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.
Monday, 31 October 2011
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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