Results are out! Find what you need...fast. Get quick advice or join the chat
Hey! Sign in to get help with your study questionsNew here? Join for free to post

The 2013 STEP Prep Thread!

Announcements Posted on
Applying to Uni? Let Universities come to you. Click here to get your perfect place 20-10-2014
    • 41 followers
    Offline

    ReputationRep:
    Damn, you guys are eager... The 2012 lot haven't even got their results yet. :p:
    • 50 followers
    Offline

    ReputationRep:
    (Original post by Umairy363)
    Hey everyone. I want to start doing some STEP work for the exams next year so that i am thoroughly prepared. Where would be a good place to start? Should i just go over past papers, because the past papers are really difficult and overwhelming when i look at them. If you've done STEP, how did you find them in the end, and where did you start?
    Finishing A-Level Maths is a good place to start so you can get started on STEP I without the limitations of not having covered the syllabus. Just keep trying with questions. It will get easier.
    • 50 followers
    Offline

    ReputationRep:
    (Original post by wcp100)
    I'm not doing STEP III officially.
    Wait wait wait wait. Was that part of a STEP III question you were giving me? :zomg:
    • 50 followers
    Offline

    ReputationRep:
    (Original post by Farhan.Hanif93)
    Damn, you guys are eager... The 2012 lot haven't even got their results yet. :p:
    We just can't restrain ourselves any longer.
    • 0 followers
    Offline

    ReputationRep:
    (Original post by wcp100)
    I'm not doing STEP III officially.
    I see -- you are doing it for fun. However, others may need it.


    Try the following. I deduced it yesterday, and then used it to derive a further result from this question.
    Spoiler:
    Show

    Simplify \displaystyle \left(\sum_{i=0}^{m} x^i \right)\prod_{n=0}^{N}\left(\sum  _{i=0}^{m}(-1)^i \cdot x^{i \cdot 2^n}\right) in your notation.
    • 50 followers
    Offline

    ReputationRep:
    (Original post by jack.hadamard)
    Try the following. I deduced it yesterday, and then used it to derive a further result from this question.
    Spoiler:
    Show

    Simplify \displaystyle \left(\sum_{i=0}^{m} x^i \right)\prod_{n=0}^{N}\left(\sum  _{i=0}^{m}(-1)^i \cdot x^{i \cdot 2^n}\right) in your notation.
    Spoiler:
    Show
    I think I keep going astray because I've ended up with zero. :sigh:


    EDIT: NVM - trivial mistake.
    • 0 followers
    Offline

    ReputationRep:
    (Original post by Maths_Lover)
    EDIT: NVM - trivial mistake.
    It wasn't for you (difficulty), but the way I deal with those is to try out simpler cases first.

    Simplify

    (1 + x + x^2)(1 - x + x^2)

    where this corresponds to m=2, N=0. More cases, any patterns?


    Have you heard of mathematical induction?
    • 11 followers
    Offline

    ReputationRep:
    (Original post by Maths_Lover)
    Spoiler:
    Show
    I think I keep going astray because I've ended up with zero. :sigh:


    EDIT: NVM - trivial mistake.
    I wouldn't do too much of the question as it's from this year paper.
    • 50 followers
    Offline

    ReputationRep:
    (Original post by jack.hadamard)
    It wasn't for you (difficulty), but the way I deal with those is to try out simpler cases first.

    Simplify

    (1 + x + x^2)(1 - x + x^2)

    where this corresponds to m=2, N=0. More cases, any patterns?


    Have you heard of mathematical induction?
    Yeah - I just thought I would try it anyway (there's no harm). That would be a good idea! I was trying to do it algebraically - simplifying each expression in the brackets then combining some bits and so on until the whole thing is done. I am now getting somewhere with the algebra but I got to a part where I'm expanding something not so nice and I've got the first and last bits and it's the bit in the middle that's annoying to work out. :lol:

    1 + x^2 + x^4 . I think I'll leave this paper. I didn't realise it was this year's.

    I have, yes.

    I see what you're getting at! I'll leave it though and then I'll have forgotten this conversation next year.
    • 50 followers
    Offline

    ReputationRep:
    (Original post by wcp100)
    I wouldn't do too much of the question as it's from this year paper.
    Ah - I didn't know. I've only seen part of this year's STEP II paper which I'm saving as a mock. In that case, I shall just leave it until next year.
    • 0 followers
    Offline

    ReputationRep:
    (Original post by Maths_Lover)
    1 + x^2 + x^4 . I'll try that tomorrow.
    You should also note that these two are geometric series.
    Do this product, but don't look at the question paper.
    • 50 followers
    Offline

    ReputationRep:
    (Original post by jack.hadamard)
    You should also note that these two are geometric series.
    Do this product, but don't look at the question paper.
    Yes, I had noticed.

    Alrighty, just this one (It's getting a bit late and my eyelids are drooping, so another day) and I won't look at the paper. Besides, there'll be 12 other questions to have a go at when I do.

    If you are interested in what I'd done before:

    Spoiler:
    Show
    \displaystyle \left(\sum_{i=0}^{m} x^i \right)\prod_{n=0}^{N}\left(\sum  _{i=0}^{m}(-1)^i \cdot x^{i \cdot 2^n}\right)

    \displaystyle \sum_{i=0}^{m} x^i \right is a geometric series with (m+1) terms, first term 1 and common ratio x, so \displaystyle S_{m+1} = \frac{(x^{m+1} -1)}{x-1} .

    \displaystyle\sum  _{i=0}^{m}(-1)^i \cdot x^{i \cdot 2^n} is another geometric series with (m+1) terms and first term 1 but common ratio \displaystyle -x^{2^n} so \displaystyle S_{m+1} = \frac{1\left(-x^{2^n})^{m+1} -1 \right)}{-x^{2^n} -1} = \frac{1-x^{2^n}\cdot (x^{2^n})^m}{x^{2^n} + 1} .

    So then

    \displaystyle \prod_{n=0}^{N}\left(\sum  _{i=0}^{m}(-1)^i \cdot x^{i \cdot 2^n}\right)

    \displaystyle = \left(\frac{1-x^{2^0}\cdot (x^{2^0})^m}{x^{2^0} + 1} \right) \left(\frac{1-x^{2^1}\cdot (x^{2^1})^m}{x^{2^1} + 1} \right)\cdots \left(\frac{1-x^{2^N}\cdot (x^{2^N})^m}{x^{2^N} + 1} \right)

    \displaystyle = \left(\frac{1-x^{m+1}}{x + 1} \right)\left(\frac{1-x^{2(m+1)}}{x^2 + 1} \right)\left(\frac{1-x^{4(m+1)}}{x^4 + 1} \right)\cdots \left(\frac{1-x^{2^N(m+1)}}{x^{2^N} + 1} \right)

    So then I split the numerator of each term to get 1/denominator - the rest/denominator in each large bracket. Now I'm trying to expand this to get sets of terms that can be summed one way or another. So far, I've noticed that all of the 1/denomintor terms multiplied together make

    \displaystyle \frac{1}{x+1} \times \frac{1}{x^2+1} \times \cdots \times \frac{1}{x^{2^N}+1} = \frac{1}{(x+1)(x^2+1)\cdots (x^{2^N}+1)}

    The denominator is just (x+1) times the geometric series we'd evaluated earlier but with N instead of n, so

    \displaystyle \frac{1}{(x+1)(x^2+1)\cdots (x^{2^N}+1)} = \frac{1}{(1+x)\left[\frac{1-x^{2^n}}{1-x^2}\right]} = \frac{1-x}{1-x^{2^N}}

    Now to deal with those pesky other terms of the expansion, simplify and then multiply by the bit at the start... hopefully it will go to plan. :lol:
    • 50 followers
    Offline

    ReputationRep:
    (Original post by jack.hadamard)
    It wasn't for you (difficulty), but the way I deal with those is to try out simpler cases first.

    Simplify

    (1 + x + x^2)(1 - x + x^2)

    where this corresponds to m=2, N=0. More cases, any patterns?


    Have you heard of mathematical induction?
    Spoiler:
    Show
    Is this what you are getting at:

    Try some more cases of m and N and try to spot a pattern. Make a conjecture then try to prove it by induction?

    It would probably be a hell of a lot quicker/neater than the method I was trying!


    This was posted from The Student Room's Android App on my GT-I9100
    • 0 followers
    Offline

    ReputationRep:
    (Original post by Maths_Lover)
    Spoiler:
    Show

    Try some more cases of m and N and try to spot a pattern. Make a conjecture then try to prove it by induction?

    Either way works, but induction would be the neater, yes.

    Spoiler:
    Show

    Notice that you have

    \displaystyle \frac{1 - x^{m+1}}{1 - x} multiplied by \displaystyle \prod_{n} \frac{1 + (-1)^m x^{(m+1)2^n}}{1+x^{2^n}}

    when m is even, this reduces to

    Spoiler:
    Show

    \displaystyle \frac{1-x^{(m+1)2^{n+1}}}{1 - x^{2^{n+1}}}

    which can be represented as a sum as well.

    • 0 followers
    Offline

    ReputationRep:
    (Original post by Maths_Lover)
    Finishing A-Level Maths is a good place to start so you can get started on STEP I without the limitations of not having covered the syllabus. Just keep trying with questions. It will get easier.
    Hey, thanks for getting back to me!

    I'm going over the C1-C4 content now :P. Is it worth doing all 3 STEP papers? Are there some things that are extra that is not in the syllabus of C1-C4 that is questioned in the exam?? Isn't STEP III Further Mths?
    • 50 followers
    Offline

    ReputationRep:
    (Original post by Umairy363)
    Hey, thanks for getting back to me!

    I'm going over the C1-C4 content now :P. Is it worth doing all 3 STEP papers? Are there some things that are extra that is not in the syllabus of C1-C4 that is questioned in the exam?? Isn't STEP III Further Mths?
    No problem!

    That is good. It depends - which degree do you plan on doing? Personally, I am going to do all three wherever I go because I'm just a bit odd like that. STEP I and STEP II are based on C1-C4, M1, M2, S1, S2 and STEP III includes the Further Maths (and Additional Further Maths!) modules. You do not have to do Additional Further Maths if you don't want to but it is rather useful for STEP III as the applied maths questions rely on it.

    There are some questions on STEP which can be done in much quicker ways if you have learned some maths outside the syllabus, for example Number Theory, Geometry, Combinatorics and Inequalities. These questions can be figured out just by thinking about them, and using the concepts covered in A-Level Maths/F.Maths but it's a lot quicker/easier if you have the experience from covering those topics.

    You may find this document useful: http://www.admissionstests.cambridge...ation_2012.pdf :cute:
    • 50 followers
    Offline

    ReputationRep:
    (Original post by jack.hadamard)
    Either way works, but induction would be the neater, yes.

    Spoiler:
    Show

    Notice that you have

    \displaystyle \frac{1 - x^{m+1}}{1 - x} multiplied by \displaystyle \prod_{n} \frac{1 + (-1)^m x^{(m+1)2^n}}{1+x^{2^n}}

    when m is even, this reduces to

    Spoiler:
    Show

    \displaystyle \frac{1-x^{(m+1)2^{n+1}}}{1 - x^{2^{n+1}}}

    which can be represented as a sum as well.

    Spoiler:
    Show
    Ah yes! Thank you for that helpful hint.

    I haven't been feeling very well for a couple of weeks (but I still had exams, my last being AEA on Tuesday, so I couldn't really rest). I'm going to chill out now for a couple of days and hopefully get better before continuing with work.
    • 5 followers
    Offline

    ReputationRep:
    Finish C3 and C4 then look at STEP? Or look at STEP first/simultaneously?
    • 0 followers
    Offline

    ReputationRep:
    (Original post by Maths_Lover)
    ...
    Yeah, I also need to take a break after all my exams. Which year are you in?
    • 50 followers
    Offline

    ReputationRep:
    (Original post by jack.hadamard)
    Yeah, I also need to take a break after all my exams. Which year are you in?
    Yeah - you deserve it!

    I am in year 12 - I take it you've just finished year 13?
Updated: June 12, 2013
New on TSR

A-level results day

Is it about making your parents proud?

Article updates
Reputation gems:
You get these gems as you gain rep from other members for making good contributions and giving helpful advice.