My course materials have now arrived for M381 and M336 and the course websites opened today and so off we go again. It is good to be in touch with people once more. It is hard doing all this preparation in advance without contact with anyone and but now that I have the actual books in front of me and the forums are open, it is much more motivating.
So how well did I do in my 6 months or so of preparation. Well, I got through NT1, 2, 3 and ML1 and 2 of M381 and IB1, 2, 3, 4 and GR1 of M336, so 10 books in all, just under a third of each course. So this will give me a head start and make the progress smoother, I hope! But I do realise that even with this advantage, I am going to have to speed up from the way I have been working so far, which has been at my usual leisurely pace.
What I have found surprising in what I have seen so far is how much more difficult the set problems are. It has been a bit of a shock to the system that there are problems that I just can't do without having a peek at the solutions. I could devote more hours to trying to solve the problems without help but I don't think this is good use of my time. So I have had a reality check about what I can and can't do and I think this is a good thing.
So now its back to the books. I am reading NT4 on Fermat's and Wilson's theorems. Strangely enough the last book I read of M336, GR1 was all about the theory of integers again, a repeat really of NT1 and 2, but actually, the way it was dealt with in GR1 was more difficult! The problems were a pain and, not like me, I couldn't be bothered to attempt the last four in the book - I felt too drained!
Another thing about these two courses is that it is all about proofs (both in the text and the set problems). So if you don't like proofs you are in for a hard time!
At some point I would like to write a summary of ML1 and 2. There was a lot of interesting stuff and writing a summary would help me to recall it all.
Good luck to you all and I look forward to meeting up with Chris and Dan again.
Tuesday, 17 September 2013
Monday, 19 August 2013
More on Generating Subgroups
I seem to be blogging more about the groups course M336 rather than the number theory and mathematical logic course M381, but I get the feeling this will all change at some point as I get further into mathematical logic.
Ok, time for an update. I finished the first mathematical logic unit of M381 without too much difficulty. I think the hardest part is dealing with a whole load of new concepts. Once you get the hang of these then it seems to be alright. I then moved onto unit 4 of groups and geometry and this was quite a time consuming book both from the length of it and the considerable number proofs that you have to wade through and also come up with yourself. A lot of the material was revision going over topics such as cosets, normal subgroups, quotient groups, isomorphisms and homomorphisms. However, there was quite a lot of new material and perhaps the most important was proving that you can generate a subgroup of a group using elements of that group (see my previous post).
The approach to this theory was different from what I expected and quite subtle. Firstly, the subgroup of a group G generated by an underlying set of elements S is defined to be the 'smallest' subgroup of G containing the set S. This is denoted <S>. The first step is to prove that there is a unique smallest subgroup of G that contains S. The trick is to consider all the subgroups of G that contain S. This collection is certainly not empty since it contains G (as G is a subgroup of itself). Now consider the intersection H of all these subgroups that contain S. The intersection of a set of subgroups is itself a subgroup (and this was proved in the unit). So H is a subgroup of G. It also contains S because our collection of subgroups each contained S. Further, H must be the 'smallest' subgroup of G containing S because H is the intersection of our set of subgroups and so is contained by those subgroups. Hence <S>=H.
Although this tells us that <S> exists it does not tell us what <S> looks like! The subgroup axioms come in handy here. <S> must contain the identity of G, e say, and the inverse of the elements of S in G. This means that <S> must also be generated by these elements (namely e, the elements of S and the inverse of the elements of S). We can call this set S'. The text then goes on to show that <S> is generated by the set of words in S', W(S'). As words in English are created by taking any finite arrangement of letters (in principle) drawn from our 26 letter alphabet, so the words in W(S') are formed by taking any finite arrangement of elements from S' (but in English we might suspect that eeeeeaaaaeeeooooaeaor isn't a word as it is in group theory!). It turns out that <S>=W(S').
Having been writing this blog over the last few days I have got further into unit 2 of mathematical logic and, surprise, surprise, I have found some of it increasingly difficult (if not impossible)! However, having said that I can't help thinking it is blooming neat and I am glad I am doing it as I think it has resonances in some of the things I have thought about number theory.
Ok, time for an update. I finished the first mathematical logic unit of M381 without too much difficulty. I think the hardest part is dealing with a whole load of new concepts. Once you get the hang of these then it seems to be alright. I then moved onto unit 4 of groups and geometry and this was quite a time consuming book both from the length of it and the considerable number proofs that you have to wade through and also come up with yourself. A lot of the material was revision going over topics such as cosets, normal subgroups, quotient groups, isomorphisms and homomorphisms. However, there was quite a lot of new material and perhaps the most important was proving that you can generate a subgroup of a group using elements of that group (see my previous post).
The approach to this theory was different from what I expected and quite subtle. Firstly, the subgroup of a group G generated by an underlying set of elements S is defined to be the 'smallest' subgroup of G containing the set S. This is denoted <S>. The first step is to prove that there is a unique smallest subgroup of G that contains S. The trick is to consider all the subgroups of G that contain S. This collection is certainly not empty since it contains G (as G is a subgroup of itself). Now consider the intersection H of all these subgroups that contain S. The intersection of a set of subgroups is itself a subgroup (and this was proved in the unit). So H is a subgroup of G. It also contains S because our collection of subgroups each contained S. Further, H must be the 'smallest' subgroup of G containing S because H is the intersection of our set of subgroups and so is contained by those subgroups. Hence <S>=H.
Although this tells us that <S> exists it does not tell us what <S> looks like! The subgroup axioms come in handy here. <S> must contain the identity of G, e say, and the inverse of the elements of S in G. This means that <S> must also be generated by these elements (namely e, the elements of S and the inverse of the elements of S). We can call this set S'. The text then goes on to show that <S> is generated by the set of words in S', W(S'). As words in English are created by taking any finite arrangement of letters (in principle) drawn from our 26 letter alphabet, so the words in W(S') are formed by taking any finite arrangement of elements from S' (but in English we might suspect that eeeeeaaaaeeeooooaeaor isn't a word as it is in group theory!). It turns out that <S>=W(S').
Having been writing this blog over the last few days I have got further into unit 2 of mathematical logic and, surprise, surprise, I have found some of it increasingly difficult (if not impossible)! However, having said that I can't help thinking it is blooming neat and I am glad I am doing it as I think it has resonances in some of the things I have thought about number theory.
Thursday, 11 July 2013
T minus 86 days and counting
I received confirmation the day before yesterday that the SAAS are paying my course fees for M381 and M336 this October and so I am all set for my next year's work. I am very grateful to the Scottish system of support for the likes of me.
I am progressing ok. I finished unit 3 of the number theory half of M381 on congruence. This was fairly straight forward and didn't offer any particular difficulties. In the mean time I have also finished unit IB3 of M336 on frieze groups and again this wasn't particularly challenging. I can now see where this course is going with its classification of friezes and wallpaper patterns. I did always wonder what Alan, our tutor for M208, was always going on about when he said that there were 17 different types of wallpaper pattern and now I have an inkling of what he may mean. I have also now got through a fair chunk of unit 1 of the mathematical logic half of M381 on computability. This is ok as it is basic linear programming.
Going back to friezes, I thought I would show you an example of a frieze that can be found in our flat.
My Dad (who knows these things because he is an architect) pointed out to me that this type of cornice is called "egg and dart" which I think is an apt description. From a mathematical point of view, if you take the decoration bit in the middle as the frieze (ignoring the fact that it obviously doesn't extend indefinitely to the left and right), then it has a number of different symmetries. For example, if you take a basic element of an egg bounded by half a dart on either side, then you can translate this element horizontally left and right by multiples of the distance between two adjacent darts. This is what defines a frieze. There are also, obviously, reflections in a vertical axis (either through a dart or through the centre of an egg). There are, however, no rotational symmetries or reflections or glide reflections in a horizontal axis.
Each frieze can be analysed according to its group of symmetries and an algorithm is given in the unit for identifying the type of frieze according to its group. This is a type 2 frieze or in international notation a pm11. There are seven types of frieze in all, as defined by their symmetry group.
I am progressing ok. I finished unit 3 of the number theory half of M381 on congruence. This was fairly straight forward and didn't offer any particular difficulties. In the mean time I have also finished unit IB3 of M336 on frieze groups and again this wasn't particularly challenging. I can now see where this course is going with its classification of friezes and wallpaper patterns. I did always wonder what Alan, our tutor for M208, was always going on about when he said that there were 17 different types of wallpaper pattern and now I have an inkling of what he may mean. I have also now got through a fair chunk of unit 1 of the mathematical logic half of M381 on computability. This is ok as it is basic linear programming.
Going back to friezes, I thought I would show you an example of a frieze that can be found in our flat.
My Dad (who knows these things because he is an architect) pointed out to me that this type of cornice is called "egg and dart" which I think is an apt description. From a mathematical point of view, if you take the decoration bit in the middle as the frieze (ignoring the fact that it obviously doesn't extend indefinitely to the left and right), then it has a number of different symmetries. For example, if you take a basic element of an egg bounded by half a dart on either side, then you can translate this element horizontally left and right by multiples of the distance between two adjacent darts. This is what defines a frieze. There are also, obviously, reflections in a vertical axis (either through a dart or through the centre of an egg). There are, however, no rotational symmetries or reflections or glide reflections in a horizontal axis.
Each frieze can be analysed according to its group of symmetries and an algorithm is given in the unit for identifying the type of frieze according to its group. This is a type 2 frieze or in international notation a pm11. There are seven types of frieze in all, as defined by their symmetry group.
Thursday, 23 May 2013
Generating subgroups
I have just finished going through book 2 of M336 Groups and Geometry and I am making reasonable progress. As usual I seem to be working at roughly half the speed that you are supposed to but I expect I will get through enough material to make life easier when the course actually starts in October. So now I have read four books in total from M336 and M381 and I am alternating between the two courses; i.e. when I finish a book from one course I start on a book from the other course. I will be looking at book 3 from M381 next which is on congruence.
Before I leave book 2 of M336, there are a few things I want to say about it. Firstly, I have noticed that the exercises in these level 3 courses require much more from the student. At level 2 it was pretty much spoon feeding - i.e. the books would give examples of how to do something and then there would be exercises where you repeat what you just learnt. Here at level 3 the questions are more demanding as the way to do an exercise is often left up to the student. I welcome this as I like thinking things through for myself, but at times it can be very time consuming. What has surprised me is that a certain amount of 'hand-waving' has appeared both in the text and in the answers to exercises; i.e. rather than thrashing out an answer exhaustively, a bit of 'well, the answer is like this because of such and such.' It can be a bit frustrating if the details are skipped over and a bit of shock after the intense rigour of M208.
I want to put down here some bits of this book which I thought ventured slightly into the hand-waving arena. This part of the book was the most difficult and the most time consuming and so it is worth going through so that later I have something to look back on for revision.
How can you generate subgroups? In M208 we have already met cyclic subgroups which are generated from integer powers of an element in a group. For example, if we consider the group of symmetries of the regular hexagon and label a rotation about its centre through π/3 as r then we can generate a cyclic subgroup from r as
$$<r>=\left\{r^{0}, r, r^{2}, r^{3}, r^{4}, r^{5}\right\}$$
Now the new idea is to expand this notion of generating subgroups to include two or more distinct elements of a group. In the text it says that "Informally, we define the subgroup generated by x,y,... to be the set obtained by forming all possible combinations of copies of x,y,.. and their inverses." In the margins it does say that a formal definition is given later in unit IB4 but I found it difficult to understand this without some formal definition. Here is my version of a formal definition for two distinct elements of a group x and y:-
$$<x,y>=\left\{uv:u,v \in \left\{ (x)^{i},(y)^{i}:i,j \in Z \right\} \right\}$$
Now it is clear that integer powers of x and integer powers of y are just the cyclic subgroups <x> and <y> and so we can rewrite this as
$$<x,y>=\left\{uv:u,v \in <x>\cup<y> \right\}$$
In other words we are taking combinations of elements taken from a union of the subgroups generated by x and y. This does not mean that the resulting subset is a subgroup. This still has to be verified by seeing if the subset satisfies the subgroup properties. One thing that becomes clear straight away is that <x,y> contains the elements of <x> and the elements of <y>. This is because if u and v are drawn from <x>, say, then because <x> is a cyclic subgroup, then the combinations of uv are also in <x>. What is new are products of elements which are not common to both sets.
Let's take an example. If we consider the regular hexagon again and call the symmetry that is a reflection in the horizontal axis s, then s is self-inverse and
$$<s>=\left\{e,s\right\}$$
If we also consider the half-turn about the centre r3 then this is also self-inverse and
$$<r^{3}>=\left\{e,r^{3}\right\}$$
But what about <r3,s>? We see from the above discussion that this is going to contain e, r3, and s but it also must contain combinations of elements which are not common to both sets, namely sr3 and r3s. These turn out to be the same element since sr3 is a reflection of some sort which is self-inverse so
$$sr^{3}=(sr^{3})^{-1}=r^{-3}s^{-1}=r^{3}s$$
Hence we have the subset
$$<r^{3},s>=\left\{e, r^{3}, s, r^{3}s\right\}$$
which does indeed turn out to be a subgroup of the symmetries of the regular hexagon and is isomorphic to the Klein group.
Before I leave book 2 of M336, there are a few things I want to say about it. Firstly, I have noticed that the exercises in these level 3 courses require much more from the student. At level 2 it was pretty much spoon feeding - i.e. the books would give examples of how to do something and then there would be exercises where you repeat what you just learnt. Here at level 3 the questions are more demanding as the way to do an exercise is often left up to the student. I welcome this as I like thinking things through for myself, but at times it can be very time consuming. What has surprised me is that a certain amount of 'hand-waving' has appeared both in the text and in the answers to exercises; i.e. rather than thrashing out an answer exhaustively, a bit of 'well, the answer is like this because of such and such.' It can be a bit frustrating if the details are skipped over and a bit of shock after the intense rigour of M208.
I want to put down here some bits of this book which I thought ventured slightly into the hand-waving arena. This part of the book was the most difficult and the most time consuming and so it is worth going through so that later I have something to look back on for revision.
How can you generate subgroups? In M208 we have already met cyclic subgroups which are generated from integer powers of an element in a group. For example, if we consider the group of symmetries of the regular hexagon and label a rotation about its centre through π/3 as r then we can generate a cyclic subgroup from r as
$$<r>=\left\{r^{0}, r, r^{2}, r^{3}, r^{4}, r^{5}\right\}$$
Now the new idea is to expand this notion of generating subgroups to include two or more distinct elements of a group. In the text it says that "Informally, we define the subgroup generated by x,y,... to be the set obtained by forming all possible combinations of copies of x,y,.. and their inverses." In the margins it does say that a formal definition is given later in unit IB4 but I found it difficult to understand this without some formal definition. Here is my version of a formal definition for two distinct elements of a group x and y:-
$$<x,y>=\left\{uv:u,v \in \left\{ (x)^{i},(y)^{i}:i,j \in Z \right\} \right\}$$
Now it is clear that integer powers of x and integer powers of y are just the cyclic subgroups <x> and <y> and so we can rewrite this as
$$<x,y>=\left\{uv:u,v \in <x>\cup<y> \right\}$$
In other words we are taking combinations of elements taken from a union of the subgroups generated by x and y. This does not mean that the resulting subset is a subgroup. This still has to be verified by seeing if the subset satisfies the subgroup properties. One thing that becomes clear straight away is that <x,y> contains the elements of <x> and the elements of <y>. This is because if u and v are drawn from <x>, say, then because <x> is a cyclic subgroup, then the combinations of uv are also in <x>. What is new are products of elements which are not common to both sets.
Let's take an example. If we consider the regular hexagon again and call the symmetry that is a reflection in the horizontal axis s, then s is self-inverse and
$$<s>=\left\{e,s\right\}$$
If we also consider the half-turn about the centre r3 then this is also self-inverse and
$$<r^{3}>=\left\{e,r^{3}\right\}$$
But what about <r3,s>? We see from the above discussion that this is going to contain e, r3, and s but it also must contain combinations of elements which are not common to both sets, namely sr3 and r3s. These turn out to be the same element since sr3 is a reflection of some sort which is self-inverse so
$$sr^{3}=(sr^{3})^{-1}=r^{-3}s^{-1}=r^{3}s$$
Hence we have the subset
$$<r^{3},s>=\left\{e, r^{3}, s, r^{3}s\right\}$$
which does indeed turn out to be a subgroup of the symmetries of the regular hexagon and is isomorphic to the Klein group.
Friday, 26 April 2013
Tilings
I am still ploughing through the first unit of M336 'Groups and Geometry' and there have been a lot of new ideas to digest concerning the topic of tilings. Sometimes I find the definitions hard to fix in my mind and it is a bit like learning a new language. For example, I found adjacency and incidence bothersome and I keep having to go back to the definitions in order to understand what they mean.
A tiling can be divided into three different parts; the tiles themselves, the vertices and the edges. Each of these parts has a particular meaning. The vertices are points on the boundary where three or more tiles meet and the edges are points on the boundary which join vertices and are where two tiles meet. Now the edges don't have to be straight lines, they can be curved. Also, if the tiles are polygons, the tiles can have corners which are not necessarily vertices and sides that can be made up of more than one edge (and vice versa). All very confusing. Some of this confusion is removed by demanding that a tiling is edge-to-edge, i.e. that each side of the polygon corresponds to exactly one edge and vice versa.
Now adjacency is a relationship between the same parts of a tiling whereas incidence is a relationship between different parts of a tiling. Now here is the definition of these two concepts.
Two distinct tiles are adjacent if they share a common edge. Two distinct vertices are adjacent if they are joined by an edge. Two distinct edges are adjacent if they share a common vertex and bound a common tile.
The edges and vertices on the boundary of a tile T in a tiling are incident with T. Similarly, T is incident with the edges and vertices on its boundary. Also, if an edge E joins vertices V and W, then E is incident with V and W and these vertices are incident with E.
Another important definition is the degree of a tile or vertex. The degree of a tile in a tiling is the number of other tiles to which it is adjacent. The degree of a vertex in a tiling is the number of tiles (or alternatively the number of edges) with which it is incident.
Some examples of Archimedean tilings can be found here. An Archimedean tiling is an edge-to-edge tiling with regular polygons that is vertex-uniform. Only 11 such tilings can be constructed (allowing for one to be a reflection of the other). The first three are the regular tilings and the other 8 are semi-regular. Under each tile is a number. For example under the snub hexagonal tiling is the number 34.6. In OU parlance this is the vertex type (3,3,3,3,6). If V is any vertex in a tiling, then the vertex type of V is given by listing the degrees of the tiles incident with V, starting with any one of them and proceeding in either direction in clockwise or anticlockwise order. So you can see how you have to understand all these definitions!
If you look at the snub hexagonal tiling you notice that all the vertices look the same. Choosing any vertex, then incident with it are four equilateral shaped tiles and one hexagonal tile. If we look at the equilateral triangle tile, then there are three tiles to which it is adjacent, that is its degree is 3 (since this is an edge-to-edge tiling there are as many adjacent tiles as there are sides of the equilateral triangle). If we look at the hexagonal tile then its degree is obviously six. Thus going round the vertex that we chose in either direction we end up with a vertex type of (3,3,3,3,6) (allowing for cyclic permutations of these numbers). A tiling is vertex-uniform if all the vertex types in the tiling are the same.
Not only can you define a vertex type but you can also define a tile type. It is similar in definition to a vertex type, but instead you list the degrees of the vertices incident with a tile in a tiling. The tile type of our snub hexagonal tiling is not uniform. The tile type for the hexagon is [5,5,5,5,5,5] (notice that square brackets are used) but the tile type for the equilateral triangle is [5,5,5].
My method for remembering how to find a vertex type or a tile type is that for vertex types it is tile-tile (i.e. you are counting tiles around tiles incident to a vertex) and for tile types it is vertex-vertex (i.e. you are counting vertices around vertices incident to a tile).
A tiling can be divided into three different parts; the tiles themselves, the vertices and the edges. Each of these parts has a particular meaning. The vertices are points on the boundary where three or more tiles meet and the edges are points on the boundary which join vertices and are where two tiles meet. Now the edges don't have to be straight lines, they can be curved. Also, if the tiles are polygons, the tiles can have corners which are not necessarily vertices and sides that can be made up of more than one edge (and vice versa). All very confusing. Some of this confusion is removed by demanding that a tiling is edge-to-edge, i.e. that each side of the polygon corresponds to exactly one edge and vice versa.
Now adjacency is a relationship between the same parts of a tiling whereas incidence is a relationship between different parts of a tiling. Now here is the definition of these two concepts.
Two distinct tiles are adjacent if they share a common edge. Two distinct vertices are adjacent if they are joined by an edge. Two distinct edges are adjacent if they share a common vertex and bound a common tile.
The edges and vertices on the boundary of a tile T in a tiling are incident with T. Similarly, T is incident with the edges and vertices on its boundary. Also, if an edge E joins vertices V and W, then E is incident with V and W and these vertices are incident with E.
Another important definition is the degree of a tile or vertex. The degree of a tile in a tiling is the number of other tiles to which it is adjacent. The degree of a vertex in a tiling is the number of tiles (or alternatively the number of edges) with which it is incident.
Some examples of Archimedean tilings can be found here. An Archimedean tiling is an edge-to-edge tiling with regular polygons that is vertex-uniform. Only 11 such tilings can be constructed (allowing for one to be a reflection of the other). The first three are the regular tilings and the other 8 are semi-regular. Under each tile is a number. For example under the snub hexagonal tiling is the number 34.6. In OU parlance this is the vertex type (3,3,3,3,6). If V is any vertex in a tiling, then the vertex type of V is given by listing the degrees of the tiles incident with V, starting with any one of them and proceeding in either direction in clockwise or anticlockwise order. So you can see how you have to understand all these definitions!
If you look at the snub hexagonal tiling you notice that all the vertices look the same. Choosing any vertex, then incident with it are four equilateral shaped tiles and one hexagonal tile. If we look at the equilateral triangle tile, then there are three tiles to which it is adjacent, that is its degree is 3 (since this is an edge-to-edge tiling there are as many adjacent tiles as there are sides of the equilateral triangle). If we look at the hexagonal tile then its degree is obviously six. Thus going round the vertex that we chose in either direction we end up with a vertex type of (3,3,3,3,6) (allowing for cyclic permutations of these numbers). A tiling is vertex-uniform if all the vertex types in the tiling are the same.
Not only can you define a vertex type but you can also define a tile type. It is similar in definition to a vertex type, but instead you list the degrees of the vertices incident with a tile in a tiling. The tile type of our snub hexagonal tiling is not uniform. The tile type for the hexagon is [5,5,5,5,5,5] (notice that square brackets are used) but the tile type for the equilateral triangle is [5,5,5].
My method for remembering how to find a vertex type or a tile type is that for vertex types it is tile-tile (i.e. you are counting tiles around tiles incident to a vertex) and for tile types it is vertex-vertex (i.e. you are counting vertices around vertices incident to a tile).
Tuesday, 9 April 2013
First books
Well, I have made a start on the course books for 'Groups and Geometry' (M336) and 'Number theory and mathematical logic' (M381). I began with M381 and the first book, which I have now gone through, 'Foundations' deals with four topics; Numbers from Patterns, Mathematical Induction, Divisibility and the Linear Diophantine Equation, much of which I have already encountered. The material is well written and the proofs are straight-forward but the exercises are a jump up in difficulty. I found myself struggling with a few and one I just couldn't see how to do at all. Still, it is to be expected in going up from level 2 to level 3.
I then started tackling the first book of M336 on Tilings and I must say that I think I like what I have read so far. It is refreshing to enter a new area of maths that I know nothing much about. This course also seems very well written and the ideas are being built up in nice easy stages.
In the mean time I have started on the second book of M381 which is on prime numbers. As this is something that interests me greatly, I will be relishing this book and am looking forward to its contents.
I have now signed up for M381 and M336 and my next aim will be to get together the funding I need to pay for the courses.
I then started tackling the first book of M336 on Tilings and I must say that I think I like what I have read so far. It is refreshing to enter a new area of maths that I know nothing much about. This course also seems very well written and the ideas are being built up in nice easy stages.
In the mean time I have started on the second book of M381 which is on prime numbers. As this is something that interests me greatly, I will be relishing this book and am looking forward to its contents.
I have now signed up for M381 and M336 and my next aim will be to get together the funding I need to pay for the courses.
Tuesday, 26 March 2013
A change of heart
Well, contrary to everything I have said before about not continuing on with my OU studies, I have had a change of heart and have decided to see if I can finish my BSc honours maths degree that I signed up for in 2009. Why this change? Well, I realise that I do have a hankering to do something in maths and I may regret it in the future if I don't try and see how far I can get. I also realise that getting ends to meet from my current work in photography is getting harder and harder and so eventually I am going to have to do something else anyway. Coming back from a week in Berlin, I realised that I just have to give it a go, regardless of the potential difficulties of approaching the time of life when most people are thinking of retirement! Still, the government wants us all to work after retirement age, so I will be trying this out. I just hope the brain cells don't wear out before I get to the point where I can contribute something to this subject.
So having made the decision, it's all kind of exciting and daunting at the same time. The first thing is to concentrate on this degree which, due to the reorganisation of these qualifications, I now have to finish by the 31/12/17. This gives me four years and, at 60 points a year, I hope to get the additional 240 points I need to pass (I already have 120 points from MST121, MS221 and M208). There are now two routes a student can follow in this degree. Route A means following what was originally prescribed and, for me, at level 2 that would mean doing the 60 point 'Mathematical methods and models' (MST209) and the 10 point summer school 'Mathematical modelling' (MSXR209). For a number of reasons, I am going to plump for route B, which means doing the new 60 point 'Mathematical methods and models' (MST210) which does not require the summer school any more. However, this does mean that I will have to wait until October 2014 to start this course and it will also mean that I will have to do the new 30 point level 1 course 'Introducing statistics' (M140).
In the mean time I need to get another 60 points under my belt starting this October and I think the best option for me having only done M208 so far, is to do the level 3 30 point courses 'Groups and Geometry' (M336) and 'Number theory and mathematical logic' (M381). This is the final presentation of these two courses before they are replaced by 'Further pure mathematics' (M303). This new course will be available from 2014 and will be an amalgamated version of M336, M381 and the now extinct 'Topology' course (M338). I think I will be pursuing a more pure maths direction in my further studies and number theory is already something I think would like to specialise in, so it all makes good sense.
As for the remaining 90 points, I have only to do 60 more points at level 3 and 30 points from a free choice of OU modules. However, I agree with Chris that I want to squeeze in the level 3 30 point 'Complex Analysis' (M337) and at the moment this is presented on alternate years (I believe), though this could change. I expect I want to get a good grounding in both pure and applied maths in case this comes in useful, so a couple of level 3 30 point applied courses may be an option.
In the mean time it is eyes down again and time to start getting ahead to alleviate future stresses of workload. So I have already got hold of some of the books for the 'Number theory and mathematical logic' course and have started on the first book. It looks like some of this will already be covered by my reading of John Stillwell's book, which I am still making some progress on, so that's all well and good.
So having made the decision, it's all kind of exciting and daunting at the same time. The first thing is to concentrate on this degree which, due to the reorganisation of these qualifications, I now have to finish by the 31/12/17. This gives me four years and, at 60 points a year, I hope to get the additional 240 points I need to pass (I already have 120 points from MST121, MS221 and M208). There are now two routes a student can follow in this degree. Route A means following what was originally prescribed and, for me, at level 2 that would mean doing the 60 point 'Mathematical methods and models' (MST209) and the 10 point summer school 'Mathematical modelling' (MSXR209). For a number of reasons, I am going to plump for route B, which means doing the new 60 point 'Mathematical methods and models' (MST210) which does not require the summer school any more. However, this does mean that I will have to wait until October 2014 to start this course and it will also mean that I will have to do the new 30 point level 1 course 'Introducing statistics' (M140).
In the mean time I need to get another 60 points under my belt starting this October and I think the best option for me having only done M208 so far, is to do the level 3 30 point courses 'Groups and Geometry' (M336) and 'Number theory and mathematical logic' (M381). This is the final presentation of these two courses before they are replaced by 'Further pure mathematics' (M303). This new course will be available from 2014 and will be an amalgamated version of M336, M381 and the now extinct 'Topology' course (M338). I think I will be pursuing a more pure maths direction in my further studies and number theory is already something I think would like to specialise in, so it all makes good sense.
As for the remaining 90 points, I have only to do 60 more points at level 3 and 30 points from a free choice of OU modules. However, I agree with Chris that I want to squeeze in the level 3 30 point 'Complex Analysis' (M337) and at the moment this is presented on alternate years (I believe), though this could change. I expect I want to get a good grounding in both pure and applied maths in case this comes in useful, so a couple of level 3 30 point applied courses may be an option.
In the mean time it is eyes down again and time to start getting ahead to alleviate future stresses of workload. So I have already got hold of some of the books for the 'Number theory and mathematical logic' course and have started on the first book. It looks like some of this will already be covered by my reading of John Stillwell's book, which I am still making some progress on, so that's all well and good.
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