If $n>1$, determine the least integer $f(n)$ such that between any $f(n)$ integers we can find $n$ of them whose sum is divisible by $n$.

I'm looking for a solution to this problem. Also, I'd like to know its original source. Thanks in advance ;)

Oh!!! I've just noticed the topic's name is wrong!!! :oops: It should be "$n$ numbers with sum divisible by $n$". Sorry :D

Erdos theorem, $f(n)=2n-1$. The proof is not simple at all

Indeed, but it appeared on the forum many times. ;)

Just look for : Erdos Ginzburg Ziv.

Well, if you wanna' prove that $f(n)>2n-2$ then you get Problem 1. from BMO 2003, if I am not wrong :)

Thank you guys :D Actually I met this problem in a test, but I didn't know it was really a known theorem :mad:


I'm just a new member with "NO"computer programming experience.How will i use the 'latex' techinique in other "Computer Rented Cafe?"

                                                    Respectfully,
                                                   A.Eddington

You can use LaTeX in this forum no matter what computer you are using. If you want to create your own documents, you need to install software on the computer itself -- see the link in the sidebar for details and instructions.


The radius of circle $ O$ is 12 inches, and $ \overline{AB}$ and $ \overline{CD}$ are tangent to the circle at $ B$ and $ D$, respectively. $ AB \equal{} 16$ inches, and $ CD\equal{}5$ inches. What is the sum of the number of inches in $ OC\plus{}OA$.

[asy]draw(Circle((0,0),12));
draw((0,0)--(8,12)--(0,12));
draw((0,0)--(-16,-12)--(0,-12));
label("$A$",(-16,-12),W);
label("$B$",(0,-12),S);
label("$C$",(8,12),E);
label("$D$",(0,12),N);
label("$O$",(0,0),S);[/asy]

Note: A radius is always perpendicular to the tangent line it intersects.

A different method:

With X being the point on the circle between C and O and extending XO to point Y on the end of the circle.

By POP $ DC^2 \equal{} CX\times{CY}$

$ 25 \equal{} x(x \plus{} 24)$

x=1 so OC=12+1=13

Letting L be the point  on AO that is on the circle, and extending AO to a point M on the circle:

By POP we have $ AB^2 \equal{} AL\times{AM}$

$ 256 \equal{} AL(AL \plus{} 24)$

AL=8 so AO=8+12=20

20+13=$ \boxed{33}$


function
floor function
inequalities
logarithms
algebra
real analysis
real analysis unsolved

Show that $ \sum_{n\geq 1} \frac {( \minus{} 1)^{E(\sqrt {n})}}{p_n}$  converges

where E(x) is floor function

$ p_n$ is n-th prime number

_______________________________________________

AMM October 2008

Nice submission to AMM Moubinool.
I can show a more general result: if $ k\in \mathbb{N}, k\geqslant 2$ and $ \alpha > 1\minus{} \frac{1}{k}$, then $ \sum_{n \equal{}1}^{\infty} \frac{(\minus{}1)^{\lfloor \sqrt[k]{n} \rfloor}}{{p_n}^{\alpha}}$ converges.

If there is no answer in few days, I'll post it here.

That's simple! Let $ u_m \equal{} \sum_{n \equal{} m^k}^{(m \plus{} 1)^k \minus{} 1}\frac 1{p_n^\alpha}$. Then $ u_{m \plus{} 1}\le (1 \plus{} \tfrac{C}{m})u_m$ (all terms in $ u_{m \plus{} 1}$ are smaller than in $ u_m$ and the number of terms is just a tiny bit greater) and $ u_m\le \frac C{m^\beta}$ with some $ \beta > 0$ (just because $ p_n > n$). Now, consider the sequences $ V_m \equal{} C\sum_{j\ge m}\frac {u_j}{j}$ and $ U_m \equal{} u_m \plus{} V_m$. The above estimates for $ u_m$ are enough to conclude that $ U_m$ and $ V_m$ are finite and decrease to $ 0$. Thus the series $ \sum_m ( \minus{} 1)^m U_m$ and $ \sum_m( \minus{} 1)^m V_m$ converge by the alternating series test. Hence $ \sum_{m}( \minus{} 1)^m u_m$ converges as well as the difference of 2 convergent series. The rest should be clear.

I said "that's simple" because initially I misread the problem and thought that it asked to prove that $ \sum_{p\in\mathbb P}\frac {( \minus{} 1)^{[\sqrt p\,]}}{p}$ converges. After an hour of thinking I realized that I know nothing about primes and returned to AoPS to ask for the idea of the solution. I almost posted something like "That is a terrific problem. Can you, please, give me some hints?" when I realized that the question was completely different from what I thought it was :roll: Why am I saying all that? Because I wonder now if there is any way to approach my misinterpretation of the question. :?

That's right fedja.
What do you think of $ \sum_{n \equal{} 1}^{\infty} \frac {( \minus{} 1)^{\lfloor \sqrt [k]{n} \rfloor}}{{p_n}^{\alpha}}$  for $ 0<\alpha \leqslant  1 \minus{} \frac {1}{k}$?

For the strict inequality case you have divergence ($ u_m\to \infty$ since $ p_n\le C n\log n$). The case $ \alpha\equal{}1\minus{}\frac 1k$ is much more interesting. On one hand, we still have $ u_m\to 0$. On the other hand, we only have $ u_m\le C(\log m)^{\minus{}\alpha}$ now, which is too weak to ensure finiteness of $ V_m$. I bet the series still converges but the proof may turn out to be much more complicated. What is clear is that you can no longer replace primes by an arbitrary sequence with the same asymptotics in this case, so you'll have to use some non-trivial facts about their distribution.


deci m-a rugat o fata sa ii fac problema asta ... si stiu sa o duc pana intr-un punct dar imi lipseste finalitatea. e asa:
Rezolvati ecuatia :
$(2x+1)(1+\sqrt{4x^2+4x+4} )+ (5x+1)(1+\sqrt{25x^2+10x+4}) =0$

rog seriozitate si promptitudine pe cat posibil . vreau solutie completa . thanks

Aratam ca ecuatia are solutia unica $x=-\frac{2}{7}$.
[hide]Ecuatia poate fi scrisa echivalent:
$(2x+1)(1+\sqrt{(2x+1)^2+3} )+ (5x+1)(1+\sqrt{(5x+1)^2+3} )=0$ si fie functia $f: R\rightarrow R,$ $f(x)=x(1+\sqrt{x^2+3})$.
Se verifica usor ca $f(-x)=-f(x),$ adica $f$ este impara.
In plus, tinand cont ca $R(x_2;x_2)=\frac{x_1(1+\sqrt{x_1^2+3})-x_2(1+\sqrt{x_2^2+3})}{x_1-x_2}>0,$ $\forall x_1, x_2\in R, x_1, x_2$distincte (dupa calcule simple; merge mai usor cu derivata), rezulta ca $f$ este strict crescatoare.
Ecuatia data se mai poate scrie echivalent $f(a)+f(b)=0,$ unde am notat $a=2x+1, b=5x+1$, inca echivalenta, prin faptul ca $f$ este impara, cu $f(a)=f(-b)$. De aici, tinand cont ca $f$ este strict crescatoare, deducem ca $a=-b$ deci $2x+1=-(5x+1)$, adica $x=-\frac{2}{7}$, solutie unica.[/hide]

superb ! mersi mult, fata va fi f recunoscatoare :P


inequalities
Putnam
inequalities proposed

Let $x_i$, $1\leq i\leq n$ be real numbers. Prove that \[ \sum_{1\leq i<j\leq n}|x_i+x_j|\geq\frac{n-2}{2}\sum_{i=1}^n|x_i|. \]
[i]Discrete version by Dan Schwarz of a Putnam problem[/i]

I wonder they have copied the problem from Iranian first TST 2006

Just look at this [url=http://www.mathlinks.ro/Forum/post-487754.html#]Problem[/url]
We know that $\sum_{i,j}|x_i+x_j|\geq n\sum_{i=1}^{n}|x_i|$. So $2\sum_{i<j}|x_i+x_j|+2\sum_{i=1}^{n}|x_i|\geq n\sum_{i=1}^{n}|x_i|$, and everything is done. And easily equality is done iff $n$ is even and half of the numbers are $k$ and the other half are eqaul to $-k$.

Also the solution is easy suppose $r$ of the numbers are positive and $n-r$ of them are negative then with the inequality $|x|+|y|\geq |x+y|$ everything is done.

[quote="Omid Hatami"]I wonder they have copied the problem from Iranian first TST 2006[/quote]Actually it wasn't copied. It was suggested from another problem (which might be the same with the one suggested in the Iranian TST).

For those wondering: no, this is not from the ISL 2005 :)

[quote="Cezar Lupu"]Valentin I think it was suggested by my problem with complex numbers.  :?[/quote]I'm not sure, but I think it was not from your problem, but rather from another problem given at Putnam ...

Ok. I see. I thought it was inspired from my problem, because I haven't seen it in shortlist. :)

In the Iranian TST problem was designed by Mohsen Jamali (designer of problem 6 IMO 2004) which hes aid he got this problem from an analytic problem.


calculus
integration
trigonometry
calculus computations

$ \int{\frac {{dx}}{{(\sin x \plus{} 2\sec x)^{2}}}}$  = $ \int\frac {\sec^{2} x dx }{ ( \tan x \plus{} 2\sec^{2} x )^{2} }$

put $ \tan x \equal{} t$

therefore
$ \int{\frac {{dx}}{{(\sin x \plus{} 2\sec x)^{2}}}}$ = $ \frac {1}{4}\int\frac {dt}{ (t^2 \plus{} \frac {t}{2} \plus{} 1 )^2} \equal{} \frac {1}{4}\int\frac {dt}{ (t \plus{} \frac {1}{4})^2 \plus{} ( \frac {\sqrt {15}}{4} )^2 }$

put $ t \plus{} \frac {1}{4} \equal{} \frac {\sqrt {15 \tan \theta}}{4}$

therefore integral becomes $ \frac {8}{\sqrt {15} 15 } \int{ (1 \plus{} \cos 2\theta ) d \theta }$

now evaluate

[u]
[b]EDITED :[/b][/u]

hello
$ I\equal{} \int{\frac{{dx}}{{(\sin x\plus{}2\sec x)^{2}}}}$
$ I\equal{} \int \ frac{cos^2(x) dx}{4\plus{}2sin(2x)\plus{}\frac{1}{4} sin^2(2x) }$
$ I\equal{}2 \int \frac{ (1\plus{}cos(2x)}{(4\plus{}sin^2(2x))^2} dx$
$ I\equal{} 2 \int \frac{1}{(4\plus{}sin^2(2x))^2} dx\plus{}2 \int\frac{cos(2x)}{(4\plus{}sin^2(2x))^2} dx$
$ a)$
For $ I_{1}\equal{}\int \frac{1}{(4\plus{}sin^2(2x))} dx$
Let $ P\equal{}\frac{cos(2x)}{4\plus{}sin(2x)}$
$ \frac{dP}{dx}\equal{}\minus{}\frac{8}{4\plus{}sin(2x)}\plus{}\frac{30}{(4\plus{}sin(2x))^2}$
integrating this
$ P \equal{}\minus{}8 \int \frac{dx}{4\plus{}sin(2x)} \plus{} 30 I_{1}$
for $ \int \frac{dx}{4\plus{}sin(2x)}$
write $ sin(2x)$ in $ tan(x)$ form and put $ tan(x)\equal{}t$
$ 30I_{1}\equal{}P\plus{} \frac{8}{\sqrt{15}}tan^{\minus{}1}(\frac{4tan(x)\plus{}1}{\sqrt{15}})$
$ b)$
for $ \int\frac{cos(2x)}{(4\plus{}sin^2(2x))^2} dx$
Put $ 4\plus{}sin(2x)\equal{}t$
$ dt\equal{}2cos(2x) dx$
Finally u will get
$ I\equal{} \frac{1}{15} \frac{cos(2x)}{4\plus{}sin(2x)}\plus{}\frac{8}{(15)^{\frac{3}{2}}} tan^{\minus{}1}(\frac{4tan(x)\plus{}1}{\sqrt{15}})\minus{}\frac{1}{4\plus{}sin(2x)}\plus{}C$

To Vijaymenon,..

$ \int \frac{\sec ^{2}xdx}{\tan x\plus{}2\sec ^{2}x}\equal{}\int \frac{dx}{(\sin x\plus{}2\sec x)^{2}}$ is not valid.

Sorry INDOCHINA it was a typo it is indeed $ \int\frac{\sec^{2}x dx }{( \tan x\plus{}2\sec^{2}x )^{2}}$


calculus
integration
inequalities
triangle inequality

In the triangle shown, what is the positive difference, between the greatest and least possible integral values of $ x$?

[asy]pair a,b,c;
b = (1,2); c = (3,0);
draw(a--b--c--a);
label("x",a--b,dir(135)); label("5",b--c,dir(45)); label("6",c--a,S);[/asy]

The triangle inequality states that the sum of two sides can not be exceeded by the third. That means the largest possible value of $ x$ is $ (5 \plus{} 6) \minus{} 1 \equal{} 10$ and the smallest value is $ (6 \minus{} 5) \plus{} 1 \equal{} 2$. Therefore, the answer is $ \boxed{8}$.


ARML Practice- The following problems I have designed are intended to find a striking relationship between combinatorics and several sequences, such as the trigonal numbers and sum of first n squares sequence. This is intended as a semi-tutorial for students that have not been exposed to this topic. Some people may find it too easy but its still good review or practice.

We will start with combinatorics, then make the connection.

1. Hockey-Stick Identity

(a)Start with an entry on the extreme left side of Pascal's Triangle (any number 1) such as $\binom{3}{0}$. Now move downwards to the right in a daigonal fashion, adding numbers along the diagonal. For example, starting at $\binom{3}{0}$, the next numbers along the diagonal are 4, 10, 20, etc. Now stop at a given number, such as 10. Compare the numbers in the diagonal (from 1 to 10) to the number below and to the left of 10 (15). Do you see a relationship? Try starting with $\binom{3}{0}$ again, going down the diagonal, and then stopping at 20. Then compare the numbers in the diagonal (from 1 to 20) to the number below and to the left of 20 (35). The relationship should be clear. Once you have a firm understanding of the relationship, summarize the result with an equation involving a summation.

(b)Now try the same thing, but this time starting at the extreme right side of the triangle. Once again, formalize the result a bit by giving an equation. Notice that for this exact relationship to hold, we must start at the a 1.

The results of the investigation are given below. If this is your first exposure to the identity, please do not look until you have tried finding it yourself.

[hide]Left Hockey-Stick Identity: $\displaystyle\sum_{k=0}^{j}\binom{n+k}{k} = \binom{n+j+1}{j}$.
Right Hockey-Stick Identity: $\displaystyle\sum_{k=0}^{j}\binom{n+k}{n} = \binom{n+j+1}{n+1}$.[/hide]

2. Proof/Derivation
Proving the hockey-stick identity is not that difficult.

Let's start by deriving the left hockey-stick identity algebraically. First off, recall pascal's identity: $\binom{n}{k-1} + \binom{n}{k} = \binom {n+1}{k}$ (basically just a more formal way of writing that the sum of two adjacent terms in a row of Pascal's triangle gives the term centered below them).

(a)Starting with the expanded form of the left-hand side of the left hockey stick identity above, use Pascal's identity to "collapse" the terms, giving you the desired result.

(b) Now do the right hockey-stick identity. Start with the side with lots of terms, and "collapse" them to get the desired result.

(c) A combinatorial-based derivation can also provide insight. First make sure you understand a key concept that defines Pascal's Triangle: the number of paths from the top most entry to the $\binom{n}{k}$ entry is $\binom{n}{k}$. This is because every path from the top to $\binom{n}{k}$ has length $n$ and exactly $k$ of the $n$ moves are to the right. It follows that there are $\binom{n}{k}$ paths to the entry since all you need to do is choose exactly $k$ of the $n$ moves to be a right move. Make sure you understand this, and test it with concrete numbers if you need to.

Back to the hockey-stick identity. Look at a specific entry in Pascal's Triangle, such as one of the 15's. Give a detailed explanation of why the hockey-stick identity holds (left and right) by considering the number of paths from the top to the entry 15.

3.
Finally we can move on to the connection with sequences.
We first concern ourselves with the question of how to derive a formula for $\displaystyle\sum_{i=1}^{n} i$ in terms of $n$. Our general strategy will be to write $i$ as a combination, and then use the hockey-stick identities to simplify.

(a) Clearly, $\binom{i}{1}=i$. Use this fact and the hockey-stick identity to derive the familiar formula for $\displaystyle\sum_{i=1}^{n} i$.

(b) The nth triangular number $t_n$ is defined by $t_n=\displaystyle\sum_{i=1}^{n} i$. The nth tetrahedral number $s_n$ is defined by $s_n=\displaystyle\sum_{i=1}^{n} t_i$. Find a formula for $s_n$ in terms of $n$.

(c) Now we move on to a more difficult problem, deriving a formula for $\displaystyle\sum_{i=1}^{n} i^2$. Once again, our strategy is the same: try hard to apply the hockeystick identity by rewriting $i^2$ in terms of binomial coefficients. Before starting, note that the binomial coefficient $\binom{i}{k}$, where $k\leq{i}$ gives an expression with degree $k$ for $i$.

(d)Now try $\displaystyle\sum_{i=1}^{n} i^2+3i+2$

(e)Find an expression for $\displaystyle\sum_{i=1}^{n} i^3$

(f)One final extension. Its a lot of algebra (and practice with manipulation), but find an expression for $\displaystyle\sum_{i=1}^{n} i^4$

I won't give answers to number 3. Most formulas are familiar and you can get an idea if your formula works by plugging-in values.

Let me know what you think about the tutorial  :) Also tell me if you find any errors.

Mark

That's pretty detailed, Didn't find any mistakes  :coolspeak:

Great job! By the way, Adam Marcus showed me another way to evaluate expressions such as

$\displaystyle\sum_{i=1}^{n} i^x$ where $x$ is an integer. I can show it to you today on the bus if you want. But I'm not sure if you'll be reading this before we get on the bus  :D

well, its too late for the bus :). just post it here.

Recall that nCr gives the number of strings of n symbols of two kinds, r of which are of one kind.  So for example, 4C6 gives the number of strings of X's and Y's of which 4 are X's (and 2 are Y's).

Then (n+j+1)C(n+1) gives the number of strings of n+j+1 X's and Y's of which n+1 are X's and j are Y's.

These can be counted by conditioning on the position of the last X.

- Last X is in the n+1st position--all previous are X's, so nC0 ways of arranging the 0 Y's before it, 1 way of arranging the j Y's after it.
- Last X is in the n+2nd position--all but one previous are X's, so (n+1)C1 ways of arranging the 1 Y before it, 1 way of arranging j-1 Y's after it.
- Last X is in the n+3rd position--all but two previous are X's, so (n+2)C2 ways of arranging the 2 Y's before it, 1 way of arranging j-2 Y's after it.
...
- Last X is in the n+j+1 position--all Y's are before it, so (n+j)Cj ways of arranging them.

== pj karafiol[/code]

Indianamath, please don't spam in other people's forums, particularly if your spam is not interesting.


$ a,b,c>0$. Show that: $ a^{b\plus{}c}\plus{}b^{c\plus{}a}\plus{}c^{a\plus{}b} \geq 1$.

[quote="hotboy12"]$ a,b,c > 0$. Show that: $ a^{b \plus{} c} \plus{} b^{c \plus{} a} \plus{} c^{a \plus{} b} \geq 1$.[/quote]
It is my problem at M&Y magazine. Moderator, please lock it. Thanks

[quote="arqady"]Nice problem, can_hang2007! :lol:[/quote]
And Please post its solution! :D


LaTeX
geometry
circumcircle
quadratics
inequalities
algebra
IMO Shortlist

this year albania got a gold medal in the JBMO!

Is there anyone who has problems given at jbmo this year? I'm very interrested.

hi! i am Andrew (MOL2) and i got gold :)
i post here the problems: i don't know the latex or how it is called, so if someone can you may "translate them"

1.
find all positive integers (nonzero) x and  y such that:
9(x^2  +  y^2 + 1) +2(3xy+2) = 2005

2.
the second was a geometry problem.
given a (acute angled)trianle ABC and the tangent at A to the circumcircle intersects BC at P.
M is the midpoint of AP. BM intersects the circumcircle again at R and PR intersects the circumcircle again at S
prove that AP || CS.

3.
(a) Prove that there exist 5 points in the plane such that there exist at least 8  right-angled triangles through the triangles with vertices in those points.   (for th (a) - 2 points)
(b) Prove that there exist 64 points in the plane such that there are at least 2005 right-angled triangles through the triangles with vertices in those points.   (for the (b) - 8 points)

4. Find all 3digit numbers abc (not as the product of a,b,c but the number with digits a,b,c.
such that     abc (not the product) = a*b*c*(a+b+c).

i think it is an ugly problem, cause almost all the solutions are by analyzing lots of cases. i analyzewd somewhere
15-20 cases, and i skipped one of them and they gave me 6 points!

the geometry has a simple solution, but few of the contestand solved it. i took five [points at it.
two from the romanian team got 10 points at it(maybe thre or four, i'm not sure)[/quote][/list][/code]

here is the official JBMO 2005 site:

http://www.hms.gr/jbmo2005/main.html

where hms means: Hellenian mathematical society

[quote="cezar lupu"]Romania got the first place?[/quote]
indeed romania got the first prize with 155 points
they got 3 golds and 3 silvers

[quote="cezar lupu"]Romania got the first place?[/quote]
congratulations to all of the romanian team, especially to Madalina Persu and Cornel Claudiu Prodesscu who got 31 (like me :) )
actually 31 was the highest mark.
and Francisc Bogzan - who is lucky as it got the last (seventh) gold medal.
and also congratulations to Padureanu Victor (silver with 22) and Matican Bogdan - silver with 23, and Stefan Stoican also silver with 23 points.
congratulations to the Moldavian team also! :) :)
i'll meet them at Iasi next year at their national MO :)

hi
i just wanted to say that it is 99% suire that next jbmo will be here, in moldova
it will be my last year (the 4th :) ) at juniors  :)

can anyone publish thee 2004 shortlist? (from jbmo)?

congratulation to the gold medals!
i think that the 4 takes a lot of time to be solved! so i'm not surprised with the fact that the highest score was 31 and that the cut-off for gold was 25!

Congratulations freemind and others who get medals especially gold one :)
     I have won silver medal with 24 points(i needed one more for gold :)).Bosnian team made big result this year,we were fifth and we almost beat Bulgarian team who was only 5 points better.Next JBMO will be held in Moldova and after that i think that JBMO 2007. will be held in Sarajevo,BiH :)

does any one have a elegant solution to the second problem!
my solution is a bit long so i think its not the proposer's one

does anyone know from where were the problems proposed :?:

[quote="silouan"][quote="freemind"]the 2nd problem was proposed by romania i think
the 4th by Albania
the 3rd by bulgaria
and the 1st by Cyprus
hope i'm not wrong

the solution at the second problem is very simple.
well consider the power point of M with respect to the circumcircle.
then MA*MA = MR*MS = MP*MP since MA=MP
thus
MP/MR=MS/MP thus triangle MPR is similar to triangle MRP. Thus angle MPR is equal to angle MSP/
but angle MSP is equal to angle ACS since BCSR is inscribed. thus angle APB = angle APC = Angle ACS thus CS||AP

congratulations to vedran6! indeed with one point more you got a gold medal

the 4th problem is somehow ugly, cause it needs lots of analyzis, and like i did, if you skip one-two cases they give you up to 6-7 points.
they gave me 6 points.[/quote]Sory freemind but i think that your solution is confused. Can you post in the site this solution better writted and i want the solutions of the other problems. please quote soon. If anyone wants i have a good solution for the first problem[/quote]

Yes his solution has some mistakes but I got the idea. It is MP*MP=MR*MB and than it is easy, just deal with angles.

silouan, why do u always say you're in agony? :)
why such a hurry? :))

It is a persuasive frase. See you. I wait the solutions.

Hi freemind where are you today? I lost you. Did you took my e-mail adress?
I send it you because i do not know well the site.I put the solution of geometry. The
file tranfered into National Contests.

Hey,Silouan what about the shortlist? You said in two days.I'm still waiting. :spam:

[quote="cezar lupu"]Although It's not right what I'm doing,here it is:

  Let $a,b,c>0$ such that $ab+bc+ca>=ab/c +bc/a +ca/b$.Prove that:

         $a^2.b+b^2.c+c^2.a>=1/a+1/b+1/c$. I think this is it. Well,if I see the shortlist I will remember

  for sure. ;)[/quote]

That condition is a very helpful one, since it gives $\sqrt{\left(abc + abc + abc\right)\left(ab/c+bc/a+ca/b\right)} \ge ab+bc+ca \ge ab/c+bc/a+ca/b$, by Cauchy, or $3abc \ge ab/c+bc/a+ca/b$. Now, the desired is weaker than

$(a^2b+b^2c+c^2a)(ab/c+bc/a+ca/b)^2\ge (3abc)^2(1/a+1/b+1/c)$

This can be shown quickly by AM-GM, since $a^2b+b^2c+c^2a \ge 3abc$ and the remainder follows by Muirhead.

Well freemind I lost you for second day. Where are you?Do you take my e-mails and my e-mail adress? See you. i hope you will post an answer as soon as possible.

Freemind to see the solution of Greek TST geometry go to Olympiad section then National olympiads and finally topics. See you

Silouan, I'm still waiting fo the shortlist. :D

I cannot give you something I have ever have. I thought it was easy to find it but the system is as for the IMO shortlist .It is not permitted to give the shortlist to someone

[quote="dule_00"]Does someone have  JBMO 2004 shorlist?
 I'm also interested in earlier shortlists.[/quote]

I am also interested in JBMO 2004 shorlist. But nobody gives it and I do not know why?

well JBMO 06 is near and im contesting in it... cya all in Moldavia and gl !


LaTeX
number theory proposed
number theory

Solve $p^n+144=m^2$ where $m,n\in \mathbb{N}$ and $p$ is a prime number.

[quote="Boll"]Solve
$p^n+144=m^2$ where $m,n\in \mathbb{N}$ and $p$ is a prime number[/quote]

We have:

$p^n=(m-12)(m+12)$ so $m-12$ must be power of $p$. Let $m-12=p^r$.

Then we have:
$p^n=p^r(p^r+24)$
$p^{n-r}=p^r+24$
$p^r(p^{n-2r}-1)=24$

$24=2^3*3$ so we have to check when $p=2, r=3$ and $p=3, r=1$.
In first case we have $2^{n-6}-1=3$ so $n=8$ and $m=2^3+12=20$.
In second: $3^{n-2}-1=8$ so $n=4$ and $m=3+12=15$

So we have solutions: $(p, n, m)=(2, 8, 20), (3, 4, 15)$

[quote="Boll"]5 points :P:P (i've done the same mistake)[/quote]

Oh.. I missed the case when $r=0$  :blush: Then we have: $(p, n, m)=(5, 2, 13)$

Is something still missing?..

An easy one, i think i have a different solution:
if p^n - 144 = m^2 then p^n = m^2 - 144 = (m-12)(m+12), but m is a positive integer so m+12 > m-12, and we know they both must be powers of p, so let m-12 = p^r and m+12 = p^s, for r<s such that r+s = n, thus
m-12 | m+12, but (m+12/m-12) = 1 + (24/m-12) wich means m-12 is a positive divider of 24, wich lead us to the solutions: {m,p,n} = {13,5,2}, {15,3,4}, and {20,2,8}
PS sorry, i didn't use Latex because i'm new and i'm just downloading it!

now i have latex :D , the solution is like this:
$p^{n}$ + 144 = $m^{2}$
$p^{n}$ = $m^{2}$ - 144 = (m-12)(m+12)
 but $m\in$ $Z^{+}$ so m+12 > m-12, and as they both must be powers of p: let m-12=$p^{r}$ and m+12 = $p^{s}$ for s>r such that r+s=n, thus $m-12 \| m+12$
but $\frac{m+12}{m-12}$ = 1 + $\frac{24}{m-12}$ wich means m-12 is a positive divider of 24, wich lead us to the solutions: {m,p,n} = {13,5,2}, {15,3,4}, and {20,2,8}


logarithms
trigonometry
function
algebra unsolved
algebra

Let $ x>1,y>1$, and $ (\log{x})^2\plus{}(\log{y})^2\equal{}\log{x^2}\plus{}\log{y^2}$, find the maximum of  $ x^{\log{y}}$[/code]

hello, do you mean with $ \log x^2\equal{}\log(x^2)$?
Sonnhard.

[quote="Dr Sonnhard Graubner"]hello, do you mean with $ \log x^2 \equal{} \log(x^2)$?
Sonnhard.[/quote]

Yes. :)

[hide="solution"]
Rewrite this:
$ \log^2x \minus{} 2\log x \plus{} \log^2y \minus{} 2\log y \equal{} 0$
$ (\log x \minus{} 1)^2 \plus{} (\log y \minus{} 1)^2 \equal{} 2$
And now let:
$ \log x \minus{} 1 \equal{} \sqrt {2}\sin\psi$
$ \log y \minus{} 1 \equal{} \sqrt {2}\cos\psi$ for $ 0\leqslant\psi\leqslant 2\pi$
And now we show find maximum value of $ x^{\log y} \equal{} e^{\log x\log y} \equal{} \exp\left[{(\sqrt {2}\sin\psi \plus{} 1)(\sqrt {2}\cos\psi \plus{} 1)}\right]$

And maximum value of $ {(\sqrt {2}\sin\psi \plus{} 1)(\sqrt {2}\cos\psi \plus{} 1)}$ is $ 4$ (easy check with derivate this function and find the extreme value).

So the answer is $ e^4$ (when $ \psi\equal{}\frac{\pi}{4}$)

[/hide]


Hi guys. I'm new in that forum. Recently I received a pretty hard task. I'm looking for some exercises from olympiads with solutions that are using complex numbers. I would be really grateful if you gave me a link or something. Thanks in advance

Well i don't know if there is a page, but i've been trying to prove these for some 2 days. Let $\omega$ be a third root of the unit. Let $P$ be a prime. Prove (or disprove) that the if factorization in $\mathbb Z[\omega]$ in irreducible numbers of $p$ is of the form $(a-\omega b)(a-\omega^{2}b)$.


MATHCOUNTS
number theory
prime factorization

What is the fastest way to find the least number with x factors?
I know that for an example:
60=$ 2^2*3*5$, so it has 3*2*2=12 factors....

but finding the least number with let's say: 20 factors would require a lot of guess and check
which would waste too much time

Edit:
I need a method for when X is not odd :)

You have to use casework to minimize the value of the prime factorization.

also:

any number with an odd number of factors MUST be a square number.

For a particular $ x$, it's essentially casework.  The general solution is related to the sequence of [url=http://mathworld.wolfram.com/HighlyCompositeNumber.html]highly composite numbers[/url], and depends strongly on the prime factorization of $ x$ itself.

I'd like to emphasize that contest problems will ask you for a particular $ x$ and that one of the reflexes Mathcounts kids might have is to memorize a formula that works for any $ x$, but many problems are not in fact that straightforward.  It's better to understand the concepts.

So let me ask you:  what [i]is[/i] the smallest number with $ 20$ factors?

well, 20=20,10*2,5*4,5*2*2, and subtravting one from each gives the sets
{19}{9.1}{4,3}{4,1,1}, which correspond to the numbers
2^19,3*2^9,3^3*2^2,2^4*3*5
2^19 is monstrously large.
3*2^9 is larger than 3^3*2^4, which is 27*16=something bigger than 240..
2^4*3*5 is 240, so 240 is the smallest.

right?

EDIT:i put 3^3*2^2 lolz... i think it's correct now..

how did you get 3^3*2^2. That number has 12 factors.

The real answer is:

[hide]
240[/hide]

[hide]

20 can be split into 5*4 or 5*2*2, which implies

2^4*3^3 or 2^4*3*5. Ofcourse the second one is lower, so the answer is 240

[/hide]


Sorry,  could  I  post  here?

I  am  using  Win  XP.       I   would  like   to   install    ConTeXt Minimals +  Texworks.

I  have   difficulty   because    I   don't   understand    path,   system path,  and  cmd.exe.

Must    I   install   Ruby?

Thanks in anticipation!

http://wiki.contextgarden.net/ConTeXt_Minimals tells you all you need. So
[list][*]http://wiki.contextgarden.net/ConTeXt_Minimals#Dependencies tells you that Ruby is required;
[*]http://wiki.contextgarden.net/ConTeXt_Minimals#GUI_Installer gives an easy to use installer. The guide at http://meeting.contextgarden.net/2008/talks/2008-08-21-vyatcheslav-win-installer/windows_installer.pdf pages 7 onwards shows there is only one decision to make.[/list]

Thank   stsvem very much!

I  have  installed   Miktex  already,  so [url=http://wiki.contextgarden.net/ConTeXt_Minimals#Command_line_method]Command  line     method[/url] is   the  only  choice.

 path, system path  ,cmd.ese?

also  ,    Integrating with texworks  is  not  easy......

You CAN use the GUI. All you need do is [i][b]not[/b][/i] tick the box "Set environment vars globally" see pages 8 ad 15 of the pdf. Also [quote]NOTE: The GUI installer asks if you want to set environmental variables globally. Click this only if you do not use MikTeX or TeXLive and do not plan to use LaTeX [/quote]

How   to      Integrate   with texworks   ?

[i][b]Always[/b][/i] read the manual. It's at http://www.leliseron.org/texworks/
[quote]So one needs to install a distribution; we will do that “before” to
 start TEXworks for the first time, this way TEXworks will automatically find what it needs.[/quote]

The Windows platform has several options for installing Ruby:

http://www.ruby-lang.org/en/downloads/

Which  one  is   better?

Many   thanks!

That site [i][b]tells you[/b][/i] which is better
[quote]If you’re unsure about how to install Ruby, the one-click installer may be the best option for you. (In addition to Ruby, the one-click installer also comes with a bunch of additional libraries built in.)[/quote]


limit
logarithms
trigonometry
calculus
calculus computations

Find L:
$ \lim_{x\rightarrow 1} \log_x(x\minus{}1) \equal{} L$

Is there a reasonable way to solve integer solutions of n?
$ \cos(n)\equal{}\sin(n)$

2:

$ \cos n= \sin n \to \tan n = 1 \to n\in \{ \frac{\pi}{4}+k\pi\}_{k\in \mathbb{Z}}$

so $ n=\frac{\pi}{4}(1+4k) \to \frac{4n}{1+4k}=\pi$ would imply that $ \pi$ is rational, which is easily disproved

hence there are no solutions

[quote="epkid08"]Find L:
$ \lim_{x\rightarrow 1} \log_x(x \minus{} 1) \equal{} L$
[/quote]
Do you mean $ \lim_{x\rightarrow 1^{\plus{}}}\log_{x}(x\minus{}1)$?

[quote="Timestopper_STG"][quote="epkid08"]Find L:
$ \lim_{x\rightarrow 1} \log_x(x \minus{} 1) \equal{} L$
[/quote]
Do you mean $ \lim_{x\rightarrow 1^{ \plus{} }}\log_{x}(x \minus{} 1)$?[/quote]
if its $ 1^{\plus{}}$, then write it as ln(x-1)/ln x. then its of the form -inf/+0. so ans should be -inf.


How do you have multiple blogs? Like Mr. Vornicu and ccy?

You can't right now. We will see if we will let that happen in the future.

But then how did ccy get it?

Edit:

http://www.artofproblemsolving.com/Forum/weblog.php?w=897
http://www.artofproblemsolving.com/Forum/weblog.php?w=305

You could do so by mimicking the form that comes up before you have a blog in an HTML file on your hard drive then submitting it, methnks.

[quote="#H34N1"]But then how did ccy get it?[/quote]The option was available at some point. It might be in the future again for certain people.

Nut ccy's iTest 2007 wasn't around until about a few days ago.

Well that means there's a hole in the system doesn't it? :) I'll take a closer look at it.


Let a convex quadrangle ABCD. The perpendicular line from A to AB,AC,AD cuts CD,BD,BC at M,N,P. Prove that M,N,P are collinear.

Hello!

$ \frac{MC}{MD}\equal{}\frac{\frac{MA\cdot sin\angle MAC}{sin\angle ACD}}{\frac{MA\cdot sin\angle MAD}{sin\angle ADC}}\equal{}\frac{sin\angle MAC}{sin\angle MAD}\cdot\frac{sin\angle ADC}{sin\angle ACD}\equal{}\frac{cos\angle BAC}{cos\angle BAD}\cdot\frac{AC}{AD}$     (*)

$ \frac{ND}{NB}\equal{}\frac{\frac{NA\cdot sin\angle NAD}{sin\angle ADB}}{\frac{NA\cdot sin\angle NAB}{sin\angle ABD}}\equal{}\frac{sin\angle NAD}{sin\angle NAB}\cdot\frac{sin\angle ABD}{sin\angle ADB}\equal{}\frac{cos\angle CAD}{cos\angle CAB}\cdot\frac{AD}{AB}$     (**)

$ \frac{PB}{PC}\equal{}\frac{\frac{PA\cdot sin\angle PAB}{sin\angle ABC}}{\frac{PA\cdot sin\angle PAC}{sin\angle ACB}}\equal{}\frac{sin\angle PAB}{sin\angle PAC}\cdot\frac{sin\angle ACB}{sin\angle ABC}\equal{}\frac{cos\angle DAB}{cos\angle DAC}\cdot\frac{AB}{AC}$     (***)

(*), (**), (***) $ \Rightarrow\frac{MC}{MD}\cdot\frac{ND}{NB}\cdot\frac{PB}{PC}\equal{}1\Rightarrow$ the points $ M$, $ N$, $ P$ are collinear. :)

The perpendicularities as the problem states, surely help us to think a solution, based on the [b][size=100]Douple ratio theory[/size][/b].

So, because of $ AM\perp AB,$ $ AN\perp AC\equiv AC',$ $ AP\equiv AP'\perp AD,$ where $ C',\ P'$ are arbitrary points on the line segments $ AC,\ AP$ respectively towards $ A,$  in the configuration of the pencils $ A.BCDN,\ A.MNP'C',$ we have that they are formed equal angles by their homologous rays $ ($ $ \angle BAC \equal{} \angle MAN,$ $ \angle CAD \equal{} \angle NAP',$ $ \angle DAN \equal{} \angle P'AC'$ $ ).$

Hence, we conclude that these pencils have equal [b][size=100]douple ratios[/size][/b] $ \Longrightarrow$ $ (A.BCDN) \equal{} (A.MNP'C')$ $ ,(1)$

The pencil $ A.BCDN$ is intersected from the line segment $ BD$ and so, we have that $ (A.BCDN) \equal{} (B,K,D,N) \equal{} (D,N,B,K) \equal{} (C.DNBK)$ $ ,(2)$ where $ K\equiv AC\cap BD.$

From $ (1),$ $ (2)$ $ \Longrightarrow$ $ (A.MNP'C') \equal{} (C.DNBK)$ $ ,(3)$

From $ (3)$ and because of the pencils $ A.MNP'C',\ C.DNBK$ have the line segment $ AC'\equiv CK\equiv AC,$ as their common ray, we conclude that the points $ M\equiv AM\cap CD$ and $ N\equiv AN\cap CN$ and $ P\equiv AP'\cap CB$ are collinear ( as the intersection points of the other pairs of their homologous rays ) and the proof is completed.

Kostas Vittas.

Yes, we easily recognise the solution which uses Menelaus's theorem (in the form of Douple ratio)  :)
We get a nicer proof with the following lemma:
Let a quadrilateral ABCD. AB meets CD at E. AD meets BC at F. Then, three circles whose diameters are the main diagonals AC,BD,EF of the complete quadrilateral are coaxial (which means that these circles have two common points).

Sorry dear all my friends for the not earlier posting of the schema, relative to the solution I presented in the post #3#.

Kostas Vittas.


geometry
geometric transformation
reflection
homothety
geometry unsolved

Given a triangle ABC and an arbitrary point P.  Let A',B',C' be the reflection of the point P wrt BC,CA,AB, and Q be the isogonal conjugate wrt triangle A'B'C'.  Let A'',B'',C'' be the reflection of the point Q wrt B'C',C'A',A'B'.
Prove that [A'B'C']²=[ABC][A''B''C''],where [ABC] is area of triangle ABC.

From the cyclic quadrilaterals $ QA''B'C'', QB''A'C''$ and $ PB'CA',$ we obtain

$ \angle A''C''Q \equal{} \angle A''B'Q \equal{} \angle PB'A' \equal{} \angle PCA'$

$ \angle B''C''Q \equal{} \angle B''A'Q \equal{} \angle PA'B' \equal{} \angle PCB' \Longrightarrow \angle ACB \equal{} \angle A''C''B''$

Mutatis mutandis $ \angle ABC \equal{} \angle A''B''C''$ $\Longrightarrow$ $ \triangle ABC \sim \triangle A''B''C''$

Perpendicular lines through $B,C$ to $ A'B'$ and $ A'C'$ meet at the isogonal congugate $ P'$ of $ P$ WRT $ \triangle ABC$ $\Longrightarrow$ $ \angle QC''B'' \equal{} \angle PCB$ and $ \angle QB''C'' \equal{} \angle PBC,$ which implies that $ \triangle QB''C''$ and $ \triangle PBC$ are homothetic, since $ PC,PB$ are parallel to $ QC'',QB''.$ By similar reasoning, we conclude that $ \triangle A''B''C''$ and $ \triangle ABC$ are homothetic. Now, according to Gergonne-Ann theorem,  $ [\triangle A'B'C']^2 \equal{} [\triangle ABC] \cdot [ \triangle A''B''C''].$


O is the circumcenter of ABC,M is on AB,N is on AC,and angleMON=180(M,O,N are on the same line).D is the midpoint of BN,E is the midpoint of CM,can we prove angleBAC=angleDOE

Let $ BO\cap (O) \equal{} \{Y\}, CO\cap (O) \equal{} \{Z\}$
Applying [b]Pascal'[/b]s theorem we get $ YN\cap ZM \equal{} \{X\}$ and $ X$ lies on $ (O)$.
Thus $ \angle MXC \equal{} \angle NXB \equal{} 90^o$.
Draw $ (E,EC)$ and $ (D,DB)$ then $ OE\perp XC$ and $ OD\perp XB$.
Therefore $ \angle DOE \equal{} 180^o \minus{} \angle BXC \equal{} \angle A$


in a group of 30 random people, what is the probability that there exists at least one month in which no one has his or her birthday?

note: assume the probability of being born in any month is the same for all months.

I think the answer is [hide](12*11^30)/12^30, because for one guy, the chance of borning in one of the 11 months is 11/12. For thirty guy is (11//12)^30. There are 11C12 number of combinations of 11 month, therefore the answer is (12*11^30)/12^30[/hide]

I'm pretty certain that's not right. [hide]First of all, I think you have a typo where you say 11C12. Do you mean 12C11? Second of all, what would happen to your method if there were say, fewer than 12 people in the group? Try your method out with smaller numbers of months and people and see what happens.[/hide]

I think this is how you do it, but the answer looks kind of large.

[hide]we can calculate the probability for there to be birthdays in all 12 months and just do 1 minus that.

There are (30-12+11)C11 = 29C11 ways to divide 30 people into 12 months with each month having at least 1 person (put 1 person into each month then putting 11 dividers in 18)

There are (30+11)C11 = 41C11 ways to divide 30 people into 12 months where each month can have any number of people (putting 11 dividers into 30)

The probability for there being birthdays in all 12 months is (29C11)/(41C11) = 70035/6395672 so the probability for not having birthdays in all 12 months is

6325637/6395672[/hide]

your method is a little too optimistic, the probability is not that high.

hint: [hide] consider the cases of no birthdays in january, no birthdays in february, no birthdays in march, etc.  These probabilities will not be mutually exclusive since there can be more than one month in which no one was born.[/hide]

zscool wrote:[hide]These probabilities will not be mutually exclusive since there can be more than one month in which no one was born.[/hide]

Sorry, maybe I'm not understanding correctly. Why is that true? Isn't it possible that [hide]say, for example, everyone was born in January[/hide]?

Yeess.. but mutually exclusive means those events could *never* both happen.. which is not true.

Oh, goodness. So sorry -- I didn't see the <b>not</b>.  :oops:


inequalities
function
inequalities solved

Guess this inequality is VERY HARD. Plese think about it.

hmm I was thinking of using this problem for the MathLinks contest starting tonight ? :D

hey .. don't panik .. i was just kiddin' :D

You can use it because I realised it's very beautiful.

yeah but some people know it ... aren't you going to participate pcalin ?

if I am not mistaking the inequality is false!!!!

if we put x=y=z=1 we get that 0>=sqrt(2) which is false... just take a look at it.. perhaps I am wrong!

cheers!

actually the equality takes place - you must be wrong somewhere Lagrangia!

oh my... indeed .. I just "saw" the ineq like so:

[(x+y+z)/(1/x+1/y+1/z)]^(1/2) >= 1+[1+[(x^2+y^2+z^2)/(1/x^2+1/y^2+1/z^2)]^(1/2)]^(1/2).

oh my... I should pay more attention when reading a problem...

sorry!! :(

cheers! :D

Yes, I have just found a solution.

Let us use notation in elementary symmetric functions: p1=x+y+z, p2=yz+zx+xy, and p3=xyz and work backward. Then the proposed ineq is equivalent to

sqrt(p1*p2/p3) >= 1 + sqrt(1+sqrt(R)/p3)

where R=p1^2*p2^2-2*p1^3*p3-2*p2^3+4*p1*p2*p3. (Yes, this is boring, but I like it!)

=> sqrt(p1*p2/p3) - 1 >= sqrt(1+sqrt(R)/p3)   ... [squaring]
=> p1*p2/p3 + 1 - 2*sqrt(p1*p2/p3) >= 1 + sqrt(R)/p3
=> p1*p2 - 2*sqrt(p1*p2*p3) >= sqrt(R)   ... [squaring]
=> p1^2*p2^2 + 4*p1*p2*p3 - 4*p1*p2*sqrt(p1*p2*p3) >= p1^2*p2^2 - 2*p1^3*p3 - 2*p2^3 + 4*p1*p2*p3

[wow, two terms are cancelled out!]

=> p1^3*p3 + p2^3 >= 2*p1*p2*sqrt(p1*p2*p3)

And this is a simple application of the AM-GM ineq. Hope there is no typo in it.

Best regards,
-Eckard


calculus
linear algebra
complex analysis

I would have to say EE requires the most mathematics. Initially, the mathematics commonly used is single variable calculus. However, in the upper level courses, multivariable calculus is a must because you will be using Maxwell's four equations in the differential form.

Linear Algebra is also quite useful in dealing when analyzing circuits using KVL and KCL because often times, you get six equations and six unkowns or ten equations and unknowns. It is also useful in control theory.

Also, nonlinear differential equations, ODE, and PDE are used when dealing with nonlinear circuits and their elements, namely transistors and diodes.

The only other engineering major that I can think of that is heavy on mathematics is probably a mechanical engineering major concentrating in HVAC systems or fluid mechanics.

I know all this because I am getting my AS in Electrical Pre-Engineering.

I hope this helps

Thanks for sharing, but please don't [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=170568]double-post[/url].

Aerospace uses a lot of upper level applied math, not just single and multivariable calculus. Lagrangian mechanics and partial differential equations are used in Orbital mechanics and complex analysis is used in some wing theory classes (i.e. Conformal mapping).


Prove that $ \boxed {\ \tan 10^{\circ}\equal{}\tan 20^{\circ}\cdot\tan 30^{\circ}\cdot\tan40^{\circ}\ }$ .

[quote="Virgil Nicula"]Prove that $ \boxed {\ \tan 10^{\circ} \equal{} \tan 20^{\circ}\cdot\tan 30^{\circ}\cdot\tan40^{\circ}\ }$ .[/quote]
[hide="My proof."]
[b][u]Proof.[/u][/b] In the my opinion, this relation is nicely and interestingly !

$ \boxed {\begin{array}{c} \\
\ \tan 10^{\circ} \equal{} \tan 20^{\circ}\cdot\tan 30^{\circ}\cdot\tan40^{\circ} \\
 \\
\sin 10\cos 20\cos 30\cos 40 \equal{} \cos 10\sin 20\sin 30\sin 40 \\
 \\
\sin 10\cos 30 (\cos 60 \plus{} \cos 20) \equal{} \sin 30\cos 10(\cos 20 \minus{} \cos 60) \\
 \\
\sin 10\cos 30(1 \plus{} 2\cos 20) \equal{} \sin 30\cos 10(2\cos 20 \minus{} 1) \\
 \\
2\cos 20\sin 10\cos 30 \plus{} \sin 10\cos 30 \equal{} 2\cos 20\sin 30\cos 10 \minus{} \sin 30\cos 10 \\
 \\
\cos 30(\sin 30 \minus{} \sin 10) \plus{} \sin 10\cos 30 \equal{} \sin 30(\cos 30 \plus{} \cos 10) \minus{} \sin 30\cos 10 \\
\ \end{array}}$[/hide]
[b]Another proof ?![/b]

[b][u]A nice geometrical interpretation.[/u][/b] Let $ ABC$ be a triangle with the orthocenter $ H$ and $ B \equal{} 50^{\circ}$ , $ C \equal{} 70^{\circ}$ . Denote

$ \{D,E,F\}\subset (BC)$ so that $ AD\perp BC$ and $ \left\|\begin{array}{c} E\in (BD)\ ,\ F\in (DC) \\
 \\
m(\widehat {BAE}) \equal{} m(\widehat {CAF}) \equal{} 10^{\circ}\end{array}\right\|$ . Prove that $ [HDE]\equal{}[ADF]$ .

[b]Notation.[/b] $ [XYZ]$ - the [u]area[/u] of $ \triangle\ XYZ$ .


[quote="AndrewTom"]Evaluate $ \cos^{4} (\frac {\pi}{24}) \minus{} \sin^{4}(\frac {\pi}{24}).$[/quote]

Or

Since these identities are true:

$ \cos^4 \alpha \equal{} \frac{1}{8}(\cos(4\alpha)\plus{}4\cos(2\alpha)\plus{}3)$

$ \sin^4 \alpha \equal{}\frac{1}{8}(\cos(4\alpha)\minus{}4\cos(2\alpha)\plus{}3)$

Therefore

$ \cos^4 \alpha\minus{}\sin^4 \alpha \equal{}\cos2\alpha \implies  \cos 2\cdot \frac{\pi}{24}\equal{}\cos({\frac{\pi}{3}\minus{}\frac{\pi}{4}})\equal{}\frac{\sqrt{2}\plus{}\sqrt{6}}{4}$


Ring Theory
superior algebra
superior algebra unsolved

Let $ F_1$ be a commutative ring with unit and $ F_2$ a ring such that $ F_1\subset F_2$. Prove that $ \operatorname{char}F_1 \equal{} \operatorname{char}F_2$ without using homomorphisms.

(Edited: $ F_1,F_2$ are not necessary fields, stupid me. I edited the problem)

Well, uhm, the unity is same in both fields  :wink:

berti's reply is rude and unnecessary

I'm not sure if this is what the professor wants but I think that if one talks about $ \ker f$ then $ f$ is a homomorphism. Right?

Clearly we need some assumptions about the inclusion.  I'm assuming that $ F_1$ is a subring of $ F_2$ of course, but what if $ F_2$ does not have a $ 1$, or if $ {1_{F_2}\neq 1_{F_1}}$?  It doesn't seem unreasonable to me that in that case we might have $ \operatorname{char} F_2 > \operatorname{char} F_1$.

Yeah, the unitary ring $ F_2 \equal{} \mathbb{F}_2\times \mathbb{F}_3$ has char $ 6$ and unitary subring $ F_1 \equal{} \mathbb{F}_2\times 0$ of char $ 2$.
And of course any ring has the unitary subring $ 0$ of char $ 1$.


Let be given a sequence of real numbers $x_1,x_2,x_3,...$ and a real number $a$ satisfying:
$0\le a\le 1,x_ {n+1}=x_n+\frac{a-x_n^2}{2}$ $\forall n\ge 0$
Prove that:$0\le \sqrt{a}-x_n\le \frac{2}{n+1}$ $\forall n\ge 0$
$($notice:It's easy to show that $0\le \sqrt{a}-x_n$ and $0\le \sqrt{a}-x_n\rightarrow$ this sequence is stricly increasing$)$

Don't spam - if somebody wants to help then he'll surely do it - don't be impatient and wait

[quote="James Potter"][u][b]Oh sorry,sorry,sorry![/b][/u] :(
Let's change "sequence of real number $x_1,x_2,x_3,..$" into "sequence of real number $x_0,x_1,x_2...$"
And [color=red]$x_0=0$[/color]
(Thank you very much Virgil for discovering this mistake!!!)[/quote]
$\sqrt{a}-x_n=(\sqrt{a}-x_{n-1})(1-\frac{\sqrt{a}+x_{n-1}}{2})\le(\sqrt{a}-x_{n-1})(1-\frac{\sqrt{a}}{2})\le...\le\sqrt{a}.(1-\frac{\sqrt{a}}{2})^n\le\frac{2}{n}.(\frac{n}{n+1})^{n+1}<\frac{2}{n+1}$
It's from [b]Viet Nam Math olympiad 1989[/b]

Thank you [color=red][b]nthd[/b][/color],a friend from [color=red][b]Vietnam[/b][/color]
Thank you very very very very$^{1001}$ much


[quote="Altariel"]Eheheh... can anyone take a few minutes to confirm my inference : nats in Nationals,  right? :lol:  :?:   I'm really sorry but I have all the intelligence of a kindergardener...some say I [b][i][u]am[/u][/i][/b] one at heart.[/quote]

Yes, Nats is national-level. It's the most fun. You get to hang out with 227 great young maths for 72 hours or so.

from past experience, many of them die during Nats, so it's more like 127.

I will most likely die when and if I got to nats...waah.  but maybe my boyfriend can save me.  He's the one I was talking about in my first post.  ;) :lol: :blush:
Me make it to nationals...what a laugh. :rotfl:

Can someone describe it to me?
I want to know what it's like.

[quote="biffanddoc"][quote="guy whosaid this"]Can someone describe it to me?
I want to know what it's like.[/quote]
Its really fun. And scary to see how fast the countdown people are :o  :o  :o.[/quote]

heh.. and I'm afraid of the countdown people in my chapter...
Even though my chapter is very likely in the top few overall in the nation...
But... its scary. Very scary.

[quote="sirorange"]It's pretty much: 228 completely crazy kids, and their equally crazy coaches, and equally crazy parent's all converge on some unsuspectin hotel, in some American city, and are forced to use an elevator that goes up 40 floors in about 40 seconds, and gives me a headache. It's pretty much just a grand ol' time, with a few hours of competition, and after that, everyone just hangs out.[/quote]

So you're saying you're crazy?  :P

I don't think the hotel has 40 floors. It's not impossible, but I think 20 floors is enough to fit all 220 competitors (unless there's others of course).

Actually it had 72 floors. But it wasn't just for competitors - it was the Marriot at the Rennaissance Center in Detroit. It's not very wide actually...

That's pretty cool. In 2004, we stayed at the Omni in Washington, D.C. Doesn't sound as interesting as where you were. :( :P

[quote="sirorange"]The hotel had around 72 floors, and it was like a giant cylinder, heading up into the air. It was pretty crazy. Our room was directly across from where GM people were working, so we'd wave to them when we felt like it. They'd actually see us, too.[/quote]

And we got to admire their nice computers ;)

There were way too many elevators - to the number of 6 per floor. Certain of those crazy people roamed the floors getting on an elevator at one floor, pushing all the buttons between that floor and the next floor, and then getting off and doing it to another elevator. How annoying. :mad:

People also stayed up way too late, impairing their judgements the next day.

They also blew us through the wind tunnel - certain people climbed the blades of the propeller until our guide warned them about the possible consequences. We got to admire their nice computers, which were stuck in the old days of Windows, but some of us didn't get to experience their 3D simulator of a car, because the computer had to reboot and that took an hour. :(

Lots of other interesting things happened too. :)

Yea SOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOO OOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOO OOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOO many people got locked in the stairs (including me) and u had to exit and go in the main door :o


probability
function
integration
limit
calculus
real analysis
real analysis unsolved

Suppose $ p(x)$ is a continuous probability density function ($ \int_{0}^{\infty}p(t)dt\equal{}1$) such that
$ \int_{0}^{\infty}p(t)tdt < \infty$, then does the following equation hold? (where n is integer)
\[ \lim_{n\rightarrow\infty}\int_{0}^{\infty}n(\int_{t}^{\infty}p(x)dx)^{n \minus{} 1}p(t)tdt \equal{} 0\]
If it holds, then how to prove? If not, please give counterexample and indicate under what conditions the equation will hold.

Thanks a lot! :)

Try $ p(x) \equal{} e^{ \minus{} x}$ as a simple test case.

[color=green]Moderator edit: It never makes sense to quote the post directly preceeding yours in full! [/color]

It is true if $ \int_0^a p(x)\,dx>0$ for all $ a>0$ (and only then). Let $ F(t)\equal{}\int_t^\infty p(x)\,dx$. The integral in question is just $ \int_0^\infty nF(t)^{n\minus{}1}tp(t)\,dt$. For every $ a>0$, we have $ \max_{t\ge a}nF(t)^{n\minus{}1}\equal{} nF(a)^{n\minus{}1}\to 0$ as $ n\to\infty$ because $ F(a)<1$. Thus, it is enough to show that $ \int_0^a$ is small for large $ n$ when $ a$ is small. But $ t\le a$ on $ (0,a)$, so the integral in question is at most $ a\int_0^a nF(t)^{n\minus{}1}p(t)\,dt\equal{}\minus{}a\int_0^a d(F^n)\equal{}a(F(0)^n\minus{}F(a)^n)\le a$.


[quote="nonie"]1.   The local election is over and a new mayor has been elected to lead Big Town. The new mayor received two votes for every vote received by her opponent. The new mayor received 1,884 votes. How many votes did her opponent receive?

2.   Emily practices figure skating an average of 18 hours per week. Amber practices figure skating an average of 1.3 hours for every hour Emily practices. How many hours a week does Amber practice?

3.   Mr. Quirk had approximately 58% of the tests graded. (a) Written as a fraction with thirty-one as a denominator, approximately what fraction of the tests did he have graded? (b) What was the approximate graded test:ungraded test ratio?[/quote]
[hide]
1. 1884/2 = 942

2. 18 x 1.3 = 23.4

3.

a) 0.58 x 31 = 17.98, 58% is apporximately 18/31

b) graded to ungraded
18 : 13
[/hide]

Those are all right except for the last one...
[hide]3b) i've also got 29:21
Because 58/2=29 and (100-58)/2=21 I can see how you got 18:13 though[/hide]

[quote="nonie"]1.   The local election is over and a new mayor has been elected to lead Big Town. The new mayor received two votes for every vote received by her opponent. The new mayor received 1,884 votes. How many votes did her opponent receive?

2.   Emily practices figure skating an average of 18 hours per week. Amber practices figure skating an average of 1.3 hours for every hour Emily practices. How many hours a week does Amber practice?

3.   Mr. Quirk had approximately 58% of the tests graded. (a) Written as a fraction with thirty-one as a denominator, approximately what fraction of the tests did he have graded? (b) What was the approximate graded test:ungraded test ratio?[/quote]

[hide="Number One"]Solution for Number One:

Ratio = $2 : 1$

$\frac{1884}{2} = 942$

[hide="Number Two"]Solution for Number Two:

$\text{Amber Ratio: }1.3 : 1.0$

$1.3 \times 18 = 23.4 \text{ hrs. per week.}$

[hide="Number Three"]Solution to Number Three:

$\boxed{A} $

$\frac{58}{100} = \frac{x}{31}$

$100x = 58\times31$

$100x = 1798$

$x = 17.98$

$17.98 \text{ Is approx. 18}$

$\boxed{\text{The fraction over 31 is approximately: }18}$

$\boxed{B} $

$\text{Graded: 58  } \text{Ungraded: 100-58 or 42}$

$\boxed{\text{The Ratio is: }58 : 42 = 29 : 21}$

[/hide]
[/hide]
[/hide]

Oh, i got my answer because i thought you wanted it to have the denominator of 31...

[quote="nonie"]Oh, in that case it was definately correct! :)[/quote]
lol thanks  :D


Is Fibonacci numbers are never squares except 1? If it is, prove it.

The question which Fibonacci numbers are perfect powers (and related questions) are completely solved. There is an article on arxiv whose title I don't remember using modular methods, but if you accept them as a black box, the proof is completely elementary.


geometry
circumcircle
analytic geometry
complex numbers

How do you show that $ OI^2 \equal{} R^2 \minus{} 2Rr$ with complex numbers? ($ O$ is the circumcenter of a triangle $ ABC$, and $ I$ is its incenter. $ R$ is the circumradius, and $ r$ is the inradius).

Note: the complex coordinates of $ I$ are $ \dfrac{\alpha a \plus{} \beta b \plus{} \gamma c}{\alpha \plus{} \beta \plus{} \gamma}$, where $ a,b,c$ are the affices of $ A,B,C$ respectively and $ AB \equal{} \gamma, BC \equal{} \alpha,$ and $ CA \equal{} \beta$.

I am not that good at finding lengths with complex numbers.

If I see this solutions hopefully I can find more lengths between triangle centers.

Thanks!

$ \alpha^2\equal{}(b\minus{}c)\cdot (b\minus{}c)\equal{}b\cdot b\minus{}2b\cdot c\plus{}c\cdot c\implies b\cdot c\equal{}R^2\minus{}\frac{\alpha^2}{2}$

Then $ OI^2\equal{}\frac{1}{(\alpha\plus{}\beta\plus{}\gamma)^2}\left(  \sum \alpha^2\cdot a\cdot a\plus{}2\sum \beta\gamma b\cdot c\right)$

$ \equal{}\frac{1}{(\alpha\plus{}\beta\plus{}\gamma)^2}\left(  R^2\sum \alpha^2\plus{}2\sum \beta\gamma \left(R^2\minus{}\frac{\alpha^2}{2}\right)\right)$

$ \equal{}\frac{1}{(\alpha\plus{}\beta\plus{}\gamma)^2}\left(  R^2\sum \alpha^2\plus{}2\beta\gamma\minus{}\sum \beta\gamma \alpha^2\right)$

$ \equal{}R^2\minus{}\frac{\alpha\beta\gamma}{\alpha\plus{}\beta\plus{}\gamma}$

And $ 2rR\equal{}2\cdot\frac{A}{s}\cdot \frac{\alpha\beta\gamma}{4A}\equal{}\frac{\alpha\beta\gamma}{\alpha\plus{}\beta\plus{}\gamma}$ so we're done.

Thanks Alex! I need to remember that $ |a|^2 \equal{} a \cdot a$...


parameterization
calculus
calculus computations

i am having troubles with these problems :D

1. A backteria culture starts with 1000 bacteria and the growth rate is proportional the the number of bactera. After 2 hours the population is 9000.
a) Find an expression for the number of bacteria after t hours
b) Find the number of bacteria after 3 hours
c) In what period of time does the number of bacteria double?

2. An isotope of strontium, Sr90, has a half life of 25 years.
a) Find the mass of Sr90 that remains from a sample of 18mg after t years.
b) How long would it take for the mass to decay to 2mg?

3. A cup of coffee has a temperature of 200F and is in a room that has a temperature of 70F. After 10 min the temperature of the coffee is 150F.
a) What is the temperature of the coffee after 15 minutes?
b) When will the coffee have cooled to 100F?

These super scientific application problems are just not my thing

These are all in the same spirit. The law governing the process is exponential in each case and you are given two points on the curve. Your task is to find the parameters on the curve and to determine a couple more points on it. Show us what you have done and we'll give you a hint as to what to do next.


This may be in the wrong forum, but does anyone know how to prove the whole AM-GM inequality thingy??? :bruce:  :heli:

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

classic proof?

[quote="wonki_noh"]$AM\Rightarrow (a+b)/2$
$GM\Rightarrow square root(ab)$

$AM-GM$ $Inequality$ $states$ $that$
$(a+b)/2 \geq square root(ab)$
$(a+b)^{2}/4 \geq (ab)$
$a^{2}+2ab+b^{2}\geq 4ab$
$a^{2}-2ab+b^{2}\geq 0$
$(a-b)^{2}\geq 0$

This is true because square of any number is equal or greater than 0.[/quote]
That's only true for the n=2 case, I'm pretty sure he is looking for the general one.


algebra polynomial induction trigonometry

Determine all polynomials $ P(x)$ such that $ P(x^{2}+1)$ = $ (P(x))^{2}+1$ and $ P(0) = 0$

Is $ P$ $ \mathbb{R}\rightarrow\mathbb{R}$?

EDIT: sorry this doesn't work  :(

[hide="If not"]
Let $ x=i$. Then, we have $ 0 = P(0) = (P(i))^{2}+1 \implies P(i) = \pm i$. So, $ P(x) = \pm x$. Checking, we see that $ P(x) =-x \implies-x^{2}-1 = x^{2}+1$, a contradiction. However, $ P(x) = x \implies x^{2}+1 = x^{2}+1$, which works, so $ P(x) = x$ is the only solution.
[/hide]

[quote="not_trig"]$ P(i) = \pm i$. So, $ P(x) = \pm x$. [/quote]

I don't think this is right, what if $ P(x)=x^{8}-x^{4}\pm x$

Yes, sorry :blush: but that doesn't satisfy the original equation does it?

[hide="New proof"]We have $ P(0^{2}+1) = (P(0))^{2}+1 \implies P(1) = 1$, $ P(1^{2}+1) = P(1)^{2}+1 = 2 \implies P(2) = 2$, and if we keep going, we see that $ P(x) = x$ at infinitely many places. Then, consider $ F(x): = P(x)-x$, which must be identically zero because EDIT: $ F$ is a polynomial with infinitely many zeroes, so $ P(x) = x$. That's much more rigorous :blush:[/hide]

[quote="not_trig"]...because $ P$ is a polynomial with...[/quote]
I think the $ P$ must be $ F$..

I really don't see how "we keep going",and in my opinion this is not rigorous at all.Even if you manage to prove that $ f(n)=n$  $ \forall n\in \mathbb N$,using induction,it won't help much,because it isn't said that $ x$ is integer.

No, because if a polynomial has infinitely many zeroes, it is identically 0.

in other words,any polynomial,from a sufficiently big $ n$ or sufficiently small $ n$,becomes strickly increasing or strickly decreasing...
so if it doesnt and it had infinitly many constant point,for example infinitly many $ n$s s.t $ p(n)=c$,then we get that $ p \equiv c$
in this problem as not-trig said $ c=0$ so $ F \equiv 0$ so $ p(x)-x \equiv 0 \Rightarrow p(x)=x$


How do you rate yourself  by appearance(I wanted to know if all intelligent people are geeky looking :D )

none of them . I cant say "i look good" or "i look dumb" , people have to judje whether how you look

no smart-alec comments from you, shady   :dry:

I can't complain... but I won't call myself 'cute'. (but I would call the number 1 voters idiot :P)
In my opinion you are still missing the middle 80% of the spectrum in your poll ;)
(unless I again don't understand the subtle meaning of those words)

PS: you guys are really friendly to eachother ;)

There should be an option named "Average"

i look like this guy:
[hide] :D.  I actually don't know what i would rate myself.
[/hide]

I am as ugly as a treausre readied to be buried

Dont worry; only people in denial of their physical traits ask such superficial and narcissistic questions!  :D

Well, I am obviously out of this world..  :roll:

Edit: Actually, I am "outta this world" (thanks for correcting me bubs)

[quote="Peter VDD"]PS: you guys are really friendly to each other ;)[/quote]

we arent that great friends anyway, thats the reason for all of shady's (and sometimes my :D ) smart remarks

ehehhehehehe  :rotfl: G-UNIT, i think you hit the bull's eye , i'll make sure shady reads that ....   ;)

[quote="Arne"]According to my girlfriend I'm cute  :P[/quote]Nice :D

The weird thing is,many people voted otta this world the most :huh: weird... :)

Nah G-Unit,some fun questions shouldn't be taken at heart,it's okay ...i know what you would vote....
And thanks eminem for telling all that sh!t(sorry),forget about the books now,that's till where yoour friendship is ,shallow.

Why watch TV when you have AoPs, this is hilarious! :rotfl:   :rotfl:

I'm the hottest AZN in the entire district  :D  :P
( it's a very true fact! I get all the girls  :D )

( because all the AZN kids in my district are either, nice but okay looking or really dumb and showing off and really ugly. ) I can get you guys their pictures too! ( the ugly people I'm referring to )

I don't think a better opportunity will take place to ask whether i should finally do something about my extremely poor cut out picture, due to incompetence (or subconscient denial?)

i voted outta this world. and i mean it... ANd im not being egotistical or narcissistic i am just simply the best.

^^^yup pfft jk
eh i get a few girls here n there. but im nothin fancy,

[quote="eminem"][quote="shadysaysurspammed"]And thanks eminem for telling all that sh!t(sorry),forget about the books now,that's till where yoour friendship is ,shallow.[/quote]

oh yeahhhhhh?????$shallow$ huh??? And what about the time in the India forum , when i was getting my pants roasted off by some guys?? that time pretty $deep$ your @#%##% friendship was, huh ??? Fat lot of a friend you were at that time, even when i hadnt said anything to you!!

ok , (look folks, what a good and forgiving soul i am :D  :P )  lets forget it. we are quits now, ok ?it was a joke neway :roll:

and ummmmmm  about those books(especially ACOPS)??? :D[/quote]

Well I didn't help you in India forum cos i too used to use the same kind of language :rotfl:  :rotfl:  :rotfl: (Wanted to get the blame on someone elses shoulder...someone as fake as u :P )

Again came to your real self,patching up cos u need ACOPS?

And guys,guys,all of you voting outta this world,be serious,you can't compete against me....really,i feel sorry for u guys.

Shouldn't-have-been-born all the way!

Come on everyone!

haha, im one of like 1000000 azns in my city...

[quote="Sunny"][quote="shinwoo"]I'm the hottest AZN in the entire district  :D  :P
[/quote]

Hahaha, I'm the hottest AZN in my town....considering I'm the only one.   :D[/quote]

  But one can also make the argument that your the ugliest "AZN" in you town....considering you're the only one.    :D

  I think we can determine the moral of this story - dont make comparative statements about single-term sets.   :lol:

I know the moral of this sotry..

many people think way to highly of themselves lol

Jeez, I would have expected the average math lover to say at best average??

You are all joking right, you do not express such adoniscomplexes irl?

Shouldn't have been born. *Darned* glasses.  :D

Especially the fact that cute, shouldn't have been born, and outta this world are almost tied.

COME ON, VOTE "shouldn't have been born"!!!!

LOL THIS TOPIC IS SO FUNNY! I voted "shouldn't have been born" although I don't think I'm that bad  :?

Wooohoooo!!!It's all tied up...wow,i thought shouldn't have....would win by a 1065 votes.

I was confused what to vote for bcos i am cute+Outtathisworld...but i settled for the latter :D

I think I'm a decent looking chap.

some people may not think so but I think so.

There aren't enough options.

Also, "cute" and "outta this world" almost don't apply to the same subject, plus the latter sounds rather dumb. :P

[quote="Treething"][quote="MysticTerminator"][quote="uberl33tmage"]Wow so many vain ppl in this thread.

I look ok in jeans and short-sleeved shirts with spiked hair. Not many males would consider themselves cute. More would go for BADDD or handsome. I fall somewhere in between.

Edit: here's a pic
[img]http://adweek.blogs.com/photos/uncategorized/michaeljackson_4.jpg[/img]

I'm not dating yet, and I'm not the best looking male in my skewl. I'm the nicest though.[/quote]

dude you look exactly like that guy who molests kids[/quote]

god i'm trying to figure out whether you're being serious or not...[/quote]

what

[quote="MysticTerminator"][/quote][quote="Treething"][quote="MysticTerminator"][quote="uberl33tmage"]Wow so many vain ppl in this thread.

I look ok in jeans and short-sleeved shirts with spiked hair. Not many males would consider themselves cute. More would go for BADDD or handsome. I fall somewhere in between.

Edit: here's a pic
[img]http://adweek.blogs.com/photos/uncategorized/michaeljackson_4.jpg[/img]

I'm not dating yet, and I'm not the best looking male in my skewl. I'm the nicest though.[/quote]

dude you look exactly like that guy who molests kids[/quote]

god i'm trying to figure out whether you're being serious or not...[/quote][quote="MysticTerminator"]

what[/quote]

I wonder if this is worthy of Babylon yet...

[quote="uberl33tmage"]Wow so many vain ppl in this thread.

I look ok in jeans and short-sleeved shirts with spiked hair. Not many males would consider themselves cute. More would go for BADDD or handsome. I fall somewhere in between.

Edit: here's a pic
[img]http://adweek.blogs.com/photos/uncategorized/michaeljackson_4.jpg[/img]

I'm not dating yet, and I'm not the best looking male in my skewl. I'm the nicest though.[/quote]

You should definitely vote for outta this world...u don't look like from my planet earth :noo:

P.S- :omighty:  almighty   M.J  for getting a skin colour not in the crayon box...I tried to draw him out,just couldn't.

[quote="Valentin Vornicu"]If I see anymore posts quoting M.J.'s picture I'll start locking threads up! :) :police:[/quote]

why ? something so wrong with MJ?

I'm wearing glasses too...  :cool: since my hair is longet than "normal" I think I look better :)

I voted "Shouldn't have been born(No offence)". (really, I shouldn't have)

I laughed a lot upon seeing the presumably child-molester's picture quoted 6 posts in row  :rotfl:

[quote="perfect_radio"]
I laughed a lot upon seeing the presumably child-molester's picture quoted 6 posts in row  :rotfl:[/quote]

I'm thinking of making him my idol..... :|

I am no beauty. I frequently dont comb my hair, and I stink a lot.
But my friends like me, and they don't take long to be my friends.


function
induction
algebra unsolved
algebra

Find all $f: R\to R$ that
$f(x+f(xy))=f(x)+xf(y)$ for all $x,y\in{R}$

[quote="mathisfun1"]Once again, $f(x)=x$ works, but I do not know how to justify my answer.[/quote]
 :o How you saw that $f(x)=x$ works? :rotfl:

I actually have fairly little skill solving functional equations, so when I approach one I first check for simple functions such as $f(x)=kx$ or the like. Many complicated functional equations in competitions yield surprisingly simple answers such as the identity function. Of course， lucky guesses would probably only earn partial credit。

[quote="AYMANE"]$f=0$ is solution so suppose that $f\neq0$
Put  $f(0)=a$
$x=y=0$ gives $f(a)=a$
$y=0$  gives $f(x+a)=f(x)+ax$
$y=1$  gives $f(x+f(x))=f(x)+xf(1)$
$x=1$ gives $f(1+f(x))=f(1)+f(x)$
$f(2a)=f(a+a)=f(a)+a^2=a+a^2=f(a+f(a))=f(a)+af(1)=a+af(1)$ then $a=f(1)$ or $a=0$
1)	Suppose that $a\neq0$ then $a=f(1)$ and $f(1+f(x))=a+f(x)$
Put $g(x)=f(x)+1$ then $g(g(x))-g(x)=a$
Now consider the sequence $x_{1}=x$ & $x_{n+1}=g(x_{n})$ gives $g(x)=x+a$ and thus $f(x)=x+a-1$ but $f(0)=a-1=a$ impossible so $a=0$
2)	Put $f(1)=b$ then $f(1+f(x))=f(x)+b$ substitute $f$ by $g$ gives $g(g(x))-g(x)=b$
Consider the same sequence as above (subtitute $a$ by $b$) gives $g(x)=x+b$ then $f(x)=x+b-1$ but $f(0)=b-1=0$ then $f(x)=x$ and its indeed solution[/quote]
First if you want to prove that $f(0)=0$,here is a shorter way.
Take $x=y$ =>$f(x+f(x^2))=f(x)(x+1)$ =>$f(-1+f(1))=0$ so there is a number $t$ that $f(t)=0$,now we have that $f(x+f(x))=f(x)+xf(1)$,if $x=t$=>$tf(1)=0$,if $f(1)=0$ then take $y=\frac{1}{x}$ you will see that $f(\frac{1}{x})=0$=>$f=0$,in the other hand we have that $t=0$=>$f(0)=0$.

Now my question,can you explain why
  [quote]Consider the same sequence as above (subtitute $a$ by $b$) gives $g(x)=x+b$[/quote]? :?:

[quote="AYMANE"]For your first point, there are many ways to prove something ;)
For the second point take it like a new problem :P[/quote]
First,I know that ther are many ways,but I said that my way is shorter :P .
Second,I think your new problem isn't correct,how I know this kind of problems can't be solved without other conditions,like that $f$ is continiu,or increasing ;) .

[quote="cefer"][quote="modeler"]It seems that you assume f(x) is surjective, which I think is difficult to prove[/quote]
 Dear, modeler, i didn't say that $f(x)$ surjective and i didn't use this fact.I only iterated by $f$ every sides of equalities. ;)[/quote]

A little explanation:(@ modeler)

[b]1[/b]For any function if $a=b$ then obviously $f(a)=f(b)$, which is the idea cefer has used.
[b]2[/b]About injetive functions:A function is called injective if from $f(a)=f(b)$ it follows  that $a=b$.
That's all. ;)


Show that the longest angle bisector of a triangle is at least as long as its shortest median

.

I think that contradiction works well in this case , any hint Ligouras ?


EF parallel to BC
Prove that EK*DC=KF*BD
KF=2cm
EK=4cm
BC=9cm
Find BD.

Refer to the picture below.

EDIT: Oops. Apparently this will be too hard for this forum. Please move.

Yeah the proof isnt' hard but proofs are just not allowed in the basics.  :D

May be too hard, but I'd like to take a shot ( 4give me ) :D
[hide]
AKE and ABD are similar so BD=4k
AKF and ACD are similar so CD=2k

So,
EK*DC=8k
and
KF*BD=8k

Therefore
EK*DC=KF*BD

BD/EK=BC/EF
BD=(9*4)/6=6 cm[/hide]

Prototype, how do you know the two pairs of similar triangles use the same scale factor k?

Another way of looking at it :

AK:AD=KF:DC=EK:BD


[quote="symmetry"]I have a room that is 20 feet by 30 feet.
I also have tiles that are 12 inches by 12 inches.
How many tiles can fit inside the room?[/quote]
[hide]
Each tile is 1 square foot.

So there can fit $20*30=600$ tiles in the room.[/hide]

20 times 30 = 600 tiles = area of floor.

Is this problem really approprate for this forum?  :? Or am I missing something here?


linear algebra
matrix
inequalities
analytic geometry
linear algebra solved

assume that $A$ is n*n invertible matrix in$M(n,\mathbb{R})$ prove that :
$\frac{1}{||A^{-1}||_{\infty}} = min \{ \ ||A-B||_{\infty}: where\ \ B\ \ is\ \ singular\ \ matrix\ \}$
($||x||_{\infty}=max_{1\leq i\leq n} |x_i|,||A||_{\infty}=sup_{x\in \mathbb{R}^n-\{ 0 \} \ }\frac{||Ax||_{\infty}}{||x||_{\infty}} $

We denote the set of singular matrices by $\mathcal S$, and by $d(A,\mathcal S)$ the distance between $A$ and $\mathcal S$ in the metric induced by the norm we're working with. We want to show that $\|A^{-1}\cdot d(A,\mathcal S)\|=1$.

In order to prove $\|A^{-1}\|\cdot d(A,\mathcal S)\ge 1$, we use $\|A^{-1}\|\cdot\|A-B\|\ge \|I_n-A^{-1}B\|,\ \forall B\in\mathcal S$, so $\|A^{-1}\|\cdot d(A,\mathcal S)\ge d(I_n,\mathcal S)$, and it's easy to prove that the RHS is equal to $1$.

For the inverse inequality we observe that $\frac 1{\|A^{-1}\|}=\inf_{\|x\|=1}\|Ax\|$, so we want to show that $\inf_{\|x\|=1}\|Ax\|\ge \inf_{B\in\mathcal S}\sup_{\|x\|=1}\|(A-B)x\|$. In order to do this, it suffices to show that given a fixed $x_0$ with $\|x_0\|=1$, we can find $B_0\in\mathcal S$ s.t. $\sup_{\|x\|=1}\|(A-B_0)x\|\le\|Ax_0\|$. We may assume WLOG that $x_0^1=1$, and we can take $B_0$ to have columns $2,\ldots,n$ identical to the corresponding ones in $A$, and column $1$ equal to $-(x_0^2C_2+\ldots+x_0^nC_n)$, where $C_i$ are the columns of $A$, and $x_0^k$ is the $k$'th coordinate of $x_0$.


Got assigned this one a few days ago in cal; I tried several methods but I kept coming short somewhere. Egh here it is:

Using the balance-scale shown, tell how many ♦ are necessary to replace the ? in the last picture. Explain your reasoning.

▼♦________=_________○

▼_________=_________♦◘

○○_________=_________◘◘◘

▼__________=_________  ?

Thanks. I did something like that, annotated and made up equations but got lost somewhere along the lines...

Also can anyone recheck this problem (same sheet) for me?

If f(x)= (2x+1)/(x+1), find all the values of x for which f(1/x) - f(-x)= 3

I got X=0.77/x=0.43 which checks out in the latter stages but I might've forgot a sign here and there and got a different final eq.

Thanks again.

Omfg I knew I took a wrong turn with this one... Lol Instead of plugging (1/x) into 2X+1/(x+1) I plugged 2X+1/(X+1) into (1/x) doh... Egh, too tired.

How do you simplify to get 2 + 1/(x-1) though? I mean how do you get the 2 by himself?

I did this -2x+1/(-x+1) then factored out -1 for 2x-1/(x-1).


trigonometry
integration
real analysis
real analysis unsolved

I have a little question: how to compute nth Cesàro mean of the Fourier series for $ \sin(kx)$ and $ \cos(kx)$.

It would be so long if we use
\[ \sigma _n \left( f, x \right) \equal{} \frac{1}
{{2\pi n}}\int_0^\pi  {\frac{{\sin ^2 \frac{{nt}}
{2}}}
{{\sin ^2 \frac{t}
{2}}}\left( {f\left( {x \plus{} t} \right) \minus{} f\left( {x \minus{} t} \right)} \right)dt} .
\]

Just think a bit of what the Fourier series for $ \sin kx$ is, what the partial sums are, and how Chesaro sums are related to partial sums ;)


这里有参加物理竞赛的人吗？

[quote="dingdongdog"]这里有参加物理竞赛的人吗？[/quote]
我是竞赛小组的, 但不参加竞赛.

澳門還未開始有物理競賽...

我参加.
但由于主要搞数学, 不允许参加实验培训.
这次理论还行, 但加上实验就萎了, 以至于没有保送资格, 挂一个空头一等(我省高考竞赛不加分), 不爽死了!
要知道我平时成绩一塌糊涂, 年级前一百有时都保不住, 搞高考定萎.

哪有呀
什麼科的物理,我讀量子的
香港都沒有

江苏数学、物理、化学、生物、计算机竞赛都有
不过从我们这一届开始取消高考加分了 :(  :(

我們這裡能在競賽得獎的大多都靠保送.

[quote="Soarer"][quote="Jefflam"]哪有呀
什麼科的物理,我讀量子的
香港都沒有[/quote]
甚麼沒有？
香港兩年前開始參加物理竞賽。
不過在香港搞物理竞賽的都不搞數學竞賽
搞數學竞賽的都進不了物理的代表隊..[/quote]

兩年前我才開始讀
oxford3定理,怎知到

[quote="shobber"]物理竞赛中有许多大学里的知识, 但大多都是直接搬过来用, 没有证明. 所以往往会越学越觉得无理.[/quote]

那得自己去尝试证明

[quote="Jefflam"]
兩年前我才開始讀
oxford3定理,怎知到[/quote]
甚麼來的? 我只知道有Newton三定律.


geometry
circumcircle
trigonometry
geometric transformation
homothety
geometry proposed

[quote]Given triangle $ ABC$ and its circumcircle $ (O)$. Denote $ M_a, M_b, M_c$ the midpoints of $ BC, CA, AB$, respectively, $ P$ an arbitrary point inside $ \Delta ABC, A',B',C'$ the second intersections of $ AP, BP, CP$ and $ (O)$. Prove that $ (A'OM_a), (B'OM_b), (C'OM_c)$ concur at two points.[/quote]

Look like we have to invert something around here too!  :P . Here is my solution

Consider the inversion through pole $ O$, power $ k \equal{} r^2$, where $ r$ is the radii of $ (O)$. We have $ \mathcal {I}(O,k)$ maps $ A',$ $ B',$ and $ C'$ into themselves, $ M_a,$ $ M_b,$ $ M_c$ is mapped into $ A''$, $ B''$ and $ C''$ respectively, where $ A''$ is the intersection of tangents from $ B$ and $ C$ of $ (O)$, so as $ B''$ and $ C''$. Thus, under the inversion through pole $ O$ power $ k$, $ (OM_aA')$ $ \mapsto$ $ A''A'$, so as $ (OM_bB')$, $ (OM_cC')$. We claim that $ A''A'$, $ B''B'$ and $ C''C'$ are concurrent. $ (*)$.

Indeed, let $ A_1,$ $ B_1,$ $ C_1$ be the pole of $ A''A'$, $ B''B'$ and $ C''C'$, respectively. Then $ A_1$ is the intersection of the tangent at $ A'$ and $ BC$, so as  $ B_1$ and $ C_1$. Also note that $ \triangle A_1A''B$ $ \cong$ $ \triangle A_1CA''$. Thus,
\[ \frac {A_1B}{A_1C} \equal{} \frac {A_1B}{A_1A'}.\frac {A_1A'}{A_1C} \equal{} {\left (\frac {A'B}{A'C}\right )}^2 \equal{} {\left (\frac {\sin \angle BAP}{\sin \angle CAP}\right )}^2\]
The same argument holds for $ B_1$ and $ C_1$. Note that $ AP$, $ BP,$ and $ CP$ have already been concurrent. Therefore,
\[ \frac {A_1B}{A_1C}.\frac {B_1C}{B_1A}.\frac {C_1A}{C_1B} \equal{} 1\]
By [i]Menelaus theorem[/i], we obtain $ A_1,B_1,C_1$ are collinear, which implies $ A''A'$, $ B''B'$ and $ C''C'$ are concurrent. We obtain $ (*)$. Thus, the result of the problem is lead as follow. Our proof is completed then. $ \square$

[img]http://i524.photobucket.com/albums/cc322/khanhto-photo/NewBitmapImage.jpg[/img]
Denote by $ D,E,F,H,T,K$ the midpoints of $ OA',OB',OC',OA,OB,OC$ .

Let  $ M,N,Q$ be  the circumcenters of triangles $ OB'C',OC'A',OA'B'$.

The homothety $ H(O,\frac {1}{2})$ maps $ AA'\to HD,BB' \to TE, CC' \to KF$

So $ HD,TE,KF$ concur at the one point $ P'$ which is the midpoint of segment $ OP$.

Based on the [b]Steinbart's theorem[/b], we have $ MH,NT,QK$ are concurrent,i.e. two triangles $ MNQ,HTK$ are perpective.

Using the [b]Desargues' theorem [/b],we have the circumcenters of $ A'OM_a,B'OM_b,C'OM_c$ lie on a traight line.

On the other hand ,$ (A'OM_a),(B'OM_b),(C'OM_c)$ have a common point which is $ O$ ,so $ (A'OM_a),(B'OM_b),(C'OM_c)$ concur at two points.
@l[b]ivetolove212[/b]: Thank you!
 :)

Dear Mathlinkers,
perhaps a shorter way...

1. consider the tangentiel triangle A*B*C* of A'B'C'
2. according to the Steinbart theorem, A*B*C* and ABC are perspective
3. the perspectrix leeds to the fact that the centers of the tree circles are collinear and we are done...

Sincerely
Jean-Louis


Determine all continuous functions $f: \mathbb R \to \mathbb R$ such that:
\[f(2009f(x) \plus{} f(y)) \equal{} 2009x \plus{} y\]
holds true for all reals $x$ and $y$.

[quote="imedeen"]Determine continuous function $ f: R\to R$ such that:

$ f(2009f(x) \plus{} f(y)) \equal{} 2009x \plus{} y$ ,$ \forall x,y \in R$[/quote]

[hide="A solution"]
Let $ P(x,y)$ be the assertion $ f(2009f(x) \plus{} f(y)) \equal{} 2009x \plus{} y$
Let $ a \equal{} 2010f(0)$

$ P(0,0)$ $ \implies$ $ f(a) \equal{} 0$
$ P(a,a)$ $ \implies$ $ f(0) \equal{} 2010a \equal{} 2010^2f(0)$ and so $ f(0) \equal{} 0$ and $ a \equal{} 0$

$ P(0,x)$ $ \implies$ $ f(f(x)) \equal{} x$
$ P(f(x),0)$ $ \implies$ $ f(2009x) \equal{} 2009f(x)$
$ P(f(x),f(y))$ $ \implies$ $ f(2009x \plus{} y) \equal{} 2009f(x) \plus{} f(y)$ $ \equal{} f(2009x) \plus{} f(y)$

And so $ f(x)$ is a continuous solution of Cauchy's equation and so $ f(x) \equal{} cx$. Plugging back in the original equation, we get $ c^2 \equal{} 1$ and so two solutions :

$ f(x) \equal{} x$ $ \forall x$
$ f(x) \equal{} \minus{} x$ $ \forall x$
[/hide]


inequalities
function
algebra unsolved
algebra

Lemma: $ \forall x \in R_ \plus{} , 4x(1 \minus{} x^2) \le (1 \plus{} x^2)^2$, and the equality occurs when $ x \equal{} \sqrt2 \minus{} 1$.

Proof: Let $ f(x) \equal{} (1 \plus{} x^2)^2 \minus{} 4x(1 \minus{} x^2)$. We have $ f'(x) \equal{} 4x^3 \plus{} 12x^2 \plus{} 4x \minus{} 4 \equal{} 4(x \plus{} 1)(x^2 \plus{} 2x \minus{} 1)$. The roots of $ x^2 \plus{} 2x \minus{} 1$ are $ \minus{} 1 \plus{} \sqrt2$ and $ \minus{} 1 \minus{} \sqrt2$. Then the function is decreasing in $ (0, \sqrt2 \minus{} 1)$ and increasing in $ (\sqrt2 \minus{} 1, \infty)$. The minimum is therefore $ f(\sqrt2 \minus{} 1) \equal{} 0$. Moreover, $ x \not \equal{} \sqrt2 \minus{} 1 \Longrightarrow f(x) > 0 \Longrightarrow 4x(1 \minus{} x^2) \le (1 \plus{} x^2)^2$.

Now, observe that $ (x^2 \plus{} 1)^2 \ge 0$, which allows us to multiplicate the equations and get

$ \prod_{cyc} 4|x|(1 \minus{} x^2) \ge \prod_{cyc} (1 \plus{} x^2)^2$

But the lemma (that can be used for $ |x|$ since $ |x| \ge 0$) implies immediatly the inequality

$ \prod_{cyc} (1 \plus{} x^2)^2 \ge \prod_{cyc} 4|x|(1 \minus{} x^2)$

So we have equality and $ |x| \equal{} |y| \equal{} |z| \equal{} \sqrt2 \minus{} 1$. It is easy to see all triples satisfying this form a solution.

Can somebody find an elementary proof for the lemma?

Ah. Actually $ f(x) \equal{} (x^2 \plus{} 2x \minus{} 1)^2$.


induction
algebra
polynomial
factorial
combinatorics
Finite Differences

Given positive integers $k, m, n$ such that $1 \leq  k \leq  m \leq  n$. Evaluate

\[\sum^{n}_{i=0} \frac{(-1)^i}{n+k+i} \cdot \frac{(m+n+i)!}{i!(n-i)!(m+i)!}.\]

[quote="suweijie"]The solution is just zero. :D
  orl,I wonder how can you find the problems of China TST?
Though I am Chinese,most of the China TST problems I can't find.
Please tell me the way that you find the problems. :)[/quote]

You can go to visit the old site of mathfan http: http://www.mathfan.net

[quote="orl"]Given positive integers $ k, m, n$ such that $ 1 \leq k \leq m \leq n$. Evaluate
\[ \sum^{n}_{i \equal{} 0}\frac {1}{n \plus{} k \plus{} i}\cdot \frac {(m \plus{} n \plus{} i)!}{i!(n \minus{} i)!(m \plus{} i)!}.\]
[/quote]

Sure that a closed form exists?

[b]EDIT:[/b] I see that there is a typo in the problem. It should be:

[color=blue][b]Corrected problem.[/b] Given integers $ k$, $ m$, $ n$ such that $ 1 \leq k \leq m \leq n$. Show that $ \sum^{n}_{i \equal{} 0}\frac {\left( \minus{} 1\right)^{i}}{n \plus{} k \plus{} i}\cdot \frac {(m \plus{} n \plus{} i)!}{i!(n \minus{} i)!(m \plus{} i)!} \equal{} 0$.[/color]

[i]Solution.[/i] We will use the following fact from the theory of finite differences (this fact can be easily shown by induction over $ n$):

[color=blue][b]Theorem 1.[/b] If $ n$ is an integer, and $ P$ is a polynomial of degree $ < n$, then

$ \sum_{i \equal{} 0}^{n}\left( \minus{} 1\right)^{i}\binom{n}{i}P\left(i\right) \equal{} 0$.[/color]

Now, since $ 1\leq k \plus{} n \minus{} m\leq n$ (in fact, $ 1\leq k \plus{} n \minus{} m$ follows from $ 1\leq k$ and $ 0\leq n \minus{} m$, the latter being a consequence of $ m\leq n$, and $ k \plus{} n \minus{} m\leq n$ follows from $ k \minus{} m\leq 0$, what is because $ k\leq m$), the polynomial $ P$ defined by $ P\left(X\right) \equal{} \frac {1}{n!}\cdot\prod_{1\leq i\leq n;\ i\neq k \plus{} n \minus{} m}\left(X \plus{} i \plus{} m\right)$ has degree $ < n$ (in fact, its degree is $ n \minus{} 1$, since it is the product of the constant factor $ \frac {1}{n!}$ and the $ n \minus{} 1$ linear factors $ X \plus{} i \plus{} m$ for $ i\in\left\{1,2,...,n\right\}\backslash\left\{k \plus{} n \minus{} m\right\}$). Hence, Theorem 1 yields

[color=green][b](1)[/b][/color] $ \sum_{i \equal{} 0}^{n}\left( \minus{} 1\right)^{i}\binom{n}{i}P\left(i\right) \equal{} 0$.

Now,

$ P\left(X\right) \equal{} \frac {1}{n!}\cdot\prod_{1\leq i\leq n;\ i\neq k \plus{} n \minus{} m}\left(X \plus{} i \plus{} m\right) \equal{} \frac {1}{n!}\cdot\frac {\prod_{1\leq i\leq n}\left(X \plus{} i \plus{} m\right)}{X \plus{} \left(k \plus{} n \minus{} m\right) \plus{} m}$
$ \equal{} \frac {1}{n!}\cdot\frac {\prod_{1\leq i\leq n}\left(X \plus{} i \plus{} m\right)}{n \plus{} k \plus{} X} \equal{} \frac {1}{n \plus{} k \plus{} X}\cdot\frac {\prod_{1\leq i\leq n}\left(X \plus{} i \plus{} m\right)}{n!}$
$ \equal{} \frac {1}{n \plus{} k \plus{} X}\cdot\frac {\prod_{1\leq i\leq n}\left(X \plus{} n \plus{} m \minus{} \left(n \minus{} i\right)\right)}{n!}$
$ \equal{} \frac {1}{n \plus{} k \plus{} X}\cdot\frac {\prod_{0\leq i\leq n \minus{} 1}\left(X \plus{} n \plus{} m \minus{} i\right)}{n!}$ (here we substituted $ n \minus{} i$ for $ i$ in the product)
$ \equal{} \frac {1}{n \plus{} k \plus{} X}\cdot\binom{X \plus{} n \plus{} m}{n}$ (since $ \frac {\prod_{0\leq i\leq n \minus{} 1}\left(X \plus{} n \plus{} m \minus{} i\right)}{n!} \equal{} \binom{X \plus{} n \plus{} m}{n}$).

Thus, [color=green][b](1)[/b][/color] becomes

$ \sum_{i \equal{} 0}^{n}\left( \minus{} 1\right)^{i}\binom{n}{i}\cdot\frac {1}{n \plus{} k \plus{} i}\cdot\binom{i \plus{} n \plus{} m}{n} \equal{} 0$.

Hence,

$ 0 \equal{} \sum_{i \equal{} 0}^{n}\left( \minus{} 1\right)^{i}\binom{n}{i}\cdot\frac {1}{n \plus{} k \plus{} i}\cdot\binom{i \plus{} n \plus{} m}{n}$
$ \equal{} \sum_{i \equal{} 0}^{n}\left( \minus{} 1\right)^{i}\cdot\frac {n!}{i!\left(n \minus{} i\right)!}\cdot\frac {1}{n \plus{} k \plus{} i}\cdot\frac {\left(i \plus{} n \plus{} m\right)!}{n!\left(\left(i \plus{} n \plus{} m\right) \minus{} n\right)!}$
$ \equal{} \sum_{i \equal{} 0}^{n}\left( \minus{} 1\right)^{i}\cdot\frac {n!}{i!\left(n \minus{} i\right)!}\cdot\frac {1}{n \plus{} k \plus{} i}\cdot\frac {\left(m \plus{} n \plus{} i\right)!}{n!\left(m \plus{} i\right)!}$
$ \equal{} \sum_{i \equal{} 0}^{n}\frac {\left( \minus{} 1\right)^{i}}{n \plus{} k \plus{} i}\cdot\frac {\left(m \plus{} n \plus{} i\right)!}{i!\left(n \minus{} i\right)!\left(m \plus{} i\right)!}$,

and the problem is solved.

[b]PS.[/b] @Orl: The transformation $ n\mapsto n!$ called factorial, not faculty. :P

  Darij

We have:

$\displaystyle S_{n,m,k}=\sum_{i=0}^n\dfrac{(-1)^i}{n+k+i}\cdot\dfrac{(m+n+i)!}{i!(n-i)!(m+i)!}=$
$=\sum_{i=0}^n\dfrac{(-1)^i(m+n+i)...(n+k+i+1). (n+k+i-1)!}{i!(n-i)!(m+i)...(k+i+1).(k+i)!}$

$\displaystyle\quad =\sum_{i=0}^n\dfrac{(m+n+i)^{\underline{m-k}}}{(m+i)^{\underline{m-k}}}\cdot \dfrac{(-1)^i{n\choose i}{n+k+i-1\choose n-1}}{n}$

*Note: $(x)^{\underline{n}}=x(x-1)...(x-n+1)$

Now, We caculate differences

$\Delta\left[\dfrac{(m+n+i)^{\underline{m-k}}}{(m+i)^{\underline{m-k}}}\right]=\dfrac{(n+k+i+2)...(m+n+i+1)}{(k+i+2)...(m+i+1)}-\dfrac{(n+k+i+1)...(m+n+i)}{(k+i+1)...(m+i)}$

$\quad =n(k-m)\cdot\dfrac{(m+n+i)^{\underline{m-k-1}}}{(m+i+1)^{\underline{m-k+1}}}$

and

$\Delta\left[(-1)^{i-1}.{n-1\choose i-1}{n+k+i-1\choose n}\right]=\dfrac{n+k}{n}(-1)^i.{n\choose i} {n+k+i-1\choose n-1}$

Therefor by [url=http://en.wikipedia.org/wiki/Summation_by_parts]Summation by parts[/url], we have:

$\displaystyle S_{n,m,k}=\dfrac{1}{n+k}\sum_{i=0}^n \dfrac{(m+n+i)^{\underline{m-k}}}{(m+i)^{\underline{m-k}}}\Delta\left[(-1)^{i-1}.{n-1\choose i-1}{n+k+i-1\choose n}\right]$

$\displaystyle = \dfrac{1}{n+k}\left.\left[\dfrac{(m+n+i)^{\underline{m-k}}}{(m+i)^{\underline{m-k}}}\cdot (-1)^{i-1}.{n-1\choose i-1}{n+k+i-1\choose n}\right]\right|_{i=0}^{n+1}$

$\quad +\dfrac{n(m-k)}{n+k}\sum_{i=0}^n (-1)^{i}.{n-1\choose i}{n+k+i\choose n} \cdot\dfrac{(m+n+i)^{\underline{m-k-1}}}{(m+i+1)^{\underline{m-k+1}}}$

$\displaystyle =\dfrac{(m-k)}{n+k}\sum_{i=0}^{n-1} \dfrac{(-1)^i(m+n+i)!}{(n+k+i+1) i!(n-1-i)!(m+i+1)!}$

$\Rightarrow S_{n,m,k}=\dfrac{m-k}{n+k}S_{n-1,m+1,k+2}=\dfrac{m-k}{n+k}\cdot\dfrac{m-k-1}{n+k+1}S_{n-2,m+2,k+4}=...=$

$=\dfrac{(m-k)(m-k-1)...(m-k-n+1)}{(n+k)(n+k+1)...(2n+k-1)}S_{0,m+n,k+2n}$

inside

$S_{0,m+n,k+2n}=\dfrac{(-1)^0 (m+n+0+0)!}{(0+k+2n+0) 0!(0-0)!(m+n+0)!}=\dfrac{1}{2n+k}$

Result: $\boxed{\displaystyle\sum_{i=0}^n\dfrac{(-1)^i}{n+k+i}\cdot\dfrac{(m+n+i)!}{i!(n-i)!(m+i)!}=\dfrac{(m-k)^{\underline{n}}}{(n+k)_{n+1}}}$

*Note: *Pochhammer symbol*: $(x)_n=x(x+1)...(x+n-1)$

Here's a *very* different (and possibly overkill) solution using calculus. We have
\begin{align*}
\sum_{i=0}^n\frac{(-1)^i}{n+k+i}\cdot\frac{(m+n+i)!}{i!(n-i)!(m+i)!} &= \sum_{i=0}^n\frac{(-1)^i(m+n+i)!}{i!(n-i)!(m+i)!}\int_0^1x^{n+k+i-1}\;dx \\
&= \int_0^1\left[\sum_{i=0}^n\binom{n}i\binom{m+n+i}n(-1)^ix^i\right]x^{n+k-1}\;dx \\
&= \int_0^1\left[\sum_{i=0}^n\binom{n}i(-1)^ix^i\cdot\frac1{n!}\frac{\partial^n}{\partial y^n}y^{m+n+i}\biggr\rvert_{y=1}\right]x^{n+k-1}\;dx \\
&= \frac1{n!}\frac{\partial^n}{\partial y^n}\left[\int_0^1\sum_{i=0}^n\binom{n}i(-1)^ix^{n+k+i-1}y^{m+n+i}\;dx\right]_{y=1} \\
&= \frac1{n!}\frac{\partial^n}{\partial y^n}\left[\int_0^1x^{n+k-1}y^{m+n}(1-xy)^n\;dx\right]_{y=1} \\
&= \frac1{n!}\frac{\partial^n}{\partial y^n}\left[y^{m-k}\int_0^yu^{n+k-1}(1-u)^n\;du\right]_{y=1} \\
&= \frac1{n!}\sum_{j=0}^n\binom{n}j\left[\frac{\partial^j}{\partial y^j}y^{m-k}\right]_{y=1}\left[\frac{\partial^{n-j}}{\partial y^{n-j}}\int_0^yu^{n+k-1}(1-u)^n\;du\right]_{y=1}
\end{align*}
Now if $j < n$, the second factor (due to the Fundamental Theorem of Calculus) is
$$\left[\frac{\partial^{n-j-1}}{\partial y^{n-j-1}}y^{n+k-1}(1-y)^n\right]_{y=1}$$
which is zero because we are differentiating at most $n-1$ times, and there is a zero of order $n$ at $y=1$. And if $j=n$, the first factor is zero because $m-k < m \le n$. So all the terms of the sum vanish, and thus the sum is zero. $\blacksquare$


probability
LaTeX
real analysis
real analysis unsolved

hi ,
 i am studying a proof of poissons limit theorem , stating :
if  m is a fixed integer, and pn is a positive sequence under one (it is a probabilty) that converges to zero if n goes to infinity

then the limit of
 [  pn^m*(1-pn)^m*binomial(n,m)   -   exp(-an)/m!*an^m  ] is n goes to infinity is zero
an =n*pn

now this theorem is quite easy if u make the natural assumption : pn=lambda/n, so an is always simply lambda

however, when working it out in general, u get :

an^m*(1-pn)^n/m!  *  (1-1/n)*....(1-(m-1)/n )    /(1-pn)^m

the right part : (1-1/n)*....(1-(m-1)/n )    /(1-pn)^m
indead goes to one, but is that a good argument to work on with simply
the left part an^m*(1-pn)^n/m!

i was never taught anything like this, compare with this sequence :

n=(1+1/n)*n*n  -n*n
1+1/n goes to one, so i get n*n-n*n=0
so i would have proven n goes to zero!! , obviously boundedness is an issue somewhere

sorry for this complicated question , i'm interested in all remarks!
 :?

fredbel, i think you consider the equality

$n=(1+\frac 1n)n^2 - n^2$

and what's happening as $n\to\infty$. now, you can't take any partial limits as $lim 1+\frac 1n=1$ if the other things do not converge! that way, the limit is clearly wrong.

indead peter that's why i gave that example
but then that error occurs too in my proof of poisson

and there, there is a subtlety : because what u get is correct in the end!

hmm is it possible that you write your proof again in latex? maybe then, i'll understand what you mean...


inequalities
induction
inequalities unsolved

If $ a_i, b_i, c_i$ are sequences of positive numbers for $ i \equal{} 1,2,\cdots,n$, prove the following inequality:

$ \sum (a_i\plus{}b_i\plus{}c_i) \sum \frac{a_ib_i \plus{} b_ic_i\plus{} c_ia_i}{a_i\plus{}b_i\plus{}c_i} \sum \frac{a_ib_ic_i}{a_ib_i \plus{} b_ic_i\plus{} c_ia_i} \leq \sum a_i \sum b_i \sum c_i$

where all summations are taken from $ i\equal{}1$ to $ i\equal{}n$.

Sorry if I'm answering my own question,
but I found that this problem becomes easy once you've proved the following lemma:

$ \sum \frac {x_i y_i}{x_i \plus{} y_i} \leq \frac {\sum x_i \sum y_i }{\sum x_i \plus{} \sum y_i}$

In order to prove that lemma, I first proved it for $ n \equal{} 2$, by expanding and cancelling terms, then you get $ (x_1y_2 \minus{} x_2y_1)^2 \geq 0$
Then induction from $ n$ to $ 2n$, and from $ n$ to $ n \minus{} 1$.

I haven't found any easier proof to the lemma.

What strikes me is that, the lemma can be rewritten as follows:
$ \sum \frac {1}{1/x_i \plus{} 1/y_i} \leq \frac {1}{1/ \sum x_i \plus{} 1/\sum y_i}$
which kinda looks like Minkowski except we can't use negative powers in Minkowski :(

[quote="hendrata01"]
What strikes me is that, the lemma can be rewritten as follows:
$ \sum \frac {1}{1/x_i \plus{} 1/y_i} \leq \frac {1}{1/ \sum x_i \plus{} 1/\sum y_i}$
which kinda looks like Minkowski except we can't use negative powers in Minkowski :([/quote]
But of course we can. (Not to say "Yes we can"  :D ) If you put $ r \equal{} 1,s \equal{} \minus{} 1$ and $ m \equal{} 2$ [url=http://www.artofproblemsolving.com/Wiki/index.php/Minkowski_Inequality]here[/url], you obtain exactly the lemma.  :)

And it is the same as Tang Zy's one [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=313265]here[/url]. Just put $ a_i \equal{} \frac1{x_i}$ and $ b_i \equal{} \frac1{y_i}.$

Yup I saw the link on Minkowski, I just never knew we could use negative powers :)

I was actually able to prove the lemma by just using Cauchy, I wrote it here: http://dharmath.thehendrata.com/2009/11/19/3-sequence-inequality/


Does anyone know how to create images, specifically on LaTeX?

Please explain what you mean with an example. Do you mean the examples at [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=122562]Clever ideas in LaTeX[/url] or do you need [[Asymptote_%28Vector_Graphics_Language%29]] or [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=53217]Ultimate guide to inserting Pictures in LaTeX[/url] or what?

As in clever ideas but it has to move around as a whole not pieces.

I still don't understand what you mean. What has to [i]move around[/i]? What are [i]pieces[/i]? Please gve an example to show exactly what you mean.

Say I want to draw a spidron in LaTeX, but then I want to be able to move it around without moving the individual pieces.

Sounds like you want animation. Make a movie with whatever program you like then include it in your pdf by using the [url=http://tug.ctan.org/cgi-bin/ctanPackageInformation.py?id=movie15]movie15 multimedia inclusion package[/url].


[b]Let[/b] $ A,X \in M_{n} \mathbb{(R)}$ , such that :
 $ \sum_{k = 1}^{\infty} {(XA)^{k}} = 0$ , $ \forall X \in M_{n} \mathbb{(R)}$ ,$ -1< a_{ij} <1$ ,$ \forall i,j \in$ {1,2,3,...,n}
[b]Prove that[/b] : [b]  A=0[/b]
"..[b]Do math to love math , love math must do math , do math , love math , math is my life [/b]!!! "
Mathlinks.ro is my friendly !!!

I'm a little confused about the condition--are we assuming that the sum always exists?  Seems like an extremely strong assumption.

Letting $ X\equal{}I$, we find that for any complex eigenvalue $ \lambda$ of $ A$, $ 0$ has an eigenvalue $ \frac{\lambda}{1\minus{}\lambda}$, so $ \lambda\equal{}0$.

If we are only allowed to assume that the sum is zero when it exists, we can do something similar: let $ X\equal{}c I$, where $ c$ is small.

eassy !!!
[b]Let[/b] $ X \equal{} E_{ij}$ here $ E_{ij} \equal{} (e_{kl})_{nxn}$
$ e_{kl} \equal{} 1$ [b]If k=i , l=j[/b],$ e_{kl} \equal{} 0$ if $ k \neq i ,l \neq j$
Since : [b] XA have row i column j is [/b] : $ (a_{j1},a_{j2},...,a_{jn})$
           [b]$ (XA)^{2}$ have row i  is [/b] : $ (a_{ji}.a_{j1},a_{ji}.a_{j2},...,a_{ji}.a_{jn})$
           .....
           [b]$ (XA)^{k}$ have row i  is [/b] : $ (a^{k \minus{} 1}_{ji}.a_{j1},a^{k \minus{} 1}_{ji}.a_{j2},...,a^{k \minus{} 1}_{ji}.a_{jn})$
          [b]$ \sum_{k \equal{} 1}^{\infty} {(XA)^{k}}$ have row i  is [/b] : $ (\frac {a_{j1}}{1 \minus{} a_{ji}},...,\frac {a_{jn}}{1 \minus{} a_{ji}})$
Loking for !!!!
[b]"..I will try to  become a good student of Jean Marie Monier ... ""[/b]


Asymptote algebra polynomial algorithm function

The asymptotes of $ y \equal{} \frac{x^4 \plus{} 3x^3 \minus{} 15x^2 \minus{} 19x \plus{} 30}{x^3 \minus{} x^2 \minus{} 30x \plus{} 72}$ are in the form $ y \equal{} a, y \equal{} b, y \equal{} cx \plus{}d$. Find the value of $ a \plus{} b \plus{} c \plus{} d$.

I keep getting the wrong answer, unless the answer key is wrong.
[hide="answer"]given answer is 11, but I keep getting 8...[/hide]

I would appreciate any help. Thanks.

$ y \equal{} \frac {x^4 \plus{} 3x^3 \minus{} 15x^2 \minus{} 19x \plus{} 30}{x^3 \minus{} x^2 \minus{} 30x \plus{} 72}$

$ \equal{} x \plus{} 4 \plus{} \frac {19x^2 \plus{} 29x \minus{} 258}{x^3 \minus{} x^2 \minus{} 30x \plus{} 72}$

$ \equal{} x \plus{} 4 \plus{} \frac {(x \minus{} 3)(19x \plus{} 86)}{(x \minus{} 3)(x \plus{} 6)(x \minus{} 4)}$, so the asymptotes are

$ y \equal{} x \plus{} 4$, $ x \equal{} \minus{} 6$, and $ x \equal{} 4$, so our answer is

$ 1 \plus{} 4 \plus{} \minus{} 6 \plus{} 4 \equal{} \boxed3$?

What did I do incorrectly?

This is what I did: I crossed out the (x+2) that you can factor from the numerator and denominator and was left with $ \frac{x^3 \plus{} x^2 \minus{} 17x \plus{} 15}{x^2 \minus{} 3x \minus{} 24}$, from which we see that $ a \plus{} b \equal{} 3$ (the sum of the roots should be the sum of the x values of the asymptotes). I also got that the horizontal asymptote is $ x \plus{} 4$ after dividing, which is how I got $ 3 \plus{} 1 \plus{}4 \equal{} \boxed 8$, which, according to the solution key (which was right for all the other questions), was wrong

Math: Let that ugly polynomial division that I am lazy to write be written in the division algorithm form $ P(x) \equal{} (x \plus{} 4)Q(x) \plus{} R(x)$. I am not that good with asymptotes, but I get the feeling that taking the remainder part $ R(x)$ and making a function out of it $ y\equal{}R(x)$ does not find the asymptotes of the original $ \frac{P(x)}{Q(x)}$.

But there is no factor of $ x \plus{} 2$ in the denominator: $ D( \minus{} 2) \equal{} 120$...?

Might you have typoed? (I have also assumed the denominator was $ x^3 \minus{} x^2 \plus{} 30x \plus{} 72$ (of which $ x \plus{} 2$ is a factor), but it doesn't give the same results as you have).

Also, why do you disagree with my method (other than arithmetic errors, etc.)? It's the same as your method in that we find all zeroes $ a$ of the denominator (that are not zeroes of the numerator as well) to deduce that $ x\equal{}a$ is an asymptote.

[quote="hurdler"]The curve $ C$ has equation $ y \equal{} \frac{x^4 \plus{} 3x^3 \minus{} 15x^2 \minus{} 19x \plus{} 30}{x^3 \minus{} x^2 \minus{} 30x \plus{} 72}$. The asymptotes of $ C$ are in the form $ y \equal{} a, y \equal{} b, y \equal{} cx \plus{} d$. Find the value of $ a \plus{} b \plus{} c \plus{} d$.

I keep getting the wrong answer, unless the answer key is wrong.
[hide="answer"]given answer is 11, but I keep getting 8...[/hide]

I would appreciate any help. Thanks.[/quote]

[hide="method"] I have added some details to the quoted text above for convenience.

This is a typical algebra problem.

$ y \equal{} cx \plus{} d$ is the equation of the oblique asymptote of the curve $ C$; to find the equation, simply divide the numerator of your curve $ C$ by your denominator and discard the remainder, leaving your quotient $ Q$. $ y\equal{}Q$ is the equation of your oblique asymptote. [You could use long division for this; just be careful about careless mistakes because those numbers look pretty awful]

I think you've typed something wrongly because since the degree of the numerator is one more than the degree of the denominator, there are no horizontal asymptotes; there exist only an oblique asymptote.

Do you mean $ x\equal{}a,x\equal{}b$? Note that $ a,b$ are the roots of your denominator.

[/hide]

Oops I typoed, I meant that $ x \equal{} a,$ and $ x \equal{} b$ are asymptotes. Sorry for the confusion.

edit: oops I made an arithmetic error. OK, I think that you are right, math. Thanks for the help, guys. The answer key is not perfect I guess.


I have found the solution, but I hope you will have too. :D  :D