The 2013 STEP Prep Thread!

But how do you know that it's some rational multiple of root 2 in the first place?

It damn sure is.

It damn sure is. I suppose you could argue something like the following:

$\cos \dfrac{\pi}{4}=\dfrac{1}{2}\left( \cdots \left( \left(4\cos^2\dfrac{\pi}{2^{n}}-2\right)^2-2\right)^2\cdots \right)^2-1$

LHS is irr. so the first bracket is irr $\to$ second is irr. $\cdots\to$ last bracket is irr.

1988 step I, q1, I drew the graph and found the first solution n=m, I'm sure there are more but I don't know how to go about it

where the hell do you find the 1988 papers or any papers earlier than 1998 in fact?

Download the Megapack in the OP.

Oh, OK. For some reason I thought it was some exercise in analysis where we couldn't assume the properties of trigonometric functions!


But how do you know that it's some rational multiple of root 2 in the first place?

Yeah simultaneous equations will work fine. Playing the exam game slightly, it's usually safe to assume that solutions will exist if they ask you to find said solutions explicitly in STEP; although I wouldn't be surprised if they expected verification of the values you found in this case - IIRC, the examiner's report for this paper pointed out that the last part of this question was completed in full by very few candidates.

Did my alternative hint make much sense? It's a lot easier to verify that the chosen a,b,c satisfy a suitable cubic than it is for that (n+1)th multivariable thing they expect you to prove.

I use injective and surjective for that matter.

You mean irrational, well sin(45) = 0.5*squrt 2. 0.5 is rational, root two is irrational so the product of a rational and an irrational is irrational.

STEP I 2007 Q4, where the two equations cannot be identical because it is given that $ac\neq b^2$ and the only way they can be identical is if $a=b=c$.

I use epic and monic

In fact, one often says "there exists a homo such that..."

This has been discussed before: people are confused by the or statement:

$x=1\Rightarrow x=\pm 1$ is a true statement, even though $x\neq 1$

inb4 ukdragon and his logic shiz

I thought I almost got a STEP III question, but it has defeated me alas. Q10 STEP III speciment paper. The 1st part was simple, but the explanation in the answers used loads of modulo stuff. When I just gave a small explanation which I thought woudl explain it enough... me being overly simplistic I suppose. Oh and the divisibility proof seems to not be able to be cracked by the an induction proof of Uk+1 - Uk. Which most divisibility proofs are normally proved by.

This has been discussed before: people are confused by the or statement:

$x=1\Rightarrow x=\pm 1$ is a true statement, even though $x\neq 1$

inb4 ukdragon and his logic shiz

I don't think that's an excuse for the poor wording of this particular question. It should simply ask: "deduce that k=1".

Hey I need to say whether this is true or false with justification

There exists a polynomial P such that

|P(x)-cos(x)|<=10^(-6) for all real x

I have no idea where to start except I think it may have something to do with the approximation for cos(x).

Any tips or hints would be appreciated.

Thanks in advance.

This has been discussed before: people are confused by the or statement:

$x=1\Rightarrow x=\pm 1$ is a true statement, even though $x\neq 1$

inb4 ukdragon and his logic shiz

Hey I need to say whether this is true or false with justification

There exists a polynomial P such that

|P(x)-cos(x)|<=10^(-6) for all real x

I have no idea where to start except I think it may have something to do with the approximation for cos(x).

Any tips or hints would be appreciated.

Thanks in advance.

Hey I need to say whether this is true or false with justification

There exists a polynomial P such that

|P(x)-cos(x)|<=10^(-6) for all real x

I have no idea where to start except I think it may have something to do with the approximation for cos(x).

Any tips or hints would be appreciated.

Thanks in advance.