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Wednesday, April 30, 2008

Displaying Cayley Tables in Mathematica



( The structure of the Ring Z[5]: addition and multiplication mod 5. )

0.x = 0 is not an axiom

The statement 0.x = 0 is not an axiom and can thus be proved.

The axioms for the integers are, for addition
(A1) a + (b + c) = (a + b) + c
(A2) a + 0 = a
(A3) a + (-a) = 0
(A4) a + b = b + a
for multiplication
(M1) a(bc) = (ab)c
(M2) 1a = a
(M3) ab = ba
and for addition and multiplication ('distributive laws')
(D1) a(b+c)=ab+ac
(D2) (a+b)c=ac+ab.

So how do mathematicians prove that 0.a=0?
They do something like this.
a = a
a = (1+0).a (by A2 and M2)
a = 1.a + 0.a (by D2)
a = a + 0.a (by M2)
0.a = 0 by (A2).

Tuesday, April 29, 2008

The field of fractions of an integral domain

V=\mathbb{Z} \times \mathbb{Z} \backslash \left\{0\right\} \\<br />R: (a,b)\equiv (c,d) \Leftrightarrow  ad=bc\\<br />\\<br />(a,b) + (c,d) = (ad + bc, bd)\\<br />(a,b) \cdot (c,d) = (ac, bd)

Create a set V of ordered pairs from {..., -2, -1, 0, 1, 2, ...} (integers) and {..., -2, -1, 1, 2, ...} (integers excluding 0). Elements of V are for example (3,1), (5,1) and (4,2).

Create an equivalence relation on elements of V. Two elements (a,b) and (c,d) are 'equivalent', 'belong to the same equivalence class' if ad=bc. For example (4,2) and (8,4) are equivalent while (4,1) and (8,4) are not.

Define addition '+' as (a,b) + (c,d) = (ad + bc, bd).

Define multiplication '.' as (a,b) . (c,d) = (ac, bd).

This is how the field of Rationals is formally constructed from the Integers.

Monday, April 28, 2008

Thursday, April 24, 2008

The differentiation matrix for arithmetic polynomials

Calculating a difference function is a straightforward process:

\\<br />f(n)=n^3-n^2+n+2\\<br />\\<br />\begin{matrix}<br />n & f(n) & \Delta f(n) \\ <br />0 & 2 & 1\\ <br />1 & 3 & 5\\ <br />2 & 8 & 15\\ <br />3 & 23 & 31\\ <br />4 & 54 & 33\\ <br />5 & 107 & \\ <br /> &  & <br />\end{matrix}<br />\\<br />\Delta f(n)=\frac{f(n+1)-f(n)}{1}=\\<br />\\<br />((n+1)^3-(n+1)^2+(n+1)+2)-(n^3-n^2+n+2)=\\<br />\\<br />3n^2+n+1\\



But it is simpler to use the differentiation matrix for arithmetic polynomials:

f(n)=2+n-n^2+n^3 \rightarrow \begin{pmatrix}<br />2 \\ <br />1 \\ <br />-1\\ <br />1\\ <br />0<br />\end{pmatrix}\\<br />\\<br />\\<br />\begin{pmatrix}<br />0 & 1 & 1 & 1 & 1\\ <br />0 & 0 & 2 & 3 & 4\\ <br />0 & 0 & 0 & 3 & 6\\ <br />0 & 0 & 0 & 0 & 4\\ <br />0 & 0 & 0 & 0 & 0<br />\end{pmatrix}<br />\begin{pmatrix}<br />2 \\ <br />1 \\ <br />-1\\ <br />1\\ <br />0<br />\end{pmatrix} =<br />\begin{pmatrix}<br />1 \\ <br />1 \\ <br />3\\ <br />0\\ <br />0<br />\end{pmatrix}\\<br />\\<br />\\<br />\begin{pmatrix}<br />1 \\ <br />1 \\ <br />3\\ <br />0\\ <br />0<br />\end{pmatrix}\rightarrow f(n)=1+n+3n^2

The 5x5 matrix above is suitable for polynomials up to degree 4. It is possible to create a (n+1)x(n+1) matrix capable of handling polynomials up to degree n.

Proof:
Exercise (hint: use falling powers).

Question: Is there a compact way ( recursive, perhaps ) of describing the matrix capable of handling polynomials up to degree n?

Wednesday, April 23, 2008

Golden Ratio

The Fibonacci numbers are well known.

\\
\begin{matrix}
n & f(n)\\ 
0 & 0\\ 
1 & 1\\ 
2 & 1\\ 
3 & 2\\ 
4 & 3\\ 
5 & 5\\ 
6 & 8\\ 
7 & 13
\end{matrix}
\\
f(n)=\frac{1}{\sqrt{5}}(\frac{1+\sqrt{5}}{2})^n-\frac{1}{\sqrt{5}}(\frac{1-\sqrt{5}}{2})^n

But did you know that the function for the Fibonacci numbers is much more elegant if we explicitly use the Golden Ratio?

\\
f(n) = \frac{\phi^n-(1-\phi)^n}{\sqrt{5}}\\\\
\\
\phi = \frac{1+\sqrt{5}}{2}\\
\\
\phi * (\phi - 1) = 1\\
\\
\phi = 1 + \frac{1}{1 + \frac{1}{1 + \frac{1}{1 + \frac{1}{\ddots}}}}

External Links
- Mathworld

Sunday, April 20, 2008

Tribonacci numbers

The Tribonacci numbers are defined as

T_n=T_{n-3}+T_{n-2}+T_{n-1}

If we set the first three numbers at 0,1 and 1, then the generating function is

T(x)=\frac{x}{1-x-x^2-x^3},

and the first 15 Tribonacci numbers are
0, 1, 1, 2, 4, 7, 13, 24, 44, 81, 149, 274, 504, 927, 1705, ...

Source: MathWorld

The Tribonacci Numbers look a lot like the Fibonacci numbers, that's how they got their name anyway. Who was Fibonacci? Fibonacci, or Leonardo of Pisa (1170-1250) was an Italian mathematician who introduced the arabic numbers ( the numbers we use today ) in Europe in his book Liber Abaci.

Saturday, April 19, 2008

Generating Function

( In mathematics ) a generating function is a formal power series whose coefficients encode information about a sequence a(n) that is indexed by the natural numbers.

( More later. )

More Mathematica Graphics

Thursday, April 17, 2008

Philosophy of Learning

I found this on a webpage from the MIT OpenCourseWare Calculus course.
Philosophy of Learning

1. Amount learned is proportional to time put in.
2. Best way to learn is to figure out ideas yourself or teach them to someone else.
3. Second best is to do so with hints from others like your friends or us.
4. Third best is to get the ideas from reading; but pause in your reading to think about them.
5. Fourth best: unacceptable: don't get them at all.
6. The object of a lecture is not so much to inform you of important facts, but rather to stimulate you to try to learn about some concept.
7. The object of the course is to empower you to use the concepts of calculus in any context.


I would like to make some comments on these points.
1. Of course. I nevertheless disagree.
- Finding the right time to study is very important. Study when you feel energetic, hungry to learn, wanting to know and agressive enough to tackle any hard problem thrown at you.
- Better study one hour each day than seven hours every Saturday. The brain somehow needs backup time to process new concepts learned.
2. Very true! The trick is to find 'things' to find out for yourself which add the knowledge required for the course you are taking. About the 'teaching', I guess he means that you can verify if you have mastered a subject by explaining it in your own words.
3, 4, 5. Yeah...
6. Lectures. Personally, I don't like lectures. They cost you a LOT of time. You either understand what's being told at a lecture ( and that's because you already mastered the subject ) or you simply don't understand what's being told which makes it all a big time-waster if not worse.
7. Yeah...

Abstract Algebra and Mathematica

I found a promising e-book on ( Explatory ) Galois Theory. The book is written at the undergraduate level and it has 'Explatory' in the title because the text uses and refers to the Mathematica ( and Maple ) packages.

What is Galois Theory anyway? Galois only wrote one ( unfinished ) paper before he died at the age of 19 in a duel, but the contribution he made to mathematics is very significant. Before Galois mathematicians were searching for a formula to find the roots of a fifth degree polynomial equation.

a_0+a_1x+a_2x^2+a_3x^3+a_4x^4+a_5x^5=0

Galois proved that it is impossible to find such a formula. It does not exist. In his proof he introduced the concept of a Group.

Abstract Algebra and computers. I learned a lot of concepts in Abstract Algebra by using the package of Hibbard and Levasseur: Exploring Abstract Algebra with Mathematica. It is a book and a software package combined as one. The package though, can be downloaded freely. Mathematica (6.0) is a requirement as the package is written in the Mathematica language.

Sunday, April 13, 2008

Aurifeuillian factorization



David Wells wrote the book Prime Numbers, The most mysterious figures in math. Well, there are some most mysterious theorems in his book...




What an insanity! Read more about Aurifeuillian factorization on MathWorld.

Saturday, April 12, 2008

Sum of consecutive cubics

Show that ...
\center{(\sum 1 + 2 + ... + n)^2 = 1^3 + 2^3 + ... + n^3\\<br />\begin{matrix}<br />n & sum\\ <br />1 & 1 = 1\\ <br />2 & 9 = 1 + 8\\ <br />3 & 36 = 1 + 8 + 27\\ <br />4 & 100 = 1 + 8 + 27 + 64\\<br />\end{matrix}}

It is of course possible to prove this identity by induction but that doesn't make you understand why the above is true. Proofs by induction generally don't contribute to understanding a problem.

( From Getaltheorie voor beginners )

Groups of Order 16

I found this article The Groups of Order Sixteen Made Easy, by Marcel Wild, from the January 2005 American Mathematical Monthly.

I am looking forward to thoroughly study this 12 page article. It's an excellent opportunity to repeat some Group Theory stuff and hopefully pick up some new insights. Group Theory is still my favourite subject.

Thursday, April 10, 2008

Problem ( Number Theory )

Let m, n be positive integers.

Show that:

If 24 / ( m * n + 1 ) Then 24 / ( m + n ).

numb3rs, 215: Running Man



Benford's Law. Very counter intuitive, but true. And explainable.

(Or: Be careful when creating a 'random' set! What is random anyway? )

One of Riemann's notes


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