User julian rosen - MathOverflow

Answer by Julian Rosen for Reals with integer powers bounded away from integers?

2013-04-10T01:59:34Z

We should be able to construct such a real number $x$. Let $\epsilon\in(0,1/2)$ be fixed. If $S\subset \mathbb{R}_{>0}$ and $r>0$, we write $S^r$ for the set of positive $r$-th powers of elements of $S$. Recursively define a sequence of intervals $I_n$ of the form $[N_n+\epsilon,N_n+1-\epsilon]$ where the $N_n$ are positive integers. Let $I_1=[N_1+\epsilon,N_1+1-\epsilon]$ where $N_1$ chosen large enough that $N_1(1-2\epsilon)>2$. Once $I_n$ has been defined, $I_n^{(n+1)/n}$ has length at least $2$, so we can find an integer $I_{n+1}$ so that $[I_{n+1}+\epsilon,I_{n+1}-\epsilon]\subset I_n^{(n+1)/n}$. Then the intervals $I_n^{1/n}$ are nested and have diameter going to 0, and we choose $x$ to be the unique element in the intersection of $I_n^{1/n}$. This $x$ is constructed in such a way that $x^n\in[I_n+\epsilon,I_n+1-\epsilon]$ for all $n$.

Answer by Julian Rosen for This might be a trivial question on Hurwitz's zeta function.

2013-04-08T17:11:18Z

Looks like there's a typo in your post. Andrews has $\left(1-\frac{sx}{n}\right)$ where you have $\left(1-\frac{x}{n}\right)$ in the rightmost term.

If you expand $\left(1+\frac{x}{n}\right)^{-s}$ as a power series in $\frac{x}{n}$, the series begins $1-s\frac{x}{n}+O(x^2/n^2)$. The first two terms are cancelled by the $-\left(1-\frac{sx}{n}\right)$, so the quantity between the square brackets is $O(n^{-2})$ as $n\to\infty$. This means the sum converges for Re$(s+2)>1$.

What is the relationship between these two notions of "period"?

2013-04-07T18:44:40Z

The motivation for this question is to understand a recent theorem of Francis Brown which implies that all periods of mixed Tate motives over $\mathbb{Z}$ lie in $\mathcal{Z}[\frac{1}{2\pi i}]$, where $\mathcal{Z}$ is the $\mathbb{Q}$-span of the set of multiple zeta values (of positive integer arguments). My picture of mixed Tate motives is not very clear, and I would like to be able to relate their periods to something I understand better.

There is a survey article of Kontsevich and Zagier which defines a period as a complex number whose real and imaginary parts are given by convergent integrals of rational functions with rational coefficients, over domains in $\mathbb{R}^n$ cut out by finitely many polynomial inequalities with rational coefficients.

What is the relationship between the set of periods of mixed Tate motives over $\mathbb{Z}$ and the set of periods in the sense of Kontsevich/Zagier? Does one of these sets contain the other?

I would be interested to see examples of periods of one kind which are not periods of the other.

Answer by Julian Rosen for A "known" tangent half-angle formula?

2012-12-13T04:30:53Z

Here's one way to derive the identity:

Suppose $\tan(\gamma)=\frac{\sin(\alpha)\sin(\beta)}{\cos(\alpha)+\cos(\beta)}$. Multiplying this equation through by $\cos(\gamma)$ gives an expression for $\sin(\gamma)$ in terms of $\cos(\gamma)$ and functions of $\alpha$, $\beta$. We now take the Pythagorean identity $\cos^2(\gamma)+\sin^2(\gamma)=1$, and replace $\sin(\gamma)$ with the expression derived above, getting $$ \left(\frac{1+\cos(\alpha)\cos(\beta)}{\cos(\alpha)+\cos(\beta)}\cos(\gamma)\right)^2=1 $$ Some simplification was done, but the only trig identity used was the Pythagorean identity.

This yields $\cos(\gamma)=\pm\frac{\cos(\alpha)+\cos(\beta)}{1+\cos(\alpha)\cos(\beta)}$, $\sin(\gamma)=\pm\frac{\sin(\alpha)\sin(\beta)}{1+\cos(\alpha)\cos(\beta)}$. We assume $\alpha,\beta,\gamma\in(0,\pi)$, so this forces the $\pm$ signs to be $+$ (it seems like this identity fails when $\alpha,\beta,\gamma<0$).

We have a tangent half-angle formula $\tan(\gamma/2)=\frac{\sin(\gamma)}{1+\cos(\gamma)}$. Combining with the formulas for $\sin(\gamma)$, $\cos(\gamma)$ gives $$ \tan(\gamma/2)=\frac{\sin(\alpha)\sin(\beta)}{(1+\cos(\alpha))(1+\cos(\beta))} $$ Using the tangent half-angle formula again gives the desired identity.

Answer by Julian Rosen for Common Zeros of Holomorphic Functions

2012-11-17T18:24:44Z

In general, $S:={(z,w,u):f=g}$ will have codimension 1. If we arrange for $S$ to be contained entirely in the fiber over $u_0$, then the desired functions $z(u)$, $w(u)$ won't exist.

In particular, if fix $z_0$, $w_0$, find $h(z,w)$ holomorphic with $h(z_0,w_0)=0$, then we can take $u_0=0$, $f(z,w,u)=h(z,w)+u$, $g(z,w,u)=h(z,w)+2u$ (we will need some condition on $h$ so that $f$ and $g$ are non-degenerate).

Answer by Julian Rosen for normalization of a bijection

2012-10-28T20:59:39Z

We should be able to construct a counterexample as follows:

Let $Y$ be an affine curve which is smooth away from a single node. We obtain $X$ from the normalization of $Y$ by removing one of the points mapping to the node of $Y$. The map from $X$ to $Y$ is a bijection, but the map $X^\nu\to Y^\nu$ is not an isomorphism.

Are there infinitely many non-Wolstenholme primes?

2012-01-11T02:38:31Z

It is known that for $p$ a prime, the following conditions are equivalent:

${2p-1\choose p-1}\equiv 1\mod{p^4}$
$p^3$ divides the numerator of $\sum_{n=1}^{p-1}\frac{1}{n}$
$p$ divides the numerator of the Bernoulli number $B_{p-3}$ (i.e., $(p,p-3)$ is an irregular pair)

A prime satisfying these conditions is called a Wolstenholme prime; only two are known. It is conjectured that there are infinitely many, and that their density is 0.

Are there known to be infinitely-many primes which are NOT Wolstenholme primes?

This would follow from the conjecture that their are infinitely many regular primes, but it seems that the property of not being a Wolstenholme prime is much weaker than being regular.

Answer by Julian Rosen for Flipping coins on a budget

2012-09-18T04:31:39Z

I want to propose a strategy in the limiting case $n=\infty$. Maybe this is better described as a limit of strategies, since I will allow a sequence of coin flips that are each assigned probability $\epsilon$ of success (where $\epsilon$ is infinitessimal). The total amount of probability we will "spend" before the next head appears will then be exponentially distributed, with mean 1.

I will denote by $f_k(x)$ the probability that my strategy results in success if we still need $k$ heads, and have $x$ probability remaining in our "budget." Here is how the strategy works: If $x\geq k$, we assign probability 1 to the next $k$ flips. This results in $f_k(x)=1$.

If $x\in (k-1,k)$, then we assign the next flip probability $x-(k-1)$. If this flip lands heads, we will win with probabilty 1. If the flip lands tails, we will win with probabilty $f_k(k-1)$. It follows that $$ f_k(x)=(x-(k-1))+(k-x)f_k(k-1) $$

Finally, if $x\leq k-1$, then we will assign probability $\epsilon$ to each subsequent flip, until we see a heads. This gives $$ f_k(x)=\int_0^x e^{-t}f_{k-1}(x-t)\,dt $$

We can recursively compute $f_k(x)$ for any $k$. Each $f_k$ is a continuous, piecewise-analytic function. The first few values (computed with the help of Mathematica; I hope they're correct) are: $$ f_1(x)=\begin{cases} x\text{ if }0\leq x\leq 1\newline 1\text{ if }x>1 \end{cases} $$

$$ f_2(x)=\begin{cases} -1+x+e^{-x}\text{ if }0\leq x\leq1\newline -1+\frac{2}{e}+(1-\frac{1}{e})t\text{ if }1\leq x\leq 2\newline 1\text{ if }x>2 \end{cases} $$

$$ f_3(x)=\begin{cases} -2+x+(x+2)e^{-x}\text{ if }0\leq x\leq 1\newline e^{-x}+\frac{3}{e}-2-\frac{x}{e}+x\text{ if }1\leq x\leq 2\newline -2+x+\frac{(1+e)(3-x)}{e^2}\text{ if }2\leq x\leq 3\newline 1\text{ if }x\geq 3 \end{cases} $$

While I don't have a proof this strategy is optimal, I've got a heuristic argument that assigning probability $\epsilon$ to each flip is a good idea. If our budget is $x$, then whatever our strategy, the expected number of heads we will have seen when we exhaust our budget is $x$. If the desired number of heads is much larger than $x$, we will need to make the variance in the number of heads large. If we assign probabilities $p_1,p_2,\ldots$ to the coin flips (with $p_1+p_2+\ldots=x$), then the variance in the number of heads is $\sum p_i(1-p_i)$, which is bounded above by $x$. We can make the variance arbitrarily close to $x$ by taking each $p_i$ as small as possible.

The argument is a little different if the next head that appears could cause our remaining budget to be larger than the number of additional heads we need to win.

Answer by Julian Rosen for Lifting identities of formal power series

2012-08-31T22:37:31Z

Take $g(x)=x-x^2\in\mathbb{Z}[[x]]$. There is an $f(x)\in x\mathbb{Z}[[x]]$ which is an inverse to $g(x)$ under composition (this is true because $g(x)$ has constant term 0 and linear term a unit; we could also write down $f(x)$ explicitly). We have $g(1)=0$, and $f(0)=0$ (both expressions converge in any topological ring, as they have only finitely many non-zero terms), so the expression $f(g(1))$ converges, and equals 0. However, $h(x):=f(g(x))=x$, and $h(1)=1\neq 0$. This construction works in every topological ring, so the statement "$h=f\circ g\Rightarrow h(r)=f(g(r))$ for all $r\in R$ for which both sides converge" does not hold in ANY non-zero topological ring $R$.

Answer by Julian Rosen for Is there a simple relation between the entropy of a matrix and its characteristic polynomial?

2012-07-18T19:21:33Z

I'm not sure if this qualifies as simple (or if this is helpful at all), but we have $$ \frac{\phi'_M(t)}{\phi_M(t)}=\sum_n\frac{1}{t-\lambda_n} $$

Using the residue theorem, we can write $$ H[M]=\frac{-1}{2\pi i}\oint\frac{\phi'_M(z)}{\phi_M(z)}z\log(z)\,dz $$ where the integral is taken over a closed contour containing all of the eigenvalues of $M$ (I guess we're either working on the Riemann surface of $\log(z)$, or we chose a branch of $\log(z)$).

Answer by Julian Rosen for Evaluating the integral $\int_{1}^{\infty}\frac{\{u\}}{u^{2}}\left(\log u\right)^{k}du.$

2011-05-11T03:29:14Z

Write $a_k$ for your integral. If we define $g(s)=\zeta(s)-\frac{1}{s-1}$, then $\left(\frac{d}{ds}\right)^n|_{s=1}g(s)=(-1)^n\gamma_n$. Your observation can be written $a_k=(-1)^k\left(\frac{d}{ds}\right)^k|_{s=1}\left(\frac{1}{s}-\frac{1}{s}g(s)\right)$. The derivative can be computed directly to give $a_k=k!-\sum_{n=0}^k \frac{k!}{n!}\gamma_n$.

Answer by Julian Rosen for Evaluating the integral $\int_0^\infty \frac{\psi(x)-x}{x^2}dx.$

2011-05-10T21:21:00Z

Consider $f(s):=\int_1^{\infty}\frac{\psi(x)-x}{x^s}dx$, which converges for $Re(s)\geq2$. For $Re(s)>2$, we can separate the numerator and integrate by parts (using the Riemann-Stieltjes integral, for convenience) to get $f(s)=\frac{1}{s-1}\int_1^{\infty}\frac{1}{x^{s-1}}d\psi(x)-\frac{1}{s-2}$. Now, $\frac{\zeta'}{\zeta}(s)=-\sum\frac{\Lambda(n)}{n^s}=-\int_1^{\infty}\frac{1}{x^s}d\psi(x)$, so we can write $f(s)=\frac{-1}{s-1}\frac{\zeta'}{\zeta}(s-1)-\frac{1}{s-2}$. $\frac{\zeta'}{\zeta}(s-1)$ has a Laurent expansion at $s=2$ of the form $\frac{\zeta'}{\zeta}(s-1)=\frac{-1}{s-2}+\gamma+O(s-2)$, so that $f(s)=\frac{-1-\gamma}{s-1}+O(s-2)$. This holds for $Re(s)>2$, but if we let $s$ decrease to 2 and use the dominated convergence theorem, we get that the value of the integral is $-\gamma-1$.

Hmm...this isn't quite the same as either value you gave. Maybe I made a mistake somewhere.

Connection between isomorphisms of algebraic topology and class field theory

2010-11-26T07:11:34Z

I am considering the following two isomorphisms:

First, if $X$ is a reasonably nice topological space, then $X$ has a normal covering space which is maximal with respect to the property of having an abelian group of deck transformations. The group of deck transformations of this covering space is isomorphic to the abelianizatin of the fundamental group of $X$, which can be identified with the singular homology group $H_1(X,\mathbb{Z})$.

Second, if $K$ is a number field, class field theory gives an isomorphism between the Galois group of the maximal abelian unramified extension of $K$ (the Hilbert class field) and the ideal class group of $K$. The ideal class group can be identified with the sheaf cohomology group $H^1(\mathrm{Spec}(\mathcal{O}_K),\mathbf{G}_m)$.

Given the apparent similarity between these two theorems, is there some more general theorem which implies both of these results as special cases?

Comment by Julian Rosen

2012-12-05T23:12:39Z

If $W$ is closed under multiplication and contains any invertible element $\varphi$, then $1\in W$: 1 can be written as a linear combination of positive powers of $\varphi$, which is follows from the fact that $\varphi$ satisfies a polynomial with non-zero constant term (namely its characteristic polynomial)

Comment by Julian Rosen

2012-11-17T17:37:23Z

Taking $f=z+w+u$, $g=z+w+2u$, $z_0=w_0=u_0=0$ gives a counterexample. Is there an extra hypothesis that should be added?

Comment by Julian Rosen

2012-08-19T00:22:36Z

In general, there are many ways to extend the valuation $v_p$ to the ring of integers of a number fields (the extension correspond to primes lying over $p$). I believe your statement that a prime $p\in\mathbb{Z}$ is prime-like in the ring of algebraic integers in a number field is false. Consider, for example, $p=5$, $R=\mathbb{Z}[i]$, $a=(2+i)^2$, $b=(2-i)^2$. We have $ab=25$, so $5^2|ab$, but $5\nmid a$, $5\nmid b$.

Comment by Julian Rosen

2012-01-11T16:18:41Z

If $p$ divides the numerator of $\frac{B_M}{M}$, then by construction $p$ cannot be irregular and non-Wolstenholme. $p$ is irregular, so $p$ must be Wolstenholme. If $0<s<p-1$ with $s\equiv M \mod{(p-1)}$, then $p|B_s$. Since $p$ is Wolstenholme, we would want to say that $s=p-3$, so that $p-1|M+2$ (this might give a bound on the numerator). However, $p$ could be Wolstenholme and also divide $B_s$ for some $s<p-3$. It seems like at best, this argument will only prove that there are infinitely many $p$ which divide $B_k$ for some $k<p-3$. Did I understand your argument correctly?

Comment by Julian Rosen

2011-05-02T08:05:33Z

In a Lie group $G$, every $g$ sufficiently close to the identity has the property that, for all positive integers $n$, there is an $h$ with $h^n=g$. The subgroup generated by all such $g$ must then contain the connected component of the identity, and so has at most countable index in $G$. For example, this shows that $\mathbb{Z}_p$ (forgetting the topology) cannot be given the structure of a Lie group. (This example is maybe not so interesting, as every torsion-free abelian Lie group is $\mathbb{R}^n$ for some $n$, but oh well)

Comment by Julian Rosen

2010-12-19T21:45:42Z

Are you claiming that $e^N E_{-N}(N) = o(N^{-1/2})$ as $N\to\infty$? Is this clear?

Comment by Julian Rosen

2010-12-15T07:49:44Z

I am a little confused by the statement of this excercise. Are we considering a single function $f(n):=1/n\log{n}\ldots\log_k{n}$, where $k$ is maximal such that $\log_k{n}\geq 1$? This question (in a slightly different form) appeared as A4 on the 2008 Putnam.