Math 100 Fall 2016: Concluding Remarks

It is that time of year. Classes are over. Campus is emptying. Soon it will be mostly emptiness, snow, and grad students (who of course never leave).

I like to take some time to reflect on the course. How did it go? What went well and what didn’t work out? And now that all the numbers are in, we can examine course trends and data.

Since numbers are direct and graphs are pretty, let’s look at the numbers first.

Math 100 grades at a glance

Let’s get an understanding of the distribution of grades in the course, all at once.

box_plots

These are classic box plots. The center line of each box denotes the median. The left and right ends of the box indicate the 1st and 3rd quartiles. As a quick reminder, the 1st quartile is the point where 25% of students received that grade or lower. The 3rd quartile is the point where 75% of students received that grade or lower. So within each box lies 50% of the course.

Each box has two arms (or “whiskers”) extending out, indicating the other grades of students. Points that are plotted separately are statistical outliers, which means that they are $1.5 \cdot (Q_3 – Q_1)$ higher than $Q_3$ or lower than $Q_1$ (where $Q_1$ denotes the first quartile and $Q_3$ indicates the third quartile).

A bit more information about the distribution itself can be seen in the following graph.

violin_plot

Within each blob, you’ll notice an embedded box-and-whisker graph. The white dots indicate the medians, and the thicker black parts indicate the central 50% of the grade. The width of the colored blobs roughly indicate how many students scored within that region. [As an aside, each blob actually has the same area, so the area is a meaningful data point].

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Computing pi with tools from Calculus

Computing $\pi$

This note was originally written in the context of my fall Math 100 class at Brown University. It is also available as a pdf note.

While investigating Taylor series, we proved that
\begin{equation}\label{eq:base}
\frac{\pi}{4} = 1 – \frac{1}{3} + \frac{1}{5} – \frac{1}{7} + \frac{1}{9} + \cdots
\end{equation}
Let’s remind ourselves how. Begin with the geometric series
\begin{equation}
\frac{1}{1 + x^2} = 1 – x^2 + x^4 – x^6 + x^8 + \cdots = \sum_{n = 0}^\infty (-1)^n x^{2n}. \notag
\end{equation}
(We showed that this has interval of convergence $\lvert x \rvert < 1$). Integrating this geometric series yields
\begin{equation}
\int_0^x \frac{1}{1 + t^2} dt = x – \frac{x^3}{3} + \frac{x^5}{5} – \frac{x^7}{7} + \cdots = \sum_{n = 0}^\infty (-1)^n \frac{x^{2n+1}}{2n+1}. \notag
\end{equation}
Note that this has interval of convergence $-1 < x \leq 1$.

We also recognize this integral as
\begin{equation}
\int_0^x \frac{1}{1 + t^2} dt = \text{arctan}(x), \notag
\end{equation}
one of the common integrals arising from trigonometric substitution. Putting these together, we find that
\begin{equation}
\text{arctan}(x) = x – \frac{x^3}{3} + \frac{x^5}{5} – \frac{x^7}{7} + \cdots = \sum_{n = 0}^\infty (-1)^n \frac{x^{2n+1}}{2n+1}. \notag
\end{equation}
As $x = 1$ is within the interval of convergence, we can substitute $x = 1$ into the series to find the representation
\begin{equation}
\text{arctan}(1) = 1 – \frac{1}{3} + \frac{1}{5} – \frac{1}{7} + \cdots = \sum_{n = 0}^\infty (-1)^n \frac{1}{2n+1}. \notag
\end{equation}
Since $\text{arctan}(1) = \frac{\pi}{4}$, this gives the representation for $\pi/4$ given in \eqref{eq:base}.

However, since $x=1$ was at the very edge of the interval of convergence, this series converges very, very slowly. For instance, using the first $50$ terms gives the approximation
\begin{equation}
\pi \approx 3.121594652591011. \notag
\end{equation}
The expansion of $\pi$ is actually
\begin{equation}
\pi = 3.141592653589793238462\ldots \notag
\end{equation}
So the first $50$ terms of \eqref{eq:base} gives two digits of accuracy. That’s not very good.

I think it is very natural to ask: can we do better? This series converges slowly — can we find one that converges more quickly?

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Series Convergence Tests with Prototypical Examples

This is a note written for my Fall 2016 Math 100 class at Brown University. We are currently learning about various tests for determining whether series converge or diverge. In this note, we collect these tests together in a single document. We give a brief description of each test, some indicators of when each test would be good to use, and give a prototypical example for each. Note that we do justify any of these tests here — we’ve discussed that extensively in class. [But if something is unclear, send me an email or head to my office hours]. This is here to remind us of the variety of the various tests of convergence.

A copy of just the statements of the tests, put together, can be found here. A pdf copy of this whole post can be found here.

In order, we discuss the following tests:

  1. The $n$th term test, also called the basic divergence test
  2. Recognizing an alternating series
  3. Recognizing a geometric series
  4. Recognizing a telescoping series
  5. The Integral Test
  6. P-series
  7. Direct (or basic) comparison
  8. Limit comparison
  9. The ratio test
  10. The root test

The $n$th term test

Statement

Suppose we are looking at $\sum_{n = 1}^\infty a_n$ and
\begin{equation}
\lim_{n \to \infty} a_n \neq 0. \notag
\end{equation}
Then $\sum_{n = 1}^\infty a_n$ does not converge.

When to use it

When applicable, the $n$th term test for divergence is usually the easiest and quickest way to confirm that a series diverges. When first considering a series, it’s a good idea to think about whether the terms go to zero or not. But remember that if the limit of the individual terms is zero, then it is necessary to think harder about whether the series converges or diverges.

Example

Each of the series
\begin{equation}
\sum_{n = 1}^\infty \frac{n+1}{2n + 4}, \quad \sum_{n = 1}^\infty \cos n, \quad \sum_{n = 1}^\infty \sqrt{n} \notag
\end{equation}
diverges since their limits are not $0$.

Recognizing alternating series

Statement

Suppose $\sum_{n = 1}^\infty (-1)^n a_n$ is a series where

  1. $a_n \geq 0$,
  2. $a_n$ is decreasing, and
  3. $\lim_{n \to \infty} a_n = 0$.

Then $\sum_{n = 1}^\infty (-1)^n a_n$ converges.

Stated differently, if the terms are alternating sign, decreasing in absolute size, and converging to zero, then the series converges.

When to use it

The key is in the name — if the series is alternating, then this is the goto idea of analysis. Note that if the terms of a series are alternating and decreasing, but the terms do not go to zero, then the series diverges by the $n$th term test.

Example

Suppose we are looking at the series
\begin{equation}
\sum_{n = 1}^\infty \frac{(-1)^n}{\log(n+1)} = \frac{-1}{\log 2} + \frac{1}{\log 3} + \frac{-1}{\log 4} + \cdots \notag
\end{equation}
The terms are alternating.
The sizes of the terms are $\frac{1}{\log (n+1)}$, and these are decreasing.
Finally,
\begin{equation}
\lim_{n \to \infty} \frac{1}{\log(n+1)} = 0. \notag
\end{equation}
Thus the alternating series test applies and shows that this series converges.

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A Notebook Preparing for a Talk at Quebec-Maine

This is a notebook containing a representative sample of the code I used to  generate the results and pictures presented at the Quebec-Maine Number Theory Conference on 9 October 2016. It was written in a Jupyter Notebook using Sage 7.3, and later converted for presentation on this site.
There is a version of the notebook available on github. Alternately, a static html version without WordPress formatting is available here. Finally, this notebook is also available in pdf form.
The slides for my talk are available here.

Testing for a Generalized Conjecture on Iterated Sums of Coefficients of Cusp Forms

Let $f$ be a weight $k$ cusp form with Fourier expansion

$$ f(z) = \sum_{n \geq 1} a(n) e(nz). $$

Deligne has shown that $a(n) \ll n^{\frac{k-1}{2} + \epsilon}$. It is conjectured that

$$ S_f^1(n) := \sum_{m \leq X} a(m) \ll X^{\frac{k-1}{2} + \frac{1}{4} + \epsilon}. $$

It is known that this holds on average, and we recently showed that this holds on average in short intervals.
(See HKLDW1, HKLDW2, and HKLDW3 for details and an overview of work in this area).
This is particularly notable, as the resulting exponent is only 1/4 higher than that of a single coefficient.
This indicates extreme cancellation, far more than what is implied merely by the signs of $a(n)$ being random.

It seems that we also have

$$ \sum_{m \leq X} S_f^1(m) \ll X^{\frac{k-1}{2} + \frac{2}{4} + \epsilon}. $$

That is, the sum of sums seems to add in only an additional 1/4 exponent.
This is unexpected and a bit mysterious.

The purpose of this notebook is to explore this and higher conjectures.
Define the $j$th iterated sum as

$$ S_f^j(X) := \sum_{m \leq X} S_f^{j-1} (m).$$

Then we numerically estimate bounds on the exponent $\delta(j)$ such that

$$ S_f^j(X) \ll X^{\frac{k-1}{2} + \delta(j) + \epsilon}. $$

In [1]:
# This was written in SageMath 7.3 through a Jupyter Notebook.
# Jupyter interfaces to sage by loading it as an extension
%load_ext sage

# sage plays strangely with ipython. This re-allows inline plotting
from IPython.display import display, Image

We first need a list of coefficients of one (or more) cusp forms.
For initial investigation, we begin with a list of 50,000 coefficients of the weight $12$ cusp form on $\text{SL}(2, \mathbb{Z})$, $\Delta(z)$, i.e. Ramanujan’s delta function.
We will use the data associated to the 50,000 coefficients for pictoral investigation as well.

We will be performing some numerical investigation as well.
For this, we will use the first 2.5 million coefficients of $\Delta(z)$

In [2]:
# Gather 10 coefficients for simple checking
check_10 = delta_qexp(11).coefficients()
print check_10

fiftyk_coeffs = delta_qexp(50000).coefficients()
print fiftyk_coeffs[:10] # these match expected

twomil_coeffs = delta_qexp(2500000).coefficients()
print twomil_coeffs[:10] # these also match expected
[1, -24, 252, -1472, 4830, -6048, -16744, 84480, -113643, -115920]
[1, -24, 252, -1472, 4830, -6048, -16744, 84480, -113643, -115920]
[1, -24, 252, -1472, 4830, -6048, -16744, 84480, -113643, -115920]
In [3]:
# Function which iterates partial sums from a list of coefficients

def partial_sum(baselist):
    ret_list = [baselist[0]]
    for b in baselist[1:]:
        ret_list.append(ret_list[-1] + b)
    return ret_list

print check_10
print partial_sum(check_10) # Should be the partial sums
[1, -24, 252, -1472, 4830, -6048, -16744, 84480, -113643, -115920]
[1, -23, 229, -1243, 3587, -2461, -19205, 65275, -48368, -164288]
In [4]:
# Calculate the first 10 iterated partial sums
# We store them in a single list list, `sums_list`
# the zeroth elelemnt of the list is the array of initial coefficients
# the first element is the array of first partial sums, S_f(n)
# the second element is the array of second iterated partial sums, S_f^2(n)

fiftyk_sums_list = []
fiftyk_sums_list.append(fiftyk_coeffs) # zeroth index contains coefficients
for j in range(10):                    # jth index contains jth iterate
    fiftyk_sums_list.append(partial_sum(fiftyk_sums_list[-1]))
    
print partial_sum(check_10)
print fiftyk_sums_list[1][:10]         # should match above
    
twomil_sums_list = []
twomil_sums_list.append(twomil_coeffs) # zeroth index contains coefficients
for j in range(10):                    # jth index contains jth iterate
    twomil_sums_list.append(partial_sum(twomil_sums_list[-1]))
    
print twomil_sums_list[1][:10]         # should match above
[1, -23, 229, -1243, 3587, -2461, -19205, 65275, -48368, -164288]
[1, -23, 229, -1243, 3587, -2461, -19205, 65275, -48368, -164288]
[1, -23, 229, -1243, 3587, -2461, -19205, 65275, -48368, -164288]

As is easily visible, the sums alternate in sign very rapidly.
For instance, we believe tha the first partial sums should change sign about once every $X^{1/4}$ terms in the interval $[X, 2X]$.
In this exploration, we are interested in the sizes of the coefficients.
But in HKLDW3, we investigated some of the sign changes of the partial sums.

Now seems like a nice time to briefly look at the data we currently have.
What do the first 50 thousand coefficients look like?
So we normalize them, getting $A(n) = a(n)/n^{5.5}$ and plot these coefficients.

In [5]:
norm_list = []
for n,e in enumerate(fiftyk_coeffs, 1):
    normalized_element = 1.0 * e / (1.0 * n**(5.5))
    norm_list.append(normalized_element)
print norm_list[:10]
1
In [6]:
# Make a quick display
normed_coeffs_plot = scatter_plot(zip(range(1,60000), norm_list), markersize=.02)
normed_coeffs_plot.save("normed_coeffs_plot.png")
display(Image("normed_coeffs_plot.png"))

Since some figures will be featuring prominently in the talk I’m giving at Quebec-Maine, let us make high-quality figures now.

 

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  1. 00000000000000, -0.530330085889911, 0.598733612492945, -0.718750000000000, 0.691213333204735, -0.317526448138560, -0.376547696558964, 0.911504835123284, -0.641518061271148, -0.366571226366719
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Math 100: Completing the partial fractions example from class

An Unfinished Example

At the end of class today, someone asked if we could do another example of a partial fractions integral involving an irreducible quadratic. We decided to look at the integral

$$ \int \frac{1}{(x^2 + 4)(x+1)}dx. $$
Notice that $latex {x^2 + 4}$ is an irreducible quadratic polynomial. So when setting up the partial fraction decomposition, we treat the $latex {x^2 + 4}$ term as a whole.

So we seek to find a decomposition of the form

$$ \frac{1}{(x^2 + 4)(x+1)} = \frac{A}{x+1} + \frac{Bx + C}{x^2 + 4}. $$
Now that we have the decomposition set up, we need to solve for $latex {A,B,}$ and $latex {C}$ using whatever methods we feel most comfortable with. Multiplying through by $latex {(x^2 + 4)(x+1)}$ leads to

$$ 1 = A(x^2 + 4) + (Bx + C)(x+1) = (A + B)x^2 + (B + C)x + (4A + C). $$
Matching up coefficients leads to the system of equations

$$\begin{align}
0 &= A + B \\
0 &= B + C \\
1 &= 4A + C.
\end{align}$$
So we learn that $latex {A = -B = C}$, and $latex {A = 1/5}$. So $latex {B = -1/5}$ and $latex {C = 1/5}$.

Together, this means that

$$ \frac{1}{(x^2 + 4)(x+1)} = \frac{1}{5}\frac{1}{x+1} + \frac{1}{5} \frac{-x + 1}{x^2 + 4}. $$
Recall that if you wanted to, you could check this decomposition by finding a common denominator and checking through.

Now that we have performed the decomposition, we can return to the integral. We now have that

$$ \int \frac{1}{(x^2 + 4)(x+1)}dx = \underbrace{\int \frac{1}{5}\frac{1}{x+1}dx}_ {\text{first integral}} + \underbrace{\int \frac{1}{5} \frac{-x + 1}{x^2 + 4} dx.}_ {\text{second integral}} $$
We can handle both of the integrals on the right hand side.

The first integral is

$$ \frac{1}{5} \int \frac{1}{x+1} dx = \frac{1}{5} \ln (x+1) + C. $$

The second integral is a bit more complicated. It’s good to see if there is a simple $latex {u}$-substition, since there is an $latex {x}$ in the numerator and an $latex {x^2}$ in the denominator. But unfortunately, this integral needs to be further broken into two pieces that we know how to handle separately.

$$ \frac{1}{5} \int \frac{-x + 1}{x^2 + 4} dx = \underbrace{\frac{-1}{5} \int \frac{x}{x^2 + 4}dx}_ {\text{first piece}} + \underbrace{\frac{1}{5} \int \frac{1}{x^2 + 4}dx.}_ {\text{second piece}} $$

The first piece is now a $latex {u}$-substitution problem with $latex {u = x^2 + 4}$. Then $latex {du = 2x dx}$, and so

$$ \frac{-1}{5} \int \frac{x}{x^2 + 4}dx = \frac{-1}{10} \int \frac{du}{u} = \frac{-1}{10} \ln u + C = \frac{-1}{10} \ln (x^2 + 4) + C. $$

The second piece is one of the classic trig substitions. So we draw a triangle.

triangle

In this triangle, thinking of the bottom-left angle as $latex {\theta}$ (sorry, I forgot to label it), then we have that $latex {2\tan \theta = x}$ so that $latex {2 \sec^2 \theta d \theta = dx}$. We can express the so-called hard part of the triangle by $latex {2\sec \theta = \sqrt{x^2 + 4}}$.

Going back to our integral, we can think of $latex {x^2 + 4}$ as $latex {(\sqrt{x^2 + 4})^2}$ so that $latex {x^2 + 4 = (2 \sec \theta)^2 = 4 \sec^2 \theta}$. We can now write our integral as

$$ \frac{1}{5} \int \frac{1}{x^2 + 4}dx = \frac{1}{5} \int \frac{1}{4 \sec^2 \theta} 2 \sec^2 \theta d \theta = \frac{1}{5} \int \frac{1}{2} d\theta = \frac{1}{10} \theta. $$
As $latex {2 \tan \theta = x}$, we have that $latex {\theta = \text{arctan}(x/2)}$. Inserting this into our expression, we have

$$ \frac{1}{10} \int \frac{1}{x^2 + 4} dx = \frac{1}{10} \text{arctan}(x/2) + C. $$

Combining the first integral and the first and second parts of the second integral together (and combining all the constants $latex {C}$ into a single constant, which we also denote by $latex {C}$), we reach the final expression

$$ \int \frac{1}{(x^2 + 4)(x + 1)} dx = \frac{1}{5} \ln (x+1) – \frac{1}{10} \ln(x^2 + 4) + \frac{1}{10} \text{arctan}(x/2) + C. $$

And this is the answer.

Other Notes

If you have any questions or concerns, please let me know. As a reminder, I have office hours on Tuesday from 9:30–11:30 (or perhaps noon) in my office, and I highly recommend attending the Math Resource Center in the Kassar House from 8pm-10pm, offered Monday-Thursday. [Especially on Tuesday and Thursdays, when there tend to be fewer people there].

On my course page, I have linked to two additional resources. One is to Paul’s Online Math notes for partial fraction decomposition (which I think is quite a good resource). The other is to the Khan Academy for some additional worked through examples on polynomial long division, in case you wanted to see more worked examples. This note can also be found on my website, or in pdf form.

Good luck, and I’ll see you in class.

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“On Functions Whose Mean Value Abscissas are Midpoints, with Connections to Harmonic Functions” (with Paul Carter)

This is joint work with Paul Carter. Humorously, we completed this while on a cross-country drive as we moved the newly minted Dr. Carter from Brown to Arizona.

I’ve had a longtime fascination with the standard mean value theorem of calculus.

Mean Value Theorem
Suppose $f$ is a differentiable function. Then there is some $c \in (a,b)$ such that
\begin{equation}
\frac{f(b) – f(a)}{b-a} = f'(c).
\end{equation}

The idea for this project started with a simple question: what happens when we interpret the mean value theorem as a differential equation and try to solve it? As stated, this is too broad. To narrow it down, we might specify some restriction on the $c$, which we refer to as the mean value abscissa, guaranteed by the Mean Value Theorem.

So I thought to try to find functions satisfying
\begin{equation}
\frac{f(b) – f(a)}{b-a} = f’ \left( \frac{a + b}{2} \right)
\end{equation}
for all $a$ and $b$ as a differential equation. In other words, let’s try to find all functions whose mean value abscissas are midpoints.

This looks like a differential equation, which I only know some things about. But my friend and colleague Paul Carter knows a lot about them, so I thought it would be fun to ask him about it.

He very quickly told me that it’s essentially impossible to solve this from the perspective of differential equations. But like a proper mathematician with applied math leanings, he thought we should explore some potential solutions in terms of their Taylor expansions. Proceeding naively in this way very quickly leads to the answer that those (assumed smooth) solutions are precisely quadratic polynomials.

It turns out that was too simple. It was later pointed out to us that verifying that quadratic polynomials satisfy the midpoint mean value property is a common exercise in calculus textbooks, including the one we use to teach from at Brown. Digging around a bit reveals that this was even known (in geometric terms) to Archimedes.

So I thought we might try to go one step higher, and see what’s up with
\begin{equation}\label{eq:original_midpoint}
\frac{f(b) – f(a)}{b-a} = f’ (\lambda a + (1-\lambda) b), \tag{1}
\end{equation}
where $\lambda \in (0,1)$ is a weight. So let’s find all functions whose mean value abscissas are weighted averages. A quick analysis with Taylor expansions show that (assumed smooth) solutions are precisely linear polynomials, except when $\lambda = \frac{1}{2}$ (in which case we’re looking back at the original question).

That’s a bit odd. It turns out that the midpoint itself is distinguished in this way. Why might that be the case?

It is beneficial to look at the mean value property as an integral property instead of a differential property,
\begin{equation}
\frac{1}{b-a} \int_a^b f'(t) dt = f’\big(c(a,b)\big).
\end{equation}
We are currently examining cases when $c = c_\lambda(a,b) = \lambda a + (1-\lambda b)$. We can see the right-hand side is differentiable by differentiating the left-hand side directly. Since any point can be a weighted midpoint, one sees that $f$ is at least twice-differentiable. One can actually iterate this argument to show that any $f$ satisfying one of the weighted mean value properties is actually smooth, justifying the Taylor expansion analysis indicated above.

An attentive eye might notice that the midpoint mean value theorem, written as the integral property
\begin{equation}
\frac{1}{b-a} \int_a^b f'(t) dt = f’ \left( \frac{a + b}{2} \right)
\end{equation}
is exactly the one-dimensional case of the harmonic mean value property, usually written
\begin{equation}
\frac{1}{\lvert B_h \rvert} = \int_{B_h(x)} g(t) dV = g(x).
\end{equation}
Here, $B_h(x)$ is the ball of radius $h$ and center $x$. Any harmonic function satisfies this mean value property, and any function satisfying this mean value property is harmonic.

From this viewpoint, functions satisfying our original midpoint mean value property~\eqref{eq:original_midpoint} have harmonic derivatives. But the only one-dimensional harmonic functions are affine functions $g(x) = cx + d$. This gives immediately that the set of solutions to~\eqref{eq:original_midpoint} are quadratic polynomials.

The weighted mean value property can also be written as an integral property. Trying to connect it similarly to harmonic functions led us to consider functions satisfying
\begin{equation}
\frac{1}{\lvert B_h \rvert} = \int_{B_h(x)} g(t) dV = g(c_\lambda(x,h)),
\end{equation}
where $c_\lambda(x,h)$ should be thought of as some distinguished point in the ball $B_h(x)$ with a weight parameter $\lambda$. More specifically,

Are there weighted harmonic functions corresponding to a weighted harmonic mean value property?
In one dimension, the answer is no, as seen above. But there are many more multivariable harmonic functions [in fact, I’ve never thought of harmonic functions on $\mathbb{R}^1$ until this project, as they’re too trivial]. So maybe there are weighted harmonic functions in higher dimensions?

This ends up being the focus of the latter half of our paper. Unexpectedly (to us), an analogous methodology to our approach in the one-dimensional case works, with only a few differences.

It turns out that no, there are no weighted harmonic functions on $\mathbb{R}^n$ other than trivial extensions of harmonic functions from $\mathbb{R}^{n-1}$.

Harmonic functions are very special, and even more special than we had thought. The paper is a fun read, and can be found on the arxiv now. It has been accepted and will appear in American Mathematical Monthly.

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Paper: Sign Changes of Coefficients and Sums of Coefficients of Cusp Forms

This is joint work with Thomas Hulse, Chan Ieong Kuan, and Alex Walker, and is a another sequel to our previous work. This is the third in a trio of papers, and completes an answer to a question posed by our advisor Jeff Hoffstein two years ago.

We have just uploaded a preprint to the arXiv giving conditions that guarantee that a sequence of numbers contains infinitely many sign changes. More generally, if the sequence consists of complex numbers, then we give conditions that guarantee sign changes in a generalized sense.

Let $\mathcal{W}(\theta_1, \theta_2) := { re^{i\theta} : r \geq 0, \theta \in [\theta_1, \theta_2]}$ denote a wedge of complex plane.

Suppose ${a(n)}$ is a sequence of complex numbers satisfying the following conditions:

  1. $a(n) \ll n^\alpha$,
  2. $\sum_{n \leq X} a(n) \ll X^\beta$,
  3. $\sum_{n \leq X} \lvert a(n) \rvert^2 = c_1 X^{\gamma_1} + O(X^{\eta_1})$,

where $\alpha, \beta, c_1, \gamma_1$, and $\eta_1$ are all real numbers $\geq 0$. Then for any $r$ satisfying $\max(\alpha+\beta, \eta_1) – (\gamma_1 – 1) < r < 1$, the sequence ${a(n)}$ has at least one term outside any wedge $\mathcal{W}(\theta_1, \theta_2)$ with $0 \theta_2 – \theta_1 < \pi$ for some $n \in [X, X+X^r)$ for all sufficiently large $X$.

These wedges can be thought of as just slightly smaller than a half-plane. For a complex number to escape a half plane is analogous to a real number changing sign. So we should think of this result as guaranteeing a sort of sign change in intervals of width $X^r$ for all sufficiently large $X$.

The intuition behind this result is very straightforward. If the sum of coefficients is small while the sum of the squares of the coefficients are large, then the sum of coefficients must experience a lot of cancellation. The fact that we can get quantitative results on the number of sign changes is merely a task of bookkeeping.

Both the statement and proof are based on very similar criteria for sign changes when ${a(n)}$ is a sequence of real numbers first noticed by Ram Murty and Jaban Meher. However, if in addition it is known that

\begin{equation}
\sum_{n \leq X} (a(n))^2 = c_2 X^{\gamma_2} + O(X^{\eta_2}),
\end{equation}

and that $\max(\alpha+\beta, \eta_1, \eta_2) – (\max(\gamma_1, \gamma_2) – 1) < r < 1$, then generically both sequences ${\text{Re} (a(n)) }$ and $latex{ \text{Im} (a(n)) }$ contain at least one sign change for some $n$ in $[X , X + X^r)$ for all sufficiently large $X$. In other words, we can detect sign changes for both the real and imaginary parts in intervals, which is a bit more special.

It is natural to ask for even more specific detection of sign changes. For instance, knowing specific information about the distribution of the arguments of $a(n)$ would be interesting, and very closely reltated to the Sato-Tate Conjectures. But we do not yet know how to investigate this distribution.

In practice, we often understand the various criteria for the application of these two sign changes results by investigating the Dirichlet series
\begin{align}
&\sum_{n \geq 1} \frac{a(n)}{n^s} \\
&\sum_{n \geq 1} \frac{S_f(n)}{n^s} \\
&\sum_{n \geq 1} \frac{\lvert S_f(n) \rvert^2}{n^s} \\
&\sum_{n \geq 1} \frac{S_f(n)^2}{n^s},
\end{align}
where
\begin{equation}
S_f(n) = \sum_{m \leq n} a(n).
\end{equation}

In the case of holomorphic cusp forms, the two previous joint projects with this group investigated exactly the Dirichlet series above. In the paper, we formulate some slightly more general criteria guaranteeing sign changes based directly on the analytic properties of the Dirichlet series involved.

In this paper, we apply our sign change results to our previous work to show that $S_f(n)$ changes sign in each interval $[X, X + X^{\frac{2}{3} + \epsilon})$ for sufficiently large $X$. Further, if there are coefficients with $\text{Im} a(n) \neq 0$, then the real and imaginary parts each change signs in those intervals.

We apply our sign change results to single coefficients of $\text{GL}(2)$ cusp forms (and specifically full integral weight holomorphic cusp forms, half-integral weight holomorphic cusp forms, and Maass forms). In large part these are minor improvements over folklore and what is known, except for the extension to complex coefficients.

We also apply our sign change results to single isolated coefficients $A(1,m)$ of $\text{GL}(3)$ Maass forms. This seems to be a novel result, and adds to the very sparse literature on sign changes of sequences associated to $\text{GL}(3)$ objects. Murty and Meher recently proved a general sign change result for $\text{GL}(n)$ objects which is similar in feel.

As a final application, we also consider sign changes of partial sums of $\nu$-normalized coefficients. Let
\begin{equation}
S_f^\nu(X) := \sum_{n \leq X} \frac{a(n)}{n^{\nu}}.
\end{equation}
As $\nu$ gets larger, the individual coefficients $a(n)n^{-\nu}$ become smaller. So one should expect that sign changes in ${S_f^\nu(n)}$ to change based on $\nu$. And in particular, as $\nu$ gets very large, the number of sign changes of $S_f^\nu$ should decrease.

Interestingly, in the case of holomorphic cusp forms of weight $k$, we are able to show that there are sign changes of $S_f^\nu(n)$ in intervals even for normalizations $\nu$ a bit above $\nu = \frac{k-1}{2}$. This is particularly interesting as $a(n) \ll n^{\frac{k-1}{2} + \epsilon}$, so for $\nu > \frac{k-1}{2}$ the coefficients are \emph{decreasing} with $n$. We are able to show that when $\nu = \frac{k-1}{2} + \frac{1}{6} – \epsilon$, the sequence ${S_f^\nu(n)}$ has at least one sign change for $n$ in $[X, 2X)$ for all sufficiently large $X$.

It may help to consider a simpler example to understand why this is surprising. Consider the classic example of a sequence of $b(n)$, where $b(n) = 1$ or $b(n) = -1$, randomly, with equal probability. Then the expected size of the sums of $b(n)$ is about $\sqrt n$. This is an example of \emph{square-root cancellation}, and such behaviour is a common point of comparison. Similarly, the number of sign changes of the partial sums of $b(n)$ is also expected to be about $\sqrt n$.

Suppose now that $b(n) = \frac{\pm 1}{\sqrt n}$. If the first term is $1$, then it takes more then the second term being negative to make the overall sum negative. And if the first two terms are positive, then it would take more then the following three terms being negative to make the overall sum negative. So sign changes of the partial sums are much rarer. In fact, they’re exceedingly rare, and one might barely detect more than a dozen through computational experiment (although one should still expect infinitely many).

This regularity, in spite of the decreasing size of the individual coefficients $a(n)n^{-\nu}$, suggests an interesting regularity in the sign changes of the individual $a(n)$. We do not know how to understand or measure this effect or its regularity, and for now it remains an entirely qualitative observation.

For more details and specific references, see the paper on the arXiv.

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Math 42 Spring 2016 Student Showcase

This spring, I taught Math 42: An Introduction to Elementary Number Theory at Brown University. An important aspect of the course was the final project. In these projects, students either followed up on topics that interested them from the semester, or chose and investigated topics related to number theory.  Projects could be done individual or in small groups.

I thought it would be nice to showcase some excellent student projects from my class. Most of the projects were quite good, and some showed extraordinary effort. Some students really dove in and used this as an opportunity to explore and digest a topic far more thoroughly than could possibly be expected from an introductory class such as this one. With the students’ permission, I’ve chosen five student projects (in no particular order) for a blog showcase (impressed by similar sorts  of showcases from Scott Aaronson).

  • Factorization Techniques, by Elvis Nunez and Chris Shaw. In this project, Elvis and Chris look at Fermat Factorization, which looks to factor $n$ by expressing $n = a^2 – b^2$. Further, they investigate improvements to Fermat’s Algorithm by Dixon and Kraitchik. Following this line of investigation leads to the development of the modern quadratic sieve and factor base methods of factorization.

  • Pseudoprimes and Carmichael Numbers, by Emily Riemer. Fermat’s Little Theorem is one of the first “big idea” theorems we encounter in the course, and we came back to it again and again throughout. Emily explored the Fermat’s Little Theorem as a primality test, leading to pseudoprimes, strong pseudoprimes, and Carmichael numbers. [As an aside, one of her references concerning Carmichael numbers were notes from an algebraic number theory class taught by Matt Baker, who first got me interested in number theory].

  • Continued Fractions and Pell’s Equation, by Max Lahn and Jonathan Spiegel. As it happened, I did not have time to teach continued fractions in the course.  So Max and Jonathan decided to look at them on their own. They explore some ideas related to the convergence of continued fractions and see how one uses continued fractions to solve Pell’s Equation.

  • Quantum Computing, by Edward Hu and Chris Long. Edward and Chris explore quantum computing with particular emphasis towards gaining some idea of how Shor’s factorization algorithm works. For some of the more complicated ideas, like the quantum Fourier transform, they make use of heuristic and analogy to purvey the main ideas.

  • Fermat’s Last Theorem, by Dylan Groos, Natalie Schudrowitz, and Kenneth Berglund. Dylan, Natalie, and Kenneth provide a historical look at attacks on Fermat’s Last Theorem. They examine proofs for $n=4$ and Sophie Germaine’s remarkable advances. They also touch on elliptic curves and modular forms, hinting at some of the deep ideas lying beneath the surface.

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Math 42: Concluding Remarks

As this semester draws to an end, it is time to reflect on what we’ve done. What worked well? What didn’t work well? What would I change if I taught this course again?

Origins of the course

This course was created by my advisor, Jeff Hoffstein, many years ago in order to offer a sort of bridge between high school math and “real math.” The problem is that in primary and secondary school, students are not exposed to the grand, modern ideas of mathematics. They are forced to drill exercises and repeat formulae. Often, the greatest and largest exposure to mathematical reasoning is hidden among statements of congruent triangles and Side-Angle-Side theorems. Most students arrive at university thinking that math is over and done with. What else could there possibly remain to do in math?

Math 42 was designed to attract nonscience majors, especially those not intending to pursue the standard calculus sequence, and to convince them to study some meaningful mathematics. Ideally, students begin to think mathematically and experience some of the thrill of independent intellectual discovery.

It is always a bit surprising to me that so many students find their way into this class each spring. This class does not have a natural lead-in, it satisfies no prerequisites, and it is not in the normal track for math concentrators. One cannot even pretend to make the argument that number theory is a useful day-to-day skill. Yet number theory has a certain appeal… there are so many immediate and natural questions. It is possible to get a hint that there is something deep going on within the first two classes.

Further, I think there is something special about the first homework assigned in a course. Homeworks send a really strong signal about the content of a course. I want this course to be more about the students exploring, asking questions, and experimenting than about repeating the same old examples and techniques from the class. So the first several questions on the first homework are dedicated to open-ended exploration.

There are side effects to this approach. Open ended exploration is uncertain, and therefore scary. I hope that it’s intriguing enough (and different enough) that students push through initial discomfort, but I’m acutely aware that this can be an intimidating time. Perhaps in another time, students were more comfortable with uncertainty — but that is a discussion for later. I’m pretty sure that many fears are assuaged after the first week, once the idea that it is okay to not know what you’re going to learn before you learn it. In fact, it’s more fun that way! (One also learns much more rapidly).

My approach to this course is strongly influenced by my experiences teaching number theory to high schoolers as part of the Summer@Brown program for the past several years. During my first summer teaching that course, I co-taught it with Jackie Anderson, who is an excellent and thoughtful instructor. I also strongly draw on the excellent textbook A Friendly Introduction to Number Theory by Joe Silverman, written specifically for this course about 20 years ago.

Trying final projects

I am still surprised each time I teach this course. I tried one major new thing in this course that I’ve thought about for a long time — students were required to do a final project on a topic of their choice. It turns out that this is a great idea, and I would absolutely do it again. The paths available to the students really opens up in the second half of the course. With a few basic tools (in particular once we’ve mastered modular arithmetic, linear congruences, and the Chinese Remainder Theorem) the number of deeply interesting and accessible topics is huge. There is some great truth here, about how a few basic structural tools allow one to explore all sorts of new playgrounds.

A final project allows students to realize this for themselves. It also fits in with the motif of experimentation and self-discovery that pervades the whole course. The structural understanding from the first half of the course is enough to pursue some really interesting, and occasionally even useful, topics. More importantly, they learn that they are capable of finding, learning, and understanding complex mathematical ideas on their own. And usually, they enjoy it, since it’s fun to learn cool things.

For various reasons, I had thought it would be a good idea to offer students an alternative to final projects, in the form of a somewhat challenging final exam. In hindsight, I now think this was not such a good idea. Students who do not perform final projects miss out on a sort of representative capstone experience in the course.

There are a few other things that I would have done differently, if given the chance. I would ask students to be on the lookout for topics and groups much earlier, perhaps about a third of the way into the semester (instead of about two thirds of the way into the course). My students did an extraordinary job at their projects this semester. But I think with some additional rumination time, more groups would pursue projects based more on their own particular interests. Or perhaps not — I’ll see next time.

Several of my students who really dove into their final projects have agreed to have their work showcased here (which is something that I’ll get back to in a later note). This means that students in later courses will have something to refer back to. [Whether this is a good thing remainds to be seen, but I suspect it’s for the better].

Interesting correlations

In these concluding notes, I often like to try to draw correlations between certain patterns of behaviour and success in the course. I’m very often interested in the question of how early on in a course one can accurately predict a final grade. In calculus courses, it seems one can very accurately predict a final grade using only the first midterm grade.

In this course, fewer such correlations are meaningful. Most notably, a large percentage of the class took the course as pass/fail. [I can’t blame them, as this course is supposed to draw students a bit out of their comfort zone into a topic they know little about]. This distorts the entire incentive structure of the course in relation to other demands of college life.

There is a very strong correlation between taking the class for a grade and receiving a high numeric grade at the end. I think this comes largely from two causes: students who are more confident with the subject material coming into the course decide to take it for a grade, and then perform in line with their expectations; and taking a class for pass/fail creates an incentive structure with high emphasis on learning enough material to pass, but not necessarily mastering all the material.

In sharp contrast to my experience in calculus courses, there is a pretty strong correlation between homework grade and with overall performance. While this may seem obvious, this has absolutely not been my experience in calculus courses. Generally poor homework grades correlate extremely strongly with poor final grades, but strong homework has had almost no correlation with strong performance.

I think that a reason why homework might be a better predicter in this course is that homework is harder. There are always open-ended problems, and every homework had at least one or two problems designed to take a lot of experimentation and thought. Students who did well on the homework put in that experimentation and thinking time, reflecting better study habits, higher commitment, and more grit (like in this TED talk).

Finally, there is an extremely high correlation with students attending office hours and strong performance in the course. It will always remain a mystery to me why more students don’t take advantage of office hours. [It might be that this is also a measure other characteristics, such as commitment, study habits, and grit].

Don’t forget the coffee

This was my favorite course I’ve taught at Brown. On the flipside, I think my students enjoyed this class more than any other class I’ve taught at Brown. This is one of those courses that rejuvenates the soul.

While returning home after one of the final classes, I flipped on NPR and listened to Innovation Hub. On the program was Steven Strogatz, a well-known mathematician and expositor, talking about his general dislike of the Calculus-Is-The-Pinnacle-Of-Mathematics style approach that is so common today in high schools and colleges. The program in particular can be found here.

He argues that standard math education is missing some important topics, especially related to financial numeracy. But he also argues that the current emphasis is not on the beauty or attraction of mathematics, but on a very particular set of applications [and in particular, towards creating rockets].

While this course isn’t perfect, I do think that it is the sort of course that Strogatz would approve of — somewhat like a Survey of Shakespeare course, but in mathematics.

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Math 420: Supplement on Gaussian Integers II

This is a secondary supplemental note on the Gaussian integers, written for my Spring 2016 Elementary Number Theory Class at Brown University. This note is also available as a pdf document.

In this note, we cover the following topics.

  1. Assumed prerequisites from other lectures.
  2. Which regular integer primes are sums of squares?
  3. How can we classify all Gaussian primes?

1. Assumed Prerequisites

Although this note comes shortly after the previous note on the Gaussian integers, we covered some material from the book in the middle. In particular, we will assume use the results from chapters 20 and 21 from the textbook.

Most importantly, for $latex {p}$ a prime and $latex {a}$ an integer not divisible by $latex {p}$, recall the Legendre symbol $latex {\left(\frac{a}{p}\right)}$, which is defined to be $latex {1}$ if $latex {a}$ is a square mod $latex {p}$ and $latex {-1}$ if $latex {a}$ is not a square mod $latex {p}$. Then we have shown Euler’s Criterion, which states that

$$ a^{\frac{p-1}{2}} \equiv \left(\frac{a}{p}\right) \pmod p, \tag{1}$$
and which gives a very efficient way of determining whether a given number $latex {a}$ is a square mod $latex {p}$.

We used Euler’s Criterion to find out exactly when $latex {-1}$ is a square mod $latex {p}$. In particular, we concluded that for each odd prime $latex {p}$, we have

$$ \left(\frac{-1}{p}\right) = \begin{cases} 1 & \text{ if } p \equiv 1 \pmod 4 \ -1 & \text{ if } p \equiv 3 \pmod 4 \end{cases}. \tag{2}$$
Finally, we assume familiarity with the notation and ideas from the previous note on the Gaussian integers.

2. Understanding When $latex {p = a^2 + b^2}$.

Throughout this section, $latex {p}$ will be a normal odd prime. The case $latex {p = 2}$ is a bit different, and we will need to handle it separately. When used, the letters $latex {a}$ and $latex {b}$ will denote normal integers, and $latex {q_1,q_2}$ will denote Gaussian integers.

We will be looking at the following four statements.

  1. $latex {p \equiv 1 \pmod 4}$
  2. $latex {\left(\frac{-1}{p}\right) = 1}$
  3. $latex {p}$ is not a Gaussian prime
  4. $latex {p = a^2 + b^2}$

Our goal will be to show that each of these statements are equivalent. In order to show this, we will show that

$$ (1) \implies (2) \implies (3) \implies (4) \implies (1). \tag{3}$$
Do you see why this means that they are all equivalent?

This naturally breaks down into four lemmas.

We have actually already shown one.

Lemma 1 $latex {(1) \implies (2)}$.

Proof: We have already proved this claim! This is exactly what we get from Euler’s Criterion applied to $latex {-1}$, as mentioned in the first section. $latex \Box$

There is one more that is somewhat straightfoward, and which does not rely on going up to the Gaussian integers.

Lemma 2 $latex {(4) \implies (1)}$.

Proof: We have an odd prime $latex {p}$ which is a sum of squares $latex {p = a^2 + b^2}$. If we look mod $latex {4}$, we are led to consider $$ p = a^2 + b^2 \pmod 4. \tag{4}$$
What are the possible values of $latex {a^2 \pmod 4}$? A quick check shows that the only possibilites are $latex {a^2 \equiv 0, 1 \pmod 4}$.

So what are the possible values of $latex {a^2 + b^2 \pmod 4}$? We must have one of $latex {p \equiv 0, 1, 2 \pmod 4}$. Clearly, we cannot have $latex {p \equiv 0 \pmod 4}$, as then $latex {4 \mid p}$. Similarly, we cannot have $latex {p \equiv 2 \pmod 4}$, as then $latex {2 \mid p}$. So we necessarily have $latex {p \equiv 1 \pmod 4}$, which is what we were trying to prove. $latex \Box$

For the remaining two pieces, we will dive into the Gaussian integers.

Lemma 3 $latex {(2) \implies (3)}$.

Proof: As $latex {\left(\frac{-1}{p}\right) = 1}$, we know there is some $latex {a}$ so that $latex {a^2 \equiv -1 \pmod p}$. Rearranging, this becomes $latex {a^2 + 1 \equiv 0 \pmod p}$.

Over the normal integers, we are at an impasse, as all this tells us is that $latex {p \mid (a^2 + 1)}$. But if we suddenly view this within the Gaussian integers, then $latex {a^2 + 1}$ factors as $latex {a^2 + 1 = (a + i)(a – i)}$.

So we have that $latex {p \mid (a+i)(a-i)}$. If $latex {p}$ were a Gaussian prime, then we would necessarily have $latex {p \mid (a+i)}$ or $latex {p \mid (a-i)}$. (Do you see why?)

But is it true that $latex {p}$ divides $latex {a + i}$ or $latex {a – i}$? For instance, does $latex {p}$ divide $latex {a + i}$? No! If so, then $latex {\frac{a}{p} + \frac{i}{p}}$ would be a Gaussian integer, which is clearly not true.

So $latex {p}$ does not divide $latex {a + i}$ or $latex {a-i}$, and we must therefore conclude that $latex {p}$ is not a Gaussian prime. $latex \Box$

Lemma 4 $latex {(3) \implies (4)}$.

Proof: We now know that $latex {p}$ is not a Gaussian prime. In particular, this means that $latex {p}$ is not irreducible, and so it has a nontrivial factorization in the Gaussian integers. (For example, $latex {5}$ is a regular prime, but it is not a Gaussian prime. It factors as $latex {5 = (1 + 2i)(1 – 2i)}$ in the Gaussian integers.)

Let’s denote this nontrivial factorization as $latex {p = q_1 q_2}$. By nontrivial, we mean that neither $latex {q_1}$ nor $latex {q_2}$ are units, i.e. $latex {N(q_1), N(q_2) > 1}$. Taking norms, we see that $latex {N(p) = N(q_1) N(q_2)}$.

We can evaluate $latex {N(p) = p^2}$, so we have that $latex {p^2 = N(q_1) N(q_2)}$. Both $latex {N(q_1)}$ and $latex {N(q_2)}$ are integers, and their product is $latex {p^2}$. Yet $latex {p^2}$ has exactly two different factorizations: $latex {p^2 = 1 \cdot p^2 = p \cdot p}$. Since $latex {N(q_1), N(q_2) > 1}$, we must have the latter.

So we see that $latex {N(q_1) = N(q_2) = p}$. As $latex {q_1, q_2}$ are Gaussian integers, we can write $latex {q_1 = a + bi}$ for some $latex {a, b}$. Then since $latex {N(q_1) = p}$, we see that $latex {N(q_1) = a^2 + b^2}$. And so $latex {p}$ is a sum of squares, ending the proof. $latex \Box$

Notice that $latex {2 = 1 + 1}$ is also a sum of squares. Then all together, we can say the following theorem.

Theorem 5 A regular prime $latex {p}$ can be written as a sum of two squares, $$ p = a^2 + b^2, \tag{5}$$
exactly when $latex {p = 2}$ or $latex {p \equiv 1 \pmod 4}$.

A remarkable aspect of this theorem is that it is entirely a statement about the behaviour of the regular integers. Yet in our proof, we used the Gaussian integers in a very fundamental way. Isn’t that strange?

You might notice that in the textbook, Dr. Silverman presents a proof that does not rely on the Gaussian integers. While interesting and clever, I find that the proof using the Gaussian integers better illustrates the deep connections between and around the structures we have been studying in this course so far. Everything connects!

Example 1 The prime $latex {5}$ is $latex {1 \pmod 4}$, and so $latex {5}$ is a sum of squares. In particular, $latex {5 = 1^2 + 2^2}$.

Example 2 The prime $latex {101}$ is $latex {1 \pmod 4}$, and so is a sum of squares. Our proof is not constructive, so a priori we do not know what squares sum to $latex {101}$. But in this case, we see that $latex {101 = 1^2 + 10^2}$.

Example 3 The prime $latex {97}$ is $latex {1 \pmod 4}$, and so it also a sum of squares. It’s less obvious what the squares are in this case. It turns out that $latex {97 = 4^2 + 9^2}$.

Example 4 The prime $latex {43}$ is $latex {3 \pmod 4}$, and so is not a sum of squares.

3. Classification of Gaussian Primes

In the previous section, we showed that each integer prime $latex {p \equiv 1 \pmod 4}$ actually splits into a product of two Gaussian numbers $latex {q_1}$ and $latex {q_2}$. In fact, since $latex {N(q_1) = p}$ is a regular prime, $latex {q_1}$ is a Gaussian irreducible and therefore a Gaussian prime (can you prove this? This is a nice midterm question.)

So in fact, $latex {p \equiv 1 \pmod 4}$ splits in to the product of two Gaussian primes $latex {q_1}$ and $latex {q_2}$.

In this way, we’ve found infinitely many Gaussian primes. Take a regular prime congruent to $latex {1 \pmod 4}$. Then we know that it splits into two Gaussian primes. Further, if we know how to write $latex {p = a^2 + b^2}$, then we know that $latex {q_1 = a + bi}$ and $latex {q_2 = a – bi}$ are those two Gaussian primes.

In general, we will find all Gaussian primes by determining their interaction with regular primes.

Suppose $latex {q}$ is a Gaussian prime. Then on the one hand, $latex {N(q) = q \overline{q}}$. On the other hand, $latex {N(q) = p_1^{a_1} p_2^{a_2} \cdots p_k^{a_k}}$ is some regular integer. Since $latex {q}$ is a Gaussian prime (and so $latex {q \mid w_1 w_2}$ means that $latex {q \mid w_1}$ or $latex {q \mid w_2}$), we know that $latex {q \mid p_j}$ for some regular integer prime $latex {p_j}$.

So one way to classify Gaussian primes is to look at every regular integer prime and see which Gaussian primes divide it. We have figured this out for all primes $latex {p \equiv 1 \pmod 4}$. We can handle $latex {2}$ by noticing that $latex {2 = (1 + i) (1-i)}$. Both $latex {(1+i)}$ and $latex {(1-i)}$ are Gaussian primes.

The only primes left are those regular primes with $latex {p \equiv 3 \pmod 4}$. We actually already covered the key idea in the previous section.

Lemma 6 If $latex {p \equiv 3 \pmod 4}$ is a regular prime, then $latex {p}$ is also a Gaussian prime.

Proof: In the previous section, we showed that if $latex {p}$ is not a Gaussian prime, then $latex {p = a^2 + b^2}$ for some integers $latex {a,b}$, and then $latex { p \equiv 1 \pmod 4}$. Since $latex {p \not \equiv 1 \pmod 4}$, we see that $latex {p}$ is a Gaussian prime. $latex \Box$

In total, we have classified all Gaussian primes.

Theorem 7 The Gaussian primes are given by

  1. $latex {(1+i), (1-i)}$
  2. Regular primes $latex {p \equiv 3 \pmod 4}$
  3. The factors $latex {q_1 q_2}$ of a regular prime $latex {p \equiv 1 \pmod 4}$. Further, these primes are given by $latex {a \pm bi}$, where $latex {p = a^2 + b^2}$.

 

4. Concluding Remarks

I hope that it’s clear that the regular integers and the Gaussian integers are deeply connected and intertwined. Number theoretic questions in one constantly lead us to investigate the other. As one dives deeper into number theory, more and different integer-like rings appear, all deeply connected.

Each time I teach the Gaussian integers, I cannot help but feel the sense that this is a hint at a deep structural understanding of what is really going on. The interplay between the Gaussian integers and the regular integers is one of my favorite aspects of elementary number theory, which is one reason why I deviated so strongly from the textbook to include it. I hope you enjoyed it too.

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