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

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DFranklin
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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.
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DFranklin
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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.
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khaixiang
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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.

\\ \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.
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Kolya
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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.
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DFranklin
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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}
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DFranklin
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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).
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generalebriety
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(Original post by 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?
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DFranklin
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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).
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DFranklin
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(Original post by 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).
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generalebriety
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(Original post by 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.
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DFranklin
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(Original post by 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).
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generalebriety
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Ah, don't see why I didn't think of it like that. Alright, thanks. I'll stop hijacking your thread now, and maybe come back with some advice of my own later. :p:
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DFranklin
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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.
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Drederick Tatum
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(Original post by 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.
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datr
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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.
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zooop
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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
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Chewwy
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(Original post by 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.
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Speleo
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In fact I would be highly surprised if anything on this page except Lusus Naturae's general advice would help with the AEA.
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insparato
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AEA only tests knowledge from C1-C4
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DFranklin
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(Original post by 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).
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