STEP Prep Thread 2016

I looked up the question and it says "You will have to explain what the LHS actually means." TBH, I have no idea what you could write during the test.

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Reply 382

Damien_Dalgaard

16

For STEP, can you answer any 6 questions, i.e. no pure questions and just mechanics and stats?

Reply 383

davros

16

Original post by Damien_Dalgaard

For STEP, can you answer any 6 questions, i.e. no pure questions and just mechanics and stats?

There's no restriction on question choice. But I think there are only 5 applied Questions now aren't there - 3 Mech and 2 Stats :smile:

Reply 384

Damien_Dalgaard

16

Original post by davros

There's no restriction on question choice. But I think there are only 5 applied Questions now aren't there - 3 Mech and 2 Stats :smile:

Yes this seems correct!

Would you ever advise anyone to do 7 questions, if say they get stuck midway of a question :smile:

Also is 45 mins the time it should take to complete one question.

Reply 385

They say it's what happens usually and it's the recommended time if you want a 1 (for which you need approx 4 well-answered questions). But if you are aiming at 120 pts, you will have to get 6 questions done in 3 hours, so 30 min per question.
Also, cambridge advices candidates to concentrte on 6 questions at most.
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(edited 8 years ago)

Reply 386

Original post by Damien_Dalgaard

Yes this seems correct!

Would you ever advise anyone to do 7 questions, if say they get stuck midway of a question :smile:

Yes; personally I'd not count questions less than about 2/3 complete when considering "how many questions?"

Also is 45 mins the time it should take to complete one question.

Not really. The "4 good questions should be a 1" advice is somewhat outdated as the grade boundaries (for STEP I/II at any rate) have crept up considerably from when that was written. And even if it wasn't, you want to be aiming to have time to spare, because otherwise if you get stuck on one question you won't have time to finish 4 other ones.

I'd say that for STEP I/II you probably want to be aiming for 30 mins or better, and for STEP III probably 36 mins (i.e. time for 5 questions).

However, if you've just started STEP I wouldn't worry too much about how long questions take. You will get faster with practice.

Reply 387

demigawdz

4

Can anybody help me find all the induction questions for step 1,2 and 3 please as it is quite hard as it is not always clear / explicit where/when a question requires induction.
I am looking for any question with any amount/ type of induction and would like to if people would at least tell me the year and step paper ex: ( step 2 /03) which means step 2, 2003 ( not sure if there is one here lol...
Thanks. :smile:

Reply 388

Original post by demigawdz

Can anybody help me find all the induction questions for step 1,2 and 3 please as it is quite hard as it is not always clear / explicit where/when a question requires induction.
I am looking for any question with any amount/ type of induction and would like to if people would at least tell me the year and step paper ex: ( step 2 /03) which means step 2, 2003 ( not sure if there is one here lol...
Thanks. :smile:

The thing is, you're explicitly asking which questions require induction, and it's very rare that a question can't be done another way (unless they tell you you must use induction, of course).

For that matter, pretty much any induction proof can easily be rewritten as "proof by contradiction" - e.g.

To prove 1+2+3+...+n = n(n+1)/2 for n = 1, 2, ... (*)
Suppose (for contradiction) this is false, then pick n to be the minimal value for which (*) fails.
We note n can't be 1, since (*) is true for n = 1. But then by minimality of n, (*) holds for n-1, and so 1+2+3+..+n-1 = n(n-1)/2. But then 1+2+3+...+n = n(n-1)/2+n = n(n+1)/2, contradicting that (*) fails for n.

Yes, this looks very similar to the induction proof. But it doesn't use induction. (It also doesn't require all the "suppose true when n = k boilerplate", which no-one ever uses beyond A-level, and drives me nuts).

Reply 389

What you wrote is exactly what induction means (you did prove it for 1 and you did have an inductive step). The difference is that induction is the conventional way of doing it.

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Reply 390

Original post by to4ka

What you wrote is exactly what induction means (you did prove it for 1 and you did have an inductive step). The difference is that induction is the conventional way of doing it.

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Note that at no point have I appealed to the principle of induction, or even mentioned induction at all. In fact, you can (under certain assumptions) *prove* the principle of induction like this.

Of course, the particular example is very "induction in disguise", but that's somewhat subjective. There are proofs where its more natural to do what I did but you *could* rephrase in terms of induction. In the end it's as much a matter of opinion as anything else. (And so asking which questions *require* induction is probably not a good idea).

On the gripping hand - asking which questions require a particular approach seems like a particularly poor way to prepare for STEP - learning to work that out for yourself is essential.

(edited 8 years ago)

Reply 391

Original post by DFranklin

Note that at no point have I appealed to the principle of induction, or even mentioned induction at all. In fact, you can (under certain assumptions) *prove* the principle of induction like this.

...

n(n-1)/2+n=n(n+1)/2
That's exactly what induction is. In fact, in introductory number theory books, I have seen induction as an axiom (check Peano's axioms). From this, they derive the well-ordering principle, using induction and contradiction only. So it's the exact same thing.

Spoiler

Reply 392

Original post by to4ka

n(n-1)/2+n=n(n+1)/2
That's exactly what induction is.I (obviously) disagree. For induction you are thinking "I want to shot it's true for 1, and if "true for n implies true for n+1", as then it's true for all n". The argument I'm making is "if it's not true for all n, what's the smallest n for which it fails, and can we derive a contradiction?". They are (usualy) quite different ways of thinking about a problem, even if the underlying principles may be equivalent.

(Note that I purposefully followed the standard inductive argument very closely. If I'd gone "well, if 1+2+...+n =/= n(n+1)/2, then 1+2+...+(n-1) =/= n(n+1) /2 - n = n(n-1)/2" the lines would already be a fair bit more blurred).

In fact, in introductory number theory books, I have seen induction as an axiom (check Peano's axioms). From this, they derive the well-ordering principle, using induction and contradiction only. So it's the exact same thing.

Yes, you can go either way. That means induction and the well ordrering principle are equivalent. It doesn't mean they are the same thing, at least in a pedagogical sense. (The Axiom of Choice is famous for having a huge number of equivalent statements, and so saying that "the proof that every vector space has a basis uses Tychonoff's theorem" might be technically correct if you squint in a particular way, but it's not really a helpful way of describing things).

But this is getting contentious, (and you can find well qualified people arguing either way) and this is in fact my point as it relates to the original question. When people will argue about whether particular solutions are "really" induction, asking people to pick out which questions require induction is not terribly sensible.

[I should also say that I intensely dislike the approach typically taken at A-level to induction, with ridiculous amounts of boilerplate written, and so I would tend to actively use proof by contradiction even where induction is "natural" for STEP. But that's just me].

(edited 8 years ago)

Reply 393

Zacken

22

Original post by to4ka

n(n-1)/2+n=n(n+1)/2

If you're preparing for the exam, then asking for "which question requires this or this technique" is detrimental. If you start on a question, knowing that you'll be needing to use induction on it - you have a massive, massive advantage. Most STEP questions require a few moments of thought before attacking it, trying different angles of attack and/or techniques.

There was a STEP III question on operators (I think raff brought it up) that I struggled through for a few hours before looking up the solution, only to see that it was a standard inductive argument - but I hadn't though of applying induction in such a question. That's one part of STEP you're going to need.

tl;dr - if you want to prepare for STEP seriously, don't give yourself detrimental (in the long run) advantages in answering questions by already knowing what angle of attack you need to use for that particular question. Approach a question with a blank mind and then start thinking about it.

Reply 394

I can feel we just have to agree to disagree. I meant that

n(n-1)/2+n...

is an inductive step, not that your proof is induction. Our discussion is not really productive btw.

I'm not studying A levels and to me, it is very badly made. If you cannot prove something, don't teach it:
● integrals - rarely do A level students know why there's a relation with area under curves
● perpendicular lines' gradients - again, you just have to memorize that they multiply to give -1, but do teachers show why?
● recently I saw someone with an A* but couldn't solve this:
6/(x-3)=5/(4x+2) or something similar
● 2nd order DEs general solutions - again, it requires uni level stuff for proof
...and much more

I cant believe how broad the syllabus is... and how you end up knowing next to nothing about many things. I really hope it's not like that at uni...

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Reply 395

Original post by to4ka

I can feel we just have to agree to disagree. I meant that

n(n-1)/2+n...

is an inductive step, not that your proof is induction. Our discussion is not really productive btw.I agree this is unproductive.

I'm not studying A levels and to me, it is very badly made. If you cannot prove something, don't teach it: ~snip~

I cant believe how broad the syllabus is... and how you end up knowing next to nothing about many things. I really hope it's not like that at uni...

Even at university, even in pure maths courses, this happens. If you exclude non-examinable proofs (that realisitically most students won't pay much attention to), it actually happens quite a lot.

Some results are so ubiquitous that you need to know/use them, even if you can't prove them. (e.g. the result that a power series can be differentiated term by term inside the radius of convergence, or Cauchy's integral formula for complex variables).

Once you get into the applied courses, it's generally a lot worse. Better than A-level, but most of the things you use you don't know how to prove. (At least in the first 2 years).

Reply 396

Original post by to4ka

I understand that, I am asking if it's valid to write it down like that.

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Original post by raff97

Its weird, i'd never seen d/dx by itself like in the question we're discussing. d/dx is an operator like you say. If dz/dx = d/dx * z then, for example wouldn't that lead to things like: dz/dx * x = d/dx * z * x = d/dx * zx = d/dx(zx) which are obviously different.
Strange

It's valid but the notation is sufficiently problematic that it's very common to avoid the

\dfrac{d}{dx}

terms by writing something like:

Write

D = \dfrac{d}{dx}

. Then (D+x)(D-x)z = (D+x)(Dz - xz) = D^2z - Dxz + xDz -x^2z = D^2z -z - xDz +xDz - x^2z = D^2z - x^2z - z.

But since this is straight substitution it obviously doesn't change the underlying mathematics, it just makes it look a little less horrible.

(If it helps, in the above calculation I don't think of D{stuff} as multiplication - D always is an operator acting on the item to the right. When you do enough of this you can often see places where you can "get away with thinking of D as a number", but generally better safe than sorry).

Reply 397

So what does it mean, the expression? Like, you know how to explain

y''=-y

so what does

(d/dx+x)(d/dx-x)z

mean? This is in the question where the expression was given.

Reply 398

Original post by to4ka

So what does it mean, the expression? Like, you know how to explain

y''=-y

so what does

(d/dx+x)(d/dx-x)z

mean? This is in the question where the expression was given.

This is literally what I've written out in the post just above.

In more detail:

(\frac{d}{dx} -x)z = (\frac{d}{dx} z) - xz = \dfrac{dz}{dx} - xz

.

Then apply

(\frac{d}{dx}+x)

to this result similarly.

Reply 399