Monday, April 16, 2018

What is a slide rule for?

I was listening to Sam Cooke's classic song "Wonderful World" and thought of the line: "Don't know what the slide rule is for". When the song was released in 1960, the line was describing how disinterested the protagonist was about math and algebra, as the slide rule was a common tool for doing calculations. Today, that line would probably describe most young adults or younger, as the slide rule has not been used for calculations for many years with the rise of the calculator. When I was in high school, we used electronic calculators in our tests and exams, but out of curiosity my brother and I bought a slide rule anyway (they were still being sold in bookstores at the time, but disappeared soon after from the shelves). I was fascinated by how you can multiply two numbers merely by aligning the start of one ruler with the first number and reading off the result off the second number.  This is due a property of logarithm: log(ab) = log(a) + log(b). Thus multiplication is reduced to addition. By printing the numbers in logarithmic scale and lining up segment end-to-end (corresponding to addition), we achieve the operation of multiplication using a slide rule. Similarly, division can be done with a slide rule as it corresponds to subtraction. And it can do a lot more, such as trigonometry and taking square and cube roots. Accuracy was a problem, and they were soon supplanted by calculators which can compute to many digits of precision. As the song is continuously being covered by many artists, I wonder if the current audience would find the lyrics strange?

Wednesday, March 14, 2018

Primes of the form H(n,-k)-1

OEIS sequence A299145 lists the primes of the form $Q(n,k) = \sum_{i=2}^n i^k$ for $n \geq 2$ and $k> 0$. It is clear that except for the case $k =1$ and $n=2$ resulting in the prime $2$, we must have $n\geq 3$.
The sum $Q(n,k)$ is equal to $H(n,-k) - 1$ where $H(n,m) = \sum_{i=1}^n \frac{1}{i^m}$ is the generalized harmonic number of order $n$ of $m$. It is well known that $H(n,-k)$ is a polynomial of $n$ of degree $k+1$. In particular, Faulhaber's formula shows that
$$H(n,-k) = \frac{1}{2} n^k + \frac{1}{k+1}\sum_{j=0}^{\lfloor k/2 \rfloor} \left(\begin{array}{c} k+1\\2j\end{array}\right) B_{2j} n^{k+1-2j}$$
where $B_i$ is the $i$-th Bernoulli number.

$H(n,k)$ and thus $Q(n,k)$ can be written as a degree $k+1$ polynomial of $n$ with rational coefficients. The smallest denominator of this polynomial is found in OEIS A064538. In 2017, Kellner and Sondow showed that this is equal to $(k+1)\prod_{p\in S}$ where $S$ is the set of primes $p \leq \frac{k+2}{2+(k \mod 2)}$ such that the sum of the base $p$ digits of $k+1$ is $p$ or larger.

Since $H(1,-k) = 1$, this implies that $Q(1,k) = 0$ and thus $n-1$ is a factor of $Q(n,k)$. Since $n\geq 3$, if $n-1$ does not get cancelled out by a factor of the denominator, we would have a nontrivial factor of $Q(n,k)$ and thus $Q(n,k)$ is not prime. This implies that $Q(n,k)$ is prime only if $n-1$ is a divisor of a(k) in sequence A064538. Thus for each $k$ there is only a finite number of values of $n$ to check. This provides an efficient algorithm to find terms of this sequence by looking only for primes in the numbers $Q(n,k)$ where $n-1$ is a divisor of A064538(k). There are 45 such numbers with $1000$ or less digits.

References
Bernd C. Kellner, Jonathan Sondow, Power-Sum Denominators, Amer. Math. Monthly 124 (2017), 695-709.

Tuesday, February 20, 2018

Chaos in music

In an earlier post, I talked about fractals mentioned in the lyrics of a song. Today, I heard the song "Jurassic Park" by Al Yankovic, which mentions chaos theory, a related branch of mathematics that studies the unpredictable and complex behavior of dynamical systems, often manifested as sensitive dependence on initial conditions. On the other hand, musical compositions themselves can exhibit fractal and chaotic behavior [1,2].

References:

[1] K J Hsü and A Hsü, "Self-similarity of the "1/f noise" called music," PNAS, vol. 88, no. 8, pp. 3507-3509, 1991.
[2] J. Beran, "Music - Chaos, Fractals, and Information," CHANCE, vol. 17, no. 4, pp. 7-16, 2004.

Monday, December 18, 2017

O, zero and eight

When my parents bought us our first computer, a Commodore VIC-20, in my early teens, I learned that to distinguish capital O with the digit zero, a diagonal line is put through the digit zero (also known as a slashed zero). This is due to the relatively low resolution computer fonts used in those days, where O is easily confused with 0 (but its use actually predates computers). On the other hand, some Scandinavian languages has something similar to a slashed zero in their alphabet, and the slashed zero can cause more problems than it solves. The use of the slashed zero is also useful when writing computer code using pen and paper, an activity that is becoming rare. Recently, I have not seen the use of slashed zero in word processing since given the number of pixels they have at their disposal, many fonts do not use it to represent the digit 0. In some fonts such as DejaVu Sans Mono and the font used in the Windows command prompt, a dotted zero is used to represent 0. Incidentally, for the Default font used in this blog, zero (0) looks almost identical to the lowercase letter "o".

The empty set (the set with zero elements) is denoted as {}, and is sometimes also denoted with a symbol similar to a slashed zero, but with the slash extending beyond the boundary of "0". This notation makes sense since in Zermelo-Fraenkel set theory the natural numbers are defined as sets with 0 being the empty set and the number n+1 defined as $n \cup \{n\}$, i.e. n+1 is the set obtained by augmenting the set n with a single element: the set consisting of the set n. In other words, 0 = {}, 1 = {{}}, 2 = {{},{{}}}, etc.

The use of a slashed zero is a good idea, but recently I found that it is causing problems for me. There is one place where I consistently see the use of a slashed zero and that is on credit card receipts. With advancing age and the onset of farsightedness, I have a hard time separating the slashed zero from the digit 8. When we go out to eat at a restaurant, the lighting is typically dim which exacerbates the problem and sometimes I have a hard time determining how much tip to leave and add it up correctly on the bill. I wish the receipt printer would not use the slash zero in their font (I don't think the dotted zero is much better either in this case).

Tuesday, October 10, 2017

The triangle inequality and its converse

The triangle inequality

The triangle inequality is a workhorse in many branches of mathematics. It expresses the subadditivity of norms and is stated as:
$$ \| A+B\| \leq \|A\| + \|B\|.$$
The name comes from its interpretation on the plane. If $A$ and $B$ are complex numbers, then we can draw a triangle with vertices at the points $0$, $A$ and $A+B$, with the resulting sides having norms equal to $|A|$, $|B|$ and $|A+B|$. Stated in another way, if $a$, $b$ and $c$ are the lengths of the $3$ sides of a triangle, then $a \leq b+c$. This is illustrated in Fig. 1.
A slightly different, but clearly equivalent statement is the following: if $a \geq b \geq c \geq 0$ are the lengths of the $3$ sides of a triangle, then $a \leq b+c$. We will use this alternative formulation as it is easier to state and prove the converse and its generalization.

Fig. 1: Illustration of the triangle inequality and its converse.

Converse of the triangle inequality on the plane

The converse of the triangle inequality is also true: If $a \geq b \geq c \geq 0$ and $a \leq b+c$, then there exists a triangle whose sides have lengths $a$, $b$ and $c$ respectively. Equivalently this can be restated as: If $a \geq b \geq c \geq 0$ and $a \leq b+c$, then there exists complex numbers $A$, $B$ and $C$ with $|A|=a$, $|B| = b$, $|C| = c$ such that $A+B+C = 0$.

The following is a simple construction of this triangle.
First we show that $d = \frac{b^2+c^2 - a^2}{2bc}$ satisfies $|d| \leq 1$. This follows from the fact that $d = \frac{(b+c)^2 -a^2-2bc}{2bc} = -1 + \frac{(b+c)^2 - a^2}{2bc} \geq -1$
and $d= \frac{(b-c)^2 -a^2+2bc}{2bc} = \frac{(b-c)^2 -a^2}{2bc} + 1\leq 1$.
Thus we can choose $\phi = \cos^{-1}(d)$ which satisfies $a^2 = b^2+c^2-2bc\cos (\phi)$ and the law of cosines shows that the triangle with sides of length $b$ and $c$ and angle $\phi$ with have the third side of length $a$. (see Fig. 1).

The generalized triangle inequality and its converse

A simple induction argument generalizes the triangle inequality to the summation of more than 2 quantities:
$$\left\| \sum_i^n A_i\right\| \leq \sum_i^n \|A_i\|$$
The geometrical interpretation is that if $r_i$ are the lengths of edges of a polygon, then $r_i \leq \sum_{j\neq i} r_j$ or equivalently, if $r_i$ are the lengths of edges of a polygon, and $r_1\geq r_i$, then $r_1\leq \sum_{i > 1}r_i$ (Fig. 2).

Fig. 2: Illustration of the generalized triangle inequality and its converse.

The converse of the generalized triangle inequality is true as well. If $r_1\geq r_i \geq 0$ and $r_1\leq \sum_{i = 2}^n r_i$, then there is a $n$-polygon with $r_i$ as the lengths of its sides (Fig. 2). Having one vertex of the polygon at the origin of the complex plane this can be reformulated as:

Lemma 1: (Ref. [1])
Let $n\geq 2$ and $r_1\geq r_2\geq \cdots r_n \geq 0$ be real sumbers such that $r_1 \leq \sum_{i=2}^n r_i$, then there exists complex numbers $c_i$ such that $|c_i| = r_i$ and $\sum_i c_i = 0$.

Proof: 
As stated in [1] this is easily proved by induction. For $n=2$ this implies that $|r_1| = |r_2|$ and thus we can set $c_1 = r_1 = -c_2$. For $n=3$ this is the converse of the triangle inequality described above. Suppose the Lemma is true for $n = k \geq 3$. Let $n = k+1$, and $s_k = r_k+r_{k+1}$. If $s_k \leq r_1$, then applying the Lemma to $r_1,\cdots r_{k-1}, s_k$ and then splitting $c_k$ into $\frac{r_k}{s_k}c_k$ and $\frac{r_{k+1}}{s_k}c_k$ would prove it for $n=k+1$. If $s_k > r_1$, then $s_k \leq r_1 + r_2 \leq \sum_{i=1}^{k-1} r_k$ and again we can apply the Lemma for $n=k$.


The degenerate case occurs when $r_1 = \sum_{i=2}^n r_i$ in which the resulting polygon must have zero area (Fig. 3). The proof of Lemma 1 also shows that the polygon can be arranged to look like a triangle, i.e. there is a partition of $\{r_i\}$ into $3$ sets $R_1$, $R_2$ and $R_3$ such that $\sum_{r_i\in R_1}r_i \geq \sum_{r_j\in R_2}r_j \geq \sum_{r_k\in R_3}r_k $ and $\sum_{r_i\in R_1}r_i \leq \sum_{r_j\in R_2\cup R_3}r_j $. This is illustrated in Fig. 4.

Fig. 3: The degenerate polygon when $r_1 = \sum_{i=2}^n r_i$.


Fig. 4: The edges of the polygon can be reordered to form a triangle.

overlapping edges

We have implicitly allowed the possibility that the edges of the polygon may overlap, i.e. the angles of the vertices are allowed to be $0$ (see for example the degenerate case in Fig. 3). The next result shows that we can have up to $n-3$ angles to be either $0$ or $\pi$.


Lemma 2: (Ref. [2])
Let $r_i$ be real numbers such that $r_1\geq r_2 \geq \cdots \geq r_n \geq 0$ and $r_1 \leq \sum_{i=2}^n r_i$, then there exists $c_i$ such that $|c_i| = r_i$ and $\sum_i c_i = 0$ and either all $c_i$'s are real or $n-2$ of the $c_i$'s are real.

Proof:
We follow the proof in [2]. Select $j\geq 3$ to be the smallest number such that $\sum_{i=3}^j r_i \geq r_1-r_2$. Such an $j$ is possible since $\sum_{i=3}^n r_i \geq r_1-r_2$. Since $r_i\leq r_2$ for $i\geq 3$, this implies that $\sum_{i=3}^j r_i < r_1$ as otherwise $\sum_{i=3}^{j-1} r_i \geq r_1-r_2$. For $k = j+1,\dots n$, if $\sum_{i=3}^{k-1} r_i \leq r_1$, then set $c_k = r_k$, otherwise set $c_k = -r_k$. Ths ensures that $r_1-r_2\leq \sum_{i=3}^k c_i \leq r_1+r_2$ for each $k$ and by the converse of the triangular inequality there exists $c_1$ and $c_2$ with $|c_1| = r_1$ and $|c_2| = r_2$ such that $\sum_i c_i = 0$. Note that in the degenerate case both $c_1$ and $c_2$ are real.


The geometric interpretation of this result is that the polygon can be folded into a triangle (possibly with some overlapping edges and angles equals to $0$) as illustrated in Fig. 5 with the edges with lengths $a$ and $b$ being the 2 longest edges.

Fig. 5: The edges of the polygon can be reordered and folded to form a triangle with the edges with lengths $a$ and $b$ being the longest 2 edges. Some of the edges may be overlapping.


Singularity of matrices

The Levy-Desplanques theorem [3,4] (which is equivalent to Gershgorin's circle criterion [5]) gives a sufficient condition for a complex matrix to be nonsingular:

Theorem 1
The matrix $A = \{a_{ij}\}$ is nonsingular if $|a_{ii}| > \sum_{j\neq i}|a_{ij}|$ for all $i$.

This is easily shown by using the generalized triangle inequality. Suppoe $A$ is singular, i.e. $Ax = 0$ for some nonzero vector $x$.
Let $i$ be such that $|x_i|\geq |x_j|$ for all $j$. Since $x\neq 0$, this implies that $|x_i| > 0$. Then applying the generalized triangle inequality to $|\sum_{j\neq i}a_{ij}x_j| = |a_{ii}x_i|$ results in:
$$ |a_{ii}||x_i| \leq \sum_{j\neq i}|a_{ij}||x_i|$$
Dividing both sides by $|x_i|$ shows that it violates the condition that $|a_{ii}| > \sum_{j\neq i}|a_{ij}|$ for all $i$.

Similarly, the converse of the triangle inequality can be used to prove the following statement:

Theorem 2: (Ref. [1])
Let $A$ be a real nonnegative matrix and let $D$ is a nonzero diagonal matrix with nonnegative diagonal elements. If $B=DA$ satisfies $b_{ij}\leq \sum_{k\neq i} b_{kj}$ for all $i$, $j$, there there exists a complex singular matrix $C = \{c_{ij}\}$ with $|c_{ij}| = a_{ij}$.

If $d_ia_{ij}\leq \sum_{k\neq i} d_ka_{kj}$ for all $i$, $j$, then Lemma 1 implies that there are $g_{ij}$ with $|g_{ij}| = d_ia_{ij}$ such that $\sum_i{g_{ij}} = 0$.
By choosing $c_{ij} = \frac{g_{ij}}{d_i}$ if $d_i \neq 0$ and $c_{ij} = a_{ij}$ if $d_i = 0$, we get a matrix such that $\sum_i d_ic_{ij} = 0$ and $|c_{ij}| = a_{ij}$. Since $D$ is not the zero matrix, this implies that the rows of $\{c_{ij}\}$ are linear dependent, hence $C$ is singular.


These results were extended by the Camion-Hoffman theorem [1] which gives necessary and sufficient conditions for a real matrix $A$ such that any complex matrix whose elements have the same norm as the corresponding elements in $A$ is nonsingular. More precisely it is stated as:

Theorem 3: (Camion-Hoffman)
Let $A = \{a_{ij}\}$ be a real matrix of nonnegative numbers. The following conditions are equivalent:

  1. If $C = \{c_{ij}\}$ is a complex matrix with $|c_{ij}| = a_{ij}$ then $C$ is nonsingular.
  2. If $D$ is a nonzero diagonal matrix with nonnegative diagonal elements, then $B = DA$ contains an entry $b_{ij}$ such that $b_{ij} >\sum_{k\neq i} b_{kj}$.
  3. There exists a permutation matrix $P$ and a diagonal matrix $D$ with positive diagonal elements such that $B = PAD$ is strictly row sum dominant, i.e. $b_{ii} >\sum_{i\neq j} b_{ij}$.


If the conditions in Theorem 3 are not satisfied, then there is a candidate matrix $C$ with $|c_{ij}| = a_{ij}$ such that $C$ is singular.
Lemma 2 can be used to show that this candidate can be chosen such that each row is real except for possibly two elements.

Lemma 3: (Ref. [2])
If $B$ is a singular complex matrix, then there exists a singular complex matrix $C$ such that each row has either $0$ or $2$ complex elements and $|b_{ij}| = |c_{ij}|$.

Proof:
Suppose $Bz = 0$. As before, we select $c_{ij} = \frac{b_{ij}z_j}{|z_j|}$ if $z_j\neq 0$ and $c_{ij} = |b_{ij}|$ otherwise, i.e. $\sum_i c_i = 0$. This means that $\sum_i c_{ij}|z_j| = 0$, i.e. $C$ is singular. By Lemma 2, we can replace up to $n-2$ of $c_{ij}$ with real numbers of the same norm.

This result is extended in [2] to show that this candidate can be chosen with at most $2$ complex elements:

Theorem 4: (Ref. [2])
If $B$ is a singular complex matrix, then there exists a singular complex matrix $C$ with either $0$ or $2$ complex elements such that $|b_{ij}| = |c_{ij}|$.

References
[1] P. Camion and A. J. Hoffman, "On the nonsingularity of complex matrices," Pacifi c Journal of Mathematics, vol. 17, no. 2, pp. 211-214, 1966.
[2] D. Coppersmith and A. J. Hoffman, "On the singularity of matrices," Linear Algebra and its Applications, vol. 411, pp. 277-280, 2005.
[3]  L. Lévy, "Sur la possibilité du l'équilibre électrique," C. R. Acad. Sci. Paris, vol. 93, pp. 706-708, 1881.
[4] J. Desplanques, "Théorème d'algèbre," J. de Math. Spec., vol. 9, pp. 12-13, 1887.
[5] S. A. Geršgorin, "Über die Abgrenzung der Eigenwerte einer Matrix," Izv. Akad. Nauk SSSR Otd. Fiz.-Mat. Nauk, vol. 7, pp. 749-754, 1931.








Friday, September 8, 2017

Numbers such that 7 is the smallest decimal digit of their squares.

9949370777987917 is the smallest number whose square has as its smallest decimal digit 7.
Note that $9949370777987917^2 = 98989978877879888789778997998889$. What is the next such number?

Sunday, July 23, 2017

Solar eclipse 2017

On August 21, 2017, there will be a solar eclipse occurring coast to coast across the continental United States. The last time this happened was almost a century ago in 1918. This has two effects on states that generate a lot of solar energy: the solar energy generated decreases during an Eclipse and people turn on their lights as the sky darkens. Both of these events stress the utility grid.  California's utilities are asking people not to turn on the lights during this event. I am surprised such a short event (and California is not even in the path of totality) can cause concerns for power utilities operators. I wonder if something similar happens when there is no wind in the summer and more people turn on their fans to cool off?

Monday, July 17, 2017

Happy Yellow Pig Day!

Today, July 17 is Yellow Pig Day. This "holiday" was created by Princeton math students Michael Spivak and David C. Kelly when they were analyzing the number 17 and wanted to have a day to celebrate this number. This year (2017) is extra special since the year also ends in the digits 17.
Michael Spivak is the author of the wonderful differential geometry texts (published by Publish-or-Perish Press which he founded) which I fondly remember reading in graduate school. He also wrote a popular typesetting book on TeX which I did not read since Leslie Lamport's book on LaTeX makes typesetting papers using LaTeX so easy that I have skipped learning TeX and went straight to using LaTeX (which sits on top of TeX).
If I remember correctly, the typesetting software I used prior to LaTeX was ChiWriter, which is quite easy to use as it is WYSIWYG, so moving to a typesetting paradigm where you relinquish control on the formatting, page breaks, etc. to the software is quite unnerving, but I soon learn to appreciate it as I can focus on the content of the document, rather than spend time improving its appearance.

Sunday, June 4, 2017

Moon and Jupiter on June 3, 2017

The Moon and Jupiter are in close proximity (not physically, but as viewed from Earth) on the night of June 3, 2017.

Sunday, May 21, 2017

The race of Cloud Computing

I wrote earlier about a news digest app returning supplemental information which is not directly related to the article. Today, I read on the app that the horse "Cloud Computing" has won the Preakness race. And sure enough, there is a box underneath the article with a definition of cloud computing from Wikipedia. What is ironic is that this news summary is probably enabled by leveraging cloud computing. "Cloud Computing" won over "Classic Empire" by coming in from behind and won the race by a head. Perhaps this is an indication how cloud computing will overtake the classical empire of on premise computing😊.