**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” 2

^{nd}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

## No comments:

## Post a Comment