Apologies for the lack of recent posts. I've actually done more reading and writing lately than I have maths.
Nonetheless, I have booked to do Pure 1 and Stats 1 on the 9th January so expect a trickle of revision related posts over the next two months!
Wednesday, 29 October 2008
Sunday, 14 September 2008
Geometry of the Line (4): Circles
Time to go back to straight line geometry for a bit.
With knowledge of how to calculate the gradient, the perpendicular gradient, the mid-point and the length of the line, you've also done quite a lot of co-ordinate geometry of the circle. This is an interesting concept. A circle on a set of cartesian co-ordinates.
You start from the premise of a circle with centre (0,0). The equation of this circle is x2 + y2 = r2. If you study a diagram of it, or even draw one yourself, the reason for this becomes clear. Like a lot of co-ordinate geometry, it's based on Pythagoras. In this case, drawing a triangle inside the circle to a point somewhere on the circumference. Say that this circle with centre (0,0) passes through the point (3,4).
The line from the centre to the point functions as the hypoteneuse of the right angled triangle, while line x measures the distance travelled horizontally to the point, and the line y measures the vertical distance to the point.
The equation of this particular circle then tells us the radius.
32 + 42 = r2
or 9 + 16 = r2
= 25
r = 5.
The radius of the circle is 5.
Things get a little more complicated if the circle's centre is not (0,0). Say we have a circle with centre (2,3). We still devise the equation of the circle based on its distance from (0,0).
This circle's centre has moved 2 units x-wards and 3 y-wards (across and up) from the origin. Any calculation of the radius needs to take into account the fact that we are already some way from the origin.
So if the circle passes through the point (5,8), the distance from the centre to the point will be 5-2 and 8-3 (otherwise we're talking about including space outside the circle).
So the equation will be: (x-2)2 + (x-3)2 = r2.
In this case: 9 + 25 = 34
So r = √34
It is perfectly acceptable to leave the answer as a surd at C1, indeed often you have to.
Usually you are given the equation, say (x-4)2 + (y-7)2 = 4
or something.
The centre of the circle is ALWAYS THE OPPOSITE OF THE NUMBERS IN BRACKETS, in this case (4,7). Worth remembering. If it is (x-a)2 + (y-b)2 = r2 the centre is always (a,b).
With knowledge of how to calculate the gradient, the perpendicular gradient, the mid-point and the length of the line, you've also done quite a lot of co-ordinate geometry of the circle. This is an interesting concept. A circle on a set of cartesian co-ordinates.
You start from the premise of a circle with centre (0,0). The equation of this circle is x2 + y2 = r2. If you study a diagram of it, or even draw one yourself, the reason for this becomes clear. Like a lot of co-ordinate geometry, it's based on Pythagoras. In this case, drawing a triangle inside the circle to a point somewhere on the circumference. Say that this circle with centre (0,0) passes through the point (3,4).
The line from the centre to the point functions as the hypoteneuse of the right angled triangle, while line x measures the distance travelled horizontally to the point, and the line y measures the vertical distance to the point.
The equation of this particular circle then tells us the radius.
32 + 42 = r2
or 9 + 16 = r2
= 25
r = 5.
The radius of the circle is 5.
Things get a little more complicated if the circle's centre is not (0,0). Say we have a circle with centre (2,3). We still devise the equation of the circle based on its distance from (0,0).
This circle's centre has moved 2 units x-wards and 3 y-wards (across and up) from the origin. Any calculation of the radius needs to take into account the fact that we are already some way from the origin.
So if the circle passes through the point (5,8), the distance from the centre to the point will be 5-2 and 8-3 (otherwise we're talking about including space outside the circle).
So the equation will be: (x-2)2 + (x-3)2 = r2.
In this case: 9 + 25 = 34
So r = √34
It is perfectly acceptable to leave the answer as a surd at C1, indeed often you have to.
Usually you are given the equation, say (x-4)2 + (y-7)2 = 4
or something.
The centre of the circle is ALWAYS THE OPPOSITE OF THE NUMBERS IN BRACKETS, in this case (4,7). Worth remembering. If it is (x-a)2 + (y-b)2 = r2 the centre is always (a,b).
Saturday, 13 September 2008
A Note on Revision
Since I've finished C1 and S1 I am sort of both doing C2 and revising C1.
If I were to give one piece of advice, it would be to treat revision as a dynamic process. Do it as soon as you finish a topic or a module. Do it randomly, answer questions out of the blue, keep coming back to it.
Revision is only boring and hard when you have piles of stuff you barely remember to do in a short space of time. If you do it regularly, treating it like another aspect of your learning, then it becomes easier and things go in better. Writing this blog has been a valuable source of revision for me as well - I hope - as some of my readers.
Keep on keeping on revising. That's the sum total of my advice.
If I were to give one piece of advice, it would be to treat revision as a dynamic process. Do it as soon as you finish a topic or a module. Do it randomly, answer questions out of the blue, keep coming back to it.
Revision is only boring and hard when you have piles of stuff you barely remember to do in a short space of time. If you do it regularly, treating it like another aspect of your learning, then it becomes easier and things go in better. Writing this blog has been a valuable source of revision for me as well - I hope - as some of my readers.
Keep on keeping on revising. That's the sum total of my advice.
Blog Purpose Update
Well I know that I don't have all that many readers here but I don't mind too much. To judge from recent Sitemeter reports some people are actually reading what I have written, even if they are not so interested in commenting.
A quick reminder. I am no expert, in anything. Just a vaguely mature student of A Level maths who is doing it in his spare time when he isn't down the pub or surfing the blogosphere with increasing frustration and disappointment. So this blog was intended to be a wholly positive space in which I could devote time and effort to something of thought, however superficial. I do have a couple of problems though, especially with graphics and symbols. I still haven't worked out how to show these in blogposts properly. I have heard of something called TeX or LaTeX but i don't know how to use it.
If you have found anything on here useful, then I'm glad. It will be the first geniune contribution a blog of mine has made to the world.
A quick reminder. I am no expert, in anything. Just a vaguely mature student of A Level maths who is doing it in his spare time when he isn't down the pub or surfing the blogosphere with increasing frustration and disappointment. So this blog was intended to be a wholly positive space in which I could devote time and effort to something of thought, however superficial. I do have a couple of problems though, especially with graphics and symbols. I still haven't worked out how to show these in blogposts properly. I have heard of something called TeX or LaTeX but i don't know how to use it.
If you have found anything on here useful, then I'm glad. It will be the first geniune contribution a blog of mine has made to the world.
Thursday, 11 September 2008
Dividing Polynomials (ii)
Alright then, let's have a crack at this properly.
We are armed with some fraction revision and a little bit of knowledge of polynomials, including factorising using the system of comparing coefficients.
We only need to divide a polynomial by a linear factor for C1.
Division has this nasty habit of not being exact, but of leaving remainders. Unfortunately, 5 does not go easily into 24 but leaves a remainder. Polynomials are also subject to this problem. A linear expression might not be a factor of P(x) as such, but might go into something else, leaving factors and a remainder when asked to be slotted into P(x).
Remember that QUADRATIC EQUATIONS sometimes factorise into TWO LINEAR expressions,
while CUBICS factorise into LINEAR + QUADRATIC (and sometimes three linear).
So if you divide a quadratic by a linear, this might happen:
x2 + 2x + 4 divided by x-2.
If we are to use the concept of comparing coefficients, which is very efficient, we first rewrite the equation into the form it would take expressed as factors and remainders:
x2 + 2x + 4 = (x+p)(x+q) + r.
Got it?
We need to divide it by x-2 though, so we already have one desired factor (THIS DOES NOT MEAN THAT x-2 WILL BE A FACTOR - just that we are trying to express the quadratic in terms of how much x-2 goes into it.
So:
x2 + 2x + 4 = (x-2)(x+q) + r.
Then we collect the different terms in this form. We are not completely, expanding the brackets, we are re-arranging them in terms of their own factors in order to look for coefficients which will tell us the other factors and remainders.
x2 + 2x + 4 = x2 + (-2 + q)x + (-2q+r)
F(x) is in this case equal to x2 and (-2+q) lots of x, and then the two constants, with no x variable, (-2q and r, the possible remainder).
Therefore 2= (-2+q) so q= 4.
4 = (-2x4 + r)
4=8+r
r = -4
x2 + 2x + 4 = (x-2)(x+4) -4.
This only expresses the quadratic as factors and a remainder. The division isn't over yet.
Now let's actually divide the quadratic by (x-2).
x2 + 2x + 4
-------------
x-2
=
(x+4) - 4
----
(x-2)
I'll explain it later. Proper work beckons.
We are armed with some fraction revision and a little bit of knowledge of polynomials, including factorising using the system of comparing coefficients.
We only need to divide a polynomial by a linear factor for C1.
Division has this nasty habit of not being exact, but of leaving remainders. Unfortunately, 5 does not go easily into 24 but leaves a remainder. Polynomials are also subject to this problem. A linear expression might not be a factor of P(x) as such, but might go into something else, leaving factors and a remainder when asked to be slotted into P(x).
Remember that QUADRATIC EQUATIONS sometimes factorise into TWO LINEAR expressions,
while CUBICS factorise into LINEAR + QUADRATIC (and sometimes three linear).
So if you divide a quadratic by a linear, this might happen:
x2 + 2x + 4 divided by x-2.
If we are to use the concept of comparing coefficients, which is very efficient, we first rewrite the equation into the form it would take expressed as factors and remainders:
x2 + 2x + 4 = (x+p)(x+q) + r.
Got it?
We need to divide it by x-2 though, so we already have one desired factor (THIS DOES NOT MEAN THAT x-2 WILL BE A FACTOR - just that we are trying to express the quadratic in terms of how much x-2 goes into it.
So:
x2 + 2x + 4 = (x-2)(x+q) + r.
Then we collect the different terms in this form. We are not completely, expanding the brackets, we are re-arranging them in terms of their own factors in order to look for coefficients which will tell us the other factors and remainders.
x2 + 2x + 4 = x2 + (-2 + q)x + (-2q+r)
F(x) is in this case equal to x2 and (-2+q) lots of x, and then the two constants, with no x variable, (-2q and r, the possible remainder).
Therefore 2= (-2+q) so q= 4.
4 = (-2x4 + r)
4=8+r
r = -4
x2 + 2x + 4 = (x-2)(x+4) -4.
This only expresses the quadratic as factors and a remainder. The division isn't over yet.
Now let's actually divide the quadratic by (x-2).
x2 + 2x + 4
-------------
x-2
=
(x+4) - 4
----
(x-2)
I'll explain it later. Proper work beckons.
Dividing Polynomials
As promised....
One of the key elements of this is the ability to split off fractions effectively.
Think of a normal, but improper fraction, such as 4/3. This could be expressed as 3/3 + 2/3 (or any other combination but this is clean and tidy and useful for us now). 3/3 is of course 1. We are left with 1 + 1/3.
The same thing can be done algebraically, so as to solve basic division questions.
Take (x + 6)/x. As above, we can think of this as x/x + 6/x. x/x = 1, so the answer is 1 + 6/x.
We use this property, making part of the numerator match the denominator, in dividing polynomials.
One of the key elements of this is the ability to split off fractions effectively.
Think of a normal, but improper fraction, such as 4/3. This could be expressed as 3/3 + 2/3 (or any other combination but this is clean and tidy and useful for us now). 3/3 is of course 1. We are left with 1 + 1/3.
The same thing can be done algebraically, so as to solve basic division questions.
Take (x + 6)/x. As above, we can think of this as x/x + 6/x. x/x = 1, so the answer is 1 + 6/x.
We use this property, making part of the numerator match the denominator, in dividing polynomials.
Sunday, 7 September 2008
Rounding Up Factorisation
...so after all that you now have the technology (all right, the skills) to factorise a polynomial completely. You first do trial and error to find the first linear factor, then you multiply it by ax2 + bx + c and compare coefficients to find the quadratic. Then you check if the quadratic factorises further.
Onto dividing polynomials proper....
Onto dividing polynomials proper....
Factor Theorem (ii)
Well that's all very neat and tidy, you might be thinking, but what good is this theorem?
It gives you a lovely and easy way of trying to find linear factors by substituting values into the polynomial you are given. Sometimes this does not work easily but it is often worth a try for a minute or so.
It also enables you to do slightly trickier things with polynomials.
You often get questions like this (we'll use the cubic from the previous post for this). A polynomial, P(x) = x3 + ax2 + bx + 6, has (x+1) and (x+2) as factors. Find the values of a and b.
This looks tricky but it is where the factor theorem comes in.
P(-1) = 0. You know this from the theorem.
that means that (-1) + a(1) + b(-1) + 6 = 0
so -1 + a - b + 6 = 0
therefore a - b = -5
P(-2) = 0.
So -8 + 4a -2b + 6 = 0
therefore 4a - 2b = 2
You now have two equations to solve SIMULTANEOUSLY!!
a - b = -5
4a - 2b = 2
Times the first equation by 2:
2a - 2b = -10
4a - 2b = 2
SUBTRACT THE TWO TO ELIMINATE b
-2a = -12
a = 6
Plug this back into one of the equations:
2(6) - 2b = -10
-2b = -22
b = 11
a = 6, b = 11
There is nothing that hard here and yet in just writing this post I have made LOADS of mistakes, which is why it's taken me an hour or so to write.
Here are the errors I made:
1) Using the wrong constant value - ie reading the cubic wrong;
2) Forgetting to cube or square the values in the cubic equation;
3) subtracting with minus signs incorrectly;
4) multiplying incorrectly.
All because I was surfing the blogs while writing this post!
It gives you a lovely and easy way of trying to find linear factors by substituting values into the polynomial you are given. Sometimes this does not work easily but it is often worth a try for a minute or so.
It also enables you to do slightly trickier things with polynomials.
You often get questions like this (we'll use the cubic from the previous post for this). A polynomial, P(x) = x3 + ax2 + bx + 6, has (x+1) and (x+2) as factors. Find the values of a and b.
This looks tricky but it is where the factor theorem comes in.
P(-1) = 0. You know this from the theorem.
that means that (-1) + a(1) + b(-1) + 6 = 0
so -1 + a - b + 6 = 0
therefore a - b = -5
P(-2) = 0.
So -8 + 4a -2b + 6 = 0
therefore 4a - 2b = 2
You now have two equations to solve SIMULTANEOUSLY!!
a - b = -5
4a - 2b = 2
Times the first equation by 2:
2a - 2b = -10
4a - 2b = 2
SUBTRACT THE TWO TO ELIMINATE b
-2a = -12
a = 6
Plug this back into one of the equations:
2(6) - 2b = -10
-2b = -22
b = 11
a = 6, b = 11
There is nothing that hard here and yet in just writing this post I have made LOADS of mistakes, which is why it's taken me an hour or so to write.
Here are the errors I made:
1) Using the wrong constant value - ie reading the cubic wrong;
2) Forgetting to cube or square the values in the cubic equation;
3) subtracting with minus signs incorrectly;
4) multiplying incorrectly.
All because I was surfing the blogs while writing this post!
Saturday, 6 September 2008
The Factor Theorem
Well that stuff in the previous post leads us nicely on to the factor theorem. Although this sounds like a low budget 70s sci-fi drama, it is in fact a key part of A Level maths.
The factor theorem is dead simple. If you have a polynomial x3 + 6x2 + 11x + 6, it has factors, namely a linear and a quadratic.
Only this time, the quadratic itself factorises, so that the cubic has 3 linear factors, namely (x+1)(x+2)(x+3).
If I now did something a bit sneaky, I could prove that these are factors of this cubic.
Watch. P(-1) = -13 + 6(-12) + 11(-1) + 6
= -1 + 6 - 11 + 6
= 0
When x=-1 the cubic equation = 0.
This proves that (x+1) is a factor of x3 + 6x2 + 11x + 6
In fact, if (x-a) is a factor of P(x) then P(a) = 0. Always and everywhere.
Why?
Well, look at (x+1)(x+2)(x+3).
These are all factors of x3 + 6x2 + 11x + 6.
So if you put x= -1 into the first factor, it will equal 0 (-1+1) and therefore the whole polynomial will equal 0. The same goes for all the other factors.
If a value, a, put into P(x) does not equal 0, then (x-a) IS NOT a factor.
The factor theorem is dead simple. If you have a polynomial x3 + 6x2 + 11x + 6, it has factors, namely a linear and a quadratic.
Only this time, the quadratic itself factorises, so that the cubic has 3 linear factors, namely (x+1)(x+2)(x+3).
If I now did something a bit sneaky, I could prove that these are factors of this cubic.
Watch. P(-1) = -13 + 6(-12) + 11(-1) + 6
= -1 + 6 - 11 + 6
= 0
When x=-1 the cubic equation = 0.
This proves that (x+1) is a factor of x3 + 6x2 + 11x + 6
In fact, if (x-a) is a factor of P(x) then P(a) = 0. Always and everywhere.
Why?
Well, look at (x+1)(x+2)(x+3).
These are all factors of x3 + 6x2 + 11x + 6.
So if you put x= -1 into the first factor, it will equal 0 (-1+1) and therefore the whole polynomial will equal 0. The same goes for all the other factors.
If a value, a, put into P(x) does not equal 0, then (x-a) IS NOT a factor.
Dividing Polynomials (1)
This can actually be a bit tricky but it's not impossible.
Any polynomial of order 3 - which is what you deal with at C1 - is the product of either linear factors, or a linear and a quadratic factor.
Finding a factor like this, you don't need to worry about remainders but these crop up soon enough, and we'll come onto them in due course.
A basic example of what I mean is: x(x2 + 2x - 4). Multiplying these two factors gives the cubic equation x3 + 2x2 - 4x.
But what do you do if you know one factor and the polynomial?
Say you know the linear factor to be (x-2) and the polynomial to be x3 + x2 - 11x + 10
The other factor must be a quadratic.
(x-2)(ax2+ bx + c) = x3 + x2 - 11x + 10
expand the two brackets as you would normally:
ax3 + bx2 + cx - 2ax2 -2bx - 2c
and then find common factors to tidy this expression up and make it usable:
= ax3 + (b-2a)x2 + (c-2b)x -2c
That bit is worth checking. You're simply collecting together the different values of x2 or whatever, as you would normally in an equation. You're only putting brackets round them because you don't know their values and you need to understand them as together making x2.
This expression is now completely identical to our original polynomial:
= ax3 + (b-2a)x2 + (c-2b)x -2c
x3 + x2 - 11x + 10
We now go through the fairly simple but again easy to cock up process of comparing the co-efficients. Co-efficients are of course the numbers in front of variables (3 is the coefficient of x in the expression 3x).
Comparing the co-efficients here means we look at our original polynomial and compare it with the unknown result - which we can see is a cubic - of our algebraic multiplication. The co-efficients in the known polynomial will tell us the missing co-efficients of our unknown quadratic.
In this case: a = 1
(b-2a) = 1
b - 2 = 1
b= 3
(c-2b) = -11
c-6 = -11
c = -5
We can check this last one by seeing -2c at the end of our algebraic expression. If c is right it will match -2 x -5 = 10, so yes it's right.
This process has given us the three co-efficients for our missing quadratic: 1, 3 and -5
so the two factors of our polynomial are (x-2) and (x2 + 3x -5).
Just be careful at this point: can the quadratic factor be factorised itself, to give three linear factors? Give it a quick b2 - 4ac test: 9 - (4x1x-5) = 9 - (-20) = 29. NO. It has to stay as a quadratic.
Often it will. You'll find that you're often - but not always - asked to find the three factors of p(x) or to factorise p(x) completely when this is the case.
Any polynomial of order 3 - which is what you deal with at C1 - is the product of either linear factors, or a linear and a quadratic factor.
Finding a factor like this, you don't need to worry about remainders but these crop up soon enough, and we'll come onto them in due course.
A basic example of what I mean is: x(x2 + 2x - 4). Multiplying these two factors gives the cubic equation x3 + 2x2 - 4x.
But what do you do if you know one factor and the polynomial?
Say you know the linear factor to be (x-2) and the polynomial to be x3 + x2 - 11x + 10
The other factor must be a quadratic.
(x-2)(ax2+ bx + c) = x3 + x2 - 11x + 10
expand the two brackets as you would normally:
ax3 + bx2 + cx - 2ax2 -2bx - 2c
and then find common factors to tidy this expression up and make it usable:
= ax3 + (b-2a)x2 + (c-2b)x -2c
That bit is worth checking. You're simply collecting together the different values of x2 or whatever, as you would normally in an equation. You're only putting brackets round them because you don't know their values and you need to understand them as together making x2.
This expression is now completely identical to our original polynomial:
= ax3 + (b-2a)x2 + (c-2b)x -2c
x3 + x2 - 11x + 10
We now go through the fairly simple but again easy to cock up process of comparing the co-efficients. Co-efficients are of course the numbers in front of variables (3 is the coefficient of x in the expression 3x).
Comparing the co-efficients here means we look at our original polynomial and compare it with the unknown result - which we can see is a cubic - of our algebraic multiplication. The co-efficients in the known polynomial will tell us the missing co-efficients of our unknown quadratic.
In this case: a = 1
(b-2a) = 1
b - 2 = 1
b= 3
(c-2b) = -11
c-6 = -11
c = -5
We can check this last one by seeing -2c at the end of our algebraic expression. If c is right it will match -2 x -5 = 10, so yes it's right.
This process has given us the three co-efficients for our missing quadratic: 1, 3 and -5
so the two factors of our polynomial are (x-2) and (x2 + 3x -5).
Just be careful at this point: can the quadratic factor be factorised itself, to give three linear factors? Give it a quick b2 - 4ac test: 9 - (4x1x-5) = 9 - (-20) = 29. NO. It has to stay as a quadratic.
Often it will. You'll find that you're often - but not always - asked to find the three factors of p(x) or to factorise p(x) completely when this is the case.
Thursday, 4 September 2008
Multiplying polynomials
This is fairly easy too: it is, at its basic level, expanding brackets, though you can make use of the distributive law of algebra to help you do it more systematically.
For example.
(x+2)(x-5) can either be expanded bit by bit (ac+ad+bc+bd) or you can write it like this: x(x-5) + 2(x-5). Instead of adding four separate products individually, you are adding two sets of them. It's basically the same thing, just a little more organised. You could also write it x(x+2) -5(x+2).
You can use this principle to help you multiply polynomials.
(x+9)(3x3 -4x2 + 3)
x(3x3 -4x2 + 3)
+ 9(3x3 -4x2 + 3)
= (3x4 - 4x3 + 3x)
+ (27x3 - 36x2 + 27)
using the knowledge of collecting polynomials from the previous post:
= 3x4 + 23x3 - 36x2 + 3x + 27.
It gets fiddlier than this but usually the problem is signs. Pay really close attention to them because they are an easy ways of losing lots of marks quickly.
For example.
(x+2)(x-5) can either be expanded bit by bit (ac+ad+bc+bd) or you can write it like this: x(x-5) + 2(x-5). Instead of adding four separate products individually, you are adding two sets of them. It's basically the same thing, just a little more organised. You could also write it x(x+2) -5(x+2).
You can use this principle to help you multiply polynomials.
(x+9)(3x3 -4x2 + 3)
x(3x3 -4x2 + 3)
+ 9(3x3 -4x2 + 3)
= (3x4 - 4x3 + 3x)
+ (27x3 - 36x2 + 27)
using the knowledge of collecting polynomials from the previous post:
= 3x4 + 23x3 - 36x2 + 3x + 27.
It gets fiddlier than this but usually the problem is signs. Pay really close attention to them because they are an easy ways of losing lots of marks quickly.
Collecting Polynomials
This is not difficult in principle, but it can become messy if you are not careful with your signs.
You can only add or subtract values with the same powers of x.
Be careful: this means that you CANNOT add 3x4 and 3x3 because they have different powers of x and are therefore completely different values.
You also CANNOT add 3x2 and 3r2 because they are different variables and so different values.
Here is a fairly basic example: add 3x3 + 5x to 2x3 -4x
It helps to collect the two expressions into brackets first:
(3x3 + 5x) + (2x3 -4x)
Now you can sort out the signs more easily:
3x3 + 5x + 2x3 -4x
As it happens there is no sort out needed here!
Now collect the like terms:
(3x3 + 2x3) + (5x - 4x)
= 5x3 + x
It does get a lot more fiddly than this sometimes:
(4x4 + 5x3 - 3x2) - (2x4 - 6x3 + x2 + 9)
But use the same principles. Here the two expressions are already in brackets so we can SORT THE SIGNS OUT:
4x4 + 5x3 - 3x2 - 2x4 + 6x3 - x2 - 9
What we've done here is simply say "all the values in the second polynomial are being subtracted and we all know what to do with a - and a + or a - and another -". We've effectively subtracted the two expressions here by reversing the signs in the second expression.
You might want to look back at that again.
OK. Now we can collect the like terms.
4x4 - 2x4 + 5x3 + 6x3 - 3x2 - x2 - 9
= 2x4 + 11x3 - 4x2 - 9
You can only add or subtract values with the same powers of x.
Be careful: this means that you CANNOT add 3x4 and 3x3 because they have different powers of x and are therefore completely different values.
You also CANNOT add 3x2 and 3r2 because they are different variables and so different values.
Here is a fairly basic example: add 3x3 + 5x to 2x3 -4x
It helps to collect the two expressions into brackets first:
(3x3 + 5x) + (2x3 -4x)
Now you can sort out the signs more easily:
3x3 + 5x + 2x3 -4x
As it happens there is no sort out needed here!
Now collect the like terms:
(3x3 + 2x3) + (5x - 4x)
= 5x3 + x
It does get a lot more fiddly than this sometimes:
(4x4 + 5x3 - 3x2) - (2x4 - 6x3 + x2 + 9)
But use the same principles. Here the two expressions are already in brackets so we can SORT THE SIGNS OUT:
4x4 + 5x3 - 3x2 - 2x4 + 6x3 - x2 - 9
What we've done here is simply say "all the values in the second polynomial are being subtracted and we all know what to do with a - and a + or a - and another -". We've effectively subtracted the two expressions here by reversing the signs in the second expression.
You might want to look back at that again.
OK. Now we can collect the like terms.
4x4 - 2x4 + 5x3 + 6x3 - 3x2 - x2 - 9
= 2x4 + 11x3 - 4x2 - 9
Polynomials - The Basics
I've got fed up with doing straight line geometry, as it needs diagrams, and I am having real difficulty drawing & importing them.
So to polynomials. Firstly, what is a polynomial? Well it is an expression involving numbers and powers of x. Such as:
2x + 1
3x2 + 4x - 6
2x4 - 4x3 + 2x2 +2
These are all polynomials. They are usually written, as here, in descending powers of x, from highest to lowest.
Sometimes they are called polynomials of degree x, depending on the highest power of x. So a cubic expression is a polynomial of degree 3, a quadratic of degree 2, a linear of degree 1 and a constant (ie just a number on its own) of degree 0 (because the highest power of x here is 0, ie x0 - ie 1).
Polynomials DO NOT have fractional or negative powers.
2x-2 + 1
is not a polynomial.
That's the basics of it.
So to polynomials. Firstly, what is a polynomial? Well it is an expression involving numbers and powers of x. Such as:
2x + 1
3x2 + 4x - 6
2x4 - 4x3 + 2x2 +2
These are all polynomials. They are usually written, as here, in descending powers of x, from highest to lowest.
Sometimes they are called polynomials of degree x, depending on the highest power of x. So a cubic expression is a polynomial of degree 3, a quadratic of degree 2, a linear of degree 1 and a constant (ie just a number on its own) of degree 0 (because the highest power of x here is 0, ie x0 - ie 1).
Polynomials DO NOT have fractional or negative powers.
2x-2 + 1
is not a polynomial.
That's the basics of it.
Friday, 29 August 2008
AS Level Update
I've finished Stats! Yay!
To be honest i've not enjoyed it all that much: too much punching numbers into a calculator for my liking. Why bother with the regression line equation at all when you can just press a button? I have found that this means the textbooks spend less time explaining the equations and their derivations, because they just want you to know it exists. The understanding - that elusive beast which is supposed to underpin school mathematics these days - is left to one side so you can get on with manipulating numbers.
I'm afraid there is no chance of any Stats posts for a while, as I intend to carry on with C1 stuff! But give me a few weeks and I might.
To be honest i've not enjoyed it all that much: too much punching numbers into a calculator for my liking. Why bother with the regression line equation at all when you can just press a button? I have found that this means the textbooks spend less time explaining the equations and their derivations, because they just want you to know it exists. The understanding - that elusive beast which is supposed to underpin school mathematics these days - is left to one side so you can get on with manipulating numbers.
I'm afraid there is no chance of any Stats posts for a while, as I intend to carry on with C1 stuff! But give me a few weeks and I might.
Tuesday, 26 August 2008
Summer Maths Reading Round-Up
Well summer is nearly over so it will be back to work soon (booo). I'm supposed to have been reading loads of children's books this summer but actually I've read a few maths books instead. I thought I'd round them up, in case anyone fancied expanding their horizon-brackets.
First off, textbooks. "A Concise Course in A Level Statistics" by Crawshaw and Chambers (4th ed). I bought this to back up my Stats module but it is far too detailed for AS students like me. I get the impression it isn't quite updated for the 2004 syllabus but it is certainly very detailed, full of useful exercises and depth. I don't think I'd advise it for AS Level but it might be handy if you were doing S2 as well.
I have a battered copy of the legendary Bostock and Chandler too, which you can get hold of very cheaply from Amazon or your local friendly maths teacher (there are loads of them around). This is very hard (later editions may be more idiot-friendly) and it progresses at light speed. Again, good for back up and for extra exercises.
And so to Popular Maths books. I get the impression that this is an infant genre, for the authors haven't really got the hang of the "popular" aspect. Popular science bedazzles its readers with accounts of billions of years of DNA multiplication, or the whizzing of quarks: the maths equivalents tend to err on the side of really hard equations and bizarre concepts that the reader is advised to just shut up and listen to and not worry about. This is probably due to the fact, which mathematicians are reluctant to admit, that maths is abstract and dry. This is why it is great, but they are desperate to prove that it is just as mind boggling as the end of the universe. Which, to be fair, it's not. That doesn't mean it's not amazing in its own right, just that it's not staggeringly mind-blowing. To fill the gaps, popular maths authors sometimes try and jazz things up with literary quotes, bullshit about poetry, or insights into the private lives of mathematicians.
Can I make a suggestion, guys? If I'm reading about maths, leave out the bloody poetry. I have tons of poetry books on my shelves and I don't expect to read garbage about cubic equations in books about TS Eliot so encountering the reverse is a pain in the arse. It really is.
Why not just focus on the austere beauty of maths? The elegance, the truthfulness, the logic?
A lot of this is an attack on Barry Mazur's book "Imagining Numbers", which is actually very well written. It just has pages of irrelevant speculation and sub-undergraduate waffle about the nature of poetry. I see what he is trying to do here but this is a guy who is one of the world's greatest mathematicians: there must be enough wonder in what he knows about maths without needing to prove that he's actually a sensitive artistic soul.
Also this book did my head in going on about imaginary numbers. I think you probably need to be brighter than me to get it, to be honest.
Having said that, the classic "Mathematics and the Imagination" by Kasner and Newman, of which I recently bought a dirt-cheap old puffin edition, explains i with much more economy and sense. This book is hard in places too, but is written with precisely the dry wit and occasional sarcastic humour (satirical even) that just needs to be sprinkled, rather than poured over this subject to make it generally entertaining. It's a great book. It rewards attention and promotes wonder, backing off at precisely the moment it knows it will lose its audience if it carries on. Genius.
"Mathematics Minus Fear" By Laurence Potter is a relaxingly easy read, aimed squarely at ordinary people cheesed off with arithmetic and probability: in short, school maths. There is even a proof of why you divide fractions by turning one of them upside down and multiplying. Frankly I can see why our teachers didn't try to explain that in primary school. This is a really classy book actually, with a wide target audience and some other stuff that is relevant and funny. Not deep enough to be a classic probably but useful and entertaining.
"E=mc2" by David Bodanis purports to be a biography of the equation but is, understandably, more about physics than maths. There are generous dollops of history too. Not bad actually and probably the first 60 or so pages are a close reading of the different items in the equation, a good explanation of the maths involved.
"one to Nine: the Inner Life of Numbers" by Andrew Hodges is a right royal pain in the arse and I would avoid it. When the hell will non-fiction writers get hold of this basic fact: WE DO NOT GIVE A TOSS ABOUT YOUR POLITICS, SO LEAVE THEM THE HELL OUT OF YOUR BOOKS. This book is "Maths the Guardian Way" and is cynical in its right-baiting irrelvances. With Labour 20 points behind in the polls it is breathtaking arrogance that leads some authors still to assume all readers are socialists.
Finally one I am still in the middle of: the memoirs of GH Hardy. Beautifully written, sensitive, thoughtful, mathematical, challenging. Lovely.
Disclaimer: My, erm "study" is a bit of a mess and so these book titles have been recovered from memory. If any are wrong i'll update later.
First off, textbooks. "A Concise Course in A Level Statistics" by Crawshaw and Chambers (4th ed). I bought this to back up my Stats module but it is far too detailed for AS students like me. I get the impression it isn't quite updated for the 2004 syllabus but it is certainly very detailed, full of useful exercises and depth. I don't think I'd advise it for AS Level but it might be handy if you were doing S2 as well.
I have a battered copy of the legendary Bostock and Chandler too, which you can get hold of very cheaply from Amazon or your local friendly maths teacher (there are loads of them around). This is very hard (later editions may be more idiot-friendly) and it progresses at light speed. Again, good for back up and for extra exercises.
And so to Popular Maths books. I get the impression that this is an infant genre, for the authors haven't really got the hang of the "popular" aspect. Popular science bedazzles its readers with accounts of billions of years of DNA multiplication, or the whizzing of quarks: the maths equivalents tend to err on the side of really hard equations and bizarre concepts that the reader is advised to just shut up and listen to and not worry about. This is probably due to the fact, which mathematicians are reluctant to admit, that maths is abstract and dry. This is why it is great, but they are desperate to prove that it is just as mind boggling as the end of the universe. Which, to be fair, it's not. That doesn't mean it's not amazing in its own right, just that it's not staggeringly mind-blowing. To fill the gaps, popular maths authors sometimes try and jazz things up with literary quotes, bullshit about poetry, or insights into the private lives of mathematicians.
Can I make a suggestion, guys? If I'm reading about maths, leave out the bloody poetry. I have tons of poetry books on my shelves and I don't expect to read garbage about cubic equations in books about TS Eliot so encountering the reverse is a pain in the arse. It really is.
Why not just focus on the austere beauty of maths? The elegance, the truthfulness, the logic?
A lot of this is an attack on Barry Mazur's book "Imagining Numbers", which is actually very well written. It just has pages of irrelevant speculation and sub-undergraduate waffle about the nature of poetry. I see what he is trying to do here but this is a guy who is one of the world's greatest mathematicians: there must be enough wonder in what he knows about maths without needing to prove that he's actually a sensitive artistic soul.
Also this book did my head in going on about imaginary numbers. I think you probably need to be brighter than me to get it, to be honest.
Having said that, the classic "Mathematics and the Imagination" by Kasner and Newman, of which I recently bought a dirt-cheap old puffin edition, explains i with much more economy and sense. This book is hard in places too, but is written with precisely the dry wit and occasional sarcastic humour (satirical even) that just needs to be sprinkled, rather than poured over this subject to make it generally entertaining. It's a great book. It rewards attention and promotes wonder, backing off at precisely the moment it knows it will lose its audience if it carries on. Genius.
"Mathematics Minus Fear" By Laurence Potter is a relaxingly easy read, aimed squarely at ordinary people cheesed off with arithmetic and probability: in short, school maths. There is even a proof of why you divide fractions by turning one of them upside down and multiplying. Frankly I can see why our teachers didn't try to explain that in primary school. This is a really classy book actually, with a wide target audience and some other stuff that is relevant and funny. Not deep enough to be a classic probably but useful and entertaining.
"E=mc2" by David Bodanis purports to be a biography of the equation but is, understandably, more about physics than maths. There are generous dollops of history too. Not bad actually and probably the first 60 or so pages are a close reading of the different items in the equation, a good explanation of the maths involved.
"one to Nine: the Inner Life of Numbers" by Andrew Hodges is a right royal pain in the arse and I would avoid it. When the hell will non-fiction writers get hold of this basic fact: WE DO NOT GIVE A TOSS ABOUT YOUR POLITICS, SO LEAVE THEM THE HELL OUT OF YOUR BOOKS. This book is "Maths the Guardian Way" and is cynical in its right-baiting irrelvances. With Labour 20 points behind in the polls it is breathtaking arrogance that leads some authors still to assume all readers are socialists.
Finally one I am still in the middle of: the memoirs of GH Hardy. Beautifully written, sensitive, thoughtful, mathematical, challenging. Lovely.
Disclaimer: My, erm "study" is a bit of a mess and so these book titles have been recovered from memory. If any are wrong i'll update later.
Monday, 25 August 2008
Gradients of Perpendicular Lines
If a line intersects another one such that the two are perpendicular,and the gradient of the other is known, it is a piece of cake to work out the unknown gradient.
The gradients of perpendicular lines have a product of -1.
Our line from the previous post has gradient 12-6/8-3 = 6/5 Its perpendicular bisector will therefore have gradient of (using the common letter m to stand for unknown gradient):
m1xm2=-1
6/5xm2 = -1
m2= -1 ÷ 6/5
At this point use the fraction division rule:
m2= -1 x 5/6
= -5/6
The gradient of the perpendicular bisector is -5/6.
The really cool bit is this. If you have two co-ordinates of the first line, say in the post below (3,6) and (8,12); then you can now easily find the equation of the perpendicular bisector.
If it bisects the first line it must pass through the mid-point: x1+x2/2, y1+y2/2
or 3+8/2, 6+12/2
which is: (5 1/2, 9)
Plug what you know of this line into y - y1 = m(x - x1).
Or: y - 9 = -5/6(x - 11/2)
I've turned 5 1/2 into 11/2 for ease of calculation.
y -9 = -1/3x + 55/12
y = -1/3x + 163/12
This is a bit messy so multiply through by 12:
12y = -4x + 163
Lovely.
NOTE: I updated this post because it was wrong: it is still a bit more tricky than most of the questions you will get at C1. Most of these type questions resolve into a line with a constant of no more than about 20. But theoretically one like this could come up.
The gradients of perpendicular lines have a product of -1.
Our line from the previous post has gradient 12-6/8-3 = 6/5 Its perpendicular bisector will therefore have gradient of (using the common letter m to stand for unknown gradient):
m1xm
6/5xm2 = -1
m2= -1 ÷ 6/5
At this point use the fraction division rule:
m2= -1 x 5/6
= -5/6
The gradient of the perpendicular bisector is -5/6.
The really cool bit is this. If you have two co-ordinates of the first line, say in the post below (3,6) and (8,12); then you can now easily find the equation of the perpendicular bisector.
If it bisects the first line it must pass through the mid-point: x1+x2/2, y1+y2/2
or 3+8/2, 6+12/2
which is: (5 1/2, 9)
Plug what you know of this line into y - y1 = m(x - x1).
Or: y - 9 = -5/6(x - 11/2)
I've turned 5 1/2 into 11/2 for ease of calculation.
y -9 = -1/3x + 55/12
y = -1/3x + 163/12
This is a bit messy so multiply through by 12:
12y = -4x + 163
Lovely.
NOTE: I updated this post because it was wrong: it is still a bit more tricky than most of the questions you will get at C1. Most of these type questions resolve into a line with a constant of no more than about 20. But theoretically one like this could come up.
Geometry of the Line (3): Length
It is a fairly simple job to calculate the length of a line. So far I've shown the mid-points, gradients, and equations. This is similar stuff, and based on the principle of Pythagoras, as for gradients.
Take a line with two points (3,6) and (8,12) lying on the line.
We can see any straight line on a graph as being equivalent to a hypotenuse of a right angled triangle.
Now Pythagoras says that the square of the hypotenuse is equal to the sum of the squares of the other two sides.
Looking at the diagram:
The line we want to measure is the hypotenuse and the other two sides are clearly horizontal and vertical. So the length of the horizontal line is easy - just x2 - x1 and the length of the vertical line the same but with ys. y2-y1
This is where Pythagoras comes in. We square the lengths of the two lines and add them together. This gives us the square of the hypotenuse.
We write it like this:
√(x2-x1)2 + (y2-y1)2
In this case:
√ (8-3)2 + (12-6)2
√ 25 +36
√ 61
This will be a surd so we can leave the answer as √61 for the purposes of C1.
Take a line with two points (3,6) and (8,12) lying on the line.
We can see any straight line on a graph as being equivalent to a hypotenuse of a right angled triangle.
Now Pythagoras says that the square of the hypotenuse is equal to the sum of the squares of the other two sides.
Looking at the diagram:
The line we want to measure is the hypotenuse and the other two sides are clearly horizontal and vertical. So the length of the horizontal line is easy - just x2 - x1 and the length of the vertical line the same but with ys. y2-y1
This is where Pythagoras comes in. We square the lengths of the two lines and add them together. This gives us the square of the hypotenuse.
We write it like this:
√(x2-x1)2 + (y2-y1)2
In this case:
√ (8-3)2 + (12-6)2
√ 25 +36
√ 61
This will be a surd so we can leave the answer as √61 for the purposes of C1.
Thursday, 21 August 2008
A Not Very Confident Interval
Ever had the feeling you just don't know what's going on?
I mean, that there you are, working and doing things, but every so often there's this tumbling sensation, like you are falling down a well, and there's a metallic taste in your mouth, and the world is slipping through your fingers and you just do not know what is going on?
No?
Oh well, this is how I feel about doing Stats. I mean plugging the data into the formulae is easy enough, and reading values off tables is alright...but as for what is really happening...
I get this odd feeling occasionally: it's a deep and throbbing pain in my nose and I fear my textbook is going to be showered in blood. But it never comes.
Still don't get it, though.
And when you're doing it for fun, what is the point if you cannot get it?
I mean, really?
I mean, that there you are, working and doing things, but every so often there's this tumbling sensation, like you are falling down a well, and there's a metallic taste in your mouth, and the world is slipping through your fingers and you just do not know what is going on?
No?
Oh well, this is how I feel about doing Stats. I mean plugging the data into the formulae is easy enough, and reading values off tables is alright...but as for what is really happening...
I get this odd feeling occasionally: it's a deep and throbbing pain in my nose and I fear my textbook is going to be showered in blood. But it never comes.
Still don't get it, though.
And when you're doing it for fun, what is the point if you cannot get it?
I mean, really?
Monday, 18 August 2008
Beware of Surds
Surds are right at the start of C1.
They are easy to forget.
As I found out tonight, trying a C1 practice paper.
Surds are basically this. They are an exact representation of irrational numbers, where a fraction or a decimal will not do. Normally we have rational numbers, like 1, 2, 3.5, 4.54, and so on. These numbers can be expressed as a fraction or as a decimal. But sometimes we get numbers like √2 or π which would go on forever if we tried to express them as a decimal (3.14159265 is only the start of π) so a decimal or a fraction would only ever give us an approximation.
1.41.... will never really describe √2 exactly, because it has an infinite number of digits after the decimal point. Apparently there is a proof of this but you don't need to prove it for C1.
So we leave these numbers expressed as a square root, or combination of square root and integer or fraction, and we call these surds (because 3 or 5 times the irrational square root of 2 will be irrational itself, usually). They enable us to do exact calculations of numbers which cannot be represented exactly as fractions or decimals.
There are a few rules of surds.
1) You can only add or subtract like surds. ie 3√7 - 2√7 = √7
2) You can multiply surds but there are a couple of rules to remember:
√5 x √6 = √30
3 x √5 = 3√5
5√3 x √4 = 5√12
3) Sometimes a surd will reveal itself to be a rational number: watch out for this:
√3 x √3 = 3
3√4 x 2√4 = 6√16 = 24
4) surds multiply out of brackets just like anything else does.
5)The trickiest thing about surds is fractions with a surd as a denominator. You need to get rid of the surd.
If you think about it, there is a way to make rational numbers out of surds that uses a rule of brackets. (a-b)(a+b) always gives you a2-b2.
So if you have 3+√5 as the denominator, multiply it by 3-√5 and you rationalise the denominator into 9-5 = 4.
Of course whatever you times the denominator by you need to times the numerator by.
This means you will often end up with your answer being a surd, but you will have eliminated the irrational denominator and so you will have been able to simplify your fraction.
I know I'm meant to be doing a series on geometry of the line but I keep getting distracted...
They are easy to forget.
As I found out tonight, trying a C1 practice paper.
Surds are basically this. They are an exact representation of irrational numbers, where a fraction or a decimal will not do. Normally we have rational numbers, like 1, 2, 3.5, 4.54, and so on. These numbers can be expressed as a fraction or as a decimal. But sometimes we get numbers like √2 or π which would go on forever if we tried to express them as a decimal (3.14159265 is only the start of π) so a decimal or a fraction would only ever give us an approximation.
1.41.... will never really describe √2 exactly, because it has an infinite number of digits after the decimal point. Apparently there is a proof of this but you don't need to prove it for C1.
So we leave these numbers expressed as a square root, or combination of square root and integer or fraction, and we call these surds (because 3 or 5 times the irrational square root of 2 will be irrational itself, usually). They enable us to do exact calculations of numbers which cannot be represented exactly as fractions or decimals.
There are a few rules of surds.
1) You can only add or subtract like surds. ie 3√7 - 2√7 = √7
2) You can multiply surds but there are a couple of rules to remember:
√5 x √6 = √30
3 x √5 = 3√5
5√3 x √4 = 5√12
3) Sometimes a surd will reveal itself to be a rational number: watch out for this:
√3 x √3 = 3
3√4 x 2√4 = 6√16 = 24
4) surds multiply out of brackets just like anything else does.
5)The trickiest thing about surds is fractions with a surd as a denominator. You need to get rid of the surd.
If you think about it, there is a way to make rational numbers out of surds that uses a rule of brackets. (a-b)(a+b) always gives you a2-b2.
So if you have 3+√5 as the denominator, multiply it by 3-√5 and you rationalise the denominator into 9-5 = 4.
Of course whatever you times the denominator by you need to times the numerator by.
This means you will often end up with your answer being a surd, but you will have eliminated the irrational denominator and so you will have been able to simplify your fraction.
I know I'm meant to be doing a series on geometry of the line but I keep getting distracted...
Sunday, 17 August 2008
Why Doesn't Completing the Square Always Work?
I was meant to be going to bed but then I saw that someone had googled the above question and come to me.
The honest answer is "it does always work" but I imagine the problem my putative reader has is that some quadratics are really inconsiderate and don't have real roots.
Well imaginary and complex numbers are Further Pure nowadays but I will hazard an answer.
A quadratic is lovely because it has its famous formula: -b +/- (sqt b2-4ac)/2a
The b2 - 4ac bit is called the discriminant and determines the kinds of solutions the equation can have.
I can't really draw a graph tonight but we see it in terms of the points the quadratic crosses the x-axis of the graph.
It goes a little bit like this:
b2 - 4ac = a square number (36, say) - quadratic factorises (ie can be reduced to (x+1)(x-2) say).
b2 - 4ac = +ve numbers, including fractions - quadratic cuts x-axis twice but can't factorise.
b2 - 4ac = 0 - quadratic hits x-axis once (x-axis is a tangent to quadratic)
b2 - 4ac = -ve number, quadratic does not touch x-axis at all.
If b2 - 4ac is less than 0 it means we need a square root of a negative number. At c1 we can't do that at all. Later on one learns that numbers which square to give -ve numbers are called imaginary numbers and have all kinds of key roles. For now we just don't really want them,
so for our A/S Level purposes the quadratic does not have real roots (ie the values would be complex numbers - an imaginary with a real number) and we simply cannot calculate it at all.
You just end up with a complex number statement like (-2+sqrt -1)/2 or something. As I say, probably great for FP1, no use for C1.
However, that does not mean that completing the square doesn't work - it can ALWAYS be used to solve a quadratic. It just means that it can produce imaginary numbers.
The honest answer is "it does always work" but I imagine the problem my putative reader has is that some quadratics are really inconsiderate and don't have real roots.
Well imaginary and complex numbers are Further Pure nowadays but I will hazard an answer.
A quadratic is lovely because it has its famous formula: -b +/- (sqt b2-4ac)/2a
The b2 - 4ac bit is called the discriminant and determines the kinds of solutions the equation can have.
I can't really draw a graph tonight but we see it in terms of the points the quadratic crosses the x-axis of the graph.
It goes a little bit like this:
b2 - 4ac = a square number (36, say) - quadratic factorises (ie can be reduced to (x+1)(x-2) say).
b2 - 4ac = +ve numbers, including fractions - quadratic cuts x-axis twice but can't factorise.
b2 - 4ac = 0 - quadratic hits x-axis once (x-axis is a tangent to quadratic)
b2 - 4ac = -ve number, quadratic does not touch x-axis at all.
If b2 - 4ac is less than 0 it means we need a square root of a negative number. At c1 we can't do that at all. Later on one learns that numbers which square to give -ve numbers are called imaginary numbers and have all kinds of key roles. For now we just don't really want them,
so for our A/S Level purposes the quadratic does not have real roots (ie the values would be complex numbers - an imaginary with a real number) and we simply cannot calculate it at all.
You just end up with a complex number statement like (-2+sqrt -1)/2 or something. As I say, probably great for FP1, no use for C1.
However, that does not mean that completing the square doesn't work - it can ALWAYS be used to solve a quadratic. It just means that it can produce imaginary numbers.
Politics Break & Completing The Square
well with the A Level results this week the annual debating on dumbing down continues. All I will say is that I can do A Level maths now and I couldn't in 1994: also that I can tell by the textbooks I have bought, borrowed and scrounged, that material has been moved, generally into higher modules over that time (complex numbers, trigonometry, series, and algebraic manipulation).
Having said that, you need, according to AQA's website, around 59/75 on C1 to get an A. This doesn't sound much but you can drop marks very easily and you don't have a lot of leeway.
I'll be signing up next month for C1 and S1 and possibly C2 in January so I can't really afford to make sarky comments about dumbing down to be honest. I don't fancy being bitten on the arse by an angry polynomial or a not very normal distribution.
By the way, for the benefit of one reader who googled completing the square "when it doesn't divide by two", it goes something like this.
3x2 + 2x -1 = 0
3[x2 + 2/3x] -1 = 0
3[(x+ 2/6)2 - 4/36] - 1 = 0
3(x + 1/3)2 -12/36 - 1 = 0
3(x + 1/3)2 = 4/3
Then you start to solve the completed quadratic:
(x + 1/3)2 = 4/9
x + 1/3 = +/- 2/3
x= 1/3
x= -1
You can check it alongside the quadratic equation if you like.
If you look at my fairly hasty working out you can see that whether the x2 term divides by anything isn't the point: you simply take out the co-efficient from the x2 term and divide the x term by it as well, as I've done here.
I will do more on completing the square soon but I wanted to see if I could answer that reader's query. I hope they have popped back to check!
Having said that, you need, according to AQA's website, around 59/75 on C1 to get an A. This doesn't sound much but you can drop marks very easily and you don't have a lot of leeway.
I'll be signing up next month for C1 and S1 and possibly C2 in January so I can't really afford to make sarky comments about dumbing down to be honest. I don't fancy being bitten on the arse by an angry polynomial or a not very normal distribution.
By the way, for the benefit of one reader who googled completing the square "when it doesn't divide by two", it goes something like this.
3x2 + 2x -1 = 0
3[x2 + 2/3x] -1 = 0
3[(x+ 2/6)2 - 4/36] - 1 = 0
3(x + 1/3)2 -12/36 - 1 = 0
3(x + 1/3)2 = 4/3
Then you start to solve the completed quadratic:
(x + 1/3)2 = 4/9
x + 1/3 = +/- 2/3
x= 1/3
x= -1
You can check it alongside the quadratic equation if you like.
If you look at my fairly hasty working out you can see that whether the x2 term divides by anything isn't the point: you simply take out the co-efficient from the x2 term and divide the x term by it as well, as I've done here.
I will do more on completing the square soon but I wanted to see if I could answer that reader's query. I hope they have popped back to check!
Thursday, 14 August 2008
Geometry of the Line (2) The Equation of the Line
The equation of the (straight) line isn't difficult. In fact it's dead easy.
The reason is this. Any curve or line has a statement that describes what happens to a y-co-ordinate when you plug in a given x-value. This is why we talk of f(x), because the y co-ordinate is basically the function of the x under certain conditions, which differ for various lines and curves. With this statement you have the potential to describe the co-ordinates of the curve or line completely knowing only the x-value.
The equation is simply the statement of what happens to the y co-ordinate when you select any x co-ordinate you like. It contains the gradient of the line within it, because clearly where the y-value is will depend on how steep the line is.
We often hear about y=mx + c , where m=gradient, and c= y-intercept (or where the line crosses the y-axis).
It is fine and dandy, to be sure.
But for MPC1 purposes the following equation is also useful, not least because solving it leads to y=mx+c.
(y - y1) = m(x - x1)
Where (x,y) is any, unknown point on the line and (x1, y1) is a known point.
For example: take a line passing through (2,4) and (6,10).
Firstly, work out m. 6/4. or 3/2.
Plug that into the equation: (y-4) = 3/2(x-6) or y-4 = 3/2x - 9 or y= 3/2x -5.
Suddenly you have your y-intercept (-5) and your y =mx+c.
The only thing is, you need the gradient, so you need to know two points on the line from the start.
The reason is this. Any curve or line has a statement that describes what happens to a y-co-ordinate when you plug in a given x-value. This is why we talk of f(x), because the y co-ordinate is basically the function of the x under certain conditions, which differ for various lines and curves. With this statement you have the potential to describe the co-ordinates of the curve or line completely knowing only the x-value.
The equation is simply the statement of what happens to the y co-ordinate when you select any x co-ordinate you like. It contains the gradient of the line within it, because clearly where the y-value is will depend on how steep the line is.
We often hear about y=mx + c , where m=gradient, and c= y-intercept (or where the line crosses the y-axis).
It is fine and dandy, to be sure.
But for MPC1 purposes the following equation is also useful, not least because solving it leads to y=mx+c.
(y - y1) = m(x - x1)
Where (x,y) is any, unknown point on the line and (x1, y1) is a known point.
For example: take a line passing through (2,4) and (6,10).
Firstly, work out m. 6/4. or 3/2.
Plug that into the equation: (y-4) = 3/2(x-6) or y-4 = 3/2x - 9 or y= 3/2x -5.
Suddenly you have your y-intercept (-5) and your y =mx+c.
The only thing is, you need the gradient, so you need to know two points on the line from the start.
Tuesday, 12 August 2008
Geometry of the Line (1) The MidPoint of the Line
Well I'm not sure that I'll bother to start with a definition of a line, although Barry Mazur, in his book that tied up my guts, Imagining Numbers, quotes Euclid: "one and only one line passes through any two distinct points on the plane". That makes sense to me.
For the purposes of this post the line will be assumed to be straight, and two dimensional. We are not going to muck about with curves and stuff.
So.
We have already seen that if you know two co-ordinates on the line you can calculate its gradient. This is very difficult to do with only one set of co-ordinates, unless you have a perpendicular line handy (which we don't, at the moment - that might come later).
If your line begins at 2,2 and ends at 8,4, as above, you need to think "How far has it gone x-wise and y-wise?" Well it has gone 6 units in the x direction and 2 in the y direction.
To find the midpoint of this line you can either just say well mid-way through a movement of 6 is 3, and mid-way through a movement of 2 is 1. So 2+3 = 5 (the x co-ordinate of your mid-point) and 2+1=3 (the y co-ordinate). Your midpoint is (5,3).
This isn't practical for a lot of lines which are (3 1/2, 5 1/2) and so on. So instead we use a simple formula:
The midpoint is, (x1 + x2)/2, (y1+y2)/2
So: (2+8)/2, (2+4)/2
Thus: 5, 3
The midpoint is (5,3), as above.
The midpoint is useful for all sorts of knotty problems involving perpendicular bisectors and stuff.
For the purposes of this post the line will be assumed to be straight, and two dimensional. We are not going to muck about with curves and stuff.
So.
We have already seen that if you know two co-ordinates on the line you can calculate its gradient. This is very difficult to do with only one set of co-ordinates, unless you have a perpendicular line handy (which we don't, at the moment - that might come later).
If your line begins at 2,2 and ends at 8,4, as above, you need to think "How far has it gone x-wise and y-wise?" Well it has gone 6 units in the x direction and 2 in the y direction.
To find the midpoint of this line you can either just say well mid-way through a movement of 6 is 3, and mid-way through a movement of 2 is 1. So 2+3 = 5 (the x co-ordinate of your mid-point) and 2+1=3 (the y co-ordinate). Your midpoint is (5,3).
This isn't practical for a lot of lines which are (3 1/2, 5 1/2) and so on. So instead we use a simple formula:
The midpoint is, (x1 + x2)/2, (y1+y2)/2
So: (2+8)/2, (2+4)/2
Thus: 5, 3
The midpoint is (5,3), as above.
The midpoint is useful for all sorts of knotty problems involving perpendicular bisectors and stuff.
Monday, 11 August 2008
Geometry of the Line
Writing yesterday's post about gradient has given me an idea for a series of posts, which really ought to have predated the dy/dx one, on the C1 basics of geometry of the line of which gradient is a part. So, although I won't do it tonight, look out for a couple of posts on: equation of a line, finding the midpoints, gradients of perpendicular lines and so on.
Yes I know it isn't that exciting but you know, it is quite interesting really.
Yes I know it isn't that exciting but you know, it is quite interesting really.
Sunday, 10 August 2008
Gradient
Hmm. I can see, from Matt's comment below, that I need to be a bit more specific. well it's early doors and though I'm giving it 110%, it's not always easy to see where I should pitch or focus this blog. I'm sure Matt will be delighted to know that I'm just going to let it evolve...
But a word on gradient.
Take a graph.
Any graph.
Put a straight line in it.
Any angle.
Take the left hand side of the line. What are its co-ordinates? In the picture above, the line begins at point (1,1). From that point, call it A, the line goes along and up. As you move along the line its co-ordinates rise in value. It moves along the x-axis and the y-axis.
Gradient is simply a measure of how y changes as you change x values. In other words, steepness.
Look at point B. It is at (5,4). From A to B you have gone 4 x-units along and 3 y-units up. On this line, wherever you are on the graph, if you go 4 x units along you will ALWAYS go 3 y units up.
Gradient, which is a general statement of this quantity, is calculated through this difference between the co-ordinates at A and B.
We divide the difference in y across the two points, by the difference in x.
y2-y1/x2-x1, meaning: the y co-ordinate of point B minus the y co-ordinate of point A, then divided by the same thing done for the x co-ordinates.
In this case,
= 4-1/5-1
= 3/4
The gradient is "three quarters".
In the picture below the line is slanting downwards. This will give it a negative gradient. Same method though.
Horizontal lines have a gradient of 0.
Vertical lines are a tricky one. If you think of a line going straight up and down, and try to apply this logic to it, you end up with a weird division.
Say you've got a vertical line through points (5,0) and up to (5, 5). If you do the y2-y1/x2-x1 calculation to find the gradient, you end up doing 5/0. You can't divide by 0, so you can't get a result. There isn't a gradient for a vertical line.
Another way of looking at it is that a vertical line cannot be a result of y = f(x) because it would be one-many - ie it's not a function.
But a word on gradient.
Take a graph.
Any graph.
Put a straight line in it.
Any angle.
Take the left hand side of the line. What are its co-ordinates? In the picture above, the line begins at point (1,1). From that point, call it A, the line goes along and up. As you move along the line its co-ordinates rise in value. It moves along the x-axis and the y-axis.
Gradient is simply a measure of how y changes as you change x values. In other words, steepness.
Look at point B. It is at (5,4). From A to B you have gone 4 x-units along and 3 y-units up. On this line, wherever you are on the graph, if you go 4 x units along you will ALWAYS go 3 y units up.
Gradient, which is a general statement of this quantity, is calculated through this difference between the co-ordinates at A and B.
We divide the difference in y across the two points, by the difference in x.
y2-y1/x2-x1, meaning: the y co-ordinate of point B minus the y co-ordinate of point A, then divided by the same thing done for the x co-ordinates.
In this case,
= 4-1/5-1
= 3/4
The gradient is "three quarters".
In the picture below the line is slanting downwards. This will give it a negative gradient. Same method though.
Horizontal lines have a gradient of 0.
Vertical lines are a tricky one. If you think of a line going straight up and down, and try to apply this logic to it, you end up with a weird division.
Say you've got a vertical line through points (5,0) and up to (5, 5). If you do the y2-y1/x2-x1 calculation to find the gradient, you end up doing 5/0. You can't divide by 0, so you can't get a result. There isn't a gradient for a vertical line.
Another way of looking at it is that a vertical line cannot be a result of y = f(x) because it would be one-many - ie it's not a function.
The dy/dx Post, Or, Differentiation for Dunderheads
Bear in mind that I know nothing at all beyond MPC1: so this post will only cover the basic basics of differentiation because that is all I know about it. Also PLEASE correct the inevitable errors. I'm not going to do the whole equation from first principles because that would just be confusing.
To kick off, a quote from Tom Freeman over at Matt's place:
In the first term, we started on calculus. I looked at the dy/dx stuff the teacher was drawing on the whiteboard, and I said...
"Don't the 'd's cancel?"
The simple answer is no. This is because when you see dy/dx you are not looking at a fraction, but a statement: the change in y with respect to the change in x. Tom knows this of course, but I didn't until a few weeks ago so it's worth pointing out!
Let us assume you have a curve. You want to find its gradient. But you can't because it's constantly changing, and besides, y2-y1/x2-x1 that you used for straight lines doesn't quite seem to cut it, what with all that mucking about going up and down and stuff.
Instead you approximate it by picking point A, which is where you want to know the gradient of the curve. Then you pick another point, B which is some distance off from A on the curve. So the coordinates of A are (x,y) and of B, (x+n, y+p) where n and p are the differences between the two sets of x and y co-ordinates. If you've started at A, and gone on to B, B's coordinates will be a + the increase, and y + the increase.
You draw a line to join A and B, a chord. You can measure the gradient of the chord, obviously (y2-y1/x2-x1). It's a bit ineaxct but it gives an approximation of the gradient of the curve between the two points and so you're nearish to the gradient at A.
But if you think about it, it would be a closer approximation if the line was closer to A...and then closer still and then closer and then closer...
Eventually you'd have such a small line that the gradient of it would be the gradient at point A plus not very much at all, ie a really good approximation of it.
So all along your gradient has been, say, 4 + f, where f is the increase in the gradient, which gets smaller and smaller as you get closer to point A. At point A your gradient is 4 (no increase). This is also the gradient of the tangent at point A.
This is differentiation! Chopping the distance between two points on a curve into smaller and smaller bits until you get invisibly close to point A, to get the gradient at Point A. It's a pain in the neck though to do this all the time, or to draw an infinite amount of tangents with gradients of their own, so we have dy/dx to do it for us.
This is the essence of the formula they give you at A Level. I said I wasn't going to do the proof, because it is a bit confusing; but it's worth looking up because it helps you understand why you are doing this stuff.
What stuff?
Well, say you have the curve y=x2. Dy/dx, the difference in y with respect to the difference in x (ie the gradient, for that is what it means), is 2x.
Reduce the power by one, and put it in front of the x. Piece of the proverbial.
What about the curve y=3x2?
Multiply the power by the number in front of the x, (6), then reduce the power as before. 6x. 6x is the gradient of the curve y=3x2.
Easy!!
But what if there's no power? And you've just got 5x? Well 5x is really 5x1. So you bring the power down as before and multiply, which makes 5x0. Stuff to the 0 is generally 1. So it's 5 times 1, so it's just 5.
If there's no power of the x, get rid of the x!! And you're done!
What if it's a constant? ie 8? Well constants always differentiate to make 0. Think of 8 as 8x0. Bring the 0 down and x it by 8, ie 0, and you get 0x-1. Ie NOWT!
But this is still a gradient formula, not the gradient itself. Well, this changes at every point on the curve, obviously. So you input the x co-ordinate of the point you want. What is the gradient at (2,3) on the curve y=3x2? Well dy/dx gives us 6x, and so put your x co-ordinate into this. 6 x 2 = 12. The gradient at (2,3) is 12.
I'll leave this for now on the question of what to do with a polynomial. The curve y= 3x4 - 2x3 + 6x2 - 3x + 4
you just differentiate bit by bit:
dy/dx = 12x3 - 6x2 + 12x - 3
Does anyone know how to do superscript in blogger? It would really help with these squares!!! (UPDATE: Yay! I've worked it out!)
Here's the proof for anyone who cares. It's come out a bit small but it's here! Click if you're bothered.
To kick off, a quote from Tom Freeman over at Matt's place:
In the first term, we started on calculus. I looked at the dy/dx stuff the teacher was drawing on the whiteboard, and I said...
"Don't the 'd's cancel?"
The simple answer is no. This is because when you see dy/dx you are not looking at a fraction, but a statement: the change in y with respect to the change in x. Tom knows this of course, but I didn't until a few weeks ago so it's worth pointing out!
Let us assume you have a curve. You want to find its gradient. But you can't because it's constantly changing, and besides, y2-y1/x2-x1 that you used for straight lines doesn't quite seem to cut it, what with all that mucking about going up and down and stuff.
Instead you approximate it by picking point A, which is where you want to know the gradient of the curve. Then you pick another point, B which is some distance off from A on the curve. So the coordinates of A are (x,y) and of B, (x+n, y+p) where n and p are the differences between the two sets of x and y co-ordinates. If you've started at A, and gone on to B, B's coordinates will be a + the increase, and y + the increase.
You draw a line to join A and B, a chord. You can measure the gradient of the chord, obviously (y2-y1/x2-x1). It's a bit ineaxct but it gives an approximation of the gradient of the curve between the two points and so you're nearish to the gradient at A.
But if you think about it, it would be a closer approximation if the line was closer to A...and then closer still and then closer and then closer...
Eventually you'd have such a small line that the gradient of it would be the gradient at point A plus not very much at all, ie a really good approximation of it.
So all along your gradient has been, say, 4 + f, where f is the increase in the gradient, which gets smaller and smaller as you get closer to point A. At point A your gradient is 4 (no increase). This is also the gradient of the tangent at point A.
This is differentiation! Chopping the distance between two points on a curve into smaller and smaller bits until you get invisibly close to point A, to get the gradient at Point A. It's a pain in the neck though to do this all the time, or to draw an infinite amount of tangents with gradients of their own, so we have dy/dx to do it for us.
This is the essence of the formula they give you at A Level. I said I wasn't going to do the proof, because it is a bit confusing; but it's worth looking up because it helps you understand why you are doing this stuff.
What stuff?
Well, say you have the curve y=x2. Dy/dx, the difference in y with respect to the difference in x (ie the gradient, for that is what it means), is 2x.
Reduce the power by one, and put it in front of the x. Piece of the proverbial.
What about the curve y=3x2?
Multiply the power by the number in front of the x, (6), then reduce the power as before. 6x. 6x is the gradient of the curve y=3x2.
Easy!!
But what if there's no power? And you've just got 5x? Well 5x is really 5x1. So you bring the power down as before and multiply, which makes 5x0. Stuff to the 0 is generally 1. So it's 5 times 1, so it's just 5.
If there's no power of the x, get rid of the x!! And you're done!
What if it's a constant? ie 8? Well constants always differentiate to make 0. Think of 8 as 8x0. Bring the 0 down and x it by 8, ie 0, and you get 0x-1. Ie NOWT!
But this is still a gradient formula, not the gradient itself. Well, this changes at every point on the curve, obviously. So you input the x co-ordinate of the point you want. What is the gradient at (2,3) on the curve y=3x2? Well dy/dx gives us 6x, and so put your x co-ordinate into this. 6 x 2 = 12. The gradient at (2,3) is 12.
I'll leave this for now on the question of what to do with a polynomial. The curve y= 3x4 - 2x3 + 6x2 - 3x + 4
you just differentiate bit by bit:
dy/dx = 12x3 - 6x2 + 12x - 3
Does anyone know how to do superscript in blogger? It would really help with these squares!!! (UPDATE: Yay! I've worked it out!)
Here's the proof for anyone who cares. It's come out a bit small but it's here! Click if you're bothered.
Thursday, 7 August 2008
Statistical Trials
As I've mentioned over at TTD, I recently finished MPC1 (the first A Level module) and I enjoyed it greatly, though some things I have a less than complete handle on (any question that begins "Determine the condition on k" for example), and I fear my algebra isn't quite up to scratch. But I am struggling a bit with S1. I find it frustrating that the textbook does not try to provide proofs for several equations and that clear, detailed explanations are hard to come by. I also find reading values off distribution tables a pain. I simply don't have the feeling that I am understanding what is going on.
Here are my thoughts on the binomial distribution: this is a distribution of probabilities, when there are only two possible outcomes, which are independent, with a fixed number of trials and a fixed probability. Like tossing a coin x number of times. I can't think, off the top of my head, of lots of applications of this. Binomial distributions are determined by combinations of probabilities (ie if you are looking for the probability of finding x widgets that fails a test if 20% of widgets fail and you have a box of 200 widgets and you pick out 35, then it is 200C35 because you could easily have loads of different ways of finding them, ie the first is ok, and the next isn't, etc etc). The 200C35 would then be multiplied by the probabilities of success and failure, each raised to the power of x and 1-x.
Or something.
The normal distribution though is for continuous data and is based around a probability curve with the peak being the mean. But then you sort of turn it into a standard normal distribution with mean 0 to gauge the probabilities more easily. I'm not really sure of the logic of this to be honest. The normal distribution gives you probabilities of things like heights across a population. The curve is deviation from the mean - standard deviation, then, yes? Or not? So what is it we're really looking for on this curve?
Here are my thoughts on the binomial distribution: this is a distribution of probabilities, when there are only two possible outcomes, which are independent, with a fixed number of trials and a fixed probability. Like tossing a coin x number of times. I can't think, off the top of my head, of lots of applications of this. Binomial distributions are determined by combinations of probabilities (ie if you are looking for the probability of finding x widgets that fails a test if 20% of widgets fail and you have a box of 200 widgets and you pick out 35, then it is 200C35 because you could easily have loads of different ways of finding them, ie the first is ok, and the next isn't, etc etc). The 200C35 would then be multiplied by the probabilities of success and failure, each raised to the power of x and 1-x.
Or something.
The normal distribution though is for continuous data and is based around a probability curve with the peak being the mean. But then you sort of turn it into a standard normal distribution with mean 0 to gauge the probabilities more easily. I'm not really sure of the logic of this to be honest. The normal distribution gives you probabilities of things like heights across a population. The curve is deviation from the mean - standard deviation, then, yes? Or not? So what is it we're really looking for on this curve?
Wednesday, 6 August 2008
Graphic
Yes I know the clumsy graphic I've chosen is nothing to do with completing the square as such...it is, of course differentiating the curve of y=f(x) (though it'll be a -x2 curve) from first principles, demonstrating how you find the gradient at A by steadily chopping smaller and smaller slices of the distance between A and B until the difference tends to 0 (ie at A). You've therefore used straight line gradients to find a pretty good approximation of the gradient at A (and therefore at any point on the curve).
Oooh, I think I am going to have to be careful here..I don't feel quite so....dogmatic as usual.
UPDATE: actually, looking at it, it seems to be a sort of cubic graph thingy in a positive x stylee. OR it might be something else. This doesn't really affect anything, you can still take dy/dx to find the gradient at any point.
I really do feel on eggshells here. Wow, this is weird.
Oooh, I think I am going to have to be careful here..I don't feel quite so....dogmatic as usual.
UPDATE: actually, looking at it, it seems to be a sort of cubic graph thingy in a positive x stylee. OR it might be something else. This doesn't really affect anything, you can still take dy/dx to find the gradient at any point.
I really do feel on eggshells here. Wow, this is weird.
Square Completed
I thought that, given that I am starting to bore the pants off my remaining reader with maths, that I would start another blog - a maths one, in order to ponder, meander, bore and generally think about maths.
As you may or may not care, I am studying A Level Maths: by profession I am an English and primary teacher. I have had "issues" with maths since around 1986, when a few doses of being ritually humilaiated in front of the class, being told to do weird things like "borrowing" and having masses of hard fractions to do cocked everything up - until then I had always been top of the class in maths and English. I had a fantastic Maths teacher for GCSE and, for a while, for AS Maths - but eventually I had to give it up as I was trying for Oxford, plus it was too hard.
So, this year giving me this year's stuff, I decided to take up the maths again and see if I could be brighter than I was at 18, or as bright as I was at 9, or just not completely crap at maths.
Please be warned that if you are coming here looking for expert comment or knowledgeable input, you are in the wrong place. I know next to nothing at all about maths and I am trying to get some sort of handle on it at A Level: I claim no authority of any kind. Anything I write here comes with the caveat "I know nothing at all about maths".
Unlike over at TTD of course...
As you may or may not care, I am studying A Level Maths: by profession I am an English and primary teacher. I have had "issues" with maths since around 1986, when a few doses of being ritually humilaiated in front of the class, being told to do weird things like "borrowing" and having masses of hard fractions to do cocked everything up - until then I had always been top of the class in maths and English. I had a fantastic Maths teacher for GCSE and, for a while, for AS Maths - but eventually I had to give it up as I was trying for Oxford, plus it was too hard.
So, this year giving me this year's stuff, I decided to take up the maths again and see if I could be brighter than I was at 18, or as bright as I was at 9, or just not completely crap at maths.
Please be warned that if you are coming here looking for expert comment or knowledgeable input, you are in the wrong place. I know next to nothing at all about maths and I am trying to get some sort of handle on it at A Level: I claim no authority of any kind. Anything I write here comes with the caveat "I know nothing at all about maths".
Unlike over at TTD of course...
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