Hey there! Sign in to join this conversationNew here? Join for free
    Offline

    2
    ReputationRep:
    (Original post by farryharnworth)
    I was not. Was there another farry there?
    Ok. Never mind then.


    Posted from TSR Mobile
    • Section Leader
    • Clearing and Applications Advisor
    Offline

    21
    ReputationRep:
    Section Leader
    Clearing and Applications Advisor
    (Original post by farryharnworth)
    I was not. Was there another farry there?
    Offline

    20
    (Original post by farryharnworth)
    In a simultaneous effort to learn a bit of LaTeX and to rub in my own face just how stupid I was to miss out some of the questions, I've typed up the questions and solutions for STEP II (credit is given, even if I've extended, abridged, corrected, or tweaked solutions).

    I figured you guys may as well have it. Give me a shout if I've messed something up.
    I think its marvelous - definitely should be attached in the OP. Thanks for sharing!

    There is one very small thing that you have missed out of question (8i) - I hesitate to mention it, since I don't think its worth much bother to add it in, and you have done a great job otherwise on the LaTex - which is that, for our succinct notation to work, we just need to clarify what happens when m=1; that is, we should define
    \sum_{i=1}^{0} f(i) = 0

    But this is a very nice resource for future STEPpers (and makes my solutions look a lot neater than they were! ), so thank you again for sharing!
    Offline

    1
    ReputationRep:
    (Original post by Mathemagicien)
    I think its marvelous - definitely should be attached in the OP. Thanks for sharing!

    There is one very small thing that you have missed out of question (8i) - I hesitate to mention it, since I don't think its worth much bother to add it in, and you have done a great job otherwise on the LaTex - which is that, for our succinct notation to work, we just need to clarify what happens when m=1; that is, we should define
    \sum_{i=1}^{0} f(i) = 0

    But this is a very nice resource for future STEPpers (and makes my solutions look a lot neater than they were! ), so thank you again for sharing!
    Thank you, you're too kind

    You're absolutely right, I did miss that. It does seem logical to me that it would be true (and a quick google turns up this, which is a bit vague, but seems to support the convention), but of course it's not worth being anything but 100% clear in a STEP exam. I've updated it - can't get it to stick in an edit, so here it is with minor correction.
    Attached Images
  1. File Type: pdf STEP II 2016.pdf (357.6 KB, 75 views)
    • Thread Starter
    Offline

    22
    ReputationRep:
    (Original post by farryharnworth)
    Give me a shout if I've messed something up.
    Q3, the LHS is the sum of non-negative terms, not positive terms.

    Edit to add: Q6, not sure about this, but I think you need to explicitly check z(1) = 1 or y_(2n)(1) = 1 or both.

    Another edit: you seem to be missing the "write down for four events" in Q12.
    Offline

    20
    (Original post by Zacken)
    Edit to add: Q6, not sure about this, but I think you need to explicitly check z(1) = 1 or y_(2n)(1) = 1 or both
    What's your reasoning?
    • Thread Starter
    Offline

    22
    ReputationRep:
    (Original post by Mathemagicien)
    What's your reasoning?
    There's a unique solution to the DE as long as the solution satisfies that y(1) = 1 thing, so if you want to claim that both z_n and y_(2n) satisfy the same DE and hence must be the same, then shouldn't you show they both satisfy the condition that z(1) = 1 for you to actually show that they are a solution?
    Offline

    20
    (Original post by Zacken)
    There's a unique solution to the DE as long as the solution satisfies that y(1) = 1 thing, so if you want to claim that both z_n and y_(2n) satisfy the same DE and hence must be the same, then shouldn't you show they both satisfy the condition that z(1) = 1 for you to actually show that they are a solution?
    Damn, I didn't read that bit

    I agree
    Offline

    18
    ReputationRep:
    (Original post by Mathemagicien)
    Damn, I didn't read that bit

    I agree
    (Original post by Zacken)
    There's a unique solution to the DE as long as the solution satisfies that y(1) = 1 thing, so if you want to claim that both z_n and y_(2n) satisfy the same DE and hence must be the same, then shouldn't you show they both satisfy the condition that z(1) = 1 for you to actually show that they are a solution?
    z(1) = 1 follows because y_n(1) = 1 and z = 2y_n^2 - 1 etc but surely that's all you'd have to say? It's given that there is one solution satisfying y_k(1) = 1 for each k, and you've provided one for k = 2n, so z = y_2n.
    • Thread Starter
    Offline

    22
    ReputationRep:
    (Original post by StrangeBanana)
    z(1) = 1 follows because y_n(1) = 1 and z = 2y_n^2 - 1 etc but surely that's all you'd have to say? It's given that there is one solution satisfying y_k(1) = 1 for each k, and you've provided one for k = 2n, so z = y_2n.
    Yeah, was just saying that there's probably a mark allocated for stating it. :yes:
    Offline

    1
    ReputationRep:
    (Original post by Zacken)
    Q3, the LHS is the sum of non-negative terms, not positive terms.

    Edit to add: Q6, not sure about this, but I think you need to explicitly check z(1) = 1 or y_(2n)(1) = 1 or both.
    Right you are; sorted.

    (Original post by Zacken)
    Another edit: you seem to be missing the "write down for four events" in Q12.
    Not sure how I managed that - whoops.

    Thank you very much.
    Attached Images
  2. File Type: pdf STEP II 2016.pdf (357.9 KB, 109 views)
    • Thread Starter
    Offline

    22
    ReputationRep:
    (Original post by farryharnworth)
    Right you are; sorted.



    Not sure how I managed that - whoops.

    Thank you very much.
    Cheers, great work!
    Offline

    2
    ReputationRep:
    Are we allowed to post Step III solutions yet?
    Offline

    3
    ReputationRep:
    (Original post by riquix)
    Are we allowed to post Step III solutions yet?
    Yes

    Edit: no
    Offline

    19
    ReputationRep:
    (Original post by riquix)
    Are we allowed to post Step III solutions yet?
    I believe you have to wait 24 hours after the start time of the exam, i.e. you have to wait until 2pm
    Offline

    2
    ReputationRep:
    (Original post by 13 1 20 8 42)
    I believe you have to wait 24 hours after the start time of the exam, i.e. you have to wait until 2pm
    Ah right
    Offline

    2
    ReputationRep:
    Grade boundary predictions?
    • Section Leader
    • Clearing and Applications Advisor
    Offline

    21
    ReputationRep:
    Section Leader
    Clearing and Applications Advisor
    (Original post by Shrek1234)
    Grade boundary predictions?
    No discussion until 2pm (BST)

    Posted from TSR Mobile
    Offline

    18
    ReputationRep:
    2pm **** sake
    Offline

    18
    ReputationRep:
    saving this space for question 7

    don't steel it :<
 
 
 
  • See more of what you like on The Student Room

    You can personalise what you see on TSR. Tell us a little about yourself to get started.

  • Poll
    Would you like to hibernate through the winter months?
  • See more of what you like on The Student Room

    You can personalise what you see on TSR. Tell us a little about yourself to get started.

  • The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

    Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

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