164

Note that if $\pi$ were rational (with even numerator), then $\sin(n)$ would equal $1$ periodically, so the series would diverge. Similarly if $\pi$ were a sufficiently strong Liouville number. Thus, to establish convergence, one must use some quantitative measure of the irrationality of $\pi$.
It is known that the irrationality measure $\mu$ of $\pi$ is ...

111

This problem turned out to be much more interesting than I originally
thought. Let me give my solution, which seems to be slightly different from
(but essentially the same as) the solution in the paper by Bremner and
MacLeod (see Allan MacLeod's answer).
Theorem. Let $a,b,c$ be positive integers. Then
$\frac{a}{b+c} + \frac{b}{c+a} + \frac{c}{a+b}$ can ...

105

I'm not a number theorist, but FWIW: I would talk, not so much about Gödel's Theorem itself, but about the wider phenomenon that Gödel's Theorem was pointing to, although the terminology didn't yet exist when the theorem was published in 1931. Namely, number theory is already a universal computer. Or more precisely: when we ask whether a given ...

96

You can evaluate this by using generating functions and integrating. The answer is $-\pi^2/16 = -0.61685 \ldots$ which is pretty close to $(1-\sqrt{5})/2=-0.61803\ldots$.
Here's a sketch: the sum is
$$
\sum_{k=1}^{\infty} \frac{(-1)^k}{k} \int_0^1 (1+x^2+ \ldots +x^{2k-2}) dx = \int_0^1 \sum_{j=0}^{\infty} x^{2j} \sum_{k=j+1}^{\infty} \frac{(-1)^k}{k} ...

95

First, Mazur's paper is arguably the first paper where the new ideas (and language) of the Grothendieck revolution in algebraic geometry were fully embraced and crucially used in pure number theory. Here are several notable examples: Mazur makes crucial use of the theory of finite flat group schemes to understand the behavior of the $p$-adic Tate modules of ...

94

My understanding of this, which is essentially cobbled together from the various news accounts, is as follows:
Let $\pi(x;q,a)$ denote the number of primes less than $x$ congruent to $a\bmod q$, and $\pi(x)$ the number of primes less than $x$. $\phi(n)$ is the number of positive integers less than or equal to n that are relatively prime to n, and is usually ...

87

We have
$$\frac{1}{\zeta(s)} = \sum_{n=1}^{\infty} \frac{\mu(n)}{n^s} \quad \mbox{for} \ Re(s)>1.$$
Taking the derivative with respect to $s$, we get the following
$$- \frac{\zeta'(s)}{\zeta(s)^2} = - \sum_{n=1}^{\infty} (\log n) \frac{\mu(n)}{n^s} \quad \mbox{for} \ Re(s)>1.$$
If we plug in $s=0$ to the second equation (which is not allowed, because ...

answered Mar 12 '13 at 15:14

David E Speyer

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85

85

### If I exchange infinitely many digits of $\pi$ and $e$, are the two resulting numbers transcendental?

If, as is commonly believed, $\pi$ and $e$ are normal numbers, then one can use a counting argument (or entropy argument) to show that no possible transposition of $\pi$ and $e$ can produce a rational number. Indeed, if there was a rational number that could be made this way, then its digit expansion would eventually be periodic with some period $q$; by ...

answered Mar 23 '17 at 21:23

Terry Tao

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The conjecture is true, in fact the equation has no solution in distinct positive real numbers. To see this, let us write the equation in the more symmetric form
$$ x^y y^z z^x = x^z y^x z^y. \tag{$\ast$}$$
We get the same equation after interchanging $x$ and $y$, or $y$ and $z$, i.e., after permuting the variables arbitrarily. Hence we can assume without ...

84

Let me answer your question "where do these involution come from" with an elementary geometric explanation. You can skip ahead to the pictures, which are somewhat self-explanatory, I hope.
The elements in the the $S$, i.e. triplets $(x,y,z)$ such that $x^2+4yz=p$, can be visualized as a square of side length $x$ together with $4$ rectangles of size $y\...

84

September 2018: There has been a back-and-forth in 2018 between Shinichi Mochizuki and Yuichiro Hoshi (MoHo) in Kyoto, and Peter Scholze and Jakob Stix (ScSt) in Germany, with ScSt spending a week in Kyoto in March 2018 to confer with MoHo.
ScSt have released a report saying they believe there is a gap in the proof of Corollary 3.12 in IUTT-3, and ...

77

Sorry I didn’t see this earlier. My memory is vague, and probably colored by subsequent events and results, but here’s how I recall things happening.
Since I had read and enjoyed Lazard’s paper on one-dimensional formal group (laws), which dealt with the case of a base field of characteristic $p$, I decided to look at formal groups over $p$-adic rings. For ...

77

It is not true anymore that a proof of this conjecture would lead to significant simplifications. Peterfalvi proved in 1984 a weaker version of this conjecture, which suffices to get rid of the chapter involving generators and relations in the original paper. Bender and Glauberman reproduce this argument in their book in one of the appendices, and it takes ...

answered Sep 13 '17 at 12:43

Jan-Christoph Schlage-Puchta

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75

Philosophically, there is essentially only one way to prove that a number is irrational/transcendental, which is to use the fact that there is no integer between 0 and 1. That is, one assumes that the number in question is rational/algebraic, and constructs some quantity that can be shown to be bounded away from 0, less than 1, and also an integer. To get ...

answered May 3 '13 at 18:21

Timothy Chow

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Jonas Meyer's comment:
Quote from arxiv.org/abs/0902.3961, Bjorn Poonen, Feb. 2009: "Harvey Friedman asked whether there exists a polynomial $f(x,y)\in Q[x,y]$ such that the induced map $Q × Q\to Q$ is injective. Heuristics suggest that most sufficiently complicated polynomials should do the trick. Don Zagier has speculated that a polynomial as simple as $x^...

73

This must have been Heinrich Kornblum (1890-1914).
[note by E. Landau in German, my translation]
$^1$ The author, born in Wohlau on August 23, 1890, had before the war
independently made the discovery that Dirichlet's classic proof of the theorem
of prime numbers in an arithmetic progression (along with the later elementary
reasons for the non-vanishing of ...

answered Nov 4 '20 at 15:02

Carlo Beenakker

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70

I don't know about analysis in general, but I think it's definitely fair to say "often" in functional analysis. My feeling is that we have a solid, thorough, elegant body of theory which usually leads to positive solutions rather quickly, when they exist. (The Kadison-Singer problem is a recent exception which required radically new tools for a positive ...

68

I have it. Mazur gave me a xerox copy off his shelf when I asked him (in grad school) if a copy exists. It's 56 pages and the first sentence is: L'objet de ce rapport est de construire la série L p-adique de Kubota-Leopold et d'établir quelques propriétés fondamentales. It was in my office and I was going to try scanning it this evening, but literally as ...

67

This is Jim (Condict) Grace, the author. I just saw this post. I'm sorry to hear that Middlebury may have lost their copy. I'm not sure where mine is, but I'll keep an eye out for it. (It may be at a family house 1,000 miles away but I'll look for it at Christmas when I visit.) If I find it, I'll scan and post it somewhere, and I'll comment again here with a ...

67

Tonight I read here [the answer by esg to another your question] that $\frac1{2\pi}\int_{-\pi}^\pi e^{-ik t}(1+e^{it})^ndt=\binom{n}{k}$, which is, well, obvious at least when both $n$ and $k$ are positive integers: just expand the binomial $(1+e^{it})^n$ and integrate. Denoting $\alpha=k/n$ we may rewrite this as $\frac1{2\pi}\int_{-\pi}^\pi (f(t))^n dt=\...

answered Nov 19 '16 at 1:18

Fedor Petrov

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66

These questions on the spacings between primes are expected to be true, but are far from being proved. They are not directly related to RH, but seem to encode other relations among zeros. The conjecture (1) follows from the Hardy-Littlewood prime $k$-tuples conjectures; this was established by Gallagher. More precise versions of Conjecture 2 were ...

66

Here is a proof of the conjecture. We shall use Hecke's theorem that the angles of the lattice points $(x_p,y_p)$ are asymptotically equidistributed in $[\pi/4,\pi/2]$, cf. this MO post.
Let $t_p\in[\pi/4,\pi/2]$ be the angle of the lattice point $(x_p,y_p)$. Let us divide the interval $[\pi/4,\pi/2]$ into $R$ subintervals of equal length, where $R$ is ...

65

"Numerology" such as you've observed is explained in the paper
Gross, B.H., and Zagier, D.: On singular moduli, J. reine angew. Math. 355 (1985), 191$-$220. MR772491 (86j:11041)
which gives more generally the factorizations of the constant terms
of the minimal polynomials of $j(\tau) - j(\tau')$ where $\tau,\tau'$
are quadratic imaginaries not ...

answered Jan 16 '14 at 6:31

Noam D. Elkies

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65

This exact problem is the subject of the paper "An Unusual Cubic Representation Problem" by Andrew Bremner (ASU) and myself. It was published in Volume 43 (2014) of Annales Mathematicae et Informaticae, pages 29-41.
It is proven that strictly positive solutions never exist for $n$ odd. They sometimes do not exist for $n$ even, and, even if they do, they can ...

63

Here is a completely different kind of answer to this question.
A perfectoid space is a term of type PerfectoidSpace in the Lean theorem prover.
Here's a quote from the source code:
structure perfectoid_ring (R : Type) [Huber_ring R] extends Tate_ring R : Prop :=
(complete : is_complete_hausdorff R)
(uniform : is_uniform R)
(ramified : ∃ ϖ : ...

answered Jul 31 '18 at 11:53

Kevin Buzzard

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61

First, recall the step's of Weil's proof, other than defining the surface:
Develop an intersection theory of curves on surfaces.
Show that the intersection of two specially chosen curves is equal to a coefficient of the zeta function.
Prove the Hodge index theorem using the Riemann-Roch theorem for surfaces (or is there another proof?).
By playing around ...

60

The limiting ratio is $\frac{1}{\zeta(2)} = \frac{6}{\pi^2}= 0.6079271018540266286632767792\ldots$
The question is easily seen to be equivalent to "What is the probability that two integers are relatively prime?" This old chestnut of elementary number theory has been addressed before on this site: see here. The idea is incredibly simple and appealing: we ...

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