The Open University online library

I was reading the note on Stirling numbers of the first kind at Matworld which has among its references Adamchik, V. "On Stirling Numbers and Euler Sums." J. Comput. Appl. Math. 79, 119-130, 1997. - I tried to search this article in the Open University  online library... And yes, got it! Access to the ( Open University ) online library is one of the great assets of being a registered student ( at the Open University ). Internet access is cool but the real gems lie at paid access sites. The mathematics in the journals is definitely source while lecture notes from some teacher are often merely squirrel. - Anyway my point is that I have discovered a vast new area of internet to surf and discover.

Generating Functions

I am revising what I have learned on generating functions. GFs are extremely cool and powerful, they are among the mathematical objects I love most.

The sequence {1,3,9,27,81,243,...} has an equivalent GF 1 / ( 1-3x).

Something else, I have this strong desire to work on a real problem, to work on mathematics of my own. But it is too early there is still so much I can pick up from courses and the literature.

M208 TMA03 Kick-off ( Revisited )

Completed M208 TMA03 - question 3 in latex format. Checked and re-checked the question. Consider it done. Five more to go on three TMA days on 4, 11, 18 may. Q3 was about finding the inverse of a matrix using row reduction. A system of three equations with three unknowns had to be solved using the just found matrix.

Chinese Remainder Theorem

The book on Discrete Mathematics by Rosen is really good. A joy to share time with. Very clear explanations. Of the Chinese Remainder Theorem for example. I have a so called "excellent" book on Number Theory from Dover publications in which the CRT seems too difficult to understand by a mortal. Rosen cuts through it right away. Mathematics is always simple once you understand it. Why not keep it simple when communicating it to students? - Anyway. The CRT is an algorithm for solving systems of linear congruence equations. Like for example.
$$
\begin{cases}
x \equiv 1 \bmod{3} \\
x \equiv 2 \bmod{5} \\
x \equiv 3 \bmod{7}
\end{cases}
$$
which has solution $x=52 \bmod{105}$ which can simply be verified like $ 52 \bmod{7} =3$ because $52 = 7.7 + 3$.

MT365 - TMA01

At times I can be so stupid! Where to begin? I wanted to typeset MT365 - TMA01 but in a course on graphs, networks and designs you can expect to do a LOT of drawing. Although it is possible to make beautifully typeset drawings in LaTeX, which requires in-depth knowledge of additional packages, it is rather time-consuming. Very time-consuming. I had to accept defeat. I sent in my work drawn and written by hand. Typesetting TMAs is not required. It is one of my perfectionistic obsessions, I suppose. No, not an obssession, more an addiciton. I like working with TeXnICCenter, LaTeX, Mathematica. It feels cool, sort of professional. As though you are producing real papers. Life is a game. Life is what you make of it. As long as it is enjoyable. Stupidity #1 is a serious planning miscalculation. I planned a LaTeX delivery which I could not deliver in time which led to the worst what can happen: passing of the cut-off date without having permission of the tutor to send in work after the cut-off date. It is all over the Open University documentation. A tutor is in its right to refuse work send in after the cut-off date and mark it with a 0. - Technically today -is- the cut-off date of MT365, I have sent in my work today by priority mail and informed my tutor. He will mark my work as it comes in. Stupidity #2 was not informing him earlier, when it was clear I couldn't ship friday. My solid advice to all ( prospective ) OU Students start with your TMAs >>now<<. I know of students who work soildly ahead. That's excellent.

About the TMA itself. There were three only three questions adding to only 60 points. Which is rather odd. But CMA41 is in the same booklet and counts for 40 in that booklet. So three questions.
One on graphs, we had to draw various types of digraphs, calculate the number of Hamiltonian cycles in a graph under different assumptions. As an application of graph theory there was a question about food webs in an eco-system. Who eats who and what are the competing species, what happens when a species disappears?
The question on networks was about calculating an optimal flow in a network with minimum and maximum capacities in the various arcs. In such a case it is important that you start from a feasible flow. As an application of network theory there was a question about the reliability of telecommunication networks.
It turns out that I like the track on designs most. It has the most pure mathematics in it. For the question on designs we had to list various polyhedra that met certain conditions. Then we had to prove mathematically that if a certain polyhedron had exactly four faces meeting at each vortex then the number pf triangles must exceed the number of pentagons by 8. The polyhedron was of type triangle / pentagon. Here we see the connection with graph theory because the proof I gave used theorems from graph theory. As an application of design theory there were questions about the planning of  a scientific research project. An area where design theory is often used for planning purposes.

I don't think it is fair to myself or realistic to strive for a distinction on MT365. I would be very satisfied with a grade 2 pass. I am confident that I will pass. IF I keep studying hard on the various topics, but MT365 is too hard to even think of a distinction. I would guess that I scored 36 / 60 or 60% and that is an inbetween .

Bell Number

$$B(n) = \sum_{k=0}^{n} S(n,k) $$

with the first five Bell numbers :
$\begin{array}{llllll}
&&&&&Total\\
1 &&&&&1 \\
0 & 1 &&&&1\\
0 & 1 & 1 & &&2\\
0 & 1 & 3 & 1 & &5\\
0 & 1 & 7 & 6 & 1&15
\end{array}$

A beautiful recurrence formula for the Bell number is
$$B(n) = \sum_{k=0}^{n-1} B(k) \cdot C(n-1,k) ,$$
for example for n=4 we get: 1* 1+ 1*3 + 2*3 + 5*1 = 15.

I now have the sixth edition of the book Discrete Mathematics and Its Applications by Kenneth Rosen.

A great book which comes with the best accompanying student website I have seen sofar for a mathematics book. It probably takes hours to just to browse through all the applets.

Stirling numbers of the first kind

$
\begin{array}{lllll}
1 &&&&\\
0 & 1 &&&\\
0 & 1 & 1 &&\\
0 & 2 & 3 & 1 &\\
0 & 6 & 11 & 6 & 1
\end{array}
$
A stirling number of the first kind is the number of permutations of {1, 2, ..., n } with k permutation cycles. Take for example the permutations of {1,2,3 }:
123 - (1)(2)(3)
132 - (1)(23)
213 - (12)(3)
231 - (132)
312 - (132)
321 - (13)(2)
each line consists of the same permutation but in different notation. After the - I noted the permutation in cycle notation. So there are 2 3-permutations of 1 cycle, 3 of 2 cycles and 1 of 3 cycles.

Stirling numbers of the second kind - continued

As we can deduce from the recurrence
 $$C(n,k) = C(n-1,k-1) + C(n-1,k)$$ that
$$C(n,k) = \frac{1}{k!} \prod_{j=0}^{k-1}(n-j),$$
where $C(n,k)$ is the number of k-subsets out of an n-set, we can similarly find a closed form for the recurrence $$S(n,k) = S(n-1,k-1) + k \cdot S(n-1,k)$$ which is
$$S(n,k) = \frac{1}{k!} \sum_{j=0}^{k} (-1)^{k-j} j^n C(k,j) ,$$ where $S(n,k)$ is the number of k-partitions of an n-set, a Stirling number of the second kind.

The tech for doing discrete sums is really spectaculair imho and therefore want to know much more about it.

V+F-E=2

Euler's celebrated polyhedron formula V+F-E=2 is part of the MT365 course. Worked on question 3 of MT365 TMA01 today ( close to the cut-off date by the way ). It was a proof which required V+F-E=2 and two Handshaking Lemma's: one for graphs and for one polyhedra.

A few years ago I read that Euler proved that the only possible -regular- polyhedra are:
- the Tetrahedron with triangle faces and V4-F4-E6,
- the Cube with square faces and V8-F6-E12,
- the Icosahedron with triangle faces and V6-F8-E12,
- the Dodecahedron with pentagon faces and V20-F12-E30 and finally,
-  the Icosahedraon with triangle faces and V12-F20-E30.

I think that I can prove this now too, if I can start from V+F-E=2 that is. I can at least reproduce the regular polyhedra. Part of question 3 was giving the names of at least three semi regular polyhedra with both triangle faces and pentagon faces.

Stirling numbers of the second kind

( Repeat / memorized ) from Aigner, A course in enumeration.

Stirling numbers of the second kind: S(n,k).
Definition. S(n,k) is the number of k-partitions of an n-set.

Example:
{A}
S(1,1)=1
A

{A,B}
S(2,1)=1
AB
S(2,2)=1
A-B

{A,B,C}
S(3,1)=1
ABC
S(3,2)=3
A-BC
B-AC
C-AB
S(3,3)=1
A-B-C

Almost similar as for the number of combinations ( C(n,k)= C(n-1,k-1) + C(n-1,k) ) we can define the Stirling numbers of the second kind recursively:
S(n,k) = S(n-1,k-1) + k * S(n-1,k).
This recursion is easy to understand as we will see when we continue with the examples for {A,B,C,D}.

/* To be continued */

Logbook

Although I will continue to write about my mathematics study at the Open University in general, I intend to use this blog more for documenting what new stuff ( entire concepts, topics, theorems, tricks, whatever ) I have learned. Since that is my personal day-to-day statistic for studying anyway. In that respect this blog becomes what "blog" suggests: a logbook. I possibly will ( and should ) focus on the Open University courses but I'll include both course topics and non-course topics, the main criterion is "mathematics".

Time for a break.

The MT365 schedule says I should take a break so I will. I am still ahead of schedule with M208. I need to refresh. Breathe in new ideas. I'll be thinking of maths I suppose. What's one or two weeks anyway? I have so many books I still want to read. Want to complete the book about Ramanujan's life for example. And I have other important things to do I tend to neglect while studying.

I watched this documentary about vinyl record collectors. Mostly guys but a few women as well. Completely into record collecting. I am developing a similar passion for mathematics. In the documentary they compare it to an addiction. Because the activity absorbs all time and energy. I consider my mathematical 'knowledge' valuable. I suppose other people will consider it useless. Like I thought about the guys in the documentary with all these records. The collectors are convinced they have a valuable collection and thus it is so. L Ron Hubbard said "What is true for you... is true for you...". He was right.

MT365 - CMA41

While redoing questions I failed I found an unneccesary error. In my answer list in OneNote I found C=4 D=3. In the assignmentbook it was C=4 E=3. Answer correct, letter wrong. That's one of the disadvantages of multiple choice. Unneccessary.
From the other two I was rather unsure of one and I had counted the other error as a deadsure correct answer. I still don't understand why I failed the deadsure one.

( This is a rather vague post, isn't it? Not very diary like and this is what the blog is supposed to be. Although I am not sure I think it is not allowed to post the actual exam question / answer pairs by the Open University. They do sell them though, so I suppose it is a copyright issue. Education is business too, nothing wrong with that at all if the product is good. )

Prime numbers

I find the following identity rather intriguing...
$\sum_{k=1}^\infty\frac{1}{k} = \prod_{p \in P}\frac{1}{1-\frac{1}{p}}$.

The LHS of the equation is a sum over all positive integers $1,2,3 \dots$, the RHS of the equation is a product over all the prime numbers.

The prime numbers is the set of all integers which are (only) divisible by 1 and itself. There is no closed formula which generates the primes 2,3,5,7,11,13,17,19,23,29, ... There are approximately x / log x primes less than x.

MST365 - CMA41 Feedback available

The results of MST365 - CMA41 are in. Much faster than was the case with MST121. We had to wait for weeks on the results of these CMA's. Anyway, my result is 88%, although I had three questions wrong. So probably not all questions had the same weight. The odd thing is that I should have studied harder on MT365, I don't 'own' the subject yet. But even if I did, I would have missed a few questions. How did the other students do? Well, 60% of the students scored in the 85-100 range, while 40% scored in the 0-84 range. Only 1% scored below the required 40% for a pass. An Open University student doing level 3 courses must be highly motivated by definition which should explain the hypothetical 99% "pass rate". - All in all, I am fairly content with this CMA result.

M208 - TMA02 Printed

Just printed M208 / TMA02, to be mailed to tutor tomorrow. TMA02 concerns Group Theory. The publication of the exam-dates triggered me to do redo my planning for the forthcoming six months. I basically want to have all TMA work done by the end of july so that I have time for revision in August and September. I have scheduled one day each week ( or three evenings ) for M208 and the weekends for MT365. I have learned that routines are very important. Studying should come very natural as a habit.

Exam-dates published

Just read the newsletter with published exam-dates.
M208: Wed 13 october - Afternoon session 14:30 - 17:30 ;
MST365: Wed 20 october - Morning session 10:00 - 13:00.

MT365 - CMA41 Done

Just submitted MT365 - CMA41. The result could be anywhere between 35% and 95% I predicted. I feel "sure" about 60% ( 12q / 20 ) but I wouldn't be surprised if it's 35%. The CMA's usually come in lower than expected in my case. Anyway, if I am lucky I may receive a nice 95%. I just don't know. - I still have M208/TMA02 to complete. Perhaps it's a better if I start fresh on that tomorrow. The weight of CMA41 is 14% of the total homework. Or 40% of the first TMA, depends how you look at it. In that case I can add 60/100 with the next TMA and my score is probably 20/100 (  if I go for 10q answered correct. ) Calculating doesn't help. Working on the next TMA does which is due the 26th of this month.

Engage

And it all started on this day ...

MT365 is hard

Although I still think I'll pass the course, MT365 is hard. For me it is a new sort of mathematics I have to get used to. It involves a lot of factual knowledge including an entirely new, rather large, nomenclature which doesn't seem to end. Completed questions 16-20 ( Design ) of CMA41 today. Still five more to go on Networks. Cut-off is Wednesday but I want to have it done tomorrow. Together with M208/TMA02 which is ready but need to be looked at one more time. So I am busy. Very busy in fact. When they are done it's time to rethink / replan.

HarmonicNumber

Let a be an odd integer, show that
$(1 + 2 + \dots + n) / (1^a + 2^a + \dots + n^a)$
While I was working on this problem I discovered again how powerful Mathematica actually is.
It is simply possible to calculate the sum $(1^a + 2^a + \dots + n^a)$ by $H(n,-a)$. Or in InputForm: HarmonicNumber(n,-a).