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    (Original post by bananarama2)
    I too want to know this
    (Original post by metaltron)
    May I ask what cyc on the bottom of a sigma means?
    http://www.gamefaqs.com/boards/57621...athon/48835430
    I think it means the sums of the combinations
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    (Original post by metaltron)
    I'm on Level 3 as well, after making your mistake yesterday. I think it might help mastering Level 3 on Combinatorics though.
    I'm on level 4 geo/combs and I still have no idea what's going on :lol: It just teaches you to think rather than master a particular topic (because you don't have solutions to compare your answers to until long after you care about the problem in most cases). I think theres an overlap aswell. What level number theory are you on?
    May I ask what cyc on the bottom of a sigma means?
    (Original post by bananarama2)
    I too want to know this
    I wondered the same thing too. I've concluded it stands for "cyclic" or something? It means like symmetrical across all the specified variables. So, for example, if you were dealing with j, k and n, \displaystyle\sum_{cyc}j=jkn means j+k+n=jkn. I just thought how funny it would be to set problems with variables j, k and n :lol: It could be like my signature :lol:
    • PS Helper
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    (Original post by Zephyr1011)
    This problem looks a lot like product rule, although I'm not sure if that has any significance.
    If you take the case x=y, then f(x^2)=2xf(x).
    If f(x)=ax+c, then f(xy)=axy+c=2axy+(x+y)c. As both x and y are positive integers greater than one, the only possible solution is if both constants are 0. Therefore a possible solution is f(x)=0
    This is about as far as I've gotten
    Could you not have just looked at it and noticed that each term contained an f(something), hence f(x) = 0 is a solution?

    (Original post by Mladenov)
    Yup, you are right, there is a flaw.
    Your first equation may not be true for y \not= y', since it is possible that f is injective.
    Secondly, continuity does not imply differentiability.
    I posed the question to my class, and made this exact point, quoting the Weierstrass function as an example :yy:
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    (Original post by Jkn)
    I'm on level 4 geo/combs and I still have no idea what's going on :lol: It just teaches you to think rather than master a particular topic (because you don't have solutions to compare your answers to until long after you care about the problem in most cases). I think theres an overlap aswell. What level number theory are you on?




    I wondered the same thing too. I've concluded it stands for "cyclic" or something? It means like symmetrical across all the specified variables. So, for example, if you were dealing with j, k and n, \displaystyle\sum_{cyc}j=jkn means j+k+n=jkn. I just thought how funny it would be to set problems with variables j, k and n :lol: It could be like my signature :lol:
    Level 3 as I didn't take the introduction questions seriously. Maybe you could post some of the Level 4/5 problems on here?
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    (Original post by metaltron)
    Level 3 as I didn't take the introduction questions seriously. Maybe you could post some of the Level 4/5 problems on here?
    What's your highest point scoring problem on the number theory section? (just type the first few words if you like, or the gist, because I want to see if its the same as some of mine), You should've listening :lol: haha

    I'm still stuck on level 4 think you've gotta do well on 2 consecutive weeks to level up and I've been a bit on-off! I've been having a go at them and I've done 4 so far on the n/t (I did a 5th but by the time I realised my mistake I'd already entered 3 different answers :lol:).

    I'll post one or two of them as problems tonight or tomorrow. The ones I've done so far are a little boring and I don't want to clog up the thread with such things.
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    (Original post by Jkn)
    What's your highest point scoring problem on the number theory section? (just type the first few words if you like, or the gist, because I want to see if its the same as some of mine), You should've listening :lol: haha

    I'm still stuck on level 4 think you've gotta do well on 2 consecutive weeks to level up and I've been a bit on-off! I've been having a go at them and I've done 4 so far on the n/t (I did a 5th but by the time I realised my mistake I'd already entered 3 different answers :lol:).

    I'll post one or two of them as problems tonight or tomorrow. The ones I've done so far are a little boring and I don't want to clog up the thread with such things.
    I got 180 points I think for a Number Theory Question, it was Q8 but it was similar to STEP I Q1 2000.

    For Combinatorics I got 180 points too, but also got three wrong answers for one so that ruined the moment for me really.

    I think you can only get 180 pts for a question in Level 3.
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    (Original post by Jkn)
    What's your highest point scoring problem on the number theory section? (just type the first few words if you like, or the gist, because I want to see if its the same as some of mine), You should've listening :lol: haha

    I'm still stuck on level 4 think you've gotta do well on 2 consecutive weeks to level up and I've been a bit on-off! I've been having a go at them and I've done 4 so far on the n/t (I did a 5th but by the time I realised my mistake I'd already entered 3 different answers :lol:).

    I'll post one or two of them as problems tonight or tomorrow. The ones I've done so far are a little boring and I don't want to clog up the thread with such things.
    Wait, are you folks talking about brilliant.org?

    Do forgive me, for I'm feeling far too indolent to scroll up to find out.
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    Solution 99

    Set f(x)=xz(x) then z(xy)=z(x)+z(y). It is a well known fact that this solves to z(x)=a\ln x,\;a\in\mathbb{R} hence f(x)=ax\ln x

    Completely forgot to reply to these, apologies:

    (Original post by Mladenov)
    I would choose MPSI and then MP* (I think, if I am good enough, they will select me to do MP* after the first year. Is this so?).
    Yes.

    (Original post by Mladenov)
    By the way, as far as I know, there is no programme which offers more mathematics, is there? .
    No there is not - I meant: be prepared to do things other than maths (which, if you're as stubborn as I am, is an issue)

    (Original post by Mladenov)
    To qualify for LLG, I ought to finish first or second on the examination, which takes place each year in May. There are roughly 50 students from my country who take the exam.
    Best of luck!

    (Original post by Jkn)
    WTAF that looks weird! So what level are the universities you go to after the prépa, like masters degrees? Are they more prestigious and exclusive than onbridge/ivy-league? Soo confused!
    Prépa stands for Classes préparatoires aux grandes écoles, so you'd usually go to a grande école afterwards (the ENS is one of these). These are not universities, see here for more information. The famous grandes écoles are very prestigious indeed, but difficult to compare with Ivies/Oxbridge since they are not universities and function somewhat differently.



    This thread is lacking integrals!

    Problem 102*

    Mladenov's functional eq. reminded me of this one.

    For 0\leq (x,y)\leq 1,\; f is a continuous function satisfying xf(y)+yf(x)\leq 1. Prove that:

    \displaystyle\int_0^1 f(x)\,dx\leq \frac{\pi}{4}


    Problem 103* (there is a slick *** solution though)

    \displaystyle \int_0^{\infty} \arctan\left( \frac{x(e-1)}{ex^2+1}\right)\frac{dx}{x}


    (Original post by Mladenov)
    It seems as though you are the only one who enjoys doing functional eqs.
    Functional equations are lovely! Here is a not-so-easy one:

    Problem 104**

    Find all continuous f: [-1,1]\to\mathbb{R} such that 2xf(x)=f(2x^2-1)

    (this one is popular on TSR so many of you may have seen it)
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    (Original post by Lord of the Flies)
    Prépa stands for Classes préparatoires aux grandes écoles, so you'd usually go to a grande école afterwards (the ENS is one of these). These are not universities, see here for more information. The famous grandes écoles are very prestigious indeed, but difficult to compare with Ivies/Oxbridge since they are not universities and function somewhat differently.
    Oh right I see (I remember reading about them requiring you to pledge to the french civil service?)

    Do you want to be a mathematician, btw?
    This thread is lacking integrals!

    Problem 102***

    \displaystyle \int_0^{\infty}\frac{1}{x}\cdot \arctan\left( \frac{x(e-1)}{ex^2+1}\right)\,dx
    "***" ?:eek: Are you sure?!
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    (Original post by Jkn)
    Oh right I see (I remember reading about them requiring you to pledge to the french civil service?)
    Yeah I think that's right :lol:

    (Original post by Jkn)
    Do you want to be a mathematician, btw?
    If I can, yes I'd love to.

    (Original post by Jkn)
    "***" ?:eek: Are you sure?!
    The solution I have in mind is ***, but on second thought I might have an idea for a * solution as well.

    Edit: yes I confirm, it is doable with A-level knowledge only.
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    Solution 101

    We prove the lower bound first.
    I claim:
    \displaystyle \sum_{cyc}x\sqrt [4]{\frac{4}{27(x+y)^2(yz+zx+xy)}} \ge \sum_{cyc}\frac{x(x+y+2z)}{x^{2}  +y^{2}+z^{2}+3(yz+zx+xy)}.
    We have
    \displaystyle 4x^{4}(x^{2}+y^{2}+z^{2}+3(yz+zx  +xy))^{4}-27x^{4}(x+y+2z)^{4}(x+y)^2(yz+zx  +xy)= (x^2+y^2-2z^2)^2(z^2+3z(x+y)+x^2+3xy+y^2)  (z^2+9z(x+y)+4x^2+9xy+4y^2) +27(x+y)^2(x+y+2z)^2(z^2-xy)^2 \ge 0 \end{aligned}.
    Since \displaystyle \sum_{cyc}\frac{x(x+y+2z)}{x^{2}  +y^{2}+z^{2}+3(yz+zx+xy)}=1 we are done.
    Let us consider the upper bound.
    Cauchy - Schwarz yields:
    \displaystyle \left(\sum_{cyc} \frac{x}{\sqrt{x+y}} \right)^{2} \le \left(\sum_{cyc} \frac{\sqrt{x}}{\sqrt{x+y}} \right)\left(\sum_{cyc} x\frac{\sqrt{x}}{\sqrt{x+y}} \right).
    We bound above each of the sums in the right-handed side.
    Without loss of generality assume xyz=1. Hence (x+y)(y+z)(z+x) \ge 8.
    Using Cauchy - Schwarz, we obtain
    \begin{aligned} \displaystyle \sum_{cyc} \frac{\sqrt{x}}{\sqrt{x+y}} & \le \sqrt{\frac{2(x+y+z)(x(y+z)+y(x+  z)+z(x+y))}{(x+y)(y+z)(z+x)}} \\&= 2\sqrt{\frac{(x+y+z)(xy+yz+zx)}{  (x+y)(y+z)(z+x)}} \\& = 2\sqrt{\frac{(x+y)(y+z)(z+x)+xyz  }{(x+y)(y+z)(z+x)}} \\& \le 2\sqrt{\frac{9}{8}} = \frac{3}{\sqrt{2}} \end{aligned}.
    Therefore, we shall have complete solution if we show that
    \begin{aligned} \displaystyle \sum_{cyc} x\frac{x}{\sqrt{x+y}} \le \sqrt{\frac{3}{2}(x^{2}+y^{2}+z^  {2})}
    Again by Cauchy - Schwarz we have:
    \begin{aligned} \displaystyle \sum_{cyc} x\frac{\sqrt{x}}{\sqrt{x+y}} \le \sqrt{\frac{(xy+yz+zx+x^{2}+y^{2  }+z^{2})(x^{2}y+y^{2}x+x^{2}z+z^  {2}x+z^{2}y+y^{2}z)}{(x+y)(y+z)(  z+x)}}.
    Let us see who likes arithmetic, and how much.
    \begin{aligned} \displaystyle \frac{(xy+yz+zx+x^{2}+y^{2}+z^{2  })(x^{2}y+y^{2}x+x^{2}z+z^{2}x+z  ^{2}y+y^{2}z)}{(x+y)(y+z)(z+x)} \le \frac{3}{2}(x^{2}+y^{2}+z^{2}) \end{aligned};
    \begin{aligned} \displaystyle 2(xy+yz+zx+x^{2}+y^{2}+z^{2})(x^  {2}y+y^{2}x+x^{2}z+z^{2}x+z^{2}y  +y^{2}z) \le 3(x^{2}+y^{2}+z^{2})(x+y)(y+z)(z  +x) \end{aligned};
    \begin{aligned} \displaystyle  4\left(\sum_{cyc} x \right)\left(\sum_{cyc} xy \right)^{2} \le \left(\sum_{cyc} x \right)^{3}\sum_{cyc} xy + 3\left(\sum_{cyc} x \right)^{2} \end{aligned};
    \begin{aligned} \displaystyle 4\left(\sum_{cyc} xy \right)^{2} \le \left(\sum_{cyc} x \right)^{2}\left(\sum_{cyc} xy \right) + 3\left(\sum_{cyc} x \right) \end{aligned}
    \begin{aligned} \displaystyle 4\left(\sum_{cyc} xy \right)^{2} \le 2\left(\sum_{cyc} xy \right)^{2}+ \left(\sum_{cyc} xy \right)\left(\sum_{cyc} x^{2} \right) + 3\left(\sum_{cyc} x \right) \end{aligned};
    \begin{aligned} \displaystyle 2\left(\sum_{cyc} x^{2}y^{2} \right) \le \left(\sum_{cyc} x^{2} \right)\left(\sum_{cyc} xy \right) - \left(\sum_{cyc} x \right) \end{aligned};
    \begin{aligned} \displaystyle 2\left( \sum_{cyc} x^{2}y^{2} \right) \le \left(\sum_{cyc} x^{3}y \right) \end{aligned};
    \begin{aligned} \displaystyle 0 \le \left(\sum_{cyc} xy(x-y)^{2} \right) \end{aligned}.

    Approximately 4 hours.
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    (Original post by MW24595)
    Wait, are you folks talking about brilliant.org?
    But of course!
    Do forgive me, for I'm feeling far too indolent to scroll up to find out.
    Hhahaa :lol: lad
    (Original post by Lord of the Flies)
    If I can, yes I'd love to.
    Me too. I'd love to lecture and write articles and stuff swell though too!

    Cannot for the life of me work out how good you need to be though... any idea?
    The solution I have in mind is ***, but on second thought I might have an idea for a * solution as well (although it slightly defies the point of the problem, in my opinion).
    How is it that you seem to know so much university-level maths?! Oh and how have you done on French maths competitions? (i.e. close to the IMO team?)
    Spoiler:
    Show
    when expanded with use of the addition formula it simply becomes a case of finding the integral of \frac{arctan{ax}}{ax} which doesn't seem too bad. I tried it though (substituted x as 1/u) and the indefinite integral had an "ax" in which doesn't work so I think I did something wrong :/
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    (Original post by Mladenov)
    Spoiler:
    Show
    Solution 101

    We prove the lower bound first.
    I claim:
    \displaystyle \sum_{cyc}x\sqrt [4]{\frac{4}{27(x+y)^2(yz+zx+xy)}} \ge \sum_{cyc}\frac{x(x+y+2z)}{x^{2}  +y^{2}+z^{2}+3(yz+zx+xy)}.
    We have
    \displaystyle 4x^{4}(x^{2}+y^{2}+z^{2}+3(yz+zx  +xy))^{4}-27x^{4}(x+y+2z)^{4}(x+y)^2(yz+zx  +xy)= (x^2+y^2-2z^2)^2(z^2+3z(x+y)+x^2+3xy+y^2)  (z^2+9z(x+y)+4x^2+9xy+4y^2) +27(x+y)^2(x+y+2z)^2(z^2-xy)^2 \ge 0 \end{aligned}.
    Since \displaystyle \sum_{cyc}\frac{x(x+y+2z)}{x^{2}  +y^{2}+z^{2}+3(yz+zx+xy)}=1 we are done.
    Let us consider the upper bound.
    Cauchy - Schwarz yields:
    \displaystyle \left(\sum_{cyc} \frac{x}{\sqrt{x+y}} \right)^{2} \le \left(\sum_{cyc} \frac{\sqrt{x}}{\sqrt{x+y}} \right)\left(\sum_{cyc} x\frac{\sqrt{x}}{\sqrt{x+y}} \right).
    We bound above each of the sums in the right-handed side.
    Without loss of generality assume xyz=1. Hence (x+y)(y+z)(z+x) \ge 8.
    Using Cauchy - Schwarz, we obtain
    \begin{aligned} \displaystyle \sum_{cyc} \frac{\sqrt{x}}{\sqrt{x+y}} & \le \sqrt{\frac{2(x+y+z)(x(y+z)+y(x+  z)+z(x+y))}{(x+y)(y+z)(z+x)}} \\&= 2\sqrt{\frac{(x+y+z)(xy+yz+zx)}{  (x+y)(y+z)(z+x)}} \\& = 2\sqrt{\frac{(x+y)(y+z)(z+x)+xyz  }{(x+y)(y+z)(z+x)}} \\& \le 2\sqrt{\frac{9}{8}} = \frac{3}{\sqrt{2}} \end{aligned}.
    Therefore, we shall have complete solution if we show that
    \begin{aligned} \displaystyle \sum_{cyc} x\frac{x}{\sqrt{x+y}} \le \sqrt{\frac{3}{2}(x^{2}+y^{2}+z^  {2})}
    Again by Cauchy - Schwarz we have:
    \begin{aligned} \displaystyle \sum_{cyc} x\frac{\sqrt{x}}{\sqrt{x+y}} \le \sqrt{\frac{(xy+yz+zx+x^{2}+y^{2  }+z^{2})(x^{2}y+y^{2}x+x^{2}z+z^  {2}x+z^{2}y+y^{2}z)}{(x+y)(y+z)(  z+x)}}.
    Let us see who likes arithmetic, and how much.
    \begin{aligned} \displaystyle \frac{(xy+yz+zx+x^{2}+y^{2}+z^{2  })(x^{2}y+y^{2}x+x^{2}z+z^{2}x+z  ^{2}y+y^{2}z)}{(x+y)(y+z)(z+x)} \le \frac{3}{2}(x^{2}+y^{2}+z^{2}) \end{aligned};
    \begin{aligned} \displaystyle 2(xy+yz+zx+x^{2}+y^{2}+z^{2})(x^  {2}y+y^{2}x+x^{2}z+z^{2}x+z^{2}y  +y^{2}z) \le 3(x^{2}+y^{2}+z^{2})(x+y)(y+z)(z  +x) \end{aligned};
    \begin{aligned} \displaystyle  4\left(\sum_{cyc} x \right)\left(\sum_{cyc} xy \right)^{2} \le \left(\sum_{cyc} x \right)^{3}\sum_{cyc} xy + 3\left(\sum_{cyc} x \right)^{2} \end{aligned};
    \begin{aligned} \displaystyle 4\left(\sum_{cyc} xy \right)^{2} \le \left(\sum_{cyc} x \right)^{2}\left(\sum_{cyc} xy \right) + 3\left(\sum_{cyc} x \right) \end{aligned}
    \begin{aligned} \displaystyle 4\left(\sum_{cyc} xy \right)^{2} \le 2\left(\sum_{cyc} xy \right)^{2}+ \left(\sum_{cyc} xy \right)\left(\sum_{cyc} x^{2} \right) + 3\left(\sum_{cyc} x \right) \end{aligned};
    \begin{aligned} \displaystyle 2\left(\sum_{cyc} x^{2}y^{2} \right) \le \left(\sum_{cyc} x^{2} \right)\left(\sum_{cyc} xy \right) - \left(\sum_{cyc} x \right) \end{aligned};
    \begin{aligned} \displaystyle 2\left( \sum_{cyc} x^{2}y^{2} \right) \le \left(\sum_{cyc} x^{3}y \right) \end{aligned};
    \begin{aligned} \displaystyle 0 \le \left(\sum_{cyc} xy(x-y)^{2} \right) \end{aligned}.

    Approximately 4 hours.
    MOTHER OF GOD YOU ACTUALLY DID IT! I ONLY POSTED IT BECAUSE IT LOOKED INSANE! :eek:

    YOU'RE A ****ING ---->:troll:

    Congratulations!

    Now to try and understand/check your proof... :bhangra:
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    (Original post by Jkn)
    Btw, have you done your team selection test yet?
    No, TST takes place on May 15 & 16.


    (Original post by Jkn)
    ..
    Problem 101 was almost impossible. I was about to give up...

    (Original post by The Polymath)
    I posed the question to my class, and made this exact point, quoting the Weierstrass function as an example :yy:
    Actually, the set of differentiable functions is a meager set in the space of continuous functions.
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    (Original post by Jkn)
    Cannot for the life of me work out how good you need to be though... any idea?
    Absolutely none!

    (Original post by Jkn)
    How is it that you seem to know so much university-level maths?! Oh and how have you done on French maths competitions? (i.e. close to the IMO team?)
    I really don't know much uni-level maths actually, just a few very basic things which I've come across in my spare time. I've never participated in any contests - my school never mentioned the Concours Général when it was time, so I missed it (thereby ruling out the possibility of doing any of the further contests as well).

    (Original post by Mladenov)
    Problem 101 was almost impossible. I was about to give up...
    That looks painful - congrats!
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    (Original post by Lord of the Flies)
    I really don't know much uni-level maths actually, just a few very basic things which I've come across in my spare time. I've never participated in any contests - my school never mentioned the Concours Général when it was time, so I missed it (thereby ruling out the possibility of doing any of the further contests as well).
    Dude that sucks! Why don't you, for the hell of it, try an IMO paper in timed conditions? Let everyone know what you get

    France vs. Bulgaria
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    (Original post by Mladenov)
    No, TST takes place on May 15 & 16.
    Mate. You're gonna destroy it.
    Problem 101 was almost impossible. I was about to give up...
    Have you done problems that hard before? :eek: Looks beyond IMO!
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    (Original post by Mladenov)
    Yup, you are right, there is a flaw.
    Your first equation may not be true for y \not= y', since it is possible that f is injective.
    Secondly, continuity does not imply differentiability.
    sigh *returns to geometry and mechanics*

    In my defence, I don't think I was saying that continuity===> differentiability (I'm not that bad at maths) merely continuity + other result in this particular case ==> existence of derivative.
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    (Original post by Lord of the Flies)
    That looks painful - congrats!
    Thanks. I believe, I am as stubborn as you are.


    Solution 102

    Note that \displaystyle \int_{0}^{1} f(x)dx = \int_{0}^{\frac{\pi}{2}} \cos y f(\sin y)dy = \int_{0}^{\frac{\pi}{2}} \sin y f(\cos y)dy.
    Hence \displaystyle 2\int_{0}^{1} f(x)dx \le \frac{\pi}{2}.
    • PS Helper
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    PS Helper
    (Original post by Mladenov)
    Problem 98**

    Let (a_{n})_{n \ge 0} be a sequence such that a_{0} > 0 and \displaystyle a_{n+1}=a_{n}+\frac{1}{a_{n}}.
    Then, the sequence (a_{n})_{n \ge 0} is divergent.
    Also, find: \displaystyle \lim_{n\to \infty} \frac{a_{n}}{\sqrt{n}}.
    No one gonna take a crack at the second part of this question? It's a pretty answer.
 
 
 
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