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The Proof is Trivial!

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Reply 700

joostan

Original post by bananarama2

I too want to know this :biggrin:

Original post by metaltron

May I ask what cyc on the bottom of a sigma means?

http://www.gamefaqs.com/boards/576216-maths-marathon/48835430
I think it means the sums of the combinations :s-smilie:

Reply 701

Jkn

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:

May I ask what cyc on the bottom of a sigma means?

Original post by bananarama2

I too want to know this :biggrin:

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:

(edited 10 years ago)

Reply 702

Doglamlico

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. :tongue:

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:

Reply 703

metaltron

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?

Reply 704

Jkn

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 :frown:

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.

Reply 705

metaltron

Original post by Jkn

haha

I'm still stuck on level 4 :frown:

).

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.

Reply 706

MW24595

Original post by Jkn

haha

I'm still stuck on level 4 :frown:

).

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? :tongue:

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

Reply 707

Lord of the Flies

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) :tongue:

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)

(edited 10 years ago)

Reply 708

Jkn

Original post by Lord of the Flies

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

"***" ?

Are you sure?!

Reply 709

Lord of the Flies

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

"***" ?

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.

(edited 10 years ago)

Reply 710

Mladenov

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

Unparseable latex formula:

\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

Unparseable latex formula:

\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:

Unparseable latex formula:

\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. :biggrin:

Reply 711

Jkn

Original post by MW24595

Wait, are you folks talking about brilliant.org? :tongue:

But of course!

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

Hhahaa

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! :smile:

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

Reply 712

Jkn

Original post by Mladenov

Spoiler

Approximately 4 hours. :biggrin:

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

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

Congratulations! :biggrin:

Now to try and understand/check your proof... :bhangra:

(edited 10 years ago)

Reply 713

Mladenov

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.

(edited 10 years ago)

Reply 714

Lord of the Flies

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! :biggrin:

(edited 10 years ago)

Reply 715

Jkn

Original post by Lord of the Flies

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 :biggrin:

France vs. Bulgaria :colone:

Reply 716

Jkn

Original post by Mladenov

No, TST takes place on May 15 & 16.

Mate. You're gonna destroy it. :colone:

Problem 101 was almost impossible. I was about to give up...

Have you done problems that hard before? :eek:

Looks beyond IMO!

Reply 717

Dirac Spinor

Original post by Mladenov

Yup, you are right, there is a flaw. :tongue:

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.