Friday, April 30, 2010
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, 119130, 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.
Wednesday, April 28, 2010
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 / ( 13x).
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.
The sequence {1,3,9,27,81,243,...} has an equivalent GF 1 / ( 13x).
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 Kickoff ( Revisited )
Completed M208 TMA03  question 3 in latex format. Checked and rechecked 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.
Tuesday, April 27, 2010
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 $52 = 7.7 + 3$.
$$
\begin{cases}
x \equiv 1 \bmod{3} \\
x \equiv 2 \bmod{5} \\
x \equiv 3 \bmod{7}
\end{cases}
$$
which has solution $x=52 \bmod{105}$ which can simply be verified like $ 52 \bmod{7} =3$ because $52 = 7.7 + 3$.
Monday, April 26, 2010
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 indepth knowledge of additional packages, it is rather timeconsuming. Very timeconsuming. 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 cutoff date without having permission of the tutor to send in work after the cutoff date. It is all over the Open University documentation. A tutor is in its right to refuse work send in after the cutoff date and mark it with a 0.  Technically today is the cutoff 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 ecosystem. 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 .
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 ecosystem. 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 .
Sunday, April 25, 2010
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
$$B(n) = \sum_{k=0}^{n1} B(k) \cdot C(n1,k) ,$$
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.
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
$$B(n) = \sum_{k=0}^{n1} B(k) \cdot C(n1,k) ,$$
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.
Saturday, April 24, 2010
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 3permutations of 1 cycle, 3 of 2 cycles and 1 of 3 cycles.
\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 3permutations of 1 cycle, 3 of 2 cycles and 1 of 3 cycles.
Friday, April 23, 2010
Stirling numbers of the second kind  continued
As we can deduce from the recurrence
$$C(n,k) = C(n1,k1) + C(n1,k)$$ that
$$C(n,k) = \frac{1}{k!} \prod_{j=0}^{k1}(nj),$$
where $C(n,k)$ is the number of ksubsets out of an nset, we can similarly find a closed form for the recurrence $$S(n,k) = S(n1,k1) + k \cdot S(n1,k)$$ which is
$$S(n,k) = \frac{1}{k!} \sum_{j=0}^{k} (1)^{kj} j^n C(k,j) ,$$ where $S(n,k)$ is the number of kpartitions of an nset, 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.
$$C(n,k) = C(n1,k1) + C(n1,k)$$ that
$$C(n,k) = \frac{1}{k!} \prod_{j=0}^{k1}(nj),$$
where $C(n,k)$ is the number of ksubsets out of an nset, we can similarly find a closed form for the recurrence $$S(n,k) = S(n1,k1) + k \cdot S(n1,k)$$ which is
$$S(n,k) = \frac{1}{k!} \sum_{j=0}^{k} (1)^{kj} j^n C(k,j) ,$$ where $S(n,k)$ is the number of kpartitions of an nset, 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.
Wednesday, April 21, 2010
V+FE=2
Euler's celebrated polyhedron formula V+FE=2 is part of the MT365 course. Worked on question 3 of MT365 TMA01 today ( close to the cutoff date by the way ). It was a proof which required V+FE=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 V4F4E6,
 the Cube with square faces and V8F6E12,
 the Icosahedron with triangle faces and V6F8E12,
 the Dodecahedron with pentagon faces and V20F12E30 and finally,
 the Icosahedraon with triangle faces and V12F20E30.
I think that I can prove this now too, if I can start from V+FE=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.
A few years ago I read that Euler proved that the only possible regular polyhedra are:
 the Tetrahedron with triangle faces and V4F4E6,
 the Cube with square faces and V8F6E12,
 the Icosahedron with triangle faces and V6F8E12,
 the Dodecahedron with pentagon faces and V20F12E30 and finally,
 the Icosahedraon with triangle faces and V12F20E30.
I think that I can prove this now too, if I can start from V+FE=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.
Tuesday, April 20, 2010
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 kpartitions of an nset.
Example:
{A}
S(1,1)=1
A
{A,B}
S(2,1)=1
AB
S(2,2)=1
AB
{A,B,C}
S(3,1)=1
ABC
S(3,2)=3
ABC
BAC
CAB
S(3,3)=1
ABC
Almost similar as for the number of combinations ( C(n,k)= C(n1,k1) + C(n1,k) ) we can define the Stirling numbers of the second kind recursively:
S(n,k) = S(n1,k1) + k * S(n1,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 */
Stirling numbers of the second kind: S(n,k).
Definition. S(n,k) is the number of kpartitions of an nset.
Example:
{A}
S(1,1)=1
A
{A,B}
S(2,1)=1
AB
S(2,2)=1
AB
{A,B,C}
S(3,1)=1
ABC
S(3,2)=3
ABC
BAC
CAB
S(3,3)=1
ABC
Almost similar as for the number of combinations ( C(n,k)= C(n1,k1) + C(n1,k) ) we can define the Stirling numbers of the second kind recursively:
S(n,k) = S(n1,k1) + k * S(n1,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 daytoday 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 noncourse topics, the main criterion is "mathematics".
Sunday, April 11, 2010
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.
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.
Saturday, April 10, 2010
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. )
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. )
Friday, April 9, 2010
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.
$\sum_{k=1}^\infty\frac{1}{k} = \prod_{p \in P}\frac{1}{1\frac{1}{p}}$.
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 85100 range, while 40% scored in the 084 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.
Wednesday, April 7, 2010
M208  TMA02 Printed
Just printed M208 / TMA02, to be mailed to tutor tomorrow. TMA02 concerns Group Theory. The publication of the examdates 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.
Examdates published
Just read the newsletter with published examdates.
M208: Wed 13 october  Afternoon session 14:30  17:30 ;
MST365: Wed 20 october  Morning session 10:00  13:00.
M208: Wed 13 october  Afternoon session 14:30  17:30 ;
MST365: Wed 20 october  Morning session 10:00  13:00.
Tuesday, April 6, 2010
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.
Monday, April 5, 2010
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 1620 ( Design ) of CMA41 today. Still five more to go on Networks. Cutoff 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).
$(1 + 2 + \dots + n) / (1^a + 2^a + \dots + n^a)$
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).
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To boldly go where no man has gone before
(Raumpatrouille – Die phantastischen Abenteuer des Raumschiffes Orion, colloquially aka Raumpatrouille Orion was the first German science fiction television series. Its seven episodes were broadcast by ARD beginning September 17, 1966. The series has since acquired cult status in Germany. Broadcast six years before Star Trek first aired in West Germany (in 1972), it became a huge success.)