Sunday, July 09, 2006

The Mathematics Genealogy Project

The intent of this project is to compile information about ALL the mathematicians of the world. We earnestly solicit information from all schools who participate in the development of research level mathematics and from all individuals who may know desired information.

Please notice: Throughout this project when we use the word "mathematics" or "mathematician" we mean that word in a very inclusive sense. Thus, all relevant data from statistics, or computer science or operations research is welcome.

In the following paragraphs we shall try to outline our goals and our underlying philosophy for the GENEALOGY PROJECT. It is our goal to list all individuals who have received a doctorate in mathematics. For each individual we plan to show the following:

  • The complete name of the degree recipient.

  • The name of the university which awarded the degree.

  • The year in which the degree was awarded.

  • The complete title of the dissertation.

  • The complete name(s) of the advisor(s).

Hilbert's 23 Problems (#10)

Definition: Given a Diophantine equation with any number of unknown quantities and with rational integral numerical coefficients: To devise a process according to which it can be determined by a finite number of operations whether the equation is solvable in rational integers.

What this basically says is that is there a way for us to give a boolean answer to whether any such equation is solvable or not without necessarily knowing the solution?

The answer turns out to be no. Get the book Hilbert's Tenth Problem by Yuri V. Matiyasevich to understand why.

Although I've started reading this book, I must say that it is a little above my level of understanding. I've already read the first chapter twice just to get some basic understanding of his definitions. However, the commentary at the end of Chapter 3 was absolutely remarkable.

In my number theory and abstract algebra classes in college (university doesn't have that nice ring to it), we learned that there does not exist any polynomial that returns only primes for all integer arguments, but rather only for some (Euler's $f(x) = x^2 + x + 41$).

Julia Robinson proved in 1952 that the binomial coefficients and the factorial are exponential Diophantine, and gave an exponential Diophantine representation for the set of all prime numbers. In 1960, Putnam noted that a Diophantine set is the positive part of the range of a polynomial. Thus, it became clear (to someone) that if exponentiation were established to be Diophantine, it would become possible to construct a polynomial such that the positive values it assumed would coincide precisely with the prime numbers. In 1971a, Matiyasevich gave the first upper bound of 24 variables in his Russian article, which was later reduced to 21 in the appendix of the English translation.

In 1976, Jones, Sato, Wada and Wiens exhibited the following polynomial:

$(k + 2){1 - [wz + h + j - q]^2 - [(gk + 2g + k + 1)(h + j) + h - z]^2$ - $[2n + p + q + z - e]^2$ - $[16(k + 1)^3(k + 2)(n + 1)^2 + 1 - f^2]^2$ - $[e^3(e + 2)(a + 1)^2 + 1 - o^2]^2$ - $[(a^2 - 1)y^2 + 1 - x^2]^2$ - $[n + l + v - y]^2$ - $[((a + u^2(u^2 - a))^2 - 1)(n + 4dy)^2 + 1 - (x + cu)^2]^2$ - $[(a^2 - 1)t^2 + 1 - m^2]^2$ - $[q + y(a - p -1) + s(2ap + 2a - p^2 - 2p - 2) - x]^2$ - $[z + pl(a - p) + t(2ap - p^2 - 1) - p\m]^2$ - $[ai + k + 1 - l -i]^2$ - $[p + l(a - n - 1) + b(2an +2a - n^2 - 2n - 2) - m]^2\}$

This polynomial contains 26 variables (all the letters of the English alphabet), and the set of its positive values is exactly the set of all prime numbers. Note: The polynomial written above, representing only primes, is itself the product of two polynomials.

Later, the bound was further reduced to 12 variables by Wada in 1975, and by Jones, Sato, Wada and Wiens in 1976. Currently, the record is at 10 variables, achieved by Matiyasevich in 1977a.

Sunday, June 11, 2006

Mathematical Poetry

Finite Simple Group (of order two)


A Klein Four original by Matt Salomone

URL: The Klein Four Group
Movie: Finite Simple Group (of Order Two)

The path of love is never smooth
But mine's continuous for you
You're the upper bound in the chains of my heart
You're my Axiom of Choice, you know it's true

But lately our relation's not so well-defined
And I just can't function without you
I'll prove my proposition and I'm sure you'll find
We're a finite simple group of order two

I'm losing my identity
I'm getting tensor every day
And without loss of generality
I will assume that you feel the same way

Since every time I see you, you just quotient out
The faithful image that I map into
But when we're one-to-one you'll see what I'm about
'Cause we're a finite simple group of order two

Our equivalence was stable,
A principal love bundle sitting deep inside
But then you drove a wedge between our two-forms
Now everything is so complexified

When we first met, we simply connected
My heart was open but too dense
Our system was already directed
To have a finite limit, in some sense

I'm living in the kernel of a rank-one map
From my domain, its image looks so blue,
'Cause all I see are zeros, it's a cruel trap
But we're a finite simple group of order two

I'm not the smoothest operator in my class,
But we're a mirror pair, me and you,
So let's apply forgetful functors to the past
And be a finite simple group, a finite simple group,
Let's be a finite simple group of order two
(Oughter: "Why not three?")

I've proved my proposition now, as you can see,
So let's both be associative and free
And by corollary, this shows you and I to be
Purely inseparable. Q. E. D.

My Poem


by Eve Andersson

There once was a number named pi
Who frequently liked to get high.
All he did every day
Was sit in his room and play
With his imaginary friend named i.

There once was a number named e
Who took way too much LSD.
She thought she was great.
But that fact we must debate;
We know she wasn't greater than 3.

There once was a log named Lynn
Whose life was devoted to sin.
She came from a tree
Whose base was shaped like an e.
She's the most natural log I've seen.

eve

A Young Don From Trinity


There was a young Don from Trinity
Who solved the square root of infinity
While counting the digits
He was seized by the fidgets
Dropped science and took up divinity

Anonymous.

A New Solution to an Old Problem


The Topologist's child was quite hyper
'Til she wore a Moebius diaper.
The mess on the inside
Was thus on the outside
And it was easy for someone to wipe her.

By Eleanor Ninestein.

An Analyst, Surname of Nero


An analyst, surname of Nero,
Is my mathematical hero.
Says he, "When in doubt,
I always start out,
'Given $\epsilon$ > 0 ...'."

By James R. Martino,
Department of Mathematics
The Johns Hopkins University
Baltimore, MD 21218.

Thursday, June 01, 2006

Continuous, Nowhere Differentiable Functions

The Weierstrass function $f_a(x) = sum_(k=1)^infty sin(pi k^a x)/{pi k^a}$ (originally defined for a = 2) is an example of a continuous function, but differentiable only on a set of points of measure zero.

The function was published by Weierstrass but, according to lectures and writings by Kronecker and Weierstrass, Riemann seems to have claimed already in 1861 that the function f(x) is not differentiable on a set dense in the reals. However, Ullrich (1997) indicates that there is insufficient evidence to decide whether Riemann actually bothered to give a detailed proof for this claim. du Bois-Reymond (1875) stated without proof that every interval of f contains points at which f does not have a finite derivative, and Hardy (1916) proved that it does not have a finite derivative at any irrational and some of the rational points. Gerver (1970) and Smith (1972) subsequently proved that f has a finite derivative (namely, 1/2) at the set of points x = ${2A+1}/{2B+1}$ where A and B are integers. Gerver (1971) then proved that f is not differentiable at any point of the form ${2A}/{2B+1}$ or ${2A+1}/{2B}$. Together with the result of Hardy that f is not differentiable at any irrational value, this completely solved the problem of the differentiability of f.

Since then, a number of mathematicians have developed examples of nowhere continuous, differentiable (CND) functions and it has been shown that these types of functions are the norm (PDF).

One thing that should be obvious is that these functions are not lines in the geometric sense. These are all limits of other continuous functions, that are mostly differentiable, which converge to a nowhere differentiable function.

In Euclidean geometry, a polygon (or straight line) should only be drawn by pencil and straighthedge. If we were to draw any line, then the straighthedge can be omitted and a person only needs to trace the pencil from the starting point to the finish using whatever path they like. Assuming that the person isn't drunk, the line should be of finite length since we are drawing a line fron start to finish. Thus, any line segment that is a subset of that line, must have an even smaller finite length.

However, with CND functions, if you are to pick a point with a pencil and try to trace it to any other point, you will not go anywhere. The reason being is that the standard definition of a line is not met with the CND function. Since at any point, the function is not differentiable, the line would not know which direction to take. Thus, it would appear to vibrate like an electron held motionless in place.

Therein lies the beauty of such functions. They exist and are continuous, but are not real in the geometric sense.

Thursday, July 28, 2005

ABC Conjecture

Conjecture

The ABC conjecture is a conjecture due to Oesterlé and Masser in 1985. For any three relatively prime numbers A, B, and C, such that A + B = C, and any infinitesimal $\epsilon$ > 0, there exists a constant c$\epsilon$ such that $sqp(ABC)^{1 + \epsilon} \ge c_{\epsilon} \times C$, where sqp(n) is the square-free product of all primes dividing n.

Modified ABC Conjecture

Given a fixed set of distinct primes $p_i$ and $q_j$.
Let C = $\prod_{i=1}^n p_i, B = \prod_{j=1}^m q_j$, where ($p_i$, $q_j$) = 1.
Let O(C) = {$C_k$} = {$\prod p_i^{c_i}, c_i \ge 1$}, be the infinite ordered set of all numbers such that C | Ck.
Let O(B) = {$B_l$} = {$\prod q_j^{b_j}, b_j \ge 1$}, be the infinite ordered set of all numbers such that B | $B_l$.
Let $A_k$ = min {Ck - Bl} > 0, ∀ l.
Let $k^'$ = {k | $A_{k^'}$ = inf($A_k$)}.

Then, $k^'$ > 0 and sqp($A_k B_l C_k$) = $B C \times sqp(A_k) \ge C_k \times \epsilon$, where $\epsilon = B C \times sqp(A_{k^'})$ / $C_(k^'}$.

The original ABC conjecture states that for a given $\epsilon$, there is a constant $c_\epsilon$ that is based only on $\epsilon$ such that the inequality holds.

The modified conjecture states that for fixed distinct primes dividing B and C, we can find a constant $c_{p,q}$ that depends only on p and q that also satisfies the inequality.

Wednesday, July 27, 2005

Fortunate Primes

Conjecture (Original)

A conjecture made by Reo F. Fortune that if $E_k$ = $p_k$# + 1, where $p_k$# is the $k^{th}$ prime and p# is primorial, and assuming $q_k$ is the next prime after $E_k$ (i.e., the smallest prime greater than $E_k$), then $F_k$ = $q_k$ - $E_k$ + 1 is prime for all k. The first values of $F_k$ are 3, 5, 7, 13, 23, 17, 19, 23, ….

Conjecture

Let S0 = $p_k$# + $p_{k+1}$. If $S_0$ is composite, let $q_0$ be the smallest prime dividing $S_0$. Let $S_i = p_k# + q_{i-1}$, where $q_{i-1}$ is the smallest prime dividing $S_i$.

Assuming $S_0 \lt M^2$, then $S_i \lt {M + 1}^2 \forall$ i. Then $S_n$ is prime for some n ≥ 0. Otherwise, $S_i$ = $S_j$ for some 0 ≤ i < j.

Tuesday, July 26, 2005

Extended Midy's Theorem

Theorem (Original)

If the period of a repeating decimal for $a/p$, where p is prime and $a/p$ is a reduced fraction, has an even number of digits, then dividing the repeating portion into halves and adding gives a string of 9s. For example, $1/7$ = 0.142857…, and 142 + 857 = 999.

Theorem (Extended)

If the period of a repeating decimal for $a/p$, where p is prime and $a/p$ is a reduced fraction, is h = $m \times n$, then dividing the repeating portion into n parts and adding gives $c_n(a,p)$ × 9 m's, where c is a constant depending on p, a and n.

Click here for my proof.