vector
superior algebra
superior algebra unsolved

First of all , how we can define the operation of (Addition)$ \plus{}$ and (Multiplication ) $ \cdot$ on $ \mathbb{R}^3$  s.t. $ (\mathbb{R}^3,\plus{},\cdot)$ is ring?

How we can define that in general for $ \mathbb{R}^n$ such to have $ (\mathbb{R}^n,\plus{},\cdot)$ Ring?

 :ninja:

Take a bijection of $ \mathbb R^n$ to $ \mathbb R$. As $ \mathbb R$ is a field, you can take your operations from there.

Give me such an example for some bijection please .

And how can I do it directly ZetaX ? I think there is some way...

Thank you

Let + be standart addition on $ \mathbb{R}^n$ as on vector space, for multiplication we can use
structure constants: if $ e_i$ is a basis
\[ e_i \cdot e_j \equal{} \sum_k C^k_{i j} e_k\]
so if we have any such constants $ C^k_{i j}$ we have multiplication on $ \mathbb{R}^n$.

not any choice of these constants works, and as I said we may put $ C_{ij}^k \equal{} \delta_{ijk}$.


Brianna is using part of the money she earned on her weekend job to buy several equally-priced CDs. She used one fifth of her money to buy one third of the CDs. What fraction of her money will she have left after she buys all the CDs?

$ \textbf{(A)}\ \frac{1}{5} \qquad
\textbf{(B)}\ \frac{1}{3} \qquad
\textbf{(C)}\ \frac{2}{5} \qquad
\textbf{(D)}\ \frac{2}{3} \qquad
\textbf{(E)}\ \frac{4}{5}$

[hide="Click for solution"]
The number of CDs that Brianna will finally buy is three times the number she has already bought.  The fraction of her money required for all purchases is $ 3 \left (\frac{1}{5} \right )\equal{}\frac{3}{5}$, so she has $ 1\minus{}\frac{3}{5}\equal{}\frac{2}{5}$ left, or $ \boxed{\textbf{(C)}}$.
[/hide]

I think all these questions would be better suited in the Wiki. I don't think this is really a question because the poster can't solve it.


function
real analysis
real analysis unsolved

Let f be a real-valued function defined and continuous on the set of real numbers R.  For the set S={f(x) : 0<x<1}

S is connected? why or why not?
S is an open subset of R? why or why not?
S is a bounded subset of R? why or why not?

If For Set S={f(x): 2<=x<=4}
For Set S={f(x): 4<x<=5}

The same responses?

For the original question: it is important that $f$ is continuous on $\mathbb{R},$ not merely continuous on $(0,1).$ In particular, $f$ is contininuous on $[0,1].$

$S$ is connected because the continuous image of a connected set is connected. (At heart, this is the Intermediate Value Theorem.)

$S$ need not be open. Consider $f(x)=\sin(10x).$ In that case, we would have $S=[-1,1].$

$S$ is bounded, because it is contained in the image of $[0,1]$ and the continuous image of a compact set is compact.

For the closed and half-open intervals of your variations on this question, the answers would be the same: connected and bounded but not necessarily open. (But the image of the closed interval would necessarily be closed.)

Could I ask why the continous image of a compact set is compact?

The lowest and simplest form of this theorem is the "Extreme Value Theorem": a continuous real-valued function on a closed bounded interval is bounded and assumes its maximum and minimum values. Have you seen a proof of that?

Yes, Yes I see. The extreme value theorem can be prooved by bounded value theorem.  Thanks a lot!


We are in topological space $X$.

$U,V$  open sets

$cl$ -  closure operator

Is it true that if $U\subset cl(V)$ then $U\subset V$


What is the number of square feet in the area of a triangle with side lengths shown?

[asy]size(101); currentpen = fontsize(10pt); draw("14 feet",(0,0)--14*(3/5,0.8),NW);
draw("15 feet",(0,0)--(15,0),S); draw("13 feet",(15,0)--14*(3/5,0.8),NE);[/asy]

use Heron's Formula.

$ S \equal{} \frac {13 \plus{} 14 \plus{} 15}{2} \equal{} 21$

therefore, the area of the triangle is $ \sqrt {21\times(21 \minus{} 15)(21 \minus{} 14)(21 \minus{} 13)} \equal{} \sqrt {21\times8\times7\times6} \equal{} 84$

answer : 84

P.S. whats an easier way?

this helped me a lot thanks? where do you get the formula?

draw altitude to the side of length 14. since this question does not mention fractions or radicals, we cheat and assume both right triangles are pythagorean triples. therefore we get that the height is 12 (we can confirm this b/c other sides are 9 and 5, which add to 14.) so area=14(12)/2=84

[asy]import olympiad;
import geometry;

pair B=(0,0);
pair C=(15,0);
pair A=intersectionpoint(Circle(B,14), Circle(C,13));
pair H=foot(C,A,B);

draw("$5$",A--H);
draw("$9$",H--B);
draw("$15$",B--C);
draw("$13$",C--A);
draw("$12$",H--C);

label("$A$",A,N);
label("$B$",B,SW);
label("$C$",C,SE);
label("$H$",H,NW);

perpendicular(H,NE,H--C);[/asy]


logarithms
floor function
function
real analysis
real analysis unsolved

I am having  little trouble with calculating/finding the big Oh and the Big Omega of log(n!).

I have tried using Stirling's formula and progressing from there, but can't get far.

Any help appreciated.

Also, having trouble solving the recurrence relation T(n)=3T(n/4) + n  where n is a positive integer.
Not sure where to start on this one.

Again help would be appreciated.

Thanks in advance.

well, ummm... do you know how to derive this (from stirling) : :(

$ \log n! \approx n\log n \minus{} n \plus{} \frac {\log n} {2} \plus{} \frac {\log(2\pi)} {2} \plus{} \frac{1}{12\,n} \minus{} \frac{1}{360\,n^3} \plus{}   \quad\ldots$

from this you can tell

$ \log n! \equal{} \mathbf O( n\log n )\qquad \because \;\log n! \le 2n\log n\quad\forall n\ge 2, n\in \mathbf Z$

also,

$ \log n! \equal{} \mathbf \Omega( n\log n )\qquad \because \;\log n! \ge \frac{1}{4}n\log n\quad\forall n\ge 2, n\in \mathbf Z$

regarding the recurrence, what you meant to say was

$ T(n)\equal{}n \plus{} 3\, T\left(\lfloor \frac{n}{4}\rfloor\right) ,\qquad \forall n\in\mathbf Z$

keep backsubstituting to get

$ T(n) \equal{} n \plus{} 3\, \left[ \lfloor \frac{n}{4}\rfloor \plus{} 3\, T\left(\lfloor \frac{n}{16}\rfloor\right)\right]$

$ T(n) \equal{} n \plus{} 3\, \lfloor \frac{n}{4}\rfloor \plus{} 9\, \left[ \lfloor \frac{n}{16}\rfloor \plus{} 3\, T\left(\lfloor \frac{n}{64}\rfloor\right)\right]$

$ T(n) \equal{} n \plus{} 3\, \lfloor \frac{n}{4}\rfloor \plus{} 9\, \lfloor \frac{n}{16}\rfloor \plus{} 27\, \left[ \lfloor \frac{n}{64}\rfloor
\plus{} 3\, T\left(\lfloor \frac{n}{256}\rfloor\right)\right]$

and so,

$ T(n) \le n\,\sum_{k\equal{}0}^\infty \;\left( \dfrac{3}{4}\right)^k \plus{} \mathbf O\left(\; n^{\ln_4\,3} \times T(1) \right)$

$ \Rightarrow T(n) \equal{} \mathbf O(n)$

Thankyou very much for your reply. I'm sure how you  the last two terms in your expression for log(n!). The others seem to come directly from the Stirling approximation, but I'll have another look at it.

Something a bit more simple:

$ n! \equal{} n\cdot(n \minus{} 1)\cdot(n \minus{} 2)\cdot\ldots\cdot2\cdot1 < n\cdot n\cdot n\cdot n\cdot\ldots\cdot n\cdot n \equal{} n^n \\
\log(n!) < \log(n^n) \equal{} n\log(n) \\
\log(n!) \equal{} \mathbf{O}(n\log(n))$

I'd be grateful if somebody could help me out with:

$ n! \equal{} \mathbf{\Omega}(2^n)$

 :)

[quote="modulo"]I'd be grateful if somebody could help me out with:

$ n! \equal{} \mathbf{\Omega}(2^n)$[/quote]

What does it mean to help you out with a false statement?

Edit: Nevermind!

Sorry!  I confused $ \Omega$ and $ \Theta$.  Bah, humbug.

[quote="Ravi B"]No, the statement is true.  Don't forget that Omega means lower bound.

Hint: $ n!$ is the product of the numbers 2, 3, 4, ..., n, each of which is at least 2.[/quote]
Can't believe it was THAT easy!

Thanks a LOT.  :)  :)  :)  :)

Hello. :)

I'm trying to figure out the big O of ${ \binom{n}{x}}$

I've found out that:

$ \binom{n}{x} = \frac {n!}{(n - x)!x!}$

and $ \log\left(\frac {n!}{(n - x)!x!}\right) = \log{\left(n!\right)} - \log{\left((n - 41)!\right)} - \log 41!$

However, I cannot figure out whether it is $ \mathbf{\Theta}\left(\binom{n}{x}\right) = \mathbf{\Theta}(n!)$ or $ \mathbf{\Theta}\left(\binom{n}{x}\right) < \mathbf{\Theta}(n!)$ :maybe:

By the way, is it true that $ \mathbf{\Theta}\left(\sqrt {n!}\right) = \mathbf{\Theta}\left(n!\right)$ ??

Since $ \log{\left(\sqrt {n!}\right)} = \frac {1}{2}\log{\left(n!\right)} = \mathbf{\Theta}(n\log n)$ and $ \log{\left(n!\right)} = \mathbf{\Theta}(n\log n)$, I assume they're of the same complexity order.

If $ x$ is fixed, $ \binom{n}{x}$ is a polynomial.  The largest possible growth rate here is certainly bounded above by $ \sum_x \binom{n}{x} \equal{} 2^n$ (and in fact you can use Stirling's to get a more exact estimate for $ \binom{2n}{n}$).

$ \Theta (\log f(n))$ is very different from $ \log (\Theta(f(n)))$ (and similarly with $ \exp$) because of the way log and exponential rules behave -- just try writing down the two expressions!  Similarly, one can show that there is no function $ f(n) \to \infty$ such that $ f(n)^2 \equal{} O(f(n))$, more or less by definition.

[quote="JBL"]If $ x$ is fixed, $ \binom{n}{x}$ is a polynomial.[/quote]

Hmmm.... why does this happen? I don't get it - or maybe I don't see it that easily... :maybe:

And how about $ \mathbf{\Theta}\left(\sqrt {n!}\right) \equal{} \mathbf{\Theta}\left(n!\right)$ ?
Is it correct?

$ \binom{n}{1} \equal{} n$.  $ \binom{n}{2} \equal{} \frac {n^2 \minus{} n}{2}$.  $ \binom{n}{3} \equal{} \frac {n(n \minus{} 1)(n \minus{} 2)}{6}$.  Etc.  $ \frac{1}{x!}$ is a constant and $ \frac{n!}{(n \minus{} x)!}$ is a product of $ x$ terms, each of which is linear in $ n$.

Please re-read my post: your question is answered in my last sentence.

[quote="JBL"]Please re-read my post: your question is answered in my last sentence.[/quote]
Aha, read it. Sorry, it had been a really hard day. :P

Does the same apply to $ \mathbf{O}$ and $ \mathbf{\Omega}$ ? I mean is $ O(\log f(n))$ different from $ \log (O(f(n)))$ too?

Since $ \sqrt {n} \equal{} \mathbf{\Omega}\left(n\right)$ can we tell that $ \sqrt {n!} \equal{} \mathbf{\Omega}\left(n!\right)$ ??

Why don't you try writing down what those statements mean and seeing for yourself?  None of these are very difficult to prove.  For example, $ \Omega(\log f(n))$ is the set of functions that are eventually greater than $ c \log f(n)$ for some $ c$, while $ \log(\Omega(f(n)))$ is the set of functions that are eventually greater than $ \log(d f(n))$ for some $ d$.  If $ f(n) \to \infty$, what is the relationship between these sets?

[quote="modulo"]Since $ \sqrt {n} \equal{} \mathbf{\Omega}\left(n\right)$ can we tell that $ \sqrt {n!} \equal{} \mathbf{\Omega}\left(n!\right)$ ??[/quote]
Where did you get $ \sqrt {n} \equal{} \mathbf{\Omega}\left(n\right)$?

[quote="Ravi B"]Where did you get $ \sqrt {n} \equal{} \mathbf{\Omega}\left(n\right)$?[/quote]

My bad, $ \sqrt {n} \equal{} \mathbf{O}\left(n\right)$ I meant. I always confuse these two.


inequalities
triangle inequality
inequalities open

[quote="pohoatza"]Let $a,b,c$ sides of a triangle. Prove the following ineq:

$\sum{\frac{1}{2a+b+c}}<\frac{7}{3(a+b+c)}$[/quote]

[hide]$\sum{\frac{1}{2a+b+c}}<\frac{7}{3(a+b+c)}$
$\Leftrightarrow \sum \frac{a}{2a+b+c}> \frac{2}{3}$

$\sum \frac{a}{2a+b+c}\ge \frac{(a+b+c)^{2}}{\sum 2a^{2}+ab+ac}$ by cauchy

$\frac{(a+b+c)^{2}}{\sum 2a^{2}+ab+ac}> \frac{2}{3}$
$\Leftrightarrow 2\sum ab >a^{2}$

which is true because $\sum (ab+ac)=\sum a(b+c)>\sum a^{2}$ by triangle inequality[/hide]

If $a,b,c$ are the sides of a  triangle, $w_{a},w_{b},w_{c}$ are the angle-bisectors, then

$\sum\frac{w_{a}}{a}\geq \frac{2}{\sqrt3}\sum\frac{a+b+c}{2a+b+c}$.


How many times does the digit 6 appear in the list of all integers from 1 to 100?

We first count the units digits.  We know that it appears in $ 6, 16, 26,\dots, 86, 96$.  It is clear that there are $ 10$.  Then, we count the tens digit.  We have $ 60, 61, 62,\dots, 68, 69$, and it is clear that there are $ 10$ of those as well (we don't count $ 66$ twice in the tens digit since we've already covered it once in the ones digit).  Therefore, we have $ 10\plus{}10\equal{}\boxed{20}$

[hide="Complementary counting"]There are $9 \cdot 9 - 1=80$(not including 0) that does not include the digit 6. There are 100 numbers, so $100-80=\boxed{20}$[/hide]

As long as we're giving different ways to do this classic...

[hide="Symmetry"]The number of sixes is the same as the number of sixes in the list of all numbers from $00$ to $99$, writing all single digit numbers with a leading zero and exchanging the number $100$ for $00$.

There are $2\cdot 100 = 200$ digits in the list.  By symmetry, every digit appears with the same frequency.  Therefore, every digit appears $\frac{200}{10} = \boxed{20}$ times.

Note that if they ask you for the number of times $0$ appears, this won't work, as we added a bunch of zeros.  It's also off for the digit $1$ because we traded $100$ for $00$.   But both of those are easy to count:  $10, 20, 30, ..., 100$ are the only numbers with $0$, and you just add one for the $100$ to count ones.[/hide]

[quote=Vapor_Biscuit23]The solution on the website is very confusing and doesn't make sense. Is it possible that you can fix that?[/quote]

I believe this post should have gone in the site support forum

[quote=DarkSun99]yeah its confusing

also the note is confusing

now don't get mad at me im not bumping it i am actually discussing[/quote]

that's still bumping but ok

There are $10$ in the column and $9$ in the row.  However, the $66$ overlaps. So the solution should be $10+9-1=\boxed{18}$.
Whats the problem???

bruh you missed 60
also 66 doesnt "overlap"

[quote=ME1231917]The AoPS solution makes no sense.[/quote]
Just report it with the flag button. No point posting it here, as the mods are unlikely to see it.


This only applys to the middle schoolers.

If you're middle schooler, you'll be given special test to benefit you to pass from Round 1. Because Round 1 involves geometry, it may not fair for some of the middle schoolers who have yet not learned geometry to attend the COP. So, I'll make a special round for middle schoolers.

After you passed this round, you'll be given same test, like the other people from Round 1.

Rule for COP Special Division (abbreviated COPSD).

There will be 10 questions on this site. What you have to do is send your full written solutions, meaning answers and how you achieved this answer. You'll be given a week to do this. The problems' categories:

Algebra
Number Theorey (Elementary)
Probability
Sequence and Series
Pattern
And some other things that might appear except geometry or any of further materials.

FAQ that I might have:

Q: I'm going to be 9th grader soon but now, I'm 8th grader. I'd like to take COPSD then Round 1, but I'm afraid that I would be 9th grader by then. Can I still allowed to take COPSD?

R: Yes, you can. Only thing is that since you're in cutoff, I'm going to ask you to solve two of the Round 1 problems plus original round questions on COPSD and to qualify, you have to get them all right. Explanation is not necessary for the two questions from Round 1.

Q: I really want to take COPSD but I'm afraid that I'll miss the contest week. What should I do?

R: You can take it early. I'll count the days from the I sent the private message and I expect the answer to come in 7 days.

Q: Why do we have to write full solutions while people in the COP Round 1 don't?

R: The reason is that for a) I want to be fair for all people that I'm not giving any special benefits to middle schoolers, b) when you go to Round 2, it would be a lot harder than Round 1. So, if you wrote the full solution to COPSD, that should definitely help you on this.

Q: Is there any special rule for middle schoolers from COPSD for say, final round? Like for freshman on AIME?

R: Nope. Younger doesn't mean you'ren't smarter. I'll simply take the highest score from the championship round to determine the winner.

Q: In my country, the systems of middle and high school is slightly different. What should I do?

R: I take that if you're under 15, you can take COPSD.

Q: Is there any possibilities that there would be total different test for younger people like junior division?

R: Afraid not. The COP is intended to start out really easy and end with very difficult so making two contests aren't necessarily good idea.

Q: Can we ask other people for help?

R: Basic rules are same as Round 1. So, the answer is NO.

Q: Calculator allowed?

R: Yes.

Any further question must be posted in here!

Thanks!

Hm...

How many questions are in Round 1? If there's less than 10, I'll do it. Of course, if there's more or an equal amount, I'll do the COPSD :P .

Why does it say Q:"Some Stuff" then R:"Some Stuff". What's the R stand for?

I'll probably do it.

Am I allowed?

I would be glad to take this test...

I guess it's the first time in my life actually feeling good about taking tests....

I live in Pittsburgh so I don't have much problems...

Yes. Anyone is allowed.

There are 5 questions to do Round 1 and 10 questions on COPSD. Only thing is that COPSD doesn't involve any geometry. So, it's very similar to AMC-8.

For taking both, it doesn't matter you take one or two. It's like going through AIME--> You can qualify in more than one ways.

Only thing is that if you take two, make sure you send the answer in a same message. So, you might want to say "COPSD answers" and blahblahblahblah... and below that (or above), "COP Round 1 answers" again blahblahblahbblahblahbl.

This is exception if you're a 8th grader who is going to be 9th grader. You can only take either one.

Hm..Since I;m a seventh grader, I'll take both of them...

Are they hard?

scale 1 to 10..

1 being middle school class tests, 10 solving like AIME problems...

I doubt if this round would get higher than 2.

COPSD is really easy round just for middle schoolers.

For AIME level, the championship round may include some questions in that level altough it's allowing calculator.

Please keep signing up.

More competitioners we have, the better the competition will get.

Note to middle-schoolers that younger doesn't mean you guys are going to be beaten by high-schoolers. RC-7th, who is a middle-schooler, placed the 2nd on the first and second COP consecutively. And he was not only competitioner there. He did very well.

So, this COPSD will benefit middle-schoolers who wants to try some challenging levels yet they're not sure they can reach it or not. This COPSD will get you there.

BIG ANNOUNCEMENT:

COP The 3rd would take place in March 7th.

Please sign up till then and if you forgot to sign up but want to participate, just send me the answers.

Thanks.

Just in case anyone wondered, I finished writing of the COPSD tonight but I'm going to hold on to this so I can make the solution packet and also I'm going to be busy for AMC.

I'll join.

And how do we access the test when it happens?

The test will be posted here and then it would be removed and placed on my website.

Although I can't get to my website now though... Must be some sort of server problem. ;)

Just note:

This topic has been nearly disappeared while so if anyone wants to sign up for COPSD, do it while it's easy to see. :D  :D

Uhh... I'll sign up for the COPSD. When is it?

[quote="Silverfalcon"]This one would be posted on this Saturday.[/quote]

Wow... So the questions are going to be posted in here. Then we have to PM you the answers with the full solutions?

[quote="Silverfalcon"]Yes.

There would be 10 questions. But if you get at least 7 right, you'll be send to Round 2.[/quote]

Ahh... Ok thanks a lot.

One more day! :P So if tomorrow is the 19th, then the answers are due on the 26th right?

[quote="Silverfalcon"]Depends.

I might give one more extra day.

I'm not sure. :)[/quote]

Lol... YES! An extra day. :P this will be my second math "competition or contest". Really excited. ;)


Let $ ABC$ be an  acute and nonisosceles triangle. Denote the midpoint $ M$ of the side $ [BC]$ . Consider the points

$ \{D,E\}\subset (AM)$ such that $ \frac {DA}{DB} \equal{} \frac {EA}{EC}$ . Denote $ F\in BD\cap CE$ . Prove that $ \widehat {AFB}\equiv\widehat {AFC}$  (see and [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=208506][color=red][b]here[/b][/color][/url]).

Apply Ptolemy's theorem to triangle BFC we have that $ \frac{DF}{DB}=\frac{EF}{EC}$ ( because M is the midpoint of BC )
$ \Rightarrow\frac{FD}{FE}=\frac{DB}{EC}=\frac{AD}{AE}$, which implies FA is the external bisector of EDF. Hence result.


Please, help me to solve the following trigonometric equation: $ \cos x \equal{} \frac{1}{2\sqrt{2}}$. NO ARC-FUNCTIONS!

The simple way to solve it, of course, would be to use inverse trig, but since that is not one of the options, I guess you can graph $ \cos x$ and analyze values that equal $ \frac {\sqrt {2}}{4}$.

I don't think $ x$ is a rational multiple of $ \pi$.  What kind of an answer are you looking for?


geometry
parallelogram
circumcircle
geometry solved

Three circles $P,P_1,P_2$ are given; $P_1$ and $P_2$ touch $P$ internally at points $A$ and $B$ respectively. It is known that the radius of $P$ is equal to the sum of the radii of $P_1$ and $P_2$. Prove that:

a) one of the intersection points of $P_1$ and $P_2$ belongs to the line segment $AB$.
b) the other one belongs to the cirucmference passing through $A$, $B$, and the center of $P$.

$P_1=(C,a)$
$P_2=(E,b)$
$P=(D,a+b)$

If $A,B,D$ are collinear, then the two small circles have one common point and the problem is trivial (the intersection point is on the line $AB$)

So, suppose that $A,B,D$ are not collinear. We have:

$AC=a, AD=a+b =>CD=b$

$BE=b, BD=a+b =>DE=a$

In the triangle $ABD$:
Bring a parallel line from $E$ to $AD$, which intersects line $AB$ at point $F$. From the parallel lines we have

$\frac{DE}{EB}=\frac{a}{b}=\frac{AF}{FB}$

Now bring a parallel line from $F$ to $BD$, which intersects line $AD$ at the point $T$.

So $\frac{AF}{FB}=\frac{AT}{TD}$ e.g. the point $T$ is the same with the point $C$.

From this construction, $CDEF$ is a parallelogram, so $CF=ED=a$, $CD=FE=b$, hence $F$ belongs to $P_1$ and to $P_2$.

The two intersection points of $P_1$ and $P_2$ are on different sides of the line $CE$. Notice that $F,D$ are on different sides of $CE$, hence the other point $H$ is on the same side with $D$.

The triangles $CDH, EHD$ are equal (they have $CD=EH=b$, $CH=ED=a$, and $DH$ common). So $\angle HCD= \angle DEH$

But $\angle HAD = \frac{1}{2} \angle HCD$ because $A,H$ are on the circle with center $C$ and $A,C,D$ are collinear.

Similarly, $\angle HBD = \frac{1}{2} \angle HED$

So $\angle HAD = \angle HBD$, that is, the quadrilateral $AHDB$ can be inscribed in a circle, e.g. $H$ lies on the circumcircle of triangle $ABD$

$Extra\ Question$

Prove that the common tangents at the points $A,B$ intersect on the circumcircle of $\triangle ABD$


Reaper
geometry
geometric transformation

I don't know is this is common in places other than in the US, but every year, near Halloween, kids dress up in costumes and beg for candy at people's houses.  I was wondering if anyone here is going to do that, and if you are, what your costume will be.

I'm going as George W. Bush.

im going as this skeleton/grim reaper thing
its mask is connected to this pump thing and if you squeeze it, blood comes down it's face. The blood gets recycled so you can squirt blood all night
it scared a lot of ppl last year :D
even my friend's gigantic dog

o yeah and a friend of mine told me they are making bush and kerry masks and which ever one gets bought the most will win the election
but i think ppl are going to buy bush masks for a different reason.......
if i went around in a bush costume i dont know how many vegetables are going to be thrown at me
 ;)

I'm just gonna stay home and throw candy at people.  It's a lot more fun to see what costumes people come up with.  For example, the highlight of last year's Halloween was when I looked up from the plastic candy cauldron to see Mario and Luigi walking up my driveway.

I'm just going to go to my friend's house and do some stuff probably. I might go trick/treat. Depends how desperate I am for sugar at that point.

[quote="cardwacko9999"]I personally dont agree with Halloween- my sources say that it originated with devil worship (I am a Christian)[/quote]
It started with some ancient Pagan holiday, but not many people know or care anymore...Halloween's been 'secularized' just like, say, 'under God' in the Pledge of Allegiance...

... the origin matters a lot to me... Im a history freak... sorry....

I probably won't be doing any real trick or treating, unless I have to go with my brother, but I will be helping at my church's carnival. It provides a place where kids can come and play all sorts of games for free, instead of being out on the streets and stuff. It also helps bring people into our church as an outreach thing. It's a lot of fun.

[quote="cardwacko9999"]I personally dont agree with Halloween- my sources say that it originated with devil worship (I am a Christian)[/quote]

Oh well. I don't mind worshipping the devil for one night if it means getting sugar high for a week.

YES YES YES YES YES!!!!! I AM SOOO GOING TRICK OR TREATING!!! probably with sum frends i eat with at lunch. and i dunno wat im gonna be...hm...probably this greek person....u no with a long toga and a laurel or sumthin. but NOT athena. it is my username, but she was also the goddess of war and i dunno were to get a helmet and a sheild thingy. and sum frends of mine are gonna be devils. SOME ppl wont need a costume for that lol. last halloween me and sum other frends tp-ed sum houses and egged one hahahaha. it was interesting. anyone seen the simpsons halloween episodes? the raven one was cool. homer goes insane lol.

You know, it's not nice to do that to people's houses.  It's a pain for the homeowners and a waste of toilet paper and eggs.  There are people starving all over the world who would kill to lick the raw eggs being splattered on vinyl siding in your neighborhood October 31.  Perhaps you'd like to come home one night after a night of mischief to find that someone (perhaps even you, if it was dark enough) had egged your own house in the dark and you had to clean it all.  People don't enjoy salmonella-flavored paint on their houses.

hey, it wasnt MY idea!!! the ppl i went with never even told me wat we were gonna do. and my jacket also got egged. there was this big yellow stain on it the next morning. and i tried 2 convince them not 2 do it. but then yea....it was still a fun nite. i got alot of candy!!!

Heh, my friend's house got egged last year. Except the people who egged his house meant to egg his neighbor's. Kind of amusing.

But still, egging houses and TPing houses is pretty dumb. Just for people who want attention.

oh yeah about egging, my sister threw an egg at me once and instead of catching it, it landed on my head :D

halloween itself is BAD BAD  BAD

trick-or-treat sounds kinda harsh but may be pretty fun iguess

like cardwacko im achristian too

dude i doubt anyone's gonna worship satan or anything.  they're just there to u know rob some candy

[quote="A+MATH"][quote="BHorseMath"]

Well I see someone did their research right.  All Hallows eve is indeed what our halloween came from but hallows eve was a Pagan holiday (if anyone has read Da Vinci Code they know this).

[/quote]

I agree with you, but I don't think you should trust a book like that, I mean he could have made that up or something to better express his point. I'm not talking about that book but just books in general. ;)

But anyways there nothing wrong with Holloween it's not like where going to worship satan on it or something.[/quote]

I wasn't saying that it was just from Da Vinci Code but it is mentioned in there and I know its true because I researched it wondering about it.

i heard of the book da vinci code. what's it about?

anyone who TP's a house or eggs one is a complete idiot worthy of an insane asylum.

And if you looked hard enough, all holidays have something evil about them
For example christmas came about as a celebrated holiday when theodosius was emperor.  He destroeyd everything pagan (sports, mathematics, science, books, learning,...) and tortured their supporters.  And so christianity itself is evil, because thats how chritianity went over big.

halloween did not come from worshipping the devil.  Ever heard of All saints day or all hallows eve? no one worships devils.  (devils may have come about because pagan gods seemed evil at the time it came about).

[quote="gameworld7"]oh yeah...the spanish don't celebrate haloween.  they celebrate 'the day of the dead" that's on november 2nd.  my spanish class will celebrate it :lol:[/quote]

I believe the tradition you speak of is from Mexico; it is indeed known as [i]El Dia de los Muertos[/i] (The Day of the Dead), and in contrast to the suppositions that Halloween is a pagan holiday, [i]El Dia de los Muertos[/i] is a day to respect one's dead relatives by placing offerings on special altars on their graves.

Well, technically, it's [i]Los Dias De Los Muertos[/i], since it's November 1 and 2.

And incidentally, "pagan god" is an oxymoron.

um i thought that the day of the dead (for u spanish people its los dias de muertos) was only on november 2nd? my spanish teacher told us that it was on november 2nd and we're having a PARTY on that date!

In Mexico, I believe it's celebrated on both days, and in large part on the night in between.

I go as yet another recourse-consuming kid in an overpopulated planet, raised to an alarming extent by Madison Avenue and Hollywood, poised with my cynical and alienated peers to take over the world when you are old and weak.

[quote=mkhaim]scared yet?[/quote]

No not a t a l l l


Sorry but I didn't understand. Could you explain me the reason for which you considered a/b two times... to be honest, I didn't understand NOTHING  :blush: Please be gentle with me, I'm really ignorant.

[quote="siuhochung"][quote="davi24"]I really can't get to a solution for this disequation. Somebody told me I can use Holder inequality but I don't even know what it is. Could you help me? Thanks!

a/b + b/c + c/a >= a+b+c  if a*b*c = 1

Thank you again![/quote]

$2a/b+b/c = a/b + a/b + b/c \ge 3 \sqrt[3]{aa/bc} = 3a$[/quote]

similarly,

$2b/c+c/a \ge 3b$
$2c/a+a/b \ge 3c$

sum up.

ps: nice soln, siuhochung


function
inequalities unsolved
inequalities

Let $ f: \mathbb R \rightarrow \mathbb R$ be a function such that

$ f(tx_1\plus{}(1\minus{}t)x_2)\leq tf(x_1)\plus{}(1\minus{}t)f(x_2)$

for all $ x_1 , x_2 \in \mathbb R$ and $ t\in (0,1)$. Show that

$ \sum_{k\equal{}1}^{2003}f(a_{k\plus{}1})a_k \geq \sum_{k\equal{}1}^{2003}f(a_k)a_{k\plus{}1}$

for all real numbers $ a_1,a_2,...,a_{2004}$ such that $ a_1\geq a_2\geq ... \geq a_{2003}$ and $ a_{2004}\equal{}a_1$

Let $ \Delta f(x,y) \equal{} \Delta f(y,x) \equal{} \frac {f(x) \minus{} f(y)}{x \minus{} y}$. Since $ f(x)$ is a convex function, then $ \Delta f(x,y)$ is an increasing function for the both variables.

$ \sum_{k \equal{} 1}^{2003}f(a_{k \plus{} 1})a_k \minus{} \sum_{k \equal{} 2}^{2004}f(a_{k \minus{} 1})a_{k} \equal{}$ $ \sum_{k \equal{} 1}^{2003}a_k(a_{k \plus{} 1} \minus{} a_{k \minus{} 1})\Delta f(a_{k \plus{} 1},a_{k \minus{} 1})$ $ \equal{} \sum_{k \equal{} 1}^{2003}a_ka_{k \plus{} 1}\Delta f(a_{k \plus{} 1},a_{k \minus{} 1}) \minus{} \sum_{k \equal{} 1}^{2003}a_ka_{k \plus{} 1}\Delta f(a_{k \plus{} 2},a_{k}) \equal{}$ $ \sum_{k \equal{} 1}^{2003}a_ka_{k \plus{} 1}\left[\Delta f(a_{k \plus{} 1},a_{k \minus{} 1}) \minus{} \Delta f(a_{k \plus{} 2},a_{k})\right]\ge 0$, since $ a_i\ge a_{i \plus{} 1}$. (That's not the complete solution, but the rest is easy :) )


Let $ a,b,c$ be the sides of a triangle.Prove that:
$ \frac{abc}{(a\plus{}b\minus{}c)(b\plus{}c\minus{}a)(c\plus{}a\minus{}b)}\ge (\frac{a^2\plus{}b^2\plus{}c^2}{ab\plus{}bc\plus{}ca})^2$

It's obviously trues because: $ LHS \ge \frac{a^3\plus{}b^3\plus{}c^3}{3abc} \ge RHS$  :maybe:
See also here: http://www.mathlinks.ro/viewtopic.php?t=299990

[quote="nguoivn"]It's obviously trues because: $ LHS \ge \frac {a^3 \plus{} b^3 \plus{} c^3}{3abc} \ge RHS$  :maybe:
See also here: http://www.mathlinks.ro/viewtopic.php?t=299990[/quote]
Find the best $ k$ that: $ \frac{abc}{(a\plus{}b\minus{}c)(b\plus{}c\minus{}a)(c\plus{}a\minus{}b)}\ge (\frac{a^2\plus{}b^2\plus{}c^2}{ab\plus{}bc\plus{}ca})^k$ still trues.


Given any three non-collinear points in a plane, two of which are fixed and one variable, what is the locus of the center of the circle through these points?

It's fairly easy, its just the perpendicular bisector of the segment of the two fixed points.   :D

Sorry....I may be dumb...but I can't see how :blush:

By definition, the center of the circle must be equidistant (the distance is all the same) from all three points, and therefore from the two fixed points.

It is also well known that the locus of all points equidistant from two points is the points' segment's perpendicular bisector.

Finally, to generate a specific point on the perpendicular bisector, just draw the circle and pick a random point on it.


Find all positive integers $n$ such that $n(n+60)$ is a perfect square.

$n=x-30$
$(x-30)*(x+30)=k^2$
$x^2-k^2=900$
$(x-k)*(x+k)=900$
x-k and x+k must be even because when we sum and divide 2 them we get x and x must be a positive integer.
x-k gets 2 from 900
x+k gets 2 from 900
then we have 225.we can give how i want the multiples of 225 to x-k and x+k.
225+1=226=x
75+3=78=x
25+9=34=x
45+5=50=x
15+15=30=x
we know that n=x-30
n's are 196,48,20,4 and 0 but n must be a positive integer then we odd 0 from them.

ok I have it this way

[hide]
Suppose $n(n+60)=k^2$ . Then we will have $(n+30)^2=k^2+30^2$ . Now we see that it is just like pytagoras theorem and we also have the pytagorean triplet satisfy
$(p^2+q^2,2pq,p^2-q^2)$ for integer $p\geq q$ .

Firstly , if all the side of the right angle triangle $a,b,c$ have $gcd(a,b,c)=1$ . Then we can say that $30=2pq$ which is equivalent to $pq=15$. WLOG , we can say $(p,q)=(15,1)$ or $(5,3)$ which gives us $n+30=226$ and $34$ . Hence , $n=196,4$

Secondly , if $gcd(a,b,c)>1$ : there are 3 such cases

(i) when $gcd(a,b,c)=3$ , divide all the three side $a,b,c$ by $3$ then we know that one side of it is $10$ . So from $2pq=10$ The only possible solution is $(p,q)=(5,1)$ .
This gives us the hypotenus is $26\times 3=78=n+30$ . Hence $n=48$

(ii)When $gcd(a,b,c)=5$ , divide the three side by $5$ , one of the side is $6$ . So $pq=3$ and $(p,q)=(3,1)$ which yields $n=20$

(iii)When $gcd(a,b,c)=2$ , one side is $15$ . So $(p+q)(p-q)=15$ and gives us $(p,q)=(4,1),(8,7)$ . But both this value gives us $n=4,196$ which is same with above .

So all the posible solution is $n=4,20,48,196$[/hide]


سلام
تا جمعه 4 روز بیشتر نمونده
حستون چیه؟
به نظرتون سوالای امسال آسونه یا سخت؟
به نظرتون قبول میشید یا نه؟
جواب بدید
بای :lol:

Man shenidam emsal asoontare.Haddeaghal computero motmaenam.Dalilesham ke hame midoonan (khode bashgaham elam karde).Be khater hamin sadegi , ghabool shodan sakht mishe (baraye kasaee ke marhale 2 mikhoondan) chon kafe ghabooli mirese be masalan $ \frac {12}{30}$ (tavajjoh konid ke Sale pish $ \frac {8.5}{30}$ bood  :!: ). Pas behtare ta akhare jalase beshinid va faghat age motmaenid test ro bezanid(nomre manfi dare !!!!)
Vali ba hameye in osaf omidvar bashid be ghaboolitoon (mesle man  :lol: ) .

slm!
man fek mikonam asun tar bashe!(har chand parsal ham khub bud!!!)
fek konam ghabul sham!
iA harkaC unghad ke talsh karde natije begire! :wink:
[i]God bless u all[/i]


$m,n$ are natural numbers, $x \in [0,1]$. Show that:

$(1-x^n)^m + (1-(1-x)^m)^n \geq 1$

It was already posted (by you!) and solved (by me!).  :lol:

http://www.mathlinks.ro/Forum/viewtopic.php?t=75884


Is there another way besides the positional algorithm (2^i) to convert a number from binary representation to the decimal one?

binnumber = d_0 d_1 ... d_n
dec=0;
for(i=0;i<=n;i++){
dec=dec*2+d_i;
}
return dec.

pretty much the same. there's yet another way too.

http://www.greatis.com/delphicb/tips/lib/mathematics-bin2hex.html

That's a neat little function.


geometry
circumcircle
projective geometry

Quadrilateral $ ABCD$ is cyclic. The tangents to the circumcircle of $ ABCD$ at $ A, B, C, D$ are, respectively, $ t_A, t_B, t_C, t_D$.

$ t_A \cap t_C \equal{} P_1$

$ t_B \cap t_D \equal{} P_2$

$ AB \cap CD \equal{} P_3$ (the lines)

$ BC \cap DA \equal{} P_4$ (the lines)

Prove $ P_1, P_2, P_3, P_4$ are collinear.

It's not too tough, but I thought it was a pretty nice property.

[hide="Solution"]

Lemma: Consider points $ LMNPQR$ on a circle. By Pascal's theorem, $ LM \cap PQ$, $ MN \cap QR$, $ NP \cap RL$ are collinear. If we force $ L$ and $ M$ to coincide and call the point $ S$, and we force $ P$ and $ Q$ to coincide and call that point $ V$, then we get $ t_S \cap t_V$, $ SN \cap VR$, $ NV \cap RS$ are collinear.

If we apply the lemma, we find that $ t_A \cap t_C, AB \cap CD, BC \cap DA$, or $ P_1, P_3, P_4$, are collinear.

Applying the lemma again, we find that $ t_B \cap t_D, AB \cap CD, BC \cap DA$, or $ P_2, P_3, P_4$, are collinear.

Since two points determine a line, $ P_1, P_2, P_3, P_4$ are all collinear, and we are done.

[/hide]

let $ w$ be the circumcircle of $ ABCD$ and let $ AC\cap BD\equal{}Q$,then $ P_1,P_2,P_3$ and $ P_4$ lie on the polar of point $ Q$ wrt circle $ w$...

Also by Pascal to the hexagon $ABCCDA$, $P_1, P_3, P_4$ are collinear, while by Pascal to the hexagon $ABBCDD$, $P_2, P_3, P_4$ are collinear, so done.

Best regards,
sunken rock


In quadrilateral $ ABCD$, sides $ \overline{AB}$ and $ \overline{BC}$
both have length 10, sides $ \overline{CD}$ and $ \overline{DA}$
both have length 17, and the measure of angle $ ADC$ is $ 60^\circ$.
What is the length of diagonal $ \overline{AC}$?
[asy]draw((0,0)--(17,0));
draw(rotate(301, (17,0))*(0,0)--(17,0));
picture p;
draw(p, (0,0)--(0,10));
draw(p, rotate(115, (0,10))*(0,0)--(0,10));
add(rotate(3)*p);

draw((0,0)--(8.25,14.5), linetype("8 8"));

label("$A$", (8.25, 14.5), N);
label("$B$", (-0.25, 10), W);
label("$C$", (0,0), SW);
label("$D$", (17, 0), E);[/asy]

think about it
if AD and CD are both 17 and angle adc=60
than that means that angle cad=60 and angle acd=60
which means that triangle acd is equilateral
hence ac=17


Hi all,

I am trying to follow the step how to write documents from the following source:

http://www.artofproblemsolving.com/LaTeX/AoPS_L_BasicFirst.php

I am stuck in the following step: " Press Ctrl-F7, or click Build, then slide the mouse over Current File, then click on Build Output. We call this process compiling the document. "

I did click BUILD and then current file, and then Build Output, but nothing seems to appear. What should happens after that???

You'll have to be a little more descriptive about what's happening.  When you build the file, what appears in the output window (at the bottom of the TeXnicCenter window)?

When nothing happens, do you mean your computer freezes, and pressing Ctrl+Alt+Delete causes Windows to tell you that TeXnic center isn't responding?

If not, you can probably ignore the rest, and maybe just try waiting a minute or so after giving the command to compile and see if it spontaneously starts compiling (this sometimes happens to me).

If so, you likely have the same problem I have.  Here's what works for me (you're compiling to PDF, right?):
First, try restarting the computer right before trying to compile.  Sometimes, restarting right before trying clears up the problem.
If this doesn't work, try compiling when Adobe Reader already has a PDF file open; if you do both, it usually works for me (don't know why, though...).

not to steal the thread, but i have trouble putting on document titles. i usually just end up centering and enlarging some text...its hard

Use

\title{Titlle goes here}
\author{your name goes here}
\maketitle

right after \begin{document}


it's really big.  2^2 is 4.  2^4 is 16.  2^16 is 65536.  I really can't go farther than that.  Also when powers are like that, it has to have parentheses.

$\displaystyle 2^{3^{4}} = 2^{81} and (2^{3)^{4}} = 2^{12}$

simple answer to a seemingly complex problem:

$\infty$

let us say it is $x$. Therefore, $2^x = x$ by substitution which is impossible except for $\infty$ and possibly complex numbers


Let $ a$, $ b$, $ c>0$ and $ abc\equal{}1$. Prove that $ \sum \frac{1}{a^2}\plus{}3\ge 2\sum a$

[quote="April"]Let $ a$, $ b$, $ c > 0$ and $ abc \equal{} 1$. Prove that $ \sum \frac {1}{a^2} \plus{} 3\ge 2\sum a$[/quote]

It's equivalent to $ \sum\left(\frac 1a\right)^2\plus{}3\left(\frac{1}{abc}\right)^{\frac 23}\ge 2\sum\frac{1}{ab}$ which is just Schur and AM-GM. :)


In the figure below, the line t is tangent to the circle of center O. Knowing that AB = $ 10\sqrt{3}$ cm, find the value of line $ \overline{BC}$. For the angle of the 70 degrees use the trigonometric table with two decimal places.

[img]http://img81.imageshack.us/img81/1310/questa.jpg[/img]

[hide="Solution"]
$ \triangle AOB$ is a 30-60-90 triangle, so $ OB$ is $ \frac{10\sqrt{3}}{\sqrt3}=10.$ We know $ \triangle OBC$ is isosceles so drop an altitude from $ O$ to $ \overline{BC}$
[asy]draw((2.5,6)--(0,0)--(5,0)--cycle,blue+linewidth(1));
draw((2.5,6)--(2.5,0),E,linewidth(1));
label("$O$",(2.5,6),N);
label("$B$",(0,0),W);
label("$C$",(5,0),E);
label("$10$",(2.5,6)--(0,0));
label("$D$",(2.5,0),S);
label("$x$",(0,0)--(2.5,0),S);[/asy]
We know $ \angle BOD=35^{\circ},$ and to get $ x$ we take the sine of $ 35^{\circ},$ which is (using half angle identities), \begin{align*}\frac{x}{10}&=\sin35^{\circ}\\x&=10\sin\frac{70^{\circ}}{2}\\&=10\sqrt{\frac{1-\cos70^{\circ}}{2}}\\&\approx10\sqrt{\frac{1-0.34}{2}}\\&=10\sqrt{0.33}\\&\approx5.74\\&\Rightarrow BC=\boxed{11.48}.\end{align*}[/hide]

I don't understand what you mean by use the trigonometric table with two decimal places.  :maybe: But anyway:

ABO is a 30-60-90 triangle so we can find that OB = 10.  OB = OC because they are both radii of the circle.  This means <OBC = <OCB = 55.  Now draw the perpendicular from O to BC call the foot of the perpendicular point P.

We have $ \cos OBC = \frac {\frac {BC}{2}}{OB}$

$ \cos 55 = \frac {BC}{20} %Error. "arrow" is a bad command.
BC \approx 11.47$

Why can't we just use the Law of Cosines?

$ OB=10\sqrt{3}\times\tan 30^\circ = 10$.
So, $ OB=OC=10$.
Thus,
\begin{align*}
BC &= \sqrt{OB^2+OC^2-2\times OB \times OC \times \cos \angle COB} \\
&= \sqrt{10^2+10^2-2\times10\times10\times0.342} \\
&= \sqrt{131.6} \\
&\approx \boxed{11.47}
\end{align*}


calculus
integration
logarithms
calculus computations

Evaluate the following definite integral.

\[\int_{e^{2}}^{e^{3}}\frac{\ln x\cdot \ln (x\ln x)\cdot \ln \{x\ln (x\ln x)\}+\ln x+1}{\ln x\cdot \ln (x\ln x)}\ dx\]

[quote="kunny"]Evaluate the following definite integral.
\[\int_{e^{2}}^{e^{3}}\frac{\ln x\cdot \ln (x\ln x)\cdot \ln \{x\ln (x\ln x)\}+\ln x+1}{\ln x\cdot \ln (x\ln x)}\ dx \]
[/quote][hide="solution?"]sorry i don't know of any other way to do this except trial-and-error for various expressions, so i tried deriving $x\ln(x\ln(x\ln{x}))$ and i got:
\begin{eqnarray*}\frac{d}{dx}(x\ln(x\ln(x\ln{x})))&=&\ln(x\ln(x\ln{x}))+\frac{x}{x\ln(x\ln{x})}\left(\ln(x\ln{x})+\frac{x}{x\ln{x}}(\ln{x}+1)\right)\\ &=&\ln(x\ln(x\ln{x}))+1+\frac{1}{\ln(x\ln{x})}+\frac{1}{\ln{x}\ln(x\ln{x})}\\ &=&\frac{\ln x\cdot \ln (x\ln x)\cdot \ln \{x\ln (x\ln x)\}+\ln x+1}{\ln x\cdot \ln (x\ln x)}+1 \end{eqnarray*}
so
\begin{eqnarray*}\int_{e^{2}}^{e^{3}}\frac{\ln x\cdot \ln (x\ln x)\cdot \ln \{x\ln (x\ln x)\}+\ln x+1}{\ln x\cdot \ln (x\ln x)}\ dx&=&x\ln(x\ln(x\ln{x}))-x|^{e^{3}}_{e^{2}}\\ &=&2e^{3}+e^{3}\ln(3+\ln{3})-e^{2}-e^{2}\ln(2+\ln{2}) \end{eqnarray*}
[/hide]

[quote="kunny"]$I=\int\frac{\ln x\cdot \ln (x\ln x)\cdot \ln \{x\ln (x\ln x)\}+\ln x+1}{\ln x\cdot \ln (x\ln x)}\ dx$.[/quote]
[color=darkblue][b]Observe that $I=J+K+L$, where $\{\begin{array}{c}J=\int \ln x\ dx\\\\ K=\int\ln(\ln(x\ln x))\ dx\\\\ L=\int\frac{1+\ln x}{\ln x\cdot\ln (x\ln x))}\ dx\end{array}$.

$\{\begin{array}{ccc}u(x)=x & \Longrightarrow & u'(x)=1\\\\ v'(x)=\frac{(x\ln x)'}{(x\ln x)\cdot\ln (x\ln x))}& \Longrightarrow & v(x)=\ln (\ln (x\ln x))\end{array}\|$ $\implies$ $L=x\cdot\ln (\ln (x\ln x))-K$.

Therefore, $I=J+K+x\ln (\ln (x\ln x))-K$ $\implies$ $\boxed{I=\int\ln x\ dx+x\ln (\ln (x\ln x))}$ a.s.o.[/b][/color]


Hello
$ FC\equal{}\frac{4}{5}AC\equal{}\frac{4}{5}(8\plus{}5)\equal{}10.4$
Thank u.

$7+8+ \frac{1}{4}AD=AD$. $AD=20$ and this allows us to set up our proportion of $\frac {20}{16}=\frac{13}{FC}$.

Hence $FC=\boxed{10.4}$


Let $x;y;z$ are positive real numbers .Prove that:
$\frac{y+z}{\sqrt{yz}}\geq \sqrt{\frac{y+z}{z+x}}+\sqrt{\frac{z+x}{y+z}}$

(8484, 21, 45) is a counterexample indeed. (152, 23, 74) isn't.

  darij

mathematica of wolfram is wrong so

lets tell this to wolfram?!


Let $G(a_{1},a_{2})$ denote the smallest positive integer which can be written as $a_{1}x_{1}+a_{2}x_{2}$, for some relatively prime integers $a_{1},a_{2}$ and such that every $b>G(a_{1},a_{2})$ has this property too. Prove that $G(a_{1},a_{2})=(a_{1}-1)(a_{2}-1)$.

If $x_{i}$ are positive integers, then this is the "Chicken McNugget Theorem." Search for it on Google, Wikipedia, or Wolfram's Mathworld.

I have the proof. I wanted to see what kind of proofs you post.


geometry
MATHCOUNTS
quadratics
number theory
algebra
quadratic formula
prime factorization

I'm a member of the organizing comiteè in Math Olympiad at San Luis Potosi, in Mexico. The educative system in Mexico isn't as good as we liked to, so when the pre-selected students come to be trained about Number Theory, Combinatorics and Geometry, they usually have only a little of knowledge about it, so, basically we have to teach them the very very basics . The problem we face is: how do we introduce them to this new form of seeing math, from math-for-fools in school to real thinking math in an Olympiad enviroment, with all it implies? Could you, more experienced people, help me on my way to make them understand, to make and forge genius, to activate the light they have not turned on?

Please, tell me some tips, strategies, entire teaching methods, everything you can!!!

1) Make math interesting to them. They don't want to learn about something boring.

2) Since you are starting out basic, introduce easy problems (and gradually build up in difficulty) to boost their morale and confidance. These problems should at first be simple application but then rise to "how can I solve this using what I know?" (MATHCOUNTS level).

3) Don't make your teachings repetitive. Make each session spontaneous and interesting.

What I also have to face with, is that as they often have no ideas, even when I have taught them theorems and formulas, still after many sessions, they continue doing problems by-foot. THis is, evaluating every case they can and solve it separately instead of finding a clever solution, or even "tabulating" a problem instead of generalizing properties of numbers we talk about. How about it? how can I prevent this from happening?

If they feel comfortable doing difficult math problems the long way (because it makes more sense to them) then that's okay. The idea of learning shortcuts, theorems, and formulas is to speed up the problem solving. Once the students have grasped the most basic concept then make them see there is an easier way.

Sometimes a math student will fear a faster method because it won't make sense. Sure they'll arrive at the correct answer, but they won't really put the thought into it. I would recommend an explanation of the faster method, why it works, and how to apply it. (For example, how the quadratic formula can be derived from completing the square).

If you want to break them of the habit of solving problems mechanically without thinking of clever solutions, try to give them a lot of problems that are REALLY HARD to do in the "standard" mechanical way, and they will be forced to try to come up with better ways. Have them work in groups; if they discuss problems that might help them practice creativity.

Give them nice problems ONLY in order they love maths  :P

A month before the national round, I came with the group with 3 difficult problems:

-IMO 2006-P4
-IMO 1990-P3
-IMO 1997-P5

For me this ordering is from easier to harder. in these problems, as you may know, you have to find specific values that satisfy the problems. My solutions to the three problems hadn't very difficult theorems implying but Fermat's Little one, the imo 2006 one was really easily solved by some factoring trick (i think this solution was easier than the official one), the second one, by prime factorization of exponents and Fermat Little theorem, and the third one, even when the solution was very long, it wasn't that difficult to get. But what they did, is to try values to see which one worked, they even didnt find one value for the IMO 1997's problem. They got stuck and did nothing, and when I solved the problems, they didn't understand what I did (I would not have said nothing if it had been only the 1997 one, but the other were EASY solutions)

-When I exposed those problems, they had already trained 6 weeks, in a 8-hour per day training. We had seen all topics that could be included in the national round, but they still kept in mechanizing and mechanizing.

What can I do? Can someone tell me where do I find nice, programmed material to teach them well, from zero?

Maybe they are too exposed to 'Class Room' Mathematics. I am merely speaking from experience. What I used to do was if I found a Number Theory problem (for example Diophantine Equation $ y^{2}\equal{}x^{3}\minus{}2$) I would simply plug values in and if I can't find anything I would conclude that no solutions exist and if I found some I would declare they are the only ones (without proofs).

I may be wrong but I think their problem may be that they do not understand the notion of a proof.

[quote="Pul de Algodoncito"]Which is the best program of learning?? How have you learned to be able to solve problems? Is there a place where I can download problems from easy to hard? Any trick in special?[/quote]

I learned my problem solving skills from AoPS books and classes. However, the books and classes are rather expensive (no offense), so I don't know how that would work out. You could take sample problems from the AoPS books, past AMC's and AIME's (because they're in increasing order of difficulty), and other stuff. To make them stop using brute force, try:

1) Making problems where brute force is near imposssible or would take an hour or so
2) Making sure they understand the concept behind the "clever" way
3) Make math interesting way for them (easy way to do this: AoPS material)
4) Let them spend time on problems

Yep, that's all I can think of right now.


calculus
integration
trigonometry
calculus computations

I would like to integrate
$ \int\frac{dx}{x^{6}\plus{}1}$

I know it somehow requires the fact that $ y^{3}\plus{}1\equal{}(y\plus{}1)(y^{2}\minus{}y\plus{}1)$ but I can't get the entire partial fraction decomposition.

Is there a more general technique to evaluate integrals of the form
$ \int\frac{dx}{x^{2n}\plus{}1}$?

I don't want some generic answer, rather a generic technique.  Thanks.

Note: This isn't for a class or anything, just a problem I've come across.

hello, note that
$ \frac{1}{x^6\plus{}1}\equal{}\frac{1}{3(x^2\plus{}1)}\plus{}\frac{\minus{}x^2\plus{}2}{3(x^4\minus{}x^2\plus{}1)}$.
Sonnhard.

$ x^{NM}\plus{}1 \equal{} (x^M\plus{}1)\sum_{k\equal{}1}^N (\minus{}1)^{k\plus{}1} x^{M(N\minus{}k)}$

but whats to say any of these guys will be easier to integrate?

The entire partial fraction decomposition is generally given by the complete factorization of the denominator into roots of unity (here $ 12^{th}$ roots), but here once you've gotten the second term in Dr Grubner's expression you can get away with writing

$ x^4 - x^2 + 1 = (x^2 + 1)^2 - 3x^2 = (x^2 - \sqrt {3} x + 1)(x^2 + \sqrt {3} x + 1)$

and completing the square.  To be as general as possible one has something like

\begin{eqnarray*}
x^{2n} + 1 &=& \prod_{k = 0}^{n - 1} \left( x^2 - 2 \cos \frac{\pi (2k + 1)}{2n} x + 1 \right) \\
 &=& \prod_{k=0}^{n-1} \left( (x - \cos \frac{\pi (2k+1)}{2n})^2 + \sin^2 \frac{\pi (2k+1)}{2n} \right). \\
\end{eqnarray*}

[quote="JRav"]
Is there a more general technique to evaluate integrals of the form
$ \int\frac {dx}{x^{2n} \plus{} 1}$?

I don't want some generic answer, rather a generic technique.  Thanks.

Note: This isn't for a class or anything, just a problem I've come across.[/quote]
Metru had once proposed this problem, and that post could be seen here [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=65639[/url].

This integral was solved (relatively) recently at an another forum by [b]bidentate[/b] and [b]Opalg[/b].  You can see their solution here [url]http://www.sosmath.com/CBB/viewtopic.php?t=19930[/url]


Ramsey Theory
combinatorics unsolved
combinatorics

Let $ n$ be the number of points on a plane. Each of these points is joined to all the other points with a line segment. (e.g. for $ n \equal{} 4$ there are $ 6$ line segments)

If there are $ k$ different colors of line segments, what is the least $ n$ so that there will have to be one triangle made up of same colored line segments?

If you mean "triangle whose vertices are among the $ n$ points we started with," you're just asking for the Ramsey number $ R(3, 3, \ldots, 3)$.  If you mean something else, I've got nothing, although there have been posts on the forum with a similar flavor (maybe a Russian problem from 1999 or 2000 or something?  try a search).

I did mean the Ramsey number thanks!
Could you explain it a little further though (or share a link that explains it further - the wiki doesn't seem to have it)

Which wiki?  :P

[url=http://en.wikipedia.org/wiki/Ramsey%27s_theorem]Ramsey's Theorem[/url]

(that was darn tricky to make this link work!!)


i am just wondering, will the national honour rolls in canadian math competitions increase my chance of getting into Cornell or Carnegie Mellon?

No, they'll actually decrease your odds.

[hide=":)"] Of [b]course[/b] math competitions help!  Why in the world wouldn't they? [/hide]

nice comment there  :D
Shreyas see the private message i have sent u.


Let $ S>0$ and $ n\ge3$ be fixed. Find the minimum value of the expression
\[ \frac{a_1(1\plus{}a_2a_3)\plus{}a_2(1\plus{}a_3a_4)\plus{}\ldots\plus{}a_n(1\plus{}a_1a_2)}
{(\sqrt[3]{a_1a_2a_3}\plus{}\sqrt[3]{a_2a_3a_4}\plus{}\ldots\plus{}\sqrt[3]{a_na_1a_2})^3},\]
where $ a_1, a_2, \ldots, a_n$ are arbitrary positive numbers such that $ a_1\plus{}a_2\plus{}\ldots\plus{}a_n\equal{}S$.


inequalities
calculus
derivative
function
ratio
inequalities proposed

[quote="darij grinberg"][quote="mecrazywong"]It is just as if a nightmare to me at the first glance :D
Transform the inequality into \[\sum(a-1)^2(a-b)(a-c)\]Then it is easy by considering the 4 cases $a\ge b\ge c\ge1$, etc.[/quote]

Well, why distinguishing cases now?[/quote]
Oh...I don't know Schur is true for $x,y,z\in\mathbb{R}$ if p is even... :blush:

[quote="darij grinberg"][quote="mecrazywong"]It is just as if a nightmare to me at the first glance :D
Transform the inequality into \[\sum(a-1)^2(a-b)(a-c)\]Then it is easy by considering the 4 cases $a\ge b\ge c\ge1$, etc.[/quote]

Well, why distinguishing cases now? You could also say that it is simply the inequality

$x^2\left(x-y\right)\left(x-z\right)+y^2\left(y-z\right)\left(y-x\right)+z^2\left(z-x\right)\left(z-y\right)\geq 0$,

applied to the numbers x = a - 1, y = b - 1 and z = c - 1. In fact, this inequality holds for [i]any reals[/i] x, y, z, as follows from http://www.mathlinks.ro/Forum/viewtopic.php?t=41284 Theorem 3 [b](b)[/b] for p = 2. Another proof of this inequality is a two-liner:

$x^2\left(x-y\right)\left(x-z\right)+y^2\left(y-z\right)\left(y-x\right)+z^2\left(z-x\right)\left(z-y\right)$
$=\frac12\left(\left(y+z-x\right)^2\left(y-z\right)^2+\left(z+x-y\right)^2\left(z-x\right)^2+\left(x+y-z\right)^2\left(x-y\right)^2\right)\geq 0$

(with cyclic sums it would become a one-liner :D ).

________________________________________________________________________________________________

By the way, Thomas, thanks a lot for the inequalities script (for the others: I mean http://web.mit.edu/~tmildorf/www/Inequalities.pdf )! Just two remarks:

- In problem 6, the convexity is clear without computing second derivatives, since the left hand side is a sum of linear and constant functions (which are trivially convex) and of the functions $\frac{a}{b+c+1}$ and $\frac{b}{c+a+1}$ which are nothing but rescaled and translated images of the function $\frac{1}{a}$ and thus also convex by AM-HM.

- And as for problem 25, I don't know whether you intend to include geometric solutions, but in case you do, here is one: Denote the triangle by ABC and its three medians by $m_a$, $m_b$, $m_c$. Then, $m_a=\frac12\sqrt{2b^2+2c^2-a^2}$, $m_b=\frac12\sqrt{2c^2+2a^2-b^2}$, $m_c=\frac12\sqrt{2a^2+2b^2-c^2}$, and thus your inequality becomes $\frac{a}{m_a}+\frac{b}{m_b}+\frac{c}{m_c}\geq 2\sqrt3$. Now, it is well-known that the medians $m_a$, $m_b$, $m_c$ of triangle ABC are, themselves, the sides of a triangle, and the medians of this new triangle are $\frac34a$, $\frac34b$, $\frac34c$. Hence, the inequality $\frac{a}{m_a}+\frac{b}{m_b}+\frac{c}{m_c}\geq 2\sqrt3$ would follow from the inequality $\frac{\frac43m_a}{a}+\frac{\frac43m_b}{b}+\frac{\frac43m_c}{c}\geq 2\sqrt3$, applied to this new triangle. How to prove the inequality $\frac{\frac43m_a}{a}+\frac{\frac43m_b}{b}+\frac{\frac43m_c}{c}\geq 2\sqrt3$ for any triangle ABC ? Well, if G is the centroid of triangle ABC, then (since the centroid of a triangle divides every median in the ratio 2 : 1) we have $AG=\frac23m_a$, so $\frac43m_a=2\cdot\frac23m_a=2\cdot AG$, and similarly $\frac43m_b=2\cdot BG$ and $\frac43m_c=2\cdot CG$, and thus this inequality rewrites as $\frac{2\cdot AG}{a}+\frac{2\cdot BG}{b}+\frac{2\cdot CG}{c}\geq 2\sqrt3$. This is equivalent to $\frac{AG}{a}+\frac{BG}{b}+\frac{CG}{c}\geq\sqrt3$. But this follows from the inequality given in http://www.mathlinks.ro/Forum/viewtopic.php?t=21016 for M = G. Sorry for the messy explanations, I hope you can understand them.

  darij[/quote]

Now I see why Theorem 3 really does seem to cover the convex version of Vornicu-Schur :) ($(a-k)-(b-k)=a-b$ allows us to shift.) Though I will continue to search. My goal in writing this one was to find an example where the convex function version would be generally necessary, but to think I could do this so quickly was foolish!

On example 6, I included that calculation which is obvious to us intentionally so that the version of convexity I was referring to was clear. For example, I seem to recall that a similar solution posted somewhere on here resulted in a brief discussion of the convexity vs. "concavity" of $abc$.

On 25, I just added your solution. It was my intention to illustrate as many different approaches as I could on each problem. "$a, b, c$ sides of a triangle..." certainly does beg for a geometric solution. :)


2/Let  a,b,c,x,y,z>0  Find  the minimal value of the expression
    P= $ \frac {( a^{3} \plus{} x^{3} \plus{} 1)( b^{3} \plus{} y^{3} \plus{} 1)( c^{3} \plus{} z^{3} \plus{} 1)}{(a \plus{} b \plus{} c)(x \plus{} y \plus{} z)}$


algebra
polynomial
inequalities
algebra proposed

Let $ f(x) \equal{} a_0 \plus{} a_1 x \plus{} a_2 x^2 \plus{} ... a_n x^n$

satisfing: $ \forall i \in [1,m] 0 \le a_i \le a_0$  (Edited, thank you bambaman for the correction)

Let :

$ f^2(x) \equal{} b_0 \plus{} b_1 x \plus{} ... b_{2n} x^{2n}$

Then prove that :

$ b_{n \plus{} 1} \le \frac {1}{2} (f(1)) ^2$

Some one knows where it was proposed, Please?

first of all, I think that you forgot to mention that the coefficients of f are non-negative (else, choose 5, -3, -2, -6 as $ a_{0},a_{1},a_{2},a_{3}$ and get a contradiction).
$ b_{n + 1} = \sum_{i + j = n + 1, i,j > 0} a_{i}a_{j} \leq$(by CS or rearrangement)$ \sum_{i = 1}^{n} a_{i}^{2}$
it's enough to show that-
$ 2\sum_{i = 1}^{n} a_{i}^{2} \leq (a_{0} + \sum_{i = 1}^{n} a_{i})^{2}$
if we look at this as an inequality in $ a_{0}$ in degree 2, it's enough to show that $ a_{0}$ is as large as the bigger root of $ (x + \sum_{i = 1}^{n} a_{i})^{2} - 2\sum_{i = 1}^{n} a_{i}^{2} = 0$:
$ a_{0} \geq - \sum_{i = 1}^{n} a_{i} + \sqrt {2\sum_{i = 1}^{n} a_{i}^2}$
which I think is true because:
$ \sqrt {2\sum_{i = 1}^{n} a_{i}^2} \leq \sqrt {2(\sum_{i = 1}^{n} a_{i})^{2}} = \sqrt {2}\sum_{i = 1}^{n} a_{i}$
and therefore RHS is negative and we're done. what did I do wrong? [EDIT - the RHS is not negative of course...]


Find a closed form for $\sum^{k}_{n=1}n!$ in terms of k.

A nice form does not exist.  See:

http://mathworld.wolfram.com/FactorialSums.html


Let's think a sequence {a_n}

"There are a_n ns ins the sequence"

find all possible {a_n}s

There are many such sequences; the most obvious is $1, 2, 2, 3, 3, 4, 4, 4, 5, 5, 5, 6, 6, 6, 6, 7, \ldots$.


geometry
rectangle
perimeter
algorithm
greatest common divisor
number theory
relatively prime

The diagram shows a rectangle that has been dissected into nine non-overlapping squares. Given that the width and the height of the rectangle are relatively prime positive integers, find the perimeter of the rectangle.

[asy]
defaultpen(linewidth(0.7));
draw((0,0)--(69,0)--(69,61)--(0,61)--(0,0));draw((36,0)--(36,36)--(0,36));
draw((36,33)--(69,33));draw((41,33)--(41,61));draw((25,36)--(25,61));
draw((34,36)--(34,45)--(25,45));
draw((36,36)--(36,38)--(34,38));
draw((36,38)--(41,38));
draw((34,45)--(41,45));[/asy]

[quote="4everwise"]The diagram shows a rectangle that has been dissected into nine non-overlapping squares. Given that the width and the height of the rectangle are relatively prime positive integers, find the perimeter of the rectangle.

[img]6600[/img][/quote]
[hide]Label the sides of the smallest two squares a and b. The sides of the next largest squares, in order, are a+b, 2a+b, 3a+2b, 5a+3b, 4a+4b, 4a+5b, 3a+6b. The two sides of the rectangle are 7a+11b and 8a+9b. Using the euclidean algorithm we find that their gcf is that of 2b-a and 25b. We can equate two values for opposite sides of the rectangle to get $7a+11b=12a+9b$, which is the same as $5a=2b$. Plugging this in, we get that their gcd is the same as that of 4a and 25b. We can make the substitution $a=2x$ and $b=5y$. x=y, and the gcf of the sides is that of 8x and 125y. Since x and y are equal, the only way these two can have a gcf of 1 is if x=y=1. So the sides are 69 and 61. So our answer is $2(69+61)=\boxed{260}$.[/hide]

the aime grind has begun <3
[center][img width = 70]https://i.imgur.com/qHYWBE2.png[/img][/center]


Hi,

I always wanted to become the best mathematician in the world when I was younger. But, I have mere feeling that there are always someone who is better than me mathematically(unfortunately)... Is it possible that I could get smarter then them, or is it all about genetics and impossible for me to get smarter then them mathematically?

A couple things:

- What do you mean by the "best" mathematician in the world? People can't agree on that one as it is
- How much you contribute to mathematics does not necessarily correlate directly with how smart you are
- It's hard to contribute anything meaningful to mathematics if your principle objective is being the "best mathematician in the world." You'll be much more able to contribute something when your view of the role mathematics plays in your life matures a bit
- There are plenty of high profile problems that you can solve if you want to contribute something to mathematics. Ex: The Riemann Hypothesis
- If you are seriously interested in getting into math that deep, know that it's a huge gamble
- If you are seriously interested in getting into math that deep, get started, if you haven't already. And there's a certain degree of urgency to this. Many meaningful results take a long time to achieve. And before that, you're going to have to spend a while studying, exposing yourself to new mathematics, thinking about mathematics, and deciding which field of mathematics interests you most

[quote="TZF"]- It's hard to contribute anything meaningful to mathematics if your principle objective is being the "best mathematician in the world." You'll be much more able to contribute something when your view of the role mathematics plays in your life matures a bit [/quote]
I'd like to emphasize this point.  If you think mathematics is primarily a competition, you've got a lot of learning to do about how the mathematical community (and the scientific community at large) actually functions and about what it means to contribute to mathematics.  Depending on where you are in life, you may benefit from reading various parts of Terence Tao's [url=http://terrytao.wordpress.com/career-advice/]career advice[/url].

In particular, I direct you to Terrence Tao's comments on the question, [url=http://terrytao.wordpress.com/career-advice/does-one-have-to-be-a-genius-to-do-maths/]"Does one have to be a genius to do maths?"[/url]


That's a little vague. You'd probably have better luck searching through the threads on one of the general problem solving forums.

i have good algebra &combinatorics&number theory questin but idontn have good geomtry question

Maybe you'd be happier in, oh say http://www.artofproblemsolving.com/Forum/index.php?f=4 ? This is the state of GA, not the nation of GA, and we haven't had IMOers in I don't know how long.


Convert $ 0.2010_3$(with a repeating bar over the 2010) to a base 10 fraction.

(i dont knwo how to do the repeating bar in latex)

[hide] This is simply $ (\frac23\plus{}\frac{1}{3^3})\plus{}(\frac{2}{3^5}\plus{}\frac{2}{3^7})\plus{}\dots$

$ \implies \frac{19}{3^3}\plus{}\frac{19}{3^7}\plus{}\dots$

$ \implies \frac{19}{27}(1\plus{}\frac{1}{81}\plus{}\frac{1}{3^8}\dots)$

Find the sum of the infinite series, multiply, and you have it.[/hide]

how did you get it down to that pattern (with the exponents and 3)?

Well, if a number ends in $ x_1x_0.b_1b_2b_3...$ in base $ b$, then that is equivalent to $ x_1 \times b^1 \plus{} x_0 \times b^0 \plus{} b_1 \times b^{\minus{}1} \plus{} b_2 \times b^{\minus{}2} \plus{} ... \plus{} b_n \times b^{\minus{}n}$.

[hide="Alternative Solution"]Let $ x$ be your desired number. $ 81x \equal{} 2010.20102010\ldots_3$, so $ 80x \equal{} 2010_3$. $ 80x \equal{} 57$ in decimal, so $ x \equal{} \boxed{\frac{57}{80}}$.
[/hide]


limit
logarithms
integration
real analysis
real analysis solved

Prove that

[tex]\lim_{n\to\infty} \frac{1}{e^n}\sum_{k=0}^n \frac{n^k}{k!}= \frac12[/tex]
where [tex]e[/tex] is base of the natural logarithm and [tex]k!= 123...k[/tex]

Using identity $e^n-\sum_{k=0}^n\frac{n^k}{k!}=\frac{1}{n!}\int_0^ne^t(n-t)^ndt$ we obtain that we need to prove that
\[\frac{1}{n!}\int_0^1n^{n+1}e^{-nt}t^ndt\stackrel{n\to\infty}{\longrightarrow}\frac{1}{2}.\]

Due to Stirling's formulae we know $\frac{n^{n+1}}{n!}\approx \frac{e^n\sqrt{n}}{\sqrt{2\pi}}$.
Therefore we need to prove $\int_0^1e^{n(1-t)}t^ndt\approx \frac{\sqrt{2\pi}}{2\sqrt{n}}$.

We can write $\int_0^1e^{n(1-t)}t^ndt=\int_0^1e^{n(1-t+\ln t)}dt$. Applying Laplace equivalence theorem we immediately obtain desired result.

I think your proof is correct but let's make things clear.

1. Is it known that $e^n-\sum_{k=0}^n\frac{n^k}{k!}=\frac{1}{n!}\int_0^ne^t(n-t)^ndt $. Just give me a link where I can find a solution for that. I am new to mathlinks.

2. What is so called "Laplace equivalence theorem". I searched in google and mathworld but found no hit.

thanks

[quote="Myth"]1. It is one of the forms of Teylor's series remainder. You can find it in any analysis book. Actually, it is integrating by parts.

2. I forgot exact english equivalent, so I write "Laplace equivalence theorem". I think, Moubinool or Kent can say exact name of this theorem.[/quote]

1. I got it.
2. Hope they answer me soon.

I know a solution using central limit theorem. Quit nice but not smart like yours.  Can you or anyone find my solution :) ?

[quote="Myth"]And what is central limit theorem?[/quote]
I think you know probability. Here we go: http://mathworld.wolfram.com/CentralLimitTheorem.html

Ok. Now i realize what you are saying about. :)


Hi,
    Where Can I purchase your books in Singapore?
Thanks
dhina

You can order them from our website, the only problem is that the shipping fees will be significantly higher than those in US (obviously.)

Click on the Bookstore link next to the top left corner and shop on your desired books.


I don't know if this inequality was posted before( I apologise if this is true :oops: ), but here it goes:

 Prove that in any triangle $ABC$ the following inequality holds:

     $3(ab+bc+ca)\leq 8(m_a m_b+m_b m_c+ m_c m_a)\leq 10(ab+bc+ca)$ ;)

It follows immediately that $2b^2+ca\geq 4m_a m_c$ and so on and by adding them we have $ 2(a^2+b^2+c^2)+ab+bc+ca\geq 4 (m_a m_b+m_b m_c+m_c m_a)$. Next use $ a^2+b^2+c^2< 2(ab+bc+ca)$ and one part is done. The second part (and if my computations are right) is an immediate consequence of the right inequality (use duality!).

The following ineq is true :
 $ \frac{4}{5}(m_am_b+m_bm_c+m_cm_a) < ab+ac+bc < \frac{20}{9}(m_am_b+m_bm_c+m_cm_a) $
  The first one can be proved by using the ineq : $ m_a +m_b+m_c < a+b+c $
  The second one can be prove by using the first one  ;)


vector
abstract algebra
superior algebra
superior algebra unsolved

Is there a field $ K$ such that $ K^*/\mathrm{Torsion}$ is not free?

$ \mathbb R^\times/\{\pm1\}$ is a (free) $ \mathbb Q$-vector space, $ \mathbb Q^\times/\{\pm1\}$ is a free abelian group. So what do you mean by "free"? owk

I meant free abelian group, so $ \mathbb{R}$ is an example.


limit
modular arithmetic
number theory unsolved
number theory

For any positive integer $ k$, let $ S(k)$ be the sum of its digits. Prove that

$ \lim_{n\rightarrow\infty}S(2^n) \equal{} \infty$

I give a try. It seems clear that $ \lim_{n\rightarrow\infty}S(2^n) \equal{} \lim_{n\rightarrow\infty}S(2^{3n})\equal{}  \lim_{n\rightarrow\infty}S(8^n)$, but $ \lim_{n\rightarrow\infty}S(8^n) \equiv \pm 1 \pmod{9}$, that is it oscillates, depending on the parity of $ n$. This means that there is no finite value to which $ S(2^n)$ tends, otherwhise $ \lim_{n\rightarrow\infty}2^n \equiv k \pmod{9}$. Very sorry for being so sketchy. Is this idea good?

[quote="geda"]I give a try. It seems clear that $ \lim_{n\rightarrow\infty}S(2^n) \equal{} \lim_{n\rightarrow\infty}S(2^{3n}) \equal{} \lim_{n\rightarrow\infty}S(8^n)$, but $ \lim_{n\rightarrow\infty}S(8^n) \equiv \pm 1 \pmod{9}$, that is it oscillates, depending on the parity of $ n$. This means that there is no finite value to which $ S(2^n)$ tends, otherwhise $ \lim_{n\rightarrow\infty}2^n \equiv k \pmod{9}$. Very sorry for being so sketchy. Is this idea good?[/quote]

Well it is NOT clear that $ \lim_{n\to \infty}S(2^{n}) \equal{} \lim_{n\to \infty}S(2^{3n})$
What is more (putting aside that every step in your proof has mistakes) the fact that it does not converge to a finite number does not mean it diverges to infinity (for example $ ( \minus{} 1)^{n}$)

It is a really nice problem and I'll post a sketch of a proof:

We shall prove a more general problem:
If $ a$ is an even natural number such that $ 5\not |a$ then $ \lim_{n\to \infty}S(a^{n}) \equal{} \plus{} \infty$
Define $ \{a_{k}\}_{k \equal{} 1}^{\infty}$ as an integer sequence consisting of digits of $ a^{n}$ (for sufficiently large k all $ a_{k}$ are zeros).
Of course $ a_{1}\neq 0$
Now we have to prove a lemma that will finish off our problem:

If $ 1\leq i \leq \frac {1}{4}n$ then among $ a_{i \plus{} 1},a_{i \plus{} 2},...,a_{4i}$ there exists such $ l\in \{i,...,4i\}$ that $ a_{l}\neq 0$ (after that the proof is quite easy)
Proof:
Assume that we have $ a_{i} \equal{} a_{i \plus{} 1} \equal{} ... \equal{} a_{4i} \equal{} 0$
Then we would have $ 10^{4i}|a^{n} \minus{} c$ where $ c$ is a number consisting of $ a_{i}a_{i \minus{} 1}...a_{2}a_{1}$ as digits
So $ 2^{4i}|a^{n} \minus{} c$ but $ a$ is even, so
$ 2^{4i}|c$ which is impossible because $ 2^{4i} > 10^{i} > c$

I'm hoping it helped a bit :wink:

It posted before. I give solution, suth work and for $ S(a^n)\to \infty$


[quote="unimpossible"]http://www.awesomemath.org/AMSP_dailysch.pdf

on page 3 it says computational geometry is the intermediate combinatorics course.

I think i should assume it means the intermediate geometry couse?[/quote]
no it means what it says, intermediate combinatorics

haha junggi...

I guess that is a typo...It obviously means an intermediate geometry course...

Everyone can make mistakes, so naturally it would be intermediate geometry. After all, I'm pretty sure there is another intermediate combinatorics course, so by the Pigeonhole Principle, one must be the fake  :) .


function
calculus
derivative
complex analysis
algebra
functional equation
real analysis

Is there a $C^{\infty}$ real function $f$ with $f(f(x))=e^x$? What about a continuous one?  :D

is this something you made up? It makes no sense

May I ask you why it makes no sense? The second question is a joke, but the first one is serious. Could you give some reasonable arguments for your assertion? And to answer you: no, I didn't make it up, it is famous problem.

The problem is not that hard, especially if one starts with the second question about existence of a continuous function. (BTW, why is it a joke?  :? ). Choose any $a<0$ and define $f$ to be any continuous strictly increasing function on $[-\infty,a]$ such that $f(-\infty)=a$ and $f(a)=0$. Then use the identity $f(x)=e^{f^{-1}(x)}$ to define $f$ inductively on the intervals $[a,0]$, $[0,e^a]$, $[e^a,1]$, and so on (each interval will be mapped onto the next one). It is not hard to check that the resulting function is continuous and satisfies $f(f(x))=e^x$. In order to make $f$ infinitely smooth, it suffices to make sure that $f'>0$ and the left derivatives of $f$ agree with the right derivatives of $e^{f^{-1}}$ at $a$. But the left derivatives of $f$ at $a$ are completely determined by the values of $f$ near $a$ and the right derivatives of $e^{f^{-1}}$ at $a$ are completely determined by the  values of $f$ near $-\infty$. Since these values can be chosen independently (you can glue any 2 functions in $C^\infty$), to make the derivatives agree is also easy.

I still don't know whether one can make $f$ real analytic. It is not hard to show that $f$ cannot be an entire function, though...  :)

Actually, I posted the second question as a hint and when I realised that it is a hint too strong, I named it a joke. Of course, with experts like you this doesn't work, but for others... I also managed to prove that $f$ cannot be entire, but the question about analytic functions seems really hard.  :(

Since it really is not hard, as fedja said, but it's pretty cute in itself :), here's a proof (a sketch of a proof, actually) that $f$ cannot be entire (the initial problem was probably too much for me, and I was one of those non-experts that harazi's "mind games" fooled :D):

We have $f(e^z)=e^{f(z)}$, so $f(z)\ne 0,\ \forall z\ne 0$. We also have $f(0)\ne 0$ from the initial functional equation, so $f$ does not vanish. This means that there is an entire function $g$ s.t. $f=e^g$, so $e^x=f(f(x))=e^{g(e^{g(x)})}$, which would make $g(e^{g(x)})$ too "slowly-increasing" (it's $x$ plus some constant of the form $2n\pi i,\ n\in\mathbb Z$) :).


inequalities
trigonometry
geometry
circumcircle
trig identities
Law of Sines
geometry solved

Let $M,N,P$ be arbitrary points on the sides $BC,CA,AB$ respectively of an acute-angled triangle $ABC$. Prove that at least one of the following inequalities is satisfied: $NP\geq \frac 12 BC,PM\geq \frac 12 CA, MN\geq \frac 12 AB$.

I hope I can explain it properly :).

Assume the contrary, and let $D,E,F$ be the midpoints of $BC,CA,AB$ respectively. It's easy to conclude that exactly one of the points $M,N,P$ must lie inside each of the triangles $AFE,FBD,NDC$. Assume WLOG that $M,N,P$ lie on the segments $BD,CE,AF$, respectively, and that $BC<AB<CA$.

If we also assume that it's impossible to move $M$ on $BD$ so that both $MN,MP$ decrease, and the same for $N$ and $P$, then we reach the conclusion that the triangles $APN,PBM,NMC$ have no obtuse angles.

Moving $M$ towards $D$ will make $MP$ increase and $MN$ decrease. We can thus move it continuously, until we reach a position where $MP=DF$ (by assumption, we had $MP<DF$). Now do the same for $N$, to reach a position where $MN=DE$ (and $NP<EF$).

We thus have a triangle $MNP$ inscribed in $ABC$, with $MN=\frac{AB}2,MP=\frac{AC}2$, and $NP<\frac{BC}2$ (and with the points $M,N,P$ as above, on the segments $BD,CE,AF$ respectively). Now start moving $M$ continuously on $BC$ towards $B$, and also move $N,P$ on $CA,AB$ continuously s.t. $MN,MP$ preserve their lengths. It's easy to see from the fact that $BC$ is the smallest side of $ABC$ that at some point $NP$ will become greater than $\frac{BC}2$, so, by continuity, it must also be equal to $\frac{BC}2$ at some point.

We now have the following situation: $MNP,DEF$ are both inscribed in $ABC$, with $M,D\in BC,\ N,E\in CA,\ P,F\in AB$ and are congruent. It's not hard to prove that this is impossible.

Can i just ask how do you get problems from serbia and montenegro olympiads because we don't have website or something like that where we  post problems from olympiads? :roll:  :roll:

My teacher gave me the questions of Serbia and Montenegro  ;)  :P

A different solution (with different notations):

[color=blue][b]Problem.[/b] Let X, Y, Z be three points on the sides BC, CA, AB of an acute-angled triangle ABC. Prove that at least one of the three inequalities $YZ\geq\frac{a}{2}$, $ZX\geq\frac{b}{2}$, $XY\geq\frac{c}{2}$ is fulfilled.[/color]

[i]Solution.[/i] Assume the contrary: assume that none of these three inequalities is fulfilled. Then, we have $YZ<\frac{a}{2}$, $ZX<\frac{b}{2}$, $XY<\frac{c}{2}$.

First, define a point P according to http://www.mathlinks.ro/Forum/viewtopic.php?t=39473 post #2. This point P will satisfy $\left(\frac{YZ}{\sin A}\right)^2\geq AP^2$, so that $\frac{YZ}{\sin A}\geq AP$. But since $YZ<\frac{a}{2}$, we have $\frac{YZ}{\sin A}<\frac{a}{2\sin A}$. Thus, $AP<\frac{a}{2\sin A}$ If R is the circumradius of triangle ABC, then the extended law of sines yields $\frac{a}{2\sin A}=R$, so this becomes AP < R. Similarly, BP < R and CP < R. Hence, $\max\left(AP;\;BP;\;CP\right)<R$. On the other hand, since the number $\min\left(AP;\;BP;\;CP\right)$ is greater or equal to each of the numbers AP, BP, CP, the circle with center P and radius $\max\left(AP;\;BP;\;CP\right)$ will contain all the three points A, B, C inside it or on its boundary, so it will contain the triangle ABC. But this contradicts with the well-known fact that a circle containing an acute-angled triangle must have its radius greater or equal to the circumradius of this triangle (in fact, as we know, our circle has its radius $\max\left(AP;\;BP;\;CP\right)$ smaller than the circumradius R of the triangle ABC).

So we have obtained a contradiction, and our assumption was wrong. Consequently, the problem is solved.

  Darij

[quote="ehsan"]My teacher gave me the questions of Serbia and Montenegro [/quote]
so would U say how your teacher get these problems  ;)  ;)

Well i didn't want to bo rood and boring but when U asked, I am joining to that question...
And why do you see your teacher for summer break.

Jesam li ja jedini iz Srbije koji je stavio svoju zastavu... izdajnici doma svoga :D :D :D

[quote]And why do you see your teacher for summer break.

[/quote]
 :lol: I have a summer classes  ;) for math,physic ,olympiad,etc.


geometry
trigonometry
function
pigeonhole principle
number theory
complex numbers

I'm planning to do an Extended Essay for IB in Math, but I have no idea what to choose as a topic. I'm an average math student, but in Math Higher Level. What is it that I could do with the knowledge I have?

My math teacher told me that it would be hard to write a paper in math unless you've done at least a bachelor's in math. But, IB does have an option for an extended essay in math. So, it can't be THAT hard. Well, the essay topic has to be very specific. For example:
1) Application of spherical geometry and trigonometry in predicted Lunar Eclipse.
2) An Analysis of the effect of alterations to the recursively defined equation of fractals.

But, I can't take those, as they've already been done. Oh, and the essay itself as the be between 2500-4000 words. So, I'll need enough to write about, and not only write, but a decent amount of mathematical content as well. Well, thank you for your input. It's truly appreciated. :D

You can do an analysis of how pure number theory affects the entire communications world. It may sound boring reading it here, but reading some of the books about it is quite exhilarating. There is also almost an endless supply of information and quotes about it. I especially enjoy Hardy's quote on how useless number theory is to the real world (obviously said before the securities boom, especially since he was dead then.)

I did an EE in math last year (received an A) heavily helped another student (who also received an A) and am currently helping two other students this year.

So I'm pretty familiar with it.

Here's (I think) a cool topic:

Applications of Complex Numbers to Geometry.

Topics which would be fairly simple and easy to write on include:

Fibonacci Numbers and their Properties.
Power Series and Generating Functions with applications to Recursive Equations.
The Pigeonhole Principle.
Coloring and Parity

Hope some of these help, if not, feel free to PM me.


[quote="math92"]How many 3 digit numbers contain the digit 5 at least once?[/quote]

[hide]In each hundred besides 500 there are 9-0+1 numbers with units digit 5. There are 9-0+1-1 with tens digit 5 not including the one with the units digit 5. 10+9=19 per hundred. There are 9-1+1-1=8 hundreds not including the 500. There are 599-500+1=100 numbers with hundreds digit 5. The total number is 19(8)+100=252.[/hide]

[quote="pianoforte"][hide]There are $9^3$ three-digit numbers that do not contain 5.
$10^3-9^3=\boxed{271}$ numbers that contain the digit 5 at least once.[/hide][/quote]

Um...there are 900 3 digit numbers ($9*10*10$) because the first digit can't be zero.  Same goes with the numbers with 5 in them $8*9*9$.  So $900-648=252$

$252$?

I probably messed up though

[hide]
There are 900 numbers.
Out of those numbers, $8\cdot9\cdot9=648$ do not have a 5. 900-648=252.[/hide]

[quote="mbabbitt2003"]well, most of the people that responded used a technique that is very handy in this problem:

How many 8-digit numbers have at least one 9 in them?

Try figuring that out.[/quote]
[hide] There are $90000000$ numbers. $8\times9^7=38263752$ numbers have no 9s. $90000000-38263752=\boxed{51736248}$ Umm... :? That doesn't look right.[/hide]
[color=orange]
[size=75][hidden by solafidefarms][/size][/color]


Source: MMPC 1995, Round 1 (#35)

On June 1, Jill earns a certain amount of money.  Each following day of June, July, and August, she earns one dollar more than she did on the previous day.  By august 31, she has been paid a total of $6670.  How much did she earn on the Fourth of July?

(A) 58
(B) 59
(C) 60
(D) 61
(E) 72

I understand this!!!!!

i hope i did this right:

[hide]days from june to august is 30+31+31=92

so set the amount of money she earned on the first day as x.

6670=92/2(2x+(92-1)1)

6670=46(2x+91)

145=2x+91

54=2x

27=x

27 dollars on june 1st

so money earned on june first would be x+(34-1)

27+33=60

C would be the answer.[/hide]

Leeyuhang, you are [hide]right as rain[/hide]!


[quote="Anaxerzia"]Bwa just bust out yo TI-89 (or even a scientific calculator) and get approximately -3.2091[/quote]

This is absolutely spam.

Number 1, you cant use a caculator
Number 2, it should be an exact value
and Number 3, its not -3.2091 because Im positive he/she means its 20 degrees and 40 degrees, because that will get you an exact value.

[hide="HINTS"]
The sum of angles for tangent $\tan \frac{\alpha}{2}= \frac{\sin \alpha}{1+\cos \alpha}= \frac{1-\cos \alpha}{ \sin \alpha}$.
Use the the relations $\cos ( \pi-\theta ) = \sin \theta$ and $\cos \theta = \sin ( \pi-\theta )$.
Then $\tan 40 = \frac{\sin 40}{\cos 40}= \frac{\cos 60}{\sin 60}$.
To get the $\tan 20$ to be simplified $\tan 20 = \frac{1-\cos 40}{ \sin 40}\equiv \frac{\sin 40}{1+\cos 40}\equiv \frac{1-\sin 60}{ \cos 60}\equiv \frac{\cos 60}{1+\sin 60}$.
Then just simplify the expression.
[/hide]

[quote="rebirth"]What is $tan20+tan40+(\sqrt{3})tan20*tan40$[/quote]
[hide="hint"]Note that $\sqrt{3}=\tan 60$.[/hide]


Find all real number $ m$ such that exits continous function $ f: R \to R$ such that
$ f(x\plus{}m)[f(x)\plus{}m]\plus{}m^2\equal{}0$

[quote="ith_power"][quote="Rust"]...or $ g(x) \equal{} \frac {a_{11}g(0) \plus{} a_{12}}{a_{21}g(0) \plus{} a_{22}}$, where
$ A \equal{} \binom{a_{11} a_{12}}{a_{21} a_{22}} \equal{} \binom{1 0}{1 1}^x$.[/quote]
Could you please explain this step  :blush: or please tell me if i can find it in a book. :blush:

by the way,is your solution true for all reals, or integers only. :maybe:[/quote]

don't worry, rust always tries to be very smart, i guess it makes him feel happy

Rust [i]is[/i] very smart, althought I also think some people here could make more detailed solutions.

Suppose by absurd there is a continuous function $ f$ such that

$ f(x \plus{} 1)[f(x) \plus{} 1] \equal{} \minus{} 1$,

If there is some $ x$ for which $ f(x) \ge 0$, we have $ f(x) \plus{} 1 > 0$, therefore $ f(x \plus{} 1) < 0$. Then, in $ [x, x \plus{} 1]$ there is some $ y$ such that $ f(y) \equal{} 0 \Rightarrow f(y \plus{} 1) \equal{} \minus{} 1 \Rightarrow \minus{} 1 \equal{} f(y \plus{} 2)[f(y \plus{} 1) \plus{} 1] \equal{} 0$, contradiction. So $ f(x) < 0 \forall x$. Let us note that $ f(x \plus{} 1) < 0 \Rightarrow f(x) \plus{} 1 > 0 \Rightarrow f(x) > \minus{} 1 \forall x$. But $ 0 > f(x \plus{} 1) > \minus{} 1$ gives $ f(x) \plus{} 1 \equal{} \frac {1}{ \minus{} f(x \plus{} 1)} \equal{} \frac {1}{|f(x \plus{} 1)|} > 1$, which implies $ f(x) > 0$. Absurd!

Then $ m \equal{} 0$ (see Rust's post).

[quote="mylove"]Find all real number $ m$ such that exits continous function $ f: R \to R$ such that
$ f(x \plus{} m)[f(x) \plus{} m] \plus{} m^2 \equal{} 0$[/quote]
[u][b]Case 1:[/b][/u] $ m\equal{}0$ then $ f(x) \equiv 0$
[u][b]Case 2:[/b][/u] $ m>0$,suppose that exits continous function $ f: R \to R$ such that
$ f(x \plus{} m)[f(x) \plus{} m] \plus{} m^2 \equal{} 0$
=>$ f(x) \neq 0$ all $ x\in R$
Note that $ f$ is a continous function,so $ f(x) >0$ with all $ x \in R$ or $ f(x) <0$ with all $ x \in R$
If $ f(x) >0$ with all $ x \in R$
then $ f(x\plus{}m) >0$ with all $ x \in R$
=>$ f(x)\plus{}m <0$ with all $ x \in R$
=>$ f(x) <\minus{}m<0$ with all $ x \in R$,false

If $ f(x)<0$ with all $ x \in R$,then $ f(x\plus{}m)<0$ with all $ x \in R$
=>$ f(x)\plus{}m>0$ with all $ x \in R$
=>$ 0>f(x)>\minus{}m$ with all $ x \in R$
We have $ 0>f(x\plus{}m)>\minus{}m;f(x)\plus{}m<m$
So $ f(x\plus{}m)(f(x)\plus{}m)>\minus{}m^2$ with all $ x \in R$,false
[u][b]Case 3: [/b][/u]$ m<0$(similar with $ m>0$)
So $ m\equal{}0$


I wasn't sure where to post this, so this'll do.

Consider the following basic example:

101 points are placed in a 1x1 square. Prove that three of them can be found which form a triangle with an area \leq 0.01.

So, my first thought was to divide the square into 50 equal rectangles, each with an area of 0.02. One of them has to have three points in it, so the triangle that they form, being completely inside of the rectangle with area 0.02, must have an area  \leq 0.01.

That's all good, but it's not true. What if those three points are in a line? For that matter, what if [b]all[/b] the points are in a line?

What would you do in such a case? State that the problem as given is impossible, because for all we know all the points might be in a line, and then go on to prove it with an additional assumption (e.g. no three points are in a line, or something weaker)?

Please advise!

Thanks,
Jonathan[/b]

We had an alike problem on last year's training stage, and there the problem said, "101 point, from which no 3 are collinear".

So indeed, the correct answer to this question is that the statement is false, and you may sometimes only get full credit if you work out the non-collinear alternative.

Ok, thanks.

Sometimes I'm not quite careful enough...

Jonathan


Sorry. Pllease go here instead. [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1156542#1156542[/url]

Well, I'm not sure if it really belongs here. How would you deal with the speed issue? It could go in the Math & Strategy Games forum.

OK. I am swiching this to the Math & Strategy Games forum. Here's a link. [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1156542#1156542[/url] SORRY!


function
trigonometry
logarithms
real analysis
real analysis unsolved

Let's say that $ f_{n}(x) \equal{} min_{m \in \mathbb{Z}} |x\minus{} \dfrac{m}{n}|$. Decide for wich $ x$ the sum $ \sum_{n \geq 1} f_{n}(x)$ converge.

Integers, and nothing else. Rewrite that as $ \frac1n s(nx)$, where $ s$ is the piecewise linear function connecting $ (0,0), (\frac12,\frac12),$ and $ (1,0)$, extended to be periodic of period $ 1$. Since $ s$ has average value $ \frac14$ and the fractional parts of $ nx$ are equidistributed on the unit interval, we expect the partial sums to be approximately $ \frac14$ those of the harmonic series. The one exception comes at integers, where $ f(nx)$ is identically zero.

For a rigorous quantitative estimate, we compare this to a series we can calculate exactly. We have that $ s(x)\ge 0.43\sin^2(\pi x)$ everywhere (sharp only at integers), so
$ S(x)\equal{}\sum_{n\equal{}1}^N \frac1n s(nx)\ge 0.43\sum_{n\equal{}1}^N \frac1n\sin^2(n\pi x)$

$ \ge 0.43\sum_{n\equal{}1}^N \frac1n\left(\frac12\minus{}\frac12\cos(2n\pi x)\right)$

$ \ge 0.43\left(\frac12 \sum_{n\equal{}1}^N\frac1n\minus{}\sum_{n\equal{}1}^N\frac{\cos(2n\pi x)}{2n}\right)$

As $ N\to\infty$, the harmonic number $ \sum_{n\equal{}1}^N\frac1n$ is approximately $ \ln N$ plus a constant. For the other term, we have the Fourier series $ \sum_{n\equal{}1}^{\infty}\frac{\cos(2n\pi x)}{2n}\equal{}\minus{}\frac12\ln(2\sin \pi x)$. At anywhere other than the integers, that's bounded. The overall sum looks like a multiple of $ \ln(N)$ plus a constant, and diverges to $ \infty$.


Find the best number $ \alpha,\beta$ so that this inequality true for all natural number $ n$:
$ (\frac{n\plus{}1}{n})^{n\plus{}\alpha}<e<(\frac{n\plus{}1}{n})^{n\plus{}\beta}$

It give $ \alpha <f(n)<\beta$, were $ f(x)\equal{}\frac{1}{ln(x\plus{}1)\minus{}ln(x)}\minus{}x$. It is not hard to show, that $ f'>0$ and $ f(1)\equal{}\frac{1}{ln 2}\minus{}1\equal{}x_0, \ f(\infty )\equal{}0.5$. Therefore $ \alpha<x_0,\beta \ge 0.5$ work.


As $n$ even clearly doesn't work we can suppose that $n=2k+1$. As $k=0$ clearly works, we might suppose that $k \geq 1$. Now:

$x^{2k+1}+4((2+x)^{2k}-....+(2-x)^{2k})=0$ so $2|x$ let $x \to 2x$ and we have:

$2^{2k-1}x^{2k+1}+2^{2k}((1+x)^{2k}-...+(1-x)^{2k})=0$ so again $2|x$ and we let $x \to 2x$. Finally:

$2^{2k}x^{2k+1}+(1+2x)^{2k}-(1-4x^2)(1+2x)^{2k-2}+....+(1-2x)^{2k} \equiv (1+2x)^{2k}-1+(1-2x)^{2k}\equiv 1 \not \equiv 0 \mod 4$ Contradiction. So we only $n=1$ satisfies given condition.

Megus please could you explain your solution...  :(

Davron

[quote="shobber"]Determine all positive integers $n$ for which the equation
\[ x^n + (2+x)^n + (2-x)^n = 0  \]
has an integer as a solution.[/quote][hide]Suppose $n$ is even. Then \[ x^n+(2+x)^n+(2-x)^n\ge0, \] with equality at $x=2+x=2-x=0$, which clearly, can not be obtained. Hence, $n$ cannot be even.

Suppose $n$ is odd. Then \[ x^n + (2+x)^n + (2-x)^n = 0  \] implies \[ x^n+(2+x)^n=(x-2)^n. \] Consider $n=1$. Then \[ x+(2+x)=(x-2) \] implies $x=-4$.

Now suppose $n\ge3$. If $x>2$, then from Fermat's Last Theorem, there are no integer solutions. The same conclusion follows if $x<-2$ and both sides are multiplied by $-1$. Furthermore, $x=2$, $x=1$, $x=0$, $x=-1$, and $x=-2$ have no solutions, upon inspection.

Therefore, the only valid $n$ for which the equation has integer solutions in $x$ is $\boxed{n=1}$.[/hide]

I think megus's solution is very easy to understand because his solution is similar to the proof of Fermat to show that the equation $ x^4+y^4=z^2 $ has no integer roots :)

[quote="shobber"]Determine all positive integers $n$ for which the equation

\[ x^n + (2+x)^n + (2-x)^n = 0 \]

has an integer as a solution.[/quote]

Trivial.Since $n$ positive integer,we see two cases.First $n$ even and the result is obvious $x=0$.now see if $n$ even,we can use fermat's last theorem

We can see that n is odd. We can also notice that x is negative so let y = -x. We have $-y^n+(2-y)^n+(2+y)^n = 0 \rightarrow (y+2)^n = y^n+(y-2)^n$.(apply fermats last theorem) Considering this equation mod y we see that $y = 2^k$. Letting 2z = y we obtain the new equation $(z+1)^n = z^n + (z-1)^n$ and considering this mod z yields z = 1, 2. y = 2 means n = 0. y = 4 means n = 1.

Amazing, this problem is APMO 1993 (this topic) and [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=352715&#p352715]IMO LongList 1988 Problem 28[/url]. :?

why does this problem remind me of fermat's last...?

if $ n $ is even then $ x=2-x=2+x=0 $ false.
if $ n $ is odd then $ (2+x)^{n}+(2-x)^{n} $ is even, $ 0 $ is even so $ x $ is even.
take $ x=2w $ so $ w^{n}+(1+w)^{n}+(1-w)^{n}=0 $ but $ (1+w)^{n} $ is congruent with 1 mod $ w $ and $ (1+w)^{n} $ is congruent with 1 mod $ w $ and $ 0 $,$ w^{n} $ are divisible by $ w $ so $ 2 $ is divisible by $ w $.
$ w $ is even, the proof of this is similar like the proof for $ x $.
1. $ w=2 $ so $ 2^{n}+3^{n}=1 $ impossible.
2. $ w=-2 $ so $ 2^{n}+1=3^{n} $ but $ 2^{n}+1<3^{n} $ for $ n>1 $ so $ n=1 $.


It is not true, Silouan. You can choose $a,b,c$ which make the equality of VASC's inequality.

hungkhtn, easier the following counterexample: $a=b=1$ and $c=0.$

Yes, Arqady. However, the following nice inequality holds

If $a,b,c$ are three side-lengths of a triangle then
\[(ab+bc+ca)(a^{2}+b^{2}+c^{2}) \le 3(a^{3}b+b^{3}c+c^{3}a).\]

I can not understand what happen here?  :blush:

[quote="hungkhtn"][quote="hungkhtn"]Yes, Arqady. However, the following nice inequality holds

If $a,b,c$ are three side-lengths of a triangle then
\[(ab+bc+ca)(a^{2}+b^{2}+c^{2}) \le 3(a^{3}b+b^{3}c+c^{3}a). \]
[/quote]

Arqady? DO you have a solution to my problem without Ravi's substitution?[/quote]
The inequality is equivalent to
\[\sum b^{2}(a-b)(a-c)+\sum a^{2}b(a-b) \ge 0 \]
Note that $a,b,c$ are the sides of a triangle then by Mildorl's Lemma (genelization of Vornicu Schur), we have
\[\sum b^{2}(a-b)(a-c) \ge 0 \]
By IMO 1983 inequality, we have
\[\sum a^{2}b(a-b) \ge 0 \]
The inequality is proved. :)


inequalities
number theory unsolved
number theory

$ 5(xy\plus{}xz\plus{}yz)\equal{}4xyz$

Of course: $ x,y,z$ are natural numbers

wlog: $ x\leq y\leq z$ and therefore $ xy\leq xz\leq yz$
so $ 5(xy \plus{} yz \plus{} xz)\leq 5(3yz) \equal{} 15yz < 4xyz$ if $ x\geq4$ therefore $ x < 4$

if $ x \equal{} 1$ we get $ 5(y \plus{} z \plus{} yz) \equal{} 4yz$ clearely no solutions

if $ x \equal{} 2$ we get $ 10(y \plus{} z) \equal{} 3yz$ and $ 10(y \plus{} z)\leq 20z < 3yz$ unless $ y < 7$ so we have posibilities $ x \equal{} 2$, $ y \equal{} 2,3,4,5,6$ due to the assumption $ x\leq y\leq z$ and by checking each of these we get following solutions for $ (x,y,z)$ $ (2,4,20)$ and $ (2,5,10)$

if $ x \equal{} 3$ we get $ 15(y \plus{} z) \equal{} 7yz$ and $ 15(y \plus{} z)\leq 30z < 7yz$ unless $ y < 5$ so we have posibilities $ x \equal{} 3$, $ y \equal{} 3,4$ both of which give no solutions

only solutions therefore are $ (2,4,20)$ and $ (2,5,10)$

[b]i)[/b]$ x\leq y\leq z$
$ \frac {1}{x} \plus{} \frac {1}{y} \plus{} \frac {1}{z} \equal{} \frac {4}{5}$
$ x$ can`t be $ 1$. So $ x\geq 2$. Then we get:
$ \frac {4}{5} \equal{} \frac {1}{x} \plus{} \frac {1}{y} \plus{} \frac {1}{z}\leq \frac {3}{x}$
So: $ 2\leq x\leq 3$. If $ x \equal{} 2$:
$ \frac {1}{y} \plus{} \frac {1}{z} \equal{} \frac {3}{10}$
Again using [b]i)[/b] we get:
$ \frac {3}{10} \equal{} \frac {1}{y} \plus{} \frac {1}{z}\leq \frac {2}{y}$
So $ 2\equal{}x\leq y\leq 6$.
Checking this we get two solutions: $ (x,y,z)\equal{}\{(2,4,20),(2,5,10)\}$.
If $ x \equal{} 3$ then:
$ \frac {1}{y} \plus{} \frac {1}{z} \equal{} \frac {7}{15}$
$ \frac {7}{15} \equal{} \frac {1}{y} \plus{} \frac {1}{z}\leq \frac {2}{y}$
$ y\leq 4$. Checking this interval we don`t find solutions.
Answer: $ (x,y,z)\equal{}\{(2,4,20),(2,5,10)\}$.

Assume without loss of generality that $ 2\leq x \leq y \leq z$. The first inequality is clear because the variables can't be equal to 1. Now we have
\[ \frac {4}{5} \equal{} \frac {1}{x} \plus{} \frac {1}{y} \plus{} \frac {1}{z} \leq \frac {3}{x},
\]
so $ \frac {3}{x} \geq \frac {4}{5}$, that is, $ x \leq 3$. So $ x \equal{} 2,\ 3$.

In the first case, we have $ \frac {1}{2} \plus{} \frac {1}{y} \plus{} \frac {1}{z} \equal{} \frac {4}{5}$, or $ \frac {1}{y} \plus{} \frac {1}{z} \equal{} \frac {3}{10}$. Then
\[ \frac {3}{10} \equal{} \frac {1}{y} \plus{} \frac {1}{z} \leq \frac {2}{y},
\]
therefore $ \frac {3}{10} \leq \frac {2}{y}$, that is, $ y \leq 5$. Now $ y \equal{} 2,\ 3,\ 4,\ 5$ and we can directly verify that $ y \equal{} 4$ and $ y \equal{} 5$ are the only values that work. We get the solutions $ (x,\, y,\, z) \equal{} (2,\, 4,\, 20)$ and $ (2,\, 5,\, 10)$.

If $ x \equal{} 3$, we get the equation $ \frac {1}{y} \plus{} \frac {1}{z} \equal{} \frac {7}{15}$. Proceeding in the same fashion we find that $ 2\leq y \leq 4$, and none of these values yields integer solutions. Hence the only solutions are $ (x,\, y,\, z) \equal{} (2,\, 4,\, 20)$ and $ (2,\, 5,\, 10)$.


algebra
polynomial
superior algebra
superior algebra solved

[quote="liyi"]Let $R$ be a ring and let $f(y)\in R[y]$ be a polynomial in one variable with coefficients in $R$. Prove that the map $R[x,y]\to R[x,y]$ defined by $x\mapsto x+f(y), y\mapsto y$ is an automorphism of $R[x,y]$.
My question is , how to prove that the map is injective?[/quote]

I think we can show that $x\mapsto x-f(y), y\mapsto y$ is the inverse of the given map; hence, an automorphism.


Se considera in spatiu punctele A,B,C,D astfel incat $AC=BD=2$ si $AB=BC=CD=DA=\sqrt{2}$.Aratati ca punctele sunt coplanare.

Asta e o problema foarte potrivita pt locala.
Se demonstreaza ca ABC, BCD, CDA, DAB sunt unghiuri drepte si se ia M mijlocul lui AC.
Se demonstreza usor ca BM+MD=BD, deci B-M-D.

Dap...asa am fkt-o si eu
da am mai aplicat th cosinusului(kre k aici mi-au skzut)