Hey there! Sign in to join this conversationNew here? Join for free
x Turn on thread page Beta
    Offline

    1
    ReputationRep:
    (Original post by Jkn)
    You're not just doing that for another shot at the IMO, are you? :/ I can picture you getting into a rather strange situation if you started university in the first year... perhaps you could take 2/3 years of exams all at once though hmm...
    Clearly not; at least, I will be twenty years old before the first day of IMO 2014..
    Roughly speaking, at our most prestigious university i.e. Sofia University, De Rham coholomogies are taught in year 4 and this is the most advanced course; this course, however, is senseless, as there is no further study of Crystalline cohomology. Not to mention that things such as Lie Groups (that is, the very basic real geometry) are only for the masters, but owing to lack of candidates such courses are taught rarely. Hence, the whole thing appears to be pointless.

    By the way, has anybody seen this before? I am gobsmacked.
    Offline

    0
    ReputationRep:
    (Original post by Mladenov)
    Clearly not; at least, I will be twenty years old before the first day of IMO 2014..
    Roughly speaking, at our most prestigious university i.e. Sofia University, De Rham coholomogies are taught in year 4 and this is the most advanced course; this course, however, is senseless, as there is no further study of Crystalline cohomology. Not to mention that things such as Lie Groups (that is, the very basic real geometry) are only for the masters, but owing to lack of candidates such courses are taught rarely. Hence, the whole thing appears to be pointless.

    By the way, has anybody seen this before? I am gobsmacked.
    Actually know a little bit about something you said. That's made my otherwise **** day.
    Offline

    1
    ReputationRep:
    (Original post by bananarama2)
    Actually know a little bit about something you said. That's made my otherwise **** day.
    I am glad then.

    Obiter, I can't say that I understand even to the least extent the recently discussed physics problems, so you are in a better position.

    Problem 244**

    Prove that for any integer a \ge 4 there exist infinitely many positive square-free integers n such that a^{n} \equiv 1 \pmod n.
    Offline

    14
    ReputationRep:
    Problem 244**

    Let  f(x) = \displaystyle\int_{0}^{sin^2(x)} \arcsin(\sqrt{t}) \ dt + \int_{0}^{cos^2(x)} \arccos(\sqrt{t}) \ dt; \quad x \in (0,\frac{\pi}{2})

    Show f(x) is constant, and find its value.


    Problem 345**/***

    Evaluate  \displaystyle\lim_{n \to \infty} \frac{ \Big{[}\displaystyle\prod_{k=1}^{n} (n+k)\Big{]} ^{\frac{1}{n} }}{n}
    Offline

    9
    ReputationRep:
    (Original post by bananarama2)
    Solve it using bananas. (Say this in Rowan Atkinsons' voice)
    Offline

    0
    ReputationRep:
    (Original post by Ateo)
    That's exactly what I was thinking of when I said it
    Offline

    1
    ReputationRep:
    Solution 245

    We differentiate f; f'(x) = x 2\sin x \cos x + x \times (-2\sin x \cos x) =0 Hence, f is constant over \displaystyle (0, \frac{\pi}{2}).

    Let x\to 0 - \displaystyle f = \int_{0}^{1} \arccos \sqrt{t} dt = .. = \int_{0}^{1} \frac{t^{2}}{\sqrt{1-t^{2}}} dt = -\int_{0}^{1} \sqrt{1-t^{2}}dt + \frac{\pi}{2} = \frac{\pi}{4}

    Solution 246

    Rewrite, \begin{aligned} \displaystyle \lim_{n \to \infty} \frac{1}{n}\left(\prod_{k=n+1}^{  2n} k \right)^{\frac{1}{n}} = \lim_{n \to \infty} \frac{1}{n}\left(\frac{(2n)!}{n!  } \right)^{\frac{1}{n}} =  \lim_{n \to \infty} \frac{(2n+2)(2n+1)n^{n}}{(n+1)^{  n+2}} = \frac{4}{e} \end{aligned}
    Offline

    1
    ReputationRep:
    (Original post by Mladenov)
    i.e. Sofia University ...
    How about universities outside Bulgaria?
    Offline

    1
    ReputationRep:
    (Original post by jack.hadamard)
    How about universities outside Bulgaria?
    I am not-so-motivated to study french - so I can't enroll at LLG although I have aced their mathematics exam.
    UK is possible, but I am not quite sure.
    I have no chance to be accepted into a university in the USA; their criteria is a bit vague for me (a friend of mine applied to several universities there last year, and he was not admitted even though he had participated in RSI).
    Offline

    18
    ReputationRep:
    Solution 246 (alternative)

    \displaystyle \frac{(2n)!^{\frac{1}{n}}}{(n!)^  {\frac{1}{n}}n}\sim\frac{(\frac{  2n}{e})^{2}\sqrt[2n]{2\pi (2n)}}{n(\frac{n}{e})\sqrt[2n]{2 \pi n}}=\frac{4\sqrt[2n]{2}}{e}\to \frac{4}{e}
    Offline

    1
    ReputationRep:
    We obviously have:
    \displaystyle \lim_{n \to \infty} \frac{1}{n} \left( \prod_{i=kn+1}^{(k+1)n} i \right)^{\frac{1}{n}} = \frac{(k+1)^{k+1}}{k^{k}e}.

    Problem 247**

    Evaluate \displaystyle \lim_{k \to \infty} \lim_{n \to \infty} \frac{1}{kn} \left( \prod_{i=kn+1}^{(k+1)n} i \right)^{\frac{1}{n}}.
    Offline

    14
    ReputationRep:
    Solution 247

     \displaystyle \lim_{k \to \infty} \dfrac{(k+1)^{k+1}}{ek^{k+1}} = \lim_{k \to \infty} \dfrac{(1+\frac{1}{k})^{k+1}}{e} = 1
    Offline

    15
    ReputationRep:
    A friend just sent me this link, are many those questions actually do-able?
    Offline

    3
    ReputationRep:
    (Original post by james22)
    A friend just sent me this link, are many those questions actually do-able?
    they're certainly doable, but some are pretty hard.
    Offline

    1
    ReputationRep:
    (Original post by Mladenov)
    I have no chance to be accepted into a university in the USA; their criteria is a bit vague for me...
    Well, admissions are usually quite individual, so don't try to predict the outcome of your potential application based on other people's experience or results. It does not cost much to apply to universities anyway, so I don't see why not try it. If you have strong grades and achievements from competitions, together with a passion for the subject (which admissions tutors can identify), then you do have a chance. It might not be as great a chance as somebody who rocked IMO (or IOI, or something), but it is still a chance. Do you just let it slip?
    Offline

    16
    ReputationRep:
    (Original post by Mladenov)
    ...
    (Original post by jack.hadamard)
    Well, admissions are usually quite individual, so don't try to predict the outcome of your potential application based on other people's experience or results. It does not cost much to apply to universities anyway, so I don't see why not try it. If you have strong grades and achievements from competitions, together with a passion for the subject (which admissions tutors can identify), then you do have a chance. It might not be as great a chance as somebody who rocked IMO (or IOI, or something), but it is still a chance. Do you just let it slip?
    I would very much agree with this. Please consider applying to a UK university (most likely, the right choice for you is Cambridge, but also look at Oxford, Warwick and Imperial too). Getting in front of an admissions tutor might be the hardest step - if you get to an interview and perform to your abilities, you should fly through the admissions process. Have you ever looked at STEP? If you feel like that's reasonable then definitely look into Cambridge.

    I also think you should look into the US too - MIT seem to be international friendly
    Offline

    0
    ReputationRep:
    (Original post by james22)
    A friend just sent me this link, are many those questions actually do-able?
    This is the opitome of why I didn't do maths. I just can't get my head around stuff like that. I can apply calculus and maths to physics amd chemistry no problem, but not stuff like that.

    Posted from TSR Mobile
    Offline

    0
    ReputationRep:
    (Original post by bananarama2)
    This is the opitome of why I didn't do maths. I just can't get my head around stuff like that. I can apply calculus and maths to physics amd chemistry no problem, but not stuff like that.

    Posted from TSR Mobile


    The real question is. Can you apply Calculus to English?
    Offline

    0
    ReputationRep:
    (Original post by Zakee)
    The real question is. Can you apply Calculus to English?
    My mistakes in that are justified by the fact I was on my phone
    Offline

    1
    ReputationRep:
    (Original post by jack.hadamard)
    Well, admissions are usually quite individual, so don't try to predict the outcome of your potential application based on other people's experience or results. It does not cost much to apply to universities anyway, so I don't see why not try it. If you have strong grades and achievements from competitions, together with a passion for the subject (which admissions tutors can identify), then you do have a chance. It might not be as great a chance as somebody who rocked IMO (or IOI, or something), but it is still a chance. Do you just let it slip?
    (Original post by shamika)
    I would very much agree with this. Please consider applying to a UK university (most likely, the right choice for you is Cambridge, but also look at Oxford, Warwick and Imperial too). Getting in front of an admissions tutor might be the hardest step - if you get to an interview and perform to your abilities, you should fly through the admissions process. Have you ever looked at STEP? If you feel like that's reasonable then definitely look into Cambridge.

    I also think you should look into the US too - MIT seem to be international friendly
    Thank you, I highly appreciate your advices and will take them into consideration.

    This thread seems abandoned.

    I am reading some number theory and here is an exercise I made.

    Problem 248**


    Let a be an arbitrary integer, p - prime number, pick random divisor n >1 of p-1, k - integer which is not divisible by n, and \displaystyle U_{a,p} = \sum_{1 \le x \le p-1} e^{2k\pi i\frac{ind  (x)}{n}}e^{2\pi i \frac{ax}{p}}.

    If \gcd(a,p)=1, show U_{a,p} = \pm \sqrt{p}.
    Next, prove that \displaystyle e^{-2k\pi i \frac{ind (a)}{n}} = \frac{U_{a,p}}{U_{1,p}}.

    Now, let p \equiv 1 \pmod 4 and \displaystyle S = \sum_{1 \le x \le p-2} e^{2\pi i \frac{ind(x^{2}+x)}{4}}. Show that p=A^{2}+B^{2}, where A and B satisfy S=A+iB.

    We next let \gcd(a,p)=1 and define x_{s} to be an arbitrary reduced residue system modulo p such that ind x_{s} \equiv s \pmod n. Set \displaystyle S_{1} = \sum_{x_{s}} e^{2\pi i \frac{ax_{s}}{p}}. Show that |S_{1}+\frac{1}{n}| < \left(1-\frac{1}{n} \right)\sqrt{p}.


    Let n and m be integers, n \geqslant 3, m \geqslant 2, \gcd(a,m)=1 (a is defined as above). Further, let R_{m} be an arbitrary complete residue system modulo m, and R'_{m} - reduced.

    Denote \displaystyle S_{a,m} = \sum_{\eta} e^{2\pi i \frac{a \eta^{n}}{m}} and \displaystyle S'_{a,m} = \sum_{\xi} e^{2\pi i \frac{a \xi^{n}}{m}}, where \eta \in R_{m}, \xi \in R'_{m}.

    Let \delta = \gcd(n,p-1)

    Show that |S_{a,p}| \leqslant (\delta -1)\sqrt{p}.

    Let \delta = 1, and s \in \mathbb{Z}, 2 \leqslant s \leqslant n. Prove that S_{a,p^{s}} = p^{s-1}, and that S'_{a,p^{s}} = 0.

    In the case s \geqslant n+1 show that S_{a,p^{s}} = p^{n-1}S_{a,p^{s-n}} and S'_{a,p^{s}}=0.

    Prove that \displaystyle |S_{a,m}| < Cm^{1-\frac{1}{n}}, where C is independent of m.


    Problem 249**

    Let p be a prime number. Show that for all k \in \mathbb{Z}, there exists integer n such that \displaystyle \left(\frac{n}{p} \right) = \left( \frac{n+k}{p} \right).
 
 
 
Reply
Submit reply
Turn on thread page Beta
Updated: February 22, 2018
Poll
Do I go to The Streets tomorrow night?
Useful resources

Make your revision easier

Maths

Maths Forum posting guidelines

Not sure where to post? Read the updated guidelines here

Equations

How to use LaTex

Writing equations the easy way

Student revising

Study habits of A* students

Top tips from students who have already aced their exams

Study Planner

Create your own Study Planner

Never miss a deadline again

Polling station sign

Thinking about a maths degree?

Chat with other maths applicants

Can you help? Study help unanswered threads

Groups associated with this forum:

View associated groups

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

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