## Recent developments in Twin Primes, Goldbach, and Open Access Tuesday, May 21 2013

It has been a busy two weeks all over the math community. Well, at least it seemed so to me. Some of my friends have defended their theses and need only to walk to receive their PhDs; I completed my topics examination, Brown’s take on an oral examination; and I’ve given a trio of math talks.

Meanwhile, there have been developments in a relative of the Twin Primes conjecture, the Goldbach conjecture, and Open Access math journals.

## 1. Twin Primes Conjecture

The Twin Primes Conjecture states that there are infinitely many primes $p$ such that $p+2$ is also a prime, and falls in the the more general Polignac’s Conjecture, which says that for any even $n$, there are infinitely many prime $p$ such that $p+n$ is also prime. This is another one of those problems that is easy to state but seems tremendously hard to solve. But recently, Dr. Yitang Zhang of the University of New Hampshire has submitted a paper to the Annals of Mathematics (one of the most respected and prestigious journals in the field). The paper is reputedly extremely clear (in contrast to other recent monumental papers in number theory, i.e. the phenomenally technical papers of Mochizuki on the ABC conjecture), and the word on the street is that it went through the entire review process in less than one month. At this time, there is no publicly available preprint, so I have not had a chance to look at the paper. But word is spreading that credible experts have already carefully reviewed the paper and found no serious flaws.

Dr. Zhang’s paper proves that there are infinitely many primes that have a corresponding prime at most $70000000$ or so away. And thus in particular there is at least one number $k$ such that there are infinitely many primes such that both $p$ and $p+k$ are prime. I did not think that this was within the reach of current techniques. But it seems that Dr. Zhang built on top of the work of Goldston, Pintz, and Yildirim to get his result. Further, it seems that optimization of the result will occur and the difference will be brought way down from $70000000$. However, as indicated by Mark Lewko on MathOverflow, this proof will probably not extend naturally to a proof of the Twin Primes conjecture itself. Optimally, it might prove the $p$ and $p+16$ – primes conjecture (which is still amazing).

One should look out for his paper in an upcoming issue of the Annals.

## 2. Goldbach Conjecture

I feel strangely tied to the Goldbach Conjecture, as I get far more traffic, emails, and spam concerning my previous post on an erroneous proof of Goldbach than on any other topic I’ve written about. About a year ago, I wrote briefly about progress that Dr. Harald Helfgott had made towards the 3-Goldbach Conjecture. This conjecture states that every odd integer greater than five can be written as the sum of three primes. (This is another easy to state problem that is not at all easy to approach).

One week ago, Helfgott posted a preprint to the arxiv that claims to complete his previous work and prove 3-Goldbach. Further, he uses the circle method and good old L-functions, so I feel like I should read over it more closely to learn a few things as it’s very close to my field. (Further still, he’s a Brandeis alum, and now that my wife will be a grad student at Brandeis I suppose I should include it in my umbrella of self-association). While I cannot say that I read the paper, understood it, and affirm its correctness, I can say that the method seems right for the task (related to the 10th and most subtle of Scott Aaronson’s list that I love to quote).

An interesting side bit to Helfgott’s proof is that it only works for numbers larger than $10^{30}$ or so. Fortunately, he’s also given a computer proof for numbers less than than on the arxiv, along with David Platt. $10^{30}$ is really, really, really big. Even that is a very slick bit.

## 3. FoM has opened

I care about open access. Fortunately, so do many of the big names. Two of the big attempts to create a good, strong set of open access math journals have just released their first articles. The Forum of Mathematics Sigma and Pi journals have each released a paper on algebraic and complex geometry. And they’re completely open! I don’t know what it takes for a journal to get off the ground, but I know that it starts with people reading its articles. So read up!

The two articles are

GENERIC VANISHING THEORY VIA MIXED HODGE MODULES

and, in Pi

-ADIC HODGE THEORY FOR RIGID-ANALYTIC VARIETIES

## Calculations with a Gauss-type Sum Wednesday, Apr 24 2013

It’s been a while since I’ve posted – I’m sorry. I’ve been busy, perhaps working on a paper (let’s hope it becomes a paper) and otherwise trying to learn things. This post is very closely related to some computations that have been coming up in what I’m currently looking at (in particular, looking at h-th coefficients of Eisenstein series of half-integral weight). I hope to write a very expository-level article on this project that I’ve been working on, outsourcing but completely providing computations behind the scenes in posts such as this one.

I’d like to add that this post took almost no time to write, as I used some vim macros and latex2wp to automatically convert a segment of something I’d written into wordpress-able html containing the latex. And that’s pretty awesome.

There is a particular calculation that I’ve had to do repeatedly recently, and that I will mention and use again. In an effort to have a readable account of this calculation, I present one such account here. Finally, I cannot help but say that this (and the next few posts, likely) are all joint work with Chan and Mehmet, also from Brown University.

Let us consider the following generalized Gauss Sum:

$\displaystyle H_m(c') : = \varepsilon_{c'} \sum_{r_1\bmod c'}\left(\frac{r_1}{c'}\right) e^{2 \pi i m\frac{r_1}{c'}} \ \ \ \ \ (1)$

where I let ${\left(\frac{\cdot}{\cdot}\right)}$ be the Legendre Symbol, and there ${\varepsilon_k}$ is the sign of the ${k}$th Gauss sum, so that it is ${1}$ if ${k \equiv 1 \mod 4}$ and it is ${i}$ if ${k \equiv 3 \mod 4}$. It is not defined for ${k}$ even.

Lemma 1 ${H_m(n)}$ is multiplicative in ${n}$.

Proof: Let ${n_1,n_2}$ be two relatively prime integers. Any integer ${a \bmod n_1n_2}$ can be written as ${a = b_2n_1 + b_1n_2}$, where ${b_1}$ runs through integers ${\bmod\, n_1}$ and ${b_2}$ runs ${\bmod\, n_2}$ with the Chinese Remainder Theorem. Then

$\displaystyle H_m(n_1n_2) = \varepsilon_{n_1n_2} \sum_{a \bmod n_1n_2} \left(\frac{a}{n_1n_2}\right) e^{2\pi i m \frac{a}{n_1n_2}}$

$\displaystyle = \varepsilon_{n_1n_2} \sum_{b_1 \bmod n_1} \sum_{b_2 \bmod n_2} \left(\frac{b_2n_1 +b_1n_2}{n_1n_2}\right) e^{2 \pi im\frac{b_2n_1 +b_1n_2}{n_1n_2}}$

$\displaystyle = \varepsilon_{n_1n_2} \sum_{b_1 \bmod n_1} \left(\frac{b_2n_1 +b_1n_2}{n_1}\right) e^{2\pi i m \frac{b_1n_2}{n_1n_2}} \sum_{b_2\bmod n_2} \left(\frac{b_2n_1 +b_1n_2}{n_2}\right) e^{2\pi im\frac{b_2n_1}{n_1n_2}}$

$\displaystyle = \varepsilon_{n_1n_2} \sum_{b_1 \bmod n_1} \left(\frac{b_1n_2}{n_1}\right) e^{2\pi i m \frac{b_1}{n_1}} \sum_{b_2\bmod n_2} \left(\frac{b_2n_1}{n_2}\right) e^{2\pi im\frac{b_2}{n_2}}$

$\displaystyle = \varepsilon_{n_1n_2}\left(\frac{n_2}{n_1}\right)\left(\frac{n_1}{n_2}\right)\sum_{b_1 \bmod n_1} \left(\frac{b_1}{n_1}\right) e^{2\pi i m \frac{b_1}{n_1}} \sum_{b_2\bmod n_2} \left(\frac{b_2}{n_2}\right) e^{2\pi im\frac{b_2}{n_2}}$

$\displaystyle = \varepsilon_{n_1n_2} \varepsilon_{n_1}^{-1} \varepsilon_{n_2}^{-1} \left(\frac{n_2}{n_1}\right)\left(\frac{n_1}{n_2}\right) H_m(n_1)H_{m}(n_2)$

Using quadratic reciprocity, we see that ${\varepsilon_{n_1n_2} \varepsilon_{n_1}^{-1} \varepsilon_{n_2}^{-1} \left(\frac{n_2}{n_1}\right)\left(\frac{n_1}{n_2}\right)= 1}$, so that ${H_m(n_1n_2) = H_m(n_1)H_m(n_2)}$. $\Box$

Let’s calculate ${H_m(p^k)}$ for prime powers ${p^k}$. Let ${\zeta = e^{2\pi i /p^k}}$ be a primitive ${p^k}$th root of unity. First we deal with the case of odd ${p}$, ${p\not |m}$. If ${k = 1}$, we have the typical quadratic Gauss sum multiplied by ${\varepsilon _p}$

$\displaystyle H_m(p) = \varepsilon_p \sum_{a \bmod p} e^{2\pi i m \frac a p}\left(\frac a p\right) = \varepsilon_p \left(\frac m p\right) \varepsilon_p \sqrt p = \left(\frac{-m} p\right) \sqrt p \ \ \ \ \ (2)$

For ${k > 1}$, we will see that ${H_m(p^k)}$ is ${0}$. We split into cases when ${k}$ is even or odd. If ${k}$ is even, then we are just summing the primitive ${p^k}$th roots of unity, which is ${0}$. If ${k>1}$ is odd,

$\displaystyle \sum_{a\bmod p^k} \zeta^a \left(\frac a {p^k}\right) = \sum_{a\bmod p^k} \zeta^a \left(\frac{a}{p}\right) = \sum_{b \bmod p}\sum_{c\bmod p^{k-1}} \zeta^{b+pc} \left(\frac b p\right)$

$\displaystyle = \sum_{b\bmod p} \zeta^b \left(\frac b p\right) \sum_{c\bmod p^{k-1}} \zeta^{pc} = 0, \ \ \ \ \ (3)$

since the inner sum is again a sum of roots of unity. Thus

$\displaystyle \left(1+ \frac{\left(\frac{-1^{k + 1/2}}{p}\right)H_m(p)}{p^{2s}} + \frac{\left(\frac{-1^{k + 1/2}}{p^2}\right)H_m(p^2)}{p^{4s}} + \cdots \right) =$

$\displaystyle = \left(1+ \frac{\left(\frac{-1^{k + 1/2}}{p}\right)H_m(p)}{p^{2s}}\right)$

$\displaystyle = \left(1+ \left(\frac {-m(-1)^{k + 1/2}}{p}\right)\frac{1}{p^{2s-\frac12}} \right)$

$\displaystyle = \left. \left(1-\frac1{p^{4s-1}}\right) \middle/ \left(1- \left(\frac{m(-1)^{k - 1/2}}{p}\right)\frac{1}{p^{2s-\frac12}}\right)\right.$

Notice that this matches up with the ${p}$th part of the Euler product for ${\displaystyle \frac{L(2s-\frac12,\left(\frac{m(-1)^{k - 1/2}}{\cdot}\right))}{\zeta(4s-1)}}$.

Now consider those odd ${p}$ such that ${p\mid m}$. Suppose ${p^l \parallel m}$. Then ${e^{2 \pi i m \ p^k} = \zeta^m}$ is a primitive ${p^{k-l}}$th root of unity (or ${1}$ if ${l \geq k}$). If ${l \geq k}$, then

$\displaystyle \sum_{a \bmod p^k} \zeta^{am} \left(\frac{a}{p^k}\right) = \sum_{a \bmod p^k} \left(\frac{a}{p^k}\right) = \begin{cases} 0 &\text{if } 2\not | k \\ \phi(p^k) &\text{if } 2 \mid k \end{cases} \ \ \ \ \ (4)$

If ${k=l+1}$ and ${k}$ is odd, then we essentially have a Gauss sum

$\displaystyle \sum_{a\bmod p^k} \zeta^{am} \left(\frac{a}{p^k}\right) = \sum_{a\bmod p^k}\zeta^{am} \left(\frac a p\right) =$

$\displaystyle = \sum_{c\bmod p^{k-1}} \zeta^{pcm} \sum_{b\bmod p} \zeta^{am} \left(\frac b p\right) = p^{k-1}\left(\frac{m/p^l}{p}\right)\varepsilon_p\sqrt p$

If ${k = l+1}$ and ${k}$ is even, noting that ${\zeta^m}$ is a ${p}$th root of unity,

$\displaystyle \sum_{a\bmod p^k} \zeta^{am}\left(\frac {a}{p^k}\right) = \sum_{\substack{a\bmod p^k\\(a,p) = 1}} \zeta^{am} =$

$\displaystyle = \sum_{a\bmod p^k}\zeta^{am} - \sum_{a\bmod p^{k-1}}\zeta^{pam} = 0 - p^{k-1} = -p^l.$

If ${k>l+1}$ then the sum will be zero. For ${k}$ even, this follows from the previous case. If ${k}$ is odd,

$\displaystyle \sum_{a\bmod p^k} \zeta^{am} \left(\frac a{p^k}\right) = \sum_{b\bmod p}\zeta^{bm} \left(\frac b p \right)\sum_{c\bmod p^{k-1}}\zeta^{pmc}= 0.$

Now, consider the Dirichlet series

$\displaystyle \sum_{c > 0, \tt odd} \frac{H_m(c)}{c^{2s}} = \prod_{p \neq 2} \left( 1 + \frac{H_m(p)}{p^{2s}} + \frac{H_m(p^2)}{p^{4s}} + \ldots\right)$.

Let us combine all these facts to construct the ${p}$th factor of the Dirichlet series in question, for ${p}$ dividing ${m}$. Assume first that ${p^l\parallel m}$ with ${l}$ even,

$\displaystyle 1 + \frac{\left(\frac{-1^{k + 1/2}}{p}\right)H_m(p)}{p^{2s}} + \frac{\left(\frac{-1^{k + 1/2}}{p^2}\right)H_m(p^2)}{p^{4s}}+ \cdots =$

$\displaystyle = \left( 1+ \varepsilon_{p^2}\frac{\phi(p^2)}{p^{4s}} + \cdots + \varepsilon_{p^l}\frac{\phi(p^l)}{p^{2ls}} + \varepsilon_{p^{l+1}}\frac{\left(\frac{(-1)^{k + 1/2}m/p^l}{p}\right)\varepsilon_p \sqrt p p^l}{p^{2(l+1)s}}\right)$

$\displaystyle = \left( 1+\frac{\phi(p^2)}{p^{4s}} + \frac{\phi(p^4)}{p^{8s}}+\cdots +\frac{\phi(p^{l})}{p^{2ls}} + \frac{\left(\frac{(-1)^{k - 1/2}m/p^l}{p}\right)p^{l+\frac12}}{p^{2(l+1)s}}\right)$

$\displaystyle = \left(1+ \frac{p^2 - p}{p^{4s}} + \cdots + \frac{p^{l}-p^{l-1}}{p^{2ls}} + \frac{\left(\frac{(-1)^{k - 1/2}m/p^l}{p}\right)p^{l+\frac12}}{p^{2(l+1)s}}\right)$

$\displaystyle = \left(1-\frac{1}{p^{4s-1}}\right)\left(1+\frac{1}{p^{4(s-\frac12)}} +\cdots + \frac{1}{p^{2(l-2)(s-\frac12)}}\right)+$

$\displaystyle +\frac{p^l}{p^{2ls}} \left(1+ \frac{\left(\frac{(-1)^{k - 1/2}m/p^l}{p}\right)}{p^{2s-\frac12}}\right)$

$\displaystyle = \left(1-\frac{1}{p^{4s-1}}\right) \left(1+ \frac{1}{p^{4(s-\frac12)}}+\cdots +\right.$

$\displaystyle + \left. \frac{1}{p^{2(l-2)(s-\frac12)}} + \frac{1}{p^{2l(s-\frac12)}}\left(1-\frac{\left(\frac{(-1)^{k - 1/2}m/p^l}{p}\right)}{p^{2s-\frac12}}\right)^{-1}\right)$

$\displaystyle = \left(1-\frac{1}{p^{4s-1}}\right) \left(\sum_{i=0}^{\lfloor \frac{l-1}{2} \rfloor} \frac{1}{p^{4(s-\frac12)i}} +\frac{1}{p^{2l(s-\frac12)}}\left(1-\frac{\left(\frac{(-1)^{k - 1/2}m/p^l}{p}\right)}{p^{2s-\frac12}}\right)^{-1} \right)$

because for even ${k}$, ${\varepsilon_{p^k} = 1}$, and for odd ${k}$, ${\varepsilon_{p^k} = \varepsilon_p}$. Similarly, for ${l}$ odd,

$\displaystyle 1+ \frac{\left(\frac{-1^{k + 1/2}}{p}\right)H_m(p)}{p^{2s}} +\frac{\left(\frac{-1^{k + 1/2}}{p^2}\right)H_m(p^2)}{p^{4s}}+ \cdots$

$\displaystyle = \left( 1+ \varepsilon_{p^2}\frac{\phi(p^2)}{p^{4s}} + \varepsilon_{p^4}\frac{\phi(p^4)}{p^{8s}} + \cdots + \varepsilon_{p^{l-1}}\frac{\phi(p^{l-1})}{p^{2(l-1)s}} + \varepsilon_{p^{l+1}}\frac{- p^l}{p^{2(l+1)s}}\right)\nonumber$

$\displaystyle = \left( 1+\frac{\phi(p^2)}{p^{4s}} + \frac{\phi(p^4)}{p^{8s}}+\cdots +\frac{\phi(p^{l-1})}{p^{2(l-1)s}} + \frac{-p^{l}}{p^{2(l+1)s}}\right) \nonumber$

$\displaystyle = \left(1+ \frac{p^2 - p}{p^{4s}} + \frac{p^4-p^3}{p^{8s}} + \cdots + \frac{p^{l-1}-p^{l-2}}{p^{2(l-1)s}} - \frac{p^l}{p^{2(l+1)s}}\right) \nonumber$

$\displaystyle = \left(1-\frac{1}{p^{4s-1}}\right)\left(\sum_{i=0}^{\frac{l-1}{2}} \frac{1}{p^{4(s-\frac12)i}}\right)$

Putting this together, we get that

$\displaystyle \prod_{p \neq 2} \left(\sum_{k=1}^\infty \frac{H_m(p)}{p^{2ks}}\right) = \frac{L_2(2s-\frac12,\left(\frac{m(-1)^{k - 1/2}}{\cdot}\right))}{\zeta_{2}(4s-1)} \times$

$\displaystyle \phantom{\sum \sum\sum\sum} \prod_{p^l \parallel m, p\neq 2} \left(\sum_{i=0}^{\lfloor \frac{l-1}{2} \rfloor} \frac{1}{p^{4(s-\frac12)i}} +\frac{\mathbf{1}_{2{\mathbb Z}}(l)}{p^{2l(s-\frac12)}}\left(1-\frac{\left(\frac{(-1)^{k - 1/2}m/p^l}{p}\right)}{p^{2s-\frac12}}\right)^{-1}\right) \ \ \ \ \ (5)$

## A Book Review of Count Down: The Race for Beautiful Solutions at the IMO Saturday, Feb 16 2013

I read a lot of popular science and math books. Scientific and mathematical exposition to the public is a fundamental task that must be done; but for some reason, it is simply not getting done well enough. One day, perhaps I’ll write expository (i.e. for non-math folk) math. But until then, I read everything I can. I then thought that if I read them all, I should share what I think.

Today, I consider the book Count Down: The Race for Beautiful Solutions at the International Mathematics Olympiad, by Steve Olson. The review itself can be found after the fold (more…)

## Hurwitz Zeta is a sum of Dirichlet L Functions, and vice-versa Friday, Feb 8 2013

At least three times now, I have needed to use that Hurwitz Zeta functions are a sum of L-functions and its converse, only to have forgotten how it goes. And unfortunately, the current wikipedia article on the Hurwitz Zeta function has a mistake, omitting the $varphi$ term (although it will soon be corrected). Instead of re-doing it each time, I write this detail here, below the fold.
(more…)

## Math journals and the fight over open access Friday, Jan 18 2013

Some may have heard me talk about this before, but I’ve caught the open source bug. At least, I’ve caught the collaboration and free-dissemination bug. And I don’t just mean software – there’s much more to open source than software (even though the term open source originated in reference to free access to source code). I use open source to refer to the idea that when someone consumes a product, they should have access to the design and details of implementation, and should be able to freely distribute the product whenever this is possible. In some ways, I’m still learning. For example, though I use linux, I do not know enough about coding to contribute actual code to the linux/unix community. But I know just enough python to contribute to Sage, and do. And I’m getting better.

I also believe in open access, which feels like a natural extension. By open access, I mean free access to peer-reviewed scholarly journals and other materials. It stuns me that the public does not generally have access to publicly-funded research. How is this acceptable? Another thing that really gets to me is how selling overpriced and overlarge calculus textbooks can allow the author to do things like spend 30+ million dollars on his home? This should not happen. At least, it shouldn’t happen now, in the internet age. All the material is freely available in at least as good of a presentation, so the cost of the textbook is a compilation cost (not worth over \$100). But these books are printed oversize, 1000+ pages, in full color and on 60-pound paper. That’s a recipe for high cost! It’s tremendously unfortunate, as it’s not as though the students even have a choice over what book they buy. But this is not the argument I want to make today, and I digress.

Recently, I was dragged down a rabbit hole. And what I saw when I emerged on the other side made me learn about a side of math journals I’d never seen before, and the fight over open access. I’d like to comment on this today – that’s after the fold.

## Are the calculus MOOCs any good: After week 1 Saturday, Jan 12 2013

This is a continuation of a previous post.

I’ve been following the two Coursera calculus MOOCs: the elementary introductory to calculus being taught by Dr. Fowler of Ohio State University, and a course designed around Taylor expansions taught by Dr. Ghrist of UPenn, meant to be taken after an introductory calculus course. I’ve completed the ‘first week’ of Dr. Fowler’s course (there are 15 total), and the ‘first unit’ of Dr. Ghrist’s course (there are 5 total), and I have a few things to say – after the fold.

## Are the calculus MOOCs any good? Tuesday, Jan 8 2013

I like the idea of massive online collaboration in math. For example, I am a big supporter of the ideas of the polymath projects. I contribute to wikis and to Sage (which I highly recommend to everyone as an alternative to the M’s: Maple, Mathematica, MatLab, Magma). Now, there are MOOCs (Massice open online courses) in many subjects, but in particular there are a growing number of math MOOCs (a more or less complete list of MOOCs can be found here). The idea of a MOOC is to give people all over the world the opportunity to a good, diverse, and free education.

I’ve looked at a few MOOCs in the past. I’ve taken a few Coursera and Udacity courses, and I have mixed reviews. Actually, I’ve been very impressed with the Udacity courses I’ve taken. They have a good polish. But there are only a couple dozen – it takes time to get quality. There are hundreds of Coursera courses, though there is some overlap. But I’ve been pretty unimpressed with most of them.

But there are two calculus courses being offered this semester (right now) through Coursera. I’ve been a teaching assistant for calculus many times, and there are things that I like and others that I don’t like about my past experiences. Perhaps the different perspective from a MOOC will lead to a better form of calculus instruction?

There will be no teaching assistant led recitation sections, as the ‘standard university model’ might suggest. Will there be textbooks? In both, there are textbooks, or at least lecture notes (I’m not certain of their format yet). And there will be lectures. But due to the sheer size of the class, it’s much more challenging for the instructors to answer individual students’ questions. There is a discussion forum which essentially means that students get to help each other (I suppose that people like me, who know calculus, can also help people through the discussion forums too). So in a few ways, this turns what I have come to think of as the traditional model of calculus instruction on its head.

And this might be a good thing! (Or it might not!) Intro calculus instruction has not really changed much in decades, since before the advent of computers and handheld calculators. It would make sense that new tools might mean that teaching methods should change. But I don’t know yet.

So I’ll be looking at the two courses this semester. The first is being offered by Dr. Jim Fowler and is associated with Ohio State University. It’s an introductory-calculus course. The second is being offered by Dr. Robert Ghrist and is associated with the University of Pennsylvania. It’s sort of a funny class – it’s designed for people who already know some calculus. In particular, students should know what derivatives and integrals are. There is a diagnostic test that involves taking a limit, computing some derivatives, and computing an integral (and some precalculus problems as well). Dr. Ghrist says that his course assumes that students have taken a high school AP Calculus AB course or the equivalent. So it’s not quite fair to compare the two classes, as they’re not on equal footing.

But I can certainly see what I think of the MOOC model for Calculus instruction.

## Math 90: Concluding Remarks Sunday, Dec 30 2012

All is said and done with Math 90 for 2012, and the year is coming to a close. I wanted to take this moment to write a few things about the course, what seemed to go well and what didn’t, and certain trends in the course. that I think are interesting and illustrative.

First, we might just say some of the numbers. Math 90 is offered only as pass/fail, with the possibility of ‘passing with distinction’ if you did exceptionally well (I’ll say what that meant here, though who knows what it means in general). We had four people fail, three people ‘pass with distinction,’ and everyone else got a passing mark. Everything else will be after the fold.

## Math 90: Week 11 and Midterm Solutions Sunday, Nov 18 2012

We had a midterm this week, and did more review during recitation. The solutions are now available below the fold

## An Application of Mobius Inversion to Certain Asymptotics I Thursday, Nov 8 2012

In this note, I consider an application of generalized Mobius Inversion to extract information of arithmetical sums with asymptotics of the form $\displaystyle \sum_{nk^j \leq x} f(n) = a_1x + O(x^{1 - \epsilon})$ for a fixed $j$ and a constant $a_1$, so that the sum is over both $n$ and $k$. We will see that $\displaystyle \sum_{nk^j \leq x} f(n) = a_1x + O(x^{1-\epsilon}) \iff \sum_{n \leq x} f(n) = \frac{a_1x}{\zeta(j)} + O(x^{1 - \epsilon})$.

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