Step tips and tricks



Identity 1

\displaystyle \left( \sum_^{\infty} a_n x^n \right) \cdot \left( \sum_^{\infty} b_n x^n \right) = \sum_^{\infty} \left( \sum_^ a_k b_ \right) x^n

Essentially \displaystyle \left( \sum_^{\infty} a_n x^n \right) \cdot \left( \sum_^{\infty} b_n x^n \right) =  a_0 b_0 + (a_0 b_1 + a_1 b_0) x + (a_0 b_2 + a_1 b_1 + a_2 b_0) x^2 + \cdots

Special case for squaring: \displaystyle \left( \sum_^{\infty} a_n x^n \right)^2 = \sum_^{\infty} \left( \sum_^ a_k a_ \right) x^n

\displaystyle \left( \sum_^{\infty} a_n x^n \right)^2 = ^2 + (2 a_0 a_1) x + (2 a_0 a_2 + ^2) x^2 + (2 a_0 a_3 + 2 a_1 a_2) x^3 + \cdots

Identity 2


Trigonometry & Complex Numbers

Identity 1

2 \cos \theta (\cos \theta + i \sin \theta) = 1 + \cos 2\theta + i \sin 2\theta

In particular, note that taking the modulus gives 2 \cos \theta = \sqrt{(1+\cos 2\theta)^2+ \sin^2 2\theta}.

Equivalently, 2 \cos \frac{\theta} = \sqrt{(1+\cos \theta)^2+ \sin^2 \theta}, 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.

\displaystyle \\ {\rm Re}{(1+e^)^m}\\={ \rm Re}{(e^)^(e^{-i\theta}+e^)^m}\\ ={ \rm Re}{(e^)(2^\cos^m{\theta}))}\\=2^\cos^m{\theta}\cos

For the case of { \rm Re}{(e^+e^+\dots+e^)}you can do similiar factorisation, which simplifies the algebra considerably.

\displaystyle \Re (\frac-1}-1}) = \displaystyle\Re [\displaystyle\frac{(e^{\frac})}{(e^{\frac})} \displaystyle\frac{(e^{\frac}-e^{\frac{-ni\theta}})}{(e^{\frac}-e^{\frac{-i\theta}})}] = \displaystyle\Re [(e^{\frac{(n-1)i\theta}})(\frac)})})] = \displaystyle \ (cos\frac)(\frac}})

This useful summation \displaystyle \sum_^{\cos} isn't difficult to memorise or derive if you want to now.

Combinatorics and Probability

Identity 1

\displaystyle \sum_^r \binom \binom = \binom

Proof: consider the coefficient of x^r in the identity (1+x)^n(1+x)^m = (1+x)^

The most famous case is when n=m=r: \displaystyle \sum_^n \binom^2 = \binom



Proof of Cauchy-Schwarz Inequality

Prove that \displaystyle  \left(\int f(x)g(x)dx\right)^2 \le \int (f(x))^2 dx \int (g(x))^2 dx.

Consider \displaystyle  I(\lambda) = \int (f(x)+\lambda g(x))^2 dx. Then I(\lambda) \ge 0 for all \lambda.

Expand out and we get

\displaystyle  I(\lambda) = \lambda^2 \underbrace{\int  (f(x))^2 dx}_A + 2\lambda \underbrace{\int f(x)g(x) dx}_B + \underbrace{\int (g(x))^2 dx}_C

\displaystyle  I(\lambda) = A\lambda^2 + 2B \lambda + C. But then since I(\lambda)\ge 0 for all \lambda, we must have (2B)^2 \le 4AC, so B^2 \le AC. Hence result.

For interest, there's equality iff I(\lambda) = 0 for some value of \lambda. In other words, g is just a multiple of f (or f = 0 and g is anything you like).

Exactly the same proof shows \displaystyle  \left(\sum a_n b_n\right)^2 \le \sum a_n^2 \sum b_n^2 etc.


Integral Calculus

The fundamental idea here is that an integral measures the area under a curve. Lots of consequences, such as if \displaystyle  f(x) \le g(x) for all \displaystyle  x \in [a,b], then \displaystyle  \int_a^b f(x) dx \le \int_a^b g(x) dx.

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 \displaystyle  \int_1^ f(x) dx \le \sum_1^n f(k) \le \int_0^n f(x) dx, 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 \displaystyle  f(x) = \frac (with a little care about what happens at x=0) to deduce results such as \displaystyle \sum_^n \frac \ge \log(n+1).

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

\displaystyle  \frac \int_a^x f(t) \,dt = f(x). 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 \displaystyle  \int_x^ f(t)\,dt \approx \delta x f(x)

Maclaurin Expansion

If you are asked to expand a function to x^2, such as

\displaystyle f(x) =  \frac

Let  f(x) = c_0 + c_1x + c_2x^2 + \cdots

 (c_0 + c_1x + c_2x^2 + \cdots)(1+\ln(1-3x)) = 2\cos3x

 (c_0 + c_1x + c_2x^2)(1 -3x - \frac + \cdots) = 2(1 - \frac + \cdots)

 (c_0 + c_1x + c_2x^2 + \cdots)(1 - 3x - \frac + \cdots) = 2 - 9x^2 + \cdots

Compare coefficients

 c_0 = 2

 -3c_0 + c_1 = 0

 c_1 = 6

 c_2 - \frac - 3c_1 = -9

 c_2 - 9 - 18 = -9

 c_2 = 18

 f(x) = 2 + 6x + 18x^2 + \cdots

Sequences & Series

Quick and dirty introduction to solving simple recurrence relations

To solve a recurrence relation of the form au_+bu_+cu_n = 0, assume a solution of form u_n = \lambda^n and deduce a\lambda^2+b\lambda+c = 0 \quad(\S)

You end up with a general solution u_n = A \lambda_1^n + B \lambda_2^n where \lambda_1, \lambda_2 are the roots of the quadratic (\S). (If the roots are repeated, the general solution takes the form u_n = (A+Bn)\lambda_1^n).

To solve au_+bu_+cu_n = f(n), first look for a particular solution, typically of the form u_n = Cf(n), and then make it general by adding solutions to au_+bu_+cu_n = 0. (All analogous to what you'd do for a linear diff equation)


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 X \sim B(n, p) and we want to prove the formulas for the mean and variance of X.

Instead of doing calculations based on finding \sum_0^n k\binom p^k(1-p)^ etc, define the indicator function I_k to be 1 if the k'th trial is a success, 0 otherwise.

Then E(I_k) = p, Var(I_k)=E(I_k^2)-E(I_k)^2 = p-p^2=p(1-p).

But X = \sum_1^n I_k, so E(X) = \sum_1^n I_k = np. (note that we can get this far without requiring the I_k to be independent, which is often very useful).

Since the I_k are independent, we also have Var(\sum I_k) = \sum Var(I_k), so Var(X) = \sum_1^n Var(I_k) = np(1-p).

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.