## Mathematics Category Archive

Below you will find the most recent posts tagged “Mathematics”, arranged in reverse chronological order.

Below you will find the most recent posts tagged “Mathematics”, arranged in reverse chronological order.

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I’m in San Diego, and it’s charming here. (It’s certainly much nicer outside than the feet of snow in Boston. I’ve apparently brought some British rain with me, though).

Today I give a talk on counting lattice points on one-sheeted hyperboloids. These are the shapes described by

$$ X_1^2 + \cdots + X_{d-1}^2 = X_d^2 + h,$$

where $h > 0$ is a positive integer. The question is: how many lattice points $x$ are on such a hyperboloid with $| x |^2 \leq R$; or equivalently, how many lattice points are on such a hyperboloid and contained within a ball of radius $\sqrt R$ centered at the origin?

I describe my general approach of transforming this into a question about the behavior of modular forms, and then using spectral techniques from the theory of modular forms to understand this behavior. This becomes a question of understanding the shifted convolution Dirichlet series

$$ \sum_{n \geq 0} \frac{r_{d-1}(n+h)r_1(n)}{(2n + h)^s}.$$

Ultimately this comes from the modular form $\theta^{d-1}(z) \overline{\theta(z)}$, where

$$ \theta(z) = \sum_{m \in \mathbb{Z}} e^{2 \pi i m^2 z}.$$

Here are the slides for this talk. Note that this talk is based on chapter 5 of my thesis, and (hopefully) soon a preprint of this chapter ready for submission will appear on the arXiv.

Posted in Math.NT, Mathematics
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This is a very short post in my collection working through this year’s Advent of Code challenges. Unlike the previous ones, this has no mathematical comments, as it was a very short exercise. This notebook is available in its original format on my github.

Given a list of strings, determine how many strings have no duplicate words.

This is a classic problem, and it’s particularly easy to solve this in python. Some might use `collections.Counter`

, but I think it’s more straightforward to use sets.

The key idea is that the set of words in a sentence will not include duplicates. So if taking the set of a sentence reduces its length, then there was a duplicate word.

In [1]:

```
with open("input.txt", "r") as f:
lines = f.readlines()
def count_lines_with_unique_words(lines):
num_pass = 0
for line in lines:
s = line.split()
if len(s) == len(set(s)):
num_pass += 1
return num_pass
count_lines_with_unique_words(lines)
```

Out[1]:

I think this is the first day where I would have had a shot at the leaderboard if I’d been gunning for it.

Let’s add in another constraint. Determine how many strings have no duplicate words, even after anagramming. Thus the string

```
abc bac
```

is not valid, since the second word is an anagram of the first. There are many ways to tackle this as well, but I will handle anagrams by sorting the letters in each word first, and then running the bit from part 1 to identify repeated words.

In [2]:

```
with open("input.txt", "r") as f:
lines = f.readlines()
sorted_lines = []
for line in lines:
sorted_line = ' '.join([''.join(l) for l in map(sorted, line.split())])
sorted_lines.append(sorted_line)
sorted_lines[:2]
```

Out[2]:

In [3]:

```
count_lines_with_unique_words(sorted_lines)
```

Out[3]:

This is the third notebook in my posts on the Advent of Code challenges. The notebook in its original format can be found on my github.

Numbers are arranged in a spiral

```
17 16 15 14 13
18 5 4 3 12
19 6 1 2 11
20 7 8 9 10
21 22 23---> ...
```

Given an integer n, what is its Manhattan Distance from the center (1) of the spiral? For instance, the distance of 3 is $2 = 1 + 1$, since it’s one space to the right and one space up from the center.

Here’s my idea. The bottom right corner of the $k$th layer is the integer $(2k+1)^2$, since that’s how many integers are contained within that square. The other three corners in that layer are $(2k+1)^2 – 2k, (2k+1)^2 – 4k$, and $(2k+1)^2 – 6k$. Finally, the closest spot on the $k$th layer to the origin is at distance $k$: these are the four “axis locations” halfway between the corners, at $(2k+1)^2 – k, (2k+1)^2 – 3k, (2k+1)^2 – 5k$, and $(2k+1)^2 – 7k$.

For instance when $k = 1$, the bottom right is $(2 + 1)^2 = 9$, and the four “axis locations” are $9 – 1, 9 – 3, 9-5$, and $9-7$. The “axis locations” are $k$ away, and the corners are $2k$ away.

So I will first find which layer the number is on. Then I’ll figure out which side it’s on, and then how far away it is from the nearest “axis location” or “corner”.

My given number happens to be 289326.

In [1]:

```
import math
def find_lowest_larger_odd_square(n):
upper = math.ceil(n**.5)
if upper %2 == 0:
upper += 1
return upper
```

In [2]:

```
assert find_lowest_larger_odd_square(39) == 7
assert find_lowest_larger_odd_square(26) == 7
assert find_lowest_larger_odd_square(25) == 5
```

In [3]:

```
find_lowest_larger_odd_square(289326)
```

Out[3]:

In [4]:

```
539**2 - 289326
```

Out[4]:

It happens to be that our integer is very close to an odd square.

The square is $539^2$, and the distance to that square is $538$ from the center.

Note that $539 = 2(269) + 1$, so this is the $269$th layer of the square.

The previous corner to $539^2$ is $539^2 – 538$, and the previous corner to that is $539^2 – 2\cdot538 = 539^2 – 1076$.

This is the nearest corner.

How far away from the square is this corner?

Posted in Expository, Programming, Python
Tagged advent of code, number spiral, python
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This is the second notebook in my posts on the Advent of Code challenges. This notebook in its original format can be found on my github.

You are given a table of integers. Find the difference between the maximum and minimum of each row, and add these differences together.

There is not a lot to say about this challenge. The plan is to read the file linewise, compute the difference on each line, and sum them up.

In [1]:

```
with open("input.txt", "r") as f:
lines = f.readlines()
lines[0]
```

Out[1]:

In [2]:

```
l = lines[0]
l = l.split()
l
```

Out[2]:

In [3]:

```
def max_minus_min(line):
'''Compute the difference between the largest and smallest integer in a line'''
line = list(map(int, line.split()))
return max(line) - min(line)
def sum_differences(lines):
'''Sum the value of `max_minus_min` for each line in `lines`'''
return sum(max_minus_min(line) for line in lines)
```

In [4]:

```
testcase = ['5 1 9 5','7 5 3', '2 4 6 8']
assert sum_differences(testcase) == 18
```

In [5]:

```
sum_differences(lines)
```

Out[5]:

In line with the first day’s challenge, I’m inclined to ask what we should “expect.” But what we should expect is not well-defined in this case. Let us rephrase the problem in a randomized sense.

Suppose we are given a table, $n$ lines long, where each line consists of $m$ elements, that are each uniformly randomly chosen integers from $1$ to $10$. We might ask what is the expected value of this operation, of summing the differences between the maxima and minima of each row, on this table. What should we expect?

As each line is independent of the others, we are really asking what is the expected value across a single row. So given $m$ integers uniformly randomly chosen from $1$ to $10$, what is the expected value of the maximum, and what is the expected value of the minimum?

Let’s begin with the minimum. The minimum is $1$ unless all the integers are greater than $2$. This has probability

$$ 1 – \left( \frac{9}{10} \right)^m = \frac{10^m – 9^m}{10^m}$$

of occurring. We rewrite it as the version on the right for reasons that will soon be clear.

The minimum is $2$ if all the integers are at least $2$ (which can occur in $9$ different ways for each integer), but not all the integers are at least $3$ (each integer has $8$ different ways of being at least $3$). Thus this has probability

$$ \frac{9^m – 8^m}{10^m}.$$

Continuing to do one more for posterity, the minimum is $3$ if all the integers are at least $3$ (each integer has $8$ different ways of being at least $3$), but not all integers are at least $4$ (each integer has $7$ different ways of being at least $4$). Thus this has probability

$$ \frac{8^m – 7^m}{10^m}.$$

And so on.

Recall that the expected value of a random variable is

$$ E[X] = \sum x_i P(X = x_i),$$

so the expected value of the minimum is

$$ \frac{1}{10^m} \big( 1(10^m – 9^m) + 2(9^m – 8^m) + 3(8^m – 7^m) + \cdots + 9(2^m – 1^m) + 10(1^m – 0^m)\big).$$

This simplifies nicely to

$$ \sum_ {k = 1}^{10} \frac{k^m}{10^m}. $$

The same style of thinking shows that the expected value of the maximum is

$$ \frac{1}{10^m} \big( 10(10^m – 9^m) + 9(9^m – 8^m) + 8(8^m – 7^m) + \cdots + 2(2^m – 1^m) + 1(1^m – 0^m)\big).$$

This simplifies to

$$ \frac{1}{10^m} \big( 10 \cdot 10^m – 9^m – 8^m – \cdots – 2^m – 1^m \big) = 10 – \sum_ {k = 1}^{9} \frac{k^m}{10^m}.$$

Subtracting, we find that the expected difference is

$$ 9 – 2\sum_ {k=1}^{9} \frac{k^m}{10^m}. $$

From this we can compute this for each list-length $m$. It is good to note that as $m \to \infty$, the expected value is $9$. Does this make sense? Yes, as when there are lots of values we should expect one to be a $10$ and one to be a $1$. It’s also pretty straightforward to see how to extend this to values of integers from $1$ to $N$.

Looking at the data, it does not appear that the integers were randomly chosen. Instead, there are very many relatively small integers and some relatively large integers. So we shouldn’t expect this toy analysis to accurately model this problem — the distribution is definitely not uniform random.

But we can try it out anyway.

Posted in Expository, Programming, Python
Tagged advent of code, Programming, python
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I thoroughly enjoyed reading through Peter Norvig’s extraordinarily clean and nice solutions to the Advent of Code challenge last year. Inspired by his clean, literate programming style and the convenience of jupyter notebook demonstrations, I will look at several of these challenges in my own jupyter notebooks.

My background and intentions aren’t the same as Peter Norvig’s: his expertise dwarfs mine. And timezones are not kind to those of us in the UK, and thus I won’t be competing for a position on the leaderboards. These are to be fun. And sometimes there are tidbits of math that want to come out of the challenges.

Enough of that. Let’s dive into the first day.

In [1]:

```
with open('input.txt', 'r') as f:
seq = f.read()
seq = seq.strip()
seq[:10]
```

Out[1]:

In [2]:

```
def sum_matched_digits(s):
"Sum of digits which match following digit, and first digit if it matches last digit"
total = 0
for a,b in zip(s, s[1:]+s[0]):
if a == b:
total += int(a)
return total
```

They provide a few test cases which we use to test our method against.

In [3]:

```
assert sum_matched_digits('1122') == 3
assert sum_matched_digits('1111') == 4
assert sum_matched_digits('1234') == 0
assert sum_matched_digits('91212129') == 9
```

For fun, this is a oneline version.

I gave an introduction to sage tutorial at the University of Warwick Computational Group seminar today, 2 November 2017. Below is a conversion of the sage/jupyter notebook I based the rest of the tutorial on. I said many things which are not included in the notebook, and during the seminar we added a few additional examples and took extra consideration to a few different calls. But for reference, the notebook is here.

The notebook itself (as a jupyter notebook) can be found and viewed on my github (link to jupyter notebook). When written, this notebook used a Sage 8.0.0.rc1 backend kernel and ran fine on the standard Sage 8.0 release , though I expect it to work fine with any recent official version of sage. The last cell requires an active notebook to be seen (or some way to export jupyter widgets to standalone javascript or something; this either doesn’t yet exist, or I am not aware of it).

I will also note that I converted the notebook for display on this website using jupyter’s nbconvert package. I have some CSS and syntax coloring set up that affects the display.

Good luck learning sage, and happy hacking.

Sage (also known as SageMath) is a general purpose computer algebra system written on top of the python language. In Mathematica, Magma, and Maple, one writes code in the mathematica-language, the magma-language, or the maple-language. Sage is python.

But no python background is necessary for the rest of today’s guided tutorial. The purpose of today’s tutorial is to give an indication about how one really *uses* sage, and what might be available to you if you want to try it out.

I will spoil the surprise by telling you upfront the two main points I hope you’ll take away from this tutorial.

- With tab-completion and documentation, you can do many things in sage without ever having done them before.
- The ecosystem of libraries and functionality available in sage is tremendous, and (usually) pretty easy to use.

Let’s first get a small feel for sage by seeing some standard operations and what typical use looks like through a series of trivial, mostly unconnected examples.

In [1]:

```
# Fundamental manipulations work as you hope
2+3
```

Out[1]:

You can also subtract, multiply, divide, exponentiate…

```
>>> 3-2
1
>>> 2*3
6
>>> 2^3
8
>>> 2**3 # (also exponentiation)
8
```

There is an order of operations, but these things work pretty much as you want them to work. You might try out several different operations.

Sage includes a lot of functionality, too. For instance,

In [2]:

```
factor(-1008)
```

Out[2]:

In [3]:

```
list(factor(1008))
```

Out[3]:

Sage knows many functions and constants, and these are accessible.

Posted in Expository, Mathematics, sage, sagemath
Tagged ipython, jupyter, notebook, sage, sagemath, tutorial
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The primary purpose of this note is to collect a few hitherto unnoticed or unpublished results concerning gaps between powers of consecutive primes. The study of gaps between primes has attracted many mathematicians and led to many deep realizations in number theory. The literature is full of conjectures, both open and closed, concerning the nature of primes.

In a series of stunning developments, Zhang, Maynard, and Tao^{1}^{2} made the first major progress towards proving the prime $k$-tuple conjecture, and successfully proved the existence of infinitely many pairs of primes differing by a fixed number. As of now, the best known result is due to the massive collaborative Polymath8 project,^{3} which showed that there are infinitely many pairs of primes of the form $p, p+246$. In the excellent expository article, ^{4} Granville describes the history and ideas leading to this breakthrough, and also discusses some of the potential impact of the results. This note should be thought of as a few more results following from the ideas of Zhang, Maynard, Tao, and the Polymath8 project.

Throughout, $p_n$ will refer to the $n$th prime number. In a paper, ^{5} Andrica conjectured that

\begin{equation}\label{eq:Andrica_conj}

\sqrt{p_{n+1}} – \sqrt{p_n} < 1

\end{equation}

holds for all $n$. This conjecture, and related statements, is described in Guy’s Unsolved Problems in Number Theory.

^{6} It is quickly checked that this holds for primes up to $4.26 \cdot 10^{8}$ in sagemath

```
# Sage version 8.0.rc1
# started with `sage -ipython`
# sage has pari/GP, which can generate primes super quickly
from sage.all import primes_first_n
# import izip since we'll be zipping a huge list, and sage uses python2 which has
# non-iterable zip by default
from itertools import izip
# The magic number 23150000 appears because pari/GP can't compute
# primes above 436273290 due to fixed precision arithmetic
ps = primes_first_n(23150000) # This is every prime up to 436006979
# Verify Andrica's Conjecture for all prime pairs = up to 436006979
gap = 0
for a,b in izip(ps[:-1], ps[1:]):
if b**.5 - a**.5 > gap:
A, B, gap = a, b, b**.5 - a**.5
print(gap)
print("")
print(A)
print(B)
```

In approximately 20 seconds on my machine (so it would not be harder to go much higher, except that I would have to go beyond pari/GP to generate primes), this completes and prints out the following output.

```
0.317837245196
0.504017169931
0.670873479291
7
11
```

Thus the largest value of $\sqrt{p_{n+1}} – \sqrt{p_n}$ was merely $0.670\ldots$, and occurred on the gap between $7$ and $11$.

So it appears very likely that the conjecture is true. However it is also likely that new, novel ideas are necessary before the conjecture is decided.

Andrica’s Conjecture can also be stated in terms of prime gaps. Let $g_n = p_{n+1} – p_n$ be the gap between the $n$th prime and the $(n+1)$st prime. Then Andrica’s Conjecture is equivalent to the claim that $g_n < 2 \sqrt{p_n} + 1$. In this direction, the best known result is due to Baker, Harman, and Pintz, ^{7} who show that $g_n \ll p_n^{0.525}$.

In 1985, Sandor ^{8} proved that \begin{equation}\label{eq:Sandor} \liminf_{n \to \infty} \sqrt[4]{p_n} (\sqrt{p_{n+1}} – \sqrt{p_n}) = 0. \end{equation} The close relation to Andrica’s Conjecture \eqref{eq:Andrica_conj} is clear. The first result of this note is to strengthen this result.

TheoremLet $\alpha, \beta \geq 0$, and $\alpha + \beta < 1$. Then

\begin{equation}\label{eq:main}

\liminf_{n \to \infty} p_n^\beta (p_{n+1}^\alpha – p_n^\alpha) = 0.

\end{equation}

We prove this theorem below. Choosing $\alpha = \frac{1}{2}, \beta = \frac{1}{4}$ verifies Sandor’s result \eqref{eq:Sandor}. But choosing $\alpha = \frac{1}{2}, \beta = \frac{1}{2} – \epsilon$ for a small $\epsilon > 0$ gives stronger results.

This theorem leads naturally to the following conjecture.

ConjectureFor any $0 \leq \alpha < 1$, there exists a constant $C(\alpha)$ such that

\begin{equation}

p_{n+1}^\alpha – p_{n}^\alpha \leq C(\alpha)

\end{equation}

for all $n$.

A simple heuristic argument, given in the last section below, shows that this Conjecture follows from Cramer’s Conjecture.

It is interesting to note that there are generalizations of Andrica’s Conjecture. One can ask what the smallest $\gamma$ is such that

\begin{equation}

p_{n+1}^{\gamma} – p_n^{\gamma} = 1

\end{equation}

has a solution. This is known as the Smarandache Conjecture, and it is believed that the smallest such $\gamma$ is approximately

\begin{equation}

\gamma \approx 0.5671481302539\ldots

\end{equation}

The digits of this constant, sometimes called “the Smarandache constant,” are the contents of sequence A038458 on the OEIS. It is possible to generalize this question as well.

Open QuestionFor any fixed constant $C$, what is the smallest $\alpha = \alpha(C)$ such that

\begin{equation}

p_{n+1}^\alpha – p_n^\alpha = C

\end{equation}

has solutions? In particular, how does $\alpha(C)$ behave as a function of $C$?

This question does not seem to have been approached in any sort of generality, aside from the case when $C = 1$.

The idea of the proof is very straightforward. We estimate \eqref{eq:main} across prime pairs $p, p+246$, relying on the recent proof from Polymath8 that infinitely many such primes exist.

Fix $\alpha, \beta \geq 0$ with $\alpha + \beta < 1$. Applying the mean value theorem of calculus on the function $x \mapsto x^\alpha$ shows that

\begin{align}

p^\beta \big( (p+246)^\alpha – p^\alpha \big) &= p^\beta \cdot 246 \alpha q^{\alpha – 1} \\\

&\leq p^\beta \cdot 246 \alpha p^{\alpha – 1} = 246 \alpha p^{\alpha + \beta – 1}, \label{eq:bound}

\end{align}

for some $q \in [p, p+246]$. Passing to the inequality in the second line is done by realizing that $q^{\alpha – 1}$ is a decreasing function in $q$. As $\alpha + \beta – 1 < 0$, as $p \to \infty$ we see that\eqref{eq:bound} goes to zero.

Therefore

\begin{equation}

\liminf_{n \to \infty} p_n^\beta (p_{n+1}^\alpha – p_n^\alpha) = 0,

\end{equation}

as was to be proved.

Cramer’s Conjecture states that there exists a constant $C$ such that for all sufficiently large $n$,

\begin{equation}

p_{n+1} – p_n < C(\log n)^2.

\end{equation}

Thus for a sufficiently large prime $p$, the subsequent prime is at most $p + C (\log p)^2$. Performing a similar estimation as above shows that

\begin{equation}

(p + C (\log p)^2)^\alpha – p^\alpha \leq C (\log p)^2 \alpha p^{\alpha – 1} =

C \alpha \frac{(\log p)^2}{p^{1 – \alpha}}.

\end{equation}

As the right hand side vanishes as $p \to \infty$, we see that it is natural to expect that the main Conjecture above is true. More generally, we should expect the following, stronger conjecture.

Conjecture’For any $\alpha, \beta \geq 0$ with $\alpha + \beta < 1$, there exists a constant $C(\alpha, \beta)$ such that

\begin{equation}

p_n^\beta (p_{n+1}^\alpha – p_n^\alpha) \leq C(\alpha, \beta).

\end{equation}

I wrote this note in between waiting in never-ending queues while I sort out my internet service and other mundane activities necessary upon moving to another country. I had just read some papers on the arXiv, and I noticed a paper which referred to unknown statuses concerning Andrica’s Conjecture. So then I sat down and wrote this up.

I am somewhat interested in qualitative information concerning the Open Question in the introduction, and I may return to this subject unless someone beats me to it.

This note is (mostly, minus the code) available as a pdf and (will shortly) appears on the arXiv. This was originally written in LaTeX and converted for display on this site using a set of tools I’ve written based around latex2jax, which is available on my github.

Posted in Math.NT, Mathematics, sage
Tagged Andrica's Conjecture, mathematics, primes, Zhang-Maynard-Tao
1 Comment

The lmfdb and sagemath are both great things, but they don’t currently talk to each other. Much of the lmfdb calls sage, but the lmfdb also includes vast amounts of data on $L$-functions and modular forms (hence the name) that is not accessible from within sage.

This is an example prototype of an interface to the lmfdb from sage. Keep in mind that this is **a prototype** and every aspect can change. But we hope to show what may be possible in the future. If you have requests, comments, or questions, **please request/comment/ask** either now, or at my email: `david@lowryduda.com`

.

Note that this notebook is available on http://davidlowryduda.com or https://gist.github.com/davidlowryduda/deb1f88cc60b6e1243df8dd8f4601cde, and the code is available at https://github.com/davidlowryduda/sage2lmfdb

Let’s dive into an example.

In [1]:

```
# These names will change
from sage.all import *
import LMFDB2sage.elliptic_curves as lmfdb_ecurve
```

In [2]:

```
lmfdb_ecurve.search(rank=1)
```

Out[2]:

This returns 10 elliptic curves of rank 1. But these are a bit different than sage’s elliptic curves.

In [3]:

```
Es = lmfdb_ecurve.search(rank=1)
E = Es[0]
print(type(E))
```

Note that the class of an elliptic curve is an lmfdb ElliptcCurve. But don’t worry, this is a subclass of a normal elliptic curve. So we can call the normal things one might call on an elliptic curve.

th

In [4]:

```
# Try autocompleting the following. It has all the things!
print(dir(E))
```

This gives quick access to some data that is not stored within the LMFDB, but which is relatively quickly computable. For example,

In [5]:

```
E.defining_ideal()
```

Out[5]:

But one of the great powers is that there are some things which are computed and stored in the LMFDB, and not in sage. We can now immediately give many examples of rank 3 elliptic curves with:

In [6]:

```
Es = lmfdb_ecurve.search(conductor=11050, torsion_order=2)
print("There are {} curves returned.".format(len(Es)))
E = Es[0]
print(E)
```

And for these curves, the lmfdb contains data on its rank, generators, regulator, and so on.

In [7]:

```
print(E.gens())
print(E.rank())
print(E.regulator())
```

In [8]:

```
res = []
%time for E in Es: res.append(E.gens()); res.append(E.rank()); res.append(E.regulator())
```

That’s pretty fast, and this is because all of this was pulled from the LMFDB when the curves were returned by the

In this case, elliptic curves over the rationals are only an okay example, as they’re really well studied and sage can compute much of the data very quickly. On the other hand, through the LMFDB there are millions of examples and corresponding data at one’s fingertips.### This is where we’re really looking for input.¶

## Now let’s describe what’s going on under the hood a little bit¶

`search()`

function.In this case, elliptic curves over the rationals are only an okay example, as they’re really well studied and sage can compute much of the data very quickly. On the other hand, through the LMFDB there are millions of examples and corresponding data at one’s fingertips.

Think of what you might want to have easy access to through an interface from sage to the LMFDB, and tell us. We’re actively seeking comments, suggestions, and requests. Elliptic curves over the rationals are a prototype, and the LMFDB has lots of (much more challenging to compute) data. There is data on the LMFDB that is simply not accessible from within sage.

**email: david@lowryduda.com, or post an issue on https://github.com/LMFDB/lmfdb/issues**

There is an API for the LMFDB at http://beta.lmfdb.org/api/. This API is a bit green, and we will change certain aspects of it to behave better in the future. A call to the API looks like

```
http://beta.lmfdb.org/api/elliptic_curves/curves/?rank=i1&conductor=i11050
```

The result is a large mess of data, which can be exported as json and parsed.

But that’s hard, and the resulting data are not sage objects. They are just strings or ints, and these require time *and thought* to parse.

So we created a module in sage that writes the API call and parses the output back into sage objects. The 22 curves given by the above API call are the same 22 curves returned by this call:

In [9]:

```
Es = lmfdb_ecurve.search(rank=1, conductor=11050, max_items=25)
print(len(Es))
E = Es[0]
```

The total functionality of this search function is visible from its current documentation.

In [10]:

```
# Execute this cell for the documentation
print(lmfdb_ecurve.search.__doc__)
```

In [11]:

```
# So, for instance, one could perform the following search, finding a unique elliptic curve
lmfdb_ecurve.search(rank=2, torsion_order=3, degree=4608)
```

Out[11]:

If there are no curves satisfying the search criteria, then a message is displayed and that’s that. These searches may take a couple of seconds to complete.

For example, no elliptic curve in the database has rank 5.

In [12]:

```
lmfdb_ecurve.search(rank=5)
```

Right now, at most 100 curves are returned in a single API call. This is the limit even from directly querying the API. But one can pass in the argument `base_item`

(the name will probably change… to `skip`

? or perhaps to `offset`

?) to start returning at the `base_item`

th element.

In [13]:

```
from pprint import pprint
pprint(lmfdb_ecurve.search(rank=1, max_items=3)) # The last item in this list
print('')
pprint(lmfdb_ecurve.search(rank=1, max_items=3, base_item=2)) # should be the first item in this list
```

Included in the documentation is also a bit of hopefulness. Right now, the LMFDB API does not actually accept

`max_conductor`

or `min_conductor`

(or arguments of that type). But it will sometime. (This introduces a few extra difficulties on the server side, and so it will take some extra time to decide how to do this).
In [14]:

```
lmfdb_ecurve.search(rank=1, min_conductor=500, max_conductor=10000) # Not implemented
```

Our

Generically, documentation and introspection on objects from this class should work. Much of sage’s documentation carries through directly.

`EllipticCurve_rational_field_lmfdb`

class constructs a sage elliptic curve from the json and overrides (somem of the) the default methods in sage if there is quicker data available on the LMFDB. In principle, this new object is just a sage object with some slightly different methods.Generically, documentation and introspection on objects from this class should work. Much of sage’s documentation carries through directly.

In [15]:

```
print(E.gens.__doc__)
```

Modified methods should have a note indicating that the data comes from the LMFDB, and then give sage’s documentation. This is not yet implemented. (So if you examine the current version, you can see some incomplete docstrings like

`regulator()`

.)
In [16]:

```
print(E.regulator.__doc__)
```

Thank you, and if you have any questions, comments, or concerns, please find me/email me/raise an issue on LMFDB’s github.

Posted in Expository, LMFDB, Math.NT, Mathematics, Programming, Python, sagemath
Tagged interface, ipython, jupyter notebook, lmfdb, sage, sage days 87, sagemath, sd87
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This is joint work with Thomas Hulse, Chan Ieong Kuan, and Alexander Walker. This is a natural successor to our previous work (see their announcements: one, two, three) concerning bounds and asymptotics for sums of coefficients of modular forms.

We now have a variety of results concerning the behavior of the partial sums

$$ S_f(X) = \sum_{n \leq X} a(n) $$

where $f(z) = \sum_{n \geq 1} a(n) e(nz)$ is a GL(2) cuspform. The primary focus of our previous work was to understand the Dirichlet series

$$ D(s, S_f \times S_f) = \sum_{n \geq 1} \frac{S_f(n)^2}{n^s} $$

completely, give its meromorphic continuation to the plane (this was the major topic of the first paper in the series), and to perform classical complex analysis on this object in order to describe the behavior of $S_f(n)$ and $S_f(n)^2$ (this was done in the first paper, and was the major topic of the second paper of the series). One motivation for studying this type of problem is that bounds for $S_f(n)$ are analogous to understanding the error term in lattice point discrepancy with circles.

That is, let $S_2(R)$ denote the number of lattice points in a circle of radius $\sqrt{R}$ centered at the origin. Then we expect that $S_2(R)$ is approximately the area of the circle, plus or minus some error term. We write this as

$$ S_2(R) = \pi R + P_2(R),$$

where $P_2(R)$ is the error term. We refer to $P_2(R)$ as the “lattice point discrepancy” — it describes the discrepancy between the number of lattice points in the circle and the area of the circle. Determining the size of $P_2(R)$ is a very famous problem called the Gauss circle problem, and it has been studied for over 200 years. We believe that $P_2(R) = O(R^{1/4 + \epsilon})$, but that is not known to be true.

The Gauss circle problem can be cast in the language of modular forms. Let $\theta(z)$ denote the standard Jacobi theta series,

$$ \theta(z) = \sum_{n \in \mathbb{Z}} e^{2\pi i n^2 z}.$$

Then

$$ \theta^2(z) = 1 + \sum_{n \geq 1} r_2(n) e^{2\pi i n z},$$

where $r_2(n)$ denotes the number of representations of $n$ as a sum of $2$ (positive or negative) squares. The function $\theta^2(z)$ is a modular form of weight $1$ on $\Gamma_0(4)$, but it is not a cuspform. However, the sum

$$ \sum_{n \leq R} r_2(n) = S_2(R),$$

and so the partial sums of the coefficients of $\theta^2(z)$ indicate the number of lattice points in the circle of radius $\sqrt R$. Thus $\theta^2(z)$ gives access to the Gauss circle problem.

More generally, one can consider the number of lattice points in a $k$-dimensional sphere of radius $\sqrt R$ centered at the origin, which should approximately be the volume of that sphere,

$$ S_k(R) = \mathrm{Vol}(B(\sqrt R)) + P_k(R) = \sum_{n \leq R} r_k(n),$$

giving a $k$-dimensional lattice point discrepancy. For large dimension $k$, one should expect that the circle problem is sufficient to give good bounds and understanding of the size and error of $S_k(R)$. For $k \geq 5$, the true order of growth for $P_k(R)$ is known (up to constants).

Therefore it happens to be that the small (meaning 2 or 3) dimensional cases are both the most interesting, given our predilection for 2 and 3 dimensional geometry, and the most enigmatic. For a variety of reasons, the three dimensional case is very challenging to understand, and is perhaps even more enigmatic than the two dimensional case.

Strong evidence for the conjectured size of the lattice point discrepancy comes in the form of mean square estimates. By looking at the square, one doesn’t need to worry about oscillation from positive to negative values. And by averaging over many radii, one hopes to smooth out some of the individual bumps. These mean square estimates take the form

$$\begin{align}

\int_0^X P_2(t)^2 dt &= C X^{3/2} + O(X \log^2 X) \\

\int_0^X P_3(t)^2 dt &= C’ X^2 \log X + O(X^2 (\sqrt{ \log X})).

\end{align}$$

These indicate that the average size of $P_2(R)$ is $R^{1/4}$. and that the average size of $P_3(R)$ is $R^{1/2}$. In the two dimensional case, notice that the error term in the mean square asymptotic has pretty significant separation. It has essentially a $\sqrt X$ power-savings over the main term. But in the three dimensional case, there is no power separation. Even with significant averaging, we are only just capable of distinguishing a main term at all.

It is also interesting, but for more complicated reasons, that the main term in the three dimensional case has a log term within it. This is unique to the three dimensional case. But that is a description for another time.

In a paper that we recently posted to the arxiv, we show that the Dirichlet series

$$ \sum_{n \geq 1} \frac{S_k(n)^2}{n^s} $$

and

$$ \sum_{n \geq 1} \frac{P_k(n)^2}{n^s} $$

for $k \geq 3$ have understandable meromorphic continuation to the plane. Of particular interest is the $k = 3$ case, of course. We then investigate smoothed and unsmoothed mean square results. In particular, we prove a result stated following.

Theorem$$\begin{align} \int_0^\infty P_k(t)^2 e^{-t/X} &= C_3 X^2 \log X + C_4 X^{5/2} \\ &\quad + C_kX^{k-1} + O(X^{k-2} \end{align}$$

In this statement, the term with $C_3$ only appears in dimension $3$, and the term with $C_4$ only appears in dimension $4$. This should really thought of as saying that we understand the Laplace transform of the square of the lattice point discrepancy as well as can be desired.

We are also able to improve the sharp second mean in the dimension 3 case, showing in particular the following.

TheoremThere exists $\lambda > 0$ such that

$$\int_0^X P_3(t)^2 dt = C X^2 \log X + D X^2 + O(X^{2 – \lambda}).$$

We do not actually compute what we might take $\lambda$ to be, but we believe (informally) that $\lambda$ can be taken as $1/5$.

The major themes behind these new results are already present in the first paper in the series. The new ingredient involves handling the behavior on non-cuspforms at the cusps on the analytic side, and handling the apparent main terms (int his case, the volume of the ball) on the combinatorial side.

There is an additional difficulty that arises in the dimension 2 case which makes it distinct. But soon I will describe a different forthcoming work in that case.

Posted in Math.NT, Mathematics
Tagged gauss circle problem, gauss sphere problem, paper
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Let’s consider a ridiculous solution to a real problem. We’re unearthing tons of carbon, burning it, and releasing it into the atmosphere.

Disclaimer: There are several greenhouse gasses, and lots of other things that we’re throwing wantonly into the environment. Considering them makes things incredibly complicated incredibly quickly, so I blithely ignore them in this note.

Such rapid changes have side effects, many of which lead to bad things. That’s why nearly 150 countries ratified the Paris Agreement on Climate Change.^{1} Even if we assume that all these countries will accomplish what they agreed to (which might be challenging for the US),^{2}

most nations and advocacy groups are focusing on *increasing efficiency* and *reducing emissions.* These are good goals! But what about all the carbon that is already in the atmosphere?^{3}

You know what else is a problem? Obesity! How are we to solve all of these problems?

Looking at this (very unscientific) graph,^{4} we see that the red isn’t keeping up! Maybe we aren’t using the valuable resource of our own bodies enough! Fat has carbon in it — often over 20% by weight. What if we took advantage of our propensity to become propense? How fat would we need to get to balance last year’s carbon emissions?

That’s what we investigate here.

Posted in Data, Mathematics, Story
Tagged carbon sequestration, data visualization, global warming, xkcd style graphics
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