Thursday, October 30, 2008

$$$

Let's count how much money i have received so far....

$4000 + $5500 + 2 years of school fees=4000 + 5500 + 13000+++
= $24000

So since my pay has already been given a lower bound, let me just use it to gauge against others who will seek to find jobs under the "brokenback" economy next May....let me gauge against the best of the best...

24000/300=80months=3 years plus plus
Hm...counting back like that, i don't think i want to complain much then...

But still complaining less doesn't mean no complain,
Somebody still deserve to get punished for robbing my Canada pie
If not i could have just spend these 2 sems doing FYP, working part-time, or working to get an economics or statistics minor

infinity vs finite

If you have a finite number of tasks to be done in a finite amount of time, can it be done? Simple arithmetic question

If you have a finite number of tasks to be done in an infinite amount of time, can it be done? trivial

If you have an infinite number of tasks to be done in a finite amount of time, can it be done? Needs some math analysis on convergence vs divergence
Example: 1 + 1/2 + 1/4 +...=2, which means in can be done under this frame.
Proof: Assume series on LHS converges to X
So X-1=1/2 + 1/4 + 1/6=0.5X
therefore X-1=0.5X
=> X=2

If you have an infinite amount of tasks to be done in an infinite amount of time, can it be done? Needs logical arguments, but arguments for both sides have been produced and recoginzed.

Morale of story: There exist infinities of different cardinality, in fact there exist an infinite amount of infinities.

Doughnut

Ok after reading more works for my FYP...which by the way is some advanced group theory known as associative schemes where elements in this space are made to permute each other and i will sum up all the possible permutations. While having uniform k-class partitions of size 2

Since i have only partitions of blocks of size 2, my eventual abstract setting is like a torus, doughnut isn't it?

Wednesday, October 22, 2008

shape of universe

After so much research done and stuff we are still trying hard to figure out the shape of the universe...
Wont it be good/funny/mocking if the shape of universe is indeed like a doughnut as predicted by Homer Simpson?

Well its a big big doughnut indeed then

Zeckendorf's theorem

Zeckendorf's theorem states that every positive integer can be represented in a unique way as the sum of one or more distinct Fibonacci numbers in such a way that the sum does not include any two consecutive Fibonacci numbers. More precisely, if N is any positive integer, there exist positive integers ci ≥ 2, with ci + 1 > ci + 1, such that

N = \sum_{i = 0}^k F_{c_i},

where Fn is the nth Fibonacci number. Such a sum is called the Zeckendorf representation of N.

For example, the Zeckendorf representation of 100 is

100 = 89 + 8 + 3

There are other ways of representing 100 as the sum of Fibonacci numbers - for example

100 = 89 + 8 + 2 + 1
100 = 55 + 34 + 8 + 3

but these are not Zeckendorf representations because 1 and 2 are consecutive Fibonacci numbers, as are 34 and 55.

For any given positive integer, a representation that satisfies the conditions of Zeckendorf's theorem can be found by using a greedy algorithm, choosing the largest possible Fibonacci number at each stage.


Proof of Zeckendorf's theorem

Zeckendorf's theorem has two parts:

  1. Existence: every positive integer n has a Zeckendorf representation.
  2. Uniqueness: no positive integer n has two different Zeckendorf representations.

The first part of Zeckendorf's theorem (existence) can be proved by induction. For n = 1, 2, 3 it is clearly true, for n = 4 we have 4 = 3 + 1. Now let each n \leq k have a Zeckendorf's representation. If k + 1 is a Fibonacci number then we're done, else there exists j such that Fj < k + 1 < Fj + 1. Now consider a = k + 1 − Fj. Since it is a < k, a has a Zeckendorf representation; moreover Fj + a < Fj + 1 then a < Fj − 1 so the Zeckendorf representation of a does not contain Fj − 1. Then k + 1 can be represented as a + Fj. Moreover, it is obvious that each Zeckendorf representation corresponds to only one integer.

The second part of Zeckendorf's theorem (uniqueness) requires the following lemma:

Lemma: The sum of any non-empty set of distinct, non-consecutive Fibonacci numbers whose largest member is Fj is strictly less than the next largest Fibonacci number Fj+1.

The lemma can be proved by induction on Fj.

Now take two non-empty sets of distinct non-consecutive Fibonacci numbers S and T which have the same sum. Eliminate common members to form two sets S' and T' with no members in common. We want to show that S' and T' are both empty i.e. that S=T.

First we show that at least one of S' and T' is empty. Assume the contrary i.e. that S' and T' are both non-empty. Let the largest member of S' be Fs and the largest member of T' be Ft. Without loss of generality, suppose Fs<Ft. Then by the lemma, the sum of S' is strictly less than Fs+1, and so is strictly less than Ft, whereas the sum of T' is clearly at least Ft. This means that S' and T' cannot have the same sum, and so S and T cannot have the same sum. So our assumption that S' and T' are both non-empty must be incorrect.

If S' is empty and T' is non-empty then S is a proper sub-set of T, and so S and T cannot have the same sum. Similarly we can eliminate the case where S' is non-empty and T' is empty. The only remaining case is that S' and T' are both empty, so S=T. We conclude that any two Zeckendorf representations that have the same sum must be identical (up to order).

[edit] Fibonacci multiplication

One can define the following operation a\circ b on natural numbers a, b: given the Zeckendorf representations a=\sum_{i=0}^kF_{c_i}\;(c_i\ge2) and b=\sum_{j=0}^lF_{d_j}\;(d_j\ge2) we define the Fibonacci product a\circ b=\sum_{i=0}^k\sum_{j=0}^lF_{c_i+d_j}.

For example, the Zeckendorf representation of 2 is F3, and the Zeckendorf representation of 4 is F4 + F2 (F1 is disallowed from representations), so 2 \circ 4 = F_{3+4} + F_{3+2} = 13 + 5 = 18.

Tuesday, October 7, 2008

on being 31

My HT supervisor is 31 years old

Has a first class honors math degree
Masters in pure mathematics, pass with distinction, his dissertation was on group theory...(yar i know grasp)
PHD in mathematics

Based in Caltech, it's hard for Asians to find a place in a ang mor university....
Has worked with world-class mathematicians...well least those people has articles on Wikipedia
Published many, gave many talks...
Works on combinatorics, algebra, graph theory...just a single one is enough to kill

My brother is 29, and his time is spend on renovating his house, wedding preps, passing ippt after failing numerous times, praying for stocks to bounce back

When i'm 31?
teaching some NT boys in some secondary school? marking scripts and attending meetings?