Category Archives: Expository

Talk on computing Maass forms

In a remarkable coincidence, I’m giving two talks on Maass forms today (after not giving any talks for 3 months). One of these was a chalk talk (or rather camera on pen on paper talk). My other talk can be found at https://davidlowryduda.com/static/Talks/ComputingMaass20/.

In this talk, I briefly describe how one goes about computing Maass forms for congruence subgroups of $\mathrm{SL}(2)$. This is a short and pointed exposition of ideas mostly found in papers of Hejhal and Fredrik Strömberg’s PhD thesis. More precise references are included at the end of the talk.

This amounts to a description of the idea of Hejhal’s algorithm on a congruence subgroup.

Side notes on revealjs

I decided to experiment a bit with this talk. This is not a TeX-Beamer talk (as is most common for math) — instead it’s a revealjs talk. I haven’t written a revealjs talk before, but it was surprisingly easy.

It took me more time than writing a beamer talk, most likely because I don’t have a good workflow with reveal and there were several times when I wanted to use nontrivial javascript capabilities. In particular, I wanted to have a few elements transition from one slide to the next (using the automatic transition capabilities).

At first, I had thought I would write in an intermediate markup format and then translate this into revealjs, but I quickly decided against that plan. The composition stage was a bit more annoying.

But I think the result is more appealing than a beamer talk, and it’s sufficiently interesting that I’ll revisit it later.

The current cover art for the Proceedings of the Royal Society

The current issue of the Proceedings of the Royal Society A1 features cover artwork made by Vikas Krishnamurthy, Miles Wheeler, Darren Crowdy, Adrian Constantin, and me.

A version of the cover pre-addition is the following.

This is based on the work in A transformation between stationary point vortex equilibria, which concerns solutions to Euler’s equation for inviscid (2D) fluid motion $$\frac{\partial \mathbf{V}}{\partial t} + (\mathbf{V} \cdot \nabla) \mathbf{V} = – \frac{\nabla p}{p_0},$$ where $\nabla = (\partial/\partial x, \partial / \partial y)$ is the 2D gradient operator. There is a notion of vortices for these systems, and the paper examines configurations of point vortices under certain idealized conditions that leads to particularly nice analysis. In the situation studied, one can sometimes begin with one configurations of point vortices and perform a transformation that yields another, bigger and more complicated configuration.

This is the situation depicted on the cover — begin with a simple configuration and iterate the process. The spiral shape was added afterwards and doesn’t describe underlying mathematical phenomena. The different colors of each vortex shows whether that vortex is a sink or a source, essentially.

I was told most of this after the fact by Miles — who researches fluid dynamics, is a friend from grad school, and was my coauthor on a paper about the mean value theorem. I do not typically think about fluid dynamics (and did not write the paper), and it’s a bit funny how I got involved in the production of this cover. But it was fun, and we produced many arresting images. In the future Miles and I intend to revisit these images and better describe how the various aspects of the image describes and reflects the underlying mathematical behavior.

As a fun aside — we didn’t only produce one image. We made many, and we made many configurations.2 In my work on visualizing modular forms, I developed a few techniques for color selection from matplotlib style colormaps, and produced several variants. I’ve collected a few of these below.

Notes from a talk at Dartmouth on the Fibonacci zeta function

I recently gave a talk “at Dartmouth”1. The focus of the talk was the (odd-indexed) Fibonacci zeta function:
$$\sum_{n \geq 1} \frac{1}{F(2n-1)^s},$$
where $F(n)$ is the nth Fibonacci number. The theme is that the Fibonacci zeta function can be recognized as coming from an inner product of automorphic forms, and the continuation of the zeta function can be understood in terms of the spectral expansion of the associated automorphic forms.

This is a talk from ongoing research. I do not yet understand “what’s really going on”. But within the talk I describe a few different generalizations; firstly, there is a generalization to other zeta functions that can be viewed as traces of units on quadratic number fields, and secondly there is a generalization to quadratic forms recognizing solutions to Pell’s equation.

I intend to describe additional ideas from this talk in the coming months, as I figure out how pieces fit together. But for now, here are the slides.

Pictures of equidistribution – the line

In my previous note, we considered equidistribution of rational points on the circle $X^2 + Y^2 = 2$. This is but one of a large family of equidistribution results that I’m not particularly familiar with.

This note is the first in a series of notes dedicated to exploring this type of equidistribution visually. In this note, we will investigate a simpler case — rational points on the line.

Points on X^2 + Y^2 = 2 equidistribute with respect to height

When you order rational points on the circle $X^2 + Y^2 = 2$ by height, these points equidistribute.

Stated differently, suppose that $I$ is an arc on the circle $X^2 + Y^2 = 2$. Then asymptotically, the number of rational points on the arc $I$ with height bounded by a number $H$ is equal to what you would expect if $\lvert I\rvert /2\sqrt{2}\pi$ of all points with height up to $H$ were on this arc. Here, $\lvert I\rvert /2\sqrt{2}\pi$ the ratio of the arclength of the arc $I$ with the total circumference of the circle.

This only makes sense if we define the height of a rational point on the circle. Given a point $(a/c, b/c)$ (written in least terms) on the circle, we define the height of this point to be $c$.

In forthcoming work with my frequent collaborators Chan Ieong Kuan, Thomas Hulse, and Alexander Walker, we count three term arithmetic progressions of squares. If $C^2 – B^2 = B^2 – A^2$, then clearly $A^2 + C^2 = 2B^2$, and thus a 3AP of squares corresponds to a rational point on the circle $X^2 + Y^2 = 2$. We compare one of our results to what you would expect from equidistribution. From general principles, we expected such equidistribution to be true. But I wasn’t sure how to prove it.

With helpful assistance from Noam Elkies, Emmanuel Peyre, and John Voight (who each immediately knew how to prove this), I learned how to prove this fact.

The rest of this note contains this proof.

Posted in Expository, Math.NT, Mathematics, sage, sagemath | | 1 Comment

Notes behind a talk: visualizing modular forms

Today, I’ll be at Bowdoin College giving a talk on visualizing modular forms. This is a talk about the actual process and choices involved in illustrating a modular form; it’s not about what little lies one might hold in their head in order to form some mental image of a modular form.1

This is a talk heavily inspired by the ICERM semester program on Illustrating Mathematics (currently wrapping up). In particular, I draw on2 conversations with Frank Farris (about using color to highlight desired features), Elias Wegert (about using logarithmically scaling contours), Ed Harriss (about the choice of colorscheme), and Brendan Hassett (about overall design choices).

There are very many pictures in the talk!

I wrote a few different complex-plotting routines for this project. At their core, they are based on sage’s complex_plot. There are two major variants that I use.

The first (currently called “ccomplex_plot”. Not a good name) overwrites how sage handles lightness in complex_plot in order to produce “contours” at spots where the magnitude is a two-power. These contours are actually a sudden jump in brightness.

The second (currently called “raw_complex_plot”, also not a good name) is even less formal. It vectorizes the computation and produces an object containing the magnitude and argument information for each pixel to be drawn. It then uses numpy and matplotlib to convert these magnitudes and phases into RGB colors according to a matplotlib-compatible colormap.

I am happy to send either of these pieces of code to anyone who wants to see them, but they are very much written for my own use at the moment. I intend to improve them for general use later, after I’ve experimented further.

In addition, I generated all the images for this talk in a single sagemath jupyter notebook (with the two .spyx cython dependencies I allude to above). This is also available here. (Note that using a service like nbviewer or nbconvert to view or convert it to html might be a reasonable idea).

As a final note, I’ll add that I mistyped several times in the preparation of the images for this talk. Included below are a few of the interesting-looking mistakes. The first two resulted from incorrectly applied conformal mappings, while the third came from incorrectly applied color correction.

Posted in Expository, Math.NT, Mathematics, sage, sagemath, sagemath | | 2 Comments

Notes from a talk at the Maine-Quebec Number Theory Conference

Today I will be giving a talk at the Maine-Quebec Number Theory conference. Each year that I attend this conference, I marvel at how friendly and inviting an environment it is — I highly recommend checking the conference out (and perhaps modelling other conferences after it).

The theme of my talk is about spectral poles and their contribution towards asymptotics (especially of error terms). I describe a few problems in which spectral poles appear in asymptotics. Unlike the nice simple cases where a single pole (or possibly a few poles) appear, in these cases infinite lines of poles appear.

For a bit over a year, I have encountered these and not known what to make of them. Could you have the pathological case that residues of these poles generically cancel? Could they combine to be larger than expected? How do we make sense of them?

The resolution came only very recently.1

I will later write a dedicated note to this new idea (involving Dirichlet integrals and Landau’s theorem in this context), but for now — here are the slides for my talk.

Posted in Expository, Math.NT, Mathematics | | 2 Comments

The Insidiousness of Mathematics

1.

a. Having a gradual and cumulative effect
b. of a disease : developing so gradually as to be well established before becoming apparent

2.
a. awaiting a chance to entrap
b. harmful but enticing

— Merriam-Webster Dictionary

In early topics in mathematics, one can often approach a topic from a combination of intution and first principles in order to deduce the desired results. In later topics, it becomes necessary to repeatedly sharpen intuition while taking advantage of the insights of the many mathematicians who came before — one sees much further by standing on the giants. Somewhere in the middle, it becomes necessary to accept the idea that there are topics and ideas that are not at all obvious. They might appear to have been plucked out of thin air. And this is a conceptual boundary.

In my experience, calculus is often the class where students primarily confront the idea that it is necessary to take advantage of the good ideas of the past. It sneaks up. The main ideas of calculus are intuitive — local rates of change can be approximated by slopes of secant lines and areas under curves can be approximated by sums of areas of boxes. That these are deeply connected is surprising.

To many students, Taylor’s Theorem is one of the first examples of a commonly-used result whose proof has some aspect which appears to have been plucked out of thin air.1 Learning Taylor’s Theorem in high school was one of the things that inspired me to begin to revisit calculus with an eye towards why each result was true.

I also began to try to prove the fundamental theorems of single and multivariable calculus with as little machinery as possible. High school me thought that topology was overcomplicated and unnecessary for something so intuitive as calculus.2

This train of thought led to my previous note, on another proof of Taylor’s Theorem. That note is a simplified version of one of the first proofs I devised on my own.

Much less obviously, this train of thought also led to the paper on the mean value theorem written with Miles. Originally I had thought that “nice” functions should clearly have continuous choices for mean value abscissae, and I thought that this could be used to provide alternate proofs for some fundamental calculus theorems. It turns out that there are very nice functions that don’t have continuous choices for mean value abscissae, and that actually using that result to prove classical calculus results is often more technical than the typical proofs.

The flow of ideas is turbulent, highly nonlinear.

I used to think that developing extra rigor early on in my mathematical education was the right way to get to deeper ideas more quickly. There is a kernel of truth to this, as transitioning from pre-rigorous mathematics to rigorous mathematics is very important. But it is also necessary to transition to post-rigorous mathematics (and more generally, to choose one’s battles) in order to organize and communicate one’s thoughts.

In hindsight, I think now that I was focused on the wrong aspect. As a high school student, I had hoped to discover the obvious, clear, intuitive proofs of every result. Of course it is great to find these proofs when they exist, but it would have been better to grasp earlier that sometimes these proofs don’t exist. And rarely does actual research proceed so cleanly — it’s messy and uncertain and full of backtracking and random exploration.

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.

Posted in Expository, Math.CA, Mathematics | | 1 Comment