I have now been working through GTB1 of M208 for the last three weeks and I still haven't finished it. I am beginning to wonder if the last two blocks of this course are tougher than the rest. I know that Christmas has slowed things down a bit, but usually I can complete a book in three weeks. One question was a corker. In Exercise 4.7 on p44, you are asked to find all the normal symmetry subgroups of regular hexagon. It sounds innocuous enough, but I ended up covering four sides of A4 with my answer.
GTB1 starts with some much needed revision of Group Theory. It is amazing what you forget when you have head full of analysis. Still, there are some things that are beginning to stick in my leaky brain.
The rest of the chapter starts to delve more deeply into conjugacy and normal subgroups. One essential idea is that conjugacy in symmetry groups represents symmetries that have a similar type. For example, when considering the square, the two reflectional symmetries which are associated with the lines of symmetry that pass through the corners of the square are of the same geometric type and are related by conjugacy. This is built up into the Fixed Point Theorem later in the chapter.
Another major section of the chapter is an exploration of the relationship between conjugacy and normal subgroups and this leads to four properties of subgroups that characterise normality. In the final section there is a look at infinite groups of 2x2 matrices.
I must say that it is tricky stuff. I find it hard to get all these abstract ideas ordered someway in my head so that they can be remembered. There is layer upon layer of ideas and whilst it is ok answering questions when you have just completed a topic, I can imagine that in an exam, it may be hard to come up with the right techniques for answering a random question.
Still, I am very glad that I have done so much advance work on this course as I will be able to spend time mastering the methods when the course actually starts. I am very much looking forward to my first assessment.
Tuesday 27 December 2011
Monday 12 December 2011
End of Analysis Block A
I have now completed the last of the books in the first analysis block of M208. This book was all about continuity in real functions, and as for the other analysis books, I really enjoyed working on it. In a nutshell, to determine whether a function f(x) is continuous at x=a, you have to ensure that, as x tends to a, f(x) tends to f(a). Of course, you need to consider the approach to x=a from both smaller and larger values of x.
I am already familiar with the idea of continuity having come across it in previous maths studies but this course deals with the subject in a much more rigorous way, using sequences in x to determine how f(x) behaves as x tends to a. One gripe that I have is that a major plank of the rest of the book, the Intermediate Value Theorem, is only proved for a special case and not proved in general. When a lot of M208 is devoted to carefully proving theorems it is sometimes surprising to find omissions, but I suppose that in some cases the proofs are too involved to reproduce for the course. The Intermediate Value Theorem is the basis of how we can define inverse functions for increasing or decreasing continuous functions.
Another gripe that I have is that it is sometimes difficult to judge the level of detail that is required in some of the answers to the exercises. Sometimes I find myself putting in too much detail and at others too little. Take Ex. 4.1 on p38 of AA4, for example. The solution says that f(n)=n2-1/n tends to infinity as n tends to infinity by the Reciprocal Rule. Is it really necessary to quote the Reciprocal Rule at this point? Isn't this obvious enough? In the examples in the text they manage fine without quoting this rule, so why suddenly do so here? There has to be some point at which you don't have to quote every theorem and rule that you have learned in M208 in order to prove something otherwise it is going to be extremely tedious/arduous. In fact, the solutions to the exercises are meant to be a guide as to how questions should be answered, but some rules, strategies and theorems are omitted in these solutions when convenient. So how are we to judge this?
I am already familiar with the idea of continuity having come across it in previous maths studies but this course deals with the subject in a much more rigorous way, using sequences in x to determine how f(x) behaves as x tends to a. One gripe that I have is that a major plank of the rest of the book, the Intermediate Value Theorem, is only proved for a special case and not proved in general. When a lot of M208 is devoted to carefully proving theorems it is sometimes surprising to find omissions, but I suppose that in some cases the proofs are too involved to reproduce for the course. The Intermediate Value Theorem is the basis of how we can define inverse functions for increasing or decreasing continuous functions.
Another gripe that I have is that it is sometimes difficult to judge the level of detail that is required in some of the answers to the exercises. Sometimes I find myself putting in too much detail and at others too little. Take Ex. 4.1 on p38 of AA4, for example. The solution says that f(n)=n2-1/n tends to infinity as n tends to infinity by the Reciprocal Rule. Is it really necessary to quote the Reciprocal Rule at this point? Isn't this obvious enough? In the examples in the text they manage fine without quoting this rule, so why suddenly do so here? There has to be some point at which you don't have to quote every theorem and rule that you have learned in M208 in order to prove something otherwise it is going to be extremely tedious/arduous. In fact, the solutions to the exercises are meant to be a guide as to how questions should be answered, but some rules, strategies and theorems are omitted in these solutions when convenient. So how are we to judge this?
Friday 25 November 2011
Infinite Series
No not Eastenders (as a tutor joked to me), but the topic of the OU's M208 book AA3. I have got through this in a fairly short space of time and I have already started the last book in this first block on analysis, continuity.
I really enjoyed AA3. I like numbers and I like sequences and series, some of which are a bit quirky at times. The good thing about this third book of analysis is that by the end of it I felt like everything was falling into place. The main emphasis is on being able to ascertain whether a series such as 1+1/2+1/3+1/4+... is convergent or not and by the end of the book all these techniques are at your fingertips. You find yourself using all the methods of AA2 on sequences as well as what is in this book, so you do feel like you are getting somewhere. I even felt that I could prove the convergence of the series for ex for x<0.
The only slight gripe I would have is that I would have liked to have seen some more work on actually determining the sum of a convergent series as I think this can be quite interesting. For example, we are told as an aside that the sum of the series 1+1/4+1/9+1/16... is π2/6, but it would be nice to know how this is obtained.
I really enjoyed AA3. I like numbers and I like sequences and series, some of which are a bit quirky at times. The good thing about this third book of analysis is that by the end of it I felt like everything was falling into place. The main emphasis is on being able to ascertain whether a series such as 1+1/2+1/3+1/4+... is convergent or not and by the end of the book all these techniques are at your fingertips. You find yourself using all the methods of AA2 on sequences as well as what is in this book, so you do feel like you are getting somewhere. I even felt that I could prove the convergence of the series for ex for x<0.
The only slight gripe I would have is that I would have liked to have seen some more work on actually determining the sum of a convergent series as I think this can be quite interesting. For example, we are told as an aside that the sum of the series 1+1/4+1/9+1/16... is π2/6, but it would be nice to know how this is obtained.
Friday 11 November 2011
Bounded, unbounded, convergent, divergent.
I am getting on ok with M208's book AA2 on sequences and I haven't come across any problems apart from one or two minor niggly points about some solutions to the exercises that I will eventually have to ask about. One of the nice things about this book is that I think I have finally got my head around the differences between the meanings of convergent/divergent, bounded/unbounded and tends to plus or minus infinity.
A sequence that tends to a limit L where L is a real number is convergent. If L is zero then this is a null sequence. If {an} is a sequence and |an|≤K for all n, for some positive number K, then the sequence is bounded. If the sequence is not bounded, then it is unbounded. If the sequence is convergent then it is bounded. If the sequence is unbounded then it is divergent. A sequence that tends to plus infinity is both unbounded and divergent. The same applies to a sequence that tends to minus infinity.
A point that should be made is that the converse of the statement "If the sequence is unbounded then it is divergent" is not true. The converse is "If the sequence is divergent then it is unbounded" and this is not true because bounded sequences such as {(-1)n} are divergent but not unbounded. Note also that some sequences such as {(-1)nn} are unbounded but do not tend to either plus infinity or minus infinity.
I hope that makes it all clear for everyone. I can bet that in a month or two it will all become opaque for me again!
A sequence that tends to a limit L where L is a real number is convergent. If L is zero then this is a null sequence. If {an} is a sequence and |an|≤K for all n, for some positive number K, then the sequence is bounded. If the sequence is not bounded, then it is unbounded. If the sequence is convergent then it is bounded. If the sequence is unbounded then it is divergent. A sequence that tends to plus infinity is both unbounded and divergent. The same applies to a sequence that tends to minus infinity.
A point that should be made is that the converse of the statement "If the sequence is unbounded then it is divergent" is not true. The converse is "If the sequence is divergent then it is unbounded" and this is not true because bounded sequences such as {(-1)n} are divergent but not unbounded. Note also that some sequences such as {(-1)nn} are unbounded but do not tend to either plus infinity or minus infinity.
I hope that makes it all clear for everyone. I can bet that in a month or two it will all become opaque for me again!
Monday 31 October 2011
Hitting the buffers with Least Upper Bound
I hit the buffers in my study of Least Upper Bounds this weekend. I was getting on ok until I looked more deeply at the strategy for determining the least upper bound of a set (Strategy 4.1, page 28 AA1).
The Strategy reads as follows:-
"Strategy 4.1. Given a subset E of ℜ, to show that M is the least upper bound, or supremum, of E, check that:
1. x≤M, for all x∈E
2. if M'<M, then there is some x∈E such that x>M' "
I had trouble with the second part.
Suppose we want to prove that the least upper bound of [0,2) is 2. The first part of the strategy is easy enough. We know that M=2 is an upper bound of [0,2) because
x≤2 for all x∈[0,2)
The second part had me bothered. The text says:-
"Suppose that M'<2. We must find an element x in [0,2) which is greater than M'. But by the Density Property, there is a real number x that is less than 2 and greater than both M' and 0. Thus x∈[0,2) and x>M', which shows that M' is not an upper bound of [0,2). Hence M=2 is the least upper bound of [0,2)."
My problem was when I started thinking about why 3 couldn't be the least upper bound according to this strategy. For example, if M'=1 (which is less than 3), then there is some x∈E such that x>M', namely x=1.5. It took ages for me to get my head round this.
The solution, of course, is that the second point of the strategy should be read as:
2. For all M'<M, then there is some x∈E such that x>M'
That 'for all' I think is implied in the fact that M' is a variable quantity. In the case where the upper bound is M=3, then step 2 isn't true when say M'=2.5 as we can't find any x in [0,2) such that x>2.5. Phew!!It definitely had me thrashing about for a few hours.
I also had to pick my way carefully through the proof of the Least Upper Bound Property of ℜ. It took some real concentration and trying it with an example before I got the gist of what was going on.
So, I am off the buffers now. AA1 is behind me and I am onto sequences in AA2.
The Strategy reads as follows:-
"Strategy 4.1. Given a subset E of ℜ, to show that M is the least upper bound, or supremum, of E, check that:
1. x≤M, for all x∈E
2. if M'<M, then there is some x∈E such that x>M' "
I had trouble with the second part.
Suppose we want to prove that the least upper bound of [0,2) is 2. The first part of the strategy is easy enough. We know that M=2 is an upper bound of [0,2) because
x≤2 for all x∈[0,2)
The second part had me bothered. The text says:-
"Suppose that M'<2. We must find an element x in [0,2) which is greater than M'. But by the Density Property, there is a real number x that is less than 2 and greater than both M' and 0. Thus x∈[0,2) and x>M', which shows that M' is not an upper bound of [0,2). Hence M=2 is the least upper bound of [0,2)."
My problem was when I started thinking about why 3 couldn't be the least upper bound according to this strategy. For example, if M'=1 (which is less than 3), then there is some x∈E such that x>M', namely x=1.5. It took ages for me to get my head round this.
The solution, of course, is that the second point of the strategy should be read as:
2. For all M'<M, then there is some x∈E such that x>M'
That 'for all' I think is implied in the fact that M' is a variable quantity. In the case where the upper bound is M=3, then step 2 isn't true when say M'=2.5 as we can't find any x in [0,2) such that x>2.5. Phew!!It definitely had me thrashing about for a few hours.
I also had to pick my way carefully through the proof of the Least Upper Bound Property of ℜ. It took some real concentration and trying it with an example before I got the gist of what was going on.
So, I am off the buffers now. AA1 is behind me and I am onto sequences in AA2.
Friday 28 October 2011
A fear of pure maths
I want to mention something which is probably at the core of why I am still slightly apprehensive about pure mathematics. When I was at school and studying for Maths 'O' level back in the late 70's, we were taught a new fangled type of maths called SMP mathematics (SMP stands for "School Mathematics Project"). I found it really difficult to understand as it was a new way of teaching mathematics that contained a high proportion of pure maths ideas.
The trouble was that these ideas were often taught without explaining why they were needed. I can still recall learning about matrices, determinants and transformations of the unit square. It all sounded like gobbledygook to me as I couldn't relate it to anything else we were learning. So, for example, we were taught how to work out the determinant of a 2x2 matrix, but there was no explanation of why we were doing it. It also didn't help that we had a maths teacher who insisted that we just listen to his explanations without being able to take notes and then work at numerous examples. I just couldn't work like this and I can remember once that I received a very poor 5/95 for a maths test!
Needless to say I didn't do well in my 'O' level and it wasn't a subject that I wanted to take for 'A' level. The trouble was that I was very interested in physics and astronomy and in the end I wanted to do an astrophysics degree. So, I ended up cramming a maths A level into a year at a different college. That's when my maths finally blossomed as I was taught the more basic 'relevant' maths which I was more comfortable with.
So since that time, I have always been somewhat fearful of pure mathematics in case I find myself floundering again.
The trouble was that these ideas were often taught without explaining why they were needed. I can still recall learning about matrices, determinants and transformations of the unit square. It all sounded like gobbledygook to me as I couldn't relate it to anything else we were learning. So, for example, we were taught how to work out the determinant of a 2x2 matrix, but there was no explanation of why we were doing it. It also didn't help that we had a maths teacher who insisted that we just listen to his explanations without being able to take notes and then work at numerous examples. I just couldn't work like this and I can remember once that I received a very poor 5/95 for a maths test!
Needless to say I didn't do well in my 'O' level and it wasn't a subject that I wanted to take for 'A' level. The trouble was that I was very interested in physics and astronomy and in the end I wanted to do an astrophysics degree. So, I ended up cramming a maths A level into a year at a different college. That's when my maths finally blossomed as I was taught the more basic 'relevant' maths which I was more comfortable with.
So since that time, I have always been somewhat fearful of pure mathematics in case I find myself floundering again.
Unequal to inequalities
The trouble with proving an inequality is that sometimes there is a very straightforward way to go about it, but for some reason it eludes you. Having gone through the audio frames for Section 3.3 of AA1 I had my head full of mathematical induction and so when asked to prove that 2n3≥(n+1)3 for n≥4 (Exercise 3.4(a)) I launched into this method without thinking. After a couple of pages of nicely worked mathematical induction I got there in the end, but the solution can actually be reached in a couple of lines.
I even think the solution in the back of the book is slightly long winded. So here is my short version:-
2n3≥(n+1)3
⇔ (3√2)n≥n+1 (Rule 5, n>0, p=3)
⇔ ((3√2)-1)n≥1 (Rule 1)
⇔ n≥1/((3√2)-1) (Rule 4, (3√2)-1>0)
1/((3√2)-1)≅3.8
Hence for integer n, 2n3≥(n+1)3 for n≥4.
The mathematical induction was at least good practice, but I wouldn't want to end up doing this in an exam. Lesson learnt!
I think I am going to enjoy analysis. It seems like we are back working with numbers and it is my sort of topic. Not that I haven't enjoyed the group theory and linear algebra - I did, but I feel like I am getting back to familiar territory.
I even think the solution in the back of the book is slightly long winded. So here is my short version:-
2n3≥(n+1)3
⇔ (3√2)n≥n+1 (Rule 5, n>0, p=3)
⇔ ((3√2)-1)n≥1 (Rule 1)
⇔ n≥1/((3√2)-1) (Rule 4, (3√2)-1>0)
1/((3√2)-1)≅3.8
Hence for integer n, 2n3≥(n+1)3 for n≥4.
The mathematical induction was at least good practice, but I wouldn't want to end up doing this in an exam. Lesson learnt!
I think I am going to enjoy analysis. It seems like we are back working with numbers and it is my sort of topic. Not that I haven't enjoyed the group theory and linear algebra - I did, but I feel like I am getting back to familiar territory.
Monday 24 October 2011
Order Properties of R - Density Property
I'ts funny how sometimes something small and innocuous turns out to be a bit of a problem. On Saturday I was looking at Exercise 1.6(b) on page 11 of AA1. The question asks:
Given positive real numbers a and b such that a<b, describe how you can find a rational number x and an irrational number y such that a<x<b and a<y<b.
The method is descibed in the text. Suppose that a=0.12333... and b=0.12345..., then the two decimals differ at the fourth digit after the decimal point. If we truncate b after this digit, then we obtain the rational number x=0.1234 which satisfies a<x<b. If we then add a sufficiently small non-recurring tail such as 010010001... to x then we obtain an irrational number y=0.1234010010001... which has a<y<b. Note that y is irrational because it is a non-recurring decimal (note also that the pattern of 010010001... is one zero 1, two zero's 1, three zero's 1 etc.).
What had me foxed was that in the solution it said:
We can arrange that a does not end in recurring 9s and b does not terminate.
I thought that this was a very odd statement and it took me a while to understand why it had been included in the solution. However, consider, for example, a=4.010... and b=4.100... Here a and b differ at the first digit after the decimal point, so we would want x to be b truncated to 4.1. However, b is already a terminating or finite decimal, so in this case you would have x=b rather than x<b. To get around this we use the fact that b is also equal to the recurring decimal 4.0999... i.e. we can arrange that b does not terminate. Now a and b differ at the second digit after the decimal point, so x would be b truncated to 4.09. Here x<b as required and now we can add a sufficiently small non-recurring tail to x to obtain y.
Consider another situation where a=3.999... and b=4.0123.. Here a and b differ at the first digit before the decimal point, so x would be b truncated to 4. The problem now is that 3.999.. i.e. 3.9 recurring is equal to 4 so now x=a, rather than x>a. The solution is that we can arrange that a does not end in recurring 9s. So if a=4.00... then x=4.01 and everything is ok again.
Given positive real numbers a and b such that a<b, describe how you can find a rational number x and an irrational number y such that a<x<b and a<y<b.
The method is descibed in the text. Suppose that a=0.12333... and b=0.12345..., then the two decimals differ at the fourth digit after the decimal point. If we truncate b after this digit, then we obtain the rational number x=0.1234 which satisfies a<x<b. If we then add a sufficiently small non-recurring tail such as 010010001... to x then we obtain an irrational number y=0.1234010010001... which has a<y<b. Note that y is irrational because it is a non-recurring decimal (note also that the pattern of 010010001... is one zero 1, two zero's 1, three zero's 1 etc.).
What had me foxed was that in the solution it said:
We can arrange that a does not end in recurring 9s and b does not terminate.
I thought that this was a very odd statement and it took me a while to understand why it had been included in the solution. However, consider, for example, a=4.010... and b=4.100... Here a and b differ at the first digit after the decimal point, so we would want x to be b truncated to 4.1. However, b is already a terminating or finite decimal, so in this case you would have x=b rather than x<b. To get around this we use the fact that b is also equal to the recurring decimal 4.0999... i.e. we can arrange that b does not terminate. Now a and b differ at the second digit after the decimal point, so x would be b truncated to 4.09. Here x<b as required and now we can add a sufficiently small non-recurring tail to x to obtain y.
Consider another situation where a=3.999... and b=4.0123.. Here a and b differ at the first digit before the decimal point, so x would be b truncated to 4. The problem now is that 3.999.. i.e. 3.9 recurring is equal to 4 so now x=a, rather than x>a. The solution is that we can arrange that a does not end in recurring 9s. So if a=4.00... then x=4.01 and everything is ok again.
Friday 21 October 2011
Diagonalising symmetric matrices and affine transformations.
I have just finished the last book in the linear algebra block of the course and I will be starting the first analysis block next. Before I do I just want to comment on a couple of things that came up in LA5.
It is stated a couple of times (see page 31 and page 33) that a symmetric nxn matrix always has an orthonormal eigenvector basis, so that it will always be orthogonally diagonalisable. I don't understand why this is not stated properly as a theorem and it makes me wonder if I have missed something. On page 33 Theorem 3.1 states that Eigenvectors corresponding to distinct eigenvalues of a symmetric matrix are orthogonal. Proof of this theorem is given on page 35 but this only goes some way to explain why a symmetric matrix always has an orthonormal eigenvector basis. What happens if the eigenvalues of a symmetric matrix are not distinct?
I actually think that this theorem may be very important because in physics, for example, symmetries play an important part in conservation laws. I wish this had been dealt with a bit better as it seems like a big result.
The other thing that is puzzling me appears in section 4 on conics and quadrics. Strategy 4.1 on page 43 uses matrices to transform the coordinates of a conic into a different system where the equation of the conic is recognised as being in 'standard form'. What puzzles me is that we are supposed to be dealing with linear transformations and yet part of the strategy involves translating the origin. If the conic is not centred at the origin, then surely we have to use affine transformations (which do not preserve the origin) rather than linear transformations (which do). It all looks rather like a trick because here we are using linear transformations to solve an affine transformation problem. I guess we aren't seeing the full details.
It is stated a couple of times (see page 31 and page 33) that a symmetric nxn matrix always has an orthonormal eigenvector basis, so that it will always be orthogonally diagonalisable. I don't understand why this is not stated properly as a theorem and it makes me wonder if I have missed something. On page 33 Theorem 3.1 states that Eigenvectors corresponding to distinct eigenvalues of a symmetric matrix are orthogonal. Proof of this theorem is given on page 35 but this only goes some way to explain why a symmetric matrix always has an orthonormal eigenvector basis. What happens if the eigenvalues of a symmetric matrix are not distinct?
I actually think that this theorem may be very important because in physics, for example, symmetries play an important part in conservation laws. I wish this had been dealt with a bit better as it seems like a big result.
The other thing that is puzzling me appears in section 4 on conics and quadrics. Strategy 4.1 on page 43 uses matrices to transform the coordinates of a conic into a different system where the equation of the conic is recognised as being in 'standard form'. What puzzles me is that we are supposed to be dealing with linear transformations and yet part of the strategy involves translating the origin. If the conic is not centred at the origin, then surely we have to use affine transformations (which do not preserve the origin) rather than linear transformations (which do). It all looks rather like a trick because here we are using linear transformations to solve an affine transformation problem. I guess we aren't seeing the full details.
Thursday 20 October 2011
All signed up
I received notification today that I am finally registered for M208. So it is all systems go, as they say. If anyone is reading this blog who is also registered for this course starting in January next year, don't be too shocked by the fact that I have already worked through half the course material already. This is not because I am some sort of super swot, but rather because I like to take things at my own rather leisurely pace!
So earlier this year I got hold of the M208 books and I have been working through them steadily in order ever since. At the moment I am just finishing LA5 on Eigenvectors. I should really have started this blog when I began work on M208 as it would have been a good place to get down some of my thoughts on the course and what I found difficult to understand. Never mind. Better late than never.
So earlier this year I got hold of the M208 books and I have been working through them steadily in order ever since. At the moment I am just finishing LA5 on Eigenvectors. I should really have started this blog when I began work on M208 as it would have been a good place to get down some of my thoughts on the course and what I found difficult to understand. Never mind. Better late than never.
How I became interested in maths again
Mathematics seems like an odd hobby to have but I remind people that it isn't so much different to doing a puzzle in a newspaper. This is how it all started again for me, because one evening I thought there must be more interesting things to puzzle over than yet another sudoku. I also blame Marcus du Sautoy and his program "The Story of Maths". His infectious interest in prime numbers has now given me the same bug and it reawakened my fascination with mathematics.
Why matrices reloaded? Well, apart from the obvious pun on the Keanu Reeves film, this is the second time I have been trying to study some maths seriously. The first time was when I was doing an astrophysics degree at Queen Mary College in London and then my priority was all about astronomy. I did go on to try the Part III mathematics course at Cambridge, but clever though I thought I was, I wasn't that clever and I failed the course. Not deterred, I did do a PhD in cosmology at Durham and subsequently a few years in research.
However, that was over 20 years ago now and this time my interest is not so much in astronomy as mathematics. Having got the prime numbers bug, I wanted to see if I could still hack it in maths and two years ago I signed up for the entry level mathematics courses MST121 and MS221 at the Open University. It seems that the old neurons are still firing ok, as I passed both courses, getting a distinction in MS221 and getting my certificate in mathematics.
So now I am signed up for a new course starting in January 2012. This is M208 Pure Mathematics. My aim at the moment is to see if I can get the diploma which consists of M208 and MST209 Mathematical Methods and Models. These two courses form the bulk of the second year BSc degree.
After this, who knows. I want to be able to strike a balance between learning new skills and having the ability to investigate things for myself. I call it 'fireside' mathematics and it is what interests me the most.
Why matrices reloaded? Well, apart from the obvious pun on the Keanu Reeves film, this is the second time I have been trying to study some maths seriously. The first time was when I was doing an astrophysics degree at Queen Mary College in London and then my priority was all about astronomy. I did go on to try the Part III mathematics course at Cambridge, but clever though I thought I was, I wasn't that clever and I failed the course. Not deterred, I did do a PhD in cosmology at Durham and subsequently a few years in research.
However, that was over 20 years ago now and this time my interest is not so much in astronomy as mathematics. Having got the prime numbers bug, I wanted to see if I could still hack it in maths and two years ago I signed up for the entry level mathematics courses MST121 and MS221 at the Open University. It seems that the old neurons are still firing ok, as I passed both courses, getting a distinction in MS221 and getting my certificate in mathematics.
So now I am signed up for a new course starting in January 2012. This is M208 Pure Mathematics. My aim at the moment is to see if I can get the diploma which consists of M208 and MST209 Mathematical Methods and Models. These two courses form the bulk of the second year BSc degree.
After this, who knows. I want to be able to strike a balance between learning new skills and having the ability to investigate things for myself. I call it 'fireside' mathematics and it is what interests me the most.
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