Can anybody help me with my problem. I've a DP solution for this problem. I submitted it on Usaco page and all cases are ok except the last one. It has timeout. Can anybody help me?
Here is my source:
[code]/*
ID:wh04mi1
TASK:money
LANG:C++
*/
#include <algorithm>
#include <iostream>
#include <fstream>
#include <vector>
using namespace std;
vector<int> units;
vector< vector<long long> > inOne;
vector<long long> comb;
int compute(int n);
int S, N;
ifstream fin("money.in");
ofstream fout("money.out");
int main()
{
      fin>>N>>S;
      units.resize(N);
      comb.resize(S+1, -1);
      inOne.resize(units.size());
      for(int i=0; i<inOne.size(); i++)
              inOne[i].resize(S+1);
      for(int i=0; i<N; i++)
              fin>>units[i];
      sort(units.begin(), units.end());
      fout<<compute(S)<<endl;
      return 0;
}
int compute(int n){
    long long res=0;
    if(n<units[0])
          res=0;
    else{
    for(int i=units.size()-1; i>=0; i--){
            if(units[i]<=n){
                   if(units[i]==1)
                          res++;
                   else {
                   for(int j=int(n/units[i]); j>=1; j--){
                           if(j*units[i]==n){
                                  res++;
                                  inOne[i][n]++;
                           }
                           else{
                                    if(n-units[i]*j==1&&units[0]==1){
                                          res++;
                                          inOne[i][n]++;
                                    }
                                    else{
                                    if(comb[n-units[i]*j]==-1)
                                      compute(n-units[i]*j);
                                        res+=comb[n-units[i]*j];
                                        inOne[i][n]+=comb[n-units[i]*j];
                                      for(int k=units.size()-1; k>=i; k--){
                                        res-=inOne[k][n-units[i]*j];
                                        inOne[i][n]-=inOne[k][n-units[i]*j];
                                      }
                                    }
                           }
                   }
                  }
              }
     }
     }
     comb[n]=res;
     return res;
}[/code]

i think ur algo is not fine enough. so i think u should describe about ur algo. actually i have problems to understand another one's code. And ur code is not aligned ezly. so..

then why wait tell me the algo! :D

Here you are

[code]#include <algorithm>
#include <iostream>
#include <fstream>
#include <vector>
using namespace std;
//GLOBAL VARIABLES
vector<int> units;  // coin types
vector< vector<long long> > inOne;    //how many combinations using coin i is in number n
vector<long long> comb;     //combinations
int compute(int n);
int S, N;     //  S-sum,  N-number of coin types
ifstream fin("money.in");
ofstream fout("money.out");
int main()
{
      //READING INPUT
      fin>>N>>S;
     //SETTING VARIABLES
      units.resize(N);
      comb.resize(S+1, -1);
      inOne.resize(units.size());
      for(int i=0; i<inOne.size(); i++)
              inOne[i].resize(S+1);
      for(int i=0; i<N; i++)
              fin>>units[i];
      sort(units.begin(), units.end());
      fout<<compute(S)<<endl;
      return 0;
}
int compute(int n){
    long long res=0;
    if(n<units[0])      //if n is smaller than smallest cointype then n can't be constructed from cointypes(that's why sorting above is needed)
          res=0;
    else{
    for(int i=units.size()-1; i>=0; i--){     //checking every unit
            if(units[i]<=n){
                   if(units[i]==1)
                          res++;
                   else {
                   for(int j=int(n/units[i]); j>=1; j--){    //checking every multiply of unit(coin type)
                           if(j*units[i]==n){   //if multiplier * cointype is equal to n then just increase result by one
                                  res++;
                                  inOne[i][n]++;      //there is one more combination of number n using coin i
                           }
                           else{
                                    if(n-units[i]*j==1&&units[0]==1){
                                          res++;
                                          inOne[i][n]++;
                                    }
                                    else{
                                    if(comb[n-units[i]*j]==-1)  //if comb are not found yet
                                      compute(n-units[i]*j) ;    //then find it
                                        res+=comb[n-units[i]*j];     //add what you found to ther result
                                        inOne[i][n]+=comb[n-units[i]*j];      //add it to the number of comb with coin i in number n
                                      for(int k=units.size()-1; k>=i; k--){   //minus number of coins with value i---currently or before checked value
                                        res-=inOne[k][n-units[i]*j];
                                        inOne[i][n]-=inOne[k][n-units[i]*j];
                                      }
                                    }
                           }
                   }
                  }
              }
     }
     }
     comb[n]=res;     //assign what you found to the array
     return res;
}[/code]


Let $ n$ be a positive integer, and let $ a,b$ be positive real numbers. Prove that $ \sum^{n}_{k\equal{}0}\frac{n!}{k!(n\minus{}k)!}log(a^{n\minus{}k}b^{k})\equal{}log((ab)^{n2^{n\minus{}1}})$


pigeonhole principle
calculus
integration
function
real analysis
real analysis solved

hi everyone ,let µ un irrational , i want to prove that for every x in ]0,1[ there existe a natural n so that {nµ}<x  where {.} designe the fractional part .
 i use the density of a subgroupe of (IR,+) but it gives a solution juste in 1 case , can any one help me with an indication ,i know that's a classic  :oops:

Pigeonhole.

Why is this in "Solved Problems"?

a romanian friend told me that i must use pigeonhole , i will try again to use it , sorry for posting this ... i dont understand the rule ? can you explain the rule to post a topic  :huh:

i use the pigeonhol, i get in the non trivial case : there is m,n so that
 [µn]-[µm]-1+x < (n-m)µ < [µn]-[µm]+1-x so [(n-m)µ] can be [µn]-[µm]-1 or [µn]-[µm]
 only the case when it equals to  [µn]-[µm] gives the solution , what can we do if it's
[(n-m)µ] = [µn]-[µm]-1 , i blok here ,can you help me ?

Here is the solution:
Of course there is an $n$ such that $\frac{1}{n} < x$ and divide the interval $[0,1]$ in intervals $[0,\frac{1}{n}],[\frac{1}{n}\frac{2}{n}],...[\frac{n-1}{n},1]$. Now consider all numbers of the form $m\cdot x$ with m integer. By pigeonhole principle there are two of these numbers such that $\{a\mu\}$ and $\{b\mu\}$ lie in the same interval that means that $0 \le|\{a\mu\}-\{bx\mu\}|\le \frac{1}{n}\Leftrightarrow 0\le a\mu-b\mu-[a\mu]+[b\mu] \le 1/n \Leftrightarrow [a\mu]-[b\mu] \le (a-b)\mu\le [a\mu]-[b\mu] +\frac{1}{n}\Rightarrow \{(a-b)\mu\} \le \frac{1}{n}<x$.
Of course that I can choose a>b.
It is also known by Kroneker Th that the set of the numbers of the form mx-n with m,n integers and x a fixed irratinal number  is dese in $\mathbb{R}$.
By [a] i meant the integral part of the number a.

it's not what i want to prove  :lol:  i was looking for other thing , i know the theorem that you have talk about but  can't solve my problem immediatly (we shold us it twice to get the result...try to do it )

Your problem is an imediat consequence of the theorem:

since the set $M=\{m\mu -n|m,n\in \mathbb{Z}\}$ is dense in $\mathbb{R}$, someone can find $a \in M\cap (0,x)$. $a=m_0 \mu -n_0 \Rightarrow n_0<n_0+a=m_0\mu<n_0+1$ so $n_0=[m_0\mu]$ and $a=\{m_0\mu\}$. Here is your number. :P

so you considere that m_0 a natural number :D  try to prove it  :P

Since $a\in M$ must have the form $m_0\mu -n_0$ with $m_0, n_0\in \mathbb{Z}$. Because the set M consists of all numbers which have that form :P  :P  :P

natural numbers is the set of IN not ZZ  :rotfl:
try to apply this theorem twice to get the result  :idea:

I've edited post #6 to answer your question. since $a>b$ $a-b \in \mathbb{N}$. Let's finish with this problem. Peace!

but the firs equivalence is not correcte ( {x} isn't an increasing function !) so can you clarify more ?


How do you find [tex]\displaystyle\int{\frac{1}{\cos{\theta}}}\;d\theta[/tex]?  There obviously has to be a way to do it since it is in the table of integrals with an exact answer...

There is a trick :

Write 1/cos(x) = cos(x) / (1-sin(x)) = 1/2 (cos(x)/(1-sin(x)) + cos(x)/(1+sin(x)).

First term is -u'/u, second is u'/u.


Do there exist an infinite number of r.v with pairwise negative correlation?
a) countable
b) uncountable

for a i can find an appropriate sequence, for b for the moment i have no good ideas...

My first thought (sorry, no time now to check whether it's true):
[b]Claim[/b] ([s]un[/s]verified) for each $ \rho < 0$ there exists $ N$ so that there cannot be more than $ N$ r.v. with pairwise correlation less than $ \rho$.

If it is true, than the answer in b) is clearly "no".

UPD: In fact, it's true (I did not need a lot of time to prove it). We can assume wlog that the variance of each r.v. is $ 1$ and consider the variance of sum, then we are done.

take an infinite number of iid random variables $ Z_k$ that are $ \mathcal N(0,1)$ and define:
\[ X_k \equal{} Z_k \minus{} \sum_{j \geq 1} \rho^j Z_{k\plus{}j}\]
then $ E[X_p \cdot X_{p\plus{}n}] \equal{} \minus{}\rho^n \plus{} \sum_{j \geq 1} \rho^j \rho^{n\plus{}j} \equal{} \rho^n \cdot \frac{2 \rho^2 \minus{} 1}{ 1 \minus{} \rho^2}$, so that if $ 2 \rho^2 \minus{} 1 < 0$, this gives an infinite sequence of negatively correlated random variables.

Thank you, zhoraster!
Thank you, alekk!

Now, everything is clear for me

[quote="zhoraster"]
If it is true, than the answer in b) is clearly "no".
[/quote]
Why? I mean, that if for some $ \rho$ you have plenty of [b]pairs[/b] with correlation $ \minus{}\rho$ or less, it still doesn't immediately mean that you have plenty of [b]variables[/b] with big negative pairwise correlations. Am I missing anything? :?

Yes, [b]fedja[/b], you are right, it does not immediately imply that.

Should it be true due to some kind of Ramsey theorem?

Let's see. You have a complete graph with uncountably many vertices with edges colored by countably many colors $ c_j$. You also have a sequence of numbers $ N_j$. You win if you can find an $ N_j$-clique of color $ c_j$. This is not a version of Ramsey I am aware of :( but it may be true after all. Let me think a bit.

Even for a bichromatic colouring of an uncountable complete graph, Ramsey does not guarantee a monochromatic clique. The classical counterexample is $ \mathbb{R}$ with some well-order $ \prec$, black edges between $ x$ and $ y$ if $ \prec$ and $ <$ coincide, and red edges if they don't.
Then only countable monochromatic cliques exist.

Sorry, it's not true, as I've found.

An easy example is $ (0,1)$ with this colouring: color $ \{x,y\}$ into $ n$th colour if the binary forms of $ x$ and $ y$ agree up to $ n$th position. Then there is no even a monochromatic $ 3$-clique.

OK. Fortunately, the original answer is still correct. What we really need is the following. If $ X$ an $ X_j$ ($ j\equal{}1,\dots,N$) are random variables such that $ EX^2\equal{}EX_j^2\equal{}1$, $ EX_iX_j\le 0$, and $ EXX_j\le\minus{}\rho$, then $ 0\le E(X\plus{}t\sum_j X_j)^2\le 1\plus{}Nt^2\minus{}2Nt\rho$, which yields $ N\le \frac 1{\rho^2}$ when $ t\equal{}\rho$.


Use the LLL algorithm to reduce the lattice with basis $ v_1\equal{}(20,16,3), v_2\equal{}(15,0,10), v_3\equal{}(0,18,9)$.

Do your own homework. Also, it's not easy to find a complete description of the algorithm- I don't have one, even though I know what it does and I'm in a course that discussed it. The best I can do is invoke it as a black box in SAGE.

I wasn't able to understand the algorithm. so i just had to post this to see if anyone can explain this algorithm.

This is a great topic! I'm not going to do your specific problem, but I will clear up what the algorithm is doing. So LLL, as you probably know, is the analog of Gram-Schmidt for lattices. Gram-Schmidt works in vector spaces because it relies on the fact that every time you subtract off the non-orthogonal component, you stay in the vector space. Thus, the span of the orthogonal vectors is the same as the span of the original linearly independent vectors. In lattices, you only have integer multiples of the basis vectors, thus Gram-Schmidt will still give you orthogonal vectors, but the lattice spanned by the orthogonal vectors is now completely different than the starting basis!

Here's the first goal of LLL: have each successive LLL-basis vector be as small as possible (ideally each successive basis vector would be the successive minima of the lattice). In other words, if $ b_1,\ldots,b_n$ is your LLL-reduced basis, then $ b_1$ is the shortest vector, $ b_2$ is the next shortest, etc. In practice you can't get there, so you have what's called $ \delta$-LLL reduced and get within $ \delta$ essentially.

The second goal of LLL: make the LLL-basis vectors as close to orthogonal as possible. Remember that we are stuck working within the confines of the lattice, not just any old vector space like we are used to. So we must subtract off integer coefficients of the basis (i.e. by rounding the usual Gram-Schmidt projection coefficients instead of using them as they are). Hopefully this high level description will help you understand your specific implementation of the algorithm -- since there are many versions.


[quote="NTAnh_K41SP"]Cho N là tập hợp các số nguyên dương có n chữ số$ \in$1;2;3;4}và số chữ số 1 bằng số chữ số 2 .Tìm lực lượng của N

minh ko biet tieng anh nen post tam bang tieng viet vay, ban nao biet thi dich giup minh voi
theanh_0808@yahoo.com[/quote]

Let $ N$ is a set positive integers have $ n$ figures $ \in\;\{1,2,3,4\}$ and the number of figures 1 equal the number of figures 2. Find $ |N|$

@The Anh em học thêm 1 chút tiếng Anh đi nha  :) lần này anh giúp, lần sau học rồi hãy post đề


Give the function f(x) and some $ a \neq$ 0. If f (x) defined at $ x \equal{} x_{0}$ so well defined at $ x \equal{} x_{0} \plus{} a$. in addition we also have: $ f(x\plus{}a)\equal{} \frac {f(x)\minus{}1}{f(x) \plus{}1}$  $ \forall x\in D$ (D is the set defined by f)

Prove that: $ f (x \plus{} 4 a) \equal{} f (x)$


find all functions $ f$ and $ g$ such that : $ f: [a, \plus{} \infty)\rightarrow \mathbb{R}$increasing and $ lim_{x\rightarrow \plus{}\infty}f(x) \in \mathbb{R}$ and $ g: (a, \plus{} \infty)\rightarrow \mathbb{R}$ increasing too.

and $ g(x) \equal{} \frac {f(x) \minus{} f(a)}{x \minus{} a}$

 :)

[quote="mr.inequalities"]find all functions $ f$ and $ g$ such that : $ f: [a, \plus{} \infty)\rightarrow \mathbb{R}$increasing and $ g: (a, \plus{} \infty)\rightarrow \mathbb{R}$ increasing too.

and $ g(x) \equal{} \frac {f(x) \minus{} f(a)}{x \minus{} a}$

 :)[/quote]

I dont understand the problem : choose any positive increasing function $ g(x)$ and take then $ f(x) \equal{} (x \minus{} a)g(x) \plus{} b$

Obviously $ f(x)$ is increasing and the pair $ (f,g)$ fit the requirements.

[quote="mr.inequalities"]find all functions $ f$ and $ g$ such that : $ f: [a, \plus{} \infty)\rightarrow \mathbb{R}$increasing and $ lim_{x\rightarrow \plus{} \infty}f(x) \in \mathbb{R}$ and $ g: (a, \plus{} \infty)\rightarrow \mathbb{R}$ increasing too.

and $ g(x) \equal{} \frac {f(x) \minus{} f(a)}{x \minus{} a}$

 :)[/quote]

Obviously no solution else $ g(x)$ would be a positive increasing function whose limit at $ \plus{}\infty$ would be $ 0$ :)

If you change "increasing" into "non decreasing", you have a unique solution $ (f(x)\equal{}c,g(x)\equal{}0)$


Let $a_{1},a_{2},\cdots$ and $b_{1},b_{2},\cdots$ be arithmetic sequences of positive integers whose common differences share no factor $>1$. Prove that every integer can be expressed in the form $a_{i}-b_{j}$ for some $i,j$.
have a look at the special case when $a_{i}=b_{j}$.

*)Let $k = a_{1}-b_{1}$; $d,e$ be the common difference of the first and the second sequence, respectively. It's clear if one of them is equal to $1$, so consider $d,e > 1$.
*)Say that we want $a_{i}-b_{j}= X$ with $a_{i}= a_{1}+(i-1) \cdot d$ and $b_{j}= b_{1}+(j-1) \cdot e$
*)Since $(d,e) = 1$, then there is some nonzero integer $a$ and $b$ so that $ad+be = 1$, and one of $a$ or $b$ must be negative.
WLOG $b < 0$. Moreover, $d \cdot (a(X-k))+e \cdot (b(X-k)) = X-k$
*)If $X \geq k$, substituting $i = a(X-k)+1$ and $j =-b(X-k)+1$ yields
$a_{i}-b_{j}=k+d \cdot (a(X-k))+e \cdot (b(X-k)) = k+(X-k) = X$ as desired.
*)If $X < k$, let $p$ be a positive integer so that $a(X-k)-ep < 0 < b(X-k)+dp$, then substituting $i = a(X-k)-ep+1$ and $j =-b(X-k)+dp+1$ gives the same result.

Just tell me if I wrote any mistakes..

the same result as me. only i have another very short proof concerning the chinese remainder theorem. let's see if you can rediscover it. :P

ps. i'm not very good at finding mistakes :D
ps.2. look at your inbox. i've sent a pm.


Any individual scores for the Australian team? Medal tally? I already know the combine score but wish to know what they did individualy.

Australian results:
Ivan Guo 32
Laurie Field 24
Alex Hua 23
Daniel Nadasi 20
Graham White 15
Kamil Khan 11
btw, who are you Templar?


well i have an assignment due on Friday and I was wondering if anyone here could help me with a few problems, im sorry if they may be kind of easy to you but i really suck in math.

6. Find the exact value of the six trigonometric functions of 45 degrees
sin45degrees=
cos45degrees=
tan45degrees=
csc45degrees=
sec45degrees=
cot45degrees=

 Describe each of the following graphs of y. Give (a) its amplitude (if applicable), (b) its maximum and minimum(if applicable) and (c) its period. If it has no amplitude, maximum or minimum write" none" on the appropriate blank line.
10. y= 3sin 1/3x
11. y=tan2A
if anyone can help me with this you will be doin me a real big favor

These are basic trig questions :o , I hope this isn't homework...

But judging by this, do you know trig identities? If not, I think it would be better if you went to mathworld.com to learn your trig identities.

[quote="suntzu17"]6. Find the exact value of the six trigonometric functions of 45 degrees
sin45degrees=
cos45degrees=
tan45degrees=
csc45degrees=
sec45degrees=
cot45degrees=
[/quote]Ok, for each of these, try drawing a right triangle with two 45 degree angles. Let the legs be 1, so the hypotenuse will have length $\sqrt{2}$.

Then use the definition of each trig function. For example, $\sin \theta=\frac{\text{opposite}}{\text{hypotenuse}}$.


Notice that we can set up a recursion (sorta). $ A_n\equal{}n\plus{}(n\minus{}1)$. We could test this on the figures we are given, and indeed it works. So the area of the 6th figure is $ 6\plus{}4(5)\equal{}6\plus{}20\equal{}\boxed{26}$.
You'll agree this is faster than drawing the 6th figure out ;)

Or......... you could just realize that each time you continue the figure u add 5...... 1+5(5). thats like 1000 times simpler


The USABO finalists, announced from CEE's website, are:

-Alexandre Boulgakov
-Johnny Hwang
-Jingkun (Ginger) Yang
-Yingyu Gao
-Jonathan Liang
-Andrew Shie
-Meng Xiao He
-Ann Ouyang
-Julia Rogers
-Mark Shteyn
-Damon Wang
-Zachary Frankel
-Ping He
-Jun Kim
-Margarita Krivitski
-Benjamin Lee
-Helen Yang
-Francois Greer
-Barry Liu
-Asad Moten
-Seungsoo Kim

Congrats to the 21 finalists!

Alexandre Boulgakov qualified for the USAMO as a 10th grader in 2006. In 2007, his high school did not even participate in the AMC.

.. and why would they be? Scores on the semifinal exam have no influence whatsoever on the next round of competition.

Some people never learn their scores after semifinals at all.

Mark Shetyn is one of them.  The others I don't know about.

Gold medalists (Team USA):
Meng Xiao He
Barry Liu
Mark Shteyn
Helen Yang

Silver medalists:
Julia Rogers
Andrew Shie
Annie Ouyang
Jonathan Liang

Bronze medalists:
Francois Greer
Seungsoo Kim
Sasha Boulgakov
Damon Wang

(No particular order within each medal category)

In case anyone was wondering, the results of the 2007 IBO in Canada can be found here:
http://ibo2007.usask.ca/results.html

Specifically, for information on how Team USA 2007 (and of all previous USABO years) did, check out the Wikipedia page:
http://en.wikipedia.org/wiki/Usabo

Could any of the finalists please explain how to study for semis?

Thanks!

just run a dictionary attack. that's what a friend of mine did.

check here

http://en.wikipedia.org/wiki/Dictionary_attack

That link is completely unhelpful. It explains WHAT a dictionary attack is (which was already obvious from the name), but gives not indication as to HOW to actually implement one in this particularly case.

umm.. could some one plz tell me how to calculate your score after semifinals?  :maybe:

Running a dictionary attack is not a good idea. I'm sure it's illegal and from what I've found can take up to 6 days to do one attack, because of the amount of words and combinations in the dictionary. Aswell as that you need a special program to do the attack which I doubt you can just buy on the internet. You probably need to know someone who has the program. Aswell as that you need to make sure you don't get caught among other things.

Lucy

Just out of curiosity why would you want to run a dictionary attack anyway?

___________
[url=http://enzytexposed.com/]Enzyte[/url]

[quote="tittu"]I downloaded the 2007 and 2008 semifinal questions and 2008 open test question papers from CEE web site but the pdf s are password protected. Can you help me out with the passwords, please?[/quote]
I can try to run the dictionary attack to you if you e-mail me.


geometry
3D geometry
sphere
real analysis
real analysis unsolved

Is it possible to divide a sphere in 15 regions of equal volume using 4 planes?

Consider a simpler problem: can one divide a disk into 7 regions of equal area using 3 lines? (Apparently not.) A proof for 2D case would have a chance of being generalized.

I still don't know the solution for this problem. Can someone post one solution??


real analysis
calculus
derivative
integration
floor function
LaTeX
complex numbers

Let $ B$ denote the Borel subsets of $ \Re$ and $ m$ denote Lebesgue measure on $ B$
For $ E\in B$ define $ vE\equal{}\sum_{n\equal{}0}^\infty \frac{m(E\bigcap[n,n\plus{}1))}{2^n}$

Show that $ (\Re, B, v)$ is finite measure space
and find the Randon-Nikodym derivative $ [\frac{dv}{dm}]$

$ \sigma$-additivity of $ v$ follows from $ \sigma$-additivity of Lebesgue measure, $ v\left( \emptyset \right) \equal{} 0$, and direct computation gives $ v\left( \mathbb{R}\right) \equal{} 2$, so it's a finite measure space.  Also for all $ E\in B$ we have $ v\left( E \right ) \equal{} \int_{E\cap \mathbb{R}^ \plus{} }{2^{ \minus{} \lfloor x \rfloor} dx}$ so
\[ \frac {dv}{dm} \equal{} \begin{cases} 2^{ \minus{} \lfloor x \rfloor} & x\geq 0 \\
0 & x < 0 \end{cases}
\]
By the way, the Fraktur $ \Re$ is typically used to denote the real part of a complex number, i.e. $ \Re (z) \equal{} \frac {z \plus{} \overline{z}}{2}$.  The symbol $ \mathbb{R}$, and less frequently $ \mathbf{R}$, is usually preferred to denote the real numbers; the relevant latex codes recognized on these fora are \mathbb{R} and \mathbf{R}, respectively.  If you have package bbm you can also use \mathbbm{R} which looks a bit nicer, but it's not used here.


solve for $x,y$ reels the system.
$\sqrt{x^{2}-2x+6}\cdot \log_{3}(6-y)=x$
$\sqrt{y^{2}-2y+6}\cdot \log_{3}(6-z)=y$
$\sqrt{z^{2}-2z+6}\cdot \log_{3}(6-x)=z$
good luck .

Let $f(a) = \frac{a}{\sqrt{a^{2}-2a+6}}$ and $g(a) = \log_{3}(6-a)$.
Note that $a > 3 \Leftrightarrow f(a) > 1, g(a) < 1$ (and vice-versa).

If $x > 3$, we have
$g(y) = f(x) > 1 \Rightarrow y < 3 \Rightarrow g(z) = f(y) < 1 \Rightarrow z > 3 \Rightarrow g(x) = f(z) > 1 \Rightarrow x < 3$.
A contradiction.
We proceed similarly if $x < 3$.

Therefore, $(x,y,z) = (3,3,3)$. It's easy to check that this solution satisfies the given condition.


probability
function
AoPS Books
number theory
conditional probability
\/closed

does anyone know any intermediate level counting/ combinatorics book that contains alot of practice problems, and is not mostly text? ( other than AoPS books??)
and also, is Number Theory, by Geroge Andrews, better, or is Fundamentals of Number Theory, by William LeVeque, better?

[quote="rrusczyk"]We'll have our Intermediate Counting & Probability text out in 3-4 months, which is probably the best fit for what you seek.  In the meantime, there are a few other less comprehensive books out there, such as Mathematics of Choice by Ivan Niven.[/quote]

i know about the interm c&p book, but are there any other good books out there??

Principles and Techniques in Combinatorics by Chen Chuan-Chong and Koh Khee-Meng

[quote="rrusczyk"][quote="rnwang2"]The Intermediate Counting & Probability text will be more comprehensive than [i]Mathematics of Choice [/i]?  :o[/quote]

For Intermediate techniques (PIE, generating functions, recursion, conditional probability, etc), absolutely.[/quote]
Yeah, I'm going to have to say that I was slightly disappointed by mathematics of choice.


What is expi (and subjects related) and how do you use it?

$ \text{expi\,}(x)\equal{}e^{ix}$ where $ x$ is real. So, expi(pi/3) is equivalent to dir(60), say. Asymptote does not distinguish between pairs and complex numbers.

sorry, but I'm not much of an asymptote person.
Whats dir?

Sorry but I dont want to sift through all 178 pages.
Where is the section on expi and dir and related?

Page 23 for dir. The entry you need reads
[quote]
pair dir(real angle)
returns a unit vector in the direction angle measured in degrees;
[/quote]
The only thing I can add to that is that the angle is measured from the positive $ x$-semiaxis counterclockwise.
[asy]size(200);
for(int i=0;i<360;i+=30)
D((0,0)--D("\mbox{dir}("+string(i)+")",dir(i),dir(i)));[/asy]

Note that if you wanted to shade the acute sector from 300 to 30 degrees, you would do -30 instead of 300.


Here is a challenge.

Can you show that $n(n+1)(n+2)(n+3) + 1$ is a perfect square for $n = 1,2,3...$?

Eh, that's tough. But I have a solution that seems really stupid...

[hide]
After expanding the expression, we get n^4+6n^3+11n^2+6n+1. Clearly this can be factored into the form of (n^2+kn+1)^2, where k is an integer. Indeed, we can factor it into (n^2+3n+1)^2, and we're done.
[/hide]

I don't think it's rigorous enough though.... Can anyone come up with a better solution/confirm mine?

Actually..

That is exactly what the answer is :D

This problem was from British Math Olympiad 1965. You just solved Olympiad level problem! :P  :P

The reason I posted this problem was because I was pretty sure someone can get the answer. :D And I was right. :ninja:

[quote="Silverfalcon"][quote="amirhtlusa"]this is a classic problem and relys on the fact that:
$ (x-1)x(x+1)(x+2)+1=(x^2+x-1)^2.$[/quote]

amir, wasn't that idea appear on one of AIME problems?[/quote]
After a quick search on Kalva, I found that problem number 1 in AIME 1989 was based on that idea.

Here was the problem:
Find the value of $\sqrt{1+(28\times29\times30\times31)}$

yes, that is true, ive seen not only in AIME, but also on IRMO(iran mathematical Olympiad) several times. :)


analytic geometry
function
calculus
calculus computations

I'm not sure if this is where to post this, so anyway.

A frictionless particle sliding down a helix r(theta)=(a cos(theta))i+(a sin(theta))j+(b*theta)k with a,b>0 under influence of gravity.  speed of a falling particle after distance z is sqrt(2gz)

a) find d(theta)/dt at theta=2pi
b) express particle's theta and z coordinates as a function of t
c) express dr/dt and d^2r/dt^2 in terms of t

the answers
a)2*sqrt(pi*g*b/(a^2+b^2))
b)theta=gbt^2/(2(a^2+b^2)), z=gb^2t^2/(2(a^2+b^2))
c) dr/dt=gbt/sqrt(a^2+b^2)T,  d^2r/dt^2=gb/sqrt(a^2+b^2)T+gbt/(a^2+b^2)N

so far, after a few hours of work

b=sqrt(2gz)
z=b*theta=2b*pi
b=2sqrt(gb*pi)
magnitude of dr/dtheta at 2pi=sqrt(a^2+b^2)
is b=dr/dt?

I am doing this for an independent study project, so any help would be appreciated as the physics and math teachers at my high school did not know what to do.

Attached are a few notes regarding the independent study project.  We're hoping a moderator or someone else would spend a few minutes to review the notes and point out any deficiencies in the theory or algebra involved.


A group of N islands are connected by bridges. Each island has bridges to at most 3 other islands.

One can travel between any 2 islands by crossing at most two bridges. What is the largest possible value of N? (Bridges are allowed to go over or under other bridges.)

u can check ur solution here:
[url]http://www.projecteureka.org/problem/question/186[/url]

An upper bound:
If Island $ a_1$ has the minimum number of bridges then it is connected to at most three other islands, $ a_2,a_3,a_4$. Each of these islands are connected to at most two other islands $ a_5...a_{10}$.
Now islands $ a_5...a_{10}$ can only reach $ a_1$ through islands $ a_2,a_3,a_4$. Any other islands we wont be able to reach $ a_1$ hence we have our maximum.
Now we must show that it exists, but that's an easy construction
[geogebra]86c7df1206953785b583affb317a463fb715604c[/geogebra]

In the case when $ N$ is maximum, all the islands are equivalent(each is connected to three others).
So, it must be possible to move around the points and stretch the lines to get a similar diagram to the original, right? But, I wasn't able to do that. Could someone help me?

Hopefully I'm understanding your comment correctly. All vertices having the same degree and being able to switch any vertices and rearrange the graph to make it look the same ([url=http://en.wikipedia.org/wiki/Vertex-transitive_graph]vertex transitive[/url]) are not equivalent statements.

ocha's example is the Petersen graph. It is vertex-transitive, so just try harder?

I didn't know anything in graph theory earlier. Thanks MellowMelon. I understand what you are saying now.
OK I am able to understand why it is the Petersen graph. So it must be vertex-transitive.
BTW, Are there Regular graphs which are not vertex-transitive?
If yes, could someone show me an example?

[b]@ocha:[/b] Your diagram is the original I was referring to.


limit
integration
real analysis
real analysis unsolved

Suppose $f\in L^{p}(E)\cap L^\infty(E)$. Does it hold that
$\lim_{r\to\infty}\|f\|_{r}= \|f\|_\infty$
if $mE=\infty$?

Yes. There is $A\subset E$ such that $m(A)<\infty$, $\int_{E\setminus A}|f|^{p}<\epsilon$, and $|f|<\epsilon$ on $E\setminus A$.


Under Karma Earned Daily on the Stats page, it shows that nobody earnd any karma, even though I know that I have earned at least a bit.

We are working on a bug with Karma - please be patient as we try to determine the problem and fix it.

Thsi reminds me of something that I'm sure a lot of people have noticed, so I'm not sure if this has been posted. When I do about 10 problems or so (for the first time in the day, and I rate the problems), sometimes the stats show that I earned 2000 something karma, even though I didn't. What does this mean?

I believe they have fixed at least that much of it.

Somehow though, it shows my karma earned to be much more than my actual karma...

karma earned - karma lost = actual karma. Your karma earned should always be higher than your actual karma.

That being said, we will be resetting karma scores some time soon due to the karma earned being off. We'll be resetting them to what should be correct values.

Will we still have what we gained so far?

[quote="levans"]We'll be resetting them to what should be correct values.[/quote]

:wink:


Place each of the digits 0, 1, 2, 3, 4, 5 and 6 in one of the boxes
below to create a true equation. What is the sum for the addition
problem?

[asy]size(50);
path box = (0,0)--(1,0)--(1,1)--(0,1)--cycle; real shifter = 1 + .2; draw(box ^^ shift(shifter*right)*box ^^ shift(shifter*down)*box ^^ shift(shifter*down + shifter*right)*box); label("$+$",shift(shifter*down + shifter*left)*(.5,.5));
draw(shift(1.15shifter*down)*((-1.5,0)--(2shifter,0)));
draw(shift((1+1.3shifter)*down + shifter*left)*box ^^ shift((1+1.3shifter)*down + shifter*right)*box ^^ shift((1+1.3shifter)*down)*box);[/asy]

[quote=Ebob]When adding 2 2-digit numbers, the hundreds place must be 1. You cannot have 6+5 in the tens place, as this would yield a 1 in the tens place, and we have already used this. Therefore the tens places are 6+4 = 10. This leaves the digits 2, 3, and 5 for the units digit, leaving an obvious choice of 2+3 = 5, making the answer

62+34 = $ \boxed{105}$[/quote]

Um I think its 62+43 instead


[quote="RohanSingh"]I want to know how the memory of computer works? I mean how the data is stored.[/quote]Do you mean on the hardware side, or how memory space is allocated to different programs?

[quote="RohanSingh"]I want the hardware part.[/quote]

Please check it out.

http://en.wikipedia.org/wiki/Dynamic_random_access_memory

http://en.wikipedia.org/wiki/Static_random_access_memory


integration
logarithms
Euler
limit
function
calculus
calculus computations

What is the Laplace transform of $ \frac{1}
{x}$ ?

Thanks!!!

$ \mathcal L\left( {\frac{1}{x}} \right) \equal{} \int_0^\infty  {\frac{{e^{ \minus{} sx} }}{x}\,dx}  \equal{} \int_0^\infty  {\int_0^\infty  {\left( {e^{ \minus{} ux}  \cdot e^{ \minus{} sx} } \right)\,du} \,dx} .$

I think this is it.

For the Laplace Transformation of $ f(x)$ to exist, $ f(x)$ must be defined on the interval $ \left[0,\plus{}\infty\right)$ and $ \frac{1}{x}$ is not!

Ohhh, I forgot that.

$ f(x) \equal{} \frac1x$ has a discontinuity at $ x \equal{} 0.$

 :ninja:

Still, Mathematica gives the solution:

$ \mathcal L\left(\frac {1}{x}\right) \equal{} \minus{} \gamma \minus{} \ln(s)$

where $ \gamma$ is the Euler-Mascheroni constant.

Evidently, this is the effect of the fact that:

$ \lim_{\tau\rightarrow 0}\int^{\infty}_{\tau}\frac {dx}{x\,e^{s x}} \equal{} \minus{} \gamma \minus{} \ln(s)$

but I don't know if this is a "proper" Laplace transformation because the conditions for the existence of one aren't fulfilled.

The problem with this being a "proper" Laplace transoform isn't so much the discontinunity - discontinuous functions can have transforms - but that $ \frac1x$ is not locally integrable in a neighborhood of zero.

I'm having a little trouble understanding that Mathematica formula; for all real $ s$, the limit in milin's post is $ \plus{}\infty.$ There's some kind of principal value-like thing going on here, but how can you have a principal value when you only have one side of the singularity?

I think Mathematica is wrong.  The only requirement for the Laplace Transform to exist (as far as I know) is that the improper integral:

$ \int_0^{\infty}f(x)e^{\minus{}sx}dx$ exists for some $ s$.  I believe the integral:

$ \int_0^{\infty}\frac{1}{x}e^{\minus{}sx}dx$ diverges for all s since:

$ \frac{1}{xe^{sx}}\sim \frac{1}{x}$ when x is close to zero and thus:

$ \int_0^1 \frac{1}{xe^{sx}}dx$ diverges.


1.  Prove the if a , b , c , and n are whole numbers with $ n > 3$, then there is one whole numer k so that none of the number $ k\plus{}a,k\plus{}b,k\plus{}c$ is divisible by n.

2. Let n be the sum of all the digits of the number $ 5555^{5555}$; let m be the sum of all the digits of n and let r be the sum of all the digits of m. Find r.

3. Find what numbers from 1 to 12 divide $ \frac{10^{601}\minus{}10}{9}$.

4. Prove the following inequality:

$ \frac{1}{2} * \frac{3}{4} * \frac{5}{6} * ...* \frac{99}{100} < \frac{1}{\sqrt{101}}$

5. Prove that there is an infinite number of primes in the form of $ 4n\plus{}3$.

[quote="modularmarc101"]1.  Prove the if a , b , c , and n are whole numbers with $ n > 3$, then there is one whole numer k so that none of the number $ k \plus{} a,k \plus{} b,k \plus{} c$ is divisible by n.

2. Let n be the sum of all the digits of the number $ 5555^{5555}$; let m be the sum of all the digits of n and let r be the sum of all the digits of m. Find r.

3. Find what numbers from 1 to 12 divide $ \frac {10^{601} \minus{} 10}{9}$.

4 Prove the following inequality:

$ \frac {1}{2} * \frac {3}{4} * \frac {5}{6} * ...* \frac {99}{100} < \frac {1}{\sqrt {101}}$

5. Prove that there is an infinite number of primes in the form of $ 4n \plus{} 3$.[/quote]

[b]1)[/b] For a given $ n > 3$, we can find a $ d$ so that $ k \plus{} d \neq k \plus{} a,k \plus{} b,k \plus{} c (mod n)$ for all $ k$. Now let $ k \equal{} \minus{} d(mod n)$ and it definitely satisfy the condition.

[b]2)[/b]We have $ 5555^{5555}\equal{}2^{5555}\equal{}2^{5}\equal{}5(mod 9)$, so we have $ r\equal{}5(mod 9)$.
Now $ 5555^{5555}$ has less than $ 4 \times 5600\equal{}22400$ digits, so $ n<201600$
$ \Rightarrow m \le 1\plus{}9\plus{}9\plus{}9\plus{}9\plus{}9\equal{}46 \Rightarrow r \le 3\plus{}9\equal{}12$
As $ r\equal{}5(mod 9)$, $ r\equal{}5$.

[b]3)[/b]$ 2,5|\frac {10(10^{600} \minus{} 1)}{9}$ is obvious. $ 10^{6 \times 100} \equal{} 1(mod 7)$ , so $ 7|\frac {10(10^{600} \minus{} 1)}{9}$ also.

As $ 10^{600} \minus{} 1$ is odd, $ 4 \nmid \frac {10(10^{600} \minus{} 1)}{9}$. Also, from $ 10^{81 \minus{} 27} \equal{} 1(mod 81)$ we have $ 10^{600} \minus{} 1 \equal{} 10^6 \minus{} 1 \equal{} (999)(1001) \equal{} 27 \times 37 \times 1001(mod 81)$, which gives $ 3|\frac {10(10^{600} \minus{} 1)}{9}$ and $ 9 \nmid \frac {10(10^{600} \minus{} 1)}{9}$.

$ 10^{600} \minus{} 1 \equal{} ( \minus{} 1)^{600} \minus{} 1 \equal{} 0(mod 11)$

Combining them gives $ 1,2,3,5,7,11$ divide $ \frac {10(10^{600} \minus{} 1)}{9}$

[b]4)[/b] Hint: $ \frac {1}{2} < \frac {2}{3}$, $ \,\,\, \frac {3}{4} < \frac {4}{5}$, ...
[b]5)[/b]Consider $ 4p_1p_2...p_n \minus{} 1$,  it's not divisible by $ p_1,p_2, ..., p_n$ and so must be divisible by a larger prime of the form $ 4k \plus{} 3$

Sorry that I didn't put my work, I don't know how to do some of them and I wanted to see how others would target them.

[quote="Arrange your tan"]modularmarc101,

five problems is not "a couple of problems."

It would be better (honest) that you split them up into
separate threads and really make them into a couple of
problems as your headline promises.

Also, show some of your own work (attempts) toward a solution.[/quote]

Stop criticizing people about showing work etc. etc.

What if they didn't know how to do it, at all??
Instead of blabbering about how many problems there should be in a post, you could help by giving a hint or solution!

[quote="Arrange your tan"]Also, show some of your own work (attempts) toward a solution.[/quote]
The only time this is asked of topic creators is when the problems look like homework. This does not look like homework.

It does look like common competition-style problems (in particular, #4 and #5 have been posted dozens of times), so most likely the OP is preparing from a book or similar and doesn't have a solution manual (and maybe not even any explanation of common techniques).  Not posting work is much more forgivable in this case.

Edit:  There is a very nice thread about #4 in which Kent, Ravi B, and I (if I recall) posted nearly the same solution within 30 seconds of each other, except that Ravi's was the cleanest.  I wonder if anyone knows what I'm talking about and can find it.


analytic geometry
rotation
conics
ellipse
symmetry
quadratics
algebra

Find the maximum value of x^2+y^2 if x^2+xy+y^2=12

Here's my solution:

We can use polar coordinates and find the maximum value of r^2:

r^2+r^2sin(t)cos(t)=12
r^2=12/(1+sin(t)cos(t))
r^2=12/(1+sin(2t)/2)
we want to minimize the denominator, whose minimum is 1+(-1)/2=1/2, so the maximum value of r^2 is 24. (I'm not sure if this is correct, but it seems so to me :P)

However, what do you all feel are the best general methods for problems where one must maximize or minimize an expression given a restriction, especially when using polar coordinates is not so simple?

Umm, are you sure there can be a maximum? The AM-GM doesn't say so...unless I'm making a mistake.

[hide]By AM-GM, we have $ \frac {x^2 \plus{} y^2}{2} \ge \sqrt {x^2y^2} \implies \frac {x^2 \plus{} y^2}{2} \ge xy.$ From $ x^2 \plus{} y^2 \plus{} xy \equal{} 12,$ we have $ xy \equal{} 12 \minus{} x^2 \minus{} y^2.$ Plugging this in, we get \[ \frac {x^2 \plus{} y^2}{2} \ge 12 \minus{} x^2 \minus{} y^2 \implies x^2 \plus{} y^2 \ge 24 \minus{} 2x^2 \minus{} 2y^2 \implies 3x^2 \plus{} 3y^2 \ge 24 \implies x^2 \plus{} y^2 \ge 8.\] So, there is no limit to their maximum. [/hide]

Do you mean minimum by any chance?

Hello.
Simply looking at the equation, we can conclude
The mininum is when xy is maximized.
xy is maximized when it equals
$ \frac {x^2\plus{}y^2} 2$ AM-GM
so x=y=2, and the minimum is 8

The maximum for positive numbers is when xy is minimized (0)
thus either x or y is 0, and the max is $ \sqrt {12}$

However, if we include the negative numbers, then there should be no limit

Yeah, the problem says maximum.

Thanks for the responses.

limac: But, I don't understand why that means there can't be a maximum.

Isn't it possible for there to be a minimum and a maximum?  [url=http://www.wolframalpha.com/input/?i=x^2%2Bxy%2By^2%3D12]Wolfram Alpha[/url] says that it is a rotated ellipse, so it makes sense that there would be a minimum and a maximum.

[quote="limac"]We end up with $ x^2 \plus{} y^2 \ge 8,$ which implies that the minimum of $ x^2 \plus{} y^2$ is 8, but it can be any value greater than 8, or am I overlooking something? :|[/quote]

You are overlooking something. What you did is find the minimum (assuming that the 8 can be achieved, which it can). However, just because you have a lower bound doesn't mean you have a maximum.

To take an extreme example, say we have f(x)=2sinx, find the max of f(x). Well, we can clearly say that $ 2sinx \geq \minus{}2$ , so the minimum is -2. Does this mean it has no maximum?

As for the problem, I have an alternate solution.

[hide]
Using Lagrange Multipliers, we have $ 2x\equal{}\lambda(2x\plus{}y)$ and $ 2y\equal{}\lambda(2y\plus{}x)$
Thus, $ \frac{2x}{2x\plus{}y}\equal{}\frac{2y}{2y\plus{}x} \implies 4xy\plus{}2x^2\equal{}4xy\plus{}2y^2 \implies x\equal{}\pm y$

If $ x\equal{}y$ we have $ x\equal{}y\equal{}\pm 2$ so $ x^2\plus{}y^2\equal{}8$

If $ x\equal{}\minus{}y$ we have $ x\equal{}\minus{}y\equal{}\sqrt{12}$ so $ x^2\plus{}y^2\equal{}24$ so $ 24$ is our maximum and $ 8$ is our minimum (note: we have no boundary cases to check)[/hide]

[b]The short-answer procedure:[/b] From looking at it, I see that you want to optimize the distance from the origin of an ellipse whose axes of symmetry are y=x and y=-x. Plug in (x,x) and (x,-x) into the equation of the ellipse to find the endpoints of the minor and major axes, respectively, and compute their square-distances from the origin.

[b]Algebraic Formalism:[/b] Let $ u \equal{} x \plus{} y$ and $ v \equal{} x \minus{} y$. We want to maximize $ F \equal{} \frac {1}{2} (u^2 \plus{} v^2)$ given $ \frac {3}{4} u^2 \plus{} \frac {1}{4} v^2 \equal{} 12$. Note we can write the latter in two helpful ways:

$ \blacktriangleright \frac {1}{2}F \plus{} \frac {1}{2} u^2 \equal{} 12 \implies \boxed{F \equal{} 24 \minus{} u^2 \leq 24}$

$ \blacktriangleright \frac {3}{2} F \minus{} \frac {1}{2} v^2 \equal{} 12 \implies \boxed{F \equal{} 8 \plus{} \frac {v^2}{3} \geq 8}$.

It should be fairly easy to demonstrate that the equality cases are attainable.

We could make that even more straightfoward by setting $ m \equal{} (x \plus{} y)/4, n \equal{} (x \minus{} y)/\sqrt {48} \Longrightarrow F \equal{} 8m^2 \plus{} 24n^2$ subject to the constraint $ m^2 \plus{} n^2 \equal{} 1$.

I think the following solution is quite different from all of the above, so it's worth for me to post it here!

[hide="Slick Solution"]

Let $ A \equal{} x^2 \plus{} y^2$. Now dividing both sides by $ 12$, we get:
$ \frac {A}{12} \equal{} \frac {x^2 \plus{} y^2}{x^2 \plus{} xy \plus{} y^2} \equal{} \frac {\frac {x^2}{y^2} \plus{} 1}{\frac {x^2}{y^2} \plus{} \frac {x}{y} \plus{} 1}$

Now, let $ \frac {x}{y} \equal{} k$, we can rewrite the $ RHS$ as:

$ \frac {A}{12} \equal{} \frac {k^2 \plus{} 1}{k^2 \plus{} k \plus{} 1}$

Clearing the denominator we end up with:

$ k^2(A\minus{}12)\plus{}kA\plus{}(A\minus{}12)\equal{}0$

This is a quadratic equation in terms of $ A$. Since our equality is true, it must have at least 1 solution, which means its $ discriminant \geq 0$:

$ (A \minus{} 12)^2 \minus{} 4(A \minus{} 12)A\geq 0$

Solving this we end up getting both upper bound and lower bound of $ A$. The rest of checking equality is all heavy computation :P

Using this method, we also can find both maximum and minimum value of $ A \equal{} ax^2 \plus{} by^2$ given that $ a,b$ are constants and $ x,y$ satisfies the following equation: $ cx^2 \plus{} dxy \plus{} ey^2 \equal{} f$, with $ c,d,e,f$ are all constants

Anyone wants to try this problem by Larange Multiplier? :P [/hide]

[quote="Mewto55555"][quote="ghjk"]
Anyone wants to try this problem by Larange Multiplier? :P[/quote]

Already did :P[/quote]

I meant my general problem :wink:


Let $ x,y,z$ positive real numbers such that $ x^{2}+y^{2}+z^{2}=1$

Prove that

$ xyz+\sqrt{\sum x^{2}y^{2}}\ge \frac{4}{3}\sqrt{xyz(x+y+z)}$

Hi manlio,
My solution is to homogenize directly from the original form that I posted there:
If $ a,b$ and $ c$ are positive reals such that $ a^{-2}+b^{-2}+c^{-2}=1$, then
$ 1+\sqrt{a^{2}+b^{2}+c^{2}}\ge\frac{4}{3}\sqrt{ab+bc+ca}$
Homogenizing:
$ 1+\sqrt{(a^{2}+b^{2}+c^{2})(a^{-2}+b^{-2}+c^{-2})}\ge\frac{4}{3}\sqrt{(ab+bc+ca)(a^{-2}+b^{-2}+c^{-2})}$
Without loss of generality, assume $ a+b+c=1$. Note that
$ \sum_{cycl}\left(\frac{2}{3}-a\right)^{2}= a^{2}+b^{2}+c^{2}$
By Cauchy-Schwarz inequality,
$ \sqrt{(a^{2}+b^{2}+c^{2})(a^{-2}+b^{-2}+c^{-2})}= \sqrt{\sum_{cycl}\left(\frac{2}{3}-a\right)^{2}\sum_{cycl}a^{-2}}= \sum_{cycl}\frac{1}{a}\left(\frac{2}{3}-a\right)$
$ = \frac{2}{3}(a+b+c)\left( \frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)-3 = \frac{2}{3}\sum_{cycl}\frac{b+c}{a}-1$
It remains to show that
$ \sum_{cycl}\frac{b+c}{a}\ge 2 \sqrt{(ab+bc+ca)(a^{-2}+b^{-2}+c^{-2})}$
which is not hard !!

But pvthuan told me that this problem is trivial. So let's ask him for his nice solution  :wink: .

Thank you very much, Ductrung.

Pvthuan, can you please post your nice solution?


But the ineuality given by you is easy argady.
Assume $xyz=1$ and use AM-GM inequality.

If $xyz=1$, it is easy too by using Schur and AM-GM :?

It is a famous (and old) inequality :
$a^{2}+b^{2}+c^{2}+3\sqrt[3]{a^{2}b^{2}c^{2}}\ge 2(ab+bc+ca)$
Using Schur.


Hello everyone,

I am having trouble on this following problem and I would appreciate some more of your help.

Thank you again!

---

1. The even integers are grouped as follows:

$ (2, 4), (6, 8, 10, 12), (14, 16, 18, 20, 22, 24), ...$

Find the value of the third term of the sequence contained in the 40th set of brackets.

---

I realise that in bracket 40, there would be (40 x 2) 80 terms.

[hide]yes there woule be 80 terms but if there are eight terms how many terms before that have there been and that would just be $ 2 \plus{} 4 \plus{} 6 \plus{} 8 \plus{}...80$ we can factor out a 2 so we have $ 1 \plus{} 2 \plus{} 3 \plus{} 4 \plus{}...40$(2) so then we have 40(41) which means that there have been 1640 numbers after the 40 group and in the 40th group we have 80 so 1640 - 80 = 1560 so the number we are looking for is the 1563 even number and that is 3126.[/hide]

Please correct if im wrong!

Hello there,

Thanks for your reply, but I don't completely understand your steps. Do you mind explaining it at a slower pace?

anther way of doing it  :)
{2,4}{6,8,10,12}....
there are 2#s in the first one,4#s in the second one, 6#s in the third one, so we can write it like this
2,4,6,8,....80 (80 #s in the 40th set)
so the sum will give us how many # are there in all the first 40 sets
(2+80)40/2=1640
the last # in the 40th set is the 1640th # in all the sets
the 3rd # in the 40th set will be (1640-77)th=1563
2+(1563-1)*2 =3126


function
calculus
inequalities
limit
derivative
real analysis
real analysis unsolved

Let $ x,y,z$ be positive real numbers, then the minimum value of $ x^z\plus{}y^z\minus{}(xy)^{z/4}$ is reached when $ x\equal{}2$, $ y\equal{}2$, and $ z\equal{}...$

hello, with the help of calculus i have got
$ x^z\plus{}y^z\minus{}(xy)^{z/4}\geq\minus{}1/8$.
Sonnhard.

[quote="Dr Sonnhard Graubner"]hello, with the help of calculus i have got
$ x^z \plus{} y^z \minus{} (xy)^{z/4}\geq \minus{} 1/8$.
Sonnhard.[/quote]

uhmm... i dont know what you mean...
does that inequality hold $ \forall x,y,z \in \mathbb{R^\plus{}}$ or
just hold for $ x\equal{}2, y\equal{}2$ and $ \forall z\in \mathbb{R^\plus{}}$ ?
but the question is the value of $ z$, isn't it?

It's well possible to reduce the number of variables: $ x^z = a^2$, $ y^z = b^2$, then we are to minimize $ a + b - \sqrt [4]{a,b}$. Further, by Cauchy, ${ a + b - \sqrt [4]{a,b}\ge a + b - \sqrt {(a + b)/2}}$ with equality iff $ a = b$. Now let $ t^2 = a + b$, so we are to minimize $ t^2 - t/\sqrt{2}$. Here the minimal value $ m$ is attained for $ t = t_m$ (you're able to find that). And it is attained iff $ a = b = \sqrt {t_m}/2$. Of course, there are infinitely many possibilities for $ x,y,z$. But I should say that $ x=y=2$ is impossible, if $ z$ has to be positive.

[hide="Now check my answer..."]
look $ x^z \plus{} y^z \minus{} (xy)^{z/4}$ as the function of $ z$, say $ g(z)$.
so the function reaches the extreme value when $ g' \equal{} 0$.
$ g'(z) \equal{} x^z \plus{} y^z \minus{} (1/4)(xy)^{z/4} \equal{} 0$, substitute $ x \equal{} 2, y \equal{} 2$
we have $ g'(z) \equal{} 2^{z \plus{} 1} \minus{} 2^{\frac {z}{2} \minus{} 2} \equal{} 0 \Leftrightarrow z \equal{} \minus{} 6$
$ \forall z > \minus{} 6, g'(z) > 0$ means the function is increasing.
So, the positive real number $ z$ satisfying is $ \lim_{z\to 0^\plus{}}z$

Please give your comment :D
[/hide]

You have big problems with differentiation, that is my comment.

hello, i have found by a numerical way
$ {x = 0.0942260227580227622, y = 0.0942260227580197785, z = 1.17380169448037218}$
inserting this in our term gives $ -\frac{1}{8}$, this should be the minimum.
Sonnhard.

This is not a good solution, Dr Sonnhard Graubner.

If you check what I've written (in fact there is a mistake: it should read $ a\equal{}x^z$ and $ b\equal{}y^z$ and $ a\equal{}b\equal{}t_m^2/2$ later), you'll be able to find a solution explicitly.

As I've written, there are infinitely many possibilities, and $ x\equal{}y\equal{}1/2, z\equal{}4$ is one of them.


calculus
integration
number theory
least common multiple
greatest common divisor
LaTeX
function

1) Prove that if an integer is of the form 6K + 5, then it is necessarily of the form 3K - 1 but not conversely.

2) Prove that square of any integer is of the form 3K or 3K + 1 but not of the form 3K +2.

3) Find positive integers a & b satisfying (a,b) = 10 and [a,b] = 100 simultaneously. Find all solutions.

 :)

hey wat spam??? :wallbash:

@slogger kindly explain ur first answer..... ]

Thanks

welcome to AoPS :)

I suppose 1st 2 are simple cases of Euclid's division Lemma(included in syllabus from our batch)

about 1,slogger just used $ n \minus{} 1$ to make things neat
you could have used $ k \equal{} n \minus{} 1$ also and then prove  :)

[hide="My attempt to 3"]
let $ a \equal{} 10m$ and $ b \equal{} 10n$ such that $ (m,n) \equal{} 1$

then $ [a,b] \equal{} 100 \equal{} 10mn$
so,$ mn \equal{} 10$
there are only two positive integral pairs with product 10
Case 1:$ (m,n) \equal{} (1,10)$
then $ (a,b) \equal{} (10\cdot1,10\cdot10)$
so,$ (a,b) \equal{} (10,10)$

Case 2:$ (m,n) \equal{} (2,5)$
then,$ (a,b) \equal{} (10\cdot2,10\cdot5)$
so,$ (a,b) \equal{} (20,50)$

similarily,we can take $ (m,n) \equal{} (10,1) or (5,2)$ and find remaining pairs to be $ (100,10)$and $ (50,20)$

so the solutions are $ (10,100),(100,10),(20,50)and(50,20)$  :P
[/hide]

@thanks skand... it was really useful.....

Here are some more.....

1) Prove that (a, b, c) = ((a, b), c).

2) Let b and g > 0 be given integers. Prove that the equations
   (x, y) = g and xy = b can be solved simultaneously if and only if
   g^2\b.

i suppose you should thanks slogger as he gave answers to 2 of your question

i'll try  :)
and no need for thanks,just help me when i am struggling  :wink:

well i guess he meant class 10 syllabi.. the chapter real number has been introdued from their batch.....

so no need to worry :lol:

[quote="Da Vinci"][quote="skand"]welcome to AoPS :)

I suppose 1st 2 are simple cases of Euclid's division Lemma([b]included in syllabus from our batch[/b])

[/quote]

what?  :o
Can you tell me where i can view the JEE 2009 syllabus?  :maybe: [/quote][/quote]

What da? Euclid's lemma is there in our 3rd stanard syllabus itself, only without the proof and in a form that is not mathematically rigorous!

and now it has been included in syllabus of class 10 in formal way with mathematical rigority and all that  :)
(I hated this chapter till i came to know how much it is helping)
and problem 2 was even in ncert mathematics book :)

[hide="my attempt"]
1)just take $ (a,b,c)\equal{}(dx,dy,dz)$ where d is H.C.F. of a,b,c and (x,y,z) are co-primes
Then the proof is straightforward

I am afraid it may be wrong(i am not too sure about my methodology)

2)g^2\b i couldn't understand :( [/hide]

Thanks a lot to all of you.....
I will keep posting my problems here :blush:

1) Let $ n \ge 2$ and $ k$ be any positive integers. Prove that $ (n\minus{}1)^ {2} divides (n^{k} \minus{} 1)$ if and only if  $ \frac {k}{(n \minus{} 1)}$

Prove that product of four consecutive numbers is divisible by 8.

Find 3 integers which are relatively prime but not relatively prime in pairs

whats up??  :(  no one posting any solution???

take this one

Prove that every positive integer is uniquely expressible in the form
$ 2^{j_{0}} \plus{} 2^{j_{1}} \plus{} 2^{j_{2}} \plus{} 2^{j_{3}} \plus{}...............\plus{} 2^{j_{m}}$
where $ m\ge0$ and $ 0 \le j_{0} < j_{1} < j_{2} < j_{3} ............. < j_{m}$ :o

Prove that any positive integer $ a$ can be uniquely expressed in form
$ a \plus{} 3^{m} \plus{} b_{m\minus{}1}3^{m\minus{}1} \plus{} b_{m\minus{}2}3^{m\minus{}2} \plus{}......................\plus{} b_{0}$

where each $ b_{i} \equal{} 0, 1 , or \minus{}1$

hey mate,you are bowling at too fast pace
calm down  :D

for multiple of consec. nos.please use Euclid's lemma

about the form $ 2^x$ or something
all I can say that every no. can be expressed as sum of $ 2^0$  :rotfl:

but i guess according to the problem we can use one power only once....... :wink:

[quote="anurags92"]Thanks a lot to all of you.....
I will keep posting my problems here :blush:

1) Let $ n \ge 2$ and $ k$ be any positive integers. Prove that $ (n \minus{} 1)^ {2} divides (n^{k} \minus{} 1)$ if and only if  $ \frac {k}{(n \minus{} 1)}$[/quote]

That does not make sense.

[quote="anurags92"]

Find 3 integers which are relatively prime but not relatively prime in pairs[/quote]

$ ab,bc,ca$, where $ (a,b)\equal{}(b,c)\equal{}(c,a)\equal{}1$

[quote="anurags92"]Prove that any positive integer $ a$ can be uniquely expressed in form
$ a \plus{} 3^{m} \plus{} b_{m \minus{} 1}3^{m \minus{} 1} \plus{} b_{m \minus{} 2}3^{m \minus{} 2} \plus{} ...................... \plus{} b_{0}$

where each $ b_{i} \equal{} 0, 1 , or \minus{} 1$[/quote]

...

[quote="anurags92"]whats up??  :(  no one posting any solution???

take this one

Prove that every positive integer is uniquely expressible in the form
$ 2^{j_{0}} \plus{} 2^{j_{1}} \plus{} 2^{j_{2}} \plus{} 2^{j_{3}} \plus{} ............... \plus{} 2^{j_{m}}$
where $ m\ge0$ and $ 0 \le j_{0} < j_{1} < j_{2} < j_{3} ............. < j_{m}$ :o[/quote]

Both these problems are direct applications of Euclid's lemma. Why dont you try them yourself and then ask if you are still not able to get them. :)

thanks ..... hell ever.....

it is that n-1 divides k

Here's my solution for the g^2/b sum[hide]
We know that the product of the two numbers is equal to the product of lcm and gcd. Denote the gcd by g and the product of the two numbers by b. Obviously the gcd of two numbers divides it's lcm  :blush: . So let lcm be some g*u. Now we get g^2*u=b. Which implies that g^2 divides b. I think the sum which involved the writing of a number as sum of powers of 2 and 3 can be done with a slight variation of the binomial theorem[/hide]

The problem was to prove that (n-1)^2 divides n^(k-1) if and only if n-1 divides k-1.[hide]
Here's my solution. (n-1)^2 divides n^(k-1) implies that n-1 divides 1+n+n^2.........+n^k-2 . Now remainder when 1 is divided by n-1 is 1. Similarly n is divided by n-1 is 1. Going on (n-1)^2 = n^2 +1 -2n which is n^2 is (n-1)^2 -1+2n which is (n-1)^2 +2(n-1) + 1 which implies the remainder when n^2 is divided by n-1 is also 1. Similarly by recursion the remainder when each of 1+n+n^2.........+n^k-2 is divided by n-1 is 1. So total remainder is k-1 and this remainder is zero only if n-1 divides k-1. I think some part of the question i copied wrong. :rotfl:  :blush:  :lol:  :D  :P  :oops: [/hide]

please use latex if u can it makes solution look clean n neat.... :wink:

dude i will try to rewrite the solution in latex but i want to know whether my answers are correct. :D
[hide] i am sure my solution for the second problem is surely wrong. :rotfl: [/hide]

didnt check for the solution but ur english is certainly wrong....  :rotfl:

hey wat clealy? :rotfl:  :rotfl:   :P

and solve this
Let f(x) be a function defined on the positive integers such that:

$ f(x)  \equal{}  x/2$if x is even
$ f(x) \equal{} (3*x\plus{}1)/2$if x is odd

Then the conjecture is: iterates of f(x) will eventually reach 1 for any
initial value of x

im talking of pure number theory only. dont involve functions here and next you challenge a question first have the capability of doing that yourself

ofcourse i can give you a solution of this problem!
but i think that it was your childishness and impatiency which let ME give you a question.so i think u should solve it first.

but if you don't have the guts of solving it or are in a grading institution that just offers its students dumb excuses to neglect the challenges coming in their way then its not my problem.

then you don't deserve to be offered a question analytic :wink:

first of all, i did not mean that u didnt hav the capability of doing that question , i just meant that when u gave the next question u dont give a question thats massively difficult for just proving that i cant solve things.but u took it too strongly and that i was insulting you but i didnt mean that in any way.

secondly it was not my childishness,it was yours , because it was it was your decision all the way to give that question cause you r just too reactive.

thirdly, never and never challenge or doubt anybody's abilities. and i was not giving any dumb excuses but still if u think i hav hurt the sentiments of any i hav edited my post
and remember to always respond things in a proactive way. giving such responses leads to tensions.i was always just taking in a light way but you i dunno y took it to your heart.

and last stop this altogether coz its going off topic . lets conc. on divisibility

The post you have edited is the height of your childishness.
because you should have remembered that it was you who had obnoxiously stated to be given a question,and secondly you called Indian's poor at number theory,without realising the fact that you might be considering other indians like yourself.

I never doubt on your skills or your talent but your irrelevant and impertinent questions seem to degrade your so called "TALENT".

I think you should end this topic here only and provide me with the solution of my problem or try solving some other problems in AoPS.

AND PLEASE REMEMBER THINK A THOUSAND TIMES BEFORE COMMENTING ON INDIANS

i dont need to argue with you as its just disturbing me. its better to ignore your posts.
@anurag: i read some of your questions and as far as i remember you posted them in the number theory section also and they have been answered and we have discussed few of them there too .
           moreover, it is better to post in other forums like high school and olympiad section on the basis of their difficulty as those forums are more active as more people come there and secondly there are very experienced solvers there , so u get a big scop of learning things there.

@rt_08
Chill buddy.... why are you getting angry......
analytic is a cool person...  He is a genius and answered many problems in the other section dedicated to number theory...... I dont know him personally but i am sure he is a imotc guy or at least has cleared rmo.....(of course if he has written these)... And something about commenting on indians.... He is himself an Indian... heheheh.... So be cool man and post questions which you want to understand... Dont challenge someone......

@analytic
I think romil misunderstood ur friendly gesture and your no nonsense attitude....
Kindly be yourself and enlighten us with your solutions....

@all forget this and carry on with the problems and their solutions....

(Saintly)

[quote="rt_08"]Let f(x) be a function defined on the positive integers such that:

$ f(x) \equal{} x/2$if x is even
$ f(x) \equal{} (3*x \plus{} 1)/2$if x is odd

Then the conjecture is: iterates of f(x) will eventually reach 1 for any
initial value of x[/quote]
[quote="rt_08"]ofcourse i can give you a solution of this problem! [/quote][size=150]
CONGRATULATIONS[/size]! [color=red]PLEASE POST IT HERE[/color]!

[quote="[url=http://en.wikipedia.org/wiki/Collatz_conjecture]Wikipedia[/url]"]The Collatz conjecture is an unsolved conjecture in mathematics.[/quote](click on blue lettered "Wikipedia" for further details)

We need to open a new topic on the "Round table". Let the mathematical community acknowledge the ground-breaking achievement.

thats why i gave the question to the over smart analytic :rotfl:  :rotfl:  :rotfl:

Cool down you two.

[quote="rt_08"]thats why i gave the question [...][/quote]
If I thought analytic was over-smart, and wanted to protest, I would *never* have done that.

It's a silly misunderstanding.
I don't see analytic acting over-smart in this thread.

[quote="analytic"]well, problems are too crowded here.can anyone give the unsolved problems till now in this topic as it has got too messy in reading.[/quote]
Most probably (like most others like me) he asked for that because he wanted problems for his own practice, that doesn't mean, the ppl responding here were stupid and he was a lot smarter (probably what rt interpreted)

And I don't know about the Indian forum, but in every other main-stream forums of the site, mods always recommend to post separate topics for separate problems, or at least the problems in the first post of the topic.

[quote="analytic"]im talking of pure number theory only. dont involve functions here[/quote]
It was pure number theory. It is generally stated in sequence form.

ok lets start with some new problems

p.s.->analytic you start with some of your high-level number theory poblems :rotfl:  :rotfl:  :rotfl:

hey someone delete this topic al together... people here are more interested in makin gothers feel low...
@rt man dont be rude and stuf..... cool down....

[quote="anurags92"]hey someone delete this topic al together... [/quote]
good idea... it should be locked.


THe solution sets of the polynomial equations f(x) = 0  and g(x) = 0 are {0,1,2,3} and {2,3,4} respectively.

 How many different number are in the solution set of the polynomial equation [f(x)][g(x)]=0?

[hide="EDITTED Possibility"]
Just combine the roots.  Yeilds $ 5$.
[/hide]

[quote="AstroPhys"][hide="Possibly?"]
Just combine the roots.  Yeilds $ 4$.
[/hide][/quote]
[hide="Why Not...?"]
$ 5$?

Knowing the solution set is $ \{0, 1, 2, 3, 4\}$ by putting together the roots?

Or did I misinterpret what the problem was asking?[/hide]

:oops: No, that is what I meant.   Just counted wrong.


Let there be $ n \ge 2$ real numbers such that none of them is greater than the arithmetic mean of the other numbers. Prove that all the numbers are equal.


category theory
abstract algebra
superior algebra
superior algebra theorems

If F,G are adjoint functors and F is exact ==> G is exact?

Do you mean, that $F$ is left-adjoint to $G$? In this case, the answer is No.

Consider the functors $F = M \otimes -$, $G = \hom(M,-)$ in $R$-Mod. Then $F$ is left-adjoint to $G$, $F$ is exact iff $M$ is flat, and $G$ is exact iff $M$ is projective. Every projective module $M$ is flat, so that $G$ exact implies $F$ exact. But the converse is not true: Take $R=\mathbb{Z}$ and $M=\mathbb{Q}$.

and does G exact ==> F exact hold in general?

No.

If $S$ is a $R$-algebra, then the functor $F : S \otimes_R - : R-\text{Mod} \to S-\text{Mod}$ is left-adjoint to the forgetful functor $U : S-\text{Mod} \to R-\text{Mod}$. Clearly, $U$ is exact, and $F$ is exact iff $U \circ F$ is exact, i.e. iff $S$ is flat over $R$. So, we only have to mention a $R$-algebra $S$ which is not flat over $R$. For $R= \mathbb{Z}$, flatness is equivalent to torsion-freeness, so that we can take $S = \mathbb{Z}/2$.

By the way, left-adjoint functors (between abelian categories) are always right-exact, and dually right-adjoint functors are always left-exact. We've seen that exactness cannot be traced back to the adjoint functor.


I already sended in my SIII form but

I don't think my school did...

Could we still participate even though that TII thing is not recevied by the AMC by the due date?

Yes, but we need the information on the T-II form, or we wouldn't have created it.  We use it to make sure your teacher, contest manager, or proctor has all the necessary information.   We also use  it to verify from whom we receive the faxed pages.    We also use it for reporting back scores and information.

The obvious thing to do is for you to talk to your Contest Manager as soon as possible.
Steve Dunbar
AMC Director

thanks ..Mr.Dunbar

I just asked my teacher
It just turned out that he had faxed it, but probably the AMC site isn't updated yet.

Anyway Thanks!


What's the difference between regular Firefox and "Safe Mode". What makes it so safe? Should I use regular mode or Safe Mode?  :?  :?  :?

i was wondering that to although i just chose regular

"When started in Safe Mode all extensions are disabled and the Default theme is used."

From the Firefox FAQ.

So if your firefox was crashing all the time by some reason you maybe do not know, then you can use the safe mode to run it and hopefully it won't crash.


[geogebra]3964a54e694f4b6be948930776fd2cd3da7d6022[/geogebra]

In the Circle with center Q, radii QD and QC form a right angle. The two shaded regions(in the diagram, the two semicircles) are tangent semicircles as shown. The radius of the circle with center Q is 14 inches. What is the radius of the smaller semicircle. Express your answer as a common fraction.

Originally, instead of point D, it was point A, and instead of point C, it was point B, but I'm bad with geogebra  :D

why is it that when ever i press "view applet", a red x appears in the top left corner?

Same with mine. If I remember the question correctly though, connect the centers of the semicircles and let the radius you want be r. Then use the Pythagorean Theorem.

[hide="Solution"]
Most of the diagram is self-explanatory; lengths denote distances between the two closest points.

So now we have $ 7^2 \plus{} (14 \minus{} r)^2 \equal{} (7 \plus{} r)^2\implies7^2 \plus{} 14^2 \minus{} 28r \plus{} r^2 \equal{} 7^2 \plus{} 14r \plus{} r^2$

$ \implies14^2 \equal{} 42r\implies r \equal{} \frac {14^2}{42} \equal{} \boxed{\frac {14}3}$.
[/hide]

Thank you math154!

If you can't see the diagram, it's because you don't have the newest version of java installed I think :)


Find the greatest constant k such that the following inequality holds for any positive real numbers a,b,c
$ a^4\plus{}b^4\plus{}c^4\plus{}k(ab\plus{}bc\plus{}ca)^2\ge (1\plus{}3k)(a^3b\plus{}b^3c\plus{}c^3a)$

[quote="lenhamquy"]Find the greatest constant k such that the following inequality holds for any positive real numbers a,b,c
$ a^4 \plus{} b^4 \plus{} c^4 \plus{} k(ab \plus{} bc \plus{} ca)^2\ge (1 \plus{} 3k)(a^3b \plus{} b^3c \plus{} c^3a)$[/quote]
The result is $ \frac{1}{3}$,It is in hungkhtn's bok

[quote="Allnames"][quote="lenhamquy"]Find the greatest constant k such that the following inequality holds for any positive real numbers a,b,c
$ a^4 \plus{} b^4 \plus{} c^4 \plus{} k(ab \plus{} bc \plus{} ca)^2\ge (1 \plus{} 3k)(a^3b \plus{} b^3c \plus{} c^3a)$[/quote]
The result is $ \frac {1}{3}$,It is in hungkhtn's bok[/quote]
I don't like slution in hungkhtn's book. i want to see a another slution
what is your sution?


[quote="bleumoose"]i just realized, going through the alabama forum, that EVERYONE here is from vestavia. is that true? or are there any non-vestavia kids that post?[/quote]

Fan has been on here some.  JHu is from Grissom.  Hopefully the community will gain broader participation as time goes on.  Get your ASFA classmates to join in!

i really should. esp the younger ones. hm. glad fan's been on some at least.

We'll let some ASFA kids here when you let some Vestavians in UAB. Just kidding... we need more people here from anywhere. I can't imagine there are a lot of ASFA math team people I didn't meet this summer, so maybe you are one of them? Who are you?

you mean you guys cant dual-enroll @uab? or what?

i havent really been active in math team over the last year, owing to outside stuff (i.e.- work). although in 9th/10th i was the only white girl on our team, and the only one to consistently place in algebra II(booya, chinese people!). *laughs* ok, that was bad racial profiling, its kind of a joke at our school, though. i try and keep up with the math but, you might have noticed this, a lot of math teamers get major burnout after 10th grade. i have been throwing myself into classroom math a lot more, though, just for kicks. so are you on the vestavia dream team?

[quote="bleumoose"]so are you on the vestavia dream team?[/quote]

?

^ Yeah, what? Is that the team in general or a specific group?

[quote="bleumoose"]you mean you guys cant dual-enroll @uab? or what?[/quote]

Well I've never heard of the opportunity from Vestavia. I only know one person who is because his dad is a professor. I don't know if he's getting any school credit for it. I also met a lot of people who were in camps, but I don't know anyone else from Vestavia who did one.

So I didn't literally mean they don't let us in, but we just don't have anyone there.

i meant in general.

ah. yeah, its a little easier for us, i guess, since we are [i]required[/i] to have the equivalent of two semester courses to graduate. although you have to be kind of careful, i got stuck in something i reeeeeeeeeeeeeeally didnt want to do, that was stinky (transportation engineering, woop woop). i am pretty sure any high schooler can go down to uab and dual enroll, with the right credits. im taking genetics this summer... you can only take one course per grading period, thats the one bad thing.

Hey, I'm not a Vestavia kid either (though they sort-of adopted me at times).

Austin High School.  Guessing you don't see them at math contests much anymore.

:lol: i knew that.  :blush: i sometimes cite you as the "see, look, you dont have to go to this super-expensive prep school to make something of yourself" example. ^_^

and you dont technically qualify as a "kid" at any rate, do you?... hey. how come you never give lectures in alabama like mr. crawford does? or did i miss them all?

I don't return to Alabama as much as MCrawford does.  I only get back there every other year or so, and I don't get down to B'ham.

And I still look young.  ish.

That's cool that your from alabama mr. rusczyk. I never would have guessed.

I'm not from Vesavtia. I'm from Mobile. We have a Math Circle run by the professors at the University of South Alabama.

wow... so r there like any MIDDLE SKOOLERS HERE??!

hey watch your yap

you're a young grasshopper too RYAN!  :rotfl:

I'm a teenager, so are you. What's the difference? Teen is a teen.

What flimsy reasoning was that?

not Long haired  Talkative Seva Tchernov

i think ryan's just bitter because he's old.
lol randolpher?
well cool tell mr. thorn i said hi.

I dont think people in alabama are interested in math

[quote="mathman440"]the more important questions is how do we get other kids from Alabama to be interested in mathematics?[/quote]

Somehow combine it with god and football.

im not from vestavia, but then again, im not from alabama.  :|

but i was at vestavia and hoover, if that counts for anything...


Its are known in the ABC triangle  AD median.\angle ADB=45°,\angle ACB=30° Find, \angle ABC . :?:

[quote="Pirkuliyev Rovsen"]Its are known in the $ ABC$ triangle  $ AD$ median.$ \angle ADB\equal{}45^\circ,\angle ACB\equal{}30^\circ$ Find, $ \angle ABC .$ :?:[/quote]

Let $ \angle ABC$ be $ \theta$
Apply m-n theorem
$ 2cot 45^\circ \equal{} cot 30^\circ\minus{}cot \theta$
$ cot \theta\equal{}\sqrt{3}\minus{}2\implies \theta\equal{}105^\circ$


Calculati:
 $\int_{0}^{u}\frac{arctgx}{\(x^{2}+ux+2}\,dx$, unde u este in $(0,2\sqrt{2})$ :)

O estimare, foarte posibil sa ma insel: $\frac{arctg^{2}u}{4}$???

Am facut mai intai prin parti, apoi o schimbare de variabila si inca o data prin parti. Iata ce mi-a iesit:
\[\frac{2}{\sqrt{8-u^{2}}}\cdot\arctan{u}\cdot\arctan{\frac{3u}{\sqrt{8-u^{2}}}}-\frac{2}{8-u^{2}}\cdot\Bigg[\Bigg(\arctan\frac{3u}{\sqrt{8-u^{2}}}\Bigg)^{2}-\Bigg(\arctan\frac{u}{\sqrt{8-u^{2}}}\Bigg)^{2}\Bigg]. \]

Nu stiu ce sa zic....raspunsul tau Damian pare asemanator cu cel corect...din pacate nu mi-au placut niciodata calculele cu fct trig inverse asa ca nu o sa vad dak coincid..sper sa am timp de solutie mai incolo :lol:


Galois Theory
superior algebra
superior algebra unsolved

Let $ K$ be a finite separable extension of a field $ k$, of prime degree $ p$. Let $ \theta \in K$ be such that $ K\equal{}k(\theta)$, and let $ \theta_1,...,\theta_p$ be the conjugates of $ \theta$ over $ k$ in some algebraic closure. Let $ \theta\equal{}\theta_1$. If $ \theta_2 \in k(\theta)$, show that $ K$ is Galois and in fact cyclic over $ k$

Well, say that there are exactly $ m$ conjugates of $ \theta$ contained in $ k(\theta)$.  How many conjugates are contained in some other $ k(\theta')$?  Is there any conjugate contained in more than one such field?


Hey everyone, im new here, so please bare with me.

While trying to figure out my HW problem on Google, i found this website.  So im wondering if anyone can help me out.
 Heres the Question...
"Find All Solutions between 0 and 2pi to the following equations:
A)5sec(x) - 3cos(x) = 2
B)3Tan^2(x) - 2tan(x)-8=0
.....
Someone Please Help me!!!
At least explain what the heck i have to do.....

Thanks
Andrew

hint for A:

$\frac{5}{\cos(x)} + 3\cos(x) = 2$
$\Rightarrow$ solve for $\cos(x)=y$
$\frac{5}{y} + 3y = 2$
 Then resubstitute your solution in $\cos(x)=y$ and solve for $x$

hint for B:

The same thing with $y=\tan(x)$

I don't think something like this belongs in pre Olympiad.

When you are done substituting in for y, always remember that your answer lies within the domain.  (There are an infinite number of solutions to the equation, but only a few satisfy the domain restriction)

This guys getting pre-olympiad homework problems???!!!

   WTF!!??!!  Where do you go to school???!!!

 :lol:

Well I guess anything easier than olympiad difficulty would be considered "pre-olympiad"


How many squares are on an 8 x 8 checker board?

[hide="Answer"]$64$ and yes I am a 4th grader[/hide]

wow, a 9 year old version of the terminator

64 squares is a good start. But there are more, of larger sizes …

When I first saw it, I thought, duh, 8*8 is 64, but then there are 2 by 2 squares, 3 by 3 squares, etc..

[quote="vishalarul"]What about squares of size $\sqrt{2}\times \sqrt{2}$, $\sqrt{5}\times \sqrt{5}$, etc...?[/quote]There really aren't any of them because there are no lines that go through there. :wink:. Also, don't try to revive year and a half old topics

[quote="Quevvy"][quote="i_like_pie"][quote="Barnacle"]how about how many rectangles?  how many non-square rectangles?[/quote]
I'm not sure about non-square rectangles, but in an $m\times n$ grid, there are $\binom{m}{2}\binom{n}{2}$ rectangles.[/quote]
Then you'd just plug it in.
$\binom{8}{2}\binom{8}{2}-204=28^{2}-204=784-204=\boxed{580}$[/quote]
I think it should be ${m+1\choose2}{n+1\choose2}$, which gives ${9\choose2}{9\choose2}=1296$. Think about it, for a 1x1 square, ${1\choose 2}{1\choose2}$ rectangles make no sense.

[quote="pianoforte"]
I think it should be ${m+1\choose2}{n+1\choose2}$, which gives ${9\choose2}{9\choose2}=1296$. Think about it, for a 1x1 square, ${1\choose 2}{1\choose2}$ rectangles make no sense.[/quote]
You are right,
[hide]${9\choose2}{9\choose2}-204=1296-204=1092$[/hide]

[hide="And the formula works becuase..."] You must choose 2 horizontal lines and 2 vertical lines to define a unique rectangle. Choosing 2 lines from the $m+1$ horizontal lines gives $\binom{m+1}{2}$. Same for $n+1$ vertical lines. [/hide]

[quote="tridecalogism_lover"]you guys are making it really complicated for a fourth grader!  :rotfl:[/quote]
That's because it was an extention and usually extentions are a couple times harder.

[quote="Quevvy"][quote="ProtestanT"]theres nothing wrong with reviving old topics[/quote]
Especially if you give an important formula or something.
Also, it might end a debate that was left in the middle...[/quote]
But usually you shouldn't unless it's to say something really really important.

ProtestanT, did you ever hear the phrase "Let lying dogs lie."?


inequalities
AMC
USA(J)MO
USAMO
trigonometry
Cauchy Inequality
inequalities proposed

I have created some nice problem. One of them which I like best is
Given $ A, B, C$ are three angles of a triangle satisfy: $ 5cosA \plus{} 6cosB \plus{} 7cosC \equal{} 9$
Prove that: $ sin^2\frac {A}{2} \plus{} sin^3\frac {B}{2} \plus{} sin^4\frac {C}{2} \geq\ \frac {7}{16}$
Althought the coefficients is diffirent but equality occurs when $ A \equal{} B \equal{} C \equal{} 60^0$. Is it nice ?
 :)

Three of my proposals for top 10.
[hide="Problem nr.1"]Let $ a,b,c\geq 0$, s.t. $ a\plus{}b\plus{}c\equal{}1$.Prove that:
$ \sqrt{a\plus{}\frac{(b\minus{}c)^2}{4}}\plus{}\sqrt{b}\plus{}\sqrt{c}\leq \sqrt{3}$ Chinese Girls MO 2007[/hide]
[hide="Problem nr.2"]Let $ a,b,c>0$.Prove that:
$ \sqrt{\frac{a}{4a\plus{}5b}}\plus{}\sqrt{\frac{b}{4b\plus{}5c}}\plus{}\sqrt{\frac{c}{4c\plus{}5a}}\leq 1$[/hide]
[hide="Problem nr.3"]Let $ a,b,c>0$.Prove that:
$ \frac{a}{\sqrt{a\plus{}b}}\plus{}\frac{b}{\sqrt{b\plus{}c}}\plus{}\frac{c}{\sqrt{c\plus{}a}}\leq \sqrt[4]{\frac{27}{4}(a^2\plus{}b^2\plus{}c^2)}$[/hide]
I would say that [hide="this problem"]Let $ a,b,c>0$.Prove that:
$ \sqrt{\frac{a}{8b\plus{}c}}\plus{}\sqrt{\frac{b}{8c\plus{}a}}\plus{}\sqrt{\frac{c}{8a\plus{}b}}\geq 1$[/hide] also deserves to be on top 10, but I don't know it's solution, so I can't say that it is nice or ugly....

[quote="Inequalities Master"]
$ \sqrt {\frac {a}{8b \plus{} c}} \plus{} \sqrt {\frac {b}{8c \plus{} a}} \plus{} \sqrt {\frac {c}{8a \plus{} b}}\geq 1$ also deserves to be on top 10, but I don't know it's solution, so I can't say that it is nice or ugly....[/quote]
See here: http://www.mathlinks.ro/viewtopic.php?t=194955 :)

In which post, can_hang2007?
Or do you want to say that:
$ \sqrt{\frac{a}{b\plus{}3c}}\plus{}\sqrt{\frac{b}{c\plus{}3a}}\plus{}\sqrt{\frac{c}{a\plus{}3b}}\geq \frac{3}{2}$ and
$ \sqrt {\frac {a}{8b \plus{} c}} \plus{} \sqrt {\frac {b}{8c \plus{} a}} \plus{} \sqrt {\frac {c}{8a \plus{} b}}\geq 1$

are same? :?:

[quote="Inequalities Master"]In which post, can_hang2007?
Or do you want to say that:
$ \sqrt {\frac {a}{b \plus{} 3c}} \plus{} \sqrt {\frac {b}{c \plus{} 3a}} \plus{} \sqrt {\frac {c}{a \plus{} 3b}}\geq \frac {3}{2}$ and
$ \sqrt {\frac {a}{8b \plus{} c}} \plus{} \sqrt {\frac {b}{8c \plus{} a}} \plus{} \sqrt {\frac {c}{8a \plus{} b}}\geq 1$

are same? :?:[/quote]
Dear friend,
It comes from my inequality
\[ \sum a\sqrt {8b^2 \plus{} c^2} \le 1
\]
with $ a,b,c \ge 0,a \plus{} b \plus{} c \equal{} 1$.
Since if we denote $ a \equal{} x^2,b \equal{} y^2,c \equal{} z^2$ $ (x,y,z \ge 0)$.
By Cauchy Schwarz Inequality, we have
\[ \sqrt {\frac {a}{8b \plus{} c}} \plus{} \sqrt {\frac {b}{8c \plus{} a}} \plus{} \sqrt {\frac {c}{8a \plus{} b}} \equal{} \sum \frac {x}{\sqrt {8y^2 \plus{} z^2}} \ge \frac {(x \plus{} y \plus{} z)^2}{\sum x\sqrt {8y^2 \plus{} z^2}}
\]
Hence, it suffices to prove that
\[ (x \plus{} y \plus{} z)^2 \ge \sum x\sqrt {8y^2 \plus{} z^2}
\]
It is proved in that link. :)

Reid Barton's proposal for IMO 2003 has an ingenious solution. See [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=6062]here[/url].

[quote="Inequalities Master"]I would like that everyone should say what are the best inequalities that you have ever seen.[/quote]
I like Jensen, Lagrange's multipliers, and the rearrangement because they are so powerful.

I am curios the opinion of our experts (like arqady, can_hang2007 and Vasc) about this question.....

[quote="Inequalities Master"]I would like that everyone should say what are the best inequalities that you have ever seen.
Of course that these inequalities must be hard, but also should have nice solutions.
I will write my proposals for top-10 later...

I didn't know where to create such a topic, so I created it here.[/quote]
I think the nice inequality, firstly it must be hard and simple in its statement. Also we dont need to use many strong theorem such as EV, AC, SID, ... And finally, it is great if we could find a proof using classial inequalities such as AM-GM, Cauchy schwarz, ...
I like too much inequalities like this. Here is one of them
\[ \sum \sqrt{ \frac{8a^2\plus{}bc}{b^2\plus{}c^2}} \ge \frac{9}{\sqrt{2}}\]
for $ a,b,c \ge 0$. :)

I agree with can.
By the way, I have a nice geometric equality with me, but it will need some inequalitic approach to solve it. So, I am posting it here. It might deserve a place in the top 100, if not in the top 10! :lol:

Let there be two triangles with sides $ a,b,c$ and $ a_{1},b_{1},c_{1}$ respectively. Prove that the triangles are similar iff
$ \sqrt{aa_{1}}\plus{}\sqrt{bb_{1}}\plus{}\sqrt{cc_{1}}\equal{}\sqrt{(a\plus{}b\plus{}c)(a_{1}\plus{}b_{1}\plus{}c_{1})}$

I also agree with you, can_hang2007.
By the way, can_hang2007, where can I see a solution to your inequality?

Do [b]Vasc [/b]and [b]arqady [/b]have anything to say about this subject?

Isn't that just a consequence of Cauchy? the RHS is $ \ge LHS$ by Cauchy, inequality iff $ (a,b,c)$ and $ (a_1,b_1,c_1)$ are linearly dependent (i.e, the triangles are similar). I might be wrong...

(sorry for double posting... I hurried in the previous post :P)

Some of my favorites:

IMO 06\3:
Determine the least M such that $ |ab(a^{2}-b^{2})+bc(b^{2}-c^{2})+ca(c^{2}-a^{2})| \leq M(a^{2}+b^{2}+c^{2})^{2}$.
Someone once told me a geometric interpretation of this gem, by I didn't follow... I will ask him to tell me and will post.

USAMO 98\3:
Let $ a_0,a_1,\cdots ,a_n$ be numbers from the interval $ (0,\pi/2)$ such that:
$ \tan (a_0 - \frac {\pi}{4}) + \tan (a_1 - \frac {\pi}{4}) + \cdots + \tan (a_n - \frac {\pi}{4})\geq n - 1.$
Prove that: $ \tan a_0\tan a_1 \cdots \tan a_n\geq n^{n + 1}.$

$ \sqrt{a^2+c^2}+\sqrt{b^2+d^2} \le \sqrt{(a+b)^2+(c+d)^2}+2\frac{|ad-bc|}{\sqrt{(a+b)^2+(c+d)^2}}$
for positive a,b,c,d

$ (1-p^a)^b+(1-q^b)^a \ge 1$, where $ p+q=1, p,q \ge 0, a,b >0$

If abc=1, a,b,c>0: $ \sum \frac{1}{1+a+b} \le 1$ (because it's so neat!)

And a gem, which possible wasn't posted in the forums yet: find the minimum of $ \frac{(1+3a)(a+8b)(b+9c)(c+8)}{abc}$, where a,b,c are positive reals.

[quote="Allnames"]How about (a hard problem )
Let a,b,c are positive number.Prove that
$ \sum \frac {a^{4}} {a^{3} \plus{} b^{3} \plus{} abc}\geq \frac {a^{3} \plus{} b^{3} \plus{} c^{3}}{a^{2} \plus{} b^{2} \plus{} c^{2}}$ (Can_hang 2007)[/quote]
It is MOLDOVA TST . I know can_hang2007 had a solution for it .
And I think it 's very hard  :)

[quote="Inequalities Master"][b]onlylove_math wrote:[/b]
[quote]It is MOLDOVA TST . I know can_hang2007 had a solution for it .
And I think it 's very hard [/quote]
What year? I haven't found it..
Are you sure?[/quote]
Yes,it is not Moldova TST,it is from Can hang 2007,
But with Moldova TST 2006,we only prove a small case of this problem.Wait a nice proof sice its authour,Can! :lol:

[quote="bambaman"]Isn't that just a consequence of Cauchy? the RHS is $ \geq L.H.S.$ by Cauchy, inequality iff $ (a,b,c)$ and $ (a_{1},b_{1},c_{1})$ are linearly dependent (i.e, the triangles are similar). I might be wrong...[/quote]
No, my friend! You are absolutely right! Thank you!
It might be an easy problem, but I liked its beauty.
 :)

can_hang2007, can you please give me the link for $ \sum \sqrt{ \frac{8a^2\plus{}bc}{b^2\plus{}c^2}} \ge \frac{9}{\sqrt{2}}$ ? :?:

[quote="nguoivn"]
Given $ A, B, C$ are three angles of a triangle satisfy: $ 5cosA \plus{} 6cosB \plus{} 7cosC \equal{} 9$
Prove that: $ sin^2\frac {A}{2} \plus{} sin^3\frac {B}{2} \plus{} sin^4\frac {C}{2} \geq\ \frac {7}{16}$
Althought the coefficients is diffirent but equality occurs when $ A \equal{} B \equal{} C \equal{} 60^0$. Is it nice ?
 :)[/quote]

And my solutuion is here:

Hey spanferkel! That's a very very nice inequality, thank u! :)
I wonder if the following inequality is true as well.... :maybe:
$ a^{\pi a} \plus{} b^{\pi b}\geq a^{\pi b} \plus{} b^{\pi a}$ for $ a,b>0$

[quote="spanferkel"][quote="Inequalities Master"]I would like that everyone should say what are the best inequalities that you have ever seen.[/quote]
I would like to see the following one included:
$ a^{ea} \plus{} b^{eb}\geq a^{eb} \plus{} b^{ea}$ for $ a,b > 0$.
As far as I can see, it is still unsolved, for almost two years now. See http://www.mathlinks.ro/viewtopic.php?p=679964#679964 :)[/quote]
Indeed, this is the best spectacular inequality in my opinion. And $ e$ is the greatest constant such that the inequality holds.

[quote="Vasc"]Indeed, this is the best spectacular inequality in my opinion. And $ e$ is the greatest constant such that the inequality holds.[/quote]
So the inequality does not hold for $ \pi$. But Vasile, how did you get that $ e$ is the largest constant satisfying the ineq.?

Can anyone post the proof for $ e\equal{}2$ please? Should I approach by Jensen?
 :)

[quote="Sunkern_sunflora"][quote="Vasc"]Indeed, this is the best spectacular inequality in my opinion. And $ e$ is the greatest constant such that the inequality holds.[/quote]
So the inequality does not hold for $ \pi$. But Vasile, how did you get that $ e$ is the largest constant satisfying the ineq.?

Can anyone post the proof for $ e \equal{} 2$ please? Should I approach by Jensen?
 :)[/quote]
the thread is longer than you think  :P
http://www.mathlinks.ro/viewtopic.php?p=720392#720392


Hi
I was wondering where can i find the results of the IPhO2005 ?
I found this :[url]http://www.ipho2005.com/[/url]
But the results are not yet up.and what is that means ?
the contest was held at 3-12 july but results are not yet up :? ?

The results are up, finally.

I have the feeling that compared to the previous IPhO's, this one was even more difficult :)

Arne , You are proffesional in Physic olympaid too? :)

Iranian results:

9 Roshan Tovrani IRAN (ISLAMIC REPUBLIC) Gold

15 Ashkan Jafar Poor Koujahi IRAN (ISLAMIC REPUBLIC) Gold

52 Seyed Ahmadreza Hosseini IRAN (ISLAMIC REPUBLIC OF) Silver

79 Ali Shahad IRAN (ISLAMIC REPUBLIC OF) Bronze

94 Seyed Peiman Nezam Abadi IRAN (ISLAMIC REPUBLIC OF) Bronze
(My schoolmate :D )

The final results:
[urlwww.helveticanet.com/ipho36/english/contenidos/documentos/resultados%20ipho%202005.pdf[/url]

Well, I was selected for IPhO last year, but I was selected for IMO too, and I couldn't do both (because of the overlap) so I chose IMO. But I did all IPhO preparations.

I definitely think this IPhO was easier than the last few ones!

I made a couple of really stupid mistakes that put me down to silver, but that wouldn't be me without stupid mistakes (for example, R=I/U and + transforming into - in the middle of calculations without any reason;)

Anyway, I think this IPhO was easy and too much of a "cooking book" - they told you almost everything you needed to do, both in theory and experiment!

Thanks!
China and Taiwan has done very well!  :)


calculus
derivative
logarithms
calculus computations

Find the first and second derivative for
a. ln(x^2)
b. (lnx)^2

I used the chain rule to obtain the following

u=x^2
du/dx=2x
y=ln(u)
dy/du=1/x
f '(x)=2/x

f ''(x)=-2/x^2

then i got that ln(x^2) = (lnx)^2

Can anyone show me some clear workings if this is incorrect.

hello, let $ f(x) = (\ln(x))^2$ then by using the chain rule we have $ f^'(x) = 2\cdot\ln(x)\cdot\frac {1}{x}$.
Do you mean in a. $ \ln(x^2)$?
Sonnhard.

hello, the second derivative of b. is by the product and chain rule
$ f^{''}(x) \equal{} 2\cdot\frac {1}{x}\frac {1}{x} \plus{} 2\ln(x)\left(\frac { \minus{} 1}{x^2}\right)$, by simplifying we have $ f^{''}(x) \equal{} \frac {2}{x^2}(1 \minus{} \ln(x))$.
Sonnhard.

PEOPLE, THIS IS NOT A CALCULUS FORUM!!!!!!!!!!

[quote="thekiterunner"]Find the first and second derivative for
a. lnx^2
b. (lnx)^2

I used the chain rule to obtain the following

u=x^2
du/dx=2x
y=ln(u)
dy/du=1/x
f '(x)=2/x

f ''(x)=-2/x^2

then i got that lnx^2 = (lnx)^2

Can anyone show me some clear workings if this is incorrect.[/quote]

$ dy/du \equal{} 1/u \implies f'(x) \equal{} 2/x$ for (a).  Note that you could have simply used the property that $ \log x^b \equal{} b \log x$ for a logarithm with any base.

(b) $ f(x) \equal{} (\ln x)^2$
By chain rule, $ f'(x) \equal{} \frac{2 \ln x}{x}$.
By quotient rule, $ f''(x) \equal{} \frac{x \cdot 2/x \minus{} 2\ln x}{x^2} \equal{} \frac{2}{x^2}(1\minus{}\ln x)$


calculus
integration
limit
logarithms
calculus computations

This problem has been published in the AMM.
Find the limit:

$\lim_{n\rightarrow \infty}\frac{1}{n}\int_{0}^{n}\frac{x\ln(1+\frac{x}{n})}{1+x}dx$

Now, A couple of generalizations:

1)
$\lim_{n\rightarrow \infty}\frac{1}{n}\int_{0}^{n}\left(\frac{x\ln(1+\frac{x}{n})}{1+x}\right)^{n}dx$

2.
$\lim_{n\rightarrow \infty}\frac{1}{n}\int_{0}^{n}\left(\frac{x\ln(1+\frac{x}{n})}{1+x}\right)^{\alpha}dx$,
where $\alpha$ is a positive real number.

Is the answer to the first problem $2\ln 2-1$?

[quote="didilica"]Yes, that is what I got. Which technique have you applied? I assume that your method is different than mine, what do you think about the problem?
Didi[/quote]
Well, for the first one, for some fixed $n$, I simply used the substitution $y = \frac{x}{n}.$ Then, we let $n$ go to $\infty ,$ obtaining an easy integral. I believe my technique is right, and if that is true, I think the problem is just too easy to appear in the AMM.


The  following   code  will    get  the  output   that   there  are   page  headers  on   the  pages   of   the   contents .  How   to   remove  the   contents
page  headers    and   the   page  number  of   the   contents   on   the   middle  of   the   page  footers  ?

Many  thanks!

 [code]
\documentclass[openany]{book}
\usepackage{fancyhdr}
\pagestyle{fancy}
\fancyhf{}
 \fancyhead[ER]{\leftmark}
\fancyhead[OL]{\rightmark}
\fancyhead[EL,OR]{\thepage}
\renewcommand{\headrulewidth}{0.4pt}
.........
[/code]

Thank  stevem  very  much!

\thispagestyle{empty}   affects   the  last  page  of  the  contents   only!
I  want  that    footer  of  the  contents   contain  page   number   centered.

In other words,    I   want   the    pagestyle  of    the   contents       is      plain   style

See the [url=http://tug.ctan.org/cgi-bin/ctanPackageInformation.py?id=fancyhdr]fancyhdr documentation[/url]
[code]\fancypagestyle{plain}{%
\fancyhf{} % clear all header and footer
\fancyfoot[C]{\bfseries \thepage} % except
\renewcommand{\headrulewidth}{0pt}
\renewcommand{\footrulewidth}{0pt}}

\begin{document}

\pagestyle{plain}
\tableofcontents
\pagestyle{fancy}[/code]

The  following  code gets what I want:
[code]\usepackage{fancyhdr}
\pagestyle{fancy}
\fancyhf{}
 \fancyhead[ER]{\leftmark}
\fancyhead[OL]{\rightmark}
\fancyhead[EL,OR]{\thepage}
\renewcommand{\headrulewidth}{0.4pt}

\begin{document}
 \frontmatter

\pagestyle{plain}
\tableofcontents
\mainmatter
\pagestyle{fancy}[/code]

[color=red]But  the  following  code will get the  output  that  all   pages  of  the  book   are   plain  style.  Why?[/color]

[code]\usepackage{fancyhdr}
\pagestyle{fancy}
\fancyhf{}
 \fancyhead[ER]{\leftmark}
\fancyhead[OL]{\rightmark}
\fancyhead[EL,OR]{\thepage}
\renewcommand{\headrulewidth}{0.4pt}
\fancypagestyle{plain}{%
\fancyhf{} % clear all header and footer
\fancyfoot[C]{\bfseries \thepage} % except
\renewcommand{\headrulewidth}{0pt}
\renewcommand{\footrulewidth}{0pt}}

\begin{document}
 \frontmatter

\pagestyle{plain}
\tableofcontents
\mainmatter
\pagestyle{fancy}[/code]

\pagestyle{fancy} should come before \mainmatter

If    pagestyle{fancy}  come before \mainmatter, the output   proved  a    disappointment:

the     pagestyle   of   the contents   is  fancy  style.

the  following code  too

[code]\usepackage{fancyhdr}
\pagestyle{fancy}
\fancyhf{}
 \fancyhead[ER]{\leftmark}
\fancyhead[OL]{\rightmark}
\fancyhead[EL,OR]{\thepage}
\renewcommand{\headrulewidth}{0.4pt}

\begin{document}
 \frontmatter

\pagestyle{plain}
\tableofcontents
 \pagestyle{fancy}
\mainmatter
[/code]

[code]\begin{document}
\frontmatter
\pagestyle{plain}
\tableofcontents
\clearpage
\pagestyle{fancy}
\mainmatter[/code]

Very  interesting!

In  that  case,     all  pages of   the   book  are   not    plain  style  or  fancy  style.

the   code affect  all pages  of  the   book(wo  want  to  affect   pages  of  the   contents  only)

The  following  code   may be  useful,  but  I  don't  know  how  to  write  the  code

[code]\makeatletter
...............
\makeatother[/code]

[quote="yzymat"]In  that  case, all  pages of the book are not plain style or fancy style.[/quote]No, all non-chapter and part pages show the fancy page style.


[quote="thanhnam2902"]Let $ (O; R)$ is circle. Let two points A, B lie on $ (O; R)$. Let $ I$ is point center of $ AB$. Let $ MN, PQ$ are two chords of $ (O; R)$ through point $ I$. $ MP$ and $ NQ$ meet $ AB$ at $ J$ and $ K$ respectively. Prove $ I$ is point center of $ JK$.

-------------------------------------------------------
Good night. I wish you have a nice dream![/quote]
This is the well-known[b] Butterfly Theorem[/b].  Please consult the following links:
http://en.wikipedia.org/wiki/Butterfly_theorem
http://www.cut-the-knot.org/pythagoras/Butterfly.shtml
http://mathworld.wolfram.com/ButterflyTheorem.html

Thank. Can you post at this topic?
Thank.

Let $ X,Y$ be midpoint of $ PM,NQ$. We have $ OX\perp PM,OY\perp NQ,OI\perp AB$. Since $ \triangle IPMand \triangle INQ$ are similar, $ \triangle IPX and \triangle INY$ are similar. Hence $ \angle IXP \equal{} \angle IYN$. But we have $ \angle IXP \equal{} \angle IOJ,\angle IYN \equal{} \angle IOK$. Hence $ \angle IOJ \equal{} \angle IOK$. This imply $ IJ \equal{} IK$.


quadratics
function
arithmetic sequence
absolute value

Let $ a,b$ and $ c$ real numbers such that the equation $ ax^2\plus{}bx\plus{}c\equal{}0$ has two distinct real solutions $ p_1,p_2$ and the equation $ cx^2\plus{}bx\plus{}a\equal{}0$ has two distinct real solutions $ q_1,q_2$. We know that the numbers $ p_1,q_1,p_2,q_2$ in that order, form an arithmetic progression. Show that $ a\plus{}c\equal{}0$.

[hide="properties between roots"]I found a couple of interesting properties. First stating the obvious:

$ p_1 \plus{} p_2 \equal{} \minus{} \frac {b}{a}$

$ p_1p_2 \equal{} \frac {c}{a}$

$ q_1 \plus{} q_2 \equal{} \minus{} \frac {b}{c}$

$ q_1q_2 \equal{} \frac {a}{c}$

Now the properties.

$ p_1p_2q_1q_2 \equal{} 1$

$ q_1 \plus{} q_2 \equal{} \frac {p_1 \plus{} p_2}{p_1p_2}$

$ p_1 \plus{} p_2 \equal{} \frac {q_1 \plus{} q_2}{q_1q_2}$

$ q_1 \minus{} p_1 \plus{} q_2 \minus{} p_2 \equal{} \frac {p_1 \plus{} p_2}{p_1p_2} \minus{} \frac {q_1 \plus{} q_2}{q_1q_2} \equal{} p_1p_2(q_1 \plus{} q_2) \minus{} q_1q_2(p_1 \plus{} p_2) \equal{} \frac {b(a \minus{} c)}{ac}$

$ ap_1 \equal{} cq_1$

$ ap_2 \equal{} cq_2$

Unfortunatley, no results :( . Maybe this will be useful to someone else..[/hide]

[hide="well"]Let $ f,g$ be the two quadratics respectively; then $ g(x) \equal{} x^2f\left(\frac 1x\right)$. If $ f(0) \equal{} c \equal{} 0$ then $ g$ does not have two distinct roots, so the roots of $ g$ are $ \frac 1{p_1}, \frac 1{p_2}$.

If $ p_1,\, \frac{1}{p_2},\, p_2,\, \frac{1}{p_1}$ is an arithmetic progression then $ p_1 \plus{} \frac{1}{p_1} \equal{} p_2 \plus{} \frac{1}{p_2}$; but $ x \plus{} \frac{1}{x}$ is a monotonically increasing function for $ x \ge 1$ and hence we have either $ p_1 \equal{} p_2$ or $ q_1 \equal{} p_2$, both of which make the progression constant and contradicting distinct roots.

If $ p_1,\, \frac {1}{p_1},\, p_2,\, \frac{1}{p_2}$ is an arithmetic progression, then $ p_1$ and $ p_2$ have opposite signs (if they have same sign then consider the distance of their absolute value from $ 1$), so $ p_1 \minus{} \frac 1{p_1} \equal{} p_2 \minus{} \frac{1}{p_2}$. Now $ x \minus{} \frac 1x$ is strictly increasing, and as $ p_1 \neq p_2$ we must have $ p_1 \equal{} \minus{}\frac{1}{p_2} \Longrightarrow p_1p_2 \equal{} \frac{\minus{}c}{a} \equal{} 1$ as desired.[/hide]