Saturday, November 8, 2014

My Morning at Honnold/Mudd Library: 2014-11-08 (Game Shows, Bravias Lattice, Prime Numbers)

My Morning at Honnold/Mudd Library

Honnold/Mudd Library - Claremont, CA


What I love about visiting the mathematics and science sections in a university library is that there is a feast of information.  Every time, there is something new to learn.   Here are some notes from today’s visit:

Game Shows and Winning

Source:  John Haigh.  “Taking Chances: Winning with Probability”  2nd Edition Oxford Press: 2003

Haigh introduces two central ideas of probability:  exclusive events and independence.  Is it impossible for two events to occur simultaneously?  Does one event have any effect on another?  Let’s explore some popular concepts of probability found in game shows.

“Find the Lady” – also known as the “Monty Hall Problem”.  You have three boxes, two are worthless and one has the prize.  After the contestant selects the box, the host shows one of the boxes that are worthless.  Statistically, if the contestant switches their initial choice, the contestant doubles the chance of winning.  That is, since the contestant initially is incorrect 2/3 of the time, and the host shows one of the incorrect boxes; the contestant who switches has a 2/3 chance of switching from a losing box to the winning box.

“Showcase Showdown” (The Price is Right) – spinning last may not be the most advantageous place to be.  It depends on the scores of the first two spinners, and the chance of getting the best score. 

“Who Wants to be a Millionaire” – In the classic structure of the game, before the time clock and the shuffle format, contestants were offered three lifelines:  Ask the Audience, 50:50, and Phone a Friend.  Haigh suggests using the lifelines in this order: 50:50, Ask the Audience, and Phone a Friend.  The risk between higher level questions increase as the game goes on.  

“Blockbusters”

A two-person team (white) goes against an individual player (blue in the UK, red in the US).  For each question the team has 2/3 chance of getting the question correctly, but will have to answer a minimum of five question correctly, while the individual can win the game as a few as four correct answers.

My calculation:  assuming that there are no blocks, the probability that the team wins with five consecutive answers is (2/3)^5 = .13169 versus (1/3)^4 = .01235. 

Ideal Crystals and Bravais Lattices

Source:  M.G. Cottam and D.R. Tilley  “Introduction to Surface and Superlattice Excitations”  Cambridge University Press:  Cambridge.  1989.

Ideal crystals are described as a basis of atoms (or ions) located on each point of a lattice, which is defined as a regular periodic array in space.  In an ideal crystal, all the lattice points are equivalent.

Each lattice is defined by the non-coplanar vector a1, a2, and a3.  That is, a1, a2, and a3 do not lie on one specific plane.  The space lattice is formed through a collection of vectors, symbolized as R, as:

R = n1*a1 + n2*a2 + n3*a3, where n1, n2, and n3 are integers. 

Since the ideal crystal has translational symmetry, the transition f(r+R)=f(r) is satisfied for all points in space, which such a point is represented by r. 

The reciprocal lattice vector of R, symbolized by Q, takes the form:

Q = v1*b1 + v2*b2 + v3*b3, where v1, v2, and v3 are integers, and

b1 = ((2*π*a2) x a3) / dot(a1, a2 x a3)
b2 = ((2*π*a3) x a1) / dot(a1, a2 x a3)
b3 = ((2*π*a1) x a2) / dot(a1, a2 x a3)

Where x represents the vector cross product and dot(v,u) represents the dot product.
Two-Dimensional Lattices

Let a1 and a2 be two translation vectors and the set R be defined as:

R = n1*a1 + n2*a2, where n1 and n2 are integers

The ends points of R from what is called the Bravais lattice.  There are five Bravais lattices which take the shape of one of these five forms:

1.    Square, |a1| = |a2|, θ = 90°
2.    Primitive Rectangle (general rectangle), |a1| ≠ |a2|, θ = 90°
3.    Centered Rectangle (rhombic), |a1| ≠ |a2|, θ = 90ׄ°
4.    Hexagonal, |a1| = |a2|, θ = 60° or 120°
5.    Oblique, all other cases

Cottam and Tilley illustrates the concept of bulk and surface excitation by using how Rayleigh’s descriptions of waves in a semi-infinite isotropic elastic medium.  In searching for plane-wave solution, in which the wave propagates parallel to the surface with vector q with velocity v and frequency w:

f’’(z) – (q^2 – w^2/v^2)*f(z) = 0

If q^2 < w^2/v^2, we have a bulk wave where

f(z) = B1*e^(i*√(w^2/v^2-q^2)*z) + B2*e^(-i*√(w^2/v^2-q^2)*z) where B1 and B2 are constants.

If q^2 > w^2/v^2, we have a decaying surface wave where

f(z) = B3*e^(√(q^2-w^2/v^2)*z)


Prime Number Goodies

Source:  Richard Carndall and Carl Pomeance.  “Prime Numbers: A Computational Perspective.  Second Edition” Springer.  2005

Twin Primes:  

Two primes that differ by 2.  Examples include 3 and 5, 29 and 31, and 311 and 313.  Is there an upper bound on twin primes, or is there an infinite set of pairs? 

V. Brun showed the upper bound for the number of twin prime sets under the integer x is:

‘π(x) = O(x*(ln ln x/ln x)^2) -> x*(ln x)^-2

Brun also showed that the sum of the reciprocal of the twin primes is finite.  That sum is denoted as the Brun constant which is:

B = (1/3 + 1/5) + (1/5 + 1/7) + (1/11 + 1/13) + …

The Brun constant is at least 1.71077 (using twin primes from 1 to 1,000,000).  As imagined, this series converges slowly.

Formulas to Produce Primes:

Euler used the polynomial x^2 + x + 41 to produce primes, which worked for all integers 0 to 39.  After x = 39, the polynomial continues produce primes at a remarkable probability. 

Mersenne Primes:

Mersenne primes, in the form of 2^q – 1 where q is a prime greater than 2.  It has been proven that if 2^q – 1 is prime, then so is q.  This form does not always produce primes (see q = 11 or q = 23).  In fact, the distribution of Mersenne primes gets scarce quickly as q increases.

Perfect numbers, integers in which the sum of its divisors except itself equals that integer (example: 6 = 1 + 2 + 3), can be characterized by Euler and Euclid:

An even number is perfect if and only if n = 2^(q-1)*(2^q-1) where 2^q-1 is prime.

Quadratic Residues:

The integer a is said to be a quadratic residue of coprime integer m if the congruence

x^2 ≡ a mod m

can be solved such as x is an integer and gcd(a, m) = 1. 

For example, 19 is a quadratic residue mod 5 since 19 mod 5 = 4 = 2^2 and gcd(19,5) = 1.

The next two basic definitions play a role in solving polynomials and equations involving prime numbers.

Legendre Symbol defined as (a/p) [normally written vertically]:

Let p be an odd prime.  Then:

(a/p) = legendresymbol(a,p) =
0 if a ≡ 0 mod p
1 if a is a quadratic residue of p
-1 if a is not a quadratic residue of p

Example:  p = 19.  The Legendre symbols of (4/19), (5/19), and (19/19) are 1, -1, and 0, respectively. (corrected on 2014-11-10 - with apologizes)

Jacobi Symbol defined as (a/m) [normally written vertically]:

Let m be any odd integer and a be any integer.  Then:

(a/m) = jacobisymbol(a, m) =  product( legendresymbol( a, prime factor of m )).

Example:  m = 27 = 3 * 9, a = 55

Jacobisymbol(55, 27) =

Legendresymbol(55, 3) * legendresymbol(55, 9) = 1 * 1 = 1

Is it shown that jacobisymbol(a, m) = 0 if and only if gcd(a, m) > 1.


Enjoy!  

Eddie

Blog Entry # 402

This blog is property of Edward Shore.  2014