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

Oxford MAT 2013/2014 Watch

Announcements
    Offline

    15
    ReputationRep:
    (Original post by MathGirl)
    I'm quite puzzled by the admissions statistics... Why is the competition for Mathematics & Statistics so much more intense than that for Maths/Maths & CompSci/Maths & Phil?:confused:
    In general there is no or little difference - there was one unusual year (the one whose figures are in the current prospectus) where fewer Maths & Stats students were admitted, I guess because the application field wasn't felt to be very strong that year.
    • Community Assistant
    Offline

    11
    ReputationRep:
    (Original post by Blazy)
    It's a shame we can't visit upstairs though



    To be honest, 'Why Maths?' and 'Why Oxford?' doesn't sound like a great question to ask in an interview when they could be spending the limited time they have assessing your ability to do maths. Needless to say, do think about it but don't get too worried about it. Enjoy it!
    Yes I will think about it and make sure I have some sort of response to the two questions. I think I am more worried about the general questions than the maths ones!

    Thanks for the advice
    Offline

    1
    ReputationRep:
    How likely is it that we'll get asked questions on A2 level vectors/mechanics etc? I'm focusing my attention on sharpening up proof/calculus/graph sketching, should I revise the rest of the syllabus aswell?
    Offline

    0
    ReputationRep:
    (Original post by Mike_Ross)
    How likely is it that we'll get asked questions on A2 level vectors/mechanics etc? I'm focusing my attention on sharpening up proof/calculus/graph sketching, should I revise the rest of the syllabus aswell?
    Doubt they will ask questions on mechanics. I'm focusing my attention on graph sketching and number theory. What kind of proof are you doing?
    Offline

    4
    ReputationRep:
    Are all the colleges supposed to have replied by now since its a week before the interview period?
    Offline

    17
    ReputationRep:
    (Original post by nahomyemane778)
    Are all the colleges supposed to have replied by now since its a week before the interview period?
    Yes.
    Offline

    4
    ReputationRep:
    (Original post by Noble.)
    Yes.
    I see- thanks.
    Offline

    4
    ReputationRep:
    (Original post by souktik)
    ...
    Holy **** Soutik - no wonder you are so good- you made it to India IMO shortlist according to stalking page- just reading through now and spotted it.
    Offline

    0
    ReputationRep:
    What are we supposed to take with us btw?
    Offline

    1
    ReputationRep:
    (Original post by nahomyemane778)
    Holy **** Soutik - no wonder you are so good- you made it to India IMO shortlist according to stalking page- just reading through now and spotted it.
    It's easier in India than in your country, trust me. India generally ranks 11-30 at the IMO, not in the top 10. Going to the selection camp here isn't half as big as going to the Trinity camp in the UK. Also, India shortlists 30+ people, much more than the UK's 20.

    Posted from TSR Mobile
    Offline

    0
    ReputationRep:
    (Original post by souktik)
    It's easier in India than in your country, trust me. India generally ranks 11-30 at the IMO, not in the top 10. Going to the selection camp here isn't half as big as going to the Trinity camp in the UK. Also, India shortlists 30+ people, much more than the UK's 20.

    Posted from TSR Mobile
    The answer to the maximum limit is (N(N-1))/2 + floor (N/2). Have you managed to prove it yet? If so I'd be interested to see your method.

    Posted from TSR Mobile
    Offline

    1
    ReputationRep:
    (Original post by NewtonsApple)
    The answer to the maximum limit is (N(N-1))/2 + floor (N/2). Have you managed to prove it yet? If so I'd be interested to see your method.

    Posted from TSR Mobile
    You could have just said floor N^2/2.
    Yes, I've proved it. I'll post the solution in LaTeX as soon as I get back home and have a proper internet connection. In very crude words, here's the main idea- when you expand V, you get each of 1 to N twice, preceded by either a + or a -. Number of + equals number of -=N. Best case is to ensure the highest N are + and the lowest N are -. Construction is easy when you get to the conditions.

    Posted from TSR Mobile
    Offline

    0
    ReputationRep:
    (Original post by souktik)
    You could have just said floor N^2/2.
    Yes, I've proved it. I'll post the solution in LaTeX as soon as I get back home and have a proper internet connection. In very crude words, here's the main idea- when you expand V, you get each of 1 to N twice, preceded by either a + or a -. Number of + equals number of -=N. Best case is to ensure the highest N are + and the lowest N are -. Construction is easy when you get to the conditions.

    Posted from TSR Mobile
    Yeah that sounds right. You can use the triangle inequality to deduce a similar argument but ultimately you construct one increasing sequence starting at the first term and one decreasing sequence starting with the second term I believe.

    Posted from TSR Mobile
    Offline

    1
    ReputationRep:
    (Original post by NewtonsApple)
    Yeah that sounds right. You can use the triangle inequality to deduce a similar argument but ultimately you construct one increasing sequence starting at the first term and one decreasing sequence starting with the second term I believe.

    Posted from TSR Mobile
    I suppose that'll work. There are many valid constructions. Can you explain how you were using the triangle inequality for the first part? You said the sum is greater than or equal to twice a. What was your a, I never figured that out.

    Posted from TSR Mobile
    Offline

    0
    ReputationRep:
    (Original post by souktik)
    I suppose that'll work. There are many valid constructions. Can you explain how you were using the triangle inequality for the first part? You said the sum is greater than or equal to twice a. What was your a, I never figured that out.

    Posted from TSR Mobile
    Okay, so write out the arrangement as the sum of differences starting at mod(A-B) and ending at mod(Z-Y). By applying the triangle inequality on each pair of successive terms we get mod(A-C)...+mod(Z-X) + (mod(A-Z)) where we simply add on the last term. When we use the triangle inequality again we get the sum of the differences is greater than or equal to two differences between two random terms. At most a difference is N-1 so we get the max is 2a or 2(N-1), we can do this since one of them is between non-successive terms. A bit convoluted but it works nonetheless for even cases. For odd cases there are some minor alterations but the method is still valid.
    Offline

    1
    ReputationRep:
    (Original post by NewtonsApple)
    Okay, so write out the arrangement as the sum of differences starting at mod(A-B) and ending at mod(Z-Y). By applying the triangle inequality on each pair of successive terms we get mod(A-C)...+mod(Z-X) + (mod(A-Z)) where we simply add on the last term. When we use the triangle inequality again we get the sum of the differences is greater than or equal to two differences between two random terms. At most a difference is N-1 so we get the max is 2a or 2(N-1), we can do this since one of them is between non-successive terms. A bit convoluted but it works nonetheless for even cases. For odd cases there are some minor alterations but the method is still valid.
    Okay, that's what I thought, but I'm not sure if this is complete. How do you deal with the cases in which a=N-1 is not achieved between two consecutive terms? Even if you deal with that case, you need to show that V=2(N-1) can actually be achieved for some configuration. The construction is trivial, but necessary.

    Posted from TSR Mobile
    Offline

    7
    ReputationRep:
    (Original post by ftball22)
    What are we supposed to take with us btw?
    The University site sums it up pretty well. I would probably add a warm coat, maybe gloves and a hat as you'll be going to other colleges for interviews. Maybe also a hot water bottle or something similar, as some of the college rooms can get pretty chilly.
    Offline

    0
    ReputationRep:
    (Original post by souktik)
    Okay, that's what I thought, but I'm not sure if this is complete. How do you deal with the cases in which a=N-1 is not achieved between two consecutive terms? Even if you deal with that case, you need to show that V=2(N-1) can actually be achieved for some configuration. The construction is trivial, but necessary.

    Posted from TSR Mobile
    Not really that difficult to fix tbh. I made a typo in my previous post which is probably why you picked up on this. It should be the sum of differences from mod(A-B) to mod(Y-Z) and we add on mod(A-Z) however applying the triangle inequality we get the differences are greater than 2*mod(A-Z)which means we only need 1 difference to be N-1 which can be easily constructed.

    Posted from TSR Mobile
    Offline

    1
    ReputationRep:
    (Original post by MathGirl)
    Doubt they will ask questions on mechanics. I'm focusing my attention on graph sketching and number theory. What kind of proof are you doing?
    Induction, contradiction and infinite descent mainly, some are just by inspection.

    Number theory as in combinatorics? Or like primes?
    Offline

    1
    ReputationRep:
    (Original post by NewtonsApple)
    Not really that difficult to fix tbh. I made a typo in my previous post which is probably why you picked up on this. It should be the sum of differences from mod(A-B) to mod(Y-Z) and we add on mod(A-Z) however applying the triangle inequality we get the differences are greater than 2*mod(A-Z)which means we only need 1 difference to be N-1 which can be easily constructed.

    Posted from TSR Mobile
    Err, I understood that you meant to say V>=2|a1-aN|. Even if |a1-aN|=N-1, a configuration with V=2(N-1) isn't guaranteed. You need to ensure that the conditions for equality of the triangle inequality are applied. And yeah, I agree that it's not difficult to fix.

    Posted from TSR Mobile
 
 
 
  • 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
    Will you be richer or poorer than your parents?
  • 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.