# Tag Archives: paper

## Choosing functions and generating figures for “When are there continuous choices for the mean value abscissa?”

In my previous note, I described some of the main ideas behind the paper “When are there continuous choices for the mean value abscissa?” that I wrote joint with Miles Wheeler. In this note, I discuss the process behind generating the functions and figures in our paper.

Our functions came in two steps: we first need to choose which functions to plot; then we need to figure out how to graphically solve their general mean value abscissae problem.

Afterwards, we can decide how to plot these functions well.

## Choosing the right functions to plot

The first goal is to find the right functions to plot. From the discussion in our paper, this amounts to specifying certain local conditions of the function. And for a first pass, we only used these prescribed local conditions.

The idea is this: to study solutions to the mean value problem, we look at the zeroes of the function $$F(b, c) = \frac{f(b) – f(a)}{b – a} – f'(c).$$ When $F(b, c) = 0$, we see that $c$ is a mean value abscissa for $f$ on the interval $(a, b)$.

By the implicit function theorem, we can solve for $c$ as a function of $b$ around a given solution $(b_0, c_0)$ if $F_c(b_0, c_0) \neq 0$. For this particular function, $F_c(b_0, c_0) = -f”(c_0)$.

More generally, it turns out that the order of vanishing of $f’$ at $b_0$ and $c_0$ governs the local behaviour of solutions in a neighborhood of $(b_0, c_0)$.

To make figures, we thus need to make functions with prescribed orders of vanishing of $f’$ at points $b_0$ and $c_0$, where $c_0$ is itself a mean value abscissa for the interval $(a_0, b_0)$.

Without loss of generality, it suffices to consider the case when $f(a_0) = f(b_0) = 0$, as otherwise we can study the function $$g(x) = f(x) – \left( \frac{f(b_0) – f(a_0)}{b_0 – a_0}(x – a_0) + f(a_0) \right),$$
which has $g(a_0) = g(b_0) = 0$, and those triples $(a, b, c)$ which solve this for $f$ also solve this for $g$.

And for consistency, we made the arbitrary decisions to have $a_0 = 0$, $b_0 = 3$, and $c_0 = 1$. This decision simplified many of the plotting decisions, as the important points were always $0$, $1$, and $3$.

## A first idea

Thus the first task is to be able to generate functions $f$ such that:

1. $f(0) = 0$,
2. $f(3) = 0$,
3. $f'(1) = 0$ (so that $1$ is a mean value abscissa), and
4. $f'(x)$ has prescribed order of vanishing at $1$, and
5. $f'(x)$ has prescribed order of vanishing at $3$.

These conditions can all be met by an appropriate interpolating polynomial. As we are setting conditions on both $f$ and its derivatives at multiple points, this amounts to the fundamental problem in Hermite interpolation. Alternatively, this amounts to using Taylor’s theorem at multiple points and then using the Chinese Remainder Theorem over $\mathbb{Z}[x]$ to combine these polynomials together.

There are clever ways of solving this, but this task is so small that it doesn’t require cleverness. In fact, this is one of the laziest solutions we could think of. We know that given $n$ Hermite conditions, there is a unique polynomial of degree $n – 1$ that interpolates these conditions. Thus we

1. determine the degree of the polynomial,
2. create a degree $n-1$ polynomial with variable coefficients in sympy,
3. have sympy symbolically compute the relations the coefficients must satisfy,
4. ask sympy to solve this symbolic system of equations.

In code, this looks like

import sympy
from sympy.abc import X, B, C, D    # Establish our variable names
def interpolate(conds):
"""
Finds the polynomial of minimal degree that solves the given Hermite conditions.

conds is a list of the form
[(x1, r1, v1), (x2, r2, v2), ...]
where the polynomial p is to satisfy p^(r_1) (x_1) = v_1, and so on.
"""
# the degree will be one less than the number of conditions
n = len(conds)

# generate a symbol for each coefficient
A = [sympy.Symbol("a[%d]" % i) for i in range(n)]

# generate the desired polynomial symbolically
P = sum([A[i] * X**i for i in range(n)])

# generate the equations the polynomial must satisfy
#
# for each (x, r, v), sympy evaluates the rth derivative of P wrt X,
# substitutes x in for X, and requires that this equals v.
EQNS = [sympy.diff(P, X, r).subs(X, x) - v for x, r, v in conds]

# solve this system for the coefficients A[n]
SOLN = sympy.solve(EQNS, A)

return P.subs(SOLN)

We note that we use the convention that a sympy symbol for something is capitalized. For example, we think of the polynomial as being represented by $$p(x) = a(0) + a(1)x + a(2)x^2 + \cdots + a(n)x^n.$$
In sympy variables, we think of this as

P = A[0] + A[1] * X + A[2] * X**2 + ... + A[n] * X**n.

With this code, we can ask for the unique degree 1 polynomial which is $1$ at $1$, and whose first derivative is $2$ at $1$.

> interpolate([(1, 0, 1), (1, 1, 2)])
2*X - 1

Indeed, $2x – 1$ is this polynomial.

### Too rigid

We have now produced a minimal Hermite solver. But there is a major downside: the unique polynomial exhibiting the necessary behaviours we required is essentially never a good didactic example. We don’t just want plots — we want beautiful, simple plots.

Firstly, we added the additional constraint that $f(1) = 1$. This is small, but it’s a small prescribed value. So now at least all three points of interest will fit within a $[0, 3] \times [0, 3]$ box.

Secondly, we also allow the choice of the value of the first nonvanishing derivatives at $1$ and $3$. In reality, we treat these as parameters to change the shape of the resulting graph. Roughly speaking, if the order of vanishing of $f(x) – f(1)$ is $k$ at $1$, then near $1$ the approximation $f(x) \approx f^{(k)}(1) x^k/k!$ is true. Morally, the larger the value of the derivative, the more the graph will resemble $x^k$ near that point.

In code, we implemented this by making functions that will add the necessary Hermite conditions to our input to interpolate.

# We fix the values of a0, b0, c0.
a0 = 0
b0 = 3
c0 = 1

# We require p(a0) = 0, p(b0) = 0, p(c0) = 1, p'(c0) = 0.
BASIC_CONDS = [(a0, 0, 0), (b0, 0, 0), (c0, 0, 1), (c0, 1, 0)]

def c_degen(n, residue):
"""
Give Hermite conditions for order of vanishing at c0 equal to n, with
first nonzero residue residue.

NOTE: the order n is in terms of f', not of f. That is, this is the amount
of additional degeneracy to add.  This may be a source of off-by-one errors.
"""
return [(c0, 1 + i, 0) for i in range(1, n + 1)] + [(c0, n + 2, residue)]

def b_degen(n, residue):
"""
Give Hermite conditions for order of vanishing at b0 equal to n, with
first nonzero residue residue.
"""
return [(b0, i, 0) for i in range(1, n + 1)] + [(b0, n + 1, residue)]

def poly_with_degens(nc=0, nb=0, residue_c=3, residue_b=3):
"""
Give unique polynomial with given degeneracies for this MVT problem.

nc is the order of vanishing of f' at c0, with first nonzero residue residue_c.
nb is the order of vanishing of f at b0, with first nonzero residue residue_b.
"""
conds = BASIC_CONDS + c_degen(nc, residue_c) + b_degen(nb, residue_b)
return interpolate(conds)

Then apparently the unique polynomial degree $5$ polynomial $f$ with $f(0) = f(3) = f'(1) = 0$, $f(1) = 1$, and $f”(1) = f'(3) = 3$ is given by

> poly_with_degens()
11*X**5/16 - 21*X**4/4 + 113*X**3/8 - 65*X**2/4 + 123*X/16

### Too many knobs

In principle, this is a great solution. And if you turn the knobs enough, you can get a really nice picture. But the problem with this system (and with many polynomial interpolation problems) is that when you add conditions, you can introduce many jagged peaks and sudden changes. These can behave somewhat unpredictably and chaotically — small changes in Hermite conditions can lead to drastic changes in resulting polynomial shape.

What we really want is for the interpolator to give a polynomial that doesn’t have sudden changes.

## Minimize change

The problem: the polynomial can have really rapid changes that makes the plots look bad.

The solution: minimize the polynomial’s change.

That is, if $f$ is our polynomial, then its rate of change at $x$ is $f'(x)$. Our idea is to “minimize” the average size of the derivative $f’$ — this should help keep the function in frame. There are many ways to do this, but we want to choose one that fits into our scheme (so that it requires as little additional work as possible) but which works well.

We decide that we want to focus our graphs on the interval $(0, 4)$. Then we can measure the average size of the derivative $f’$ by its L2 norm on $(0, 4)$: $$L2(f) = \int_0^4 (f'(x))^2 dx.$$

We add an additional Hermite condition of the form (pt, order, VAL) and think of VAL as an unknown symbol. We arbitrarily decided to start with $pt = 2$ (so that now behavior at the points $0, 1, 2, 3$ are all being controlled in some way) and $order = 1$. The point itself doesn’t matter very much, since we’re going to minimize over the family of polynomials that interpolate the other Hermite conditions with one degree of freedom.

In other words, we are adding in the condition that $f'(2) = VAL$ for an unknown VAL.

We will have sympy compute the interpolating polynomial through its normal set of (explicit) conditions as well as the symbolic condition (2, 1, VAL). Then $f = f(\mathrm{VAL}; x)$.

Then we have sympy compute the (symbolic) L2 norm of the derivative of this polynomial with respect to VAL over the interval $(0, 4)$, $$L2(\mathrm{VAL}) = \int_0^x f'(\mathrm{VAL}; x)^2 dx.$$

Finally, to minize the L2 norm, we have sympy compute the derivative of $L2(\mathrm{VAL})$ with respect to VAL and find the critical points, when the derivative is equal to $0$. We choose the first one to give our value of VAL.1

In code, this looks like

def smoother_interpolate(conds, ctrl_point=2, order=1, interval=(0,4)):
"""
Find the polynomial of minimal degree that interpolates the Hermite
conditions in conds, and whose behavior at ctrl_point minimizes the L2
norm on interval of its derivative.
"""
# Add the symbolic point to the conditions.
# Recall that D is a sympy variable
new_conds = conds + [(ctrl_point, order, D)]

# Find the polynomial interpolating new_conds, symbolic in X *and* D
P = interpolate(new_conds)

# Compute L2 norm of the derivative on interval
L2 = sympy.integrate(sympy.diff(P, X)**2, (X, *interval))

# Take the first critical point of the L2 norm with respect to D
SOLN = sympy.solve(sympy.diff(L2, D), D)[0]

# Substitute the minimizing solution in for D and return
return P.subs(D, SOLN)

def smoother_poly_with_degens(nc=0, nb=0, residue_c=3, residue_b=3):
"""
Give unique polynomial with given degeneracies for this MVT problem whose
derivative on (0, 4) has minimal L2 norm.

nc is the order of vanishing of f' at c0, with first nonzero residue residue_c.
nb is the order of vanishing of f at b0, with first nonzero residue residue_b.

"""
conds = BASIC_CONDS + c_degen(nc, residue_c) + b_degen(nb, residue_b)
return smoother_interpolate(conds)

Then apparently the polynomial degree $6$ polynomial $f$ with $f(0) = f(3) = f'(1) = 0$, $f(1) = 1$, and $f”(1) = f'(3) = 3$, and with minimal L2 derivative norm on $(0, 4)$ is given by

> smoother_poly_with_degens()
-9660585*X**6/33224848 + 27446837*X**5/8306212 - 232124001*X**4/16612424
+ 57105493*X**3/2076553 - 858703085*X**2/33224848 + 85590321*X/8306212

> sympy.N(smoother_poly_with_degens())
-0.290763858423069*X**6 + 3.30437472580762*X**5 - 13.9729157526921*X**4
+ 27.5001374874612*X**3 - 25.8452073279613*X**2 + 10.3043747258076*X

Is it much better? Let’s compute the L2 norms.

> interval = (0, 4)
> sympy.N(sympy.integrate(sympy.diff(poly_with_degens(), X)**2, (X, *interval)))
1865.15411706349

> sympy.N(sympy.integrate(sympy.diff(smoother_poly_with_degens(), X)**2, (X, *interval)))
41.1612799050325

That’s beautiful. And you know what’s better? Sympy did all the hard work.

For comparison, we can produce a basic plot using numpy and matplotlib.

import matplotlib.pyplot as plt
import numpy as np

def basic_plot(F, n=300):
fig = plt.figure(figsize=(6, 2.5))
b1d = np.linspace(-.5, 4.5, n)
f = sympy.lambdify(X, F)(b1d)
ax.plot(b1d,f,'k')
ax.set_aspect('equal')
ax.grid(True)
ax.set_xlim([-.5, 4.5])
ax.set_ylim([-1, 5])
ax.plot([0, c0, b0],[0, F.subs(X,c0),F.subs(X,b0)],'ko')
fig.savefig("basic_plot.pdf")

Then the plot of poly_with_degens() is given by

The polynomial jumps upwards immediately and strongly for $x > 3$.

On the other hand, the plot of smoother_poly_with_degens() is given by

This stays in frame between $0$ and $4$, as desired.

### Choose data to highlight and make the functions

This was enough to generate the functions for our paper. Actually, the three functions (in a total of six plots) in figures 1, 2, and 5 in our paper were hand chosen and hand-crafted for didactic purposes: the first two functions are simply a cubic and a quadratic with certain points labelled. The last function was the non-analytic-but-smooth semi-pathological counterexample, and so cannot be created through polynomial interpolation.

But the four functions highlighting different degenerate conditions in figures 3 and 4 were each created using this L2-minimizing interpolation system.

In particular, the function in figure 3 comes is

F3 = smoother_poly_with_degens(nc=1, residue_b=-3)

which is one of the simplest L2 minimizing polynomials with the typical Hermite conditions, $f”(c_0) = 0$, and opposite-default sign of $f'(b_0)$.

The three functions in figure 4 are (from left to right)

F_bmin = smoother_poly_with_degens(nc=1, nb=1, residue_c=10, residue_b=10)
F_bzero = smoother_poly_with_degens(nc=1, nb=2, residue_c=-20, residue_b=20)
F_bmax = smoother_poly_with_degens(nc=1, nb=1, residue_c=20, residue_b=-10)

We chose much larger residues because the goal of the figure is to highlight how the local behavior at those points corresponds to the behavior of the mean value abscissae, and larger residues makes those local behaviors more dominating.

## Plotting all possible mean value abscissae

Now that we can choose our functions, we want to figure out how to find all solutions of the mean value condition $$F(b, c) = \frac{f(b) – f(a_0)}{b – a_0} – f'(c).$$
Here I write $a_0$ as it’s fixed, while both $b$ and $c$ vary.

Our primary interest in these solutions is to facilitate graphical experimentation and exploration of the problem — we want these pictures to help build intuition and provide examples.

Although this may seem harder, it is actually a much simpler problem. The function $F(b, c)$ is continuous (and roughly as smooth as $f$ is).

Our general idea is a common approach for this sort of problem:

1. Compute the values of $F(b, c)$ on a tight mesh (or grid) of points.
2. Restrict attention to the domain where solutions are meaningful.
3. Plot the contour of the $0$-level set.

Contours can be well-approximated from a tight mesh. In short, if there is a small positive number and a small negative number next to each other in the mesh of computed values, then necessarily $F(b, c) = 0$ between them. For a tight enough mesh, good plots can be made.

To solve this, we again have sympy create and compute the function for us. We use numpy to generate the mesh (and to vectorize the computations, although this isn’t particularly important in this application), and matplotlib to plot the resulting contour.

Before giving code, note that the symbol F in the sympy code below stands for what we have been mathematically referring to as $f$, and not $F$. This is a potential confusion from our sympy-capitalization convention. It is still necessary to have sympy compute $F$ from $f$.

In code, this looks like

import sympy
import scipy
import numpy as np
import matplotlib.pyplot as plt

def abscissa_plot(F, n=300):
# Compute the derivative of f
DF = sympy.diff(F,X)

# Define CAP_F --- "capital F"
#
# this is (f(b) - f(0))/(b - 0) - f'(c).
CAP_F = (F.subs(X, B) - F.subs(X, 0)) / (B - 0) - DF.subs(X, C)

# build the mesh
b1d = np.linspace(-.5, 4.5, n)
b2d, c2d = np.meshgrid(b1d, b1d)

# compute CAP_F within the mesh
cap_f_mesh = sympy.lambdify((B, C), CAP_F)(b2d, c2d)

# restrict attention to below the diagonal --- we require c < b
# (although the mas inequality looks reversed in this perspective)

# Set up plot basics
fig = plt.figure(figsize=(6, 2.5))
ax.set_aspect('equal')
ax.grid(True)
ax.set_xlim([-.5, 4.5])
ax.set_ylim([-.5, 4.5])

# plot the contour
ax.contour(b2d, c2d, valid_cap_f_mesh, [0], colors='k')

# plot a diagonal line representing the boundary
ax.plot(b1d,b1d,'k--')

# plot the guaranteed point
ax.plot(b0,c0,'ko')

fig.savefig("abscissa_plot.pdf")

Then plots of solutions to $F(b, c) = 0$ for our basic polynomials are given by

for poly_with_degens(), while for smoother_poly_with_degens() we get

And for comparison, we can now create a (slightly worse looking) version of the plots in figure 3.

F3 = smoother_poly_with_degens(nc=1, residue_b=-3)
basic_plot(F3)
abscissa_plot(F3)

This produces the two plots

For comparison, a (slightly scaled) version of the actual figure appearing in the paper is

## Copy of the code

A copy of the code used in this note (and correspondingly the code used to generate the functions for the paper) is available on my github as an ipython notebook.

Posted in Expository, Math.CA, Mathematics, Programming, Python, sagemath | | 3 Comments

# When are there continuous choices for the Mean Value Abscissa?

Miles Wheeler and I have recently uploaded a paper to the arXiv called “When are there continuous choices for the mean value abscissa?”, which we have submitted to an expository journal. The underlying question is simple but nontrivial.

The mean value theorem of calculus states that, given a differentiable function $f$ on an interval $[a, b]$, then there exists a $c \in (a, b)$ such that
$$\frac{f(b) – f(a)}{b – a} = f'(c).$$
We call $c$ the mean value abscissa.
Our question concerns potential behavior of this abscissa when we fix the left endpoint $a$ of the interval and vary $b$. For each $b$, there is at least one abscissa $c_b$ such that the mean value theorem holds with that abscissa. But generically there may be more than one choice of abscissa for each interval. When can we choose $c_b$ as a continuous function of $b$? That is, when can we write $c = c(b)$ such that
$$\frac{f(b) – f(a)}{b – a} = f'(c(b))$$
for all $b$ in some interval?
We think of this as a continuous choice for the mean value abscissa.

This is a great question. It’s widely understandable — even to students with only one semester of calculus. Further it encourages a proper understanding of what a function is, as thinking of $c$ as potentially a function of $b$ is atypical and interesting.

But I also like this question because the answer is not as simple as you might think, and there are a few nice ideas that get to the answer.

Should you find yourself reading this without knowing the answer, I encourage you to consider it right now. Should continuous choices of abscissas exist? What if the function is really well-behaved? What if it’s smooth? Or analytic?

Let’s focus on the smooth question. Suppose that $f$ is smooth — that it is infinitely differentiable. These are a distinguished class of functions. But it turns out that being smooth is not sufficient: here is a counterexample.

In this figure, there are points $b$ arbitrarily near $b_0$ such that the secant line from $a_0$ to $b$ have positive slope, and points arbitrarily near such that the secant lines have negative slope. There are infinitely many mean value abscissae with $f'(c_0) = 0$, but all of them are either far from a point $c$ where $f'(c) > 0$ or far from a point $c$ where $f'(c) < 0$. And thus there is no continuous choice. From a theorem oriented point of view, our main theorem is that if $f$ is analytic, then there is always a locally continuous choice. That is, for every interval $[a_0, b_0]$, there exists a mean value abscissa $c$ such that $c = c(b)$ for some interval $B$ containing $b_0$. But the purpose of this article isn’t simply to prove this theorem. The purpose is to exposit how the ideas that are used to study this problem and to prove these results are fundamentally based only on a couple of central ideas covered in introductory single and multivariable calculus. All of this paper is completely accessible to a student having studied only single variable calculus (and who is willing to believe that partial derivatives exist are a reasonable object). We prove and use simple-but-nontrivial versions of the contraction mapping theorem, the implicit function theorem, and Morse’s lemma. The implicit function theorem is enough to say that any abscissa $c_0$ such that $f”(c_0) \neq 0$ has a unique continuous extension. Thus immediately for “most” intervals on “most” reasonable functions, we answer in the affirmative. Morse’s lemma allows us to say a bit more about the case when $f”(c_0) = 0$ but $f'{}'{}'(c_0) \neq 0$. In this case there are either multiple continuous extensions or none. And a few small ingredients and the idea behind Morse’s lemma, combined with the implicit function theorem again, is enough to prove the main result. ## Student projects A calculus student looking for a project to dive into and sharpen their calculus skills could find ideas here to sink their teeth into. Beginning by understanding this paper is a great start. A good motivating question would be to carry on one additional step, and to study explicitly the behavior of a function near a point where $f”(c_0) = f'{}'{}'(c_0) = 0$, but $f^{(4)}(c_0) \neq 0$. A slightly more open question that we lightly touch on (but leave largely implicit) is ther inverse question: when can one find a mean value abscissa $c$ such that the right endpoint $b$ can be written as a continuous function $b(c)$ for some neighborhood $C$ containing the initial point $c_0$? Much of the analysis is the same, but figuring it out would require some attention. A much deeper question is to consider the abscissa as a function of both the left endpoint $a$ and the right endpoint $b$. The guiding question here could be to decide when one can write the abscissa as a continuous function $c(a, b)$ in a neighborhood of $(a_0, b_0)$. I would be interested to see a graphical description of the possible shapes of these functions — I’m not quite sure what they might look like. There is also a nice computational problem. In the paper, we include several plots of solution curves in $(b, c)$ space. But we did this with a meshed implicit function theorem solver. A computationally inclined student could devise an explicit way of constructing solutions. On the one hand, this is guaranteed to work since one can apply contraction mappings explicitly to make the resulting function from the implicit function theorem explicit. But on the other hand, many (most?) applications of the implicit function theorem are in more complicated high dimensional spaces, whereas the situation in this paper is the smallest nontrivial example. ## Producing the graphs We made 13 graphs in 5 figures for this article. These pictures were created using matplotlib. The data was created using numpy, scipy, and sympy from within the scipy/numpy python stack, and the actual creation was done interactively within a jupyter notebook. The actual notebook is available here, (along with other relatively raw jupyter notebooks). The most complicated graph is this one.

This figure has graphs of three functions along the top. In each graph, the interval $[0, 3]$ is considered in the mean value theorem, and the point $c_0 = 1$ is a mean value abscissa. In each, we also have $f”(c_0) = 0$, and the point is that the behavior of $f”(b_0)$ has a large impact on the nature of the implicit functions. The three graphs along the bottom are in $(b, c)$ space and present all mean value abscissa for each $b$. This is not a function, but the local structure of the graphs are interesting and visually distinct.

The process of making these examples and making these figures is interesting in itself. We did not make these figures explicitly, but instead chose certain points and certain values of derivatives at those points, and used Hermite interpolation find polynomials with those points.1

In the future I plan on writing a note on the creation of these figures.

## “Second Moments in the Generalized Gauss Circle Problem” (with T. Hulse, C. Ieong Kuan, and A. Walker)

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.

Theorem

There 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.

## The Collatz Conjecture – recent development?

On his site, John D. Cook recently proliferated a paper by Gerhard Opfer that claimed to solve the Collatz Conjecture. The Collatz Conjecture is simple to state:

Collatz (or the 3n + 1 conjecture):
Starting at any number do the following: if n is even, divide by 2; if n is odd, multiply by 3 and add 1.
The conjecture states that no matter what positive integer you start at, you will end up at 1 (the so-called 1-4-2 loop).

At first, I had high hopes for the paper. It seems relatively well-written and was submitted to the Mathematics of Computation, a very respectable journal. I even sent out a brief email about the paper. But the paper is flawed. The problem, I think, can be succinctly summarized by the following: he relies on the assumption that starting with any number $n_0$, one will eventually hit a number that is less than $n_0$. When stated like this, it seems obvious that there is a problem, but he only relied on that one number (rather than the apparent infinite descent that could follow). The exact problem occurs with his ‘annihilation argument’ on page 11 of the pdf above. He more or less states that one can start at 1 and reach every number by doing a sort of reverse Collatz function (he’s actually a bit wittier than that), but does not prove it.

More commentary can be found on reddit, reddit again, and on math.SE (a question protected by Qiaocho Yuan – go him).

I use this as an intro to a sort of joke that goes around mathematician’s circles. A while back, Sean Carroll wrote up ‘The Alternative-Science Respectability Checklist,’ and it’s awesome. Find it here. It turns out that Scott Aaronson wrote up a similar article, inspired by Sean Carroll, that is titled “Ten Signs a Claimed Mathematical Breakthrough is Wrong.”

His inspiration was the time-old problem that simply stated problems encourage generations up people to attack them, and frequently to think that they have made progress. So he asks :

Suppose someone sends you a complicated solution to a famous decades-old math problem, like P vs. NP. How can you decide, in ten minutes or less, whether the solution is worth reading?

And thus his 10 signs were created. I happen to have heard a few people say that this most recent paper on the Collatz Conjecture only failed three: #6 (The paper jumps into technicalities without presenting a new idea), #8 (The paper wastes lots of space on standard material), and #10 (The techniques just seem too wimpy for the problem at hand). {though perhaps #8 is debateable – some say it’s related to a different convention of writing papers, but I don’t know about any of that}

In my experience, I rely mostly on #1 (it’s not written in $TeX$), #4 (it conflicts with some impossibility result), and #7 (it doesn’t build on any previous work). But both of these articles are very funny, though not exactly precise nor entirely true.

Posted in Expository, Humor, Mathematics, Open | | 2 Comments