INTERMAJOR MATHS OLYMPIAD - IMO

Today's challenge is to prove that every [math] m \times n [/math] rectangle, where m

this seems trivial unless you put some restrictions on what the extended square looks like

how is this not obvious?

>how is this not obvious?
addendum: if i understand the problem correctly then just start with 1, 2, ..., n and for every new line push all numbers to the right and put the last number on the first spot.

No matter what the [math] m \times n [/math\ rectangle looks like, you have to prove that you can expand it into a square.

do you mean expand it so that every column and row is the set {1,2,..,n}?

yes, that's what i mean. You have m lines that each line is a permutation of the set {1, 2, ... n} and want to add n-m more lines to make it a square such that each column is the set {1, 2, .. n} as well.

>yes, that's what i mean. You have m lines that each line is a permutation of the set {1, 2, ... n} and want to add n-m more lines to make it a square such that each column is the set {1, 2, .. n} as well.
impossible e.g.
(1,2,3,...,n)
(1,2,3,...,n)
(1,2,3,...,n)
...
(1,2,3,...,n)

with n-1 lines cannot be extended that way despite each line being a permutation of (1,2,3,...,n)

i agree with the other replies, this appears to be obvious.

Jesus fucking Christ you dumb niggers how fucking stupid can you be? WE CALL AN m x n RECTANGLE WITH m

try not wording the question like a brainlet next time

so for each column, you already have m unique numbers, which leaves you n-m unique numbers with which to construct your extra rows. you can fill out the matrix however you want. it's trivial

Nope, this problem is far from easy. For each position on the new line you have n-m possible numbers to choose from, yes, but what guarantee do you have that they can be chosen in such a way that the new line you have constructed is a permutation of {1, 2,... n} ? i.e. isn't it possible that randomly choosing suitable elements for each position will make you run out of valid choices by the end of the line?from my point of view the question was worded properly and you're the brainlet for not understanding it, sorry :^

read the OP again brainlet

'extended by adding more lines until we obtain an nxn square' is completely trivial

you will not "run out of numbers", because you have n-m rows to fill out, and for each column, n-m unique numbers to choose from.

I give up, you guys are fucking retarded. Kill yourselves please.

also, let's call it a matrix like everyone else

>I give up, you guys are fucking retarded. Kill yourselves please.
lel brainlet

What did you fucking expect? This is a fucking hole full of losers, of course basically no one is going to be able to solve these.

consider latin rectangle

a{11} a{12} ... a{1n}
a{21} a{22} ... a{2n}
...
a{m1} a{m2} ... a{mn}

let j be any integer between 1 and n. there are n-m columns missing a j, number them j_1, ..., j_{n-m}

so set a{m+r,j_r}=j, for r=1,...,n-m and you get a latin square

confirmed for undergrad brainlet

once you get past 1st year you'll realize the importance of wording things properly

i hope you realize this in time for your calculus exams or you'll likely lose a lot of marks and go crying to the prof for not understanding your gibberish

what's this even about anyway?

This seems to work

why are you replying to yourself? embarrassing

Fuck OP don't let this discourage you from making these threads in the future. These threads are exactly what Veeky Forums needs.

we dont need people making threads asking questions they dont even understand themselves well enough to word it properly, thats essentially shitposting

You're fucking retarded though, the question is perfectly clear : "how do you extend a latin rectangle to a latin sqaure".

there's like 12 people saying its trivial or obvious, and you still think you worded it clearly?

the way wolfram defines it, no row or column can contain duplicates. that's kind of important.

heres another difficult question to solve, seeing that nobody really understands OPs question.

this is pretty much the same actually. these terrible wiki pages are throwing me off.

0, upperbounding of a positive sequence

your rows are always going to be a permutation of {1...n}, so the rows are unique by default. derp.

its some kind of autistic savant code, i havent deciphered it yet

brb my hot indian gf needs help studying for mcat

-t. biologist

>hot
>indian
user just pick one

ah, i see why it looks harder this way.

you can use a proof by contradiction: if you don't have enough unique numbers to choose from when constructing the next row, then the previous rows do not make a latin square.

i don't feel like putting the whole proof together though.

*latin rectangle.

Could you elaborate? I'm not the guy who posted it but I am trying to understand it and I see that you get to a form of infinity over infinity.

How did you get to that evaluating to 0?

Ok, I was always to prove inductively that this limit is equal to

[math]\lim_{n\to\infty} \frac{1}{n} \sum_{j=1}^{n} \frac{j \pi coth(j \pi) - 1}{2j} [/math]

After that I don't know what the fuck to do so this is as far as Calc 2 will take me.

I did try to use wolframalpha to compute it numerically up to 9000 and it seems to be a number between 1.57 and 1.58 or at least around that.

So it is not 0... but I don't know how to go on analytically.

Now that I think about it, I've never worked worked with the hyperbolic function of cotangent, interesting.

You haven't? It's fairly common. I love when it comes up and I'm helping out students. I say it like "cotch" and pretend to not know what's funny when they snicker.

[math]\pi/2[/math]

HOLY FUCK that fits with my estimations in How did you do it? Can you describe the procedure?

Please don't Cleo us like this.

It's a fun, hard problem. I'm not about to rob you of the enjoyment you'll get when you solve it yourself.

I'm not gonna solve it though. I have exhausted my tools. Best I can do with my knowledge is estimate it, which I did.

Just tell me what you did man. Fuck.

Keep it on the back burner and when you learn the theory of residues, you might want to bring it back up. You can solve it. You're definitely smart enough, you might just not have enough knowledge quite yet.

>theory of residues

And what is that? When do I learn it? Calc 3?

That's complex analysis. Once you have calc 3 down, you can pick this up pretty easily. You might even be able to pick it up now, but that would be a little harder since you haven't done line integrals.

question 2

[eqn] \lim_{n \to \infty} \frac{1}{n} \sum_{j=1}^n \sum_{k=1}^n \frac{j}{j^2 + k^2} [/eqn]
[eqn]= \lim_{n \to \infty} \sum_{j=1}^n \sum_{k=1}^n \frac{ \left( \frac{j}{n} \right) }{ \left( \frac{j}{n} \right)^2 + \left( \frac{k}{n} \right)^2 } \frac{1}{n^2}[/eqn]
[eqn] = \int_0^1 \int_0^1 \frac{x}{x^2 + y^2} dy dx [/eqn]
[eqn] = \frac{\pi}{4} + \frac{1}{2} \log(2) [/eqn]

and this is actually the correct answer

Just how the fuck did you figure that out?

Proof is trivial and left to the reader

im the one who gave the question and even i dont know how he figured it... it seems so random and random...

not the guy you're replying to, but he used riemann sums to get the answer, which might seem random at first but isn't that hard after you've done it a few times, we covered it in analysis 1

what do you mean by every line?

nevermind you could've called it a matrix instead of only a rectangle

what do you expect from the 1st year brainlets who clog up this board?

Call me a brainlet bou won't get very far in math if you can't translate ideas into plain english. Go ahead and tell people about a rectangle of numbers. Even secondary school students know better.

i'm calling OP a brainlet not you

he probably hasnt even heard of matrices yet

Also I think that it's trivial if there is no restriction on rows.

no it's not. think again. there's a restriction on both rows and columns.
i'm not even going to bother explaining how retarded you are
>ITT: What is reading comprehension?
I've already reformulated the problem in more formal terms once. I'm not going to spoonfeed you explanations on how to even read a maths problem.
Jeses fuck guys just try thinking about the problem for one moment, I know it appears trivial at first but it's not. The proof I know of uses Graph theory and Hall's Marriage theorem.
t. maths olympiad participant who isn't even in uni yet :^

> isn't even in uni yet
oh it shows

how does it feel, wasting time on a tibetan birdwatching mailing list and making fun of a high schooler and future Cambridge student because he tried to state a problem in such simple terms that you could understand it, but you're still too dumb to actually understand the problem, let alone think about it? keep circlejerking about "muh calculus 3" and how freshmen are "hurr durr polluting this board" while i go to undergraduate courses while still in high school and score first in my country in maths olympiads. I bet you're one of the guys making new stupid question threads daily and posting shit like "why is the jordan normal form so hard to understand? my meanie uni prof said it was easy but i don't get it" and then shitpost about how simple the collatz conjecture is. Now, which one of us is polluting this board, mongrel? Not me, that's for sure :^

Why the fuck are you here if you're so fucking intelligent? Go be immensely better than me somewhere else please, I feel inadequate enough just by existing.

thanks for the laugh kid, am i supposed to be impressed?

I'm here because this is an imageboard dedicated to advanced science and mathematics and I thought this would be a nice place to hang out and learn some cool maths stuff. I also tried to contribute to the wellbeing of this board by making this thread but I see it wasn't thd right choice. I'll maybe make one or 2 more threads and if they don't go too well I'll just give up and leave this board. There's just too much shitposting here for my liking. I understand this is Veeky Forums, I've been on other boards for some time now but I thought this board would be a bit more serious. Also, I too am miserable and depressed and that is why I came here, thinking I could find other smart people to learn cool stuff with and from. I suppose this board isn't for me, after all.
Let's just try one more time. I'll make another thread today with a problem taken directly frm an IMO shortlist and I'll be posting a picture of the text so you can't hate on me any more. I'll be glad to have meaningful coversations about that problem and others like it in that thread. I admit the problem posted here is probably not interesting for maths students (others than those studying combinatorics, which probably already know this problem). The next one will also be about combinatorics but with a more algorithmic feel to it.
Fuck off, i only listed my acomplishments so that the other faggot making fun of me would stop. I never said i'm too intelligent for this board, or that i'm smarter than anyone here. I just said that he shouldn't treat me with disdain just for still being in HS and argued my point. Congrats on the admission to cambridge, but i'm not impressed either.

>the other faggot
there is no other faggot, there's only me

>I too am miserable and depressed
>I too
also, stop projecting

>high schooler and future Cambridge student
>i go to undergraduate courses while still in high school and score first in my country in maths olympiads
>I too am miserable and depressed
I don't get it.

>I never said i'm too intelligent for this board, or that i'm smarter than anyone here.
You are and you probably are (or at least could be).

There's gotta be tons of maths communities out there, so I don't know why you chose the worst one.

I don't think anyone here considers Veeky Forums to be a math community instead of a "mostly popsci shitposting but sometimes also science/math related homework" community.

Can't you just plug in the Taylor series of log(1-x) and then integrate each term by parts?

we calc2 now

^12 {p{e+12%c)p=-)
you guys need to retire before I get involved

>we calc2 now

>We smug bastards now

No one has to change sums to integrals in Calc 2. The only time I ever saw that was when the professor wanted to prove the theorems for volume of revolution problems.

>No one has to change sums to integrals in Calc 2
Isn't this how you DEFINE the integral in calc2?

Yeah, I have that. Where delta is a regular partition of the inteval [a.b] given by a = x0 < x1 < ... < xn = b with deltaxi being the length of the ith interval and ci being any point inside the ith subinterval then the integral from a to b is equal to the limit as the norm of delta approaches 0 of the sum from i=1 to n of f(ci)*deltaxi

And that was it. But we enver did it directly.

>:(

And we never did it with double integrals which are not covered in calc 2

>:(

It's also wrong

is this a troll? the function isnt even defined from 0 to 1

nevermind, i read x-1 instead of 1-x

Is this a troll?

It's called limits.

try doing this with change of variables

its not wrong, its correct.

...

heres the solution.

Kurisu is a useless whore who should kill herself.

question 3

which book ?

It's 2. Just pull the limit inside the integral.

2 is the answer but your reasoning does not make sense.

[eqn] \lim_{y \to 0} \frac{1}{y} \int_0^\pi \tan(y \sin(x)) dx \\
= \int_0^\pi \left(\lim_{y \to 0} \frac{ \tan(y \sin(x)) }{y \sin(x)} \right) \sin(x) dx \\
= \int_0^\pi \sin(x) dx \\
= 2 [/eqn]

What makes no sense?

alright user... you got away this time... but im watching u. question 4

Where are those exercises from ?

analysis is fucking terrifying

try subtracting each equation from the next one
add them all together, too

you will see some neat simplifications that may help

question 5

The book, faggot
Can't you answer a simple fucking question

Amen

not anymore i can't

what's the name of the book?