STEP / AEA Revision Thread (i.e. useful tips and tricks)

The purpose of this thread is to put some of the more useful tips and tricks people have found doing STEP / AEA questions (or elsewhere) into one place.

If enough gets posted to make it worthwhile, I'll try to make this post into some kind of "index", similarly to how others have done the solutions threads.

Reply 1

DFranklin

OP

18

Trig: If you only learn one identity for STEP, make it this one:

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}{2} = \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.

Reply 2

khaixiang

5

Great idea for a thread David.

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.

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

For the case of

\Re{(e^{i\theta}+e^{2i\theta}+\dots+e^{ni\theta})}

you can do similiar factorisation, which simplifies the algebra considerably.

Unparseable latex formula:

\\ \displaystyle \Re{\frac{e^{ni\theta}-1}{e^{i\theta}-1}}\\ =\Re{\frac{e^{\frac{ni\theta}{2}}(e^{\frac{ni \theta}{2}}-e^{\frac{-ni\theta}{2}})}{e^{\frac{i\theta}{2}}(e^{\frac{i \theta}{2}}-e^{\frac{-i\theta}{2}})}}\\ =\Re{(e^{(\frac{n-1}{2})i\theta}(\frac{2i\sin{\frac{n\theta}{2}}}{2i\sin{\frac{\theta}{2}}}))}\\ =\frac{\cos{(\frac{n-1}{2}\theta)}\sin{(\frac{n\theta}{2})}}{\sin{ (\frac{\theta}{2})}}}

This useful summation

\displaystyle \sum_{k=1}^{n}{\cos{k\theta}}

isn't difficult to memorise or derive if you want to now. This trick seems to be used quite often in the solutions provided by Siklos.

Reply 3

Kolya

17

I am not sure what they are what David is looking for, but they may provide a small help, so here are my notes which I have taken from the mark schemes and my own practise:

General Advice
- The examination requires intensity and persistence.
- 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 slightly 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.

Reply 4

DFranklin

OP

18

Something different, that I've seen on the old CCE but not sure if I have in STEP:

\displaystyle \sum_{k=0}^r \binom{n}{k} \binom{m}{r-k} = \binom{n+m}{r}

Proof: consider the coefficient of

x^r

in the identity

(1+x)^n(1+x)^m = (1+x)^{n+m}

The most famous case is when n=m=r:

\displaystyle \sum_{k=0}^n \binom{n}{k}^2 = \binom{2n}{n}

Reply 5

DFranklin

OP

18

Quick and dirty introduction to solving simple recurrence relations:

To solve a recurrence relation of the form

au_{n+2}+bu_{n+1}+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_{n+2}+bu_{n+1}+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_{n+2}+bu_{n+1}+cu_n = 0

. (All analogous to what you'd do for a linear diff equation).

Reply 6

generalebriety

16

DFranklin

Quick and dirty introduction to solving simple recurrence relations:

To solve a recurrence relation of the form

au_{n+2}+bu_{n+1}+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_{n+2}+bu_{n+1}+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_{n+2}+bu_{n+1}+cu_n = 0

. (All analogous to what you'd do for a linear diff equation).

Interesting, I did mean to look this up some time. Anyone know where I can find any practice questions on this? Also, can we legitimately assume there's a solution in the form u_n = (lambda)^n in STEP?

Reply 7

DFranklin

OP

18

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

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{n}{k} p^k(1-p)^{n-k}

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)

.

Reply 8

DFranklin

OP

18

generalebriety

Interesting, I did mean to look this up some time. Anyone know where I can find any practice questions on this?

Don't know, sorry.

Also, can we legitimately assume there's a solution in the form u_n = (lambda)^n in STEP?

Yes, because you're not "assuming" there's a solution of that form, you're actually providing a solution of that form. (In other words, you've produced a solution that works, and they can't really ask for anything else).

The more difficult problem is proving that not only have you found a valid solution, but that your solution is in fact the only possible one. Under exam conditions, I wouldn't bother about this (unless you are in the situation where you can't find anything better to do or the question seems particularly strict in its language).

But if you do have to prove it, suppose you have 2 solutions u_n, v_n with u_0 = v_0, u_1 = v_1. Put w = u-v, then w also satisfies the difference equation, but w_0 = w_1 = 0. But then

w_2 = (bw_1+cw_0)/a = 0

and then

w_3 = (bw_2+cw_1)/a = 0

, etc... (use induction if you want to be really formal).

Reply 9

generalebriety

16

DFranklin

The more difficult problem is proving that not only have you found a valid solution, but that your solution is in fact the only possible one. Under exam conditions, I wouldn't bother about this (unless you are in the situation where you can't find anything better to do or the question seems particularly strict in its language).
I suppose my question is more along the lines of: why does there always exist such a solution? Sorry, I don't mean to hijack this thread or anything, it's just something I've wondered for a while. What if there isn't such a solution? Then you'd still get an answer out of it which would satisfy the first couple of terms in the recurrence relation but would inevitably fail elsewhere.

Reply 10

DFranklin

OP

18

generalebriety

I suppose my question is more along the lines of: why does there always exist such a solution? Sorry, I don't mean to hijack this thread or anything, it's just something I've wondered for a while. What if there isn't such a solution? Then you'd still get an answer out of it which would satisfy the first couple of terms in the recurrence relation but would inevitably fail elsewhere.

There's always a valid solution because we can always find roots to the auxiliary quadratic equation, and if

\lambda

is a root it is immediate that

u_n = A\lambda^n

is a valid solution (since for all n we have

au_{n+2}+bu_{n+1}+cu_n = A(a\lambda^2+b\lambda+c)\lambda^n = 0

). Since you can always add two solutions, this justifies the general solution for the standard case with no repeated roots.

It does get a lot more dodgy when you have a repeated root and want to claim there's a solution of the form

(A+Bn)\lambda^n

. This is tricky to justify (more tricky than I think is justified discussing in detail here - I can't see it being expected for STEP!). But you can do it by basically "factoring" the difference equation into linear terms, and solving each term in sequence, with the solution to each term becoming the inhomogenous part on the right hand side of the next.

Edit: decided I might as well do an example for the repeated root case; I did get asked it at interview (for a differential equation, not a difference equation), so it is the kind of question that could happen (though I'm certain they'd give some hints. I certainly got stuck on it!).

Suppose we want the general solution of

u_{n+2} -4u_{n+1}+4u_n = 0

. Factor the aux eqn as

x^2-4x+4 = (x-2)(x-2)

. Consider

v_n = u_{n+1}-2u_n

(corresponding to (x-2)). Then

v_{n+1}-2v_n = 0

, from which we deduce

v_n = A 2^n

. Then

u_{n+1}-2u_n = A 2^n

and we deduce

u_n = B^n + (nA2^n)/2

(where there's still a certain amount of "pulling a rabbit out of a hat" in terms of guessing the form of the solution, but it is much easier to justify for this case where

u_{n+1}

depends only on

u_n

and n).

Reply 11

generalebriety

16

Ah, don't see why I didn't think of it like that. :s-smilie:

Alright, thanks. I'll stop hijacking your thread now, and maybe come back with some advice of my own later. :p:

Reply 12

DFranklin

OP

18

Proof of Cauchy-Schwarz Inequality

There are various forms of this, I'll just do one:

Prove that

\left(\int f(x)g(x)dx\right)^2 \le \int f^2(x) dx \int g^2(x) dx

.

Consider

I(\lambda) = \int (f(x)+\lambda g(x))^2 dx

. Then

I(\lambda) \ge 0

for all

\lambda

.

Expand out and we get

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

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

\left(\sum a_n b_n\right)^2 \le \sum a_n^2 \sum b_n^2

etc.

Reply 13

Drederick Tatum

12

DFranklin

Something different, that I've seen on the old CCE but not sure if I have in STEP:

\displaystyle \sum_{k=0}^r \binom{n}{k} \binom{m}{r-k} = \binom{n+m}{r}

Proof: consider the coefficient of

x^r

in the identity

(1+x)^n(1+x)^m = (1+x)^{n+m}

The most famous case is when n=m=r:

\displaystyle \sum_{k=0}^n \binom{n}{k}^2 = \binom{2n}{n}

Quite a nice way of thinking about these questions is by noting that the binomial coefficient "n choose k" is exactly that - the number of ways of choosing k objects from n objects.
So we can prove the first identity above by noting that the RHS is the number of ways of picking r players from 2 teams (one of size m, the other of size n).

But we can do this by picking 0 players from the n-team, and r from the m-team,
or by picking 1 player from the n-team and the other r-1 from the m-team,
...
or by picking r players from the n team and none from the m-team.

This is exactly the LHS. Infact, this kind of thinking might be helpful on other questions, particularly in the probability section.

Reply 14

datr

12

I made notes of all the advice given at the STEP summer school this year, when i get a chance I'll write them up if anyone is interested. Although they are much more elementary than what has already been mentioned.

Reply 15

zooop

1

I am sitting the AEA this time round and I haven't seem to have studied half the things that have been mentioned above! It makes me a little worried! Will I need to know half of what has been previously said? Thanks

Reply 16

Chewwy

17

zooop

I am sitting the AEA this time round and I haven't seem to have studied half the things that have been mentioned above! It makes me a little worried! Will I need to know half of what has been previously said? Thanks

no.

Reply 17

Speleo

7

In fact I would be highly surprised if anything on this page except Lusus Naturae's general advice would help with the AEA.

Reply 18

insparato

15

AEA only tests knowledge from C1-C4

Reply 19

DFranklin

OP

18

zooop

I am sitting the AEA this time round and I haven't seem to have studied half the things that have been mentioned above! It makes me a little worried! Will I need to know half of what has been previously said? Thanks

As others have said, no. Personally, I don't really know what would be useful for AEA. Even for STEP, I'm aiming somewhat at "stuff you do at university that you could make into a STEP question", because those are the things where you get a big advantage from having seen something like it before. (E.g. the "diamond-derivative" question, which you do formally in algebra at university. And is a 20 marks in 2 minutes question if you're familiar with how to go about it).

STEP / AEA Revision Thread (i.e. useful tips and tricks)

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