Happy 2011

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                                            -=*  POP! *=-
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Looking back on 2010 - 1

The number of readers of this blog has been more or less constant during the first two years of collecting stats. Somewhere in October both the number of unique visitors and pages viewed doubled. ( Visits were high in the exam-period. ) Visitors stabilized on this new level in November and December. I don't think there is one explanation for the doubling but attributing factors might be regular 'on-topic' posting and tweets about new posts. It is nice to know that someone is reading your scribblings.

Somewhere during 2010 I implemented MathJax ( LaTeX for HTML ). Because Blogger does not officially support MathJax ( yet, I hope ), I am leeching resources elsewhere. This slows down MathJax somewhat. Recently I installed the AMS extensions which further slowed down MathJax performance. One way or the other I'll fix this issue.

A conjecture about perfect numbers

Perfect number

In number theory the sum of the divisors is denoted as $\sigma$: $$\sigma(n) = \Sigma_{d/n} d$$ and $s(n)=\sigma(n) - n$ is the sum of the proper divisors. A perfect number is equal to the sum of its proper divisors. All known perfect numbers are even, it is unknown if odd perfect numbers exist. The number $2^{p-1}(2^p-1)$ is perfect if and only if $(2^p-1)$ is prime.

Conjecture

Show that: if $p$ is odd then
$$ 2^{p-1}(2^p-1) = \sum_{k=1}^{\frac{p+1}{2}-1} (2k-1)^3$$
( Notice that $2^{p-1}(2^p-1)$ yields a perfect number if $(2^p-1)$ is prime. )

Example

$6$ is perfect, since $6 = 1 + 2 + 3.$
$28$ is perfect, since $28= 1 + 2 + 4 + 7 + 14.$

Any perfect number ( except 6 ) can be represented as a sum of cubes.
$\begin{array}{ccc}
\underline{p} & \underline{Pf} &\underline{s}\\
3 & 28 & 1^3 + 3^3 \\
5 & 496 & 1^3 + 3^3 + 5^3 + 7^3 \\
7 & 8128 & 1^3 + 3^3 + ... + 15^3 \\
13 & 33550336 & 1^3 + 3^3 + ... + 127^3
\end{array}$

Proof

My exercise for New Year's Day. ( You may have noticed that I like doing 'sums'. ) Later...

( Source:
- A primer of analytic number theory, From Pythagoras to Riemann by Jeffrey Stopple
)

Stirling numbers in Discrete Calculus

Definition

Stirling numbers of the second kind represent the number of k-partitions of an n-set and are recursively defined as $\left\{ 0,0 \right\} = 1$, and $\left\{ n,k \right\} = \left\{ n-1,k-1 \right\} + k \cdot \left\{ n-1,k \right\}$. They are used in the Discrete Calculus to convert powers to factorial powers, i.e. $n^2 = n^{\underline{1}} + n^{\underline{2}}$.

The matrix below shows the Stirling numbers for $n=0$, to $n=5$.
$\left(
\begin{array}{cccccc}
1 & 0 & 0 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 & 0 & 0 \\
0 & 1 & 1 & 0 & 0 & 0 \\
0 & 1 & 3 & 1 & 0 & 0 \\
0 & 1 & 7 & 6 & 1 & 0 \\
0 & 1 & 15 & 25 & 10 & 1
\end{array}
\right)$

Example

Calculate $\sum_{k=1}^{n} k^5$.

$\Sigma \Delta k^5$
$=\Sigma ((n+1)^{\underline{1}} + 15(n+1)^{\underline{2}} + 25(n+1)^{\underline{3}} + 10(n+1)^{\underline{4}} + (n+1)^{\underline{5}} )$
$=\frac{1}{2}(n+1)^{\underline{2}} + 5(n+1)^{\underline{3}} + \frac{25}{4}(n+1)^{\underline{4}} + 2(n+1)^{\underline{5}} + \frac{1}{6}(n+1)^{\underline{6}}$
$=-\frac{n^2}{12}+\frac{5 n^4}{12}+\frac{n^5}{2}+\frac{n^6}{6}$
$=\frac{1}{12} n^2 (1+n)^2 \left(-1+2 n+2 n^2\right)$

\begin{array}{lll}
\underline{n} & \underline{n^5} & \underline{\frac{1}{12} n^2 (1+n)^2 \left(-1+2 n+2 n^2\right)} \\
1 & 1 & 1 \\
2 & 32 & 33 \\
3 & 243 & 276 \\
4 & 1024 & 1300 \\
5 & 3125 & 4425
\end{array}

Continued Fractions

I started studying continued fractions..., a vast subject. ( 'new land!' )

Here I calculate the Extended GCD of $9976$ and $6961$ using Blankinship's matrix method. A by-product of the calculation ( column 1 ) are the numbers for expressing $\frac{9976}{6961}$ as a finite continued fraction.
$\left(
\begin{array}{cccc}
- & - & 9976 & 6961 \\
- & 9976 & 1 & 0 \\
1 & 6961 & 0 & 1 \\
2 & 3015 & 1 & -1 \\
3 & 931 & -2 & 3 \\
4 & 222 & 7 & -10 \\
5 & 43 & -30 & 43 \\
6 & 7 & 157 & -225 \\
7 & 1 & -972 & 1393\\
& 0 & 6961 & -9976 \\
\end{array}
\right)$

$\frac{9976}{6961}=1+\frac{1}{2+\frac{1}{3+\frac{1}{4+\frac{1}{5+\frac{1}{6+\frac{1}{7}}}}}}$

Five proofs for the sum-formula of 1+2+3+ ... +n

The running totals of 1,2,3 ... are called the triangular numbers. We will show that $1 + 2 + 3 + \cdots + n = \frac{n(n+1)}{2}$

Proof-1 Gauss's proof

( Gauss supposedly came up with this proof when he was 8 years old. On this page you will find more than 100 different tellings of this story. )
$s = 1 + 2 + 3 + \cdots + n$
$\underline{s = n + (n-1) + (n-2) + \cdots + 1}$
$2s = (n + 1) + ((n-1)+2) + ((n-2)+3) + \cdots + (1+n) \Leftrightarrow $
$2s = n \cdot (n + 1) \Leftrightarrow $
$s = \frac{n(n+1)}{2}$

Proof-2 By induction

Let $S=\left\{ n \in \mathbf{N} | \sum_{k=1}^{n} k = \frac{n(n+1)}{2} \right\}$
Clearly $1 \in S$
Assume, $n \in S$:
$\sum_{k=1}^{n+1} k = \frac{n(n+1)}{2} + (n+1) = \frac{(n+1)(n+2)}{2}$, or $n \in S \Rightarrow n+1 \in S$
Now, since $(S \subset \mathbf{N} \wedge 1 \in S \wedge n \in S \Rightarrow n+1 \in S ) \Rightarrow S=\mathbf{N}.$

Proof-3 With the Pascal Triangle

Because $n^k$ is in the PT for any $k \in \mathbf{N}$, sums of polynomials with integer coefficients can be read from the PT. ( $n={n \choose 1}$, $n^2={n \choose 1} + 2{n \choose 2}$, and so forth. )
$\underline{n}$
0: 1
1: 1 - 1
2: 1 - 2 - 1
3: 1 - 3 - 3 - 1
4: 1 - 4 - 6 - 4 - 1
5: 1 - 5 - 10-10 - 5 - 1
From $n$ we seek the second column, one row down or ${n+1 \choose 2}= \frac{n(n+1)}{2}$

Proof-4 Geometric

Look at the pattern
X
and
X---Y

X
X-X
and
X---Y-Y
X-X---Y

X
X-X
X-X-X
and
X---Y-Y-Y
X-X---Y-Y
X-X-X---Y

The number of X's and Y's are equal. The triangle X-pattern with base of n X's is replaced by a rectangular shape of n+1 by n X's OR Y's.

Proof-5 With Discrete Calculus

The discrete analog of solving a differential equation.
$\Delta f(n) = n+1$
$\Sigma \Delta f(n) = \Sigma (n+1) $
$f(n) = \frac{1}{2}(n+1)^{\underline{2}} + C$
$f(n) = \frac{1}{2}(n+1)n + C$
Since $f(1) = 1, C=0$
$f(n) = \frac{n(n+1)}{2}$

Divisibility by 7

Let n be an integer with last digit d.
Repeat until divisibility of n by 7 has been decided:
- Set m to n with last digit removed.
- Set n to m - 2d
- Determine if n is divisible by 7.

Proof:
m = (n - d)/10
m = (n - d)/10 - 2d
m = ( n - 21d ) / 10
If 7 / n then 7 / ( n - 21d ) / 10 since (7,10)=1 and 7/21.

Example:
n = 8 641 969
m = 864 196
n = 864 196 - 18 = 864 178
m = 86 417
n = 86 417- 16 = 86 401
m = 8 640
n = 8640 - 2 = 8638
m = 863
n = 863 - 16 = 847
m = 84
n = 84 - 14 = 70
Divisible by 7.

I suppose trivia like this are only interesting for math(s)(*) enthusiasts and number geeks. Computational number theorists working with numbers of several million digits may actually use methods like this.

P.S.
(*) Just learned from Math is Fun that "Mathematics is commonly called Math in the US and Maths in the UK and in many other countries.".

Merry Christmas

*★Merry★* 。 • ˚ ˚ ˛ ˚ ˛ •
•。★Christmas★ 。* 。
° 。 ° ˛˚˛ * _Π_____*。*˚
˚ ˛ •˛•˚ */______/~＼。˚ ˚ ˛
*° •˛• ☃｜ 田田 ｜門｜ ☃˚╰☆╮
...To ALL readers :) ♥♥♥

[Books] - Problem-Solving and Selected Topics in Number Theory

Provides a self-contained introduction to classical number theory;
Includes step-by-step proofs of theorems and solutions to exercises;
Designed for undergraduate students ( ... );
( Olympiad-caliber problems )

My Christmas Present from myself ( who else? )... Do you mind? ;-) - This boook symbolizes my commitment to Number Theory. Besides M373 / M381 ( 2011 ) and MST209 / M336 ( 2012 ) I will focus my self-study on Number Theory. Besides Elementary ( or Classical ) Number Theory which is covered by M381 I have a shortlist of books on Analytical-, Algebraic- and Computational Number Theory.

Oh, and I think I'll ask Caz ( who has a column on Platform ) to blog about why people think they are successful in their studies. It could lead to some interesting discussion and hopefully valuable ideas. Where do they come from: Study Results <- Progress <- Concentration <- Focus <- Discipline ??

OU final registrration date extended until 6/Jan-'11

Received the following e-mail.

The final date for reserving a place on an undergraduate course starting in Feb 2011 has now been extended until 6th January 2011, due to the adverse weather conditions.

( I have quite a 'registration-story' forthcoming too. )

[Sign of the times] To: Richard Branson

Richard Branson has criticised the British education system for "overeducating" students and failing to prepare them for the business world, according to leaked diplomatic cables. Branson is by no means the only successful businessman to choose experience over education. - The Virgin tycoon, who is dyslexic and left school at the age of 15, has previously said that he gained valuable business experience while his peers were at university. (From The Telegraph)

Dear Mr. Richard Branson,
All the money you made can't buy you a single pass of a 10 point level 1 course. I suppose 90% of the 16-year olds are smarter than you because you were so stupid to leave school at the age of 15. You may need brainless morons to obey your orders in exchange for a few coins, society needs some very smart people to solve the current global problems which you are partly responsible for.
Kind regards.

M373

Just registered for M373 - Optimization per Feb 2011. ( More later... )

M381

Just registered for M381 - Number Theory and Logic per Feb 2011. ( More later... )

P.S. $$\sum_{p \in \mathbf{P}} \frac{1}{p}$$
... diverges.

M208 result is in

Grade 2 pass. - More later...

Two days and counting ...

... before the M208 result comes in. I wish that we all get the results we have hoped for and deserve considering the work put in.

Anyway, I am definitely going to do M381 Number Theory and Logic next year / month. I'll decide tomorrow if I add M337, MST209 or leave it with "just" M381.

[Video] - Why Pi?

Why Pi? is the title of Don Knuth’s 16th Annual Christmas Tree Lecture. Don Knuth is a famous computer scientist and the designer and programmer of TeX. The first book I seriously self-studied was Concrete Mathematics by Knuth, Ron Graham and Patashnik with beautiful stuff on the Fibonacci series, the Pascal triangle, combinatorial identities, generating functions and numbers and divisibility. It's a beautiful book. I haven't watched the ( entire ) lecture yet, i parked it on my watch-queue.

[Exercise] - 3

Calculate, or otherwise determine $$\sum_{k=1}^{500}k! \mod{189}$$

Reminder 2011 course registration

Ref: 47911183 Reminder - changes to registration dates for undergraduate courses starting in February 2011 - 22 Dec 2010.

I still haven't decided yet... I really don't know at this moment. Tired, tomorrow...

[Exercise] - 2a: The coconut congruence equation

The coconut problem is what is called an algebra word problem. In a course in elementary number theory we could skip the words and ask to solve the following equation:

$$\frac{4}{5}(\frac{4}{5} (\frac{4}{5} (\frac{4}{5} (x-1) - 1) -1) -1) \equiv 1 \mod{5}$$

[Sign of the times] - Wikileaks: Yes, USA: No

Often heard: "What can I do about it, anyway?"

Well, you can close your MasterCard account for example. Or stop buying books from Amazon. Don't buy American if you have the option. You can also support Wikileaks directly of course. I think they need it and deserve it. - They belong to the good guys on this planet. Wikileaks is more than just Assange, although he is important.

Just my 0,02c.

[Exercise] - 1a : Hint

Let $n \in \mathbb{N}$, show that $$f(n) = \frac{(2+\sqrt{3})^{1+2n}+(2-\sqrt{3})^{1+2n}+2}{6}$$ is a square.

Hint: MST121 / MS221 math suffices to solve this one, we need to find a function $A(n)$ such that $${A(n)}^2 = \frac{(2+\sqrt{3})^{1+2n}+(2-\sqrt{3})^{1+2n}+2}{6}$$
and
$$f(n) = A(n) \in \mathbb{N}$$
( To be continued. )

[Exercise] - 2

This is a famous problem. I found it on the Internet by searching for "mathematics, monkey, coconut, problem". - My version is in a Dutch book called Algebra by M. Riemersma.

Five men and a monkey were shipwrecked on a desert island, and they spent the first day gathering coconuts for food. Piled them all up together and then went to sleep for the night. But when they were all asleep one man woke up, and he thought there might be a row about dividing the coconuts in the morning, so he decided to take his share. So he divided the coconuts into five piles. He had one coconut left over, and he gave that to the monkey, and he hid his pile and put the rest all back together.

By and by the next man woke up and did the same thing. And he had one left over, and he gave it to the monkey. And all five of the men did the same thing, one after the other; each one taking a fifth of the coconuts in the pile when he woke up, and each one having one left over for the monkey. And in the morning they divided what coconuts were left, and they came out in five equal shares. Of course each one must have known there were coconuts missing ; but each one was guilty as the others, so they did not say anything.

How many coconuts were there in the beginning?”

To be continued ( i.e. answer and comment )

[Exercise] - 1

Let $n \in \mathbb{N}$, show that $$f(n) = \frac{(2+\sqrt{3})^{1+2n}+(2-\sqrt{3})^{1+2n}+2}{6}$$ is a square.

For $n=1$ to $5$ we have ($n, \ \sqrt{f(n)}, \ f(n)$):
$\begin{array}{lll}
1. & 3. & 9. \\
2. & 11. & 121. \\
3. & 41. & 1681. \\
4. & 153. & 23409. \\
5. & 571. & 326041.
\end{array}$

[News] - Math and industry

The European Science Foundation published the ESF Forward Outlook "Mathematics and Industry". Download a PDF copy from this page.

[Sign of the times] - New York /12-2010

From an article in the New York Times: LAPTOPISTAN

credit: Piotr Redlinski for The New York Times

 

( A new series of posts [Sign of the times] with mainly pictures. Common theme is of course the theme of this blog: "mathematics and / or study". ) Let me know if you think this series doesn't belong here.

Comment:
Take your laptop with you and you can go out on your own. Have coffee, work, study and enjoy the company of people around you. In the Netherlands public libraries are ( or already have ) transformed themselves into embassies of Laptopistan with free power-points and WiFi.

[Sign of the times] - Abacus still in use

The Corporation is optimistic about the potential of the abacus system of mental arithmetic, which it introduced in two of its primary schools on Friday. The system is aimed at improving the comfort of young students with numbers and the mathematical functions.

Source: express buzz

Link: Abacus ( see the comment )

Link: An introduction to the abacus

Link: Chinese Abacus + Manual

Video Lectures on Number Theory

Among lectures on Calculus I,II and III, ( Introduction to ) Linear Algebra and ( Introduction to ) Differential Equations from the UCCS ( University of Colorado and Colorado Springs ) Department of Mathematics you will find video lectures on Math 311 Number Theory by Professor Dr. Seung Son here. I have watched most of them earlier this year. This week I watched some of them again.

While watching a video on mathematical induction something amazing happened, not sure if I would call it a cognition, but it's close. Since I was able to do proofs by mathematical induction and thus understood it, I thought I was done studying mathematical induction. ( I mean both the MS221 and M208 exams included questions on induction). Well, I close-to-cognited that I didn't understand proofs by mathematical induction -at all-.

Do you? If so:
- state the first principle ( of mathematical induction ) using symbols only,
- state the second principle using symbols only,
- re-formulate: "Show that: ... $$\sum_{k=1}^{n}k = \frac{n(n+1)}{2}$$ ..." using Set Terminology,
- can you explain the difference between the first and second principle?
- give an example of a proof using the first principle,
- give an example of a statement which can only be proved with the second principle.

I failed ( note: past tense ) all answers to the questions above. Post is To Be Continued ...