Completing The Square

Housekeeping

2011-05-02T11:07:00.001+01:00

I've been tidying and revising some of my older posts. Some errors have been corrected, some bad writing made less awful.

However, although it's blog etiquette to notify your readers of when and how you have updated your posts, I'm not doing that because it will take too long and I cannot be bothered.

Do feel free to point out errors that are still obscuring the light of truth...

Completing the Square Redux

2011-05-01T22:54:00.002+01:00

An amazing site which explains completing the square much better than I ever could:

http://www.mathsisfun.com/algebra/completing-square.html

The Modulus Function

2011-05-01T16:07:00.002+01:00

Put very basically, the modulus function finds the absolute value of a number.

It is indicated by a pair of straight brackets around an x or a part of an equation.

It just means that -x when operated on by the modulus function becomes x while x stays as x.

It ignores negative values of x, or of the statement in x, such as (x-2).

Transformation of Graphs, contd

2011-05-01T16:02:00.002+01:00

When you are doing composite transformation of graphs, as you will at C3 and presumably C4 level, you do need to remember that sometimes the order of transformation counts. I have to admit to struggling with this (I don't have a tutor right now) so forgive me if I fail to offer a convincing explanation. But it's something like this:

If you are stretching in the x-direction, that comes BEFORE any translation.

If you are stretching in the y-direction, that comes AFTER any translation.

That's what I've picked up from getting lots of answers wrong, anyhow. I think it's to do with the way that a y-stretch affects the whole equation, while an x-stretch affects anything directly attached to the x part of the equation.

Some websites do suggest that you start at the innermost point of the equation and then work out - I don't know if this works for all composite transformations. I doubt it if what I've written above is right.

More LaTeX Experiments: Transformation of Graphs

2011-04-26T17:06:00.004+01:00

If you take a graph and transform it in various ways, here is what you get:

A reflection in the y-axis turns a graph of y=f(x) into the graph of y= f(-x).

A reflection in the x-axis turns a graph of y=f(x) into the graph of y= -f(x)

- note the difference -

A stretch of factor s in the y- direction turns the graph of y= f(x) into the graph of

y = s f(x).

A stretch of factor m in the x-direction turns the graph of y = f(x) into the graph of \[y = f(\frac{x}{m})\]

A translation by \[ \left( \frac{k}{l} \right) \] turns the graph of y = f(x) into the graph of y = f(x-k) + l

Another Test

2011-04-26T14:07:00.003+01:00

I'm trying to see if I can use LaTeX to write formulae onto the blog, but I've always been a bit befuddled by it. Instead I've installed a javascript thing from www.watchmath.com and that's supposed to recognise code and turn it into proper maths.

Well here goes.

Let's have a look at the function f(x) = 1/x. By rights in LaTeX that should be \[ \frac{1}{x} \]

Test

2011-04-26T14:04:00.001+01:00

\[ E(\mathbf{x}) = \sum_{i \in \mathcal{V}} \theta_i(x_i) + \sum_{ij \in \mathcal{E}} \theta_{ij}(x_i, x_j) \]

C3 Functions Round-Up

2011-04-26T12:09:00.002+01:00

Reminder: this is really about AQA C3, as OCR and Edexcel are slightly different. They don't change the truths of mathematics though.

After quite a long break, I've started doing maths again. I've forgotten chunks of it, as you would expect. Here are some things I've forgotten (on the assumption that they're generally forgettable):

1. Inverse functions are simply functions reversed. The domain becomes the range and vice versa. However, not all functions can have an inverse. If a function is many-one (as quadratic functions generally are) then they can't have an inverse - it would be one-many, which is not a function at all.

2. To get an inverse function you can do two things. If x appears only once in the original function, then you can do a reverse flow chart, reversing the order and the operations. BUT BEWARE OF SELF INVERSE FUNCTIONS - if you meet a subtraction from or a divide into, those operations stay the same because they're self-inverse functions. IF x appears more than once you need to do some equation-fiddling. On a function, x is a/the domain value(s). y is the same as f(x) - ie it's the range of the function. So by re-arranging the equation to isolate x, you're turning the function INSIDE OUT and making x the f(x) instead of the variable at the beginning (the domain).

3. composite functions are two functions put together. They're usually written as fg(x). What this means is you do both operations. You write out g(x) first, say it's x² + 4. Then you put it into f(x), which might be 2x - 3. The resulting composite function is 2(x²+4)-3.

4. On some functions, there are values that cannot be defined and that affects the range as well as the domain. Take f(x) = 1/x for example. X cannot be 0. But this means there are going to be impossible values of f(x) as well. A good way of estimating this is to draw the graph of f(x) on your calculator!! Always use the calculator wherever possible, even if you need to sketch the graph by more formal methods.

Polynomials

2011-04-06T15:38:00.000+01:00

These are a very common sight in almost any study of mathematics. It’s worth taking a moment to define them properly and think about how they can be used.

A polynomial is a mathematical expression involving a whole, positive power of x. The type of polynomial depends on the highest power of x in the expression. The different bits of the expression are known as terms, and the polynomial as a whole is the sum of the terms it contains.

Take a quadratic expression: 4x² + 2x - 2. It has three powers of x: 2, 1 and 0. These are whole numbers and positive (except 0). This is therefore a polynomial. The same applies to cubic expressions: x³ + 4x² + 5x + 2 is a polynomial too.

In fact, the same applies, whatever the power of x, as long as it is not a fractional or negative number. A polynomial does not need to have several terms, like most cubics and quadratics do. So, 1 is also a polynomial, being a term in x raised to the power of zero. The same applies to any constant (ie any number). 4x + 1 is a polynomial too, as x is x raised to the power of one.

Polynomials are written in descending powers of x, though they do not have to be.

The numbers in front of the x variables are called coefficients, and can take any value, including fractional or negative values, as can constants.

The highest power of x in a polynomial defines the degree of that polynomial. So 1 is a polynomial of degree 0, because the highest power of x here is 0. 4x + 1 is a polynomial of degree 1, and quadratics are polynomials of degree 2.

Polynomials can be created by expanding brackets. (x+2)² gives x² + 4x + 4 when expanded. In fact, both expressions here are polynomials. If the brackets were to give us an expression with a fractional power, then of course this would not be the case.

When you make a polynomial like a quadratic equal to 0, then you have a polynomial equation. In the case of a quadratic, such as 4x² + 2x – 2, the resulting equation is usually written as: 4x² + 2x – 2 = 0. As an equation, it is now a description of a parabolic curve which intersects the x-axis at (-1,0) and (½, 0).

Polynomials can be added, subtracted, multiplied and divided. The Factor and Remainder Theorems can be used to find out additional information about them. For example, if a number assigned to the variable x causes the polynomial to equal zero, that number is a factor of the polynomial (a root, a value of x where the curve of the equation crosses the x-axis). So we can demonstrate that -1 is a factor of 4x² + 2x – 2 by putting it in place of x. 4(-1)² + (-1x2) – 2 = 0. Therefore -1 is a factor of this polynomial. This is the essence of the Factor Theorem. The Remainder Theorem is a little more involved but states that if we divide the polynomial by x-a then the remainder is f(a) – in other words, the remainder is the sum of the polynomial when a is put in place of the variable x.

These are the absolute basics. Of course it does get a lot more complicated! But it is worth familiarising yourself with these principles before moving on.

What is Multiplication?

2010-01-28T11:47:00.003Z

Can be a tricky question, this.

Primary schoolchildren are taught that it is repeated addition, which makes a lot of sense.

4 x 4 =

4 + 4 + 4 + 4 =

16

Four lots of four/groups of/sets of.

But multiplication is also a scaling quantity. It ratchets things up massively quickly. If you type 2x2 into your calculator, then keep multiplying every answer by 2, the calculator will quickly run out of digits. To compare this with addition misses the point that multiplication is the centre of all geometric operations.

It does through the squaring and cubing and so on effect. If you take four centimetres, and for each of these four, you add another four, you get 4 x 4. It is known as squaring because the Greeks used to see it geometrically as the way to find the area of a square. For each centimetre across, there are four up (and vice versa).

x² is a powerful, recurring idea, which has its role in pretty much the entire universe. (E = mc²)

It's also worth noting the effect of multiplication by a fractional quantity.

4 x 1/4 will obviously increase the fraction, but it will have a decreasing effect on the whole number. Unlike multiplication by two whole numbers, where the answer is greater than either, in this operation, the answer will always be smaller than one of the inputted values.

Tips For Revising for a Module

2010-01-19T21:30:00.003Z

It's exam season: in fact, it's just gone. So many of us will be turning our thoughts to retakes - or in my case, to reapplying for exams we had to drop out of.

So how do you study for an A Level maths exam?

1) Don't necessarily do all the questions. For some, like me, it's essential to do every question in the book and all the past papers. For many, that's just repetition. You need to do the first few and the last few. Past papers are good to study as well as to do, to work out which questions often come up.

2) Make sure your calculator has new batteries. Yes - graphical calculator batteries do NOT last for years!

3) MAke sure you know the formulae which are NOT in the formula book for the module you are doing. There are usually several really common ones - like trig identities for example. Check each chapter in the textbook to see which formulae you need. It is easy to forget these, especially if you have been *ahem* working on the same module a while.

4) DON'T do anything the night before. The formulae will start to jump around in your head.

5) Revise the previous modules a little. You will find some identities and formulae you studied before that are not explicitly repeated come in handy. For example, the cosine and sine rules from C2 can be very handy when doing the questions on reciprocal trig functions in C3. Also the rules and techniques of radians are very useful indeed, though quite a bit of practice on this is done through C3.

6)If you do past papers and you can't get a tutor to mark them for you, you can do it yourself via the online answer papers which are published. It's not an exact science because you are not amazing at maths, unlike the markers, which is why you're studying it now. So BE HARSH on yourself and be conservative. Markers will often be generous with follow-through marks so you can at least be prepared. That's the pessimistic side of me coming through.

e (2.71828)

2010-01-18T19:45:00.004Z

No, this is not a post for luvved-up early 90s teenagers.

e is a bizarrely cool number, sometimes known as the natural logarithm.

Like π it is an irrational number - ie it cannot be expressed exactly as the ratio of two whole numbers (and thus as a fraction).

How do we get it then?

Imagine an exponential curve. A curve that is n^x. Say, y= 3^x.

As usual, the gradient will generally differ as x differs.

Is there a curve which has a gradient of 1, where x=0? There is, and, coolly, this curve has a gradient of 1 where x=0 and ALSO passes through y=1, ie the value of the function is 1.

This curve is y = e ^x. It's not just at y=1 where the value of the function equals the gradient, it's every point on the curve.

As you can imagine, that makes differentiating with e easy...

There is an excellent page here which shows you some interesting examples of e in action.

Reciprocal Trig Functions

2010-01-18T19:32:00.003Z

Sine, cosine and tangent are functions. You input a value - an angle in radians, or degrees, and for each angle there is a value. They are repeating functions.

They also have reciprocal functions.

The reciprocal of sine x is cosec x and is (1/sin x)

The reciprocal of cosine x is secant x and is (1/cos x)

The reciprocal of tan x is cotangent x (cot x) and is (1/tan x).

The Self-Inverse Function

2009-11-30T20:48:00.003Z

This sounds really weird or complex but it's not.

Basically a self-inverse function is just a function that gives you the same answer when you do the function to the answer.

Wot?

Alright. Say you have a function f(x) = 4-x

Say x = 2

4-2 = 2.

Now apply the answer to the function

4-2 = 2.

Same answer.

Try again f(x) - 7-x

Say x - 5.

7-5=2.

7-2=5

We have x again.

And f(x) - 10-x

x=4

10-4 = 6

Put 6 into it:

10-6=4.

Brings us back to our starting point with x.

When you do the operation twice, finding the function of the answer, you get your starting value of x.
Reciprocals always give self-inverses too.

When Can an Inverse Function Exist?

2009-11-30T20:41:00.002Z

You can't always have an inverse function (See previous post or previous but one or thereabouts).

A function is defined as any mapping which is one-one or many-one. This means that for any input value a unique output value is generated.

Something like y= x ² is a many-one function. 2² and -2² give the same value.

You can always draw a graph of a function.

A function CAN'T be a one-many mapping. It doesn't make a lot of sense to most of us to have some kind of operation which could generate loads of different answers for exactly the same input.

So.

If a function would generate a one-many correspondence, it follows that it's not a function. So a many-one function CANNOT have an inverse, because its inverse would be one-many (ie not a function at all).

Only one-one type functions have inverse functions.

C1 Advice

2009-11-19T14:35:00.002Z

Hmmm.

I was bored so I thought I'd offer a little advice for anyone revising for C1. As it is a non-calculator paper, you need to consider a few things.

1) Your mental arithmetic should be good
2) You should know square and cube numbers up to and including 5³
3) You should be totally au fait handling surds. This is an early part of C1 and you might have forgotten it by the time of the exam. There are some fiddly rules surrounding eliminating surds from equations so learn 'em.
4) You should remember to x stuff by -1 to get rid of unwanted negative numbers in your answers.
5) You need to be good at fractions, including cancelling down.
6) You also should be confident expanding brackets (and other GCSE stuff). This might be annoying if you did GCSE some years ago, like myself. It's worth buying a GCSE revision textbook and looking the Higher level stuff up.
7) Because of the non-calculator thing, the answers to C1 questions are generally nice, like 3 and 5 and 2. If you work something out not in surd form and it is a bizarre fraction a good rule of thumb is to go back and check your working.

Functions II - The Inverse

2009-11-02T15:40:00.005Z

You can reverse a function - not always, to give you what you started with.

This is known as the inverse function

Take y=x²

This function, or y=f(x), will give 4 for the input of 2.

So what will give us 2, from an input of 4?

It would be the opposite, the inverse - x = √y

This is sometimes written as f^-1.

It does get more complicated but that is the idiot's version.

A Recommendation

2009-11-02T13:37:00.003Z

Normally I would not advise students of maths under degree level to go anywhere near Wikipedia. It's not the reliability issue as such, it's that it gets very complicated very quickly.

An exception is the page on Trig. It gives some useful basics and some really handy animations which illustrate the key trigonmetric functions.

Functions

2009-11-01T08:09:00.002Z

A simple definition of a function is an operation which, for every value put in, will give a single value. It can give the same answer for different inputs. y=x² will give the answer 4 when both 2 and -2 are put in, but is still a function.

What a function can't do is give multiple answers to the same inputted value. A graph that appears to show this is not the graph of a function.

Geometric Series Part II

2009-10-31T13:10:00.003Z

Because timesing is a funny operation, a scaling up or down operation - repeated addition or however you like to look at it, it does strange things to series.

A common ratio of 2, like in the previous post, gives you a series which just keeps getting more and more massive, into infinity. It therefore has no sum - because it just keeps going on. You can find the sum of the first n terms of course.

But what if you had a common ratio of 1? Well then your series stays exactly the same

2,2,2,2,2,2,2,2,2,.....into infinity. It's the same sort of thing as above. Finding a complete sum of the series isnot going to be possible, because there will simply be an infinite load of 2s.

But if your common ratio is a fraction:

1/2, 1/4, 1/8, 1/16

Here the common ratio is 1/2. The terms of this sequence get smaller all the time, but will never reach 0.

When we think of the sequence properly as a series:

1/2 + 1/4 + 1/16 + 1/32 + 1/64

Then we can see that it is an infinite series, and will go on halving ad infinitum.
If we add the first few terms, we get 1/2, 3/4, 13/16, 27/32....

The sum is getting larger each time but by progressively smaller amounts.

The sum is getting closer and closer to 1, but without ever quite reaching it. If it had an infinite number of terms, then its sum would indeed reach 1. Its sum to infinity is 1.

It converges on 1, and is therefore called a convergent series.

This happens with certain types of common ratio, which we will come back to.

Geometric Series

2009-10-31T13:00:00.002Z

These are a bit like arithmetic series but instead of there being a common difference between terms there is a common power-type difference.

Consider 2,4,8,16,32,64...

This series appears to double each time, which means that there is no common number which gives you the nth term when added to the n-1th term.

But the common difference is "doubling" - or rather, powers of 2.

The series goes 2¹, 2², 2³, 2⁴.....

With geometric series, therefore, we don't talk about a common difference, but a common ratio - the thing you times each term by to get the next one. So this series has common ratio 2.

Sum of an Arithmetic Series

2009-10-30T10:01:00.005Z

This was supposedly demonstrated by Gauss, aged eight or something, when he was set a problem by a teacher desperate to get on with something more interesting to add together all the numbers from 1-100. He's supposed to have realised that if you paired up the numbers, they all had the same sum ie 101. It was then a matter of spotting the number of pairs and multiplying the two (5050).

The sum of an arithmetic series is done in a similar way. Think of a series as being:

a₁ + (a₁ + d) + (a₁ + 2d) +......

since d is a constant...

You want to find the sum of the first n terms.

S_n = a₁ + (a₁ + d) + (a₁ + 2d) +....

....(a₁ + (n-1)d)

This is probably quite a few terms (otherwise you wouldn't bother trying to find the sum, would you?).

So the wisest option would be to pair up the terms, like Gauss did, and find the sum of each pair.

The formula is derived from writing the above series out from beginning to n, and the opposite way, and adding the terms.

However, just as Gauss's addition always gave him 101, in our case we will always get 2a + (n-1)d

Since we want the sum to n of the series, there will be n of these pairs. So the sum is n(2a + (n-1)d)

BUT -

We have just added TWO series! Because we wrote it out twice to make adding the terms easier.

Therefore, the final formula will be: n(2a + (n-1)d)/2.

Quite fiddly to prove, but easy enough to use.

Series and Sequences 1 - nth terms

2009-10-29T10:17:00.005Z

There are a few of these.

Let's just start with a definition. A sequence is a load of numbers in a list, with there being a common difference between the numbers.

A series is the same thing, but added together.

Here is an arithmetic sequence: 0,2,4,6,8,10....

Here is an arithmetic series: 0+2+4+6+8+10.....

In an arithmetic series or sequence, there is always a common difference between the numbers. In the ones above, the common difference is 2. It is always a constant (and not anything weird like n² - that's more like geometric series).

So finding the nth term of an arithmetic series or sequence is easy enough.

You need the first term a₁ and the common difference d.

a_n = a₁ + (n-1)d

So say I wanted to find the 42nd term of the above sequence.

a₄₂ = 0 + 41d
= 0 + 41x2
= 82

The 42nd term of the sequence 0,2,4,6,8...is 82

Because the difference is a constant with arithmetic sequences, this formula is simple enough to grasp. To find the nth term you need the first term, and then the number of terms before the nth one timesed by the common difference because there are that many lots of the common difference.

Slowness

2009-10-28T10:55:00.002Z

Hmm, 50 posts in over a year isn't much is it? Well, I get bored easily.

You should see the other place...

I will try to post more, as I come back to maths after a few months' absence and try to tackle C3.

Radians

2009-10-28T10:34:00.002Z

These are little blighters used to describe the angles of circles. Like degrees, they can be used to express arc widths, or can be used in equations (especially trig equations). Basically they're just like degrees, but are based on π and the radius of the circle (hence radian). Imagine a line from the centre to any point on the circle. That's a radius. Then, in any direction, draw an arc around the surface of the circle with the same length as the radius. Stop. Draw a radius from this point back to the centre.

The angle subtended is one radian.

There are 360 degrees in a circle and one degree really isn't very large.

A radian is much bigger (around 57 degrees). Every circle has 2π radians, like every circle has 360°. 180° is therefore π radians.

Most calculations involving radians will also involve fractions. Using π enables you to be exact without lots of fiddly decimals. The same is true of fractions. In some ways then, using radians is satisfyingly precise.

It is also easy to convert one to the other.

360° = 2π radians

Therefore

1° = 2π radians/360.

and

say

12° = (2π/360) x 12

= 24π/360

= π/15

and it's usually sufficient to leave it like this for C2. At other levels you might want to do the calculation to n dps or whatever.

Similarly

2π radians = 360°

1 radian = 360/2π

and

so

3 radians = (360/2π) x 3.

Easy.

Get In!

2009-08-20T20:23:00.000+01:00

C2 = 92%

AS Level = A

TTD = Maths Genius

New Maths Specifications

2009-05-23T14:23:00.004+01:00

As it's a mere five years since the last new A Level maths specifications came out the QCA are clearly in dire need of updating their syllabus. Accordingly they're now in the consultation stage for a new curriculum to begin in 2012. This means of course that uncompleted A Levels will come to the end of their shelf-life at this time. Hopefully that won't affect me. But I was interested to see some of their recommendations.

1) They want to abolish the non-calculator paper. I think C1 is great. You learn how to do a lot of manipulation of formulae and a lot of arithmetic in your head. Differentiation, integration, surds, quadratics (factorising) can all be easily done without recourse to a calculator. This, surely is good for fast, flexible thinking and for confidence. The down side is that it does mean C1 can't be that challenging - you can't really put radians into it for example, even though conceptually radians is a piece of the proverbial. The same goes for basic trig. Trig would be better introduced in C1 rather than stuffed into C2 (which is I think around 35-40% trig, all told).

2) They are trying to bring it down to four modules again. This is not a bad idea. The current A Level maths six unit system is complicated and contradictory, though it is flexible for people with different specialisms. It also throws up anomalies. For example as it currently stands maths is the only A Level where you can do 4 AS modules and two A2 ones - which I am doing, because in addition to C1,C2,C3 and C4 I took S1 for AS and am doing M1 for A2. M1 is an AS module. So I'm only doing C3 and C4 as actual A2 Levels. The situation can be reversed - Pure Maths AS is two AS modules and one Further! And you can also do AS modules in Further Maths (you could do FP1-4 and S1 if you hadn't done it). This leads to anomalies in the awarding of grades - the A* grade is only awarded on the basis of C3 and C4.

Four modules would simply take us back to where we started in the 90s with the equivalent of P1 and P2 - ie the four Core modules collapsed into two again. This is probably where this idea dovetails with the abolition of the calculator paper.

3) There are going to be no formulae to learn. This strikes me as disastrous. It makes an expecation of zero knowledge on the part of candidates - and this is an advanced qualification we are talking aobut here. The authorities do not view knowledge as important, but expect candidates to problem solve. Well you can only solve problems if you know how to approach them - which strategies you know of to use. I think the motivation for this is clear dumbing down, or as the QCA write, to "provide students with equality of opportunity and a common basis for progression.". Hmmm. An advanced qualification should be for people who are advanced in their knowledge and understanding of that subject - it should not pander to political ideas of equality, and believe me the QCA does just that. In the questionnaire I answered yesterday I was asked whether the new specifications promote gender and race diversity, or whether they were discriminatory against disabled people. For crying out loud, why is even pure knowledge infected with this leftist rot?

4) They want slightly to change the balance between pure and applied. This probably isn't a bad idea. Right now applied is 33% of AS and A2, but they would like it to be more flexible, up to 40%. With fewer papers I don't know how they will do this, except by weighting, although the QCA do suggest that they might allow certain exams to be longer - this would be an excellent idea. At the same time they think the pure content should be kept the same. So without changing weightings or lengths of exams or indeed mixing up applied content into pure modules, I'm not sure how this will work.

5)There is a lot of guff on the QCA site about being more stretching. They conceive it as using longer, less structured questions. This is clearly an excellent idea, where possible and where practicable. I don't know if they just mean in Further or in Core maths as well.

You can go and look all this up for yourself at the QCA website.

Long Time, No Blog

2009-05-23T14:13:00.002+01:00

It's been a tricky old year so far, like a particularly knotty equation, so I haven't found the time for maths blogging, or even for maths.

So it was with some trepidation that I looked at my calendar about a month ago and saw the date for the AQA C2 exam. At that point I knew nothing of radians, logarithms I couldn't even spell, and geometric series I thought were just lots of pretty pictures (lots of geometry type stuff). So I spent a month fairly hectically doing the last four chapters of C2, missing out a few questions on the way and also, disappointingly, not really bothering to look at the proofs for the various formulae but just learning them.

C2 was yesterday. It was ok, though exams are always hard to assess until you get the marks. I answered all qs, but I think the differentiation with fractional indices one might have cost me a few %.

As before, the exam was thrilling. No I really mean this. I had bags of adrenalin, lots of excitement, masses of determination to show what I could do. I am like this with things that don't matter. Only important things have me quivering in the corner like a wobbly jelly who's just been made professor of wobbling at Oxford (ok ok that's a Blackadder joke sort of). Lining up beforehand with fifty sixth formers all pointing at me and whispering was a bit strange but it just reminded me of how utterly uninterested in it I was at their age. You do it for reasons, to get to uni, or because you happen to be good it, but you rarely do it just because you love it - you learn that later. Sometimes.

Integration, Differentiation

2009-02-16T09:10:00.002Z

This is quite tricky to get your head round, I think.

Differentiation is what you do to a curve to obtain its gradient. It works by splitting the curve into infinitesimally tiny chunks and tending towards a limit (ie where you started from). I think I put the proof of it on an earlier post.

Basically it gives you the rate of change in the curve as a formula. Because curves are always changing, the gradient will be different at each point. You plug the x-coordinate into the differentiated gradient formula and there's your gradient at x=whatever.

But if you integrate that curve you get the area under the curve. It works in a similar way to differentiation: it takes the space under the curve and chops it up into tiny rectangles to approximate the area.

It seems to work out that in terms of a formula, integration is the opposite of differentiation (ie raise the power by one and divide by the value of the new power, instead of dropping the power by one and timesing by the new power).

The curve bounding an area with the x-axis is now the value of the area formula differentiated - ie it is the rate of change of the area itself.

Differentiation and integration are inverse processes: that's the theorem of calculus (which means stones used for counting, or something).

If you have a derivative, you can integrate it to get the indefinite integral - which you can then find the full equation for if you know a set of co-ordinates. So by knowing the derivative, you can find the equation of that curve.

If you have a derivative, often you can differentiate again to find a stationary point on the curve - the second derivative test.

I'm still not sure I get this, you know. I still find the connection between the two loose and hard to actually articulate. I can do it with the formulae no problem at all - but understanding the linkage and how rates of change relate to curves is much harder.

More Basic Trig

2009-02-08T12:10:00.002Z

Every angle has a sin or cosine or tan (well, not quite every angle, in the case of tan). You can use this value to find the angle, doing it easily on a calculator.

You just press sin^-1 .456 or whatever it is, and that gives you the angle (if you have the calc. set to degrees that is).

Amen Trig Corner

2009-02-07T21:33:00.002Z

Well I guess I don't get it, but then maths is like that. Someone tells you to do x, you do x, you get it right or you get it wrong: no one gives a toss. No one cares that you've just described the entire universe (less the supernovae or the black holes), no one cares that you've inscribed patterns across space - patterns which remain, whether you like them or not; whether you think the guy who made them matters or not: they remain.

They are not contingent like the semi colon, or culturally derived like the feminism or the theism: they are not subject to the whims of tenured professors: there they are and you can just something off if you don't like it, tbh.

But to just do x, if you have been told to: that defies and defeats the entire point of mathematics. They tell us this because they know we are thick and also because in a pyramidal structure like maths, you just cannot *understand* things that you need to know how to do, when you are little.

No-one tells an 8 year old about subjects, objects, predicates, the passive voice or the genitive case. Equally, you shouldn't need to tell an 8 year old about the mechanics of division in order to teach them how to do it. But so paranoid are we about maths, so in hoc to an educational vision based more in socialism than in intellectualism, that we cannot believe we can teach these skills without sending 8 year olds into A Level territory, and in the meantime, depriving them of the skills of actually dividing one number by another.

Division by repeated subtraction: for crying out loud are you REALLY going to stop bright 11 year olds dividing by decimals by using this infantile but ridiculously complex method? Are you REALLY going to prevent tough calculation by saying "well, just take away and take away and take away"?

More to the point: stop insisting on an "understanding" that defies most adults, let alone children. You do not need to know that a sentence consists of subject, object and predicate to write one. Why are we baffling kids with mathematical jargon? Why are we constructing entirely false notions of "understanding" which exclude the simple explanation "well, I am timesing x by y" or "well, I guess I am seeing how often z goes into a". - and hamstringing bright kids, and stopping them calculating?

It could'nt be because socialist academics who write the curriculum don't want kids to be able to do these things...could it?

A Proper Go At Explaining Trig

2009-02-07T16:52:00.002Z

Alright, well, clearly it is the study of angles and triangles and by use of the unit circle, of circles too.

We talk about sin, cosine, and tan a lot in the basic stuff.

Although we learned SohCahToa at school, it really only applies to right angled triangles. But by using the unit circle (radius 1, centre at the origin) and drawing a right angled triangle inside it, we can calculate basic values of sine, cosine and tan, which hold for any angle and can be used in calculations involving any triangle.

The sine of an angle is a measurement of its distance from the horizontal axis, which can be negative or positive and which can be between 1 and -1 inclusive. If you think of an angle sweeping up around the circle, it reaches its furthest point from the axis (ie sin of 1) at 90 degrees.

The cosine of an angle is a meaurement of the distance from the vertical y axis of a point on the end of the angle; again, between 1 and -1.

The tan of an angle is a measurement of whether a line of angle x would meet a tangent to the circle drawn at right angles to the radius (the x axis) near, far, or not at all. The tan can have any value. Calculators really don't like being asked tan 90 or tan 270 because the lines of these angles will never meet a tangent.

There are some cool diagrams on the wikipedia page that show these things in action.

More Trig

2009-02-07T11:46:00.001Z

I can't say I get it! But I can use the formulae.

I loathe doing stuff I don't get.

General AS Advice

2009-02-05T16:16:00.000Z

Know your volumes and areas from GCSE.

I don't.

The Really Basic Basics of Trig

2009-02-05T16:05:00.002Z

Trig is basically to do with triangles and circles. It works from the properties of angles relative to the sides of triangles and how these can be seen in circles (unit circles - with a radius of 1). It's used to tell you the size of angles and sides using other information you already know about the triangle.

The trig identities, I reckon, are probably best defined as ratios.

All fractions are ratios, as well as being processes.

For a right angled triangle (where we tend to begin with these)

SohCahToa

Sin =opposite/hypotenuse

Cos = adjacent/hypotenuse

Tan = opposite/adjacent

For each angle, there will be a different value for each one of these. So sin 42 will be the same whatever triangle you have. It regulates what the sizes of the opposite and hypotenuse will be only in terms of the ratio between them.

Sin and Cos (not sure about Tan) will always give a value between 1 and -1 inclusive.

On Fractions

2009-01-31T10:03:00.005Z

I think these are great: more adaptable and more accurate than decimals.

But it's the division thing which always bugs me. You know: "turn the divisor upside down and multiply" - what the hell? Well, ok, I'll do it....

Well, doodling during a lesson the other day (yes I know I am supposed to be the teacher), I worked out a kind of rationalisation.

Take 3/6 divided by 1/2.

Flip over the 1/2 to make 2.

3/6 x 2 = 6/6 = 1.

So a half goes into 3/6 once. Makes sense, as 3/6 is a half.

But think of it without fractions.

6 / 3 = 2, right?

but 6 x 1/3 also = 2.

Because division is the inverse of multiplication, you are basically just doing the inverse. Also because multiplication is a scaling process, when you times by a number less than 1, the answer will be less than the number you were trying to scale. It scales downwards. 4x2 = 8 but 4x1/2 = 2.

And reciprocals are essential here. A reciprocal is any number you multiply any other number by to make 1. In practice this means that the reciprocal of a whole number is 1/that number.

1/6 x 6 = 1

But the reciprocal of any fraction is that fraction turned upside down.

3/4 - 4/3.

This is because you can imagine any whole number as n/1.

I guess you could say the reciprocal of a number is the inverse of that number...

So, 10/2 = 5. Ie 2 goes into 10 five times.

And 10 x 1/2 = 5. ie 10 lots of 1/2 make 5.

And 10 / 1/2 = 10 x2 = 20. ie a half goes into 10 20 times, which is the same as multiplying ten by two.

The fact that this holds makes it a lot easier to divide fractions, so it is a trick, but a useful one. There are proper proofs of it, but this isn't supposed to be a proof, only a sort of meditation on the subject.

On The Importance Of Factorising

2009-01-22T13:38:00.002Z

Without it, even linear equations become extremely difficult to re-arrange.

Take 6xy + 3x - 2y = 7

Re-arrange this for x.

MOVE the non x bit.

6xy + 3x = 7- 2y.

SPOT the common factor

x(6y + 3) = 7 - 2y.

DIVIDE by (6y+3)

x = 7-2y
----
6y+3

PIECE of the proverbial.

Indices And Differentiation

2009-01-22T13:33:00.002Z

Yesterday, courtesy of my beautiful and awesomely intelligent maths tutor, I suddenly worked out a way of making differentiation with negative indices slightly easier.

When inputting the x value to find the gradient at that point, to make the final stage of adding it all up much easier, turn your x^-2 into 1/x². It makes calculations a lot, lot easier.

Laws of Indices

2009-01-17T08:10:00.002Z

Are pretty simple.

a^m x aⁿ = a^m+n

a^m / aⁿ = a^m-n

(a^m)ⁿ = a^mxn

But a^m + aⁿ DOES NOT equal anything other than what it says.
There isn't much you can do to collect the terms if the indices are different, or if the constant is different (ie x^m + y^m)

a^m + a^m = 2(a^m), of course.

Fractional Indices (2)

2009-01-14T21:55:00.002Z

Well yes it gets a bit harder.

You see...a ^n/m

means ^m√aⁿ

...but I will try to explain this later.

Fractional Indices (1)

2009-01-14T21:39:00.005Z

Think about it.

Think about it.

x is not just x. x is x¹. x is x raised to the power of one: x multiplied by itself no times at all; x just being x. So then x ¹ is x on its own. x is, therefore, x.

So then, x^1/2 is going to be a number that, multiplied by itself, makes x.

In other words, the square root of x.

√x = x^1/2

It does get a bit more complicated than this....

Negative Indices

2009-01-11T17:51:00.001Z

Negative Indices

Now these seem simple enough: you raise a number to a power and it usually means multiplying a number by itself n times.

a³ = a x a x a

All numbers to the power of 1 are themselves, and all numbers to the power 0 are one.

a⁰ = 1

Easy enough.

But then you can also have negative powers.

a^-3

What? You can’t multiply a number by itself a negative number of times!

Well, no, clearly. But you can see how negative powers come about.

It’s to do with the powers rule.

aⁿ x a^m = a^n+m

You can see this if you write out aⁿ and a^m in full.

aⁿ x a^-n = a^{n+ -n}

= a^n-n
=a¹
=a

Therefore, the negative powers are used to denote reciprocals (the number you multiply n by to get 1 – so 1/6 is the reciprocal of 6 – and it is always 1/n.

So.

3³ =27

3^-3 = 1/27 (or 1/33 – the reciprocal).

Negative powers are easy to manipulate, they just seem a bit weird until you think that positive powers are going up by multiplying the number by itself:

2⁴ is 2 x 2 x 2 x 2

But if you go down, towards 0, the same process is division by two.

64...32...16..8...4..2..

This continues as you go down below zero:

......2, 1, ½, ¼, 1/8, 1/16......

(2¹, 2⁰, 2^-1, 2^-2, 2^-3, 2^-4)

And you can see that these are reciprocals of the powers of 2.

The same applies for the powers of each number.

C1 and S1 taken!

2009-01-11T09:24:00.004Z

Apologies for the lack of posting. I sort of became involved in my actual revision!

But I took C1 and S1 on Friday. C1 was ok but S1 was really tricky. Not sure how I did there.

Anyhow. I'm doing C2 now but I will try to post some more pure-related posts. I'm not sure I want to think about stats again!

Exams Booked & Paid For

2008-10-29T09:59:00.001Z

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!

Geometry of the Line (4): Circles

2008-09-14T12:25:00.001+01:00

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).

A Note on Revision

2008-09-13T16:13:00.002+01:00

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.

Blog Purpose Update

2008-09-13T16:10:00.000+01:00

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.

Dividing Polynomials (ii)

2008-09-11T15:33:00.000+01:00

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.

Dividing Polynomials

2008-09-11T15:13:00.000+01:00

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.

Rounding Up Factorisation

2008-09-07T13:01:00.000+01:00

...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....

Factor Theorem (ii)

2008-09-07T11:24:00.000+01:00

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!

The Factor Theorem

2008-09-06T16:09:00.000+01:00

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.

Dividing Polynomials (1)

2008-09-06T14:45:00.000+01:00

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.

Multiplying polynomials

2008-09-04T14:50:00.000+01:00

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.

Collecting Polynomials

2008-09-04T13:02:00.000+01:00

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

Polynomials - The Basics

2008-09-04T12:32:00.000+01:00

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.

Bit Brighter

2008-09-04T11:54:00.001+01:00

Thought it was a bit dark and dismal around here! This looks better to me.

AS Level Update

2008-08-29T14:41:00.000+01:00

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.

Summer Maths Reading Round-Up

2008-08-26T13:15:00.000+01:00

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.

Gradients of Perpendicular Lines

2008-08-25T17:12:00.000+01:00

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.

Geometry of the Line (3): Length

2008-08-25T16:32:00.000+01:00

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.

A Not Very Confident Interval

2008-08-21T22:13:00.000+01:00

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?

Beware of Surds

2008-08-18T21:40:00.001+01:00

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...

Why Doesn't Completing the Square Always Work?

2008-08-17T21:35:00.001+01:00

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.

Politics Break & Completing The Square

2008-08-17T15:01:00.000+01:00

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!

Geometry of the Line (2) The Equation of the Line

2008-08-14T21:46:00.001+01:00

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.

Geometry of the Line (1) The MidPoint of the Line

2008-08-12T10:22:00.000+01:00

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.

Geometry of the Line

2008-08-11T21:52:00.000+01:00

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.

Gradient

2008-08-10T15:55:00.001+01:00

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.

The dy/dx Post, Or, Differentiation for Dunderheads

2008-08-10T11:36:00.000+01:00

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.

Statistical Trials

2008-08-07T14:20:00.001+01:00

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?

Graphic

2008-08-06T23:05:00.000+01:00

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.

Square Completed

2008-08-06T22:08:00.000+01:00

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...