M208
 TMA01 85% ( part 1 77% , part 2 89% )
 TMA02 88%,
 TMA03 Awaiting result. Delivered after cutoff ;) Expecting 78 (72  84)
 Studying TMA04. Somewhat behind schedule )
MT365
 CMA41 88%
 TMA01 79%
Problems with MT365. Will/must concentrate on CMA42 with cutoff day after tomorrow first. Then evaluate situation. I simply haven't studied enough on MT365. And I know why. More when evaluation has been done.  My mainfocus is M208 because it is mainline mathematics, it's required for B31 Pure Mathematics and represents 60 points. Focus will remain on M208.
Monday, May 31, 2010
Sunday, May 30, 2010
Generating functions demonstrated
Yes! Solved a problem ( well, just an exercise from Aigner's book, A course in Enumeration, which I particularly like ). Let me explain.
The problem was as stated as folows. Find $a_n$ such that
$$\sum_{k=0}^n {a_k * a_{nk}} =1, \text{ for all } n$$.
In the case of a sequence of only two items this means that the following two statements must be true:
$a_1 * a_1 = 1$,
$a_1 * a_2 + a_2 * a_1 = 1$
We can simply see that the first two items of the sequence must be $(1,\frac{1}{2})$ since $1 * 1 = 1$, and $1 * \frac{1}{2} + \frac{1}{2} * 1 = 1$.
But we are looking for an infinite sequence! Would such a sequence at all exist? Well, it exists and there are probably several ways to find it. I used the theory of generating functions. As I explained earlier in my blog generating functions are formal power series representing infinite sequences.
Since the generating function of the sequence $(1,1,1 \cdots)$ is $\frac{1}{1z}$, we know,if $g(z)$ is the generating function of the sequence we are looking for, that :
$${g(z)}^2 = \frac{1}{1z} \Leftrightarrow $$
$${g(z)} = \sqrt{\frac{1}{1z}} \Leftrightarrow $$
$$g(z) = (1z)^{\frac{1}{2}}.$$
But this is simply Newtons binomial formula, so the sequence we are looking for is: $$a(n) = C(\frac{2n1}{2}, n)$$ with first five items:
$$(C(\frac{1}{2},0), C(\frac{1}{2},1), C(\frac{3}{2},2), C(\frac{5}{2},3), C(\frac{7}{2},4), \cdots) = $$
$$(1,\frac{1}{2},\frac{3}{8},\frac{5}{16},\frac{35}{128}, \cdots)$$.
This is why I find generating functions beautiful and incredibly powerful.
The problem was as stated as folows. Find $a_n$ such that
$$\sum_{k=0}^n {a_k * a_{nk}} =1, \text{ for all } n$$.
In the case of a sequence of only two items this means that the following two statements must be true:
$a_1 * a_1 = 1$,
$a_1 * a_2 + a_2 * a_1 = 1$
We can simply see that the first two items of the sequence must be $(1,\frac{1}{2})$ since $1 * 1 = 1$, and $1 * \frac{1}{2} + \frac{1}{2} * 1 = 1$.
But we are looking for an infinite sequence! Would such a sequence at all exist? Well, it exists and there are probably several ways to find it. I used the theory of generating functions. As I explained earlier in my blog generating functions are formal power series representing infinite sequences.
Since the generating function of the sequence $(1,1,1 \cdots)$ is $\frac{1}{1z}$, we know,if $g(z)$ is the generating function of the sequence we are looking for, that :
$${g(z)}^2 = \frac{1}{1z} \Leftrightarrow $$
$${g(z)} = \sqrt{\frac{1}{1z}} \Leftrightarrow $$
$$g(z) = (1z)^{\frac{1}{2}}.$$
But this is simply Newtons binomial formula, so the sequence we are looking for is: $$a(n) = C(\frac{2n1}{2}, n)$$ with first five items:
$$(C(\frac{1}{2},0), C(\frac{1}{2},1), C(\frac{3}{2},2), C(\frac{5}{2},3), C(\frac{7}{2},4), \cdots) = $$
$$(1,\frac{1}{2},\frac{3}{8},\frac{5}{16},\frac{35}{128}, \cdots)$$.
This is why I find generating functions beautiful and incredibly powerful.
Thursday, May 27, 2010
Visual mathematics ( step by step... )
Another video.
As (internet) consumers, I suppose, we take the technology and underlying mathematics of 3D graphics for granted, like most other technology we simply consume it. It takes a huge effort to impress.
My latest mathematics video is called 'Small sphere circling around large(r) sphere in elliptic orbit.'  Simple motions, it seems. It took me nevertheless quite a while to find the equations of the motions I initially perceived in my mind. The satisfaction of doing this is seeing the perceived become reality, it's euphoric, addictive, like most math.
Step by step ...
As (internet) consumers, I suppose, we take the technology and underlying mathematics of 3D graphics for granted, like most other technology we simply consume it. It takes a huge effort to impress.
My latest mathematics video is called 'Small sphere circling around large(r) sphere in elliptic orbit.'  Simple motions, it seems. It took me nevertheless quite a while to find the equations of the motions I initially perceived in my mind. The satisfaction of doing this is seeing the perceived become reality, it's euphoric, addictive, like most math.
Step by step ...
Tuesday, May 25, 2010
M208  TMA03
Done. At last. :(  I really like the M208 topics though. TMA03 was definitely a case of serious bad planning. I hope the tutor accepts and will mark the TMA. The paradox is that I think I might get somewhere between 7590 points for this TMA. Have to see and wait. ( Next urgency is MST365 CMA42. )
Sunday, May 23, 2010
Mathematica Cookbook
It took O'Reilly a while but now Mathematica programmers have a Cookbook too.
(Update:)
First impression: not sure yet, could be good, could be cut and pasted from the Wolfram documentation. ( Aren't most computer books written like that? )
As is the case with all programming books, once you are profound in a language, books seem redundant because you can basically dream the manual anyway. The Manual. Oddly enough, books, ( including manuals ) aren't popular among those who are still learning a language ( or should I say platform ).
How did I learn Mathematica myself? Initially, by trial and error, thus not making much progress although I thought otherwise at the time, I could differentiate, integrate, solve equations, table and plot graphs so I thought I was quite good at it actually ). Truth is that I used Mathematica for years without even understanding pure functions, regular expressions or dynamic variables. I started making progress when I sat down and started to read Mathematica books like 'Introduction to Programming with Mathematica', and more recently 'Mathematica Navigator'. The Mathematica Navigator is a book but it can also be installed as an AddOn package to Mathematica in which case it neatly integrates with the existing documentation.  It's like mathematics itself really, doing exercises comes after the reading of new topics.
The CookBook doesn't integrate with Mathematica, neither has it been written in Mathematica. Which is odd because Mathematica is a premier technical publishing platform. My motto is: 'Practice what you preach", what the CookBook authors certainly did not.
(Update:)
First impression: not sure yet, could be good, could be cut and pasted from the Wolfram documentation. ( Aren't most computer books written like that? )
As is the case with all programming books, once you are profound in a language, books seem redundant because you can basically dream the manual anyway. The Manual. Oddly enough, books, ( including manuals ) aren't popular among those who are still learning a language ( or should I say platform ).
How did I learn Mathematica myself? Initially, by trial and error, thus not making much progress although I thought otherwise at the time, I could differentiate, integrate, solve equations, table and plot graphs so I thought I was quite good at it actually ). Truth is that I used Mathematica for years without even understanding pure functions, regular expressions or dynamic variables. I started making progress when I sat down and started to read Mathematica books like 'Introduction to Programming with Mathematica', and more recently 'Mathematica Navigator'. The Mathematica Navigator is a book but it can also be installed as an AddOn package to Mathematica in which case it neatly integrates with the existing documentation.  It's like mathematics itself really, doing exercises comes after the reading of new topics.
The CookBook doesn't integrate with Mathematica, neither has it been written in Mathematica. Which is odd because Mathematica is a premier technical publishing platform. My motto is: 'Practice what you preach", what the CookBook authors certainly did not.
Another YouTube video
Although the motions and objects ( let alone the texture ) are simple in these videos I nevertheless have a framework for experimenting and visualizing a range of mathematical concepts.  I tend to call it a Virtual Clay Demo Table, I hope Source will not object to that though ;)  I can add geometric objects and give them motion by simply adding functions for {x(t), y(t), z(t)}.  Simple indeed, as it is intended by the developers of Mathematica.
Saturday, May 22, 2010
Friday, May 21, 2010
M208  TMA03
I have asked for a last minute extension of the cutoff date. While reviewing the draft TMA I found several issues that need to be addressed further.  And in the meantime the cutoff for MT365CMA42 is dooming up on the horizon.
Tuesday, May 18, 2010
MT365 TMA01 Result is in
MT365  TMA01 Result is in: 79. The lowest score ever on a OU TMA sofar. I can and must do better next time. It is not a bad result as I don't own the subject as I do the M208 materials, have been studying those for years. A TMA result is an important statistic and I am currently downstat.
But: this is an M208 TMA03 cutoff date week, I'll have to deal with MST365 later.
But: this is an M208 TMA03 cutoff date week, I'll have to deal with MST365 later.
Saturday, May 15, 2010
M208  TMA03 now in draft
M208  TMA03 has entered draft status which means that all six questions have been answered, all graphics done, all LaTeX done. What remains to be done are three full QA cycles.
P.S. How would you mark this?
If you have read a previous post about marking I am sure you thought I am paranoid with regards to receiving marks, I wrote something like "as low as possible on the tutor's interpretation band". But let me give you an example. Modified, of course. But you'll get the idea. Suppose we have a set with elements V = {1,2,3}. A function f: V>N has been defined as follows: v mapsto v^2. The question is give the function values of the elements of V. Answer: {1,4,9}. Is this answer correct? Well, I got 0 marks ( for similar question of course ). My question is: do you agree / disagree with the tutor? Or is it perhaps fair to give 25%, 75%, 90%, 99%? Mark it at M208 level. Or is the answer bloody well right?  What do you think?  I think you will be very surprised by the tutor's opinion.
Update.
{1,4,9} is a set. In a set the items have no order. So answer should be {1^2=1,2^2=4,3^2=9}. Right?
P.S. How would you mark this?
If you have read a previous post about marking I am sure you thought I am paranoid with regards to receiving marks, I wrote something like "as low as possible on the tutor's interpretation band". But let me give you an example. Modified, of course. But you'll get the idea. Suppose we have a set with elements V = {1,2,3}. A function f: V>N has been defined as follows: v mapsto v^2. The question is give the function values of the elements of V. Answer: {1,4,9}. Is this answer correct? Well, I got 0 marks ( for similar question of course ). My question is: do you agree / disagree with the tutor? Or is it perhaps fair to give 25%, 75%, 90%, 99%? Mark it at M208 level. Or is the answer bloody well right?  What do you think?  I think you will be very surprised by the tutor's opinion.
Update.
{1,4,9} is a set. In a set the items have no order. So answer should be {1^2=1,2^2=4,3^2=9}. Right?
Friday, May 14, 2010
How to turn a sphere inside out.
Mindblowing, fascinating, must see video. I think stuff like this is studied in Topology. The actual transformation what would it be like? Most likely complex analytic stuff. Hope to discover rather sooner than later.
Thursday, May 13, 2010
M208TMA03 Continued
M208  TMA03, question 5ad done, e in daft, f to do. And Q6 of course.
It may look as though I am terribly struggling with this TMA. I am not. Linear Algebra is the first topic I started to study in my preOU days while mathematics was a hobby without deadlines and TMA's. I am just being perfectionistic. I have to since my marks on the previous two were as low as possible on the tutor's interpretation band. I would like to think of a tutor happily sitting behind his desk, marking... jabadabada, marking, tralala, being happy. Taking all the time answering questions of students via email, phone and other media. And of course ambitiuously preparing extremely interesting lectures ( called tutorials in OUspeak ), besides that the tutor is doing his / her own research.  We wouldn't think of a tutor viciously marking thinking of any means to destroy a student's motivation, pleasure, ambitions, dreams regarding any mathematical future they dream of because they haven't quite made it as far as Stephen Hawking did.  The truth is probably somewhere in the middle, although I hope for all of us I merely had a black thought and the first option is, of course, true. I wish I was in the Truman Show.
Anyway. Since a very long time I am thinking of writing a program for myself again. Programming can be fun, but it all depends on what the code does. I spent some time learning Mathematica features I didn't know yet, like NotebookCreate, Manipulate and so in. And Grid which is nice too.
It may look as though I am terribly struggling with this TMA. I am not. Linear Algebra is the first topic I started to study in my preOU days while mathematics was a hobby without deadlines and TMA's. I am just being perfectionistic. I have to since my marks on the previous two were as low as possible on the tutor's interpretation band. I would like to think of a tutor happily sitting behind his desk, marking... jabadabada, marking, tralala, being happy. Taking all the time answering questions of students via email, phone and other media. And of course ambitiuously preparing extremely interesting lectures ( called tutorials in OUspeak ), besides that the tutor is doing his / her own research.  We wouldn't think of a tutor viciously marking thinking of any means to destroy a student's motivation, pleasure, ambitions, dreams regarding any mathematical future they dream of because they haven't quite made it as far as Stephen Hawking did.  The truth is probably somewhere in the middle, although I hope for all of us I merely had a black thought and the first option is, of course, true. I wish I was in the Truman Show.
Anyway. Since a very long time I am thinking of writing a program for myself again. Programming can be fun, but it all depends on what the code does. I spent some time learning Mathematica features I didn't know yet, like NotebookCreate, Manipulate and so in. And Grid which is nice too.
Wednesday, May 12, 2010
M208TMA03 Revisited
Completed question 4, parts 4d and e. Applied an interesting strategy to express vectors as linear combinations of vectors in an orthogonal basis. Gettng closer, but question 5 consists of six (6!) parts.
Fibonacci Trivia
A while back I wrote that the number of ordered partitions ( called compositions in the Handbook of Discrete Mathematics ) of n of length k is C(n1,k1). It seems that the case where no 1's are used reduces that number to F(n1) ( where F means the Fibonacci sequence ). I haven't been able to figure out a proof yet though. I just find it fascinating how often this number actually pops up.
Example.
The partitions of 6 are.
6
51
42
411
33
321
3111
222
2211
21111
111111
Or, a total of 11.
Ordered partitions with no 1s used are:
6
42
24
33
222
Or a total of 5, which is equal to, as expected F(5) = 5.
Partitions of length three are:
411
321
222
Or, a total of 3.
Ordered partitions of length three are:
411
141
114
321
312
213
231
123
132
222
Or a total of 10, which is equal to, as expected C(5,2) = 10.
Example.
The partitions of 6 are.
6
51
42
411
33
321
3111
222
2211
21111
111111
Or, a total of 11.
Ordered partitions with no 1s used are:
6
42
24
33
222
Or a total of 5, which is equal to, as expected F(5) = 5.
Partitions of length three are:
411
321
222
Or, a total of 3.
Ordered partitions of length three are:
411
141
114
321
312
213
231
123
132
222
Or a total of 10, which is equal to, as expected C(5,2) = 10.
Tuesday, May 11, 2010
M208TMA03
Revised chapter Linear Algebra about vector spaces, subspaces, linear (in)dependency, spanning sets, bases, dimensions and orthogonal bases. Completed parts a,b,c of question 4. Parts 4d,e to do, but I made them in draft. Besides that I have a bit more than a week left for questions 5 and 6. I planned to work on 5 thursday and on 6 tuesday next week. The situation has been worse. Where does the time go?
Monday, May 10, 2010
Mathematica and OOP
Mathematica has features Java programmers can only dream about. But Java is OO, and if you are used to that paradigm ( like me ) switching is hard. Switching may be no problem in the case of simple stored calculation type of programs, it is hard for any large data processing programs or games. Thankfully there are solutions which add OOP features to Mathematica like Objectica. There is however a free option which is called An OO System for Mathematica, Version 3. I have tried it and it works. See how it compares to Java, Ruby and others on this comparison matrix for OOP languages.
What does it do? It adds the possibility to ( externally ) define ( and store ) classes. Classess basically have constructors, properties and methods. In Mathematica through the OO package objects can be created from a class. There is only one featur that is available in Java and not in Mathematica OO which is Access Control. Through Access Control certain properties are made public, default or private. Maybe this will be added in some new version, I don't know.
What does it do? It adds the possibility to ( externally ) define ( and store ) classes. Classess basically have constructors, properties and methods. In Mathematica through the OO package objects can be created from a class. There is only one featur that is available in Java and not in Mathematica OO which is Access Control. Through Access Control certain properties are made public, default or private. Maybe this will be added in some new version, I don't know.
Sunday, May 9, 2010
Recurrence equations
Back in MST121 I learned that the solution of the following first order recurrence equation
$$a_n  5 a_{n1}=0 $$
is
$$a_n = a_0 * 5^n.$$
I have been playing around with Mathematica and digging in Discrete Mathematics books and I am now able to solve equations of the type $a_n  c_1 \cdot a_{n1}=f(n) $, for example if $a_0=1$ the solution of
$$a_n  5 a_{n1}=n^2 $$
is
$$a_n = \frac{1}{32}(8n^220n15+47 \cdot 5^n).$$
Mathematica has a very nice function for solving recurrence equations ( of any order ) which is called RSolve which I only used to verify my own solution. The key to solving recurrence equations of the first order is always finding some sort of sum. Like the sum of the first $n$ integers, which is $\frac{1}{2} n(n+1)$, is in fact the solution of $a_n  a_{n1} = n$, with $a_0=0$.
$$a_n  5 a_{n1}=0 $$
is
$$a_n = a_0 * 5^n.$$
I have been playing around with Mathematica and digging in Discrete Mathematics books and I am now able to solve equations of the type $a_n  c_1 \cdot a_{n1}=f(n) $, for example if $a_0=1$ the solution of
$$a_n  5 a_{n1}=n^2 $$
is
$$a_n = \frac{1}{32}(8n^220n15+47 \cdot 5^n).$$
Mathematica has a very nice function for solving recurrence equations ( of any order ) which is called RSolve which I only used to verify my own solution. The key to solving recurrence equations of the first order is always finding some sort of sum. Like the sum of the first $n$ integers, which is $\frac{1}{2} n(n+1)$, is in fact the solution of $a_n  a_{n1} = n$, with $a_0=0$.
Saturday, May 8, 2010
About students and tutors.
TMA Machine is another OU ( Mathematics ) student with a blog on http://catbear.wordpress.com/. In his right sidebar he maintains a list of OU Students with a blog. I thought about copying them to my bar but since the whole idea of the WorldWideWEB is linking ( and thus not copying ) I simply provide the link.
While reading some other blogs I realized that it is quite a handicap that Englush is not my native language. And worse I don't see much improvement either while reading back some of the posts I have written over the years.  ( Making a thought jump here. )  Which proves my point that selfstudying mathematics is nice but having a tutor looking at your work once in a while is the only guarantee that you will make any progress. If some English teacher would comment on the English in my blog entries I would have made some progress.
Ramanujan, considered to be one of the greatest mathematicians who ever lived, was mainly a selftaught mathematician when he came to England to work with Hardy. It showed immediately: Ramanujan wasn't able to construct a simple formal mathematical proof. He could only progress in that area with the help of Hardy.
While reading some other blogs I realized that it is quite a handicap that Englush is not my native language. And worse I don't see much improvement either while reading back some of the posts I have written over the years.  ( Making a thought jump here. )  Which proves my point that selfstudying mathematics is nice but having a tutor looking at your work once in a while is the only guarantee that you will make any progress. If some English teacher would comment on the English in my blog entries I would have made some progress.
Ramanujan, considered to be one of the greatest mathematicians who ever lived, was mainly a selftaught mathematician when he came to England to work with Hardy. It showed immediately: Ramanujan wasn't able to construct a simple formal mathematical proof. He could only progress in that area with the help of Hardy.
Friday, May 7, 2010
Travelling Salesman Problem
The Travelling Salesman Problem ( TSP ) is broadly known. Most people seem to have heard about it and are able to describe it in laymen's terms. In MT365  Graphs  2 the TSP is described mathematically. Not very deep, I suppose, the TSP is introduced. The formal definition of the TSP is as follows: "Given a weighted complete graph, find a minimumweight Hamiltonian cycle." In Graphs  1 all the words in this definition have been given a precise meaning.
Thursday, May 6, 2010
Latex in HTML
Until 2008 or so texts like $\zeta(2) = \int_0^1 \int_0^1 \frac{dx dy}{1x y} = \frac{\pi^2}{6}$ where produced by inserting image files in the HTML. Thanks to Ajax technology we can simply type Latex in HTML. Amazing as current technology seemed impossible only a decade ago.  What I still can't do  in a flexible, simple, manner is producing graphs, dwawings and the likes.
I wanted to write about graph theory as I am studying Graphs2 of MST365, but even drawing one nice and clear graphs image is close to impossible. Even in Mathematica it takes tons of code.  The topic I wanted to discuss is really, really interesting. About some deep connection between graph theory and number theory via unlabelled trees and integerpartitions. Well, maybe another time.
I wanted to write about graph theory as I am studying Graphs2 of MST365, but even drawing one nice and clear graphs image is close to impossible. Even in Mathematica it takes tons of code.  The topic I wanted to discuss is really, really interesting. About some deep connection between graph theory and number theory via unlabelled trees and integerpartitions. Well, maybe another time.
M208  TMA02 Result is in
I scored 88% on M208  TMA02 ( Group Theory ). Just read the result on my student page. Haven't seen the marked work yet though.
OFFTOPIC. I suppose the mail is a bit behind. Yesterday was a national holiday in our tiny country. According to our government we should celebrate every five years that we ( ... ) defeated the Germans and were liberated by the US in 1945. You know that government that killed 3000 of its own people on 9/11 to win support for a murder, rape and loot trip to Iraq.  A good thing that came out of the war in Iraq though is the movie The Hurt Locker. Must have seen it maybe ten times now.
OFFTOPIC. I suppose the mail is a bit behind. Yesterday was a national holiday in our tiny country. According to our government we should celebrate every five years that we ( ... ) defeated the Germans and were liberated by the US in 1945. You know that government that killed 3000 of its own people on 9/11 to win support for a murder, rape and loot trip to Iraq.  A good thing that came out of the war in Iraq though is the movie The Hurt Locker. Must have seen it maybe ten times now.
Wednesday, May 5, 2010
M208 TMA03 Question 4
Done in draft. Question about finding an orthogonal base of some abstract vector space. Not really difficult considering the marks assigned ( 20 ).
In MST121 / MS221 we learned to find the closed form for the Fibonacci sequence. Or solving the recurrence equation $a_{n}  a_{n1}  a_{n2} = 0$, equations of this type are called homogenous linear recurrence equations ( of the second order ). Inhomogenous equations have a function of n on the RHS. An inhomogenous recurrence equation of the first order is for example $a_{n} = 2 a_{n1} + n$. The solutions of this equation consist of a homonegenous part and a particular part. ( More later. )
In MST121 / MS221 we learned to find the closed form for the Fibonacci sequence. Or solving the recurrence equation $a_{n}  a_{n1}  a_{n2} = 0$, equations of this type are called homogenous linear recurrence equations ( of the second order ). Inhomogenous equations have a function of n on the RHS. An inhomogenous recurrence equation of the first order is for example $a_{n} = 2 a_{n1} + n$. The solutions of this equation consist of a homonegenous part and a particular part. ( More later. )
Tuesday, May 4, 2010
Exercise  comments
A reader ( Paddy ) surprised me with an excellent answer to the exercise I posted. The series I posted was in fact the running total of the squares of the Fibonacci numbers. Let me put that in a table.
$ \begin{matrix}
n & F(n) & (F(n))^2 & \Sigma\\
1 & 1 & 1 & 1\\
2 & 1 & 1 & 2\\
3 & 2 & 4 & 6\\
4 & 3 & 9 & 15\\
5 & 5 & 25 & 40\\
6 & 8 & 64 & 104\\
7 & 13 & 169 & 273
\end{matrix} $
Paddy's answer...
$ \begin{matrix}
n & F(n) & F(n) * F(n+1)\\
1 & 1 & 1 \\
2 & 1 & 2 \\
3 & 2 & 6 \\
4 & 3 & 15 \\
5 & 5 & 40 \\
6 & 8 & 104 \\
7 & 13 & 273
\end{matrix} $
So $ \sum_{k=1}^{n} F_n^{2} = F_n * F_{n+1} $
An interesting conjecture to prove formally.
$ \begin{matrix}
n & F(n) & (F(n))^2 & \Sigma\\
1 & 1 & 1 & 1\\
2 & 1 & 1 & 2\\
3 & 2 & 4 & 6\\
4 & 3 & 9 & 15\\
5 & 5 & 25 & 40\\
6 & 8 & 64 & 104\\
7 & 13 & 169 & 273
\end{matrix} $
Paddy's answer...
$ \begin{matrix}
n & F(n) & F(n) * F(n+1)\\
1 & 1 & 1 \\
2 & 1 & 2 \\
3 & 2 & 6 \\
4 & 3 & 15 \\
5 & 5 & 40 \\
6 & 8 & 104 \\
7 & 13 & 273
\end{matrix} $
So $ \sum_{k=1}^{n} F_n^{2} = F_n * F_{n+1} $
An interesting conjecture to prove formally.
M208TMA03
M208 TMA03 consists of 6 questions. I had already finished question 3. Today I completed questions 1 and 2. The three questions I have done sofar still only amount to 30% of the marks. None of them was really about linear algebra though, questions 46 ( 70% ) definitely are. As mentioned before q3 was about solving 3 simultaneous eqns with 3 unknowns using the inverse of a 3by3 matrix. Question 1 was about 2D geometry with vectors, simple but extremely important. Question 2 was about conics. A parabola this time. The conic stuff in M208 is very easy compared to what I had to learn in MS221. Or maybe it is because of MS221 that conics don't scare me anymore. That's how it usually goes. All math looks horrifying difficult at first sight. And then slowly the mist disappears and then it dawns, things get clearer and clearer.
Monday, May 3, 2010
Generating functions
See how multiplication with $\frac{1}{x}$ summates the sequence upto n.
$
\begin{matrix}
\text{sequence} & \text{generating function} & \text{formula}\\
1,1,1, \cdots & \frac{1}{1x} & f(n)=1\\
1,2,3, \cdots & \frac{x}{(1x)^2} & f(n)=n\\
1,3,6, \cdots & \frac{x}{(1x)^3} & f(n)=\frac{n(n+1)}{2}\\
1,4,10, \cdots & \frac{x}{(1x)^4} & f(n)=C(n+3,n)
\end{matrix}
$
$
\begin{matrix}
\text{sequence} & \text{generating function} & \text{formula}\\
1,1,1, \cdots & \frac{1}{1x} & f(n)=1\\
1,2,3, \cdots & \frac{x}{(1x)^2} & f(n)=n\\
1,3,6, \cdots & \frac{x}{(1x)^3} & f(n)=\frac{n(n+1)}{2}\\
1,4,10, \cdots & \frac{x}{(1x)^4} & f(n)=C(n+3,n)
\end{matrix}
$
Sunday, May 2, 2010
How am I doing  (2) April
M208
 TMA01 85% ( part 1 77% , part 2 89% )
 TMA02 done, waiting for result Exp.75%
 On schedule for TMA03 cutoff 21/5
MT365
 CMA41 88%
 TMA01 done, waiting for result Exp.60%
 On schedule for CMA42 cutoff 2/6
 On schedule for TMA02 cutoff 14/6
( I am still experimenting with my study workflow. Some changes I introduced did not work as expected. I'll undo some changes I implemented. as the coming month ( as of 3/5 ) will be very busy not just studywise. I do spend time learning other mathematics than that in M208 and MT365. I guess that is sort of inevitable. In general topics that are in Concrete Mathematics. In a way that's the first math book I picked up in years and I still don't have the feeling that I own the material in it. Before that I won't be ready with that book. )
 TMA01 85% ( part 1 77% , part 2 89% )
 TMA02 done, waiting for result Exp.75%
 On schedule for TMA03 cutoff 21/5
MT365
 CMA41 88%
 TMA01 done, waiting for result Exp.60%
 On schedule for CMA42 cutoff 2/6
 On schedule for TMA02 cutoff 14/6
( I am still experimenting with my study workflow. Some changes I introduced did not work as expected. I'll undo some changes I implemented. as the coming month ( as of 3/5 ) will be very busy not just studywise. I do spend time learning other mathematics than that in M208 and MT365. I guess that is sort of inevitable. In general topics that are in Concrete Mathematics. In a way that's the first math book I picked up in years and I still don't have the feeling that I own the material in it. Before that I won't be ready with that book. )
We are alone in the universe.
The number of planets in the universe is approximately 3*10^21. A large enough number for scientists to speculate about alien life in the universe. Stephen Hawking as his new series, Into the Universe, proves, has a rich imagination too. ( The series promises to be excellent, more episodes follow. )
Personally, I am pessimistic though. It is my belief that we are utterly alone. To be precise: With life I mean intelligent life with a sense of purpose. Simply because I know that a number like 3*10^21 is not big at all, it is small, tiny in fact.  Would we be here if some 65 million years ago the earth was not hit by some huge meteor that supposedly killed all the dinosaurs on Earth? What are the odds of a meteor hitting Earth? How many other events simply had to happen so that the human species could evolve to what it is today? Suppose the chances of a species like us evolving on some planet somewhere in the Universe is equal to throwing 1000 sixes in a row with one dice. Onethousand oneoutofsix events in a period of say a billion years? Not so hard to imagine is it?
Anyway, if that would be true then there would be life in 1 out of 10^(778) planets which is way beyond the total number of planets in the universe.  My guess is that we are alone. Once you have accepted that we are alone the belief in ( a ) God is simple, almost automatic as a logical conclusion.
Personally, I am pessimistic though. It is my belief that we are utterly alone. To be precise: With life I mean intelligent life with a sense of purpose. Simply because I know that a number like 3*10^21 is not big at all, it is small, tiny in fact.  Would we be here if some 65 million years ago the earth was not hit by some huge meteor that supposedly killed all the dinosaurs on Earth? What are the odds of a meteor hitting Earth? How many other events simply had to happen so that the human species could evolve to what it is today? Suppose the chances of a species like us evolving on some planet somewhere in the Universe is equal to throwing 1000 sixes in a row with one dice. Onethousand oneoutofsix events in a period of say a billion years? Not so hard to imagine is it?
Anyway, if that would be true then there would be life in 1 out of 10^(778) planets which is way beyond the total number of planets in the universe.  My guess is that we are alone. Once you have accepted that we are alone the belief in ( a ) God is simple, almost automatic as a logical conclusion.
Saturday, May 1, 2010
Pascal Triangle  Ordered kpartitions
The number of integer solutions to
x1 + x2 + x3 = 6
with x1,x2,x3 >= 1 is equal to the number of integer solutions to
x1 + x2 + x3 = 3
with x1,x2,x3 >=0.
It is like distributing three similar balls into three different buckets, solutions are:
b b b   ( x1=3, x2=0, x3=0 )
b b  b  ( x1=2, x2=1, x3=0 ).
A pattern appears. We are in fact looking at the number of different arrangements of three b's and 2 's, which is ( 3 + 2 )! / 3! 2! = C(5,2) = 10.
Let's list the solutions anyway.
3  0  0
2  1  0
2  0  1
1  2  0
1  1  1
1  0  2
0  3  0
0  2  1
0  1  2
0  0  3
or translated to the original question:
4  1  1
3  2  1
3  1  2
2  3  1
2  2  2
2  1  3
1  4  1
1  3  2
1  2  3
1  1  4.
They are in fact all the ordered 3partitions of 6.
So we can interpret C(n,k) as the ordered (k+1)partitions of (n+1).
x1 + x2 + x3 = 6
with x1,x2,x3 >= 1 is equal to the number of integer solutions to
x1 + x2 + x3 = 3
with x1,x2,x3 >=0.
It is like distributing three similar balls into three different buckets, solutions are:
b b b   ( x1=3, x2=0, x3=0 )
b b  b  ( x1=2, x2=1, x3=0 ).
A pattern appears. We are in fact looking at the number of different arrangements of three b's and 2 's, which is ( 3 + 2 )! / 3! 2! = C(5,2) = 10.
Let's list the solutions anyway.
3  0  0
2  1  0
2  0  1
1  2  0
1  1  1
1  0  2
0  3  0
0  2  1
0  1  2
0  0  3
or translated to the original question:
4  1  1
3  2  1
3  1  2
2  3  1
2  2  2
2  1  3
1  4  1
1  3  2
1  2  3
1  1  4.
They are in fact all the ordered 3partitions of 6.
So we can interpret C(n,k) as the ordered (k+1)partitions of (n+1).
Pascal Triangle
I suppose we can all dream the Pascal Triangle.
1
11
121
1331
14641
15101051
1615201561
etc.
Until today I did not know that we could simply read the following sum from the triangle.
$$ s(n) = \sum_{i=1}^{n} i \cdot ( ni ) $$
Like for example 100 * 1 + 100 * 2 + 100 * 3 + 100 * 4 + ... + 1 * 100 = ? The answer is in the triangle. Amazing. Or perhaps "fascinating" is a better word.
1
11
121
1331
14641
15101051
1615201561
etc.
Until today I did not know that we could simply read the following sum from the triangle.
$$ s(n) = \sum_{i=1}^{n} i \cdot ( ni ) $$
Like for example 100 * 1 + 100 * 2 + 100 * 3 + 100 * 4 + ... + 1 * 100 = ? The answer is in the triangle. Amazing. Or perhaps "fascinating" is a better word.
Exercise
Readers,
I would like to propose a challenge.
What follows ...
a) 1  2  6  15  40  104  273  ?
b) If the sequence in a) is generated by a function f: N>R then what is f[n] ? Or... what is the closed form for the sequence in a).
c) Modify the polynomial involved in creating f such that at least one power of ten becomes a member of the image of f.
Remarks:
a) Should be simple for those with a pass on MS221;
b) Not solvable with the standard MS221 / M208 tools but I'll expect 221+ folk will accept the challenge;
c) No hints as it is the bonus question;
Solvers are nominated for the gallery of excellence.
Update:
( 1/5'10)
What sort of a person are you? Do you give it a try? Or don't you think it isn't worth the time? Then may I ask, why read my blog in the first place? Do you only like the " TMAsortofexercices " which are easy to solve and get marked? Then... mathematics is not for you. Think about it. It concerns: you.
Anyway, I designed parts b) and c) of the exercise for myself. I haven't find the answer yet. It's more difficult than I thought. But I am learning... I'll solve it.l
I would like to propose a challenge.
What follows ...
a) 1  2  6  15  40  104  273  ?
b) If the sequence in a) is generated by a function f: N>R then what is f[n] ? Or... what is the closed form for the sequence in a).
c) Modify the polynomial involved in creating f such that at least one power of ten becomes a member of the image of f.
Remarks:
a) Should be simple for those with a pass on MS221;
b) Not solvable with the standard MS221 / M208 tools but I'll expect 221+ folk will accept the challenge;
c) No hints as it is the bonus question;
Solvers are nominated for the gallery of excellence.
Update:
( 1/5'10)
What sort of a person are you? Do you give it a try? Or don't you think it isn't worth the time? Then may I ask, why read my blog in the first place? Do you only like the " TMAsortofexercices " which are easy to solve and get marked? Then... mathematics is not for you. Think about it. It concerns: you.
Anyway, I designed parts b) and c) of the exercise for myself. I haven't find the answer yet. It's more difficult than I thought. But I am learning... I'll solve it.l
Fixedpointfree permutations
The permuations of the elements of {1,2,3} are followed by the permutioncycle(s)
123 = (1)(2)(3)
132 = (1)(23)
213 = (12)(3)
231 = (123)
312 = (132)
321 = (13)(2).
Only two of them are so called "fixedpointfree", i.e. they have no cycles of length 1: 231 and 312. There is a symbol for this number D for derangement. So D(3)=2. We can easily find D(4) by thinking in permutation cycles: those that do not include length 1 cycles are fixedpointfree. The possible cycle structures of permutations of length 4 are 1to1 related to the partitions of 4 ( number of permutations ).
4 = xxxx ( 6 = 4*3*2*1 / 4 )
31= xxxx ( 8 = 4*3*2 / 3 )
22 = xxxx ( 3 = 4*3 / 2 * 2 )
211 = xxxx ( 6 = 4*3 /2)
1111 = xxxx ( 1 = 1 )
All basic M208/GTA1 material, I suppose. This formula is however not in the course.
$$D(n) = n! \cdot \sum_{k=0}^{n} \frac{(1)^k}{k!}$$
With the formula for D we can simply check the result for 4.
$$D(4) = 4! \cdot (
\frac{(1)^0}{0!} +
\frac{(1)^1}{1!} +
\frac{(1)^2}{2!} +
\frac{(1)^3}{3!} +
\frac{(1)^4}{4!} ) =9.
$$
123 = (1)(2)(3)
132 = (1)(23)
213 = (12)(3)
231 = (123)
312 = (132)
321 = (13)(2).
Only two of them are so called "fixedpointfree", i.e. they have no cycles of length 1: 231 and 312. There is a symbol for this number D for derangement. So D(3)=2. We can easily find D(4) by thinking in permutation cycles: those that do not include length 1 cycles are fixedpointfree. The possible cycle structures of permutations of length 4 are 1to1 related to the partitions of 4 ( number of permutations ).
4 = xxxx ( 6 = 4*3*2*1 / 4 )
31= xxxx ( 8 = 4*3*2 / 3 )
22 = xxxx ( 3 = 4*3 / 2 * 2 )
211 = xxxx ( 6 = 4*3 /2)
1111 = xxxx ( 1 = 1 )
All basic M208/GTA1 material, I suppose. This formula is however not in the course.
$$D(n) = n! \cdot \sum_{k=0}^{n} \frac{(1)^k}{k!}$$
With the formula for D we can simply check the result for 4.
$$D(4) = 4! \cdot (
\frac{(1)^0}{0!} +
\frac{(1)^1}{1!} +
\frac{(1)^2}{2!} +
\frac{(1)^3}{3!} +
\frac{(1)^4}{4!} ) =9.
$$
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(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.)