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/math/ - Mathematics


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25 Dec 2021Mathchan is launched into public

12 / 3 / 11 / ?

File: question18.png ( 11.57 KB , 693x61 , 1690377094530.png )

Image
Prove you are not a Midwit.
>>
>>456
sure
ill start with the case where n is between 1 and 6
in this case, ceil(n(n-6)/19) <= 0.
at least 0 elements being unrepresentable is always true.
>>
>>457
Pre-School level proof.

Let
m:=n(n1)2m:=\frac{n(n-1)}{2}
,
M:={1,,m}M:=\{1,\ldots,m\}
and
D:={aiaj:ai>aj,1i,jn}={d1,,dm}D:=\{a_i-a_j: a_i>a_j, 1\leq i,j\leq n\}=\{d_1,\ldots,d_m\}
be the multiset of differences.
Then let
A:=MDA:=M\cap D
,
B:=M\A={b1,,bt}B:=M\backslash A=\{b_1,\ldots,b_t\}
,
C:=D\A={c1,,ct}C:=D\backslash A=\{c_1,\ldots,c_t\}
. Here
tt
is the number wanted to be larger than
n(n6)19\lceil \frac{n(n-6)}{19} \rceil
.
Consider the generating function
F(z):=za1++zanF(z):=z^{a_1}+\cdots+z^{a_n}
,
G(z):=F(z)F(1z)=n+(zdk+zdk)G(z):=F(z)F(\frac{1}{z})=n+\sum (z^{d_k}+z^{-d_k})
,
H(z):=zm+zm+1++1+z++zmH(z):=z^{-m}+z^{-m+1}+\cdots+1+z+\cdots+z^m
. We have
G(z)=n1+H(z)(zbl+zbl)+(zcl+zcl)G(z)=n-1+H(z)-\sum (z^{b_l}+z^{-b_l}) + \sum (z^{c_l}+z^{-c_l})
.
Take
z=1|z|=1
, then
G(z)=F(z)20G(z)=|F(z)|^2\geq 0
and
G(z)(n1)H(z)4t|G(z)-(n-1)-H(z)|\leq 4t
. Write
zz
as
e2iθe^{2i\theta}
then
H(z)=sin(2m+1)θsinθH(z)=\frac{\sin(2m+1)\theta}{\sin\theta}
. Take
(2m+1)θ=32π(2m+1)\theta=\frac{3}{2}\pi
, we have
H(z)=1sinθ<n2n+132πH(z)=-\frac{1}{\sin\theta}<-\frac{n^2-n+1}{\frac{3}{2}\pi}
. Hence
G(z)(n1)H(z)23π(n2(1+3π2)n)|G(z)-(n-1)-H(z)|\geq \frac{2}{3\pi}(n^2-(1+\frac{3\pi}{2})n)
. So we can get
t16πn(n6)t\geq \frac{1}{6\pi}n(n-6)
.

Great problem.
>>
>>456
Isn't the captha enough?
>>

File: IMG_20230803_204442.jpg ( 434.51 KB , 975x937 , 1691085059640.jpg )

Image
No one in this site can do these two 1/2
>>

File: IMG_20230803_204731.jpg ( 217.67 KB , 939x486 , 1691085174123.jpg )

Image
2/2
>>
>>456 Where did you get these questions from? Just curious. I could solve the first one with ease, the other ones are a little harder for me.
>>
>>482
All these questions are from China Team Selection Test for IMO.
And also if you actually can solve first one you're pretty above average.
>>
>>460
Can be done in a couple of ways; summing
akmbkm\lfloor \frac{ak}{m} \rfloor \lfloor \frac{bk}{m} \rfloor
or computing
01{ax}{bx}\int_0^1 \{ax\}\{bx\}
works.

For(1),We will show that
λ=01{ax}{bx}dx\lambda = \int_{0}^{1}\left \{ ax \right \} \left \{ bx \right \}dx
satisfied. In fact,we would show that
[ab(k1)m,abkm]Z=\left [ \frac{ab\left ( k-1 \right )}{m} ,\frac{abk}{m}\right ]\cap\mathbb{Z}= \emptyset

then we are done. This
{akm}{bkm}mk1mkm{ax}{bx}dx=O(1m)\Longleftrightarrow\left \{ \frac{ak}{m} \right \} \left \{ \frac{bk}{m} \right \} -m\int_{\frac{k-1}{m} }^{\frac{k}{m} }\left \{ ax \right \}\left \{ bx \right \} dx=\mathrm {O}\left ( \frac{1}{m} \right )
and
supx[k1m,km]{ax}{bx}infx[k1m,km]{ax}{bx}=O(1m)\Longleftrightarrow \sup_{x\in \left [ \frac{k-1}{m} ,\frac{k}{m} \right ]}\left \{ ax \right \} \left \{ bx \right \} -\inf_{x\in \left [ \frac{k-1}{m} ,\frac{k}{m} \right ]}\left \{ ax \right \} \left \{ bx \right \} =\mathrm {O} \left ( \frac{1}{m} \right )
It's obviously right.

\blacksquare
>>
>>458
>>514
>mathchan
>can't even render weblatex
this is sad
>>
>>533
You have to use
\[ ... \]
instead of dollars.
>>
>>533
Exactly just what I was thinking lmao
>>
>>534
>he did not read the sticky