the Stirling formula? That $n!$ is rather close'' to $n^n$ but you need to adjust it a bit via division by $e^n$. </p> <p>Also, if he knows logarithms, the Prime Number Theorem is a good example why $e$ is the correct'' base of logarithms. </p> http://mathoverflow.net/questions/40103/a-question-regarding-polynomials-whose-roots-satisfy-certain-algebraic-relation/40116#40116 Answer by Nikita Sidorov for A question regarding polynomials whose roots satisfy certain algebraic relation Nikita Sidorov 2010-09-27T09:39:04Z 2010-09-27T09:39:04Z <p>This is rather vague but regarding relations between roots of a polynomial you may try some of Chris Smyth's papers as a starting point. For instance, this one:</p> <p>C. J. Smyth, Conjugate algebraic numbers on conics, Acta Arith. 40 (1982), 333–346.</p> http://mathoverflow.net/questions/39159/isodiametric-hull Isodiametric hull Nikita Sidorov 2010-09-17T22:55:45Z 2010-09-18T03:49:53Z <p>Let A be a convex compact set in the plane (with a piecewise smooth boundary, say). We want to inflate' it in such a way that the diameter does not increase. </p> <p>More accurately, we are looking for all sets C such that </p> <p>a) A is a subset of C; b) diam(A)=diam(C)</p> <p>Let now B is the largest possible set C which satisfies these two properties. </p> <p>By largest' I mean either that it m(B) = max m(C), where m is the Lebesgue measure; or that B actually contains any C with these properties. Let us call B the <em>isodiametric hull</em> of A.</p> <p>The simplest example of A is of course the square: here B is the superscribed disc, and it is the isodiametric hull of A in the strong sense.</p> <p>Another example is the equilateral triangle, for which B is the <a href="http://mathworld.wolfram.com/ReuleauxTriangle.html" rel="nofollow">Reuleaux triangle</a>. Similarly, for any regular 2n-gon we have the disc, and for any regular (2n+1)-gon its isodiametric hull is a Reuleaux polygon. </p> <p>The first non-trivial example that comes to mind is an isosceles triangle that isn't equilateral. It is clear that the hull is always a set of constant diameter but how does one actually obtain it? It seems that its boundary - a curve of constant width - is not a finite union of circular arcs.</p> <p>I wonder if all this is well known (being such a natural question!). In particular, does the isodiametric hull of a set always exists in the strong sense? </p> <p><strong>Added:</strong> of course, if there is no IDH in the strong sense, B may not be unique. Its area <em>is</em> unique, though. How does one find it? </p> http://mathoverflow.net/questions/35986/measure-0-sets-on-the-line-with-hausdorff-dimension-1/36012#36012 Answer by Nikita Sidorov for Measure 0 sets on the line with Hausdorff dimension 1 Nikita Sidorov 2010-08-18T19:14:03Z 2010-08-18T19:14:03Z <p>It is a very common phenomenon in ergodic theory when the set of points which do <strong>not</strong> satisfy the Birkhoff ergodic theorem (i.e., a set of zero measure) has full Hausdorff dimension.</p> <p>See, for instance, <a href="http://www.math.psu.edu/pesin/papers_www/birk.pdf" rel="nofollow">http://www.math.psu.edu/pesin/papers_www/birk.pdf</a></p> http://mathoverflow.net/questions/121915/linear-numeration-systems/121918#121918 Comment by Nikita Sidorov Nikita Sidorov 2013-02-16T22:41:09Z 2013-02-16T22:41:09Z No problem. Hope it'll help. http://mathoverflow.net/questions/119241/harmonacci-recurrence-and-identities-for-pi Comment by Nikita Sidorov Nikita Sidorov 2013-01-18T08:59:18Z 2013-01-18T08:59:18Z So, in terms of continued fractions, we have $a_{n+1}=[a_{n-1}; a_{n-2},\dots, a_2, a_1, a_0, a_1]$. http://mathoverflow.net/questions/119216/fibonnaci-sequence-and-series-limits Comment by Nikita Sidorov Nikita Sidorov 2013-01-18T00:06:09Z 2013-01-18T00:06:09Z Try Binet's formula (<a href="http://mathworld.wolfram.com/BinetsFibonacciNumberFormula.html" rel="nofollow">mathworld.wolfram.com/&hellip;</a>). Btw, I don't understand how both series can be equal to 1 when one is clearly larger? Something's not right. http://mathoverflow.net/questions/118811/hutchinsons-formula-for-asymptotically-homogeneous-cantor-sets/118818#118818 Comment by Nikita Sidorov Nikita Sidorov 2013-01-13T17:38:55Z 2013-01-13T17:38:55Z Thanks. Yes, introducing measures looks like a good approach here. http://mathoverflow.net/questions/118811/hutchinsons-formula-for-asymptotically-homogeneous-cantor-sets/118826#118826 Comment by Nikita Sidorov Nikita Sidorov 2013-01-13T17:37:57Z 2013-01-13T17:37:57Z This paper is about lacunary sets which are generalizations of self-similar sets. I am interested in perturbed self-similar sets, so I don't quite see how his result could possibly help me. http://mathoverflow.net/questions/118320/leaving-academia Comment by Nikita Sidorov Nikita Sidorov 2013-01-08T01:19:23Z 2013-01-08T01:19:23Z A school teacher? http://mathoverflow.net/questions/117633/a-question-about-the-limit-of-a-sequence-of-pointwise-convergent-analytic-funtion Comment by Nikita Sidorov Nikita Sidorov 2012-12-30T14:47:53Z 2012-12-30T14:47:53Z Of course, not. The set of analytic functions is dense in $C(\Delta)$, so for any continuous non-analytic function ($f(z)=\overline z$, say) there exists a sequence of analytic functions which converges to $f$ uniformly and hence, pointwise. http://mathoverflow.net/questions/117571/evaluating-publication-lists-how-to-use-the-new-internet-tools Comment by Nikita Sidorov Nikita Sidorov 2012-12-30T00:58:15Z 2012-12-30T00:58:15Z Yes, MathSciNet is less generous and indeed more accurate. However, it is too old-fashioned in the sense that it only counts <i>published</i> papers. So, if you upload a preprint and then it takes a few years for this paper to appear, say (which often happens given low speed of refereeing and massive backlogs in some journals), then all the citations you preprint gathers during this unfortunate limbo will be ignored by MathSciNet - but not by Google Scholar. Also, there are borderline mathematical' or just non-mathematical journals MathSciNet doesn't list, which may affect your H-index as well. http://mathoverflow.net/questions/115509/when-we-use-bernstein-polynomials-in-application/115510#115510 Comment by Nikita Sidorov Nikita Sidorov 2012-12-06T10:16:46Z 2012-12-06T10:16:46Z Margaret: Thanks. Yes, this was the reason, and Bernstein saw it mainly as a result in Probability. http://mathoverflow.net/questions/115509/when-we-use-bernstein-polynomials-in-application/115510#115510 Comment by Nikita Sidorov Nikita Sidorov 2012-12-06T00:16:28Z 2012-12-06T00:16:28Z Fredrik: not really. Take, for instance, $f(x)=x^2$; then $B_n(f;x)=x^2+x(1-x)/n$, so $B_n(f;x)-f(x)\asymp 1/n$. http://mathoverflow.net/questions/115509/when-we-use-bernstein-polynomials-in-application/115510#115510 Comment by Nikita Sidorov Nikita Sidorov 2012-12-05T23:55:53Z 2012-12-05T23:55:53Z Thanks! What's a Tonelli polynomial? I have to admit I've never heard of them and Google is no help either... http://mathoverflow.net/questions/112419/an-identity-which-involves-eulers-totient-function Comment by Nikita Sidorov Nikita Sidorov 2012-11-15T12:51:03Z 2012-11-15T12:51:03Z Sinai - No, I haven't. I guess I'm not sure how... fedja - Thanks, but this constant is not that important. It's just a part of a technical argument, nothing more. http://mathoverflow.net/questions/112050/logistic-map-periodic-point/112054#112054 Comment by Nikita Sidorov Nikita Sidorov 2012-11-11T13:13:42Z 2012-11-11T13:13:42Z Right. Note however that for this word we know for sure that it is of period $m$ and not of any $k&lt;m$. http://mathoverflow.net/questions/112050/logistic-map-periodic-point/112054#112054 Comment by Nikita Sidorov Nikita Sidorov 2012-11-11T12:56:53Z 2012-11-11T12:56:53Z The word $(10^{m-1})^\infty$ is $m$-periodic for $\sigma$, whence $x=0.5/(1-2^{-m})$ is $m$-periodic for $T$. If your formula is correct I haven't checked it), then $\sin^2(\pi/(1-2^{-m}))$ is your guy. http://mathoverflow.net/questions/111454/pure-greedy-algorithm Comment by Nikita Sidorov Nikita Sidorov 2012-11-04T15:27:10Z 2012-11-04T15:27:10Z What's $G_m(f,D)$?