C1 Advice

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

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

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

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

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

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

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.