A Summer of Maths

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  • View Poll Results: Have you studied any Group Theory already?
    Yes, I did some Group Theory during/at my A-level.
    		24 34.78%
    No, but I plan to study some before Uni.
    		15 21.74%
    No, I haven't.
    		30 43.48%
    Voters: 69. You may not vote on this poll

  1. DamoclesAustria's Avatar
    401 12-07-2012  16:34
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    Re: A Summer of Maths
    (Original post by Rahul.S)
    thanks for the stuff. Ive done little IMO stuff before, but some of the questions do seem interesting. Would you recommend getting the books?
    Which books?

    I recommend Titu Andreescu's "Problems from the Book". (There is a draft version online)
    I met Titu at the IMO last year and had my copy signed
    Post rating:
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  2. Rahul.S's Avatar
    402 12-07-2012  16:50
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    Re: A Summer of Maths
    (Original post by DamoclesAustria)
    Which books?

    I recommend Titu Andreescu's "Problems from the Book". (There is a draft version online)
    I met Titu at the IMO last year and had my copy signed
    http://www.artofproblemsolving.com/index.php?mode=books this stuff?

    Titu sounds jokes guessing Romanian? do you do the IMO last year?
  3. TheMagicMan's Avatar
    403 12-07-2012  18:21
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    (Original post by Rahul.S)
    is this IMO stuff any good for uni shizz?
    None of the material is useful. The two main things you get out of training from the IMO are problem solving skills and a forced early introduction to 'advanced' maths.... I wouldn't have done half the Uni maths I have if I hadn't been dragged through the AMS for four years


    This was posted from The Student Room's iPhone/iPad App
  4. Tobedotty's Avatar
    404 12-07-2012  19:10
    • Exalted and Worshipped Member
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    Re: A Summer of Maths
    (Original post by TheMagicMan)
    None of the material is useful. The two main things you get out of training from the IMO are problem solving skills and a forced early introduction to 'advanced' maths.... I wouldn't have done half the Uni maths I have if I hadn't been dragged through the AMS for four years


    This was posted from The Student Room's iPhone/iPad App
    Did you want to do the AMS? And are you glad that you did?
  5. Rahul.S's Avatar
    405 12-07-2012  20:34
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    Re: A Summer of Maths
    (Original post by TheMagicMan)
    None of the material is useful. The two main things you get out of training from the IMO are problem solving skills and a forced early introduction to 'advanced' maths.... I wouldn't have done half the Uni maths I have if I hadn't been dragged through the AMS for four years


    This was posted from The Student Room's iPhone/iPad App
    ah I see. I might still just look at some questions- I get little time nowadays to do much- and there are IMO questions which I can dive into without actually reading any new material my mate was telling me there are quite alot of Trinity 2012 applicants in the IMO team- they must of raised the STEP boundary
  6. jack.hadamard's Avatar
    406 12-07-2012  20:47
    • Benevolent Member
    • Posts: 696
    Re: A Summer of Maths
    (Original post by Rahul.S)
    http://www.artofproblemsolving.com/index.php?mode=books this stuff?

    Titu sounds jokes guessing Romanian? do you do the IMO last year?
    I don't know how it sounds, but this guy's team was the first to obtain a perfect score in the IMO.
  7. jack.hadamard's Avatar
    407 12-07-2012  20:57
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    • Posts: 696
    Re: A Summer of Maths
    (Original post by DamoclesAustria)
    Does anybody want to post their solutions to any pending problems?

    Here's a new one:
    Let p_k(x)=1+x+\ldots+x^{k-1}. Given that gcd(k,n)=1, show that p_k(x) divides p_k(x^n).
    Spoiler:
    Show

    I would bash it quickly before dinner.

    \displaystyle p_k(x) = \frac{1 - x^k}{1 - x} and \displaystyle p_k(x^n) = \frac{1 - x^{kn}}{1 - x^n}

    Hence,

    \displaystyle \frac{p_k(x^n)}{p_k(x)} = \frac{(1 - x^{kn})(1 - x)}{(1 - x^n)(1 - x^k)}

    and, considering subgroups, on top we have C_{kn} with C_1, and on the bottom subgroups C_n \cap C_k = C_1, where C_{nk} the group of kn-th roots of unity.
  8. jack.hadamard's Avatar
    408 12-07-2012  20:59
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    • Posts: 696
    Re: A Summer of Maths
    (Original post by Lord of the Flies)
    I fixed what I had overlooked with the fct. eq. and now have a flawless proof/solution (or so it seems)
    Didn't have time for it; being busy all day. Will have a look at it again tonight.

    (Original post by ben-smith)
    Where do you think the quote in my sig comes from?
    Well, I didn't know people had signatures -- they are rather hidden.
  9. jack.hadamard's Avatar
    409 12-07-2012  21:03
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    Re: A Summer of Maths
    I think Groups will be more interesting than Analysis?
  10. Tobedotty's Avatar
    410 12-07-2012  21:26
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    Re: A Summer of Maths
    (Original post by jack.hadamard)
    I think Groups will be more interesting than Analysis?
    I agree
  11. DamoclesAustria's Avatar
    411 12-07-2012  21:27
    • Full Member
    • Posts: 90
    Re: A Summer of Maths
    (Original post by jack.hadamard)
    I think Groups will be more interesting than Analysis?
    Groups are interesting as well but I don't have as many exercises to share...

    Here's one:
    Let G be a group. The conjugation c:G\to G of an element x with y is c_y(x)=yxy^{-1}, where x,y\in G. (Show that c is a group homomorphism.)

    Define the function t: G\to Aut(G) that maps an element of G to the conjugation with this element, so y\mapsto c_y. Show that t is also a group homomorphism.
  12. Dog4444's Avatar
    412 12-07-2012  21:35
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    Re: A Summer of Maths
    (Original post by DamoclesAustria)
    ...
    Is it true that a lot of IMO-level/type problems (and olympiad problems in general) can be solved easily by advanced (undergrad and beyond) maths?
  13. Rahul.S's Avatar
    413 12-07-2012  21:36
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    Re: A Summer of Maths
    (Original post by jack.hadamard)
    I don't know how it sounds, but this guy's team was the first to obtain a perfect score in the IMO.
    is that a big ting? because thats the first im hearing of it
    Post rating:
    1
  14. jack.hadamard's Avatar
    414 12-07-2012  21:41
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    • Posts: 696
    Re: A Summer of Maths
    (Original post by DamoclesAustria)
    Groups are interesting as well but I don't have as many exercises to share...
    I have a lot of exercises, but we have to get started from somewhere; i.e. conjugation and group homomorphisms is quite a jump.

    Does everyone know what a group is? Can we get some axioms and examples?
  15. jack.hadamard's Avatar
    415 12-07-2012  21:51
    • Benevolent Member
    • Posts: 696
    Re: A Summer of Maths
    (Original post by Rahul.S)
    is that a big ting? because thats the first im hearing of it
    Well, unless this year there is a team with a perfect score, it will be also the only one.
  16. Rahul.S's Avatar
    416 12-07-2012  21:55
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    Re: A Summer of Maths
    (Original post by jack.hadamard)
    Well, unless this year there is a team with a perfect score, it will be also the only one.
    guess that is sik?! I dont even know a lot about IMO etc. :lol:
  17. DamoclesAustria's Avatar
    417 12-07-2012  21:55
    • Full Member
    • Posts: 90
    Re: A Summer of Maths
    (Original post by Dog4444)
    Is it true that a lot of IMO-level/type problems (and olympiad problems in general) can be solved easily by advanced (undergrad and beyond) maths?
    No, definitely not. Even if you disregard elementary geometry, hard olmypiad problems require a lot of experience and training both of which you don't get simply by studying university maths.
  18. DamoclesAustria's Avatar
    418 12-07-2012  22:08
    • Full Member
    • Posts: 90
    Re: A Summer of Maths
    (Original post by jack.hadamard)
    I have a lot of exercises, but we have to get started from somewhere; i.e. conjugation and group homomorphisms is quite a jump.

    Does everyone know what a group is? Can we get some axioms and examples?
    So you suggest that we all do the same stuff at the same time? That won't work.

    Anyway, here's a playlist of a course on group theory (on youtube).

    The first chapter here gives an introduction to groups as well.
  19. TheMagicMan's Avatar
    419 12-07-2012  22:18
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    • Posts: 2,940
    (Original post by Tobedotty)
    Did you want to do the AMS? And are you glad that you did?
    I had no clue what it was...I was a little kid when I was asked to do it so I wasn't going to say no

    The unfortunate thing was was that by the time I was mature enough (I wasn't nearlytalented enough to be challenging for the IMO as an early teen like a Tao) to challenge for the IMO team I was completely jaded with that kind of maths


    This was posted from The Student Room's iPhone/iPad App
  20. jack.hadamard's Avatar
    420 12-07-2012  22:43
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    • Posts: 696
    Re: A Summer of Maths
    (Original post by jack.hadamard)
    Question:

    Let f: \mathbb{R} \to \mathbb{R} be a function satisfying \displaystyle f\left(\frac{x_1 + x_2}{2}\right) = \frac{f(x_1) + f(x_2)}{2} for any x_1, x_2.

    Prove that

    \displaystyle f\left(\frac{x_1 + x_2 + \cdots + x_n}{n}\right) = \frac{f(x_1) + f(x_2) + \cdots + f(x_n)}{n}

    for any x_1, x_2, ..., x_n.
    Solution:

    Spoiler:
    Show

    Notice that the case n = 2 is contained in the hypothesis.


    Lets assume n = 2^k, and we show, by induction on k, that n = 2^{k+1} must be true.


    \displaystyle f\left(\frac{x_1 + \cdots x_{2^k} + x_{2^k + 1} + \cdots + x_{2^{k+1}}}{2^{k+1}}\right)\ =\ \frac{\displaystyle f\left(\frac{x_1 + \cdots + x_{2^k}}{2^k}\right) + f\left(\frac{x_{2^k + 1} + \cdots + x_{2^{k+1}}}{2^k}\right)}{2}


    \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\ \frac{\displaystyle \frac{f(x_1) + \cdots + f(x_{2^k})}{2^k} + \frac{f(x_{2^k + 1}) + \cdots + f(x_{2^{k+1}})}{2^k}}{2}


    \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\ \frac{f(x_1) + \cdots + f(x_{2^k}) + f(x_{2^k + 1}) + \cdots + f(x_{2^{k+1}})}{2^{k+1}}



    Now, we show that if true for n, then it is true for n - 1.

    Consider x_1, x_2, ..., x_{n-1} and \displaystyle x_n = \frac{x_1 + x_2 + \cdots + x_{n-1}}{n-1}, then by hypothesis we obtain


    \displaystyle f\left(\frac{x_1 + \cdots + x_{n-1} + \frac{x_1 + \cdots + x_{n-1}}{n-1}}{n}\right) = \frac{f(x_1) + \cdots + f(x_{n-1}) + f\left(\frac{x_1 + \cdots + x_{n-1}}{n-1}\right)}{n}

    which is the same as

    \displaystyle f\left(\frac{x_1 + \cdots + x_{n-1}}{n - 1} \right) = \frac{f(x_1) + \cdots + f(x_{n-1})}{n} + \frac{1}{n} f\left(\frac{x_1 + \cdots + x_{n-1}}{n-1} \right)

    and it is easily rearranged to

    \displaystyle f\left( \frac{x_1 + x_2 + \cdots + x_{n-1}}{n-1} \right) = \frac{f(x_1) + f(x_2) + \cdots + f(x_{n-1})}{n-1}

    and the proof is complete.
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