you should be able to do this at Tue, 9 Apr 2024 03:03:04 UTC No. 16120271

Posting two more problems from an anon in /mg/, which were the motivation for this general

you should be able to do this at Tue, 9 Apr 2024 03:04:12 UTC No. 16120274

you should be able to do this at Tue, 9 Apr 2024 03:07:08 UTC No. 16120282

a nice but perhaps standard calculus problem

you should be able to do this at Tue, 9 Apr 2024 03:21:54 UTC No. 16120299

a number theory prob to make things more spicy!
(not as easy as it looks as this is a contest problem, won't reveal which contest yet as it may scare people off unnecessarily)

Anonymous at Tue, 9 Apr 2024 04:35:11 UTC No. 16120369

>- Napkin (freely available on Evan Chen's site) to learn some ug math at a level accessible to talented high schoolers(this a good resource for learning calc theory, starting from the ground up with metric spaces, topological spaces etc)
Nothing evan chen has ever written is a good resource. Even his lecture notes are somehow mostly terrible. The napkin especially massively lacks in rigor, which is terrible if you're new to proof-based math.
Whatever makes you kool-aid drinking kids think notes written by some young olympiad coach are superior to books by experienced professors of mathematics is beyond me. And even if you had to learn topology (fast) from evan chen out of all people, you should read his notes on, say, M55b as he's mostly just copying Gaitsgory, you know, someone who knows his shit. Not to mention, just about every book on undergrad math is accessible to high schoolers.

If you want some really nice problems, check out Lang's book on algebra. For something less terse and more linear algebra focused (i.e. more undergrad-level), Gorodentsev is pretty nice. Zorich, Rudin, Fichtenholz, and Amann-Escher have great problems on elementary analysis (what you call "calc theory"). For measure theory specifically, Bogachev's book is full of very interesting ones. Work through those and you can continue with fun anal etc. None of them are inaccessible and refusing to read them just because they have no "for math olympiads" plastered all over the cover will be your bane.

Cult of Passion at Tue, 9 Apr 2024 04:58:08 UTC No. 16120389

You need BOOKS... to find problems to solve...

You will NEVER be a mathematician

you should be able to do this at Tue, 9 Apr 2024 07:25:57 UTC No. 16120496

>>16120369
obviously napkin is not directed to math majors, just high schoolers, he clearly mentions in the beginning that examples and intuitions are prioritised, not formal proofs, although most important results mentioned do have proofs, at least thats what i gathered from reading chapters 2, 6 and 7(currently reading)...can you point out where it lacks rigor in these chapters apart from examples?
also, obviously I won't refuse to read books because they are not catered towards olympiads and i never said evan's books are superior to those by profs(although i am willing to die on the hill that EGMO is the best book ever written about euclidean geometry), i just believe that at this stage, building strong intuition is very important and i like solving problems more than reading literature, i will have plenty of that in my later education but now i just want to have fun

you seem to be under the impression that somehow people from olympiad backgrounds do not understand rigor, have you never seen the shortlists of imo and the proofs presented?
>>16120389
post problems or solve them, or gtfo

Anonymous at Tue, 9 Apr 2024 07:34:12 UTC No. 16120506

>>16120266
Inequalities by Korovkin is a nice book too

Anonymous at Tue, 9 Apr 2024 07:40:05 UTC No. 16120515

Anonymous at Tue, 9 Apr 2024 08:08:33 UTC No. 16120548

>>16120515
I think this can be solved by induction

Anonymous at Tue, 9 Apr 2024 16:00:20 UTC No. 16120959

>>16120515
[math]|a-b|=\max\{a,b\}-\min\{a,b\}[/math] for all real [math]a[/math]and [math]b[/math].[math]E=\sum_{k=1}^{n}\max\{a_k,b_k\}-\sum_{k=1}^{n}\min\{a_k,b_k\}[/math]

The main idea is to prove that actually [math]\max\{a_1,b_1\},\max\{a_2,b_2\},\ldots,\max\{a_n,b_n\}[/math] is a permutation of [math]n+1,n+2,\ldots,2n[/math] and similarly [math]\min\{a_1,b_1\},\min\{a_2,b_2\},\ldots,\min\{a_n,b_n\}[/math] is a permutation of [math]1,2, \ldots ,n[/math] (clearly, it's enough to prove the first statement, the second follows immediately).

Anonymous at Tue, 9 Apr 2024 16:53:04 UTC No. 16121021

>>16120369
I do agree that "Olympiads" themselves shouldn't be the focus, Math should be the focus.
We want teachers who teach Maths and students who study Maths.

Anonymous at Tue, 9 Apr 2024 17:14:22 UTC No. 16121066

>>16120271
a) looks like pretty standard stars and bars

b) This is one of the gazillion standard identities on Pascal's Triangle, you could also expand it out but there's a much easier combinatorial argument.
The LHS sum is the number of subsets (of any size) of [n] that don't contain consecutive integers.
Count the subsets that don't contain n, it's the number of such subsets of [n - 1], and count the subsets that do contain n, it's the number of such subsets of [n - 2]. Therefore the recurrence follows the Fibonacci recurrence and you just need to show the initial values are the same.

you should be able to do this at Tue, 9 Apr 2024 18:58:00 UTC No. 16121207

>>16120506
checked it out and it looks nice at first glance, i will read it as soon as i get the chance
>>16120515
>the unreasonable effectiveness of simplicity
nice approach anon, let me fill in the blanks for others
suppose [math]\max\{a_k,b_k\}=a_k[/math], we then have [math]k-1[/math] elements in [math]\{a_n\}[/math] which are less than [math]a_k[/math], [math]b_k[/math] itself also less than [math]a_k[/math] and [math]b_k[/math] being greater than [math](n-k)[/math] elements in [math]\{b_n\}[/math], we have a total of [math]n[/math] elements smaller than [math]a_k[/math] and we are done. the other case is similar.
>>16121021
i agree, the resources mostly are books related to olympiads only because they are a good source of problems
>>16121066
well done anon, i had the same approach to part b

Anonymous at Tue, 9 Apr 2024 20:05:43 UTC No. 16121333

>>16120282
Lim x->c f(x)-f(c) = lim x->c (x-c) f(x)-f(c)/x-c

Anonymous at Wed, 10 Apr 2024 01:32:47 UTC No. 16121747

everything
[spoiler]
>>16120266
greedy algorithm starting at the biggest, show that can make choices so the difference is at most a_i after using a_i, final is difference <=1. difference is even so =0
>>16120271
a) balls and urns
b) recurrence depending if n in the set
>>16120274
a) strictly increasing 1. nondecreasing 2n choose n, lattice path in square or balls and urns
b) Catalan, lattice path in square under diagonal
>>16120282
continuous in complement of countable set. proof is noncontinuous point have a jump on one or two sides gives open interval in the range which contains a rational
>>16120299
n=1 and 3. proof is n is odd and if p is smallest prime divisor then p^2 divides 2^(2n)-1, the order of 2 in the multiplicative group mod p divides gcd(2n,p-1)=2, now p divides 2^2-1, p=3 then if 3^k divides n but not 3^(k+1) it's a problem, 2^n+1 has k+1 power of 3 not 2k. so k=1, n=3m and q next smaller, order of 2 mod q divides gcd(6m,q-1)=6 so q=7 but 2^n+1=2 mod 7
>>16120515
if ai and bj are nearby, you can switch and the sum is same (i<j then ai<bi and bj<aj)
for 123...n and 2n...n+1 it is sum of odds n^2
[/spoiler]

Anonymous at Wed, 10 Apr 2024 01:40:53 UTC No. 16121756

Get absorbed in to /mg/ already. Ive seen dozens of new attempted generals happen that easily fit into the other established ones, yours is no different.

🗑️ Anonymous at Wed, 10 Apr 2024 02:43:25 UTC No. 16121810

>muh vanity thread!!
>I'm such a special snowflake, everyone pay attention to me!!
>gibes me all your internet social media dopamine!!

you should be able to do this at Wed, 10 Apr 2024 03:39:51 UTC No. 16121847

>>16121756
yeah i was thinking the same thing, this is way too niche and only mg would be a good place to post these...there are not many responses as well so its time for it to die kek

you should be able to do this at Wed, 10 Apr 2024 03:43:51 UTC No. 16121852

>>16121747
looks good, i just saw the number 7 in your sol to the nt and knew it had to be correct lol
except for the last one tho, switching things sometimes does not preserve order(you have to take a few examples to understand this), i don't know if you are trying to do that, because i tried just that and failed

Anonymous at Wed, 10 Apr 2024 06:31:53 UTC No. 16122005

>>16121852
123456
ABAABB
I can switch adjacent A and B preserve order
easy "bubble" As to the front

Anonymous at Wed, 10 Apr 2024 19:51:07 UTC No. 16122926

>>16120266
Did anyone solve the first one?

Anonymous at Wed, 10 Apr 2024 23:22:08 UTC No. 16123277

Prove that if a,b,c,d,m are integers such that m^2<a<b<c<d<(m+1)^2 then ad != bc.

Anonymous at Thu, 11 Apr 2024 01:08:44 UTC No. 16123429

Color each point of a circle one of 1998 colors. Show there are 4 distinct points of the same color which are the vertices of a trapezoid.

🗑️ Anonymous at Thu, 11 Apr 2024 01:17:25 UTC No. 16123435

>>16120282
You mean Holder continuous right?

Anonymous at Thu, 11 Apr 2024 10:17:39 UTC No. 16123903

>>16120266
>Simple statements
The ordered [from least to greatest] tuple [math] t [/math] is something like [math] T^1 = (0,1,1) [/math] or [math] T^2 = (0,1,2,3) [/math], where [math] t_{-1} [/math] is the last and largest value.
The sum of the values for [math] t [/math] is [math] S(t) [/math], so [math] S(T^1) = 2 [/math] and [math] S(T^2) = 6 [/math]. Oddly enough, [math] S(T^1) = 2T^1_{-1} [/math] and [math] S(T^2) = 2T^2_{-1} [/math].
[math] t [/math] can be partitioned into two ordered tuples [math] x(t), y(t): S(t) = S(x) + S(y) [/math].
[math] t [/math] can be appended by the non-trivial value [math] a': 1 \leq a' \leq 2t_{-1} [/math] with [math] (t,a') [/math], which orders the values.

For all relevant [math] t [/math] we are considering, it turns out is always true that
>[math]0.)\ \ a' \leq S(t) [/math].
I'll talk about it later. Just assume it's true.

There's the funny thing about [math] T^1 [/math] and [math] T^2 [/math] is that
>[math] 1.)\ \ \forall i: 0 \leq i \leq S(t) \ , \ \exists x(t): S(x) = i[/math],
meaning for [math] T^2 [/math], that [math] S(0) = 0,\ S(0,1) = 1,\ S(0,2)=2,\ S(0,3) = 3,\ S(1,3) = 4,\ S(2,3) = 5, [/math] and [math] S(1,2,3) = 6 [/math]. The same can also be said for [math] y[t] [/math].

Anonymous at Thu, 11 Apr 2024 10:18:40 UTC No. 16123904

Suppose this is true for all [math] t [/math] we are considering. Then for all [math] (t,a') [/math], obviously 1.) holds true for [math] i \leq S(t) [/math], and there also obviously is an [math] x(t,a') [/math] for [math] i = S(t)+a' = S(t,a') [/math]. Turns out, this is actually true for all [math] i [/math] in between, or that
>[math] 2.)\ \ \forall i: 0 \leq i \leq S(t,a') \ , \ \exists x(t,a'): S(x) = i[/math].
To see this, first set [math] x(t) = t [/math], meaning [math] y(t) = (0) [/math]. Then let [math] x' = (x, a') [/math] with [math] S(x') = S(x) + a' [/math] initally. From 1.), we can alter [math] y [/math] to increase [math] S(y) [/math] to any integer between [math] 0 [/math] and [math] S(t) [/math], which in turn decreases [math] S(x,a') [/math] by the same amount. Since [math] a' \leq S(t) [/math], at most we would only need to increase [math] S(y) [/math] by [math] a' [/math], and therefore we can ALWAYS reduce [math] S(x,a') [/math] to a number between [math] S(t) \leq i \leq S(t) + a' [/math].

>OP's problem
For OP's problem, if only one integer, [math] a' [/math], is appended to the list, notice that it has to be an even number between [math] 2 [/math] and [math] 2t_{-1} [/math]. From 2.), we can always construct an [math] x(t,a') [/math] for any given [math] i = S(x) [/math] within the right range, so why not set it to [math] S(x) = S(t,a')/2 [/math]. Since [math] a' [/math] and [math] S(t) [/math] are both even, then [math] S(x) = S(y) [/math].

Therefore, even numbers within the range can always be added to an already correct [math] t [/math]. For [math] T^2 [/math], you can add as many 2s, 4s, and 6's as u want, one at a time.

Anonymous at Thu, 11 Apr 2024 10:19:50 UTC No. 16123908

Odd numbers need to be added in pairs, [math] (a',b') [/math]. Here, three cases are needed to be considered, but theyre all pretty easy:
>A.) [math] a' \leq b' \leq t_{-1} [/math]
>B.) [math] a' \leq t_{-1} \leq b' \leq 2t_{-1} [/math]
>C.) [math] t_{-1} \leq a' \leq b' \leq 2a' [/math]
For A.), [math] a' [/math] and [math] b' [/math] can be added individually and still satisfy 2.). But since two odd numbers add to an even number, [math] S(t,a',b') [/math] is also even, so we can set [math] S(x) = S(t,a',b')/2 = S(y) [/math]. B.) is pretty similar, im tired of typing, fill in the blanks urself. C.) needs more words and thinking but the logic is all here.

>the 0.) part.
Ugh, i just wanna be done. Ok, notice that for [math] T^1 [/math] and [math] T^2 [/math], all possible even [math] a' [/math] and odd [math] (a',b') [/math] will keep 0.) true. Think about (1,1), (2), (4), (6), (3,3), (3,5), (5,5). It's set to equality when [math] a' = 2t_{-1} [/math]. Adding anything else will increase [math] S(t) [/math] and [math] t_{-1} [/math] by a different amount where [math] S(t) [/math] is always greater or equal. When [math] t [/math] satisfies OP's problem, we create new solutions by adding more numbers. Adding more numbers can only keep equality or set inequal but always maintaining 0.)

Yall can fill in the blanks.

Anonymous at Thu, 11 Apr 2024 14:51:40 UTC No. 16124241

Have fun

Anonymous at Thu, 11 Apr 2024 16:38:37 UTC No. 16124393

>>16123908
If you start with something like [math] T^{3} = (0,1,2) [/math], you still get 1.) even though it doesn't satisfy OP's problem. 2.) still holds for any applicable [math] a' [/math], so you can now build up to any [math] t [/math] you want.

I actually think the odd ordered pair thing is a little too much. Just literally add any [math] a' [/math] to [math] T^1 [/math] or [math] T^3 [/math] and beyond that satisfies
> 4.) [math] 0 \leq a' \leq S(t) + 1 [/math]
so that
> 5.) [math] \forall i : 1 \leq i \leq S(t,a') \ , \ \exists x(t,a'): S(x) = i [/math]
is always true.

If you know that
>6.) [math] a' \leq 2t_{-1} [/math],
then for OP's problem, there's always an [math] S(x) [/math] that is half of [math] S(t) [/math] since the sum is given to be even. This could be described better but yall get the gist.

The big point i guess is how does 6.) lead to 4.). Use induction. Start off with
>7.) [math] t_{-1} \leq S(t \setminus t_{-1}) + 1 = S(t) - t_{-1} + 1 [/math],
which means [math] a' \leq 2t_{-1} \leq S(t) + 1 [/math]. [math] T^1 [/math] and [math] T^3 [/math] satisfy 7.), so any [math] t [/math] that builds off em with [math] a' [/math] satisfying 6.) works fine.

I think i got most of it.

Anonymous at Fri, 12 Apr 2024 03:43:36 UTC No. 16125054

>>16124241
2*sum = sum mn/(3^m3^n) because switching m,n and adding symmetric
2*sum=X^2, X=sum m/3^m=(1/3)/(1-1/3)^2=3/4
sum=9/32

Anonymous at Fri, 12 Apr 2024 09:38:15 UTC No. 16125360

>>16125054
Yep

Anonymous at Fri, 12 Apr 2024 09:47:31 UTC No. 16125364

>>16120496
You only have to compare his treatment of group theory with, say, Gerritzen's. Even in his rather incomplete M55a notes, the treatment of linear algebra is imo better than in the napkin. For more advanced topics (algebraic geometry, etc.) there's no reason at all to use the napkin. At best, you can get an overview of how things are connected, but the actual pedagogical value is akin to skimming through undergraduate books, especially skipping the proofs.
>obviously napkin is not directed to math majors, just high schoolers
Any book written for undergraduates could just as easily be written for interested high school students.
Nothing you learn in your final years of high school is a strict prerequisite for undergraduate mathematics per se.
>examples and intuitions are prioritized
Well-written elementary textbooks do all this, while treating the theory rigorously. Eventually, you'll be mature enough in mathematics to construct your own examples, which in turn will allow you to read more concise graduate-level books.
Another problem with avoiding proofs is that no matter how many examples you work through while doing the former, they won't help you solve problems or go on to anything more advanced. For example, try constructing a regular 17-gon using a compass and a ruler. This may seem stupidly difficult now, but with Gorodentsev's treatment of Galois groups, this is just another doable problem. That really applies to nearly all olympiad problems: tricky with little math knowledge but trivial, the more you have.

Anonymous at Fri, 12 Apr 2024 12:06:33 UTC No. 16125492

bump for potential

Anonymous at Fri, 12 Apr 2024 14:41:19 UTC No. 16125645

>>16125364
>That really applies to nearly all olympiad problems: tricky with little math knowledge but trivial, the more you have.
Could you stop larping?
https://terrytao.wordpress.com/2009/07/20/imo-2009-q6-as-a-mini-polymath-project/
https://gowers.wordpress.com/2014/07/19/mini-monomath/
We get it, you didn't win any contests. No need to larp as some kind of hardcore math major.

Anonymous at Fri, 12 Apr 2024 16:45:12 UTC No. 16125880

>>16123277
>>16123429
>paraphrase contest problems so no one can google the answers
>no one can solve them

🗑️ Anonymous at Fri, 12 Apr 2024 17:56:12 UTC No. 16125981

>>16123277
log is strictly concave. the result is trivial. ad<bc
>>16125880
happy now?

you should be able to do this at Fri, 12 Apr 2024 17:57:43 UTC No. 16125983

>>16125645
it was so stupid what he said, i didn't even bother giving him anymore (You)s

Anonymous at Sat, 13 Apr 2024 02:08:52 UTC No. 16126612

>>16123277
[math]19\sqrt{2}[/math]

Hehehe... Are you even trying?

Anonymous at Sat, 13 Apr 2024 02:18:21 UTC No. 16126626

>>16126612
Well you best start thinking about reversing that for two reasons. One its stupid. And 2. All it takes it a blue and this all goes at once the other way and since you're getting aggressive with me I now seek that resolution and it could be any second now you receive a wrathful cancellation for your stupidity.

Yes I was being framed. I don't care how long it was for you.