Sorry, not really been keeping up to date with this blog (bad dragon! naughty dragon!) I will try harder during MST221 I promise.
Anyway over Xmas I got a little letter saying I passed MST121 with a rather credible 93% which I was pleased with. I was especially pleased to see I got 95% on the final end of course assessment which was a surprise as I thought I had royally screwed up on the calculus question and I wasn't too happy with my answer to the crocodile modelling question either - turns out they were my best 2!
Anyway more to come soon.
Wednesday, 30 December 2009
Monday, 17 August 2009
One down, lots to go.
Well I got my first Open University result back, this one from "Easy Maths Even You Can Do" or "MU120" as they call it. To be honest I probably could have skipped that course as it became pretty obvious early on that I could easily handle the maths involved but it was useful for revision, especially some of the later on trig as that was rather rusty and some of the stats stuff like boxplots was new to me (they didn't teach those back when I was a hatchling) and most of all it was useful for getting into doing things the OU way so I was not stressing about the maths and the admin.
I actually completed the course well ahead of the alloted time too and so it was a bit of a surprise when the results actually turned up and a very nice surprise to find I'd got 96% which is a first (or "distinction" as they call them) in anyones language except MU120 is a straight pass/fail.
It was even more surprising that what I call the "wibble" questions I got full marks for. The "wibble" questions are the non-mathematical ones that ask you to pick a bt of the course and say what you found hard, how you coped and what your "learning outcomes" were. I guess having lived with a social worker for the last 15 years helps here as I just came up with a stream of wibbly nonsense with the right kind of buzzwords in it and bingo, 16 marks out of 16. I don't mind this sort of question and I can see that in a low level course with people just starting out it makes sense but I don't agree with them being a marked part of the course and final exam work as, well, it's not really maths is it?
Still, bumped my marks up!
I actually completed the course well ahead of the alloted time too and so it was a bit of a surprise when the results actually turned up and a very nice surprise to find I'd got 96% which is a first (or "distinction" as they call them) in anyones language except MU120 is a straight pass/fail.
It was even more surprising that what I call the "wibble" questions I got full marks for. The "wibble" questions are the non-mathematical ones that ask you to pick a bt of the course and say what you found hard, how you coped and what your "learning outcomes" were. I guess having lived with a social worker for the last 15 years helps here as I just came up with a stream of wibbly nonsense with the right kind of buzzwords in it and bingo, 16 marks out of 16. I don't mind this sort of question and I can see that in a low level course with people just starting out it makes sense but I don't agree with them being a marked part of the course and final exam work as, well, it's not really maths is it?
Still, bumped my marks up!
Thursday, 4 June 2009
Integration.
MST121 has moved onto integration now. Integration conceptually seems pretty simple, it's like differentiation only seen from the back.
It's not too hard to do either: if you remember down there when I differentiated X^2 and ended up with 2x you can do a reverse operation and it turns out the formula, yielded by some relatively straightforward arseing about with algebra or geometry, to integrate a simple function of the form
is
which happily yields x^2
However the devil in the detail is that the result of integration needs to have an arbitrary constant added on as differentiating x^2 + a (where a is any constant) gives 2x so the reverse operation has to add this "constant of integration" back in. Apparently this makes these "Indefinite Integrals" but luckily for me you can get them to be a bit more definite when you subtract one from the other.
And subtracting the results of integrating a function for different values of the dependent variable is the one bit of calculus I vaguely remember being shown at school (I never did A-levels as I loathed the two maths teachers with a passion so I'm guessing this got touched on somewhere in AO level1 pure maths) probably for finding areas under curves.
Oh and I got a book on Calculus as well (see - still a swotty dragon, reading around the course now!) called "Calculus made Simple (for scaly green firebreathing creatures like you)2" written by some guy in the early 1900's and apparently still the definitive text on the subject today. I will let you know how I get on
1 This was an old UK exam that I think got dropped sometime in the 80's. I, being a swotty hatchling, was in an "advanced" stream at secondary school and we took our O levels (called GCSEs now) a year early at 15 and we also did a couple of these "midway between an O level and A level". I had a google and although I couldn't find a AO level paper I did come across a marking scheme for one here: www.edexcel-international.org/VirtualContent/49349/7362_PURE_MATHEMATICS_F.pdf and it does have a little bit of calculus in it.
2 I think I might have imagined the subtitle.
It's not too hard to do either: if you remember down there when I differentiated X^2 and ended up with 2x you can do a reverse operation and it turns out the formula, yielded by some relatively straightforward arseing about with algebra or geometry, to integrate a simple function of the form
n
ax
is
a 1 n+1
--- x
n+1
which happily yields x^2
However the devil in the detail is that the result of integration needs to have an arbitrary constant added on as differentiating x^2 + a (where a is any constant) gives 2x so the reverse operation has to add this "constant of integration" back in. Apparently this makes these "Indefinite Integrals" but luckily for me you can get them to be a bit more definite when you subtract one from the other.
And subtracting the results of integrating a function for different values of the dependent variable is the one bit of calculus I vaguely remember being shown at school (I never did A-levels as I loathed the two maths teachers with a passion so I'm guessing this got touched on somewhere in AO level1 pure maths) probably for finding areas under curves.
Oh and I got a book on Calculus as well (see - still a swotty dragon, reading around the course now!) called "Calculus made Simple (for scaly green firebreathing creatures like you)2" written by some guy in the early 1900's and apparently still the definitive text on the subject today. I will let you know how I get on
1 This was an old UK exam that I think got dropped sometime in the 80's. I, being a swotty hatchling, was in an "advanced" stream at secondary school and we took our O levels (called GCSEs now) a year early at 15 and we also did a couple of these "midway between an O level and A level". I had a google and although I couldn't find a AO level paper I did come across a marking scheme for one here: www.edexcel-international.org/VirtualContent/49349/7362_PURE_MATHEMATICS_F.pdf and it does have a little bit of calculus in it.
2 I think I might have imagined the subtitle.
Sunday, 10 May 2009
Zero as a limit
Well MST121, which is the Open University maths course I'm doing, has now reached the dreaded "Chapter C" in which we get to study "The Calculus".
Quite why calculus gets a definite article is a mystery to me, I mean I never studied "The Geometry" or "The Algebra" that I can remember.
Well the first thing that's presented is a demonstration of differentiation which, in essence, is finding out about the rate of how things change.
A nice way to think about this and one that helped me get my head around it is to think of a graph like this which shows distance travelled (on the y axis) versus time taken (on the x axis).
In this example the distance travelled in metres is the square of time taken in seconds so in maths-speak we say that distance (d) is a function of time (t) and that in this case the function is t squared; which you write like this
OK so from this you can tell how far you have gone in how many seconds. But how fast are you going at any time? Well as speed is distance divided by time you can work out an average speed between two times, t and t + h by joining the points on the curve at t and t+h and dividing the distance covered by the time taken.
Now the time taken is (t+h) - t which is just h and the distance covered is the difference between the values of distance d at times t and t + h; but as we know that d = t^2 we can express these distances as (t+h)^2 and t^2.
So that gives us
which simplifies to
and more simply:
Now what we do is start to reduce h and keep reducing it to a limit of 0
Ah, you can't divide by zero though, we went through that on the last post didn't we, you get a black hole if you do.
"Yes", says my OU tutor who has a brain so large there's nothing left for dress sense, but we don't go to zero, we TEND to the LIMIT of zero so think of h getting infinitessimally small, so small it doesn't count. So small you can just get rid of them leaving:
2t
And there you go, the first derivative of t^2. So after 5 seconds you've travelled 25 metres and you're going at 10m/s
Wel it does work but there's something inside of me that doesn't quite like it. I'm just a bit concerned that it's a bit of a fiddle throwing away these very tiny numbers. Apart from that at least I'm feeling pretty comfy with differentiation so far.
By the way if you type "Zero as a limit" into Google you get a load of song lyrics to a Human League tune from the 80's. Was the floppy haired one a mathematician I wonder?
Quite why calculus gets a definite article is a mystery to me, I mean I never studied "The Geometry" or "The Algebra" that I can remember.
Well the first thing that's presented is a demonstration of differentiation which, in essence, is finding out about the rate of how things change.
A nice way to think about this and one that helped me get my head around it is to think of a graph like this which shows distance travelled (on the y axis) versus time taken (on the x axis).
In this example the distance travelled in metres is the square of time taken in seconds so in maths-speak we say that distance (d) is a function of time (t) and that in this case the function is t squared; which you write like this
d = f(t) = t^2
OK so from this you can tell how far you have gone in how many seconds. But how fast are you going at any time? Well as speed is distance divided by time you can work out an average speed between two times, t and t + h by joining the points on the curve at t and t+h and dividing the distance covered by the time taken.
Now the time taken is (t+h) - t which is just h and the distance covered is the difference between the values of distance d at times t and t + h; but as we know that d = t^2 we can express these distances as (t+h)^2 and t^2.
So that gives us
(t+h)^2 - t^2
-------------
h
which simplifies to
t^2 + 2th + h^2
---------------
h
and more simply:
2th
---
h
Now what we do is start to reduce h and keep reducing it to a limit of 0
Ah, you can't divide by zero though, we went through that on the last post didn't we, you get a black hole if you do.
"Yes", says my OU tutor who has a brain so large there's nothing left for dress sense, but we don't go to zero, we TEND to the LIMIT of zero so think of h getting infinitessimally small, so small it doesn't count. So small you can just get rid of them leaving:
2t
And there you go, the first derivative of t^2. So after 5 seconds you've travelled 25 metres and you're going at 10m/s
Wel it does work but there's something inside of me that doesn't quite like it. I'm just a bit concerned that it's a bit of a fiddle throwing away these very tiny numbers. Apart from that at least I'm feeling pretty comfy with differentiation so far.
By the way if you type "Zero as a limit" into Google you get a load of song lyrics to a Human League tune from the 80's. Was the floppy haired one a mathematician I wonder?
Tuesday, 28 April 2009
Divide By Zero
Despite what you might have read on teh interwebs you don't get a black hole if you divide by zero. But what do you get?
If you try it on a calculator you get an error or "can't divide by zero" message, although oddly if you are writing a computer program like this one...
double DivByZero(double n)
{
return n/0;
}
...you may well get a special value passed back which is either "Positive Infinity" or "Negative Infinity" which is a plausable number, but mathematically speaking wrong.
Lets look at the infinity business first. Imagine you divide 1 by a half (0.5), you get 2. Now divide 1 by a tenth (0.1), you get 10, now divide 1 by a hundredth (0.01), you get 100. So as the number you divide by gets progressively smaller and smaller the result gets larger and larger and this is true regardless of the number that is being divided.
In maths-speak what they say is that as the divisor (the number on the bottom) tends to zero then the result tends to infinity. Of course if the number being divided is negative then the result tends to a larger and larger negative value, "negative infinity".
The important thing to notice here is the "tends to zero" bit. The number is approaching zero, becoming exceedingly small but it is never actually zero. This works fine for computers where the rules say that all floating point operations have to have a defined result but we still have not actually divided by zero.
Why you can't actually divide by a real zero is that it makes maths itself break.
To see why you need to look at what division is. Basically put it is multiplication backwards. If you take a number, say 6 and divide it by 2 you get 3, now if you multiply it by the same number 2 you get back to 6. We say that division is the inverse function of multiplication.
Right, so let's for now pretend that you can divide by zero and you get this number "infinity" as a result: so 1 divided by 0 is infinity and infinity multiplied by 0 must be 1. Except it isn't as anything multiplied by zero is zero.
You could do the same thing with 2 divided by zero, three divided by zero, in fact any number divided by zero Now the same calculation can't have different results as the only way this could happen is if the result of a divide by zero as being every possible number simultaneously; so the correct answer to what do you get when you divide by zero is "the result is undefined".
"Simples" as that irritating meerkat on the telly says.
Mathematics For Dragons
Hello everyone (all two of you who follow my witterings anyway).
This is going to be an offshoot of my normal blog Grumpy Dragon (Warning, adult language, swearing, cursing and stuff) where I'm going to make the occasional post about adding up as I wander through my Open University maths degree.
If you're at the Open University, a student or just interested in sums you might like to have a read. The rest of you will probably be bored witless.
This is going to be an offshoot of my normal blog Grumpy Dragon (Warning, adult language, swearing, cursing and stuff) where I'm going to make the occasional post about adding up as I wander through my Open University maths degree.
If you're at the Open University, a student or just interested in sums you might like to have a read. The rest of you will probably be bored witless.
Subscribe to:
Posts (Atom)
