Matrices and the Pacal Triangle

Create a matrix from the first n rows of the Pascal Triangle like

1 0 0 0
1 1 0 0
1 2 1 0
1 3 3 1

Then the inverse of this matrix is...

1 0 0 0
-1 1 0 0
1 -2 1 0
-1 3 -3 1

Discrete Mathematics Using Latin Squares

A truly beautiful mathematics book. A book that demands to be read, studied, worked! I will. I love it already. Happiness is easy.

In the past two decades, researchers have discovered a range of uses for Latin squares that go beyond standard mathematics. People working in the fields of science, engineering, statistics, and even computer science all stand to benefit from a working knowledge of Latin squares. Discrete Mathematics Using Latin Squares is the only upper-level college textbook/professional reference that fully engages the subject and its many important applications.

Fibonacci sequence in the Pascal triangle

Where is the Fibonacci sequence in the Pascal triangle?

1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
1 6 15 20 15 6 1
1 7 21 35 35 21 7 1
1 8 28 56 70 56 28 8 1
1 9 36 84 126 126 84 36 9 1

Count as follows.
Start left with a 1.
While on a number:
- Move right
- Move up
- Add to running total.

Let's go.
First row.
1.
Stop.

Second row.
1
Stop.

Third row.
1, right, up
1
Stop.

Fourth row
1, right, up
2
Stop.

Fifth row
1, right, up
3, right, up
1
Stop.

Sixth row
1, right, up
4, right, up
3
Stop.

Totals were 1, 1, 2, 3, 5, 8.

Fibonacci sequence, obvious.

Now what's the proof?

Discrete Mathematics video

I watched video 11-15-00: Combinations and permutations. A lecture by Shai Simonson. A lecture in the Discrete Mathematics series on the ADUni.org website.

Topics:
Counting principles.
- Multiplication
- Addition
- Complement
- Counting double
- 'When are things the same'?
Permutation
Combination
n Choose k.
Binomial Theorem
Pascal Triangle
Proofs
- by formula
- by induction
- by combinatorial argument

Doing proofs by combinatorial argument is a powerful technique. The basic identity from the Pascal triangle
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
1 6 15 20 15 6 1
1 7 21 35 35 21 7 1
...
[n,k]=[n-1,k-1]+[n-1,k]
for example
[6,3]=[5,2]+[5,3]=20=10+10
can easily proven by a combinatorial argument.

Such an argument could be the following.
Let n be the number of employees of a small company with one director. We have to select a k-member team from all employees. The number of teams is [n,k]
Exclude the director from selection and select a k-member team. The number of teams is [n-1,k].
Now select a (k-1)-team, the number of teams is [n-1,k-1] and add the director to it which makes it a k-member team including the director.
Adding the number of k-member teams without a director and k-member teams with a director is the total number of k-member teams or [n-1,k] + [n-1, k-1].

This proves [n,k]=[n-1,k-1]+[n-1,k].

1935

From: Alan Turing, the Enigma by Andrew Hodges, page 94.

He was still only twenty-two. The fellowship carried with it GBP 300 a year for three years, which would normally be extended to six, and no explicit duties. He was entitled to room and board when he chose to reside at Cambridge, and to dine at High Table.

Alan Turing's biography reads like a thriller, situated in the era of Brideshead Rivisited. I am now close to the point where he starts thinking about his (Turing) machines.

Pythagoras' Theorem

a^2 + b^2 = c^2

Play a few times with [ this applet ] and I promise you'll never forget the proof of Pythagoras' Theorem again. Ok, just one of the dozens of proofs, but the proof in the applet is the original one. The requirements for understanding the proof are 1) the formula for the area of a triangle: area = ( base * height )/2 and 2) the area of a triangle does not change if the triangle is rotated.

Conic Sections (2)

I have installed JavaView. The integration with Mathematica (5.2) is flawless. No further comment on that. The available methods and settings for displaying graphics seem infinite. These pictures are the result of my first try of displaying a Mathematica graphics object with JavaView. The major difference between Mathematica and JavaView is the possibility to interactively modify practically all the object's properties and of course rotating and resizing the object. Go to the JavaView site to get a feel for that.

Conic sections

The circle, ellips, parabola and hyperbola are examples of conic sections:



A conic section is an affine variety like V(x^2 + y^2 - z^2, ax + by + cz). Where x^2 + y^2 - z^2 = 0 is the equation of a cone and ax + by + cz = 0 is the equation of a plane in three dimensional affine space. Depending on the values of a, b and c the affine variety takes the form of a circle, ellips, parabola or hyperbola.

Interesting, not? Let's experiment in Mathematica.

Show[
  ContourPlot3D[x^2 + y^2 - z^2, {x, -2, 2}, {y, -2, 2}, {z, -2, 2}],
  ContourPlot3D[-2x + y - z, {x, -2, 2}, {y, -1, 1}, {z, -2, 2}]
  ]

This images almost asks to be rotated in 3D for further visual inspection. This is where the superb JavaView comes in but that is something for another time.

ImplicitPlot

ImplicitPlot[{y^2\[Equal]x^3-3x+1},{x,-2,2}]

( Elliptic Curve )

Educational software for Abstract Algebra

Exploring Abstract Algebra with Mathematica supports groups, rings and morphisms.


( Example of a visualization of the morphism Z12 -> Z6 )

How to construct this image in Mathematica ?

( What do you think ? )

Quadrifolium (2)

Mathematica plots a quadrifolium if you enter

PolarPlot[Sin[2 theta], {theta, 0, 2Pi}]