TSR Wiki > Study Help > Subjects and Revision > Revision Notes > Mathematics > STEP Tips and Tricks
These notes are currently being updated from this thread in the forum; STEP / AEA Revision Thread
Identities
Algebra
Identity 1
Essentially 
Special case for squaring:
Identity 2
Trigonometry & Complex Numbers
Identity 1
In particular, note that taking the modulus gives 
.
Equivalently, 
, which comes up a lot when calculating the distance between two points on the circumference of a circle
Identity 2
This trick is particularly useful in summing trigonometric series or any other situations which involve extracting real or imaginary parts out of a complex expression.
For the case of 
you can do similiar factorisation, which simplifies the algebra considerably.
= 
![\displaystyle\frac{(e^{\frac{ni\theta}{2}}-e^{\frac{-ni\theta}{2}})}{(e^{\frac{i\theta}{2}}-e^{\frac{-i\theta}{2}})}]](http://www.thestudentroom.co.uk/latexrender/pictures/c8/c8454042ade5f14aa994b51a634c351b.png "\displaystyle\frac{(e^{\frac{ni\theta}{2}}-e^{\frac{-ni\theta}{2}})}{(e^{\frac{i\theta}{2}}-e^{\frac{-i\theta}{2}})}]")
= ![\displaystyle\Re [(e^{\frac{(n-1)i\theta}{2}})(\frac{2isin(\frac{n\theta}{2})}{2isin(\frac{\theta}{2})})]](http://www.thestudentroom.co.uk/latexrender/pictures/3a/3a966b9ea9fc6df8ff67824d60f832bf.png "\displaystyle\Re [(e^{\frac{(n-1)i\theta}{2}})(\frac{2isin(\frac{n\theta}{2})}{2isin(\frac{\theta}{2})})]")
=
This useful summation 
isn't difficult to memorise or derive if you want to now.
Combinatorics and Probability
Identity 1
Proof: consider the coefficient of 
in the identity 
The most famous case is when n=m=r: 
Methods
Algebra
Proof of Cauchy-Schwarz Inequality
Prove that 
.
Consider 
. Then 
for all 
.
Expand out and we get
. But then since 
for all , we must have 
, so 
. Hence result.
For interest, there's equality iff 
for some value of . In other words, g is just a multiple of f (or f = 0 and g is anything you like).
Exactly the same proof shows 
etc.
Calculus
Integral Calculus
The fundamental idea here is that an integral measures the area under a curve. Lots of consequences, such as if 
for all ![\displaystyle x \in [a,b]](http://www.thestudentroom.co.uk/latexrender/pictures/93/934ae3eab7fba07dd34b4f1e09145b93.png "\displaystyle x \in [a,b]")
, then 
.
In general, when they ask you to prove something like this, you just have to draw a vague sketch to justify yourself!
A particular case that comes up often is that if f(x) is a decreasing function, then 
, as can be seen by sketching the graph of f and drawing "staircase" functions that are constant on each interval [k,k+1].
A particular case of the particular case(!) is to consider 
(with a little care about what happens at x=0) to deduce results such as 
.
Note I haven't explained this terribly well, because you really need the diagrams, but this concept of relating a sum to an integral is a pretty important one, and so you should spend a little time going over this.
Fundamental theorem of calculus
. This is isn't necessarily true if f isn't continuous at x, but don't worry too much about that. Again, if you need to prove it, you just need to draw a vague sketch justifying that 
Maclaurin Expansion
If you are asked to expand a function to x^2, such as
Let 
Compare coefficients
Sequences & Series
Quick and dirty introduction to solving simple recurrence relations
To solve a recurrence relation of the form 
, assume a solution of form 
and deduce 
You end up with a general solution 
where 
are the roots of the quadratic 
. (If the roots are repeated, the general solution takes the form 
).
To solve 
, first look for a particular solution, typically of the form 
, and then make it general by adding solutions to . (All analogous to what you'd do for a linear diff equation)
Statistics
Indicator functions and Expectation
All an indicator function is is a function that is 1 when an event happens and 0 when it doesn't. What's neat about indicator functions is that you can often use them to break a more complicated function down into something almost trivial.
E.g. Suppose 
and we want to prove the formulas for the mean and variance of X.
Instead of doing calculations based on finding 
etc, define the indicator function 
to be 1 if the k'th trial is a success, 0 otherwise.
Then 
.
But 
, so 
. (note that we can get this far without requiring the to be independent, which is often very useful).
Since the are independent, we also have 
, so 
.
General Advice
- The examination requires intensity and persistence.
- Expect to take a long time answering questions at first.
- Answering a question yourself then checking is much more pleasing than giving up and looking at the answer.
- Questions require a systematic approach.
- Checking will improve the work of many candidates.
- The fluent, confident and correct handling of mathematical symbols is necessary and expected.
- Set out a well-structured answer.
- Sometimes a fresh start to a question is needed.
- Sound technique is necessary, and checking required.
- Working to be legible.
- Aim for thoughtful and well set-out work.
- Arithmetic and algebraic accuracy would most improve marks.
- It is not a good idea to plunge into the algebra without thinking about alternative methods.
Specific Advice
- Using abbreviations can save a great deal of writing
- The parts of a question are often linked together, but sometimes with slight modifications.
- To show a statement is true, give a formal proof; to show one is false, give a (if possible, simple) counterexample.
- It doesn't matter if you start from the given answer and work backwards - it is still a mathematical proof and any proof will get the marks.
- A geometric understanding of modulus questions can help when examining the different cases.
- If you are unsure what to do some way into a question, examine what you have already demonstrated. STEP often teaches small tricks in the first part of the question then gets you to use this method by yourself.
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