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The Proof is Trivial!

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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.
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.
Original post by bananarama2
Actually know a little bit about something you said. That's made my otherwise **** day.


I am glad then.:tongue:

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.:biggrin:

Problem 244**

Prove that for any integer a≥4a \ge 4 there exist infinitely many positive square-free integers nn such that an≡1(modn)a^{n} \equiv 1 \pmod n.
Problem 244**

Let f(x)=∫0sin2(x)arcsin⁡(t) dt+∫0cos2(x)arccos⁡(t) dt;x∈(0,π2) 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
Unparseable latex formula:

\displaystyle\lim_{n \to \infty} \frac{ \Big{[}\displaystyle\prod_{k]} ^{\frac{1}{n} }}{n}

(edited 10 years ago)
Reply 1724
Avatar for Ateo
Ateo
9
Original post by bananarama2
Solve it using bananas. (Say this in Rowan Atkinsons' voice)


Original post by Ateo


That's exactly what I was thinking of when I said it :biggrin:
Solution 245

We differentiate ff; f′(x)=x2sin⁡xcos⁡x+x×(−2sin⁡xcos⁡x)=0f'(x) = x 2\sin x \cos x + x \times (-2\sin x \cos x) =0 Hence, ff is constant over (0,π2)\displaystyle (0, \frac{\pi}{2}).

Let x→0x\to 0 - f=∫01arccos⁡tdt=..=∫01t21−t2dt=−∫011−t2dt+π2=π4\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, lim⁡n→∞1n(∏k=n+12nk)1n=lim⁡n→∞1n((2n)!n!)1n=lim⁡n→∞(2n+2)(2n+1)nn(n+1)n+2=4e\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}
Original post by Mladenov
i.e. Sofia University ...


How about universities outside Bulgaria?
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).
Solution 246 (alternative)

(2n)!1n(n!)1nn∼(2ne)22π(2n)2nn(ne)2πn2n=422ne→4e\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}
(edited 10 years ago)
We obviously have:
lim⁡n→∞1n(∏i=kn+1(k+1)ni)1n=(k+1)k+1kke\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 lim⁡k→∞lim⁡n→∞1kn(∏i=kn+1(k+1)ni)1n\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}}.
Solution 247

lim⁡k→∞(k+1)k+1ekk+1=lim⁡k→∞(1+1k)k+1e=1 \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
A friend just sent me this link, are many those questions actually do-able?
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.
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?
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
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
Reply 1737
Avatar for Zakee
Zakee
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? :wink:
Original post by Zakee
The real question is. Can you apply Calculus to English? :wink:


My mistakes in that are justified by the fact I was on my phone :wink:
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 aa be an arbitrary integer, pp - prime number, pick random divisor n>1n >1 of p−1p-1, kk - integer which is not divisible by nn, and Ua,p=∑1≤x≤p−1e2kπiind(x)ne2πiaxp\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\gcd(a,p)=1, show Ua,p=±pU_{a,p} = \pm \sqrt{p}.
Next, prove that e−2kπiind(a)n=Ua,pU1,p\displaystyle e^{-2k\pi i \frac{ind (a)}{n}} = \frac{U_{a,p}}{U_{1,p}}.

Now, let p≡1(mod4)p \equiv 1 \pmod 4 and S=∑1≤x≤p−2e2πiind(x2+x)4\displaystyle S = \sum_{1 \le x \le p-2} e^{2\pi i \frac{ind(x^{2}+x)}{4}}. Show that p=A2+B2p=A^{2}+B^{2}, where AA and BB satisfy S=A+iBS=A+iB.

We next let gcd⁡(a,p)=1\gcd(a,p)=1 and define xsx_{s} to be an arbitrary reduced residue system modulo pp such that indxs≡s(modn)ind x_{s} \equiv s \pmod n. Set S1=∑xse2πiaxsp\displaystyle S_{1} = \sum_{x_{s}} e^{2\pi i \frac{ax_{s}}{p}}. Show that ∣S1+1n∣<(1−1n)p|S_{1}+\frac{1}{n}| < \left(1-\frac{1}{n} \right)\sqrt{p}.


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

Denote Sa,m=∑ηe2πiaηnm\displaystyle S_{a,m} = \sum_{\eta} e^{2\pi i \frac{a \eta^{n}}{m}} and Sa,m′=∑ξe2πiaξnm\displaystyle S'_{a,m} = \sum_{\xi} e^{2\pi i \frac{a \xi^{n}}{m}}, where η∈Rm\eta \in R_{m}, ξ∈Rm′\xi \in R'_{m}.

Let δ=gcd⁡(n,p−1)\delta = \gcd(n,p-1)

Show that ∣Sa,p∣⩽(δ−1)p|S_{a,p}| \leqslant (\delta -1)\sqrt{p}.

Let δ=1\delta = 1, and s∈Zs \in \mathbb{Z}, 2⩽s⩽n2 \leqslant s \leqslant n. Prove that Sa,ps=ps−1S_{a,p^{s}} = p^{s-1}, and that Sa,ps′=0S'_{a,p^{s}} = 0.

In the case s⩾n+1s \geqslant n+1 show that Sa,ps=pn−1Sa,ps−nS_{a,p^{s}} = p^{n-1}S_{a,p^{s-n}} and Sa,ps′=0S'_{a,p^{s}}=0.

Prove that ∣Sa,m∣<Cm1−1n\displaystyle |S_{a,m}| < Cm^{1-\frac{1}{n}}, where CC is independent of mm.


Problem 249**

Let pp be a prime number. Show that for all k∈Zk \in \mathbb{Z}, there exists integer nn such that (np)=(n+kp)\displaystyle \left(\frac{n}{p} \right) = \left( \frac{n+k}{p} \right).
(edited 10 years ago)

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