text:
### Oh, these Mathematicians!

There are three kinds of mathematicians: those who can count and those who cannot.

My geometry teacher was sometimes acute, and sometimes obtuse, but always, he was right.

Old mathematicians never die; they just lose some of their functions.

The great logician Betrand Russell (or was it A.N. Whitehead?) once claimed that he could prove anything if given that 1 +1 = 1.
So one day, some smartypants asked him, “Ok. Prove that you’re the Pope.”
He thought for a while and proclaimed, “I am one. The Pope is one. Therefore, the Pope and I are one.”

Asked how his pet parrot died, the the mathematician answered “Polynomial. Polygon.”

A physicist and a mathematician are sitting in a faculty lounge. Suddenly, the coffee machine catches fire. The physicist grabs a bucket, leaps toward the sink, fills the bucket with water and puts out the fire. Second day, the same two sit in the same lounge. Again, the coffee machine catches fire. This time, the mathematician gets a bucket and hands the bucket to the physicist, thus reducing the problem to a previously solved one.

Three men are in a hot-air balloon. Soon, they find themselves lost in a canyon somewhere. One of the three men says “I’ve got an idea. We can call for help in this canyon and the echo will carry our voices far.”
So he leans over the basket and yells out, “HELLLLLLOOOOO! Where are we?”
(They hear the echo several times).
15 minutes later, they hear this echoing voice: “HELLLLLLOOOOO! you’re lost!!”
One of the men says, “That must have been a mathematician.”
Puzzled, one of the other asks, “Why do you say that?”
The reply : “For three reasons. (1) he took a long time to answer, (2) he was absolutely correct, and (3) his answer was absolutely useless.”

A mathematician wandered home at 3 A.M. His wife angrily told him “You are late! You said you’d be home by a 11:45″. The mathematician replied “I am right on time. I said I’d be home by a quarter of twelve”.

An Engineer, a Physicist and a Mathematician are asked to find the value of 2 +2.
Engineer (after 3 minutes, with a slide rule) : “The answer is precisely 3.9974″.
Physicist (after 6 hours of experiments) : “The value is approximately 4.002, with an error of plus or minus 0.005″.
Mathematician (after a week of calculation) : “Well, I haven’t found an answer yet but I can prove that an answer exists.”

A Mathematician, an engineer and a physicist are given an identical problem:

“Prove that all odd numbers greater than 2 are prime number.”

They proceed:

Mathematician : 3 is a prime, 5 is a prime, 7 is a prime, 9 is not a prime – counterexample -claim is false.
Physicist : 3 is a prime, 5 is a prime, 7 is a prime, 9 is an experimental error, 11 is a prime,….
Engineer : 3 is a prime, 5 is a prime, 7 is a prime, 9 is a prime, 11 is a prime,….

A Mathematician, an engineer and a physicist were traveling through Scotland when they saw a black sheep through the window of the train.
“Aha”, says the engineer, “I see that Scottish sheep are black.”
“Hmm”, says the physician, “You mean that some Scottish sheep are black”.
“No”, says the mathematician, “All we know is that there is at least one sheep in Scotland, and that at least one side of that one sheep is black!”

A statistician can have his head in an oven and his feet in ice, and he will say that on the average he feels fine.

Q : What does a mathematician do when he is constipated?
A : He works it out with pencil.

Q : Did you hear about the murderous mathematician?
A : He went on a killing spree with a pair of axis!

A Mathematician is someone who is ready to assume everything except responsibility

### Numbers and Counting

” What’s one and one and one and one and one and one and one and one and one and one and one and one?”
” I don’t know” said Alice. “I lost count.”
“She can’t do addition.” said the Red Queen.
-Lewis Carroll (Through the looking glass)

Prof : How much is 7 x 24?
Student : It’s 168.
Prof : Prove it.
Student : 16 +8 = 24.
Prof : and 7 x 27?
Student : 189, proof 18 +9 = 27.
Prof : and 7 x 21?
Student : 147, proof 14 +7 = 21.
Prof : and 7 x 18?
Student : 126, proof 12 +6 = 18.

Q : Divide 14 sugar cubes into 3 cups of coffee so that each cup has an odd number of sugar cubes.
A : 1, 1, 12
Riposte : 12 isn’t odd!
A : It’s an odd number of cubes to put in a cup of coffee! (groan)

Complete the next two terms of this sequence:
O, T, T, F, F, S, S, E,….

Likewise here:
3 3 5 4 4 3 5 5
(Answer: 4 3 -number of letters in the words “Nine” and “Ten”)

Theorem : A cat has nine tails.
Proof : No cat has eight tails. A cat has one more tail than no cat. Therefore, a cat has nine tails.

Mathematics is made up of 50 percent formulas, 50 percent proofs and 50 percent imagination.

A child defined infinity as the place where things happen that shouldn’t.

### Higher Math Jokes

#### Can you math these?

 What did the acorn say when he grew up? M – bisects A dead parrot P – center What should you do when it rains? S – hypotenuse A guy who has been to the beach recently G- -Coincide The set of card is missing E – Polygon The boy has a speech defect V – secant How they schedule gym classes H – tangent What did he do when his mom-in-law wanted o go home? K – ellipse A tall kettle boiling on the stove B – geometry Why doesn’t the girl run a 4-minute mile? J – Decagon

### de horse and de cart

There was once a very smart horse. Anything that was shown it, it mastered easily, until one day, its teachers tried to teach it about rectangular coordinates and it couldn’t understand them. All the horse’s acquaintances and friends tried to figure out what was the matter and couldn’t. Then a new guy (what the heck, a computer engineer) looked at the problem and said, “Of course he can’t do it. Why, you’re putting Descartes before the horse!”

A guy gets on a bus and starts threatening everybody : “I’ll integrate you! I’ll differentiate you!!!”
So everybody gets scared and runs away. Only one person stays. The guy comes up to him and says: “Aren’t you scared, I’ll integrate you! I’ll differentiate you!!!”
And the other person says; “No, I am not scared, I am $e^x$

### B-Complex Capsules

Q : Why did the mathematician name his dog “Cauchy”?
A : Because he left a residue on every pole.

A group of Polish tourists is flying on a small airplane through the Grand Canyon on a sightseeing tour. The tour guide announces: “On the right of the airplane, you can see the famous Bright Angle Falls.” The tourists leap out of their seats and crowd to the windows on the right side. This causes a dynamic imbalance, and the plane violently rolls to the side and crashes into the canyon wall. All aboard are lost. The moral to this episode is:
Always keep your poles off the right side of the plane.

Q : What’s the contour integral around Western Europe?
A : Zero, because all the poles are in Eastern Europe!
Addendum : Actually, there ARE some Poles are in Western Europe, but they are removable!

The limit as n goes to infinity of sin(x)/n is 6.
Proof : Cancel the n in the numerator and denominator.

1 +1 = 3, for larger values of 1

lim
8->9 Sqrt(8)=3

### Random jokes

Why did the girl eat her maths home work?
Because the teacher said it was a piece of cake!

Who invented fractions?
Henry the eighth!

Theorem: 45 – 45 = 45
PROOF:

```   987654321 = 45(SUM OF DIGITS)
-123456789 = 45(SUM OF DIGITS)
864197532 = 45(SUM OF DIGITS)
```

With 9999 how can you get 100?
Solution: 99+9/9=99+1=100.

A somewhat advanced society has figured how to package basic knowledge in pill form. A student, needing some learning, goes to the pharmacy and asks what kind of knowledge pills are available. The pharmacist says “Here’s a pill for English literature.” The student takes the pills, swallows it and has new knowledge about English literature!
“What else do you have?” asks the student.
“Well, I have pills for art history, biology, and world history,” replies the pharmacist.
The student asks for these, swallows them and has new knowledge about those subjects.
Then the student asks “Do you have a pill for math?”
The pharmacist says “Wait just a moment”, goes back into the storeroom and brings back a whopper of a pill and plunks it on the counter.
“I have to take that huge pill for math?” inquires the student.
The pharmacist replies “Well, you know math always was a little hard to swallow.”

If A = B and B = C then A = C. It is nice except in practice. If Julie Loves Nick, and Nick loves Sara, then Julie loves Sara. Is that right?

Math problems?
Call 1 -800 -[(10x)^(13i)^2 ] -[ sin (x y) / 2.362x].


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## CV limbo [Regulatives / Guidelines]

Dear Helmut,

Could you please take a look at my questions above. Many thanks.

format: plain


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Laplacian spectrum of small graphs

This is a collection of entirely unoriginal remarks about Laplacian spectrum of graphs. For an accessible overview of the subject I recommend the M.S. thesis The Laplacian Spectrum of Graphs by Michael William Newman. It also includes a large table of graphs with their spectra. Here I will avoid introducing matrices and enumerating vertices.

Let ${V}$ be the vertex set of a graph. Write ${u\sim v}$ if ${u, v}$ are adjacent vertices. Given a function ${f\colon V\to \mathbb R}$, define ${L f(v) = \sum_{u\colon u\sim v}(f(v)-f(u))}$.
This is a linear operator (the graph Laplacian) on the Euclidean space ${\ell^2(V)}$ of all functions ${f\colon V\to \mathbb R}$ with the norm ${\|f\|^2 = \sum_{v\in V} f(v)^2}$. It is symmetric: ${\langle L f, g\rangle = \langle f, L g\rangle }$ and positive semidefinite: ${\langle L f, f\rangle = \frac12 \sum_{u\sim v}(f(u)-f(v))^2\ge 0}$. Since equality is attained for constant ${f}$, 0 is always an eigenvalue of ${L}$.

This is the standard setup, but I prefer to change things a little and replace ${\ell^2(V)}$ by the smaller space ${\ell^2_0(V)}$ of functions with zero mean: ${\sum_{v\in V}f(v)=0}$. Indeed, ${L}$ maps ${\ell^2(V)}$ to ${\ell^2_0(V)}$ anyway, and since it kills the constants, it makes sense to focus on ${\ell^2_0(V)}$. It is a vector space of dimension ${n-1}$ where ${n=|V|}$.

One advantage is that the smallest eigenvalue is 0 if and only if the graph is disconnected: indeed, ${\langle L f, f\rangle=0}$ is equivalent to ${f}$ being constant on each connected component. We also gain better symmetry between ${L}$ and the Laplacian of the graph complement, denoted ${L'}$. Indeed, since ${L' f(v) = \sum_{u\colon u\not \sim v}(f(v)-f(u))}$, it follows that ${(L+L')f(v) = \sum_{u\colon u\ne v} (f(v)-f(u)) = n f(v)}$ for every ${f\in \ell^2_0(V)}$. So, the identity ${L+L' = nI}$ holds on ${\ell^2_0(V)}$ (it does not hold on ${\ell^2(V)}$). Hence the eigenvalues of ${L'}$ are obtained by subtracting the eigenvalues of ${L}$ from ${n}$. As a corollary, the largest eigenvalue of ${L}$ is at most ${n}$, with equality if and only if the graph complement is disconnected. More precisely, the multiplicity of eigenvalue ${n}$ is one less than the number of connected components of the graph complement.

Let ${D}$ denote the diameter of the graph. Then the number of distinct Laplacian eigenvalues is at least ${D}$. Indeed, let ${u, v}$ be two vertices at distance ${D}$ from each other. Define ${f_0(u) = 1}$ and ${f_0=0}$ elsewhere. Also let ${f_k=L^k f_0}$ for ${k=1, 2, \dots}$. Note that ${f_k\in \ell_0^2(V)}$ for all ${k\ge 1}$. One can prove by induction that ${f_k(w)=0}$ when the distance from ${w}$ to ${u}$ is greater than ${k}$, and ${(-1)^k f_k(w) > 0}$ when the distance from ${w}$ to ${u}$ is equal to ${k}$. In particular, ${f_k(v) = 0}$ when ${k and ${f_D(v)\ne 0}$. This shows that ${f_D}$ is not a linear combination of ${f_1, \dots, f_{D-1}}$. Since ${f_k = L^{k-1}f_1}$, it follows that ${L^{D-1}}$ is not a linear combination of ${L^0, L^1, \dots, L^{D-2}}$. Hence the minimal polynomial of ${L}$ has degree at least ${D}$, which implies the claim.

Let’s consider a few examples of connected graphs.

3 vertices

There are two connected graphs: the 3-path (D=2) and the 3-cycle (D=1). In both cases we get D distinct eigenvalues. The spectra are [1, 3] and [3, 3], respectively.

4 vertices

• One graph of diameter 3, the path. Its spectrum is ${[2-\sqrt{2}, 2, 2+\sqrt{2}]}$.
• One graph of diameter 1, the complete graph. Its spectrum is ${[4, 4, 4]}$. This pattern continues for other complete graphs: since the complement is the empty graph (${n}$ components), all ${n-1}$ eigenvalues are equal to ${n}$.
• Four graphs of diameter 2, which are shown below, with each caption being the spectrum.

Remarks:

• The graph [1, 3, 4] has more distinct eigenvalues than its diameter.
• The graph [2, 2, 4] is regular (all vertices have the same degree).
• The smallest eigenvalue of graphs [1, 1, 4] and [2, 2, 4] is multiple, due to the graphs having a large group of automorphisms (here rotations); applying some of these automorphisms to an eigenfunctions for the smallest eigenvalue yields another eigenfunction.
• [1, 3, 4] and [2, 4, 4] also have automorphisms, but their automorphisms preserve the eigenfunction for the lowest eigenvalue, up to a constant factor.

5 vertices

• One graph of diameter 4, the path. Its spectrum is related to the golden ratio: it consists of ${(3\pm \sqrt{5})/2, (5\pm \sqrt{5})/2}$.
• One graph of diameter 1, the complete one: [5, 5, 5, 5]
• Five graphs of diameter 3. All have connected complement, with the highest eigenvalue strictly between 4 and 5. None are regular. Each has 4 distinct eigenvalues.
• 14 graphs of diameter 2. Some of these are noted below.

Two have connected complement, so their eigenvalues are less than 5 (spectrum shown on hover):

One has both integers and non-integers in its spectrum, the smallest such graph. Its eigenvalues are ${3\pm \sqrt{2}, 3, 5}$.

Two have eigenvalues of multiplicity 3, indicating a high degree of symmetry (spectrum shown on hover).

Two have all eigenvalues integer and distinct:

The 5-cycle and the complete graph are the only regular graphs on 5 vertices.

6 vertices

This is where we first encounter isospectral graphs: the Laplacian spectrum cannot tell them apart.

Both of these have spectrum ${3\pm \sqrt{5}, 2, 3, 3}$ but they are obviously non-isomorphic (consider the vertex degrees):

Both have these have spectrum ${3\pm \sqrt{5}, 3, 3, 4}$ and are non-isomorphic.

Indeed, the second pair is obtained from the first by taking graph complement.

Also notable are regular graphs on 6 vertices, all of which have integer spectrum.

Here [3, 3, 3, 3, 6] (complete bipartite) and [2, 3, 3, 5, 5] (prism) are both regular of degree 3, but the spectrum allows us to tell them apart.

The prism is the smallest regular graph for which the first eigenvalue is a simple one. It has plenty of automorphisms, but the relevant eigenfunction (1 on one face of the prism, -1 on the other face) is compatible with all of them.

7 vertices

There are four regular graphs on 7 vertices. Two of them are by now familiar: 7-cycle and complete graph. Here are the other two, both regular of degree 4 but with different spectra.

There are lots of isospectral pairs of graphs on 7 vertices, so I will list only the isospectral triples, of which there are five.

Spectrum 0.676596, 2, 3, 3.642074, 5, 5.681331:

Spectrum 0.726927, 2, 3.140435, 4, 4, 6.132637:

Spectrum 0.867363, 3, 3, 3.859565, 5, 6.273073:

Spectrum 1.318669, 2, 3.357926, 4, 5, 6.323404:

All of the triples mentioned so far have connected complement: for example, taking the complement of the triple with the spectrum [0.676596, 2, 3, 3.642074, 5, 5.681331] turns it into the triple with the spectrum [1.318669, 2, 3.357926, 4, 5, 6.323404].

Last but not least, an isospectral triple with an integer spectrum: 3, 4, 4, 6, 6, 7. This one has no counterpart since the complement of each of these graphs is disconnected.

8 vertices

Regular graphs, excluding the cycle (spectrum 0.585786, 0.585786, 2, 2, 3.414214, 3.414214, 4) and the complete one.

Degree 6 regular

Credits: the list of graphs by Brendan McKay and NetworkX library specifically the methods read_graph6 (to read the files provided by Prof. McKay), laplacian_spectrum, diameter, degree, and draw.

Extremal Taylor polynomials

Suppose ${f(z)=a_0+a_1z+a_2z^2+\cdots}$ is a holomorphic function in the unit disk ${|z|<1}$ such that ${|f|\le 1}$ in the disk. How large can its Taylor polynomial ${T_n(z)=a_0+a_1z+\cdots +a_n z^n}$ be in the disk?

We should not expect ${T_n}$ to be bounded by 1 as well. Indeed, the Möbius transformation ${f(z)=(z+1/2)/(1+z/2)}$ has Taylor expansion ${(z+1/2)(1-z/2+O(z^2)) = 1/2 + (3/4)z + O(z^2)}$, so ${T_1(1)=5/4}$ in this case. This turns out to be the worst case: in general ${T_1}$ is bounded by 5/4 in the disk.

For the second-degree polynomial ${T_2}$ the sharp bound is ${89/64}$, attained when ${f(z) = (8z^2 + 4z + 3)/(3z^2 + 4z + 8)}$; the image of the unit circle under the extremal ${T_2}$ is shown below. Clearly, there is something nontrivial going on.

Edmund Landau established the sharp bound for ${|T_n|}$ in his paper Abschätzung der Koeffizientensumme einer Potenzreihe, published in Archiv der Mathematik und Physik (3) 21 in 1913. Confusingly, there are two papers with the same title in the same issue of the journal: one on pages 42-50, the other on pages 250-255, and they appear in different volumes of Landau’s Collected Works. The sharp bound is in the second paper.

First steps

By rotation, it suffices to bound ${|T_n(1)|}$, which is ${|a_0+\cdots +a_n|}$. As is often done, we rescale ${f}$ a bit so that it’s holomorphic in a slightly larger disk, enabling the use of the Cauchy integral formula on the unit circle ${\mathbb T}$. The Cauchy formula says ${2\pi i a_k = \int_{\mathbb T} z^{-k-1} f(z) \,dz}$. Hence

${\displaystyle 2\pi |T_n(1)| = \left| \int_{\mathbb T} z^{-n-1}(1+z+\dots+z^n) f(z) \,dz \right|}$

It is natural to use ${|f(z)|\le 1}$ now, which leads to

${\displaystyle 2\pi |T_n(1)| \le \int_{\mathbb T} |1+z+\dots+z^n|\, |dz| }$

Here we can use the geometric sum formula and try to estimate the integral of ${|(1-z^{n+1})/(1-z)|}$ on the unit circle. This is what Landau does in the first of two papers; the result is ${O(\log n)}$ which is the correct rate of growth (this is essentially the Dirichlet kernel estimate from the theory of Fourier series). But there is a way to do better and get the sharp bound.

Key ideas

First idea: the factor ${1+z+\dots+z^n}$ could be replaced by any polynomial ${Q}$ as long as the coefficients of powers up to ${n}$ stay the same. Higher powers contribute nothing to the integral that evaluates ${T_n(1)}$, but they might reduce the integral of ${|Q|}$.

Second idea: we should choose ${Q}$ to be the square of some polynomial, ${Q=P^2}$, because ${(2\pi)^{-1}\int_{\mathbb T} |P(z)|^2\, |dz|}$ can be computed exactly: it is just the sum of squares of the coefficients of ${P}$, by Parseval’s formula.

Implementation

Since ${1+z+\dots+z^n}$ is the ${n}$-th degree Taylor polynomial of ${(1-z)^{-1}}$, it is natural to choose ${P}$ to be the ${n}$-th degree Taylor polynomial of ${(1-z)^{-1/2}}$. Indeed, if ${P_n(z) = (1-z)^{-1/2} + O(z^{n+1})}$, then ${P_n(z)^2 = (1-z)^{-1} + O(z^{n+1}) = 1+z+\dots+z^n + O(z^{n+1})}$ as desired (asymptotics as ${z\to 0}$). The binomial formula tells us that
${\displaystyle P_n(z)=\sum_{k=0}^n (-1)^k\binom{-1/2}{k}z^k }$

The coefficient of ${z^k}$ here can be written out as ${(2k-1)!!/(2k)!!}$ or rewritten as ${4^{-k}\binom{2k}{k}}$ which shows that in lowest terms, its denominator is a power of 2. To summarize, ${|T_n(1)|}$ is bounded by the sum of squares of the coefficients of ${P_n}$. Such sums are referred to as the Landau constants,

${\displaystyle G_n = 1+ \left(\frac{1}{2}\right)^2 + \left(\frac{1\cdot 3}{2\cdot 4}\right)^2 + \cdots + \left(\frac{(2n-1)!!}{(2n)!!}\right)^2 }$

A number of asymptotic and non-asymptotic formulas have been derived for ${G_n}$, for example Brutman (1982) shows that ${G_n - (1/\pi)\log(n+1)}$ is between 1 and 1.0663.

Sharpness

To demonstrate the sharpness of the bound ${|T_n|\le G_n}$, we want ${|f|\equiv 1}$ and ${P_n(z)^2f(z)/z^n\ge 0}$ on the unit circle. Both are arranged by taking ${f(z) = z^n P_n(1/z) / P_n(z)}$ which is a Blaschke product of degree ${n}$. Note that the term ${P_n(1/z)}$ can also be written as ${\overline{P_n(1/\bar z)}}$. Hence ${P_n(z)^2f(z)/z^n = P_n(z) \overline{P_n(1/\bar z)}}$ which is simply ${|P_n(z)|^2}$ when ${|z|=1}$. Equality holds in all the estimates above, so they are sharp.

Here are the images of the unit circle under extremal Taylor polynomials ${T_5}$ and ${T_{20}}$.

These polynomials attain large values only on a short subarc of the circle; most of the time they oscillate at levels less than 1. Indeed, the mean value of ${|T_n|^2}$ cannot exceed the mean of ${|f|^2}$ which is at most 1. Here is the plot of the roots of extremal ${T_n}$:  they are nearly uniform around the circle, except for a gap near 1.

But we are not done…

Wait a moment. Does ${f(z) = z^n P_n(1/z) / P_n(z)}$ define a holomorphic function in the unit disk? We are dividing by ${P_n}$ here. Fortunately, ${P_n}$ has no zeros in the unit disk, because its coefficients are positive and decreasing as the exponent ${k}$ increases. Indeed, if ${p(z)=c_0+c_1z+\cdots + c_nz^n}$ with ${c_0>c_1>\dots>c_n > 0}$, then ${(1-z)p(z)}$ has constant term ${c_0}$ and other coefficients ${c_1-c_0}$, ${c_2-c_1}$, … ${c_n-c_{n-1}}$, ${-c_n}$. Summing the absolute values of the coefficients of nonconstant terms we get ${c_0}$. So, when these coefficients are attached to ${z^k}$ with ${|z|<1}$, the sum of nonconstant terms is strictly less than ${c_0}$ in absolute value. This proves ${P_n\ne 0}$ in the unit disk. Landau credits Adolf Hurwitz with this proof.

In fact, the zeros of ${P_n}$ (Taylor polynomials of ${(1-z)^{-1/2}}$) lie just outside of the unit disk.

The zeros of the Blaschke products formed from ${P_n}$ are the reciprocals of the zeros of  ${P_n}$, so they lie just inside the unit circle, much like the zeros of ${T_n}$ (though they are different).

The distribution of pow(2, n, n)

For a positive integer ${n}$ let ${f(n)}$ be the remainder of the division of ${2^n}$ by ${n}$. This is conveniently computed in Python as pow(2, n, n). For example, the first 20 values are returned by

[pow(2, n, n) for n in range(1, 21)]

and they are:

[0, 0, 2, 0, 2, 4, 2, 0, 8, 4, 2, 4, 2, 4, 8, 0, 2, 10, 2, 16]

Obviously, ${f(n)=0}$ if and only if ${n}$ is a power of ${2}$. It also looks like ${f}$ is always even, with powers of 2 dominating the list of its values. But ${f}$ does take on odd values, although this does not happen often: only 6 times in the first 100 integers.

[(n, pow(2, n, n)) for n in range(1, 101) if pow(2, n, n) % 2]

returns

[(25, 7), (45, 17), (55, 43), (91, 37), (95, 13), (99, 17)]

It is conjectured that the range of ${f}$ consists of all nonnegative integers except 1. Here is a proof that ${f(n)\ne 1}$, provided by Max Alexeyev on the OEIS page for A036236

Suppose ${2^n \equiv 1 \bmod n}$ for some ${n>1}$. Let ${p}$ be the smallest prime divisor of ${n}$. Then ${2^n\equiv 1 \bmod p}$. This means that the order of ${2}$ in the multiplicative group ${(\mathbb Z/ p \mathbb Z)^*}$ is a divisor of ${n}$. But this group has ${p-1}$ elements, and the only divisor of ${n}$ that is smaller than ${p}$ is ${1}$. Thus, the order of ${2}$ is ${1}$, which means ${2\equiv 1 \bmod p}$, which is absurd.

The fact that the smallest ${n}$ with ${f(n) = 3}$ is 4700063497 deserves to be mentioned here. But I would like to consider the most frequent values of ${f}$ instead. Experimentally, they are found like this:

from collections import Counter
freq = Counter(pow(2, n, n) for n in range(1, 100000000 + 1))
print(freq.most_common(20))

which prints 20 pairs (value, frequency):

(2, 5763518),
(4, 3004047),
(8, 2054167),
(16, 1569725),
(64, 1076293),
(256, 824438),
(512, 737450),
(1024, 668970),
(32, 638063),
(4096, 569705),
(128, 466015),
(65536, 435306),
(262144, 389305),
(1048576, 351723),
(2097152, 326926),
(16384, 246533),
(16777216, 243841),
(32768, 232037),
(2048, 153537),
(8192, 131614)

These are all powers of 2… but not exactly in the order one might expect. I repeated the experiment for the first ${10^8k}$ integers with ${k=1, 2, 3, 4, 5, 6}$. The frequency order remained stable at the top. These are the most common 11 values, with their frequency (in %) listed for the aforementioned six ranges.

2, [5.8, 5.5, 5.4, 5.3, 5.3, 5.2]
4, [3.0, 2.9, 2.8, 2.8, 2.7, 2.7]
8, [2.1, 2.0, 1.9, 1.9, 1.9, 1.8]
16, [1.6, 1.5, 1.5, 1.4, 1.4, 1.4]
64, [1.1, 1.0, 1.0, 1.0, 1.0, 1.0]
256, [0.8, 0.8, 0.8, 0.8, 0.7, 0.7]
512, [0.7, 0.7, 0.7, 0.7, 0.7, 0.7]
1024, [0.7, 0.6, 0.6, 0.6, 0.6, 0.6]
32, [0.6, 0.6, 0.6, 0.6, 0.6, 0.6]
4096, [0.6, 0.5, 0.5, 0.5, 0.5, 0.5]
128, [0.5, 0.4, 0.4, 0.4, 0.4, 0.4]

It seems that 32 and 128 are consistently less common than one might expect.

The most common value, 2, is contributed by primes and pseudoprimes to base 2. The value of 4 appears when ${n = 2p}$ with ${p}$ prime (but not only then). Still, the primes having zero density makes it difficult to explain the frequency pattern based on primality. A more convincing reason could be: when ${n=2k}$ is even, we are computing ${4^k \bmod 2k}$ and that is likely to be a power of ${4}$. This boosts the frequency of the even (generally, composite) powers of 2 in the sequence.

Magic angles

The function ${u(x, y, z) = 2z^2 - x^2 - y^2}$ is notable for the following combination of properties:

1. It is a harmonic function: ${\Delta u = -2 - 2 + 4 = 0}$
2. It vanishes at the origin ${(0, 0, 0)}$ together with its gradient.
3. It is positive on the cone ${C=\{(x, y, z) : z > \sqrt{(x^2+y^2)/2}\}}$

The cone C has the opening angle ${\theta_3 = \cos^{-1}(1/\sqrt{3}) \approx 57.4^\circ}$ which is known as the magic angle in the context of NMR spectroscopy. Let us consider the mathematical side of its magic.

If C is replaced by any larger cone, the properties 1-2-3 cannot be satisfied by a harmonic function in a neighborhood of the origin. That is, C is the largest cone such that a harmonic function can have a critical point at its vertex which is also a point of its extremum on the cone. Why is that?

Let ${u}$ be a harmonic function in some neighborhood of ${(0,0,0)}$ and suppose ${u(0,0,0)=0}$, ${\nabla u(0,0,0) = 0}$, and ${u>0}$ on some cone ${C = \{(x, y, z) \colon z>c \sqrt{x^2+y^2+z^2} \}}$. Expand ${u}$ into a sum of polynomials ${p_2+p_3+\cdots }$ where each ${p_k}$ is a harmonic homogeneous polynomial of degree ${k}$. Let ${m}$ be the smallest integer such that ${p_m}$ is not identically zero. Then ${p_m}$ has the same properties as ${u}$ itself, since it dominates the other terms near the origin. We may as well replace ${u}$ by ${p_m}$: that is, ${u}$ is a spherical harmonic from now on.

Rotating ${u}$ around the ${z}$-axis preserves all the properties of interest: harmonic, positive on the cone, zero gradient at the origin. Averaging over all these rotations we get a rotationally symmetric function known as a zonal spherical harmonic. Up to a constant factor, such a function is given by ${P_m(\cos \phi)}$ where ${\phi}$ is a spherical coordinate (angle with the ${z}$-axis) and ${P_m}$ is the Legendre polynomial of degree ${m}$.

The positivity condition requires ${P_m(t) > 0}$ for ${t>c}$. In other words, the bound on ${\theta}$ comes from the greatest zero of the Legendre polynomial. As is true for orthogonal polynomials in general, the zeros are interlaced: that is, the zeros of ${P_m=0}$ appear strictly between any two consecutive zeros of ${P_{m+1}}$. It follows that the value of the greatest zero grows with ${m}$. Thus, it is smallest when ${m=2}$. Since ${P_2(t) = (3t^2-1)/2}$, the zero of interest is ${1/\sqrt{3}}$, and we conclude that ${c \ge 1/\sqrt{3}}$. Hence the magic angle.

The magic angle is easy to visualize: it is formed by an edge of a cube and its space diagonal. So, the magic cone with vertex at (0,0,0) is the one that passes through (1, 1, 1), as shown above.

In other dimensions the zonal harmonics are expressed in terms of Gegenbauer polynomials (which reduce to Chebyshev polynomials in dimensions 2 and 4). The above argument applies to them just as well. The relevant Gegenbauer polynomial of degree ${2}$ is ${nx^2-1}$ up to a constant. Thus, in ${\mathbb R^n}$ the magic angle is ${\cos^{-1}(1/\sqrt{n})}$, illustrated by the harmonic function ${u(x)=nx_n^2 - |x|^2}$.

This analysis is reminiscent of the Hopf Lemma which asserts that if a positive harmonic function in a smooth domain has boundary value 0 at some point, the normal derivative cannot vanish at that point. The smoothness requirement is typically expressed as the interior ball condition: one can touch every boundary point by a ball contained in the domain. The consideration of magic angles shows that if the function is also harmonic in a larger domain, the interior ball condition can be replaced by the interior cone condition, provided that the opening angle of the cone is greater than ${\cos^{-1}(1/\sqrt{n})}$.

Measurability of Banach space valued functions

There is only an indirect proof of the existence of a function ${f\colon [0, 1]\to \mathbb R}$ that is not Lebesgue measurable. But it’s easy to give an explicit example when the codomain of ${f}$ is a Banach space: just let ${b(t)}$ be the sequence of the binary digits of ${t}$, considered as an element of the sequence space ${\ell_\infty}$.

Why is ${b}$ not measurable? Recall that a Banach space-valued function ${f}$ is (Bochner) measurable iff there is a sequence of simple functions ${\sum v_k \chi_{A_k}}$ (finite sum, measurable ${A_k}$, arbitrary vectors ${v_k}$) that converges to ${f}$ almost everywhere. This property implies that, with an exception of a null set, the range of ${f}$ lies in the separable subspace spanned by all the vectors ${v_k}$ used in the sequence of simple functions. But ${b}$ has the property ${\|b(t)-b(s)\|=1}$ whenever ${t\ne s}$, so the image of any uncountable set under ${b}$ is nonseparable.

Another way to look at this: on the interval [0, 1) the function ${b}$ is injective and its range has discrete topology, which implies that every subset of [0, 1) is the preimage of some open subset of ${\ell_\infty}$ under ${b}$.

The binary-digits functions can also be used to illustrate an issue with the duality of Lebesgue-Bochner spaces ${L_p(0, 1; X)}$ where ${X}$ is a Banach space and ${1\le p<\infty}$. (So, ${f}$ belongs to this space iff it is Bochner measurable and the ${L^p}$ norm of ${\|f\|\colon [0, 1]\to [0, \infty)}$ is finite.) In general we do not have the expected relation ${L_p(0, 1; X)^* = L_q(0, 1; X^*)}$ with ${1/p+1/q=1}$. The natural isometric embedding of ${L_q(0, 1; X^*)}$ into ${L_p(0, 1; X)^*}$ is still there: any ${g\in L_q(0, 1; X^*)}$ acts on ${L_p(0, 1; X)}$ by ${f\mapsto \int \langle f(t), g(t) \rangle\, dt}$. But unless ${X^*}$ has the Radon–Nikodym property, these are more bounded linear functionals on ${L_p(0, 1; X)}$.

To construct such a functional, let ${b_n(t)}$ be the ${n}$-th binary digit of ${t}$. Given ${f\in L_1(0, 1; \ell_1)}$, write it in coordinates as ${(f_1, f_2, \dots)}$ and define ${\varphi(f) = \sum_n \int_0^1 f_n b_n}$. This is a bounded linear functional, since ${|\varphi(f)|\le \sum_n \int_0^1 |f_n| = \|f\|}$. But there is no function ${g\in L_\infty(0, 1; \ell_\infty)}$ that represents it, i.e., ${\varphi(f) = \int_0^1 \langle f(t), g(t)\rangle \,dt = \sum_n \int_0^1 f_n g_n }$. Indeed, if such ${g}$ existed then by considering ${f}$ with only one nonzero coordinate, we find that ${g_n}$ must be ${b_n}$, using the duality ${L_1^* = L_\infty}$ in the scalar case. But the function ${[0, 1]\to \ell_\infty}$ with the components ${(b_1, b_2, \dots)}$ is not measurable, as shown above.

This example, which applies to all ${1\le p<\infty}$, also serves as a reminder that the duality relation ${L_p(0, 1; X)^* = L_q(0, 1; X^*)}$ depends on the dual space ${X^*}$ having the Radon-Nikodym property (RNP), not ${X}$ itself. Indeed, ${X=\ell_1}$ has the RNP; its dual does not.

The importance of ${X^*}$ having the RNP becomes clear once one tries to follow the usual proof of ${L_p^*=L_q}$. Given ${\varphi\in L_p(0,1;X)^*}$, we can define an ${X^*}$-valued measure ${\tau}$ on ${[0, 1]}$ by ${\tau(A)(v) = \varphi( v\chi_A)}$ where ${A\subset [0, 1]}$ is Lebesgue measurable and ${v\in X}$. This measure has reasonable finiteness and continuity properties coming from ${\varphi}$ being a bounded functional. Still, the existence of a density ${g\colon [0, 1]\to X^*}$ of the measure ${\tau}$ depends on the structure of the Banach space ${X^*}$.

Immaculate perfection

— I am so lucky. I was born on April 4th 1944, that’s 4/4/44. If you add that up, it comes to 16, one six. One plus six is seven: luckiest number of all.
— It’s more than math, Mike, it’s… immaculate perfection.

The value of 7 for the the digital root indeed seems preferable, at least on this blog. But looking at the dates formed by repeated digits is too anthropocentric (not to mention xkcd1179). Besides, mathematics already has a concept of perfect numbers: being equal to the sum of proper divisors.

So, are there any perfect numbers with digital root equal to 7, equivalently ${n\equiv 7\bmod 9}$? For even perfect numbers, one can expect that the Euclid-Euler theorem ${n = 2^{p-1} (2^p-1)}$ will help with the answer, but even so, the solution turns out to be surprisingly simple.

A lucky coincidence is that ${2^6\equiv 1\bmod 9}$ (${64 = 1 + 7\cdot 9}$). Since all primes greater than 3 have remainder 1 or 5 mod 6, there are only a few cases to check:

• ${p\equiv 1\bmod 6}$, hence ${2^{p-1}\equiv 2^{1-1} \equiv 1 \bmod 9}$ and ${2^p-1\equiv 2^1-1 \equiv 1\bmod 9}$. In this case ${n\equiv 1\cdot 1 \equiv 1 \bmod 9}$
• ${p\equiv 5\bmod 6}$, hence ${2^{p-1}\equiv 2^{5-1} \equiv 7 \bmod 9}$ and ${2^p-1\equiv 2^5-1 \equiv 4\bmod 9}$. In this case ${n\equiv 7\cdot 4 \equiv 1 \bmod 9}$
• Exceptional cases p=2, 3 correspond to n = 6, 28. The latter is congruent to 1 mod 9, matching the preceding cases.

So, every even perfect number except 6 is congruent to 1 mod 9. Should a perfect number be congruent to 7 mod 9, it has to be odd. An odd perfect number, should it exist, must be either divisible by 9 (not good for us) or congruent to 1 mod 12. (Reference: On the form of an odd perfect number). The condition ${n\equiv 1 \bmod 12}$ is consistent with ${n\equiv 7 \bmod 9}$: for example, 25 satisfies both. Come to think of it, the sum of proper divisors of 25 is… not 25. But there might still be a 7 mod 9 perfect number out there.

What numbers k have the property that the sequence {n mod k : n is perfect} stabilizes? Assuming odd perfect numbers do not exist, every power of 2 has this property. So do 3 and 9 by the above (and therefore, their products with powers of 2). I don’t know any other such k, in particular n mod 27 does not appear to stabilize. Of course, it is impossible to prove any negative result here since we don’t know if there are infinitely many perfect numbers.

Bipartite metric spaces and positive kernels

Bipartite metric spaces provide simple examples of highly non-Euclidean metric spaces. Let ${A}$ and ${B}$ be two disjoint abstract sets, say, with ${m}$ and ${n}$ points respectively. Define a metric on the union ${A\cup B}$ as follows, where ${a, b, c}$ are some positive numbers:

• The distance between distinct elements of ${A}$ is ${a}$.
• The distance between distinct elements of ${B}$ is ${b}$.
• The distance between two elements from different sets is ${c}$.

The triangle inequality is satisfied provided that ${a, b\le 2c}$. When ${c}$ is large, the space is not very interesting: it is approximately the union of two simplexes placed at some large distance from each other. On the other hand, when ${a=b=2c}$ we encounter a bizarre situation: every point of ${B}$ is a “midpoint” between any two distinct points of ${A}$, and vice versa.

A function ${K\colon X\times X \to \mathbb R}$ is called a positive kernel if the matrix ${K(x_i, x_j)_{i, j=1}^n}$ is positive semidefinite for any choice of ${n}$ and ${x_1, \dots, x_n\in X}$. A classical theorem of Schoenberg says that a metric space ${(X, d)}$ is isometric to a subset of a Hilbert space if and only if ${\exp(-\gamma d^2)}$ is a positive kernel for all ${\gamma >0}$. One may ask: is there any function ${f\colon [0, \infty)\to\mathbb R}$ such that ${f\circ d}$ is a positive kernel on ${X}$ for every metric space ${(X, d)}$?

Suppose ${f}$ is such a function. Applying it to the bipartite metric defined above, we find that the following matrix must be positive semidefinite (the case ${m=3}$, ${n=2}$ shown; in general the diagonal blocks have sizes ${m\times m}$ and ${n\times n}$):

${\displaystyle \begin{pmatrix} f(0) & f(a) & f(a) & f(c) & f(c) \\ f(a) & f(0) & f(a) & f(c) & f(c) \\ f(a) & f(a) & f(0) & f(c) & f(c) \\ f(c) & f(c) & f(c) & f(0) & f(b) \\ f(c) & f(c) & f(c) & f(b) & f(0) \end{pmatrix} }$

Test the PSD property against vectors of the form ${(s, s, s, t, t)}$, that is ${m}$ entries equal to some number ${s}$ followed by ${n}$ entries equal to some number ${t}$. The result is the quadratic form ${As^2 + Bt^2 + 2C^2st }$ where ${A = (m^2-m) f(a)+mf(0)}$, ${B = (n^2-n)f(b)+nf(0)}$, and ${C = mnf(c)}$.

For this quadratic form to be positive semidefinite, we need ${A\ge 0}$ and ${C^2\le AB}$ to hold. The inequality ${A\ge 0}$ amounts to ${f(a)\ge 0}$ and ${f(0)\ge 0}$, because ${m}$ can be any positive integer. Simply put, ${f}$ must be nonnegative.

The inequality ${C^2\le AB}$ simplifies to

${ mn f(c)^2 \le ( (m-1) f(a)+f(0)) \cdot ((n-1)f(b)+ f(0)) }$

When ${m=n=1}$ and ${a=b}$, we obtain ${f(c)\le f(0)}$; so, ${f}$ is maximized at ${0}$. Letting ${m, n\to \infty}$ yields ${f(c)^2 \le f(a)f(b)}$. What does this mean?

If ${c=(a+b)/2}$, the inequality ${f(c)^2 \le f(a)f(b)}$ amounts to log-convexity of ${f}$, and there are lots of log-convex functions. But crucially, our only constraint on ${c}$ in the construction of a bipartite metric space is ${c\ge \max(a, b)/2}$. In the special case ${a=b}$ we find that ${f(c)\le f(a)}$ whenever ${c\ge a/2}$ (and ${a>0}$). On one hand, ${f}$ must be nonincreasing because all ${c\ge a}$ are allowed. On the other hand, ${f(a/2)\le f(a)}$ implies that the sequence ${k\mapsto f(2^k)}$ is nondecreasing for ${k\in\mathbb Z}$. These conditions can be satisfied together only if ${f}$ is constant on ${(0, \infty)}$.

And there you have it, a complete description of functions ${f\colon [0, \infty)\to\mathbb R}$ that transform every distance matrix into a positive semidefinite matrix: ${f\equiv c\ge 0}$ on ${(0, \infty)}$, and ${f(0)\ge c}$. It is immediate that such ${f}$ indeed works for all metric spaces, not just bipartite ones.

Noncontracting Jordan curves

A simple closed curve ${\Gamma}$ on the plane can be parameterized by a homeomorphism ${f\colon \mathbb T\to \Gamma}$ in infinitely many ways. It is natural to look for “nice” parameterizations, like nonexpanding ones in the previous post. This time, let us look for noncontracting parameterizations: ${|f(a)-f(b)|\ge |a-b|}$ for all ${a, b\in \mathbb T}$. Note that Euclidean distance is used here, not arclength. And we still want ${f}$ to be a homeomorphism. Its noncontracting property simply means that the inverse ${f^{-1}}$ is nonexpanding aka 1-Lipschitz.

What are some necessary conditions for the existence of a noncontracting parameterization? We can mimic the three from the earlier post Nonexpanding Jordan curves, with similar proofs:

1. The curve must have length at least ${2\pi}$.
2. The curve must have diameter at least 2.
3. The curve must enclose some disk of radius 1. (Apply Kirszbraun’s theorem to ${f^{-1}}$ and note that the resulting Lipschitz map G of the plane will attain 0 somewhere, by a topological degree / winding number argument. Any point where G = 0 works as the center of such a disk.)

This time, the 3rd item supersedes both 1 and 2. Yet, the condition it presents is not sufficient for the existence of a noncontracting parameterization. A counterexample is a disk with a “comb-over”:

Indeed, suppose that ${g=f^{-1}}$ is a nonexpanding map from this curve onto the unit circle. Let ${1 = A < B < C}$ be the three points at which the curve meets the positive x-axis. Since every point of the curve is within small distance from its arc AB, it follows that ${g(AB)}$ is a large subarc of ${\mathbb T}$ that covers almost all of the circle. But the same argument applies to the arcs ${g(BC)}$ and ${g(CA)}$, a contradiction.

No matter how large the enclosed disk is, a tight combover around it can force arbitrarily large values of the Lipschitz constant of ${f^{-1}}$. Sad!

1. It is sufficient for the curve to be star-shaped with respect to the origin, in addition to enclosing the unit disk. (Equivalent statement: ${\Gamma}$ has a polar equation ${r = r(\theta)}$ in which ${r \ge 1}$.) Indeed, the nearest-point projection onto a convex set is a nonexpanding map, and projecting ${\Gamma}$ onto the unit circle in this way gives the desired parameterization.
2. It is sufficient for the curve to have curvature bounded by 1. I am not going into this further, because second derivatives should not be needed to control a Lipschitz constant.

Recall that for nonexpanding parameterizations, the length of the curve was a single quantity that mostly answered the existence question (length at most 4 — yes; length greater than ${2\pi}$ — no). I do not know of such a quantity for the noncontracting case.

Extreme values of a reproducing kernel for polynomials

For every nonnegative integer ${n}$ there exists a (unique) polynomial ${K_n(x, y)}$ of degree ${n}$ in ${x}$ and ${y}$ separately with the following reproducing property:

${\displaystyle p(x) = \int_{-1}^1 K_n(x, y)p(y)\,dy}$

for every polynomial ${p}$ of degree at most ${n}$, and for every ${x}$. For example, ${K_1(x, y)= (3xy+1)/2}$; other examples are found in the post Polynomial delta function.

This fact gives an explicit pointwise bound on a polynomial in terms of its integral on an interval:

${\displaystyle |p(x)| \le M_n(x) \int_{-1}^1 |p(y)|\,dy}$

where ${M_n(x) = \sup\{|K(x, y)| \colon y\in [-1, 1]\}}$. For example, ${M_1(x) = (3|x|+1)/2}$.

Although in principle ${x}$ could be any real or complex number, it makes sense to restrict attention to ${x\in [-1, 1]}$, where integration takes place. This leads to the search for extreme values of ${K}$ on the square ${Q=[-1, 1]\times [-1, 1]}$. Here is how this function looks for ${n=1, 2, 3}$:

The symmetries ${K(x, y)=K(-x, -y) = K(y, x)}$ are evident here.

Explicitly,

${\displaystyle K_n(x, y) = \sum_{k=0}^n \frac{2k+1}{2} P_k(x)P_k(y)}$

where ${P_k}$ is the Legendre polynomial of degree ${k}$ and the factor ${(2k+1)/2}$ is included to make the polynomials an orthonormal set in ${L^2(-1, 1)}$. Since ${P_k}$ oscillates between ${-1}$ and ${1}$, it follows that

${\displaystyle |K_n(x, y)|\le \sum_{k=0}^n \frac{2k+1}{2} = \frac{(n+1)^2}{2}}$

and this bound is attained at ${K(1, 1)=K(-1,-1)=(n+1)^2/2}$ because ${P_k(1)=1}$ and ${P_k(-1)=(-1)^k}$.

Is

${\displaystyle K_n(-1, 1) =\sum_{k=0}^n (-1)^k\frac{2k+1}{2} = (-1)^n \frac{n+1}{2}}$

the minimum value of ${K}$ on the square ${Q}$? Certainly not for even ${n}$. Indeed, differentiating the sum

${\displaystyle S_n(x) = K_n(x, 1) = \sum_{k=0}^n \frac{2k+1}{2} P_k(x)}$

with respect to ${x}$ and using ${P_k'(-1) =(-1)^{k-1}k(k+1)/2}$, we arrive at

${\displaystyle S_n'(-1) = (-1)^{n-1} \frac{n(n^2+3n+2)}{4}}$

which is negative if ${n}$ is even, ruling out this point as a minimum.

What about odd ${n}$, then: is it true that ${K_n \ge -(n+1)/2}$ on the square ${Q}$?

${n=1}$: yes, ${K_1(x, y) = (3xy+1)/2 \ge (-3+1)/2 = -1}$ is clear enough.

${n=3}$: the inequality ${K_3\ge -2}$ is also true… at least numerically. It can be simplified to ${35(xy)^3 + 9(xy)^2 + 15xy \ge (21x+21y+3)(x^2+y^2)}$ but I do not see a way forward from there. At least on the boundary of ${Q}$ it can be shown without much work:

${\displaystyle K_3(x, 1) + 2 = \frac{5}{4}(x+1)(7x^2-4x+1)}$

The quadratic term has no real roots, which is easy to check.

${n=5}$: similar story, the inequality ${K_5\ge -3}$ appears to be true but I can only prove it on the boundary, using

${\displaystyle K_5(x, 1)+3 = \frac{21}{16}(x + 1)(33 x^4 - 18x^3 - 12x^2 + 2x + 3)}$

The quartic term has no real roots, which is not so easy to check.

${n=7}$: surprisingly, ${K_7(4/5, 1) = -2229959/500000}$ which is about ${-4.46}$, disproving the conjectural bound ${K_7\ge -4}$. This fact is not at all obvious from the plot.

Questions:

• Is ${K_n \ge -Cn}$ on the square ${Q = [-1, 1]\times [-1, 1]}$ with some universal constant ${C}$?
• Is the minimum of ${K_n}$ on ${Q}$ always attained on the boundary of ${Q}$?
• Can ${M_n(x) = \sup\{|K(x, y)| \colon y\in [-1, 1]\}}$ be expressed in closed form, at least for small degrees ${n}$?

Institutions ranked by the number of AMS Fellows, 2019 edition

Only those with the count of 3 or greater are included. Not counting the deceased. Considering CUNY as a single institution. Source of data.

1. Rutgers The State University of New Jersey New Brunswick : 44
2. University of California, Los Angeles : 38
3. Massachusetts Institute of Technology : 34
4. University of California, Berkeley : 34
5. University of Michigan : 32
6. Cornell University : 26
7. Princeton University : 26
8. University of Illinois, Urbana-Champaign : 24
9. University of Wisconsin, Madison : 24
10. Brown University : 23
11. Stanford University : 22
12. University of Texas at Austin : 22
13. New York University, Courant Institute : 21
14. The City University of New York : 21
15. University of California, San Diego : 21
16. University of Illinois at Chicago : 21
17. University of Chicago : 20
18. University of Washington : 20
19. Stony Brook University : 19
20. Texas A&M University : 19
21. University of California, Santa Barbara : 19
22. University of Minnesota-Twin Cities : 18
23. University of Pennsylvania : 17
24. Duke University : 16
25. Indiana University, Bloomington : 16
26. Purdue University : 16
27. University of Maryland : 16
28. Georgia Institute of Technology : 15
29. Northwestern University : 15
30. Ohio State University, Columbus : 15
31. Pennsylvania State University : 15
32. University of California, Irvine : 13
33. University of Utah : 13
34. Johns Hopkins University, Baltimore : 12
35. University of British Columbia : 12
36. Boston University : 11
37. Harvard University : 11
38. University of Notre Dame : 11
39. University of Toronto : 11
40. Eidgenössische Technische Hochschule Zürich (ETH Zürich) : 10
41. University of North Carolina at Chapel Hill : 10
42. University of Virginia : 10
43. Vanderbilt University : 10
44. Brandeis University : 9
45. University of California, Davis : 9
46. University of Georgia : 9
47. Columbia University : 8
48. Institute for Advanced Study : 8
49. Rice University : 8
50. Tel Aviv University : 8
51. University of Oregon : 8
52. California Institute of Technology : 7
53. Ecole Polytechnique Fédérale de Lausanne (EPFL) : 7
54. Michigan State University : 7
55. Microsoft Research : 7
56. North Carolina State University : 7
57. University of Nebraska-Lincoln : 7
58. University of Oxford : 7
59. University of Southern California : 7
60. Williams College : 7
61. Carnegie Mellon University : 6
62. The Hebrew University of Jerusalem : 6
63. Université Pierre et Marie Curie (Paris VI) : 6
64. University of Arizona : 6
65. University of Rochester : 6
66. Harvey Mudd College : 5
67. Northeastern University : 5
68. Temple University : 5
69. Université Paris-Diderot : 5
70. University of California, Riverside : 5
71. University of Colorado, Boulder : 5
72. Virginia Polytechnic Institute and State University : 5
73. Boston College : 4
74. Florida State University : 4
75. Louisiana State University, Baton Rouge : 4
76. NYU Polytechnic School of Engineering : 4
77. Norwegian University of Science and Technology : 4
78. Rutgers The State University of New Jersey Newark : 4
79. University of Cambridge : 4
80. University of Connecticut, Storrs : 4
81. University of Missouri-Columbia : 4
82. University of Tennessee, Knoxville : 4
83. University of Warwick : 4
84. Washington University : 4
85. Weizmann Institute of Science : 4
86. Yale University : 4
87. Alfréd Rényi Institute of Mathematics, Hungarian Academy of Sciences : 3
88. Australian National University : 3
89. Barnard College, Columbia University : 3
90. Emory University : 3
91. Imperial College : 3
92. Københavns Universitet : 3
93. KTH Royal Institute of Technology : 3
94. Mathematical Institute, Oxford University : 3
95. McGill University : 3
96. Pomona College : 3
97. Shanghai Jiao Tong University : 3
98. Tsinghua University : 3
99. Tufts University : 3
100. Università degli Studi di Milano : 3
101. Université Paris-Sud (Paris XI) : 3
102. Université de Montréal : 3
103. University of Iowa : 3
104. University of Melbourne : 3
105. University of Memphis : 3


text:
# History of the classification of mathematical subjects

I would like to know if there are sources on the history of the classification of mathematical subjects. Gérard Lang

• I like this question, but I think it could benefit from more context. E.g. are there particular instances of classification, or changes of classification over time, which you're interested in? Feb 14 '18 at 15:21
• Two early examples I know of in which the whole of mathematics was essentially classified are: (1) Jahrbuch über die Fortschritte der Mathematik (late 1860s); (2) Royal Society of London Catalogue of Scientific Papers 1800-1900. Volume I, Pure Mathematics (1908) (Bull. AMS review by G. A. Miller). Feb 14 '18 at 15:26
• Thank you. These are two very interesting sources, that are completing the article of Craig Fraser. The review of the second source in the AMS is not too optimistic about the quality of the catalogue of the Royal Society. Feb 14 '18 at 21:35
• Trying to answer to Noah Schweber, I would have been specially interested to have a study on this question on the long term, beginning in the Antiquity and going on until now, specially concerning the possible changes of status regarding Logic and Geometry. Feb 14 '18 at 21:44
• "The review of the second source in the AMS is not too optimistic about the quality of the catalogue $\ldots$" That review was by G. A. Miller who was/is well known for highly critical reviews and short papers criticizing various (often minor, but not always) issues about someone's paper(s) and/or book(s). I've read quite a few such reviews and critical commentaries by him (probably over 50) in the past 20 some years, and for what it's worth, this particular review is extremely high praise for him. Feb 14 '18 at 21:49


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The rapid increase of research in nanoscale devices and nanotechnology in the past few decades has revealed that nanomaterials may possess exceptional properties that are significantly different from the bulk counterpart, due to local rearrangements of the atoms at surfaces and defects. Transmission electron microscopy (TEM) is an indispensable tool when it comes to the characterization of nanomaterials, primarily due to its ability to resolve the local-structure of materials at the atomic-scale. To study dynamic...

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### Biogeomorphology of Bedrock Fluvial Systems

Tasnuba Jerin
The dynamic interactions between fluvial processes and vegetation vary in different environments and are uncertain in bedrock settings. Bedrock streams are much less studied than alluvial in all aspects, and in many respects act in qualitatively different ways. This research seeks to fill this lacuna by studying bedrock streams from a biogeomorphic perspective. The first part of this research aims to identify the impacts of woody vegetation that may be common to fluvial systems and...

### Scale modeling of dust capture through a flooded-bed dust scrubber integrated within a longwall shearer

Sampurna Arya, Ashish R. Kumar, Kozo Saito, Thomas Novak & Adam Levy
Meeting federal dust standards at a longwall mine face is among the more difficult challenges for a longwall mine operator. With recent changes in federal dust regulations requiring lower worker exposure, maintaining compliance has become increasingly difficult. The authors recommend the concept of controlling respirable and float dust, which is inherent in longwall mining, through the application of a flooded-bed dust scrubber to a longwall shearer. A full-scale physical model of a longwall shearer, modified...

### THE ONCOGENIC ROLE OF THE PROTEIN TYROSINE PHOSPHATASE 4A3 (PTP4A3 OR PRL-3) IN T-CELL ACUTE LYMPHOBLASTIC LEUKEMIA

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T-cell Acute Lymphoblastic Leukemia (T-ALL) is an aggressive blood cancer. There are no immunotherapies and few molecularly targeted therapeutics available for the treatment of this malignancy. The identification and characterization of genes and pathways that drive T-ALL progression is critical for the development of new therapies for T-ALL. The protein tyrosine phosphatase PTP4A3, also known as PRL-3, has been extensively reported to play a causative role in numerous cancers, including several types of blood malignancies....

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Jack Swab
Geographic representation and thinking has a long history in the American business world. This thesis examines the role of geographic representation and thinking in the fire insurance industry in the late nineteenth and early twentieth centuries through the Sanborn Map Company and in the development of site selection as a concept in the mid-twentieth century through the biography of William Applebaum. Through these case studies, I explore the relevance applied cartographic representations to the business...

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Quasicrystals have been shown to exhibit physical properties that are dramatically different from their periodic counterparts. A limited number of magnetic quasicrystals have been fabricated and measured, and they do not exhibit long-range magnetic order, which is in direct conflict with simulations that indicate such a state should be accessible. This dissertation adopts a metamaterials approach in which artificial quasicrystals are fabricated and studied with the specific goal of identifying how aperiodicity affects magnetic long-range...

### AN EXPLORATION OF SMETANA’S Z DOMOVINY (FROM MY HOMELAND), TWO PIECES FOR VIOLIN AND PIANO, JB 1:118

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Maryam Ali Al Ghafli
The inverse eigenvalue problem for a given operator is to determine the coefficients by using knowledge of its eigenfunctions and eigenvalues. These are determined by the behavior of the solutions on the domain boundaries. In our problem, the Schrödinger operator acting on functions defined on the unit ball of $\mathbb{R}^3$ has a radial potential taken from $L^2_{\mathbb{R}}[0,1].$ Hence the set of the eigenvalues of this problem is the union of the eigenvalues of infinitely many...

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Research has indicated accredited, U.S. higher education-governed intensive English programs (IEPs) often struggle financially due to a scarcity of resources (namely students) because of political and global economic factors and increased competition (ICEF Monitor, 2017; IIE, 2017; Ladika, 2018; Soppelsa, 2015). However, few IEPs advertise online language acquisition (OLA) courses despite the increase in online study methods at the higher education institutes governing the programs and its use by competitors. The purpose of this study...

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Jane Samuel
Despite calls for increased Adoption Communication Openness (ACO) within the adoptive family, research indicates that families still struggle to accomplish the recommended elements and levels of openness. What could be holding families back from this key process? Three focus groups comprised of 17 adoptive parents of early adolescents (aged 10-14) who were age 0-2 at the time of placement were thematically coded. This inductive analysis revealed the complexity rooted in being —sometimes successfully and sometimes...

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Charlie Sun
The Kentucky Transportation Cabinet’s current practice for determining the compressive strength of structural concrete consists of two methods. Both methods require that samples be delivered to an off-site facility for testing. Accordingly, affected parties on the project site must wait for the delivery of samples to the off-site facility, the performance of tests and analysis of test data, and the return of test results to the project site. Analysis of existing structural concrete requires core...

### BARRIERS AND FACILITORS OF HEALTHCARE USE AMONG PEOPLE WHO INJECT DRUGS

Candice Kuns-Adkins
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Tracy Gastineau-Stevens
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text:
## CryptoDB

### Paper: The Curse of Small Domains: New Attacks on Format-Preserving Encryption

Authors: Viet Tung Hoang Stefano Tessaro Ni Trieu DOI: 10.1007/978-3-319-96884-1_8 (login may be required) Search ePrint Search Google Slides CRYPTO 2018 Format-preserving encryption (FPE) produces ciphertexts which have the same format as the plaintexts. Building secure FPE is very challenging, and recent attacks (Bellare, Hoang, Tessaro, CCS ’16; Durak and Vaudenay, CRYPTO ’17) have highlighted security deficiencies in the recent NIST SP800-38G standard. This has left the question open of whether practical schemes with high security exist.In this paper, we continue the investigation of attacks against FPE schemes. Our first contribution are new known-plaintext message recovery attacks against Feistel-based FPEs (such as FF1/FF3 from the NIST SP800-38G standard) which improve upon previous work in terms of amortized complexity in multi-target scenarios, where multiple ciphertexts are to be decrypted. Our attacks are also qualitatively better in that they make no assumptions on the correlation between the targets to be decrypted and the known plaintexts. We also surface a new vulnerability specific to FF3 and how it handles odd length domains, which leads to a substantial speedup in our attacks.We also show the first attacks against non-Feistel based FPEs. Specifically, we show a strong message-recovery attack for FNR, a construction proposed by Cisco which replaces two rounds in the Feistel construction with a pairwise-independent permutation, following the paradigm by Naor and Reingold (JoC, ’99). We also provide a strong ciphertext-only attack against a variant of the DTP construction by Brightwell and Smith, which is deployed by Protegrity within commercial applications. All of our attacks show that existing constructions fall short of achieving desirable security levels. For Feistel and the FNR schemes, our attacks become feasible on small domains, e.g., 8 bits, for suggested round numbers. Our attack against the DTP construction is practical even for large domains. We provide proof-of-concept implementations of our attacks that verify our theoretical findings.
##### BibTeX
@inproceedings{crypto-2018-28865,
title={The Curse of Small Domains: New Attacks on Format-Preserving Encryption},
booktitle={Advances in Cryptology – CRYPTO 2018},
series={Lecture Notes in Computer Science},
publisher={Springer},
volume={10991},
pages={221-251},
doi={10.1007/978-3-319-96884-1_8},
author={Viet Tung Hoang and Stefano Tessaro and Ni Trieu},
year=2018
}


text:
# Physics problems related to green function ?

1. May 27, 2013

### lofaif

hello all !

my teacher told me to do a research on examples of problems that has connection with green function on solving differential equations (with programmed numerical solutions) in my final year project , can you give me such problems to work on as an undergraduate ? , thank you !

Last edited: May 27, 2013
2. May 27, 2013

### SteamKing

Staff Emeritus
Just about any problem governed by Laplace's equation, Del^2 Phi = 0. This equation is used in 2-D elasticity and fluid flow.

3. May 28, 2013

### sudu.ghonge

Try simulating a classical damped, driven harmomic oscillator. I'm assuming you're working with Fourier transforms, yes?

4. May 28, 2013

### lofaif

yes , how about its connection with quantum mechanics ? , can i do that or is it hard for non physics student ?

5. May 28, 2013

### NegativeDept

The Schrödinger equation* is a linear PDE with a $\nabla^2$ in it:

$\imath \hbar \partial_t \Psi(\mathbf{r},t) = \frac{-\hbar^2}{2m}\nabla^2 \Psi(\mathbf{r},t) + V(\mathbf{r},t) \Psi(\mathbf{r},t)$

So it can be useful in QM to know Green's functions for the Helmholtz equation, which is closely related to what SteamKing mentioned. Probably the simplest example would be to look up a 1-dimensional "particle in a box" problem. Pick some initial wavefunction $\Psi(x,0)$ and use Green's functions and convolution to find future wavefunctions $\Psi(x,t)$.

If you want to really show off, combine this with sudu.ghonge's idea and do the same for a 1-dimensional quantum harmonic oscillator instead of a particle-in-a-box. But it might be a good idea to practice on something other than quantum mechanics because QM is often counterintuitive and hard to visualize.

* in position representation, if anyone asks.


text:
## Area Bounded By Non Negative Function

The Definite Integration of a non negative function f(x) with respect to ‘x’ from ‘a’ to ‘b’ gives the area of the bounded region abBAa (given in above figure). Now let f(x) is a countinuous non negative function on an interval [a, b], such that f(x)≥0. Then the area of the closed region bounded by the function f(x) from ‘a’ to ‘b’ with respect to ‘x’ is equal to the definite integral $\int ^{b}_{a}f\left( x\right) dx$.

Now for the best understanding of the topic you are highly recommended to attempt at least five questions based on the theory that you have learned.

Example: find the area of the region bounded by the curve x²-2x+4, x-axis and line segments x = -1 and x = 3.
Solution:
\begin{aligned}\int ^{3}_{-1}\left( x^{2}-2x+4\right) dx=\int ^{3}_{-1}x^{2}dx-2\int ^{3}_{-1}xdx+4\int ^{3}_{-1}dx\\ \Rightarrow \left( \dfrac{x^{3}}{3}\right) _{-1}^{3}-2\left( \dfrac{x^{2}}{2}\right) _{-1}^{3}+4\left( x\right) _{-1}^{3}=\dfrac{52}{3}\end{aligned}

Now You People are Advised to Attempt the following unsolved problems for better understanding this topic.

Question: Find the area of the region bounded by the function Cos-1(cos x), x-axis on the interval [0, 2π].

Question: Let f(x)=cos-1(cos x)-sin-1(sin x) is a continuous function. Find the area of the region bounded by the function f(x) and x-axis in [0, 2π].

Question: Find the area of the region bounded by the function f(x)=x|sin x|and x-axis in the interval [0, 2π].

Question: Find the area of the region bounded by the curve f(x)=tan-1(x)+ln(x), x-axis, and the straight line segments x=1 and x=2.

Question: Let f(x) be a function defined as f(x)=max{|x-2|, (4-x2), (x-2)1/3}, x belongs to [-2, 4]. Now find the area bounded by the curve and x-axis.

So this is it from this tutorial. Hoping you people have attempted all the unsolved problems that have written above. In the next tutorial, we will discuss the topic “area of the region bounded by the non-positive continuous function”.


text:
# Deriving transfer equations of active filters with Wolfram Language

Posted 3 years ago
7831 Views
|
5 Replies
|
7 Total Likes
|

## Introduction

Today I had the opportunity to revisit an old topic for my Bachelor's degree (Op Amps) and felt the urge to see just how much can Mathematica help in the classroom for deriving equations for an application of op amps known as active filters.

For those who do not know what active filters are, as their name suggest, a filter that permits the separation of useful frequencies in a signal to pass through a system from one stage to another. They are classified according to the type of filtering that they execute: Low Pass, High Pass, Band Pass, and Band Rejection. There is an extensive amount of literature that can greatly expand on my explanations such as Electronic Devices by Floyd.

## Getting started

To start off, I chose to derive the transfer equation for a second order low pass active filter. The circuit diagram is shown below.

From the field of Electronics and Electrical engineering there are great tools available to analyze the system without much difficulty. The concepts which are used are Nodal Voltage Analysis of circuits, Laplace transformation to convert from the time domain to the frequency domain and a bit of Control Theory to study the system after obtaining the equation.

## Deriving the equations

For the case of op amps and active filters a couple of assumptions must be made for the analysis.

• Input Impedance is close to infinity
• Output Impedance is close to zero
• The voltage difference between input terminals (V+ and V-) is zero

We start by converting the capacitors to impedances (Z) in the frequency domain.

Z1= 1 / (s c1);
Z2 = 1/ (s c2);

Establish the equations for each node (V1, V2, V3)

eq01 = (v1 - vin)/R1 + (v1 - v2)/R2 + (v1 - vout)/Z2 == 0;
eq02 = (v2 - v1)/R2 + v2/Z1 == 0;
eq03 = (v3 - vout)/R3 + v3/R4 == 0;

As mentioned earlier, the input voltage difference accross voltage terminals is zero. Thus

v2 = v3;

We now solve the system of equations

{solutionLP} = Solve[{eq01, eq02, eq03}, {v1, v2, vout}]

We obtain the transfer function for the system which describes the relation between output and input

H(s) = Voutput / Vinput (* Transfer function equation in control theory*)
LP = vout/vin /. {solutionLP} (* Transfer function for the Low Pass Filter *)

Let us now give values to the capacitors and resistors to simulate its behavior

parameters = {R1 -> 1000, R2 -> 1000, R3 -> 100, R4 -> 100, c1 -> 0.000001, c2 -> 0.000001}

Use the appropriate functions in Mathematica to get a transfer function model

tfLP = TransferFunctionModel[vout/vin /. {solutionLP}, s] /. parameters

Let us now see how the system responds to different input signals and frequencies

out01 = OutputResponse[tfLP, Sin[5000 t], t];
Plot[out01, {t, 0, 1/100}]

out02 = OutputResponse[tfLP, UnitStep[t], t]
Plot[out02, {t, 0, 0.02}, PlotRange -> All]

We can get a more useful visualization by plotting the dB gain of the output depending on the frequency

LogLinearPlot[20 Log[Abs[tfLP[I frequency]]], {frequency, 0, 10000}]

A smarter way to do the same is with a built-in function

BodePlot[tfLP]

## Conclusion

Overall, Mathematica proves that what could have taken me about three hours of work by hand can be done in less than ten minutes. This speeds up dramatically the process of learning about a new topic in Electronics and implementing a design for a specific problem. One can easily make several iterations of the system, coupled with other equations to obtain proper values for the passive components of the filters and simulate its response in mere seconds.

## To Do

I'd love to expand a bit more on this topic, adding the equations for the others filtered mentioned. Of course more useful examples would be provided from my own notes and real world comparisons. The example shown above was created just for this demonstration and can obviously be improved as well.

5 Replies
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Posted 3 years ago
 Hi Sam, I did not know about the existence of those functions. I'll play around with them later. I still think you work has educational value as you derive things from scratch I'm glad the point got accross, I made the post with the intention of showing how Mathematica can be integrated at an introductory level for Electronics. And sure, the automated functions are always faster to implement in code; my focus was to test development from the physical implementation perspective with limited values available for components (standard capacitor and resistor values).The next step with the code is to set a Manipulate[] function which could solve the system of equations for the missing quantities and compare that to the standard equations used in Electronics to define resistor and capacitor values.
Posted 3 years ago
 Issac have you seen these built-in filters: Linear and Nonlinear Filters and specifically LowpassFilter ? I still think you work has educational value as you derive things from scratch. Perhaps it would also be interesting to compare your filters to built-in ones.
Posted 3 years ago
 - Congratulations! This post is now a Staff Pick as distinguished on your profile! Thank you, keep it coming! In future it is a good idea to add some meaningful outputs (as we did for formulas in your post) and possibly attach the notebook. We could not understand if meant this as a valid Wolfram Language syntax: H(s) = Voutput / Vinput (* Transfer function equation in control theory*) LP = vout/vin /. {solutionLP} (* Transfer function for the Low Pass Filter *)


text:
# string and subscript in label

s = Import[
"Desktop/crack.xlsx", {"Data", 2, Range[1, 546], {1, 2, 4}}];
ListContourPlot[s, AspectRatio -> Automatic,
ColorFunction -> "Pastel",
Contours -> Range[0.054, 0.27, (0.27 - 0.054)/10],
FrameLabel -> {Subscript[X, 1], Subscript[X, 2]},
PlotLegends -> Placed[Automatic, Right], ColorFunction -> "Pastel",
PlotLabel -> Style["MLS" Subscript[u, 2]
OverTilde[\[Eta]] "=0.048%", FontSize -> 18]]

The label of this ouput has subscript and string. But I want the label show like the below. How to modify the Plotlabel command??

You can try using Row to layout the text, and use Spacer command to adjust the spacing as needed

Plot[Sin[x], {x, -Pi, Pi},
PlotLabel ->
Text@Style[
Row[{"MLS", Spacer[5], Subscript[u, 2], Spacer[20],
OverTilde[\[Eta]] , Spacer[5], "= 0.048%"}]], FontSize -> 18],

Or, the way I would actually do it, is use Szhorvat nice MaTex package, like this

<< MaTeX
label = MaTeX["\\text{MLS}\\, u_2 \\qquad \\tilde\\eta = 0.048\%",
Magnification -> 2];
Plot[Sin[x], {x, -Pi, Pi}, PlotLabel -> label, ImagePadding -> 20]

• I personally prefer to just maintain labels as Strings like so: TraditionalForm["MLS \!$$\*SubscriptBox[\(u$$, $$2$$]\) \ \!$$\*OverscriptBox[\(\[Eta]$$, $$~$$]\)=0.048%"]. The InputForm is quite ugly, but within the notebook one can use typesetting and it becomes a one-liner along the lines of TraditionalForm["MLS u_2 ~\Eta=0.048%"] – LLlAMnYP Apr 13 '15 at 7:23

Your output is formatted wrongly, because what you gave to Style is not a series of things to style, but a multiplication of its elements. Just evaluate the arguments on their own in your notebook to see the effect:

"MLS" Subscript[u, 2] OverTilde[\[Eta]] "=0.048%"
(* "=0.048%" "MLS" \!$$\*OverscriptBox[\(\[Eta]$$, $$~$$]\) Subscript[u, 2] *)

To get what you wanted, you could modify PlotLabel to

PlotLabel->TraditionalForm@
Style[Row[{"MLS ", Subscript[u, 2], "\t", OverTilde[\[Eta]],
"=0.048%"}], FontSize -> 18]

• fyi, I am not sure that using tab "\t" is recommend for spacing, as it might generate different space amount depending on the platform or some other system configuration. I think Spacer[n] is supposed to generate the exact space on all systems. (mac, linux, windows) – Nasser Apr 13 '15 at 8:02
• @Nasser: Well, \t` works for me. ;) But you are right, of course. – Jinxed Apr 13 '15 at 8:05


text:
# zbMATH — the first resource for mathematics

Homogenization limits on diffusion equations in thin domains. (English) Zbl 0672.73009
The problem deals with a composite material in a plate-like domain in the limit as the plate thickness tends to zero. the related mathematical study is to derive a consistent two-dimensional model for this structure as the limit of some three-dimensional problem. The paper studies a simple problem: a linear diffusion equation for a composite medium in a thin plate-like domain of $$R^ n$$, $$\Omega_{\epsilon}=\omega X(- \epsilon /2,\epsilon /2)$$, where $$\omega \subset R^{n-1}$$ is smooth and bounded. The diffusivity matrices are spatially varying.
Two main results are proved. The first theorem shows that, as the thickness tends to zero, the limit solution must satisfy a corresponding effective diffusion equation in the mid section $$\omega$$. This analysis does not need any information about the geometry of the inhomogenities. In the particular case of a mixture of two isotropic components distributed with a horizontal periodicity (in the first n-1 variables), optimal bounds on the effective conductivity of the composite, as the thickness tends to zero, are given by the second theorem. These bounds depend only on the diffusivity coefficients of the two components and on their volume fraction, they are independent of the geometry.
This study extends results originally obtained by D. Caillerie [RAIRO, Anal. Numer. 15, 295-319 (1981; Zbl 0483.35003)].
Reviewer: Th.Lévy

##### MSC:
 74E05 Inhomogeneity in solid mechanics 74E30 Composite and mixture properties 35A35 Theoretical approximation in context of PDEs 74K20 Plates 74S30 Other numerical methods in solid mechanics (MSC2010) 46S30 Constructive functional analysis
Full Text:
##### References:
 [1] D CAILLERIE, Homogénéisation des équations de la diffusion stationnaire dans les domaines cylindriques aplatis R A I R O Analyse Numérique, 15 (1981), pp 295-319 Zbl0483.35003 MR642495 · Zbl 0483.35003 · eudml:193384 [2] D CAILLERIE, Thin elastic and periodic plates Math Methods Appl Sci , 6 (1984), pp 159-191 Zbl0543.73073 MR751739 · Zbl 0543.73073 · doi:10.1002/mma.1670060112 [3] A DAMLAMIAN, M VOGELIUS, Homogenization limits of the equations of elasticity in thin domains SIAM J Math Anal , 18 (2) (1987), pp 435-451 Zbl0614.73012 MR876283 · Zbl 0614.73012 · doi:10.1137/0518034 [4] E DE GIORGI, Quelques problèmes de $$\Gamma$$-convergence, in proceedings of the Conference Computing Methods in Applied Sciences and Engineering, Versailles 1979 (R Glowinski and J L Lions eds ), North Holland, 1980, pp 637-643 Zbl0445.49018 MR584059 · Zbl 0445.49018 [5] R KOHN - G W MILTON, On bounding the effective conductivity of anisotropic composites, in Homogenization and Effective Moduli of Materials and Media (J R Ericksen, D Kinderlehrer, R Kohn, J L Lions eds ) IMA Volumes in Math Appls, 1, Springer, 1986, pp 97-125 Zbl0631.73012 MR859413 · Zbl 0631.73012 [6] R KOHN, M VOGELIUS, A new model for thin plates with rapidly varying thickness II a convergence proof Quart Appl Math , 43 (1985) pp 1-22 Zbl0565.73046 MR782253 · Zbl 0565.73046 [7] K A LURIE, A V CHERKAEV, Exact estimates of the conductivity of composites formed by two isotropically conducting media taken in prescribed proportions Proc Royal Soc Edinburgh, 99A (1984), pp 71-87 Zbl0564.73079 MR781086 · Zbl 0564.73079 · doi:10.1017/S030821050002597X [8] F MURAT, H-convergence, Séminaire d’Analyse Fonctionnelle et Numérique 1977/1978 Département de Mathématiques, Université d’Alger [9] F MURAT, G FRANCFORT, Homogenization and optimal bounds in linear elasticity Arch Rat Mech Anal , 94 (1986), pp 307-334 Zbl0604.73013 MR846892 · Zbl 0604.73013 · doi:10.1007/BF00280908 [10] L TARTAR, Cours Peccot, Collège de France, Paris, 1977 [11] L TARTAR, Estimations fines des coefficients homogénéises, in Ennio De Giorgi’s Colloquium (P Kree ed), Pitman Research Notes in Math, Pitman, 1985 Zbl0586.35004 MR909716 · Zbl 0586.35004
This reference list is based on information provided by the publisher or from digital mathematics libraries. Its items are heuristically matched to zbMATH identifiers and may contain data conversion errors. It attempts to reflect the references listed in the original paper as accurately as possible without claiming the completeness or perfect precision of the matching.


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unable to evaluate the function to numeric values - Maple Help

Home : Support : Online Help : System : Error Message Guide : unable to evaluate the function to numeric values

Warning, unable to evaluate the function to numeric values in the region; see the plotting command's help page to ensure the calling sequence is correct

Error, empty plot

 Description Note: For users of Maple 13 and earlier versions, this warning occurs when you create a Maple plot without points, curves, or surfaces.  In the warning message, "function" refers to the function being plotted (expressed as an algebraic expression or procedure) and "region" refers to the plotting domain.   Verify that you are using the correct syntax for the plotting command and that the function values are real numbers in the specified region.

Examples

Example 1: 3-D plot of a function

 >
 ${h}{:=}\left({x}{,}{y}\right){→}{{x}}^{{2}}{}{\mathrm{cos}}{}\left({y}\right)$ (2.1)
 >

Solution 1: Specify rather than $h$ in the plot3d calling sequence.  This solution uses an expression () as the argument to the plot3d command.

 >

Solution 2: To plot the function $h$, leave $h$ in operator form and use the second calling sequence listed on the plot3d help page.

 >

Example 2: Expression with non-real values

 >

In the range given, the solutions for the function $\sqrt{x}$ are all complex values that cannot be plotted with the plot command.

Solution:  Change the range to one for which the expression is real-valued.

Alternatively, to plot in the complex plane, use plots[complexplot].

Example 3: Additional parameter not given numerical value

 >

The expression to be plotted includes two unknowns, but plot expects only one unknown..

Solution 1: To plot this expression, replace $y$ with a numeric value.

 >

Solution 2: To create a 3-D plot of this expression, use plot3d and specify a range for $y$.

 >

Example 4: Independent variable not specified

 >

Solution: To plot this function, explicitly assign the range to $x$.

 >

Important: Previously in Maple 10, you had to specify a range when entering a plot.
Example 5 (Maple 10): Function has not been assigned to anything

 > $\mathrm{plot}\mathit{}\left(f\right)$
 Warning, unable to evaluate the function to numeric values in the region; see the plotting command's help page to ensure the calling sequence is correct
 Error, empty plot

Here,  plot(f) is interpreted as operator-form and a default range of -10..10 is assumed. The error occurs because $f$ has not been assigned to anything.

Example 5 (Maple 11 and later): You can enter a plot without having to specify a range.

Here, $f$ is interpreted as the independent variable.  A default range of -10..10 is assumed.

 > $\mathrm{plot}\left(f\right)$


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# The inside dimensions of a rectangular wooden box are 6 inches by 8 in

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The inside dimensions of a rectangular wooden box are 6 inches by 8 in [#permalink]

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The inside dimensions of a rectangular wooden box are 6 inches by 8 inches by 10 inches. A cylindrical cannister is to be placed inside the box so that it stands upright when the closed box rests on one of its six faces. Of all such cannisters that could be used, what is the radius, in inches, of one that has the maximum volume?

(A) 3
(B) 4
(C) 5
(D) 6
(E) 8
[Reveal] Spoiler: OA

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The inside dimensions of a rectangular wooden box are 6 inches by 8 in [#permalink]

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04 Dec 2017, 23:31
Bunuel wrote:
The inside dimensions of a rectangular wooden box are 6 inches by 8 inches by 10 inches. A cylindrical cannister is to be placed inside the box so that it stands upright when the closed box rests on one of its six faces. Of all such cannisters that could be used, what is the radius, in inches, of one that has the maximum volume?

(A) 3
(B) 4
(C) 5
(D) 6
(E) 8

Let's assume that the bottom side of the box has dimensions 8*10 when the canister is placed in it. Obviously, the diameter can't be more than eight so the radius is 4 Height is 6
$$V = \pi * r^2 * h = 16\pi * 6 = 96\pi$$

Let's assume that the bottom side of the box has dimensions 8*6 when the canister is placed in it. Obviously, the diameter can't be more than 6 so the radius is 3 Height is 10
$$V = \pi * r^2 * h = 9\pi * 10 = 90\pi$$

Last edited by Vorovski on 05 Dec 2017, 08:17, edited 2 times in total.
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The inside dimensions of a rectangular wooden box are 6 inches by 8 in [#permalink]

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05 Dec 2017, 04:35
Bunuel wrote:
The inside dimensions of a rectangular wooden box are 6 inches by 8 inches by 10 inches. A cylindrical cannister is to be placed inside the box so that it stands upright when the closed box rests on one of its six faces. Of all such cannisters that could be used, what is the radius, in inches, of one that has the maximum volume?

(A) 3
(B) 4
(C) 5
(D) 6
(E) 8

Volume of a cylinder: $$pi*r^2*h$$
Radius = $$\frac{Diameter}{2}$$

The maximum diameter of a cylinder that fits inside the rectangular box is restricted to the length and width of the box.
So, if we have a length =8, width = 10 and height = 6, then max. diameter = 8 and radius = 4;
And, the volume of the cylinder: $$pi*4^2*6=pi*96$$

Let's try another, with length =8, width = 6 and height = 10, then max. diameter = 6 and radius = 3;
And, the volume of the cylinder: $$pi*3^2*10=pi*90$$

Largest radius we can have is 4.

Last edited by exc4libur on 07 Dec 2017, 15:41, edited 1 time in total.
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Re: The inside dimensions of a rectangular wooden box are 6 inches by 8 in [#permalink]

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05 Dec 2017, 06:34
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exc4libur wrote:
Bunuel wrote:
The inside dimensions of a rectangular wooden box are 6 inches by 8 inches by 10 inches. A cylindrical cannister is to be placed inside the box so that it stands upright when the closed box rests on one of its six faces. Of all such cannisters that could be used, what is the radius, in inches, of one that has the maximum volume?

(A) 3
(B) 4
(C) 5
(D) 6
(E) 8

Volume of a cylinder: $$pi*r^2*h$$

Let's test the wooden box dimensions for the bottom and the height.
If we have a bottom of 6 and a height of 10 then the radius would be 3 and volume of cylinder = $$pi*3^2*10 = pi*90$$
If we have a bottom of 8 and a height of 10 then the radius would be 4 and volume of cylinder = $$pi*4^2*10 = pi*160$$
If we have a bottom of 10 and a height of 8 then the radius would be 5 and volume of cylinder = $$pi*5^2*8 = pi*200$$

So the maximum volume of the cylinder would be $$pi*200$$ with a radius of 5.

Might i check if you have considered how the cylinder could fit in?

For example, the length of the bottom is 10 and the height is 8, that means the width the bottom is 6. How do you fit in a cylinder with a radius of 5 (i.e., diameter = 10) while the width is only 6?

Re: The inside dimensions of a rectangular wooden box are 6 inches by 8 in   [#permalink] 05 Dec 2017, 06:34
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# Bipolar coordinates

Bipolar coordinate system

Bipolar coordinates are a two-dimensional orthogonal coordinate system. There are two commonly defined types of bipolar coordinates.[1] The first is based on the Apollonian circles. The curves of constant σ and of τ are circles that intersect at right angles. The coordinates have two foci F1 and F2, which are generally taken to be fixed at (−a, 0) and (a, 0), respectively, on the x-axis of a Cartesian coordinate system. The second system is two-center bipolar coordinates. There is also a third coordinate system that is based on two poles (biangular coordinates).

The term "bipolar" is sometimes used to describe other curves having two singular points (foci), such as ellipses, hyperbolas, and Cassini ovals. However, the term bipolar coordinates is reserved for the coordinates described here, and never used to describe coordinates associated with those other curves, such as elliptic coordinates.

Geometric interpretation of the bipolar coordinates. The angle σ is formed by the two foci and the point P, whereas τ is the logarithm of the ratio of distances to the foci. The corresponding circles of constant σ and τ are shown in red and blue, respectively, and meet at right angles (magenta box); they are orthogonal.

## Definition

The most common definition of bipolar coordinates (στ) is

$x = a \ \frac{\sinh \tau}{\cosh \tau - \cos \sigma}$
$y = a \ \frac{\sin \sigma}{\cosh \tau - \cos \sigma}$

where the σ-coordinate of a point P equals the angle F1 P F2 and the τ-coordinate equals the natural logarithm of the ratio of the distances d1 and d2 to the foci

$\tau = \ln \frac{d_1}{d_2}$

(Recall that F1 and F2 are located at (−a, 0) and (a, 0), respectively.) Equivalently

$x + i y = a i \cot\left( \frac{\sigma + i \tau}{2}\right)$[2][3]

## Curves of constant σ and τ

The curves of constant σ correspond to non-concentric circles

$x^2 + \left( y - a \cot \sigma \right)^2 = \frac{a^{2}}{\sin^2 \sigma}$

that intersect at the two foci. The centers of the constant-σ circles lie on the y-axis. Circles of positive σ are centered above the x-axis, whereas those of negative σ lie below the axis. As the magnitude |σ| increases, the radius of the circles decreases and the center approaches the origin (0, 0), which is reached when |σ| = π/2, its maximum value.

The curves of constant $\tau$ are non-intersecting circles of different radii

$y^2 + \left( x - a \coth \tau \right)^2 = \frac{a^2}{\sinh^2 \tau}$

that surround the foci but again are not concentric. The centers of the constant-τ circles lie on the x-axis. The circles of positive τ lie in the right-hand side of the plane (x > 0), whereas the circles of negative τ lie in the left-hand side of the plane (x < 0). The τ = 0 curve corresponds to the y-axis (x = 0). As the magnitude of τ increases, the radius of the circles decreases and their centers approach the foci.

## Reciprocal relations

The passage from the Cartesian coordinates towards the bipolar coordinates can be done via the following formulas:

$\tau = \tanh^{-1} T_\tau,$

and

$\sigma = \tan^{-1} T_\sigma,$

where

$T_\tau = \frac{2 a y}{x^2 + y^2 + a^2},$

and

$T_\sigma = \frac{2 a x}{x^2 + y^2 - a^2}.$

## Scale factors

The scale factors for the bipolar coordinates (στ) are equal

$h_\sigma = h_\tau = \frac{a}{\cosh \tau - \cos\sigma}$

Thus, the infinitesimal area element equals

$dA = \frac{a^2}{\left( \cosh \tau - \cos\sigma \right)^2} \, d\sigma\, d\tau$

and the Laplacian is given by

$\nabla^2 \Phi = \frac{1}{a^2} \left( \cosh \tau - \cos\sigma \right)^2 \left( \frac{\partial^2 \Phi}{\partial \sigma^2} + \frac{\partial^2 \Phi}{\partial \tau^2} \right)$

Other differential operators such as $\nabla \cdot \mathbf{F}$ and $\nabla \times \mathbf{F}$ can be expressed in the coordinates (στ) by substituting the scale factors into the general formulae found in orthogonal coordinates.

## Applications

The classic applications of bipolar coordinates are in solving partial differential equations, e.g., Laplace's equation or the Helmholtz equation, for which bipolar coordinates allow a separation of variables. A typical example would be the electric field surrounding two parallel cylindrical conductors.

## Extension to 3-dimensions

Bipolar coordinates form the basis for several sets of three-dimensional orthogonal coordinates. The bipolar cylindrical coordinates are produced by projecting in the z-direction. The bispherical coordinates are produced by rotating the bipolar coordinates about the $x$-axis, i.e., the axis connecting the foci, whereas the toroidal coordinates are produced by rotating the bipolar coordinates about the y-axis, i.e., the axis separating the foci.

## References

• H. Bateman "Spheroidal and bipolar coordinates", Duke Mathematical Journal 4 (1938), no. 1, 39–50
• Hazewinkel, Michiel, ed. (2001), "Bipolar coordinates", Encyclopedia of Mathematics, Springer, ISBN 978-1-55608-010-4
• Lockwood, E. H. "Bipolar Coordinates." Chapter 25 in A Book of Curves. Cambridge, England: Cambridge University Press, pp. 186–190, 1967.
• Korn GA and Korn TM. (1961) Mathematical Handbook for Scientists and Engineers, McGraw-Hill.