… Data aequatione quotcunque fluentes quantitates involvente fluxiones invenire et vice versa …

## Category: Maths

### A better estimate of Kempner’s series

The Kempner series recently regained some notoriety due to a Saturday Morning Breakfast Cereal comic (the last panel). The observation first appeared in a 1914 American Mathematical Monthly article, in which it was shown that the series consisting of the usual harmonic series

$\displaystyle \sum_{n = 1}^{\infty} \frac{1}{n}$

but with all the terms, whose decimal expansion includes the digit ‘9’, removed, in fact converges to some number below 80. The original proof is given in the Wikipedia article linked above, so I will not repeat it. But to make it easier to see the idea: let us first think about the case where the number is expressed in base 2. In base 2, all the positive integers has the leading binary bit being 1 (since it cannot be zero). Therefore there are no binary positive numbers without the bit ‘1’ in its expansion. So the corresponding series converges trivially to zero. How about the case of the bit ‘0’? The only binary numbers without any ‘0’ bits are

$1, 3 = (11)_2, 7 = (111)_2, 15 = (1111)_2, \ldots, 2^n - 1$.

So the corresponding series actually becomes

$\displaystyle \sum_{n = 1}^\infty \frac{1}{2^n - 1} \leq \sum_{n = 1}^\infty \frac{1}{2^{n-1}} = 2$

So somewhere from the heavily divergent harmonic series, we pick up a rapidly converging geometric series. So what’s at work here? Among all the n-bit binary numbers, exactly 1 has all bits not being 0. So the density of these kinds of numbers decays rather quickly: in base 2, there are $2^{n-1}$ numbers that are exactly n-bit long. So if a number $x$ has a binary representation that is exactly n bits long (which means that $2^{n} \leq x < 2^{n+1}$), the chances that it is one of the special type of numbers is $\frac{1}{2^{n-1}} \approx \frac{2}{x}$. This probability we can treat then as a density: replacing the discrete sum $\sum \frac{1}{n}$ by the integral $\int \frac{1}{x}\mathrm{d}x$ (calculus students may recognize this as the germ of the “integral test”) and replacing the $\mathrm{d}x$ by the density $\frac{2}{x} \mathrm{d}x$, we get the estimate

$\displaystyle \text{Binary Kempner series} \approx \int_1^\infty \frac{2}{x^2} = 2$.

Doing the same thing with the original Kempner series gives that the chances a n-digit number does not contain the digit nine to be

$\displaystyle \left(\frac89\right)\left(\frac9{10}\right)^{n-1} \approx \left( \frac{9}{10}\right)^{n}$

The length of the decimal expansion of a natural number $x$ is basically $1 + \log x$. So the density we are interested in becomes

$\displaystyle \left( \frac{9}{10}\right)^{1+\log x} ~\mathrm{d}x$

From this we can do an integral estimate

$\displaystyle \text{Kempner series} \approx 0.9 \times \int_1^\infty \left( \frac{9}{100}\right)^{\log x} ~\mathrm{d}x$

The integral can be computed using that

$\displaystyle a^{\log b} = b^{\log a}$

to get

$\displaystyle 0.9 \times \int_1^{\infty} \left( \frac{9}{100}\right)^{\log x} ~\mathrm{d}X = 0.9\times \int_1^\infty x^{\log 9 - 2} ~\mathrm{d}x = \frac{0.9}{1 - \log 9} \approx 19.66$

Notice that this estimate is much closer to the currently known value of roughly 22.92 than to the original upper bound of 80 computed by Kempner.

Kempner’s estimate is a heavy overestimate because he performed a summation replacing every n-digit long number that does not contain the digit 9 by $10^{n-1}$; this number can be many times (up to 9) times smaller than the original number. Our estimate is low because among the n-digit long numbers, the numbers that do not contain the digit 9 are not evenly distributed: they tend to crowd in the front rather than in the back (in fact, we do not allow them to crowd in the back because none of the numbers that start with the digit 9 is admissible). So if in the original question we had asked for numbers that do not contain the digit 1, then our computation will give an overestimate instead since these numbers tend to crowd to the back.

### Simulating closed cosmic strings (or yet another adventure in Julia)

One of my current research interests is in the geometry of constant-mean-curvature time-like submanifolds of Minkowski space. A special case of this is the study of cosmic strings under the Nambu-Goto action. Mathematically the classical (as opposed to quantum) behavior of such strings are quite well understood, by combining the works of theoretical physicists since the 80s (especially those of Jens Hoppe and collaborators) together with recent mathematical developments (such as the work of Nguyen and Tian and that of Jerrard, Novaga, and Orlandi). To get a better handle on what is going on with these results, and especially to provide some visual aids, I’ve tried to produce some simulations of the dynamics of closed cosmic strings. (This was also an opportunity for me to practice code-writing in Julia and learn a bit about the best practices for performance and optimization in that language.)

The code

After some false starts, here are some reasonably stable code.

function MC_corr3!(position::Array{Float64,2}, prev_vel::Array{Float64,2}, next_vel::Array{Float64,2}, result::Array{Float64,2}, dt::Float64)
# We will assume that data is stored in the format point(coord,number), so as a 3x1000 array or something.
num_points = size(position,2)
num_dims = size(position,1)

curr_vel = zeros(num_dims)
curr_vs = zeros(num_dims)
curr_ps = zeros(num_dims)
curr_pss = zeros(num_dims)

pred_vel = zeros(num_dims)
agreement = true

for col = 1:num_points  #Outer loop is column
if col == 1
prev_col = num_points
next_col = 2
elseif col == num_points
prev_col = num_points - 1
next_col = 1
else
prev_col = col -1
next_col = col + 1
end

for row = 1:num_dims
curr_vel[row] = (next_vel[row,col] + prev_vel[row,col])/2
curr_vs[row] = (next_vel[row,next_col] + prev_vel[row,next_col] - next_vel[row,prev_col] - prev_vel[row,prev_col])/4
curr_ps[row] = (position[row,next_col] - position[row,prev_col])/2
curr_pss[row] = position[row,next_col] + position[row,prev_col] - 2*position[row,col]
end

beta = (1 + dot(curr_vel,curr_vel))^(1/2)
sigma = dot(curr_ps,curr_ps)
psvs = dot(curr_ps,curr_vs)
bvvs = dot(curr_vs,curr_vel) / (beta^2)
pssps = dot(curr_pss,curr_ps)

for row in 1:num_dims
result[row,col] = curr_pss[row] / (sigma * beta) - curr_ps[row] * pssps / (sigma^2 * beta) - curr_vel[row] * psvs / (sigma * beta) - curr_ps[row] * bvvs / (sigma * beta)
pred_vel[row] = prev_vel[row,col] + dt * result[row,col]
end

agreement = agreement && isapprox(next_vel[:,col], pred_vel, rtol=sqrt(eps(Float64)))
end

return agreement
end

function find_next_vel!(position::Array{Float64,2}, prev_vel::Array{Float64,2}, next_vel::Array{Float64,2}, dt::Float64; max_tries::Int64=50)
tries = 1
result = zeros(next_vel)
agreement = MC_corr3!(position,prev_vel,next_vel,result,dt)
for j in 1:size(next_vel,2), i in 1:size(next_vel,1)
next_vel[i,j] = prev_vel[i,j] + result[i,j]*dt
end
while !agreement && tries < max_tries
agreement = MC_corr3!(position,prev_vel,next_vel,result,dt)
for j in 1:size(next_vel,2), i in 1:size(next_vel,1)
next_vel[i,j] = prev_vel[i,j] + result[i,j]*dt
end
tries +=1
end
return tries, agreement
end


This first file does the heavy lifting of solving the evolution equation. The scheme is a semi-implicit finite difference scheme. The function MC_Corr3 takes as input the current position, the previous velocity, and the next velocity, and computes the correct current acceleration. The function find_next_vel iterates MC_Corr3 until the computed acceleration agrees (up to numerical errors) with the input previous and next velocities.

Or, in notations:
MC_Corr3: ( x[t], v[t-1], v[t+1] ) --> Delta-v[t]
and find_next_vel iterates MC_Corr3 until
 Delta-v[t] == (v[t+1] - v[t-1]) / 2

The code in this file is also where the performance matters the most, and I spent quite some time experimenting with different algorithms to find one with most reasonable speed.

function make_ellipse(a::Float64,b::Float64, n::Int64, extra_dims::Int64=1)  # a,b are relative lengths of x and y axes
s = linspace(0,2π * (n-1)/n, n)
if extra_dims == 0
return vcat(transpose(a*cos(s)), transpose(b*sin(s)))
elseif extra_dims > 0
return vcat(transpose(a*cos(s)), transpose(b*sin(s)), zeros(extra_dims,n))
else
error("extra_dims must be non-negative")
end
end

function perturb_data!(data::Array{Float64,2}, coeff::Vector{Float64}, num_modes::Int64)
# num_modes is the number of modes
# coeff are the relative sizes of the perturbations

numpts = size(data,2)

for j in 2:num_modes
rcoeff = rand(length(coeff),2)

for pt in 1:numpts
theta = 2j * π * pt / numpts
for d in 1:length(coeff)
data[d,pt] += ( (rcoeff[d,1] - 0.5) *  cos(theta) + (rcoeff[d,2] - 0.5) * sin(theta)) * coeff[d] / j^2
end
end
end

nothing
end


This file just sets up the initial data. Note that in principle the number of ambient spatial dimensions is arbitrary.

using Plots

pyplot(size=(1920,1080), reuse=true)

function plot_data2D(filename_prefix::ASCIIString, filename_offset::Int64, titlestring::ASCIIString, data::Array{Float64,2}, additional_data...)
x_max = 1.5
y_max = 1.5
plot(transpose(data)[:,1], transpose(data)[:,2] , xlims=(-x_max,x_max), ylims=(-y_max,y_max), title=titlestring)

end
end

png(filename_prefix*dec(filename_offset,5)*".png")
nothing
end

function plot_data3D(filename_prefix::ASCIIString, filename_offset::Int64, titlestring::ASCIIString, data::Array{Float64,2}, additional_data...)
x_max = 1.5
y_max = 1.5
z_max = 0.9
tdata = transpose(data)
plot(tdata[:,1], tdata[:,2],tdata[:,3], xlims=(-x_max,x_max), ylims=(-y_max,y_max),zlims=(-z_max,z_max), title=titlestring)

plot!(tdata[:,1], tdata[:,2], tdata[:,3])
end
end

png(filename_prefix*dec(filename_offset,5)*".png")
nothing
end


This file provides some wrapper commands for generating the plots.

include("InitialData3.jl")
include("MeanCurvature3.jl")
include("GraphCode3.jl")

num_pts = 3000
default_timestep = 0.01 / num_pts
max_time = 3
plot_every_ts = 1500

my_data = make_ellipse(1.0,1.0,num_pts,0)
perturb_data!(my_data, [1.0,1.0], 15)
this_vel = zeros(my_data)
next_vel = zeros(my_data)

for t = 0:floor(Int64,max_time / default_timestep)
num_tries, agreement = find_next_vel!(my_data, this_vel,next_vel,default_timestep)

if !agreement
warn("Time $(t*default_timestep): failed to converge when finding next_vel.") warn("Dumping information:") max_beta = 1.0 max_col = 1 for col in 1:size(my_data,2) beta = (1 + dot(next_vel[:,col], next_vel[:,col]))^(1/2) if beta > max_beta max_beta = beta max_col = col end end warn(" Beta attains maximum at position$max_col")
warn("   Beta = $max_beta") warn(" Position = ", my_data[:,max_col]) prevcol = max_col - 1 nextcol = max_col + 1 if max_col == 1 prevcol = size(my_data,2) elseif max_col == size(my_data,2) nextcol = 1 end warn(" Deltas") warn(" Left: ", my_data[:,max_col] - my_data[:,prevcol]) warn(" Right: ", my_data[:,nextcol] - my_data[:,max_col]) warn(" Previous velocity: ", this_vel[:,max_col]) warn(" Putative next velocity: ", next_vel[:,max_col]) warn("Quitting...") break end for col in 1:size(my_data,2) beta = (1 + dot(next_vel[:,col], next_vel[:,col]))^(1/2) for row in 1:size(my_data,1) my_data[row,col] += next_vel[row,col] * default_timestep / beta this_vel[row,col] = next_vel[row,col] end if beta > 1e7 warn("time: ", t * default_timestep) warn("Almost null... beta = ", beta) warn("current position = ", my_data[:,col]) warn("current Deltas") prevcol = col - 1 nextcol = col + 1 if col == 1 prevcol = size(my_data,2) elseif col == size(my_data,2) nextcol = 1 end warn(" Left: ", my_data[:,col] - my_data[:,prevcol]) warn(" Right: ", my_data[:,nextcol] - my_data[:,col]) end end if t % plot_every_ts ==0 plot_data2D("3Dtest", div(t,plot_every_ts), @sprintf("elapsed: %0.4f",t*default_timestep), my_data, make_ellipse(cos(t*default_timestep), cos(t*default_timestep),100,0)) info("Frame$(t/plot_every_ts):  used \$num_tries tries.")
end
end


And finally the main file. Mostly it just ties the other files together to produce the plots using the simulation code; there are some diagnostics included for me to keep an eye on the output.

The results

First thing to do is to run a sanity check against explicit solutions. In rotational symmetry, the solution to the cosmic string equations can be found analytically. As you can see below the simulation closely replicates the explicit solution in this case.

The video ends when the simulation stopped. The simulation stopped because a singularity has formed; in this video the singularity can be seen as the collapse of the string to a single point.

Next we can play around with a more complicated initial configuration.

In this video the blue curve is the closed cosmic string, which starts out as a random perturbation of the circle with zero initial speed. The string contracts with acceleration determined by the Nambu-Goto action. The simulation ends when a singularity has formed. It is perhaps a bit hard to see directly where the singularity happened. The diagnostic messages, however, help in this regard. From it we know that the onset of singularity can be seen in the final frame:

The highlighted region is getting quite pointy. In fact, that is accompanied with the “corner” picking up infinite acceleration (in other words, experiencing an infinite force). The mathematical singularity corresponds to something unreasonable happening in the physics.

To make it easier to see the “speed” at which the curve is moving, the following videos show the string along with its “trail”. This first one again shows how a singularity can happen as the curve gets gradually more bent, eventually forming a corner.

This next one does a good job emphasizing the “wave” nature of the motion.

The closed cosmic strings behave like a elastic band. The string, overall, wants to contract to a point. Small undulations along the string however are propagated like traveling waves. Both of these tendencies can be seen quite clearly in the above video. That the numerical solver can solve “past” the singular point is a happy accident; while theoretically the solutions can in fact be analytically continued past the singular points, the renormalization process involved in this continuation is numerically unstable and we shouldn’t be able to see it on the computer most of the time.

The next video also emphasizes the wave nature of the motion. In addition to the traveling waves, pay attention to the bottom left of the video. Initially the string is almost straight there. This total lack of curvature is a stationary configuration for the string, and so initially there is absolutely no acceleration of that segment of the string. The curvature from the left and right of that segment slowly intrudes on the quiescent piece until the whole thing starts moving.

The last video for this post is a simulation when the ambient space is 3 dimensional. The motion of the string, as you can see, becomes somewhat more complicated. When the ambient space is 2 dimensional a point either accelerates or decelerates based on the local (signed) curvature of the string. But when the ambient space is 3 dimensional, the curvature is now a vector and this additional degree of freedom introduces complications into the behavior. For example, when the ambient space is 2 dimensional it is known that all closed cosmic strings become singular in finite time. But in 3 dimensions there are many closed cosmic strings that vibrate in place without every becoming singular. The video below is one that does however become singular. In addition to a fading trail to help visualize the speed of the curve, this plot also includes the shadows: projections of the curve onto the three coordinate planes.

### Riemann-, Generalized-Riemann-, and Darboux-Stieltjes integrals

(The following is somewhat rough and may have typos.)

Let us begin by setting the notations and recalling what happens without the Stieltjes part.

Defn (Partition)
Let $I$ be a closed interval. A partition $P$ is a finite collection of closed subintervals $\{I_\alpha\}$ such that

1. $P$ is finite;
2. $P$ covers $I$, i.e. $\cup P = I$;
3. $P$ is pairwise almost disjoint, i.e. for $I_\alpha, I_\beta$ distinct elements of $P$, their intersection contains at most one point.

We write $\mathscr{P}$ for the set of all partitions of $I$.

Defn (Refinement)
Fix $I$ a closed interval, and $P, Q$ two partitions. We say that $P$ refines $Q$ or that $P \preceq Q$ if for every $I_\alpha\in P$ there exists $J_\beta \in Q$ such that $I_\alpha \subseteq J_\beta$.

Defn (Selection)
Given $I$ a closed interval and $P$ a partition, a selection $\sigma: P \to I$ is a mapping that satisfies $\sigma(I_\alpha) \in I_\alpha$.

Defn (Size)
Given $I$ a closed interval and $P$ a partition, the size of $P$ is defined as $|P| = \sup_{I_\alpha \in P} |I_\alpha|$, where $|I_\alpha|$ is the length of the closed interval $I_\alpha$.

Remark In the above we have defined two different preorders on the set $\mathscr{P}$ of all partitions. One is induced by the size: we say that $P \leq Q$ if $|P| \leq |Q|$. The other is given by the refinement $P\preceq Q$. Note that neither are partial orders. (But that the preorder given by refinement can be made into a partial order if we disallow zero-length degenerate closed intervals.) Note also that if $P\preceq Q$ we must have $P \leq Q$.

Now we can define the notions of integrability.

Defn (Integrability)
Let $I$ be a closed, bounded interval and $f:I \to \mathbb{R}$ be a bounded function. We say that $f$ is integrable with integral $s$ in the sense of

• Riemann if for every $\epsilon > 0$ there exists $P_0\in \mathcal{P}$ such that for every $P \leq P_0$ and every selection $\sigma:P \to I$ we have
$\displaystyle \left| \sum_{I' \in P} f(\sigma(I')) |I'| - s \right| < \epsilon$

• Generalised-Riemann if for every $\epsilon > 0$ there exists $P_0 \in \mathcal{P}$ such that for every $P \preceq P_0$ and every selection $\sigma: P\to I$ we have
$\displaystyle \left| \sum_{I' \in P} f(\sigma(I')) |I'| - s \right| < \epsilon$

• Darboux if
$\displaystyle \inf_{P\in\mathscr{P}} \sum_{I' \in P} (\sup_{I'} f )|I'| = \sup_{P\in\mathscr{P}} \sum_{I' \in P} (\inf_{I'} f )|I'| = s$

From the definition it is clear that “Riemann integrable” implies “Generalised-Riemann integrable”. Furthermore, we have clearly that for a fixed $P$
$\displaystyle \sum_{I' \in P} (\inf_{I'} f) |I'| \leq \sum_{I' \in P} f(\sigma(I')) |I'| \leq \sum_{I' \in P} (\sup_{I'} f) |I'|$
and that if $P \preceq Q$ we have
$\displaystyle \sum_{I' \in Q} (\inf_{I'} f) |I'| \leq \sum_{I' \in P} (\inf_{I'} f) |I'| \leq \sum_{I' \in P} (\sup_{I'} f) |I'| \leq \sum_{I' \in Q} (\inf_{I'} f) |I'|$
so “Darboux integrable” also implies “Generalised-Riemann integrable”. A little bit more work shows that “Generalised-Riemann integrable” also implies “Darboux integrable” (if the suprema and infima are obtained on the intervals $I'$, this would follow immediately; using the boundedness of the intervals we can find $\sigma$ such that the Riemann sum approximates the upper or lower Darboux sums arbitrarily well.

The interesting part is the following
Theorem
Darboux integrable functions are Riemann integrable. Thus all three notions are equivalent.

Proof. Let $P, Q$ be partitions. Let $|P| \leq \inf_{I'\in Q, |I'| \neq 0} |I'|$, and let $m$ be the number of non-degenerate subintervals in $Q$. We have the following estimate
$\displaystyle \sum_{I'\in Q} (\inf_{I'} f) |I'| - (m-1) |P| (\sup_I 2|f|) \leq \sum_{J'\in P} f(\sigma(J')) |J'| \leq \sum_{I'\in Q} (\sup_{I'} f) |I'| + (m-1) |P| (\sup_I 2|f|)$
The estimate follows by noting that “most” of the $J'\in P$ will be proper subsets of $I'\in Q$, and there can be at most $m-1$ of the $J'$ that straddles between two different non-degenerate sub-intervals of $Q$. To prove the theorem it suffices to choose first a $Q$ such that the upper and lower Darboux sums well-approximates the integral. Then we can conclude for all $P$ with $|P|$ sufficiently small the Riemann sum is almost controlled by the $Q$-Darboux sums. Q.E.D.

Now that we have recalled the case of the usual integrability. Let us consider the case of the Stieltjes integrals: instead of integrating against $\mathrm{d}x$, we integrate against $\mathrm{d}\rho$, where $\rho$ is roughly speaking a “cumulative distribution function”: we assume that $\rho:I \to \mathbb{R}$ is a bounded monotonically increasing function.

The definition of the integrals are largely the same, except that at every step we replace the width of the interval $|I'|$ by the diameter of $\rho(I')$, i.e. $\sup_{I'} \rho - \inf_{I'} \rho$. The arguments above immediately also imply that

• “Riemann-Stieltjes integrable” implies “Generalised-Riemann-Stieltjes integrable”
• “Darboux-Stieltjes integrable” implies “Generalised-Riemann-Stieltjes integrable”
• “Generalised-Riemann-Stieltjes integrable” implies “Darboux-Stientjes integrable”

However, Darboux-Stieltjes integrable functions need not be Riemann-Stieltjes integrable. The possibility of failure can be seen in the proof of the theorem above, where we used the fact that $|P|$ is allow to be made arbitrarily small. The same estimate, in the case of the Stieltjes version of the integrals, has $|P|$ replaced by $\sup_{J'\in P} (\sup_{J'} \rho - \inf_{J'} \rho)$, which for arbitrary partitions need to shrink to zero. To have a concrete illustration, we give the following:

Example
Let $I = [0,1]$. Let $\rho(x) = 0$ if $x < \frac12$ and $1$ otherwise. Let $f(x) = 0$ if $x \leq \frac12$ and $1$ otherwise. Let $Q_0$ be the partition $\{ [0,\frac12], [\frac12,1]\}$. We have that
$\displaystyle \sum_{I'\in Q_0} (\sup_{I'} f) (\sup_{I'} \rho - \inf_{I'} \rho) = 0 \cdot (1 - 0) + 1\cdot (1 - 1) = 0$
while
$\displaystyle \sum_{I'\in Q_0} (\inf_{I'} f) (\sup_{I'} \rho - \inf_{I'} \rho) = 0 \cdot (1-0) + 0 \cdot(1-1) = 0$
so we have that in particular the pair $(f,\rho)$ is Darboux-Stieltjes integrable with integral 0. However, let $k$ be any odd integer, consider the partition $P_k$ of $[0,1]$ into $k$ equal portions. Depending on the choice of the selection $\sigma$, we see that the sum can take the values
$\displaystyle \sum_{I'\in P_k} f(\sigma(I')) (\sup_{I'} \rho - \inf_{I'}\rho) = f(\sigma([\frac12 - \frac1{2k},\frac12 + \frac1{2k}])) (1 - 0) \in \{0,1\}$
which shows that the Riemann-Stieltjes condition can never be satisfied.

The example above where both $f$ and $\rho$ are discontinuous at the same point is essentially sharp. A easy modification of the previous theorem shows that
Prop
If at least one of $f,\rho$ is continuous, then Darboux-Stieltjes integrability is equivalent to Riemann-Stieltjes integrability.

Remark The nonexistence of Riemann-Stieltjes integral when $f$ and $g$ has shared discontinuity points is similar in spirit to the idea in distribution theory where whether the product of two distributions is well-defined (as a distribution) depends on their wave-front sets.

### Bouncing a quantum particle back and forth

If you have not seen my previous two posts, you should read them first.

In the two previous posts, I shot particles (okay, simulated the shooting on a computer) at a single potential barrier and looked at what happens. What happens when we have more than one barrier? In the classical case the picture is easy to understand: a particle with insufficient energy to escape will be trapped in the local potential well for ever, while a particle with sufficiently high energy will gain freedom and never come back. But what happens in the quantum case?

If the intuition we developed from scattering a quantum particle against a potential barrier, where we see that depending on the frequency (energy) of the particle, some portion gets transmitted and some portion gets reflected, is indeed correct, what we may expect to see is that the quantum particle bounces between the two barriers, each time losing some amplitude due to tunneling.

But we also saw that the higher frequency components of the quantum particle have higher transmission amplitudes. So we may expect that the high frequency components to decay more rapidly than the low frequency ones, so the frequency of the “left over” parts will continue to decay in time. This however, would be wrong, because we would be overlooking one simple fact: by the uncertainty principle again, very low frequency waves cannot be confined to a small physical region. So when we are faced with two potential barriers, the distance between them gives a characteristic frequency. Below this frequency (energy) it is actually not possible to fit a (half) wave between the barriers, and so the low frequency waves must have significant physical extent beyond the barriers, which means that large portions of these low frequency waves will just radiate away freely. Much above the characteristic frequency, however, the waves have large transmission coefficients and will not be confined.

So the net result is that we should expect for each double barrier a characteristic frequency at which the wave can remain “mostly” stuck between the two barriers, losing a little bit of amplitude at each bounce. This will look like a slowly, but exponentially, decaying standing wave. And I have some videos to show for that!

In the video we start with the same random initial data and evolve it under the linear wave equation with different potentials: the equations look like

$\displaystyle - \partial^2_{tt} u + \partial^2_{xx} u - V u = 0$

where $V$ is a non-negative potential taken in the form

$\displaystyle V(x) = a_1 \exp( - x^2 / b_1) - a_2 \exp( -x^2 / b_2)$

which is a difference of two Gaussians. For the five waves shown the values of $a_1, b_1$ are the same throughout. The coefficients $a_2$ (taken to be $\leq a_1$) and $b_2$ (taken to be $< b_1$) increases from top to bottom, resulting in more and more-widely separated double barriers. Qualitatively we see, as we expected,

• The shallower and narrower the dip the faster the solution decays.
• The shallower and narrower the dip the higher the “characteristic frequency”.

As an aside: the video shown above is generated using Python, in particular NumPy and MatPlotLib; the code took significantly longer to run (20+hours) than to write (not counting the HPDE solver I wrote before for a different project, coding and debugging this simulation took about 3 hours or less). On the other hand, this only uses one core of my quad-core machine, and leaves the computer responsive in the mean time for other things. Compare that to Auto-QCM: the last time I ran it to grade a stack of 400+ multiple choice exams it locked up all four cores of my desktop computer for almost an entire day.

As a further aside, this post is related somewhat to my MathOverflow question to which I have not received a satisfactory answer.

### … and scattering a quantum particle

In the previous post we shot a classical particle at a potential barrier. In this post we shoot a quantum particle.

Whereas the behaviour of the classical particle is governed by Newton’s laws (where the external force providing the acceleration is given as minus the gradient of the potential), we allow our quantum particle to be governed by the Klein-Gordon equations.

• Mathematically, the Klein-Gordon equation is a partial differential equation, whereas Newton’s laws form ordinary differential equations. A typical physical interpretation is that the state space of quantum particles are infinite dimensional, whereas the phase space of physics has finite dimensions.
• Note that physically the Klein-Gordon equation was designed to model a relativistic particle, while in the previous post we used the non-relativistic Newton’s laws. In some ways it would’ve been better to model the quantum particle using Schroedinger’s equation. I plead here however that (a) qualitatively there is not a big difference in terms of the simulated outcomes and (b) it is more convenient for me to use the Klein-Gordon model as I already have a finite-difference solver for hyperbolic PDEs coded in Python on my computer.

To model a particle, we set the initial data to be a moving wave packet, such that at the initial time the solution is strongly localized and satisfies $\partial_t u + \partial_x u = 0$. Absent the mass and potential energy terms in the Klein-Gordon equation (so under the evolution of the free wave equation), this wave packet will stay coherent and just translate to the right as time goes along. The addition of the mass term causes some small dispersion, but the mass is chosen small so that this is not a large effect. The main change to the evolution is the potential barrier, which you can see illustrated in the simulation.

The video shows 8 runs of the simulation with different initial data. Whereas in the classical picture the initial kinetic energy is captured by the initial speed at which the particle is moving, in the quantum/wave picture the kinetic energy is related to the central frequency of your wave packet. So each of the 8 runs have increasing frequency offset that represents increasing kinetic energy. The simulation has two plots, the top shows the square of the solution itself, which gives a good indication of where physically the wave packet is located. The bottom shows a normalized kinetic energy density (I have to include a normalization since the kinetic energy of the first and last particles differ roughly 10 fold).

One notices that in the first two runs, the kinetic energy is sufficiently small that the particle mostly bounces back to the left after hitting the potential.

For the third and fourth runs (frequency shift $\sqrt{2}$ and $\sqrt{3}$ respectively) we see that while a significant portion of the particle bounces back, a noticeable portion “tunnels through” the barrier: this caused by a combination of the quantum tunneling phenomenon and the wave packet form of the initial data.

The phenomenon of quantum tunneling manifests in that all non-zero energy waves will penetrate a finite potential barrier a little bit. But the amount of penetration decays to zero as the energy of the wave goes to zero: this is known as the semiclassical regime. In the semiclassical limit it is known that quantum mechanics converge toward classical mechanics, and so in the low-energy limit we expect our particle to behave like a classical particle and bounce off. So we see that naturally increasing the energy (frequency) of our wave packet we expect more of the tunneling to happen.

Further, observe that by shaping our data into a wave packet it necessarily contains some high frequency components (due to Heisenberg uncertainty principle); high frequency, and hence high energy components do not “see” the potential barrier. Even in the classical picture high energy particles would fly over the potential barrier. So for wave packets there will always be some (perhaps not noticeable due to the resolution of our computation) leakage of energy through the potential barrier. The quantum effect on these high energy waves is that they back-scatter. Whereas the classical high energy particles just fly directly over the barrier, a high energy quantum particle will leave some parts of itself behind the barrier always. We see this in the sixth and seventh runs of the simulation, where the particle mostly passes through the barrier, but a noticeable amount bounces off in the opposite direction.

In between during the fifth run, where the frequency shift is 2, we see that the barrier basically split the particle in two and send one half flying to the right and the other half flying to the left. Classically this is the turning point between particles that go over the bump and particles that bounces back, and would be the case (hard to show numerically!) where a classical particle comes in from afar with just enough energy that it comes to a half at the top of the potential barrier!

And further increasing the energy after the seventh run, we see in the final run a situation where only a negligible amount of the particle scatters backward with almost all of it passing through the barrier unchanged. One interesting thing to note however is that just like the case of the classical particle, the wave packet appears to “slow down” a tiny bit as it goes over the potential barrier.

### Shooting a classical particle…

Here’s a small animation of what happens when you try to shoot a classical particle when there’s a potential barrier. For small initial kinetic energies, the particle bounces back. For large initial kinetic energies, the particle goes over the hump, first decelerating and then accelerating in the process.

(It may be best to watch this full screen with HD if the network supports it.)

(The NumPy code is pretty simple to write for this; and it runs relatively fast. The one for my next post is a bit more complicated and takes rather much longer to run. Stay tuned!)

### An optimization problem: variation

Examining the theorem proven in the previous post, we are led naturally to ask whether there are higher order generalizations.

Question: Let $f \in C^{k}([-1,1])$ with $f^{(k)} > 0$. What can we say about the minimizer of $C = \int_{-1}^1 |f(x) - p(x)|~\mathrm{d}x$ where $p$ ranges over degree $k-1$ polynomials?

It is pretty easy to see that we expect $p$ to intersect $f$ at the maximum number of points, which is $k$. We label those points $x_1, \ldots, x_k$ and call $x_0 = -1$ and $x_{k+1}= 1$. Then the cost function can be written as
$\displaystyle C = \sum_{j = 0}^k (-1)^j \int_{x_j}^{x_{j+1}} f(x) - p(x; x_1, \ldots, x_k) ~\mathrm{d}x$
Since we know that values of $p$ at the points $x_1, \ldots, x_k$ we can write down the interpolation polynomial explicitly using Sylvester’s formula:
$\displaystyle p = \sum_{j = 1}^k \left( \prod_{1 \leq m \leq k, m\neq j} \frac{x - x_m}{x_j - x_m} \right) f(x_j) = \sum L_j(x; x_1, \ldots, x_k) f(x_j)$

The partial derivatives are now
$\displaystyle \partial_n C = \sum_{j = 0}^k (-1)^{j+1} \int_{x_j}^{x_{j+1}} \partial_n p(x; x_1, \ldots, x_k) ~\mathrm{d}x$
It remains to compute $\partial_n p$ for $1 \leq n \leq k$. We observe that when $n \neq j$
$\displaystyle \partial_n L_j = - \frac{1}{x - x_n} L_j + \frac{1}{x_j - x_n} L_j$
and also
$\displaystyle \partial_n L_n = - \left( \sum_{1\leq m \leq k, m\neq n} \frac{1}{x_n - x_m} \right) L_n$
So
$\displaystyle \partial_n p = \sum_{j \neq n} \frac{x-x_j}{(x_j - x_n)(x - x_n)} L_j f(x_j) + L_n f'(x_n) - \left( \sum_{1\leq m \leq k, m\neq n} \frac{1}{x_n - x_m} \right) L_n f(x_n)$
Now, we observe that
$\displaystyle \frac{x - x_j}{x - x_n} L_j = - \left( \prod_{m \neq n,j} \frac{x_n - x_m}{x_j - x_m} \right) L_n$
so after some computation we arrive at
$\displaystyle \partial_n p = L_n(x) \cdot \left[ f'(x_n) - \sum_{j \neq n} \frac{1}{x_j - x_n} \left(\left( \prod_{m \neq j,n}\frac{x_n - x_m}{x_j - x_m}\right)f(x_j) - f(x_n) \right)\right]$
which we can further simplify to
$\displaystyle \partial_n p = L_n(x) \cdot \left( f'(x_n) - p'(x_n)\right)$
Now, since $f$ and $p$ cross transversely at $x_n$, the difference of their derivatives is non-zero. (This harks back to our assumption that $f^{(k)} > 0$.) So we are down, as in the case where $k = 2$, to equations entirely independent of $f$.

More precisely, we see that the stationarity condition becomes the choice of $x_1, \ldots, x_k$ such that the integrals
$\displaystyle \sum_{j = 0}^k (-1)^{j} \int_{x_j}^{x_{j+1}} L_n(x) ~\mathrm{d}x = 0$
for each $n$. Since $L_n$ form a basis for the polynomials of degree at most $k-1$, we have that the function
$\chi(x) = (-1)^j \qquad x \in (x_j, x_{j+1})$
is $L^2$ orthogonal to every polynomial of degree at most $k-1$. So in particular the $x_j$ are solutions to the following system of equations
$x_0 = -1, \qquad x_{k+1} = 1$
$\sum_{j = 0}^k (-1)^j \left[ x_{j+1}^d - x_{j}^d \right] = 0 \qquad \forall d \in \{1, \ldots, k\}$

From symmetry considerations we have that $x_j = - x_{k+1 - j}$. This also kills about half of the equations. For the low $k$ we have

1. $\{ 0\}$
2. $\{ -1/2, 1/2\}$
3. $\{-1/2, 0, 1/2\}$
4. $\{ (\pm 1 \pm \sqrt{5})/4 \}$
5. $\{ 0, \pm\frac12, \pm \frac{\sqrt{3}}2 \}$

### An optimization problem: theme

Let’s start simple:

Question 1: What is the linear function $\ell(x)$ that minimizes the integral $\int_{-1}^1 |x^2 + x - \ell(x)| ~\mathrm{d}x$? In other words, what is the best linear approximation of $x^2 + x$ in the $L^1([-1,1])$ sense?

This is something that can in principle be solved by a high schooler with some calculus training. Here’s how one solution may go:

Solution: All linear functions take the form $\ell(x) = ax + b$. The integrand is equal to $x^2 + x - \ell(x)$ when $x^2 + x \geq \ell(x)$, and $\ell(x) - x - x^2$ otherwise. So we need to find the points of intersection. This requires solve $x^2 + x - ax - b = 0$, which we can solve by the quadratic formula. In the case where $x^2 + x - a x - b$ is signed, we see that changing $b$ we can make the integrand strictly smaller, and hence we cannot attain a minimizer. So we know that at the minimizer there must exist at least one root.

Consider the case where there is only one root in the interval (counted with multiplicity), call the root $x_0$. We have that the integral to minimize is equal to
$\displaystyle \left| \int_{-1}^{x_0} x^2 + x - ax - b~\mathrm{d}x \right| + \left| \int_{x_0}^1 x^2 + x - a x - b ~\mathrm{d}x \right|$
each part of which can be computed explicitly to obtain
$\displaystyle \left| \frac13 x_0^3 + \frac13 + \frac{1-a}{2} x_0^2 - \frac{1-a}{2} - b x_0 - b \right| + \left| \frac13 - \frac13 x_0^3 + \frac{1-a}{2} - \frac{1-a}{2} x_0^2 - b + b x_0\right|$
Since we know that the two terms comes from integrands with different signs, we can combine to get
$\displaystyle \left| \frac23 x_0^3 + (1-a) x_0^2 - (1-a) - 2b x_0 \right|$
as the integrand. Now, we cannot just take the partial derivatives of the above expression with respect to $a,b$ and set that to zero and see what we get: the root $x_0$ depends also on the parameters. So what we would do then is to plug in $x_0$ using the expression derived from the quadratic formula, $x_0 = \frac{1}{2} \left( a - 1 \pm \sqrt{ (1-a)^2 + 4b}\right)$, and then take the partial derivatives. Before that, though, we can simplify a little bit: since $x_0^3 + (1-a) x_0^2 - b x_0 = 0$ from the constraint, the quantity to minimize is now
$\displaystyle \left| - \frac13 x_0^3 - (1-a) - b x_0 \right|$
A long computation taking the $\partial_b$ now shows that necessarily $x_0 = 0$, which implies that $b = 0$. But for the range of $a$ where there is only one root in the interval, the quantity does not achieve a minimum. (The formal minimizer happens at $a = 1$ but we see for this case the integrand of the original cost function is signed.

So we are down to the case where there are two roots in the interval. Now we call the roots $x_+$ and $x_-$, and split the integral into
$\displaystyle \left| \int_{-1}^{x_-} x^2 + x - ax - b~\mathrm{d}x - \int_{x_-}^{x_+} x^2 + x - ax - b~\mathrm{d}x + \int_{x_+}^1 x^2 + x - ax - b~\mathrm{d}x \right|$
and proceed as before. (The remainder of the proof is omitted; the reader is encouraged to carry out this computation out by hand to see how tedious it is.) Read the rest of this entry »

### Bessaga’s converse to the contraction mapping theorem

In preparing some lecture notes for the implicit function theorem, I took a look at Schechter’s delightfully comprehensive Handbook of Analysis and its Foundations (which you can also find on his website), and I learned something new about the Banach fixed point theorem. To quote Schechter:

… although Banach’s theorem is quite easy to prove, a longer proof cannot yield stronger results.

I will write a little bit here about a “converse” to the Banach theorem due to Bessaga, which uses a little bit of help from the Axiom of Choice.

### Compactifying (p,q)-Minkowski space

In a previous post I described a method of thinking about conformal compactifications, and I mentioned in passing that, in principle, the method should also apply to arbitrary signature pseudo-Euclidean space $\mathbb{R}^{p,q}$. A few days ago while visiting Oxford I had a conversation with Sergiu Klainerman where this came up, and we realised that we don’t actually know what the conformal compactifications are! So let me write down here the computations in case I need to think about it again in the future. Read the rest of this entry »