This is my first game, and these are the rules.
The first person will post a word in BOLD in a certain topic, like board games (I will post the topic), and the second player will post a word with the 1st letter being the same as the last letter of the previous post.

Example:
Player 1: Monopoly
Player 2: Yahtzee
Player 3: Ummm.....
So since Player 2 got 3 stumped, he/she gains 2 points.
If Player 3 has any points (which is not the case), he/she loses 3 points because he/she couldn't think of a word fitting the topic. There is no penalty if the player has 0 points. If you post with a word, then you get 1 point.

Scoreboard:
Player 1: 1
Player 2: 2

Please post the scoreboard.  Are the rules clear enough?

I don't get any points for starting us off.

The topic is..... food. (Every once in a while I'll change the topic.)

Ok.....
[b]Hamburger[/b]

[u]Scoreboard[/u]
Nobody yet...

Like this?

[b]ribs[/b]

[hide="Scoreboard"]
PowerOfPi - 1
[/hide]

[b]sausage[/b]

[hide="Scoreboard"]
PowerOfPi - 1
ytrewq - 1
[/hide]

Oh, and also, don't do any repeating, please. Don't count as a point if you can find a repeat. And no really random things, like Zebra Meat. Some thing that most people can eat a resturant or something. Finally, scratch out everything I said about the being stumped part. If nobody can find something to post for..... a week, then the person who posted last gets 2 points instead of 1, and a new topic begins.
[/b]

[b]eggs[/b]

[hide="Scoreboard"]
PowerOfPi - 2
ytrewq - 1
[/hide]

[b]sushi[/b]

[hide] [u]Scoreboard[/u]
PowerOfPi - 2
ytrewq - 1
louiethechewy123- 1[/hide]

[b]ice cream[/b]

[hide="Scoreboard"]
PowerOfPi - 3
ytrewq - 1
louiethechewy123- 1[/hide]

[b]mahi-mahi[/b]

[hide][u]Scoreboard[/u]
PowerOfPi - 3
louiethechewy123- 2
ytrewq - 1
[/hide]

EDIT: wait, is this too wierd? If so, go with [b]meat[/b].

[b]apricot[/b]

[hide="in here"]PowerOfPi - 4
louiethechewy123- 2
ytrewq - 1
1=2 - 1[/hide]

[b]taco[/b]

[hide="Scoreboard"]
PowerOfPi - 5
louiethechewy123- 2
ytrewq - 1
1=2 - 1
[/hide]

[quote="PowerOfPi"][b]Tilapia[/b]

[hide="ScoreBoard"]
PowerOfPi - 4
louiethechewy123- 2
ytrewq - 1
[/hide]

There are like no foods starting with "i".[/quote]

How does Tilapia start with an "i"?

[b]orange[/b]

[hide="Scoreboard"]
PowerOfPi - 5
ytrewq - 2
louiethechewy123- 2
1=2 - 1
[/hide]

[b]eggplant[/b]

[hide="Scoreboard"]
PowerOfPi - 6
ytrewq - 2
louiethechewy123- 2
1=2 - 1
[/hide]

[b]tomato[/b]
[hide="in here"]
PowerOfPi - 6
ytrewq - 2
louiethechewy123- 2
1=2 - 2[/hide]

[b]olive[/b]

[hide="Scoreboard"]
PowerOfPi - 7
ytrewq - 2
louiethechewy123- 2
1=2 - 2
[/hide]

It's been 3 days since I posted. Is this game dying?

I think it needs a new topic, candy has too many plurals.

Or maybe plants. We could start with [b]birch tree[/b]

[b]elm[/b]

[hide="scoreboard"]tennis123 - 18
gohuskers - 15
PowerOfPi - 14
louiethechewy123 - 10
ytrewq - 6
1=2 - 5
bbgun34 - 4
james4l - 4
jjfun1-3
pinkmuskrat - 2
isabella2296 - 1
MATTH - 1
FlyAgaric - 1
meewhee009 - 1
Sabien5 - 1[/hide]

[b]eucalyptus[/b]
[hide="scoreboard"]tennis123 - 18
gohuskers - 15
PowerOfPi - 14
louiethechewy123 - 10
ytrewq - 6
1=2 - 5
bbgun34 - 4
james4l - 4
jjfun1-4
pinkmuskrat - 2
isabella2296 - 1
MATTH - 1
FlyAgaric - 1
meewhee009 - 1
Sabien5 - 1
mathfreak77-1[/hide]

[b]Sugar Maple[/b]
[hide="Scoreboard"]
tennis123 - 19
gohuskers - 15
PowerOfPi - 14
louiethechewy123 - 10
ytrewq - 6
1=2 - 5
bbgun34 - 4
james4l - 4
jjfun1-4
pinkmuskrat - 2
isabella2296 - 1
MATTH - 1
FlyAgaric - 1
meewhee009 - 1
Sabien5 - 1
mathfreak77-1[/hide]
p.s. do i get 3 points for winning the last round?

[b]daffodils[/b]

[hide="scoreboard"]tennis123 - 22
gohuskers - 15
PowerOfPi - 14
louiethechewy123 - 10
ytrewq - 6
1=2 - 5
jjfun1-5
bbgun34 - 4
james4l - 4
pinkmuskrat - 2
isabella2296 - 1
MATTH - 1
FlyAgaric - 1
meewhee009 - 1
Sabien5 - 1
mathfreak77-1[/hide]

@tennis: i suppose, i added 3 to ur score for u

yew tree

is there such thing as that?? :huh:

[b]ebony tree[/b]
[hide="Scoreboard"]
tennis123 - 23
gohuskers - 15
PowerOfPi - 14
louiethechewy123 - 10
ytrewq - 6
jjfun1 - 6
1=2 - 5
bbgun34 - 4
james4l - 4
pinkmuskrat - 2
mathfreak77 - 2
meewhee009 - 2
isabella2296 - 1
MATTH - 1
FlyAgaric - 1
Sabien5 - 1 [/hide]
Also, could you try to post the scoreboard every post so less updating is needed per person?


Find the product of all real numbers $ n$ which satisfy $ |{n^2\minus{}9n\plus{}20}| \equal{} |{16\minus{}n^2}|$

[hide="Answer"]2[/hide]

Square both sides of the equation to eliminate the abs value signs.  Then, I guess you could just solve the really convoluted equation using synthetic division, though I'm not sure if that's the easiest method.

[hide]
$ |{n^2 \minus{} 9n \plus{} 20}| \equal{} |{16 \minus{} n^2}|$ means either

$ n^2 \minus{} 9n \plus{} 20 \equal{} 16 \minus{} n^2$ or $ n^2 \minus{} 9n \plus{} 20 \equal{} \minus{} (16 \minus{} n^2)$.

The first one is therefore $ 2n^2 \minus{} 9n \plus{} 4 \equal{} 0$, while the second is $ \minus{} 9n \equal{} \minus{} 36$.

The product of the roots of a quadratic is $ \frac {c}{a}$, so the product of the roots of the first equation is 2, while the second equation has solution $ 4$. But note that 4 is a repeated solution, because it also satisfies the quadratic. Our answer is therefore $ \fbox{2}$.[/hide]

FIXED!

[hide="My Solution"]

Factor both sides and distribute the absolute value:
$ |{n - 5}| |{n - 4}| = |{4 + n}| |{4 - n}|$

Note that $ |{n - 4}|$ = $ |{4 - n}|$, so factor $ |{n - 4}|$ out of the equation. Note that that takes out a solution, $ n = 4$, because both sides would evaluate to zero, that we have to remember to add in our final solution.

\begin{align*} |{n - 5}| & = |{n + 4}| \\
n^2 - 10n + 25 & = n^2 + 8n + 16 \\
9 & = 18n \\
n &= \frac {1}{2} \end{align*}

So we have two solutions, $ 4$ and $ \frac {1}{2}$. The product of these is $ \boxed{2}$.
[/hide]

[quote="Ihatepie"]distracted has the wrong solution but the right answer
why can't that happen to me?
 :wallbash:[/quote]
What do you mean? I think it's right.

I found that solution to be a bit hard to understand, but I figured it out in the end :)

Yeah, I meant that 4 is also a solution to the quadratic. I'll change it I guess. Incidentally, my solution [b]was[/b] wrong at first, but I fixed it within a couple of minutes.

[quote="myspacesucks"]Distracted, yours is still wrong because it can be -(n^2-9n+20)=16-n^2 if you have n=3.  However, it doesn't work anyway, so you got lucky.  At NATS, that won't happen.[/quote]

that quadratic is equivalent to $ n^2\minus{}9n\plus{}20\equal{}\minus{}(16\minus{}n^2)$ if you multiply both sides by $ \minus{}1$.  Also, the solution to that is $ n\equal{}4$.

That's the same equation as my second one. You just multiplied it by -1. And the solution to that is 4  :lol: . What sort of nationals experience do you have?

*gets beat.


abstract algebra
Ring Theory
superior algebra
superior algebra unsolved

What is an example of a $ k[x,y]$-module $ F$ and submodule $ U$ of $ F$ such that $ F$ is free and has finite rank $ n$ over $ k[x,y]$ yet the rank of $ U$ over $ k[x,y]$ is $ >n$?

I think there is no example because $ k(X,Y)$ is flat as a $ k[X,Y]$-module.

There is a theorem stating that for $ R$ a PID, $ F$ a free $ R$-module of finite rank $ n$, and $ U$ a submodule of $ F$, then $ U$ has rank $ \leq n$.

I am wondering of counterexamples to this theorem when $ R$ is not a PID. To me, $ k[x,y]$ is the simplest example of a non-PID.


I can't go, I have to do english homework.

U know what this is the Georgia forum transplanted. I just had to tell you. We need a Georgia/Alabama forum.


Let $ f_n$ be the number of permutations of the set $ S_n\equal{}\{ 1,2,...,n \}$ having no fixed points, and $ g_n$ be the number with exactly one fixed point. Show that $ |f_n\minus{}g_n|\equal{}1.$

Clearly $ g_n \equal{} n f_{n\minus{}1}$.  On the other hand, it is well-known that $ f_n \equal{} n f_{n\minus{}1} \plus{} (\minus{}1)^n$.

A short proof is as follows:  inclusion-exclusion gives the well-known identity

$ \sum_{n \ge 0} \frac{f_n}{n!} x^n \equal{} \frac{e^{\minus{}x}}{1 \minus{} x}$

and multiplying by $ x$ we obtain

$ \sum_{n \ge 1} \frac{n f_{n\minus{}1}}{n!} x^n \equal{} \frac{e^{\minus{}x}}{1 \minus{} x} \minus{} e^{\minus{}x}$.


A positive integer is called a perfect number if it is equal to the sum of all its positive
factors, excluding itself. For example, the number 6 is a perfect number, because its
factors (excluding itself) are 1, 2, and 3, and 1 + 2 + 3 = 6. Which of the following is a
perfect number?
A. 24
B. 28
C. 30
D. 36
E. 42

[hide]
28 b/c its factors are 1,2,4,7,14[/hide]


geometry
rectangle
limit
algebra proposed
algebra

A parcel of land has dimensions $n \times n$ meters and is divided into strips (rectangles $1\times m$ or $m \times 1$, $m$ is an integer).
A man can choose an integer $k$ and take the land on all the strips that have length $k$.
Let $l(n)$ be the largest integer such that he may take at least $l(n)$ square meters of land, no matter how the parcel is divided.

Compute $\lim_{n \rightarrow \infty} \frac{l(n)}{n}$

(I got $\sqrt{\frac{3}{12-\pi^2}}$)

iura,

can you illustrate with an example from a 3 by 3 or a 4 by 4 square what the largest l(n) is/can be to help make this clearer to me?  Thanks.

Today in the classroom we found another limit:
$\sqrt{\frac{6}{6-\pi^2}}$.

@ a cute angle : for $3$x$3$ you have $l(n)=4$, because use it can't be divided
in 3 groups with 3 cells in each one(land with length 2 doesn't allow to do this).
Analogously for $4$x$4$ - $l(n)=5 (4+3+2\cdot2+5\cdot1)$

Actually, I think [b]iura[/b]'s limit is the good one, isn't it?

Because, [b]prowler[/b], yours is an imaginary limit...  :?


Can anybody prove the Fano's axiom (Diagonal points of a complete quadrilateral are never collinear)

an axiom is a proposition which is obvious but that you cannot prove.
So...

:|  I had known this but I thought that it's possible to prove it... I think that it's not necessary... Is it true??

Firstly, I am not sure what you call the "diagonal points of a complete quadrilateral", so I will simply assume that, by Fano's axiom, you mean the following fact:

[color=blue][b]Fano's axiom.[/b] Let A, C, D, F be four points in the plane, no three of them being collinear. Then, the points $AD\cap CF$, $AC\cap DF$ and $CD\cap FA$ are not collinear.[/color]

Note that I have chosen the letters A, C, D, F since I will later introduce two others B and E.

Whether Fano's axiom is an axiom or not depends on how you define geometry, i. e. whether you use Hilbert's axioms, or whether you define the Euclidean plane as the vector space $\mathbb{R}^2$, or whether you introduce geometry in some other way. Anyway, if you define the Euclidean plane as the vector space $\mathbb{R}^2$, then Fano's axiom is surely not an axiom, since it can be derived from the theorems of Ceva and Menelaos (which both can be algebraically verified if you define the Euclidean plane as the vector space $\mathbb{R}^2$):

We will use directed segments.

Assume, for the sake of contradiction, that the points $AD\cap CF$, $AC\cap DF$ and $CD\cap FA$ are collinear. Let $E=AC\cap DF$ and $B=CD\cap FA$; then, we obtain that the points $AD\cap CF$, E and B are collinear. In other words, the point $AD\cap CF$ lies on the line BE; equivalently, the lines AD, BE and CF concur at one point. Hence, by the Ceva theorem, applied to the triangle ABC, we get $\frac{BD}{DC}\cdot\frac{CE}{EA}\cdot\frac{AF}{FB}=1$.

On the other hand, since the points D, E, F lie on one line, the Menelaos theorem, applied to the triangle ABC, yields $\frac{BD}{DC}\cdot\frac{CE}{EA}\cdot\frac{AF}{FB}=-1$. Hence, 1 = -1, what is a contradiction. Thus, our assumption is wrong, and the points $AD\cap CF$, $AC\cap DF$ and $CD\cap FA$ cannot be collinear. This proves Fano's axiom.

  Darij


[quote="Albanian Eagle"][quote]Albanian Eagle wrote:
Most probably I will be there too.
See you guys there.

You are a contestant? I never imagine that?

[/quote]

yes! :blush:
For me and Tiks it's the fifth time to participate.
At least we can meet there, right ?

PS:Maybe we will have the chance to meet also in CA.[/quote]
Weren't you supposed to be at ISEF (International Science and Engineering Fair) earlier this month? I thought I saw your name in one of the team projects, then checked that project to see that no one had arrived there :(

[quote="Albanian Eagle"]@ me@home:
no I wasn't at the Intel ISEF because of lack of Sponsor :blush:

and well I have to inform you that even I did qualify on the TST for Vietnam our leader didn't register my name by breaking the rules of the olympiad here and entering in conflict with The Ministry of Education. I really don't know why I am so unlucky...
The leader says that now everything is over because the registration can not be changed. (Can anyone confirm that? I mean maybe I still do have some chance... :roll: )
The leader here declared:
It is not our job to send the best students but just to send some students...

This is really very strange :huh:  :o[/quote]

I can't believe that this kind of things happen---  :|

[quote="Albanian Eagle"]The leader says that now everything is over because the registration can not be changed. (Can anyone confirm that? I mean maybe I still do have some chance... :roll: )
The leader here declared:
It is not our job to send the best students but just to send some students...

This is really very strange :huh:  :o[/quote]

You may contact your ministry of education and people from the [url=http://www.imo2007.edu.vn/index.php?module=ViewCatOpenFirst&CatId=8]organization[/url] of the VN IMO 2007. As you can see under deadlines the [url=http://www.imo2007.edu.vn/index.php?module=login]students registration[/url] ends today. But I think the VN IMO organization won't really mind if a few selected changes of participants are issued after the deadline. Good luck to you!!!

[quote="hsiljak"]I can't believe that this kind of things happen---  :|[/quote]I remember 2 years ago in Mexico Rust did not make it because the Russian embassy did not release his team's passports in time. So actually these things do happen.

In this case I'm not sure what's going on though  :huh:

The case is that I have graduated from high school last year but didn't enroll in the university thus am able to participate in IMO.
To enter the TST I had to take a confirmation from the Albanian Mathematical Society and The Ministry of Education, which I both did. Our leader in IMO is the vice president of the society, but the strange thing he did here was that he organised the TST on his own and even the president was not informed on the issue. At the end he decided (quite arrogantly) not to register me ( he had no reason do do this except the fact that the new students registered had some acquiatances...).
Now I am trying to clear this thing out but the leader says everything is over ?! (the registration)
I really hope for the best.(I doubt it)
Thank to all of you guys for the support. :oops:

What registration are you talking about? TST registration? Or IMO registration? For the IMO, I'm sure it's not done yet (although the deadline is 1 June) as for example the Romanian IMO team hasn't been selected yet.

[quote="Albanian Eagle"]Sorry fore not explaining myself clearly.
I entered the TST and qualified but our leader didn't include my name in the team when he made the registration to IMO. (It's very ridiculous I know)
Now I'm asking if changes to the team are allowed. (Here they aree telling me that it's not.)[/quote]

I agree with Valentin's remark. The question is whether they actually want you to take part into it or they are just too lazy to make any changes to the list of registered IMO contestants of Albania.

[quote="Albanian Eagle"]The ministry and the society wants me to take part in IMO, it's just the leader who decided not to take me eventhough I qualified in the TST! After this was discussed he said that he had already made the registration and could not make any changes![/quote]

What can I say? I only can repeat what I said above. You should tell the leader that you know that the entries are still subject to change. And if he denies your request you need to talk to the ministry such that they put pressure upon him to select the strongest team. And see if you are lucky.

[quote="Tiks"]How has this ended Eagle? Are you coming to Vietnam? :roll:[/quote]

I guess not: http://imo2007.edu.vn/index.php?module=ViewInforTableForm.php&ct=ALB

Apparently I have no chance!
They agreed on signing a new rule at the ministry of education that only students in high school will be eligible to participate to the IMO  :)
This was done a few days ago just to stop me from complaining  :rotfl:
So I wish all of you good luck there and will be watching the IMO from ML here  :ninja:

Oh, don't worry about it. As one of my friends once said "IMOs are important but it is more important to get into a good school". I know that you got into Caltech and it is a great school and I am very glad for you.

Good luck my friend!!! :wink:


1)Find all triples of integer numbers $ a;b;c$ such that for any positive integer $ n$; $ a.2^n\plus{}b$ is a divisor of $ c^n\plus{}1$
2) Find all integers $ a;b;c;d$ such that for any positive integer $ n$; $ a.4^n\plus{}b.9^n\plus{}c.6^n$ is a divisor of $ d^n\plus{}1$
(2) is mine.

[quote="Allnames"]1)Find all triples of integer numbers $ a;b;c$ such that for any positive integer $ n$; $ a.2^n \plus{} b$ is a divisor of $ c^n \plus{} 1$
2) Find all integers $ a;b;c;d$ such that for any positive integer $ n$; $ a.4^n \plus{} b.9^n \plus{} c.6^n$ is a divisor of $ d^n \plus{} 1$
(2) is mine.[/quote]
No solution?. First, please post your nicest solution for 1)!.
I will come back 3 days later !

Dear Allnames and mathlinkers, I have tried to solve these nice problems, but I failed.

Now I'm waiting for solution from all. Everyone, please give me solution or just some hints. Thanks

[quote="forevercontinue"]
Now I'm waiting for solution from all. Everyone, please give me solution or [b]just some hints[/b]. [/quote]
For 1) You should find a suitable integer sequence $ x_n$ (for example $ x_n \equal{} \frac {c^n \plus{} 1}{a.2^n \plus{} b}$; it is just an example  :) ).And then; when $ n$ tenses to the infinity we will find out $ a; b; c$.( some analysis knowledges work here)
It is surprising that no one interested my posts!.Two months ago!


modular arithmetic
algebra unsolved
algebra

$ 1\plus{}5^n\plus{}6^n\plus{}11^n \equiv 3 \pmod{5}$ for all positive integers $ n$.
But for all $ k \in \mathbb{N}$, $ k^2\equal{}0,1,4 \pmod{5}$, so $ 1\plus{}5^n\plus{}6^n\plus{}11^n$ is never a perfect square.


Let $ a,b$ be reals such that $ 0 < a,b < 1$
$ e \equal{} 2.71828...$ is a well-known constant.
Prove or disprove that
$ a^{ea} \plus{} b^{eb} \geq a^{eb} \plus{} b^{ea}$
Please help!

who can help me ?
the problem which $ e$ is replaced by $ 1$ or $ 2$ was proven long time ago in this Forum.
So i really need someone who can help me to prove this hard problem ^^.
Please help.

Help me pliz...
Please help me.
Does anyone have an idea ?

Very soon an article concerning the inequality a^(ea)+b^(eb) >= a^(eb)+b^(ea) will appear in JIPAM Journal.
In this article I used a single reference
[1] A. ZEIKII, V. CIRTOAJE and W. BERNDT, Mathlinks Forum, Nov. 2006, [ONLINE:http://www.mathlinks.ro/Forum/viewtopic.php?t=118722].

I sent six months ago a message to Zeikii to say me his actual complete name but I received no answer. I have only three days to change "A. Zeikii" with the true name.


function
number theory
Diophantine equation
number theory open

Hi,I have one question which seems simple but I don't know what to do :
suppose we have the diophantine equation $ \alpha x\plus{}\beta y \equal{}\gamma$,
.with $ \alpha,\beta,\gamma>0$.Can we find the number of the positive x that satisfy
this equation?(I couldn't find it on the internet.)

In order for there to be [i]any[/i] solutions (that make sense), we need the following assumptions:

- $ \alpha, \beta, \gamma,x, y \in \mathbb{N}$
- $ \gcd(\alpha, \beta) | \gamma$ (we can WLOG take $ \gcd(\alpha, \beta) \equal{} 1$ in this case)

Even then, half of the values of $ \gamma$ such that $ \gamma \le \alpha \beta \minus{} \alpha \minus{} \beta$ turn out to have [i]no[/i] solutions (see [url=http://mathworld.wolfram.com/CoinProblem.html]coin problem[/url]).  For the ones that do, the number of solutions depends on the nature of the solution with smallest $ x$.  The best we can do is find upper and lower bounds, I think.

When $ \gamma\equal{}\gcd(\alpha,\beta)$ there exist a nice formula for the number of non-negative solutions.
See http://www.mathlinks.ro/Forum/viewtopic.php?t=191692

I don't follow.  $ \gcd(\alpha, \beta) \le \min(\alpha, \beta)$ so there aren't [i]any[/i] non-negative solutions.

Actually, I've addressed the case when $ \alpha$ and $ \beta$ can be any integers, including negative ones, while the variable $ x$ and $ y$ are non-negative (or positive).
When $ \alpha$ and $ \beta$ are positive, since the restriction on the sign $ y$ is not specified, we can assume that $ y < 0$ and change the sing of $ \beta$ and $ y$ together, and we get the same case again.

When all $ \alpha,\ \beta,\ x,\ y$ are positive, the problem becomes harder. Still, in this case it is possible to find out an ordinary generating function for the number of positive solutions. Namely, the number of positive solutions equals to the coefficient of $ z^{\gamma}$ in
\[ (z^{\alpha} \plus{} z^{2\alpha} \plus{} z^{3\alpha} \plus{} \dots)(z^{\beta} \plus{} z^{2\beta} \plus{} z^{3\beta} \plus{} \dots) \equal{} \frac {z^{\alpha \plus{} \beta}}{(1 \minus{} z^{\alpha})(1 \minus{} z^{\beta})}.
\]


number theory
prime numbers
number theory proposed

[quote="Johan Gunardi"]Find integer $ n$ with $ 8001 < n < 8200$ such that $ 2^n \minus{} 1$ divides $ 2^{k(n \minus{} 1)! \plus{} k^n} \minus{} 1$ for all integers $ k > n$.[/quote]
From condition we have $ n|k(n\minus{}1)!\plus{}k^n,\forall k>n$ . Choose $ k\equal{}n\plus{}1$ ,it is easy to check that $ n$ must be a prime and from Fermat's and Wilson's theorem , it is clear that all prime numbers satisfy .
Now only need to find out all prime between $ (8001,8200)$ .

may ...
$ 2^n \minus{} 1$ divides $ 2^{k(n \minus{} 1)! \plus{} k^n} \minus{} 1$  iff $ n|k(n \minus{} 1)! \plus{} k^n$  ?

[quote=jgnr]Find integer $ n$ with $ 8001 < n < 8200$ such that $ 2^n \minus{} 1$ divides $ 2^{k(n \minus{} 1)! \plus{} k^n} \minus{} 1$ for all integers $ k > n$.[/quote]

We need $n|k(n - 1)! + k^n$ or $n|k^n$ for composites and $n|0$ for primes(which is clearly true)
So if n was composite choose any $k \equiv -1 \pmod n$ and it won't work,so for this condition to hold we need n to be prime so the answer is $\boxed{21}$


Να βρείτε όλε τις συναρτήσεις $ f: N\rightarrow [1,\infty)$ τέτοιες ώστε:
1) $ f(2)\equal{}2$
2)$ f(mn)\equal{}f(m)f(n)$ για όλα τα $ m,n$ φυσικούς
3)$ m<n\iff f(m)<f(n)$

Ωραία άσκηση Δημήτρη!! Mήπως όμως το σύνολο τιμών είναι το $ [1,\plus{}\infty]\cap \mathbb{N}$ ? Μήπως εξασφαλίζεται αυτό με κάποιο τρόπο από τα δεδομένα ή δεν χρειάζεται καθόλου? Στην παρακάτω λύση έχω πάρει ως δεδομένο ότι $ f(n) \in \mathbb{N}$ για κάθε $ n\in \mathbb{N}$.

Λοιπόν ξεκινάμε με μερικές παρατηρήσεις. Παρατηρούμε ότι $ f(2)\equal{}f(1\cdot 2)\equal{}f(1)f(2) \Rightarrow f(1)\equal{}1$.

Επίσης για $ k\geq 2$ έχουμε $ f(2^k)\equal{}\underbrace{f(2)\cdots f(2)}_{k\minus{}\phi o \rho \epsilon \varsigma}\equal{}\left[f(2)\right]^k\equal{}2^k$.

$ 2<3<4 \Rightarrow 2<f(3)<4 \Rightarrow f(3)\equal{}3$.

$ 4<5<6 \Rightarrow 4<f(5)<f(6)\equal{}f(2)f(3)\equal{}6 \Rightarrow f(5)\equal{}5$.

Θα δείξουμε με κατάλληλη ισχυρή επαγωγή ότι $ f(n)\equal{}n$ για κάθε $ n$ φυσικό.

Παρατηρήστε ότι αρκεί να αποδείξουμε ότι $ f(p)\equal{}p$ για κάθε περιττό πρώτο $ p$, διότι τότε λόγω της πολλαπλασιαστικότητας της συνάρτησης (Ιδιότητα (2)), εαν $ n\equal{}2^kp_1^{a_1}\cdot p_k^{a_k}$ η ανάλυση του $ n$ σε πρώτους παράγοντες, τότε $ f(n)\equal{}\ldots\equal{}2^kf^{a_1}(p_1)\cdots f^{a_k}(p_k)$.

Για $ p\equal{}3$ ισχύει.

Έστω ότι ισχύει για όλους τους πρώτους $ q$ τους μικρότερους του αριθμού $ p$. Τότε σίγουρα θα ισχύει και για όλους τους αριθμούς (ανεξάρτητα αν είναι πρώτοι ή όχι) τους μικρότερους του $ p$ (Απλά αναλύουμε τον κάθε αριθμό ο οποίος ως μικρότερος του $ p$ και κατ' επέκταση πρώτος προς το $ p$ θα έχει μικρότερους από το $ p$ πρώτους παράγοντες, άρα  λόγω της πολλαπλασιαστικότητας της συνάρτησης και της επαγωγικής υπόθεσης, θα πάρουμε το αποτέλεσμα για όλους τους αριθμούς τους μικρότερους του $ p$).

Θα το αποδείξουμε για τον $ p$. Όμως $ p\minus{}1<p<p\plus{}1  \Rightarrow f(p\minus{}1)<f(p)<f(p\plus{}1)$ και το $ p\minus{}1$ είναι αριθμός μικρότερος του $ p$, άρα λόγω του παραπάνω βήματος θα έχουμε $ f(p\minus{}1)\equal{}p\minus{}1$. Τέλος επειδή το $ p$ είναι περιττός, άρα το $ p\plus{}1$ είναι άρτιος άρα $ f(p\plus{}1)\equal{}f\left(2\cdot \frac{p\plus{}1}{2}\right)\equal{}2f\left(\frac{p\plus{}1}{2}\right)$, και επειδή και πάλι ο αριθμός $ \frac{p\plus{}1}{2}$ είναι μικρότερος του $ p$, έχουμε $ f\left(\frac{p\plus{}1}{2}\right)\equal{}\frac{p\plus{}1}{2}$ οπότε τελικά $ f(p\plus{}1)\equal{}p\plus{}1$. Δηλαδή $ p\minus{}1<f(p)<p\plus{}1$ άρα $ f(p)\equal{}p$ για κάθε περιττό πρώτο $ p$ που ολοκληρώνει την απόδειξη.

Αλέξανδρος

έστω g η συνάρτηση της εκφώνησης τότε
αν f μια συνάρτηση ίδια με την g αλλά με πεδίο ορισμού τους πραγματικούς (οι υπόλοιπες ιδιότητες και στοιχεία ίδια) έχουμε:

$ f(0 \plus{} 0) \equal{} f(0)f(0) \Leftrightarrow f(0)(f(0) \minus{} 1) \equal{} 0 \Leftrightarrow$
ή $ f(0) \equal{} 0 < 1 \Leftrightarrow$ άτοπο
ή $ f(0) \equal{} 1$

ομοίως $ f(1) \equal{} 1$

θα δείξουμε ότι για κάθε $ n$ μη μηδενικό φυσικό είναι $ f(n) \equal{} n$

για n = 2 ισχύει

έστω ότι ισχύει για $ n \le p \minus{} 1$

τότε για $ n \equal{} p$ είναι

$ 2f(\frac {p}{2}) \equal{} p \Leftrightarrow f(\frac {p}{2}) \equal{} \frac {p}{2}$

και

$ \frac {p \minus{} 1}{2} \plus{} \frac {1}{2} \le \frac {p}{2} \le \frac {p \plus{} 1}{2} \minus{} \frac {1}{2} \Leftrightarrow$

$ \frac {p}{2} \le f(\frac {p}{2}) \le \frac {p}{2} \Leftrightarrow$

$ p \le 2f(\frac {p}{2}) \le p \Leftrightarrow p \le f(p) \le p \Leftrightarrow$

$ f(p) \equal{} p$

άρα είναι $ f(n) \equal{} 1$ αν $ n \equal{} 0$ και $ f(n) \equal{} n$ αν $ n \ge 1$
άρα $ g(n) \equal{} 1$ αν $ n \equal{} 0$ και $ g(n) \equal{} n$ αν $ n \ge 1$

τώρα που το ξαναβλέπω, η παραπάνω λογική χάνει κάπως γιατί στους περιτούς το $ \frac{p}{2}$ δεν εξετάζεται από την επαγωγή και άρα πρέπει να θεωρήσουμε την επαγωγή για τα μισά ακεραίων,
έπειτα πάλι για τον ίδιο λόγο πρέπει να πάρουμε τα μισά των μισών και πάει λέγοντας...
άρα μάλλον χρειάζεται ο περιορισμός η συνάρτηση να έχει σύνολο τιμών τους φυσικούς
ή απλά δεν γίνεται ο υπολογισμός με επαγωγή (τουλάχιστον με αυτό τον τρόπο)

πάντως πρέπει να εξεταστεί ξεχωριστά η περίπτωση $ n \equal{} 0$

[quote="bilot4"]Να βρείτε όλε τις συναρτήσεις $ f: N\rightarrow [1,\infty)$ τέτοιες ώστε:
1) $ f(2) \equal{} 2$
2)$ f(mn) \equal{} f(m)f(n)$ για όλα τα $ m,n$ φυσικούς
3)$ m < n\iff f(m) < f(n)$[/quote]

thn phres apo vivlio me ble eksofyllo?

Κατ' αρχάς να διευκρινίσω ότι η άσκηση έλεγε ότι $ f: N\rightarrow (1,\infty)$ (στο 1 είναι είναι κλειστο το διάστημα αλλά όταν βάζω το κατάλληλο σύμβολο για κάποιο λόγο δεν μου βγαίνει καλά το latex, μάλλον λόγω του quote που κάνω μετά...) και δεν μιλούσε μόνο για φυσικούς (υπήρχε μια προηγούμενη άσκηση η οποία αναφερόταν σε σύνολο τιμών τους φυσικούς οπότε αυτή η άσκηση είναι ξεχωριστή και αναφέρεται σε αυτό που προανέφερα, αποτελώντας μια δύσκολη γενίκευση!) Το πρόβλημα γαι φυσικούς εξάλλου λύνεται επίσης με strong induction υποθέτωντας ότι ισχύει γαι όλα τα $ n\leq 2k$ και αποδεικνύοντας ότι αυτό ισχύει και γαι $ 2k \plus{} 1$ (είναι αρκετά εύκολο αυτό αφού το $ 2k \equal{} f(2k)\leq f(2k \plus{} 1)\leq f(2k \plus{} 2) \equal{} f(2)f(k \plus{} 1) \equal{} 2k \plus{} 2$ λόγω της επαγωγής).

Το πρόβλημα όμως όπως είπα είναι πιο δύσκολο από αυτό :wink:

[quote="maniopas"]
$ \frac {p \minus{} 1}{2} \plus{} \frac {1}{2} \le \frac {p}{2} \le \frac {p \plus{} 1}{2} \minus{} \frac {1}{2} \Leftrightarrow$

$ \frac {p}{2} \le f(\frac {p}{2}) \le \frac {p}{2} \Leftrightarrow$

$ p \le 2f(\frac {p}{2}) \le p \Leftrightarrow p \le f(p) \le p \Leftrightarrow$

$ f(p) \equal{} p$

[/quote]
Maniopas μπορείς plz να εξηγήσεις αυτό το σημείο της λύσης σου. Καταλαβαίνω ότι κάνεις κάτι σαν την επαγωγή που προανέφερα αλλά δεν καταλαβαίνω σε αυτές τις γραμμές πώς πας απ' το ένα στο άλλο. Εξάλλου αυτό που γράφεις στην πρώτη σειρά ισχύει σαν ισότητα κι όχι σαν ανισότητα... Σε κάθε περίπτωση θα παρακαλούσα να το εξηγήσεις λίγο τι εννοείς (μήπως είναι κανα typo?). Α και btw το 0 δεν συμπεριλαμβάνεται στους φυσικούς $ N$ :wink: .

@rapanos:
Νίκο το βιβλίο απ' το οποίο πήρα την άσκηση λέγεται "functional equations- A problem solving approach" αλλά δεν το έχω στο πρωτότυπο οπότε δεν μπορώ να σου πω τι χρώμα έχει το εξώφυλλο :P

Έχεις δίκιο Δημήτρη! Δεν χρειάζεται η προϋπόθεση για το σύνολο τιμών. Με λίγο ψάξιμο βρήκα ότι το θέμα που έβαλες, χωρίς την πρώτη συνθήκη είναι ένα θεώρημα του Erdos που δημοσιεύθηκε το 1946. Η λύση που σας δίνω δεν είναι εκείνη που δημοσιεύθηκε από τον Erdos (η οποία δεν είναι και τόσο απλή) αλλά μία αρκετά μεταγενέστερα, το 1986 από τον Everett Howe και η οποία δημοσιεύθηκε στο American Mathematical Monthly. Είναι αρκετά απλή στην κατανόησή της. Σας επισυνάπτω ολόκληρο το συγκεκριμένο άρθρο (με 1 επιπλέον θεώρημα πιο ισχυρό).

Αποδεικνύει ότι εαν ισχύουν οι συνθήκες 2), 3) του Δημήτρη τότε υπάρχει σταθερά $ c$ τέτοια ώστε $ f(n) \equal{} n^c,$ για κάθε $ n\geq 1$ (Οπότε με τη συνθήκη 1) παίρνουμε $ c \equal{} 1$ κι έτσι $ f(n) \equal{} n$, για κάθε $ n\geq 1$.

Τέλος, μπορείτε να δείτε ένα σχετικό θέμα στο 2003 TST της Κίνας στη σελίδα http://www.mathlinks.ro/Forum/viewtopic.php?p=346551#346551

Αλέξανδρος

Η αρχική λύση μου ούτως ή άλλως είναι λανθασμένη.

Μετά όμως από πολλές προσπάθειες κατάφερα να βρω μια απόδειξη επαγωγής, την οποία θα παραθέσω αναλυτικά.

θα δείξουμε με ισχυρή επαγωγή στους πραγματικούς ότι ισχύει $ f(n) \equal{} n$ για κάθε $ n \ge 1$

εύκολα f(n) = συνεχής στο $ [1, \plus{}\infty)$

για $ n \equal{} 1$ είναι $ f(n) \equal{} n$ όπως έδειξα στο προγούμενό μου post

έστω ότι ισχύει για κάθε $ 1 < n < p$ ότι $ f(n) \equal{} n$

τότε για $ n \equal{} p$ είναι

$ 1 < p \Rightarrow 0 < p \Leftrightarrow p < 2p \Leftrightarrow \frac {p}{2} < p \Leftrightarrow$ (από την υπόθεση)

$ f(\frac {p}{2}) \equal{} \frac {p}{2} \Leftrightarrow f(2)f(\frac {p}{2}) \equal{} 2\frac {p}{2} \Leftrightarrow$

$ f(p) \equal{} p$

maniopas δεν είμαι σιγουρος αλλά νομίζω ότι με αυτό που έκανες αποδεκνύεις μόνο την περίπτωση που το σύνολο τιμών είναι οι φυσικοί. Ο λόγος είναι ότι δεν μπορείς να κάνεις επαγωγή σοτυς πραγματικούς (τουλάχιστον έτσι όπως την κάνεις). Δηλαδή γαι να κάνεις την ισχυρή επαγωγή που έκανες δεν φτάνει το $ f(1)\equal{}1, f(2)\equal{}2$. Αυτο θα αρκούσε στους ακέραιους...Το πρόβλημα δηλαδή είναι ότι αν σου πω απλώς να αποδείξεις ότι $ f(\frac{3}{2})\equal{}\frac{3}{2}$ δεν αποδεικνύεται αυτό με την επαγωγή που χρησιμοποιείς. Τελος πάντων νομίζω ότι γενικά δεν μπορείς να κάνεις τέτοιου είδους επαγωγή στους πραγματικούς (σκέψου απλά για παράδειγμα πώς θα πας απ' το $ 1$ στο $ 1.1$...)

Γι' αυτό αποδικνύω τη συνέχεια της συνάρτησης.
Αποδεικνύω για n=1
έπειτα, λόγω της συνέχειας, σε μια πολύ μικρή περιοχή δεξιά του 1 (αν θέλεις βάλε την να έχει πλάτος $ lim_{(x \to \plus{} \infty)}\frac{1}{2^x}$) ισχύει.
Μετά εφαρμόζω επαγωγή. (Δηλαδή: για κάθε αριθμό που έχει το μισό του στη μικρή αυτή περιοχή του 1 θα ισχύει η σχέση. Έτσι διπλασιάζω το διάστημα που αυτή ισχύει. Συνεχίζω αυτή τη διαδικασία $ 2^{2x}$ φορές. To πλάτος του διαστήματος για το οποίο θα ισχύει μετά από $ x \to \plus{} \infty$ φορές εκτέλεσης του παραπάνω θα είναι:
$ lim_{(x \to \plus{} \infty)}\frac {2^{2x}}{2^x} \equal{} lim_{(x \to \plus{} \infty)}2^x \equal{} \plus{} \infty$)

Είναι ποτέ δυνατόν μια συνάρτηση με πεδίο ορισμού το $ N$ να είναι συνεχής???????? :?:  :blush:
Γιατί μου φαίνεται ότι αυτό το πορισμα είναι λίγο βλακεία???? :blush:

Λέγαμε για την ίδια άσκηση όταν η συνάρτηση ορίζεται στους πραγματικούς.

[quote="maniopas"]

θα δείξουμε με ισχυρή επαγωγή στους πραγματικούς ότι ισχύει $ f(n) \equal{} n$ για κάθε $ n \ge 1$

εύκολα f(n) = συνεχής στο $ [1, \plus{} \infty)$

για $ n \equal{} 1$ είναι $ f(n) \equal{} n$ όπως έδειξα στο προγούμενό μου post

έστω ότι ισχύει για κάθε $ 1 < n < p$ ότι $ f(n) \equal{} n$

τότε για $ n \equal{} p$ είναι

$ 1 < p \Rightarrow 0 < p \Leftrightarrow p < 2p \Leftrightarrow \frac {p}{2} < p \Leftrightarrow$ (από την υπόθεση)

$ f(\frac {p}{2}) \equal{} \frac {p}{2} \Leftrightarrow f(2)f(\frac {p}{2}) \equal{} 2\frac {p}{2} \Leftrightarrow$

$ f(p) \equal{} p$[/quote]

Κατ' αρχάς απολογούμαι γιατί στο προηγούμενο post μου έκανα λάθος. Η άσκηση ήταν πράγματι όπως λέει ο meinstein $ f: N\rightarrow [1,\infty)$ (κι οπότε αυτο που λέει με την συνέχεια έχει απόλυτο δίκιο...) , και αυτό που είπα για $ f(\frac {3}{2})$ δεν μπορεί να υφίσταται... Απλά μπερδεύτηκα με αυτο΄που γράφεις maniopas για "επαγωγή σοτυς πραγματικούς" η οποία αφενός μεν επαναλαμβάνω ότι νομίζω ότι δεν υφίσταται (ίσως κάνω και λάθος βέβαια, δεν είμαι σίγουρος :blush: ) και αφετέρου δεν έχει κάποια σχέση αφού το πεδίο ορισμού είναι το $ N$. Σε μια τέτοια περίπτωση η λύση σου παραπάνω είναι λάθος κι ο λόγος είναι ότι γαι $ p$ περιττό δεν υφίσταται το $ f(\frac {p}{2})$ αφού το $ \frac{p}{2}$ δεν ανήκει στο $ N$ . Η διαφορά με το να έχουμε σύνολο τιμών τους φυσικούς ειναι ότι τότε το $ f(2k \plus{} 1)$ το βρίσκαμε βάσει του: $ 2k \equal{} f(2k) < f(2k \plus{} 1) < f(2k \plus{} 2) \equal{} 2k \plus{} 2$ . Εδώ το πρόβλημα ξεκινάει ήδη απ' την εύρεση του $ f(3)$ (Αν βρεις ένα τρόπο να βρεις αυτό λογικά θα μπορείς να το αποδείξεις για όλα τα $ n$).

Ευχομαι να κατάλαβες τι θέλω να πω... και συγγνώμη ξανά για το λάθος :blush:

Η βασική ιδέα είναι ότι ορίζω μια άλλη συνάρτηση με ακριβώς τον ίδιο τρόπο στους πραγματικούς και αποδεικνύω τη σχέση γι' αυτήν. Έπειτα, αφού η σχέση ισχύει στο πεδίο ορισμού της, θα ισχύει και για κάθε υποσύνολο του πεδίου ορισμού της άρα και για τους φυσικούς. Δυστυχώς όμως, θυμόμουν λάθος τη θεωρία. Δεν υπάρχει "επαγωγή στους πραγματικούς", με την αυστηρή έννοια τουλάχιστον.

Να πως θα ήταν το σωστό:
Έπρεπε να αποδείξω ότι ιδιότητα αυτή ισχύει για κάθε y που βρίσκεται σε μια περιοχή του 0 που να περιέχει πάνω από ένα στοιχεία.
Έπειτα, για κάθε x θετικό θα υπήρχε k τέτοιο ώστε το $ y \equal{} x\cdot2^{ \minus{} k}$ να βρίσκεται σε αυτή την περιοχή του 0 και άρα $ f(y) \equal{} y \Leftrightarrow f^k(2)f(y) \equal{} 2^ky \Leftrightarrow f(x) \equal{} x$.

Το λάθος μου βρισκόταν στο ότι:
1. η περιοχή που θεώρησα ήταν γύρω από το 1 αντί το 0
2. η περιοχή που το απέδειξα, περιείχε μόνο ένα στοιχείο (βασικά δεν έπρεπε να την πάρω απείρως μικρή)

Ευχαριστώ πάντως για την υπομονή σου να μου δείξεις το λάθος μου (αν προσπαθούσα μόνος μου να καταλάβω ότι έκανα λάθος, το πολύ - πολύ να κατέληγα στο ότι
1 + 1 = !@#%^&**&^#@%^ & ^%^^*&#  :rotfl: ).

Παιδιά, δύο σχολιάκια:

1. [u]κάθε[/u] συνάρτηση με πεδίο ορισμού τους φυσικούς $ \mathbb{N}$ είναι/θεωρείται συνεχής, αλλά αυτό δε σημαίνει ότι έχει τις συνήθεις ιδιότητες των συνεχών συναρτήσεων ;)

2. Για να κάνετε επαγωγή στους πραγματικούς χρειάζεστε την έννοια της transfinite επαγωγής (sorry, δεν ξέρω πώς μεταφράζεται το transfinite) και αυτό απαιτεί να έχει το $ \mathbb{R}$ μία καλή διάταξη. Η συνήθης διάταξη $ <$ του $ \mathbb{R}$ δεν είναι καλή, επειδή υπάρχουν υποσύνολα (πχ το $ \mathbb{R}$ το ίδιο) που δεν έχουν κάποιο ελάχιστο στοιχείο. Εν συντομία, καλύτερα να ξεχάσετε την επαγωγή στους πραγματικούς... αν ίσχυε έτσι απλά, θα το είχαμε κλείσει το μαγαζί προ πολλού ;)

Slow down guys, μη γράφετε πολλά μαζεμένα και βιαστικά πράγματα με αναφορές σε διαφορετικά posts, γιατί μπερδευόμαστε όλοι και δημιουργείται χάος...

Cheerio,

Durandal 1707

[quote="Durandal"]

1. [u]κάθε[/u] συνάρτηση με πεδίο ορισμού τους φυσικούς $ \mathbb{N}$ είναι/θεωρείται συνεχής, αλλά αυτό δε σημαίνει ότι έχει τις συνήθεις ιδιότητες των συνεχών συναρτήσεων ;)

Cheerio,

Durandal 1707[/quote]

 :?:

Μία συνάρτηση $ f: X\to Y$ λέγεται συνεχής όταν η αντίστροφη εικόνα $ f^{\minus{}1}(U)\subseteq X$ είναι ανοικτή στον $ X$ για κάθε ανοικτό υποσύνολο $ U\subseteq Y$.

Όταν πεδίο ορισμού είναι το $ \mathbb{R}$, ο ορισμός αυτός ταυτίζεται με τον ορισμό του σχολικού βιβλίου (btw, αυτό είναι πολύ διδακτική ασκησούλα πάνω στον ορισμό των ορίων!). Ωστόσο, στη (διακριτή) τοπολογία των φυσικών, κάθε υποσύνολο θεωρείται ανοικτό, οπότε και κάθε συνάρτηση με πεδίο ορισμού τους φυσικούς θα είναι συνεχής. Αυτό όμως δε σημαίνει ότι πρέπει να έχεις την ίδια διαίσθηση για μία ακολουθία και για μία συνεχή πραγματική συνάρτηση.

Περισσότερο εγκυκλοπαιδικά το ανέφερα (κυρίως σε αντιπαραβολή με το σχόλιο του meinstein) για να μη δημιουργούνται παρερμηνείες και να μην εκπλαγούν όσοι το βρουν μπροστά τους κάποια στιγμή. Αν σ'ενδιαφέρουν περισσότερες λεπτομέρειες για το θέμα, κάθε εισαγωγικό βιβλίο τοπολογίας η ανάλυσης θα σε καλύψει.

Cheerio,

Durandal 1707


how do i proove that there is a infinity of pairs $(a,b,t)\in Z^3$
with the property that $a^2+b^2+4=t^2$ ?
it's with pytagoreic numbers i think but i don't know really how to proove this...

could you at least proove that...please
i want to learn about how to solve such problems

I would also be interesed in the solution in this problem..

This problem is solved; just substitute those values and calculate.

James! They want to know how to guess such an answer.

yes .. you are right how do I get to have such numbers....I want the demonstration please

Look: we know that $a^2+4a+4=(a+2)^2$ is a perfect square, so it is sufficient to find $b$ s.t. $b^2=4a$. From here we immediately obtain the blang's answer. I am sure blang thought in the same way.

[quote="Myth"]$a^2+4a+4=(a+2)^2$ is a perfect square, so it is sufficient to find $b$ s.t. $b^2=4a$ (...) I am sure blang thought in the same way.[/quote]
You're right Myth  ;)

so $a=w^2$ a perfect square and then $b^2=4w^2$ so $b=2w$ and $t^2=(w^2+2)^2$ so $t=w^2+2$

so we have triplets of form : $(w^2,2w,w^2+2)$


$(0.\overline{3})^2=0.\overline{x}$, where $x$ is a digit.  What is the value of $x$?

[hide]We know that: $0.\overline{3}=\frac{1}{3}$, so therefore, the given equation equals:
$(\frac{1}{3})^2=\frac{1}{9}=0.\overline{1}$

Therefore, the digit $x$ equals $\boxed{1}$.[/hide]


integration
calculus
advanced fields
advanced fields unsolved

Let $A$ be a symmetric positive definite $n \times n$ matrix.

Show that

$\int \,dx_{1} \ldots \,dx_{n} \text{exp}\left(-\sum_{i,j} x_{i}A_{ij}x_{j}\right) = \pi ^{n/2} (\text{Det}A)^{-1/2} = \pi ^{n/2} \prod_{i} \lambda_{i}^{-1/2}$

where $\lambda_{i}$'s are the eigenvalues of $A$.

This is a generalization of the fact that $\int_{\mathbb{R}^{d}} \text{exp}\left(-\pi |x|^{2}\right) \,dx = 1$. (this is where $A$ is the identity).

The integral can be evaluated by a simple change of variables. Let $B$ be the unique symmetric positive definite square root of $A$; then $\exp(-\sum x_iA_{ij}x_j)=\exp(-X^TAX)=exp(-Y^TY)$ where $Y=BX$. The Jacobian for this transformation is $\det B=\sqrt{\det A}$, so $\int_{\mathbb{R}^n} \exp(-\sum x_iA_{ij}x_j)\, dx_1dx_2\cdots dx_n=\frac1{\det B}\int_{\mathbb{R}^n}\exp(-Y^TY)\, dy_1dy_2\cdots dy_n$
$=\frac1{\sqrt{\det A}}\int_{\mathbb{R}^n}\exp(-y_1^2-y_2^2-\cdots-y_n^2)\, dy_1dy_2\cdots dy_n=\frac1{\sqrt{\det A}}\left(\int_{-\infty}^\infty e^{-y^2}\, dy\right)^n.$

That Gaussian has integral $\sqrt{\pi}$, so the original integral is $\frac1{\sqrt{\det A}}\cdot\pi^{\frac n2}.$


Dear everyone

For a triangle [b]ABC[/b], three points [b]A'[/b], [b]B'[/b], [b]C'[/b] are on [b]BC[/b], [b]CA[/b], [b]AB[/b] respectively. The three circumcircles of triangles [b]AB'C'[/b], [b]BC'A'[/b], [b]CA'B'[/b] meet at a point [b]K[/b]. Let [b]A"[/b] be the intersection of [b]BC[/b] and the circumcircle of triangle [b]KAA'[/b]. In the same way, define [b]B"[/b], [b]C"[/b].

Prove the following

(1) [b]A"[/b], [b]B"[/b], [b]C"[/b] are collinear if and only if [b]AA'[/b], [b]BB'[/b], [b]CC'[/b] are concurrent.

(2) if [b]AA'[/b], [b]BB'[/b], [b]CC'[/b] concur at a point [b]P[/b], the line [b]A"B"C"[/b] is the polar of [b]P'[/b] w.r.t. the circumcircle of triangle [b]ABC[/b], where [b]P'[/b] is the isogonal conjugate of [b]P[/b].

I got this from my friend but don't know where it was from. May I ask you a favor?
Your advices help me a lot. Thanks.

Dear Mathlinkers,
this problem comes from a French periodical Journal de Mathematiques Elementaires 65 (1940-41), p. 22, #13023 with solution...
Sincerely
Jean-Louis


Let $a_{i}\in R,i=\overline{1,n}$ be constant and
$f(x)=\sum_{i=1}^{n}\frac{\cos(a_{i}+x)}{2^{i-1}}$ prove that
if $f(a)=f(b)$ then $a-b=k\pi$

You are not wrong but I'm not wrong too because
Mine say if $f(a)=f(b)\Rightarrow a-b=k\pi$ it isn't contradictory with yours, you say $a+b=\frac{\pi}{2}\Rightarrow f(a)=f(b)\Rightarrow a-b=k\pi$
Thus we can infer $a=\frac{\pi}{4}+k\frac{\pi}{2},a=\frac{\pi}{4}-k\frac{\pi}{2}$
There isn't contradiction here!

Oh, come on.
${\pi\over3}+{\pi\over 6}={\pi\over 2}$, so $f({\pi\over3})=f({\pi\over 6})$, but ${\pi\over3}-{\pi\over 6}={\pi\over3}\not =k\pi$ for any $k\in\mathbb{Z}$.

ok ok you are right, sorry I lost zero number in my problem :D ,my problem must be
Let $a_{i}\in R,i=\overline{1,n}$ be constant and
$f(x)=\sum_{i=1}^{n}\frac{\cos(a_{i}+x)}{2^{i-1}}$ prove that
if $f(a)=f(b)=0$ then $a-b=k\pi$
Thanks

See also [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=80905[/url] for a slightly different approach.


Suppose log base 4n of 40 root 3 = log base 3n of 45, find n^3.

Let $ x \equal{} \log_{4n}{40\sqrt3} \equal{} \log_{3n}{45}$.

Then $ 4^xn^x \equal{} 40\sqrt3$, and $ 3^xn^x \equal{} 45$, so dividing gives

$ \frac {4^xn^x}{3^xn^x} \equal{} \left(\frac {2^2}{3^1}\right)^x \equal{} \frac {40\sqrt3}{45} \equal{} \frac {8\sqrt3}9 \equal{} \frac {2^33^{\frac12}}{3^2} \equal{} \frac {2^3}{3^{\frac32}}$

$ \implies x \equal{} \frac32\implies(3n)^{\frac32} \equal{} 45\implies(3n)^3 \equal{} 3^3n^3 \equal{} 45^2 \equal{} 3^4\cdot5^2$

$ \implies n^3 \equal{} 3\cdot5^2 \equal{} \boxed{75}$.


The figure below shows two parallel lines, $ \ell$ and $ m$, that are distance $ 12$ apart:
[asy]unitsize(7);
  draw((-7, 0) -- (12, 0));
  draw((-7, 12) -- (12, 12));
  real r = 169 / 48;
  draw(circle((0, r), r));
  draw(circle((5, 12 - r), r));
  pair A = (0, 0);
  pair B = (5, 12);
  dot(A);
  dot(B);
  label("$A$", A, plain.S);
  label("$B$", B, plain.N);
  label("$\ell$", (12, 0), plain.E);
  label("$m$", (12, 12), plain.E);[/asy]
A circle is tangent to line $ \ell$ at point $ A$.  Another circle is tangent to line $ m$ at point $ B$.  The two circles are congruent and tangent to each other as shown.  The distance between $ A$ and $ B$ is $ 13$.  What is the radius of each circle?

[hide]Let $ P$ be the center of the circle containing $ A$. We have $ \cos \angle PAB\equal{}\frac{12}{13}$ by the problem conditions. Letting $ C$ be the common point of tangency for the two circles, and $ M$ be the midpoint of $ AC$, we have that $ AM\equal{}\frac{13}{4}$. By looking at the right triangle $ AMP$, we find $ AP\cos\theta \equal{} \frac{13}{4} \iff AP\equal{}\frac{13}{4}\cdot\frac{13}{12}\equal{}\frac{169}{48}$.[/hide]

[hide="a little better maybe?"]
Let the point of intersection be X. A,X,B are collinear, and AX=BX. Drop the perpendicular from A to m, and let it intersect the circle containing A a second time at Y, and m at A'. AYX is right, and AYX is similar to ABA'. AX=13/2, and AY/AX=AB/AA'=13/12, so AY=169/24, and the radius of either circle is 169/48.[/hide]

[hide="Yay I got a geometry for once"]

Connect the centers of the two circles. Drop the perpendicular from B to line l, intersecting at point C. Draw the perpendicular from the center of circle A to the line just drawn. Then we have a right triangle with side lengths unknown, 12-2x, and 2x as the hypotenuse. We find the unknown side to be $ \sqrt{48x \minus{} 144}$ by pythag.

we see another right triangle, ACB with lengths already 12 and 13 as the hypotenuse. Thus, we must have $ AC \equal{} \sqrt{48x \minus{} 144} \equal{} 5$

Solving, we find $ x \equal{} \boxed{\dfrac{169}{48}}$

[/hide]

I have tried it almost 10 times but not get the answer and I am too excited now to get the answer and will try whole night.

There are three good solutions above.  The "official" solution is here:
[url]http://mathprize.atfoundation.org/archive/2009/index[/url].

No, that's not it, but if you explain your reasoning, we can figure out where you went astray.


Let a, b>0 and n>1. Show that A is true (see file attached)

[/code]


An intersting identity with an intersting consequence


a1=k

an+1= an - 1 (if an can be divided by 2)
an+1= (an - 1)/2 (if an can't be divided by 2)

what has to be the minimal value of k so that a2005 is the first to turn 0.

$k=2^2003$, because recurent equation $a_{n+1}=f(a_n)$ equavalent to "$a_{n+1}=a_n>>1$" in C++, $a_{n+1}$ had one digits less than  $a_n$ in basic 2.

k=2^2 003?
Is there a problem with my web browser? does anyone else see this number?
I got k=2^1003-3
is it correct?

No, there's nothing wrong with your browser. i wonder what he means by $2^2003$ I think  he means$2{}^2{}^0{}^0{}^3$. It's a common error.

No, there's nothing wrong with your browser. i wonder what he means by $2^2003$ I think  he means$2{}^2{}^0{}^0{}^3$. It's a common$LaTex$ error.


$x^8-x^5+x^2-x+1$

prove that this expression is always positive for x being real

Assuming x is defined for all real values (and not for complex... test x=i gets -2i+1>0 which is just wrong because it implies i<0.5), clearly if x negative, then it is really quite trivial (the odd powers simply become positive).  So, let us only consider the positive cases,

rearranging,
x^8+x^2+1>x^5+x
now this should be obvious for all x>1,

So proving for all 0<x<1,
letting x=1/a such that a>1 and then multiplying through with a^8 and rearranging should be pretty obvious again ( i don't know how to use latex so i won't bother trying to write it all out)

of course testing x=0,1 will show that it is positive

[quote="zeus_three"]Assuming x is defined for all real values (and not for complex... test x=i gets -2i+1>0 which is just wrong because it implies i<0.5), clearly if x negative, then it is really quite trivial (the odd powers simply become positive).  So, let us only consider the positive cases,
[/quote]
-2i + 1 > 0 means nothing. Inequalities don't exist for complex numbers, because they don't have any order like real numbers do. If a question deals with an inequality or "positive/negative" it is understood that it's dealing with real numbers.

I know, i just thought it looked a little blank without any reference to exactly what x is

[quote] If a question deals with an inequality or "positive/negative" it is understood that it's dealing with real numbers.[/quote]
Unless one  states  that $x$   is real, the  originial statement  is [b] NOT   true[/b], The expression can be poistive, negative, or even complex.  Nothing trivial or "picky"  about  about zeus_three's statement .  What could be "understood" here is the  problem is askling to prove that "[b]expression[/b]"  is positive (and thus real),   which,   of course is NOT the same thing as  stating as "$x$ is positive (or real)"    (and is infact  the statement may not be  true when x is not real)

Eminem - You may like to edit the problem  and remove this  this serious ommision.

Also, FWIW the expression can also be written as
 $x^8-x^5+x^2/4 \ \ \ + x^2/2+ \ \ \ x^2/4-x+1\\ = (x^4-(x/2))^2+ (x^2/2)+((x/2)-1)^2$
so, as long as x is real, it (the expression)  is positive.

[hide="Sol"]
this can be written as,
$x(x-1)[x^4(x^2+1+x)+1]+1$
Now for all REAL values $x^2+x+1$ = $(x+1)^2-x$ which is positive,
And then if $x$ is negative then $x(x-1)$ is also negative,
If positive and greater than $1$ then the whole expression is positive,
If $0$ then whole expression=$1$

I know this is not perfect sol but if there is anything wrong please tell me.... OKAY sachin?
[/hide]

[quote="Gyan"]
Also, FWIW the expression can also be written as $(x^4-(x/2))^2+((x/2)-1)^2 +(x^2/2)$ so, as long as x is real, it is positive.[/quote]
How did you get that final expression? :?  :(

[quote]How did you get that final expression? [/quote]
Way of thinking which makes this kind of work  easier ...
Trivial stuff but read it  if you are curious --
 [hide]Point 1:   Square of anything (real number that is) is always non negative...
Point 2:     $x^2 (- \ or \ +) Bx(something)$  would be a square if you add  $(B/2)^2$ to it
So when you see $(-X^5$  you add   ((1/2)(whatever square  is needed  etc) to it and see ...if you do end on the right (or correct ) side  of the  positive/negative universe..... same thing you do witn $-x^3$ and fortunately  you still have half of $x^2$  to spare .... Hope this helps. [/hide]

Gyan- Then how come this question only works for REAL no.s? if your answer was coming that way?

:blush:  :blush:  :blush:  :blush:  :oops:  :oops:  :oops:  sorry guys , i guess i made a bloomer , it is real , i forgot to give that . i 've edited it now

thanx a lot gyan , and cool soln!!!


Is there an easier way to label a line?

I do it like:

[asy]size(200);
draw((0,0)--(500,0));
label("$l$",(250,0),S);[/asy]

Any better way than manually finding the midpoint?

You can label paths. It always finds the center point of the path.

[asy]size(200);
draw((0,0)--(500,0));
label("$l$",(0,0)--(500,0),S);[/asy]

(Oh, and this is posted in the wrong forum.)

You could also use the midpoint() function passing in the path as the argument.

Optimized code:
[asy]draw("$l$", (0,0)--(200,0));[/asy]

You can simply specify the position, the color etc... like this:
[asy]draw(Label("$l$",position=Relative(0.75), red), (0,0)--(200,0));[/asy]

If you want to rotate...

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=220794


search
ratio
algorithm
calculus
probability
number theory

I am in a strange situation.I want to find a book that I once read in a library,but to make it short- it's impossible to visit that library any more.

I tried google book search/web search,but it didn't help.

Since many of you read a lot of math books, I will give a description of that book on the basis of what I remember.See if anything you read fits the description and tell me it's name/author.

If i remember correctly,
-the book was mainly on discrete mathematics (and probably the title also implied that)
-the book had more than one author,most likely three(not necessarily western,the names had vowels-like I)
-towards the beginning of the book,there was discussion on sets ,saying how everything can be in a universal set or something like that
-[b]most importantly,there was a lot of discussion on the game of nim,at least 4 pages,and an entire chapter was dedicated to similar simple games.[/b]
-the chapter on games was most likely chapter 6(otherwise it was chapter 7 or 8),if I am not mistaken.
-later ,there was a chapter discussing generating functions/recurrence relations/difference equation.
-somewhere in that chapter,the value of golden ratio(the limiting value of fibonacci ratio) was calculated.

So,can you help me?I really want to find that book,as I have some unfinished material to read,especially on nim.I know it's difficult from my description,but any help will be deeply appreciated.Thanks

[i]Selections from Modern Abstract Algebra[/i] by Andree pages 33 to 35 covers Nim.
[i]Elementary Number Theory[/i] by Upensky and Heaslet, pages 15 to 19.

Thanks for the guesses and suggestions.

I think there is a chance that the boook might be:
[b]Introduction to Discrete Mathematics [i]by[/i] R.J.McEliece,R.B.Ash and C.Ash.[/b]
But I can't confirm my suspect without knowing at least the names of the chapters in this book.Googling did not help me find out the chapter names.If anyone has access to this book,can you please tell me the names of the chapters in this book?

The game of NIM is also discussed and the strategy proven in Yaglom and Yaglom - Non-elementary problems treated in an Elmentary way (approximate title).

I just ordered a copy for 9 cents plus shipping. I'll let you know in a few weeks.

Chapter One Preliminaries
Chapter Two Combinatorics: The Theory of Counting
Chapter Three Graphs and Algorithms
Chapter Four Graphs and Algorithms Part II
Chapter Five Propositional Calculus and Boolean Algebra
Chapter Six Mathematical Models for Computing Machines
Chapter Seven Formal Languages and Algorithms
Chapter Eight Mathematical Games
Chapter Nine Discrete Probability Theory
Chapter Ten Finite Difference Equations


\[%Error. "eqalign" is a bad command.
{ \&{\rm{Let 0 < b}}\le{\rm{a}}{\rm{. Prove that}}\cr \&{1 \over 8}{{\left({a-b}\right)^{2}}\over a}\le{{a+b}\over 2}-\sqrt{ab}\le{1 \over 8}{{\left({a-b}\right)^{2}}\over b}\cr}\]

We transform the inequality equivalently:
$\frac{(a-b)^{2}}{8a}< \frac{a+b}{2}-\sqrt{ab}<\frac{(a-b)^{2}}{8b}\iff$
$\frac{(a-b)^{2}}{4a}<a+b-2\sqrt{ab}<\frac{(a-b)^{2}}{4b}\iff$
$\frac{(a-b)^{2}}{4a}<\left(\sqrt{a}-\sqrt{b}\right)^{2}<\frac{(a-b)^{2}}{4b}\iff$
$\frac{a-b}{2\sqrt{a}}<\sqrt{a}-\sqrt{b}<\frac{a-b}{2\sqrt{b}}\iff$
$a\sqrt{b}-b\sqrt{b}<2a\sqrt{b}-2b\sqrt{a}<a\sqrt{a}-b\sqrt{a}$.
For the left part: $2a\sqrt{b}-2b\sqrt{a}-a\sqrt{b}+b\sqrt{b}= a\sqrt{b}-2b\sqrt{a}+b\sqrt{b}= \left(a^{1/2}b^{1/4}-b^{3/4}\right)^{2}>0$, for the right part: $a\sqrt{a}-b\sqrt{a}-2a\sqrt{b}+2b\sqrt{a}= b\sqrt{a}-2a\sqrt{b}+a\sqrt{a}= \left(b^{1/2}a^{1/4}-a^{3/4}\right)^{2}$ which ends the proof.

starts the same as Rzeszut did :)
$\frac{(a-b)^{2}}{4a}<(\sqrt{a}-\sqrt{b})^{2}<\frac{(a-b)^{2}}{4b}$
$\iff4a\leq (\sqrt{a}+\sqrt{b})^{2}\leq 4b$


Find $f: \mathbb{R}\rightarrow\mathbb{R}$ satifies conditions:
a) $f(x+y)\leq f(x)+f(y)\forall x,y\in\mathbb{R}$
b) $\lim_{x\to\infty}\frac{f(x)}{x}=1$

Has anyone tried this problem?

I haven't solved it yet, but I got strange results on the "road".

The problem can solve by uesing the fact inscreasing function


algebra
polynomial
function
analytic geometry
graphing lines
slope
calculus

let P be polynomial of odd degree.
show that from any point of the plan passes a tangent line to the curve of P

I would do this:

let $P(x)=a_{2k+1}x^{2k+1}+\cdots+a_{0}$ be your polynomial.

Fix (a,b) a point in the plane and you have to prove that there is a point located on the graph of $P$, namely $(\alpha, P(\alpha))$, such that the tangent line passes through (a,b).

The equation of the tangent line is

$y=P(\alpha)+P'(\alpha)(x-\alpha)$,

you solve the equation

$b=P(\alpha)-P'(\alpha)(a-\alpha)$ for $\alpha$ , more precisely you have to show that the preceding equation has at least a solution , which has since $P(\alpha)-P'(\alpha)*(a-\alpha)$ is a polynomial of odd degree which has at least a real solution.
Didi

The following question is natural.

Find all polynomials of even degree for which a similar statement holds or prove that none exists.

Didi

I would conjecture that none exist.

I don't feel like writing out a proof, but here's the intuition. For the function $f(x)=x^{2k},$ one cannot write a tangent line to the curve from any point "inside" the graph. (That is, $(x,y)$ such that $y>x^{2k).}$)

For a general even polynomial, there's a bounded region outside which the graph is (closely enough) approximated by the graph of $f(x)=ax^{2k}.$ We may be able to reach some of the points "inside" the graph by lines tangent to the curve within that bounded region. However, $f'(x)$ is bounded within that bounded region, so the slopes of the tangent lines are bounded, which limits how "high" the can reach into the "inside" of the graph.

Someone with enough patience should be able to use that intuition to construct a proof.

Firstly, let me point out a little mistake:
[quote="didilica"]The equation of the tangent line is

$y=P(\alpha)+P'(\alpha)(x-\alpha)$,

you solve the equation

$b=P(\alpha)-P'(\alpha)(a-\alpha)$ for $\alpha$, more precisely you have to show that the preceding equation has at least a solution , which has since $P(\alpha)-P'(\alpha)*(a-\alpha)$ is a polynomial of odd degree which has at least a real solution.
Didi[/quote]
The equation is $b = P \left( \alpha \right)+P^\prime \left( \alpha \right) \left( a-\alpha \right)$.

[quote="didilica"]Find all polynomials of even degree for which a similar statement holds or prove that none exists.[/quote]
$P \left( \alpha \right)+P^\prime \left( \alpha \right) \left( a-\alpha \right)$ is a polynomial of even degree, hence its image is not the whole $\mathbb R$.

Let`s conclude with the following result:

Lemma: Let P be a polynomial. The following statements are equivalent:

1) P is a polynomial of odd degree

2) From any point of the plane passes a tangent line to the curve of P.

Solution.
1) implies 2) has been proven in a previous post. Assume 2) holds, then we have to show that P is a polynomial of odd degree. By contradiction if the degree of P is even then using Perfec Ratio`s argument we get the contradiction.


A carton contains milk that is $ 2\%$ fat, and amount that is $ 40\%$ less fat than the amount contained in a carton of whole milk. What is the percentage of fat in whole milk?

$ \textbf{(A)}\ \frac{12}{5} \qquad
\textbf{(B)}\ 3 \qquad
\textbf{(C)}\ \frac{10}{3} \qquad
\textbf{(D)}\ 38 \qquad
\textbf{(E)}\ 42$

[hide="Click for solution"]
Let $ x$ denote the percent of milk-fat in whole milk.  Then:
\begin{align*} x - 0.4x & = 2 \\
0.6x & = 2 \\
x & = \frac {10}{3} \end{align*}

This corresponds to answer choice $ \boxed{\textbf{(C)}}$.
[/hide]

Does anyone else think that this question could've been worded better, as in "an amount that is 40% less [b]fatty[/b] than the amount contained in a carton of whole milk"?


Solve the system of equations:

$x+xy+xyz = 12$
$y+yz+yzx = 21$
$z+zx+zxy = 30$

Obviously neither variable can be zero ($x(1+y+yz)=12\neq 0$ etc.). Multiplying the second equation by $x$ and subtracting from the first we get

$x-yzx^{2}=12-21x\iff 22x-12=yzx^{2}\iff xyz=22-{12\over x}$

Multiplying the third by $y$ and subtracting from the second, and multiplying the first by $z$ and subtracting from the third, we similarly find

$xyz=31-{21\over y}, xyz=13-{30\over z}$

Hence $22-{12\over x}=31-{21\over y}=13-{30\over z}$, which yields

$y={7x\over 4+3x}, z={10x\over 4-3x}\qquad(*)$

Now the first equation becomes

$x+{7x^{2}\over 4+3x}+{70x^{3}\over 16-9x^{2}}=12$

or, after simplifying,

$5x^{3}+17x^{2}+2x-24=0$

This factors to $(x-1)(x+2)(5x+12)=0$.

For $x=1$ from $(*)$ we find $y=1,z=10$

For $x=-2$ we get $y=7,z=-2$

For $x=-{12\over 5}$ we get $y={21\over 4},z=-{15\over 7}$

Hence all the solutions are $(1,1,10),(-2,7,-2),(-{12\over 5},{21\over 4},-{15\over 7})$


Suppose $A\cup B=\mathbb{N}$ and $A\cap B=\varnothing$. Prove that for any $c>0$ there are increasing sequences $a_n\in A$, $b_n\in B$ s.t. $\frac{a_}{b_n}\to c$ as $n\to \infty$.

I think, something similar was on the forum. But how to find it now?

I guess it was grobber who immediately deleted his reply?  :?

I also seem to remember something very similar, if not the same, but I have no idea how one could start looking for it: I can think of no good keywords, and I don't remember any of the details (like forum members involved in the discussion) :).

I suppose both $A,B$ must be infinite, right? For $c=1$ it's obvious, so assume WLOG that $c>1$. Assume the contrary to  what we have to prove. This translates to the fact that we can find $\varepsilon$ s.t. for all $b_n\in B$ (with $n$ sufficiently large) and all $a_m\in A,a_m>b_n$, we have $\frac{a_m}{b_n}>c+\varepsilon$ or $\frac{a_m}{b_n}<c-\varepsilon$, so $\mathbb N\cap [b_n\cdot (c-\varepsilon),b_n\cdot (c+\varepsilon)]\subset B$ if $n$ is large enough. Then the naturals in $[b_{n_1}\cdot(c-\varepsilon)^2,b_{n_2}\cdot(c+\varepsilon)^2]$ also belong to $B$, where $b_{n_1},b_{n_2}$ are the smallest and largest naturals in the first interval we considered, respectively, and so on. However, if we reiterate this procedure indefinitely, then the intervals will cover all of $\mathbb N$, so we will find that all sufficiently large naturals belong to $B$, contradicting the infinitude of $A$.

Is it Ok?

[Edit: yes, Myth, there were some pretty stupid mistakes in my first reply, as usual; I hope it's not the case now, but who knows? :)]


Hi guys:

has any of you taken an extra year at high school?

well, i'm planning to do so, since i want to have more science credit and improve my english bit =)

so, here s the thing--can i still write the gr 12 contests even if i stay one more year? i will be 18~19 during that extra year. COMC, Euclid, AMC 12 AIME, ? all of them or none of them?

thanks!

I'm not sure if u can participate in Euclid officially. (It depends on what exactly is ur birthday, because u need to be born on or after Jan 1, 1989 to write Euclid officially)

Other contest, I don't know ^^

Hope this helps

Well, I guess you can take Gr12 contests if you plan to stay for an extra year (at least I haven't heard that you can't). But I still suggest you talk to your teacher about that.

i was born may 12 1990

what contests can i write?

thanks!

i'll take an extra yr...
i think u can write most of those contests except....ur too old

i think on the Euclid under the grade thing there is a bubble that says Grade 13..w/e im sure u can write all the contests

EDIT: whoops i thought zhuyihan was in grade 12.. :blush: (not sure if zhuyihan is in grade 12..)

hi Sma4t

im in grade 11 now =)

to vincent-five:

too old?
what contests am i too  old for?

thanks!

i didn't see ur post ...
u r eligible for all the contests


What is the value of $ \frac{14!}{5!9!}$ ?


Can you find a tetrahedron T contained inside a tetrahedron W such that the sum of the lengths of the edges of T is bigger than the sum of the lengths of the edges of W?

I don't know the answer to this one myself, but it sounds interesting, and I heard it is from an old issue of Crux.

My guess would be that the answer is "no."  It definitely is true for lower-dimensional variants (a segment and a triangle), but I don't really know how to prove it.  I mean, I want to make an argument like "moving it closer to a vertex will always increase the lengths, so putting each corner at the vertices of the original tetrahedron is best-case, but of course that is just a greedy argument and not necessarily true.  What I need to be able to say "moving it away from a vertex always makes the lengths shorter," but that definitely isn't true.  So I don't know.

This was, of course, just a long-winded BUMP.

Wow, this is the second bump for this one.


I haven't seen this solution yet on the forums and I was surprised I saw it myself during the test.

Recognize that the first row represents the difference of consecutive squares and therefore the second row represents squares of integers 2 apart and so on... The first row Starting with 1^2 - 0^2 and Ending with 50^2 - 49^2
Hence we are looking for the difference of 2 squares that is a multiple of 67

x^2 - y^2 = 67k  where x>y
(x+y)(x-y) = 67k

because 67 is prime one of two above must be 67 and it can't be (x-y) because the max of x is 50.
so...
(x+y) =
(50+17)
(49+18)
(48+19)
...
(34+33) and stop here because x cannot be less than y

There are 17 values here so it occurs 17 times!

It's a nice solution, but can you not make a new thread next time?

hmm i counted differentlyu at the end but same answer (although my solution looks similar)

basically I got down to

n+2k=67 (clearly it has to be 67*1, or k is too big) for 0<=k<=50-n
(there was also some 2^n-1 attatched to it but thats irrelevant)
where n is the row and k is the term

counting same way you did by just listing for rows and term

(1, 33)
(3, 32)
...
(33,17)

Thats 33-17+1=017 for number6.


Given a partition $ \lambda \equal{} (\lambda_1 \geq \lambda_2 \geq \ldots \geq \lambda_k > 0)$ with $ \lambda_1 \plus{} \ldots \plus{} \lambda_k \equal{} n$ ($ n$ fixed, $ k$ variable), we associated the [i]Young diagram[/i] that consists of $ k$ rows of boxes, the $ i$th row containing $ \lambda_i$ boxes, all justified to the left.

A [i]standard tableau[/i] of shape $ \lambda$ is a filling of the boxes of the diagram of $ \lambda$ with the numbers $ 1, \ldots, n$, each occuring once, so that the result is increasing along rows and columns.  The number of such fillings is denoted $ f^\lambda$.

A [i]semistandard tableau[/i] of shape $ \lambda$ by $ [m]$ is a filling of the boxes of the diagram of $ \lambda$ with the numbers $ 1, \ldots, m$, so that the result is weakly increasing along rows and strictly increasing along columns.  The number of such fillings is denoted $ f(\lambda; m)$.

Question 1: We get a partial ordering $ P$ on the set of partitions of $ n$ by placing $ \lambda$ above $ \mu$ exactly when $ f^\lambda > f^\mu$, and similarly we get a poset $ P_m$ for each $ m$ by placing $ \lambda$ above $ \mu$ exactly when $ f(\lambda; m) > f(\mu; m)$.  Show that eventually $ P_m$ is constant (i.e., for large enough $ m$) and that the limit poset is a refinement of $ P$.

Question 2 (for which I do not have a solution): What can you say about $ P_n$?  For example, what are the maximal and minimal elements?

A hint for the first one: [hide]Let $ g(\lambda; k)$ be the number of semistandard tableaux of shape $ \lambda$ in which exactly $ k$ distinct values appear.  Express $ f(\lambda; m)$ in terms of these numbers.[/hide]

The second problem is probably very hard (at least, I see no reason for it to be easy, and similar problems are hard), but here's one quite tractable special case:

Question 3: Among the [i]hooks[/i] (that is, partitions with one part of size $ n \minus{} k$ and $ k$ parts of size 1 for some $ k$), which one maximizes the number of semistandard tableaux filled with $ [n]$?

Equivalently, let $ f(n, k)$ be the number of tuples $ (a, b_1, \ldots, b_{n \minus{} k}, c_1, \ldots, c_{k})$ of integers such that $ 1 \leq a \leq b_1 \leq \ldots \leq b_{n \minus{} k} \leq n$ and $ 1 \leq a < c_1 < \ldots < c_k \leq n$.  For fixed $ n$, which of $ f(n, 0), f(n, 1), \ldots, f(n, n)$ is maximal?

I should definitely know something about this.

The minimal element is easy: it's the one-column partition $ 1^n$, which admits exactly one tableau. This diagram has zero tableaux for smaller $ m$, and is minimal for larger $ m$ as well. [Sentence deleted. I looked at the hint, and saw I was about to repeat it]. Given a standard tableau on $ \lambda$, we get an injective map from $ m$-tableaux on $ 1^n$ to semistandard $ m$-tableaux on $ \lambda$ taking $ n$ distinct values by rearranging cells. The images are even disjoint if we take other standard tableaux.

The maximum is harder.


[quote="vangelismak"]If  a,b are possitive integers such a,b > 4 prove that $ ab > 2(a \plus{} b)$[/quote]

Thats odd, the latex has show it differently for me, still

$ ab > 2(a\plus{}b) \Rightarrow ab\minus{}2a\minus{}2b > 0 \Rightarrow ( a\minus{}2 ) ( b\minus{}2 ) \minus{} 4 > 0 \Rightarrow ( a\minus{}2 ) ( b\minus{}2 ) > 4$

The inequality shows that the point $ (2,\ 2)$ lies in the region with the line $ \frac{x}{a}\plus{}\frac{y}{b}<1.$ for $ a>4,\ b>4.$ Considering the $ x$ intercept and $ y$ intercept gives...

My solution.

We have to prove , ... $ ab > 2(a \plus{} b)$ <=> $ a(b \minus{} 2) > b \plus{} b$

let  $ s_1 \equal{} s_2 \equal{} s_3 \equal{} ... \equal{} s_{b \minus{} 2} \equal{} 1 \plus{} 1 \plus{} ... \plus{} 1 , ( a \minus{} aces )$ ,... then  LHS  is equivalent to $ s_1 \plus{} s_2 \plus{} s_3 \plus{} ... \plus{} s_{b \minus{} 2}$

so we have to prove  $ s_1 \plus{} s_2 \plus{} s_3 \plus{} ... \plus{} s_{b \minus{} 2} > b \plus{} b$

Taking 2 aces of each $ s_i , i \equal{} 1,2,... (b \minus{} 2) ,$ and adding, we have $ (1\plus{}1)\plus{}(1\plus{}1) \plus{}...\plus{}(1\plus{}1) \equal{} 2(b\minus{}2)$

 ,now the inequality is  $ 1\plus{}1\plus{}...\plus{}1 \plus{} 2b\minus{}4 > b\plus{}b$. The sum of aces at the LHS is > 4 because the condition b>4 show that there exist at least three $ s_i$ and the condition a>4 show that $ s_i$ have at least 3 aces, so we are done!

[quote="vangelismak"]If  a,b are possitive integers such a,b > 4 prove that $ ab > 2(a \plus{} b)$[/quote]

The same holds true for a, b reals.

$ a > 4 \Leftrightarrow ab > 4b$ since $ b > 4 > 0$
similarly $ ab > 4a$

thus,
$ ab \plus{} ab > 4a \plus{} 4a \Leftrightarrow 2ab > 4(a \plus{} b) \Leftrightarrow ab > 2(a \plus{} b)$

While solving the above problem, it occured to me that the following holds true, too:

for $ a_1, a_2, ... a_n > k \ge 0$ it is
$ a_1a_2...a_n > \frac {k}{n}\sum_{syc} a_1a_2...a_{n \minus{} 1} > \frac {k}{n}(n \minus{} 1)\frac {k}{n \minus{} 1}\sum_{syc}a_1a_2...a_{n \minus{} 2} > ... \frac {k^{n \minus{} 1}}{n}(a_1 \plus{} a_2 \plus{} ... a_n)$

The proof is similar to the one I outlined above.

Yes i Know that  $ ab > 2(b\plus{}a)$ , if a,b are reals ( a,b > 4 ) but the solution above is true only if a,b are integers ...  :)

Any way, ther is no big deal whith this inequality ... :|


What should I look out for as I fill out the common application?

For a school that gives you the option to use the common application or their own application, should I just use the common app?

Make sure to use optional parts of the form to self-report AP scores and other distinctions not specifically asked for on the form. The same goes for AMC scores and the like. I think the Common Application is a good idea, and it is the only form this year for Caltech, as it has long been for Harvard.

[quote="Iversonfan2007"]I think the Common App asks for too many things (for example, an extra essay). Also, it takes a while to do in my opinion.

And it's not very "user friendly" compared to some apps i've done on the college websites (for ex. Princeton app and Brown app).[/quote]

yeah i noticed that. i had to spend about 30 minutes reading "instructions" before I could even begin anything. MIT had a much nicer format.

Sorry to bring up an old topic, but where should I incorporate AMC scores?

And if they were only borderline good (ie I did quite well in 10th grade, did not take the exam in 9th grade, and barely qualified for AIMEs in 11th), should I report them?

Thanks!

EDIT:  Forgot to say thank you  :oops:

Yes, you should report them. It shows that you care about math. It depends on the school.

MIT asks for an AMC score on the application I believe.

In the common, it goes on the resume part.


Solve the equation: $ 3^{33} \cdot x \minus{} [2^{51}\minus{}9^{17}]\equal{}8^{17},$ where $ n \in \mathbb{Z}.$

$ x\equal{}\frac{2.8^{17}\plus{}9^{17}}{3^{33}}$

how can i simplify it further?

[quote="moldovan"]Solve the equation: $ 3^{33} \cdot x \minus{} [2^{51} \minus{} 9^{17}] \equal{} 8^{17},$ where $ n \in \mathbb{Z}.$[/quote]His equation would have an easy simple answer if it is $ 3^{33} \cdot x \plus{} [2^{51} \minus{} 9^{17}] \equal{} 8^{17}$, where $ x\in\mathbb{Z}$, and not $ n$ :P


Suppose f is defined in an interval (a,b) and f is concave (-f is convex).  Suppose f>0 in (a,b) and g:(0,infinity) ---> Real Numbers is decreasing and convex.  Show g o f( (g o f)(x)=g(f(x)) ) is convex.

Let's replace $ f(x)$ with $ \minus{}f(x)$ and $ g(t)$ with $ g(\minus{}t)$. This way, we are to prove that a convex function remains convex after composition with an increasing convex function.


I am having trouble understanding a couple problems:

11.5.4: A [b]unit fraction[/b] is a fraction of the form [i]1/n[/i] for some nonzero integer [i]n[/i]. Compute the number of ways to write 1/6 as the sum of two distinct positive unit fractions. (The order of  the fractions in the sum does not matter, so 1/2+1/3 would be considered the same sum as 1/3+1/2.)

11.5.5: Find all pairs of integers ([i]p,q[/i]) such that [i]pq-[/i]3[i]p[/i]+5[i]q[/i]=0.

I looked at both of the solutions, but they didn't make sense. Could someone please explain the solutions in a different way? :read:

It might help if you say what parts of the solution are confusing, so that we don't make the same mistake that the solution manual does.

[hide="Hint for 2"]Try factoring using SFFT (Simon's Favorite Factoring Trick)[/hide]

[hide="Solution"] Factor to $ (p \plus{} 5)(q \minus{} 3) \equal{} \minus{} 15$
Taking the factors of $ \minus{} 15$ - $ ( \minus{} 1, 15),(15, \minus{} 1),( \minus{} 3, 5),$ and $ ( \minus{} 5, 3)$
$ p \plus{} 5 \equal{} \minus{} 1$
$ q \minus{} 3 \equal{} 15$
$ p \equal{} \minus{} 6$
$ q \equal{} 18$
and so on for the rest of the factors...
[/hide]

[quote="Temperal"]For the second one, you just factor it into two products (too lazy to do it right now), adding an integer to both sides if necessary. Then, you take the constant on the RHS and list its' factors, which the two factors on the LHS would have to match up with. You take all cases of matching the factors, solve for the variables, and those are your solutions.[/quote]
In other words, SFFT.

EDIT:Someone already posted.

I hardly think that a student taking Introduction to Algebra is likely to be familiar with SFFT; don't you think the poster would already have solved the problem if she knew what you were talking about?  Especially since she's already stated that she was confused by the given solution, which is likely to be clearer than yours? :mad:

[b]pianogirl:[/b]  The first thing to do is to make sure that you understand the FOIL method.  The distributive property of multiplication over addition tells us that

$ a(c \plus{} d) \equal{} ac \plus{} ad$

and similarly that

$ b(c \plus{} d) \equal{} bc \plus{} bd$.

Now suppose we want to evaluate the product $ (a \plus{} b)(c \plus{} d)$.  Well, distributing the $ (c \plus{} d)$ first, we have

$ (a \plus{} b)(c \plus{} d) \equal{} a(c \plus{} d) \plus{} b(c \plus{} d)$

and then distributing $ a$ and $ b$ gives

$ (a \plus{} b)(c \plus{} d) \equal{} ac \plus{} ad \plus{} bc \plus{} bd$.

The basis of both of these problems is to FOIL in reverse.  (This is sometimes known as "Simon's Favorite Factoring Trick," which is what the other posters are trying to tell you.)  To start you off:

[hide="11.5.4"] We want to write $ \frac{1}{6} \equal{} \frac{1}{a} \plus{} \frac{1}{b}$.  Clearing denominators, this is equivalent to

$ ab \equal{} 6a \plus{} 6b \Leftrightarrow$
$ ab \minus{} 6a \minus{} 6b \equal{} 0$

which is more or less the same form as the next problem.  Now, note that

$ (a \minus{} 6)(b \minus{} 6) \equal{} ab \minus{} 6a \minus{} 6b \plus{} 36$.

How does this help you? [/hide]
[hide="11.5.5"] As above, note that

$ (p \plus{} 5)(q \minus{} 3) \equal{} pq \minus{} 3p \plus{} 5q \minus{} 15$.

Again, how does this help you? [/hide]

I thought a person taking Intro to Algebra would know FOIL... it seems to me that taking the factors and solving was the tricky and hard part (which I couldn't get myself when I first saw these type of problems) instead of SFFT.

[quote="pianogirl"][quote="t0rajir0u"]I hardly think that a student taking Introduction to Algebra is likely to be familiar with SFFT; don't you think the poster would already have solved the problem if she knew what you were talking about?  Especially since she's already stated that she was confused by the given solution, which is likely to be clearer than yours? :mad:

[b]pianogirl:[/b]  The first thing to do is to make sure that you understand the FOIL method.  The distributive property of multiplication over addition tells us that

$ a(c \plus{} d) \equal{} ac \plus{} ad$

and similarly that

$ b(c \plus{} d) \equal{} bc \plus{} bd$.

Now suppose we want to evaluate the product $ (a \plus{} b)(c \plus{} d)$.  Well, distributing the $ (c \plus{} d)$ first, we have

$ (a \plus{} b)(c \plus{} d) \equal{} a(c \plus{} d) \plus{} b(c \plus{} d)$

and then distributing $ a$ and $ b$ gives

$ (a \plus{} b)(c \plus{} d) \equal{} ac \plus{} ad \plus{} bc \plus{} bd$.

The basis of both of these problems is to FOIL in reverse.  (This is sometimes known as "Simon's Favorite Factoring Trick," which is what the other posters are trying to tell you.)  To start you off:

[hide="11.5.4"] We want to write $ \frac {1}{6} \equal{} \frac {1}{a} \plus{} \frac {1}{b}$.  Clearing denominators, this is equivalent to

$ ab \equal{} 6a \plus{} 6b \Leftrightarrow$
$ ab \minus{} 6a \minus{} 6b \equal{} 0$

which is more or less the same form as the next problem.  Now, note that

$ (a \minus{} 6)(b \minus{} 6) \equal{} ab \minus{} 6a \minus{} 6b \plus{} 36$.

How does this help you? [/hide]
[hide="11.5.5"] As above, note that

$ (p \plus{} 5)(q \minus{} 3) \equal{} pq \minus{} 3p \plus{} 5q \minus{} 15$.

Again, how does this help you? [/hide][/quote]

I actually sorta got lost after 1/6=1/a+1/b. The rest is a bit of a blur. Is it possible to break it down a tiny bit more? :maybe:[/quote]
He multiplied both sides by $ 6ab$ to eliminate fractions (or "clear denominators") and then just moved things over.

Not really completely relevant, but an easy way to look at unit fractions, (which would do more harm than good here) would be thinking of them as smaller than the one on the left.

[hide="Hence"]
$ \frac{1}{n}\equal{}\frac{1}{a}\plus{}\frac{1}{b}$
Can be thought of as
$ \frac{1}{n}\equal{}\frac{1}{6\plus{}x}\plus{}\frac{1}{6\plus{}y}$
[/hide]

[quote="pianogirl"]I actually sorta got lost after 1/6=1/a+1/b. The rest is a bit of a blur. Is it possible to break it down a tiny bit more? :maybe:[/quote]

I'll try to explain it in detail:

[hide]So you understand up to the point of $ \frac{1}{6} \equal{} \frac{1}{a} \plus{} \frac{1}{b}$. Now, we multiply both sides by the quantity $ 6 a b$, so we can get rid of the fractions. We get:

$ a b \equal{} 6 b \plus{} 6 a$

Now, we move everything to the other side:

$ a b \minus{} 6 b \minus{} 6 a \equal{} 0$

We can factor out a $ b$ from the first two terms:

$ b(a \minus{} 6) \minus{} 6 a \equal{} 0$

(This is the tricky/clever part) We add 36 to both sides, and watch how I factor this:

$ b(a \minus{} 6) \minus{} 6 a \plus{} 36 \equal{} 36$

$ b(a \minus{} 6) \minus{} 6(a \minus{} 6) \equal{} 36$

Now, we can factor out the quantity $ a \minus{} 6$:

$ (b \minus{} 6)(a \minus{} 6) \equal{} 36$

Do you understand everything up to this point? If you are still confused, try expanding out the left hand side using the FOIL method (you get $ b a \minus{} 6 a \minus{} 6 b \plus{} 36$).

Now comes the hard part. Since $ a$ and $ b$ are integers, so are $ a \minus{} 6$ and $ b \minus{} 6$. So we are looking for two integers who have a product of $ 36$. One example is $ 2$ and $ 18$. They are two integers whose product is $ 36$. So,

$ b \minus{} 6 \equal{} 2$ AND $ a \minus{} 6 \equal{} 18$

Moving everything to the other side,

$ b \equal{} 8$ and $ a \equal{} 24$

This works:

$ \frac{1}{8} \plus{} \frac{1}{24} \equal{} \frac{3}{24} \plus{} \frac{1}{24} \equal{} \frac{4}{24} \equal{} \frac{1}{6}$

Try to find some other examples yourself. Hint: $ 3\times 12 \equal{} 36$, and let $ a\minus{}6\equal{}3$ and $ b\minus{}6\equal{}12$.

For the second problem, we subtract 15 from both sides to get:

$ p q \minus{} 3 p \plus{} 5 q \minus{} 15 \equal{} \minus{}15$

$ p(q \minus{} 3) \plus{} 5(q \minus{} 3) \equal{} \minus{}15$

$ (p \plus{} 5)(q \minus{} 3) \equal{} \minus{}15$

Now you can use the same method as we did in the previous problem.[/hide]

Do you get it now? If you get stuck, tell us where.


Two positive integers are chosen. The sum is revealed to logician A, and the sum of the squares is revealed to logician B. Both A and B are given this information and the information contained in this sentence. The conversation between A and B goes as follows: B starts.

B: "I can't tell what the two numbers are."
A: "I can't tell what the two numbers are."
B: "I can't tell what the two numbers are."
A: "I can't tell what the two numbers are."
B: "I can't tell what the two numbers are."
A: "I can't tell what the two numbers are."
B: "Now I can tell what the two numbers are."

What are the two numbers?

it's in PEN
before u post sth search wether it has been posted before or not :)
http://www.mathlinks.ro/viewtopic.php?t=150971


These are two conceptual physics problems that I can't figure out.  Maybe the answer is really obvious....I dont know....

(1) If you have a beaker filled with water and you drop an 100 g mass in the water, the water level rises 15 mL.  However, if you place the mass in a styrofoam cup and place the styrofoam cup in the water, the mass in cup displaces 100 mL of water.  Explain this in terms of density.

(2) How come battleships that are made of steel can float in the ocean? (EDITED: Nevermind, I got it, dont answer this one.)

(3) A cargo ship in a huge tub of water looses a few cartons overboard.  Will the water level in the tub increase, decrease, or stay the same after this event?  (Is the answer obvious b/c of #1?)

Thanks for the help!

(NOTE: I realize that this is a math website but I am hoping someone here can help me.)

Try to answer number 3 first. What determines the amount of water displaced when an object floats on top of it? In other words, the amount of water displaced (by weight) is equal to _________.

Once you answer that question, think about how it pertains to question 3 (It doesn't say whether the cartons that fall overboard float or sink, but I assume they float).

Now for question 1, think about what determines the amount of water displaced when an object sinks (i.e. is totally submerged in the water). Now compare this with your answer for how much water is displaced when something floats on top of the water. (incidentally, why does the cup + weight float on the water, but the weight by itself sinks?)

TO clarify, the cartons sink (not float).

Well the amount of water displaced (its weight) is equal to the buoyant force. However I dont see how this helps....  Just know that the buoyant force is less than the weight of the mass b/c the mass sinks.

For question 1, the weight of liquid displaced by an object that sinks = weight of the object.  So 100g ( 1 kg / 10^3 g)(9.8 m/s^2) = 0.98 N of water is displaced by the mass.  However, for an object that floats, it is equal to buoyant force.  F_b=V*p*g = (105 mL)(1.0 g/mL)(9.8 m/s^2)( 1 kg/ 10^3 g) = 1.03 N of water is displaced.  Therefore mathematically it works out.  My problem is that I am not sure if Im allowed to use the 105 mL in the equation b/c that's the # I am trying to show...

I'm not in physics but I'll try #3.  Wouldn't the water level rise?  The water level is determined by how much is submerged in the water.  Since it sinks I believe that more would be submerged raising the water level.  But also since the ship is lighter wouldn't it float more making the water displaced by it less?  I'm pretty sure that the water displaced by the crates would outweigh the decrease of the water displaced by the ship.


ARML
Duke
college
AMC
AMC 8
limit
integration

hey guys, I heard duke lost KIND OF SURPRISING HUH

or is it. wait no it's not. cause duke sucks?

I suppose I might as well give y'all a problem here:

you have a heavy particle with some charge sitting at the origin of the plane; some other (light) particle comes in on a horizontal line with some specified charge, energy, and angular momentum. It scatters so that it eventually leaves along some line with inclination theta. Solve for theta in terms of anything you need to.

the primary purpose of this thread should be to note the immense amount of suckage that duke possesses, though

1. Thanks for at least trying to revive the forum.
2. What did duke lose this time? Or are you just referring to the situation in general?
3. Go find a physics forum.

EDIT: Ok, maybe I was feeling a bit harsh this morning. The title is pretty ridiculous, though.

eh; I think nontrivial physics problems are just as good as math problems for training problem-solving, and in the past, I've certainly posted problems on here that could be posted elsewhere (linear algebra, advanced fields, etc)

pssh we all know duke>pittsburgh. they just somehow managed to luckily sneak in that 1 point lead.

arnav - duke losing is spam, duh. UNC losing, however, should be rated 6.

Duke losing was total crap. Argh.

anyway, I know little or no physics

and my brother only tied for 5th in the state in his grade (!) on AMC8 which makes me sad...

argh

In the spirit of problems that do not even remotely relate to ARML, I present the following. I am seriously interested in the answer, though. Knowing me, it will probably turn out to be trivial, but bleh.

Let $ M$ be a metric space with metric $ d$. Given $ A,B\in M$, say that $ \ell$ is a curve from $ A$ to $ B$ if $ \ell : [t_0,t_1]\rightarrow M$, $ \ell (t_0) \equal{} A$, and $ \ell (t_1) \equal{} B$. Let also $ \ell^{\prime}(t) \equal{} \lim_{s\rightarrow 0}\frac {d(\ell (t \plus{} s),\ell (t))}{|s|}$. Over all curves $ \ell$ from $ A$ to $ B$, define $ d^{\prime}(A,B) \equal{} \inf{\displaystyle\int^{t_1}_{t_0}\ell^{\prime}(t)\,dt}$. Certainly $ d^{\prime}$ is a metric on $ M$. But when is it the same as $ d$? When it isn't the same as $ d$, what are its properties?

Note when $ M$ is Euclidean n-space, $ \displaystyle\int^{t_1}_{t_0}\ell^{\prime}(t)\,dt$ is the arc length of the curve $ \ell$ between $ A$ and $ B$, so $ d^{\prime}$ is just the shortest distance between $ A$ and $ B$, which is coincidentally the definition of $ d$. This is the motivation for the question. Also, $ d^{\prime} \equal{} d$ for some nonstandard metrics in Euclidean space, but in a concave subspace of a Euclidean space, $ d^{\prime}\neq d$.

Haha... open ended questions like this can't be "trivial"

By the way, how do you make that fancy l instead of $ l$?

I will think more on this later, but I can make a few remarks off the top of my head.

There exist injective continuous curves in $ \mathbb{R}^n$ for which a good notion of arc length does not exist (the integral in your statement would blow up to infinity).  Imagine for example that your metric space is the trace (image) of one such curve.  Any curve connecting the two endpoints would have infinite length, so even taking the infimum won't rescue you :(...  $ d'$ would not be a well-defined metric.

Curves for which arc length exists are called rectifiable, and they are relatively rare amongst all continuous curves.  I am not sure but quite possibly by defining arc length in terms of $ l'$ you have imposed additional restrictions on $ l$ beyond rectifiablity.

So the first point is that $ d'$ is not well-defined at all for a general metric space  :(.  Usually one needs smoothness of some sort to guarantee rectifiablity, but in that case we can't work in a general metric space.

If we look at open sets in $ \mathbb{R}^n$, then $ d'$ corresponds to "distance within the set", and $ d$ corresponds to "distance in $ \mathbb{R}^n$", as I'm sure you know based on your example.  In general it seems to me like $ d'$ is the more useful measure of distance, and $ d$ depends too much on the particular embedding of the open set in $ \mathbb{R}^n$.  The important point about $ d$ is that it induces the same topology and smoothness structures as $ d'$, so locally $ d$ and $ d'$ are the same.

Please clarify if you are interested in "nice spaces" (say, smooth manifolds) as your metric spaces, or if the main point of the problem was to investigate the behavior of d' on wild spaces, where it might be difficult to define.

If you like this stuff, you should check out a book on Riemmanian (spelling?) geometry.

[quote="n^4+4"]
By the way, how do you make that fancy l instead of $ l$?
[/quote]
\mathcal{l}

you can also hover over (or click) his LaTeX.

[quote="n^4+4"]
If you like this stuff, you should check out a book on Riemmanian (spelling?) geometry.[/quote]
Riemann :D

ok that was kind of silly but I answered two questions

[quote="n^4+4"]By the way, how do you make that fancy l instead of $ l$?[/quote]

ell

[quote="n^4+4"]Any curve connecting the two endpoints would have infinite length, so even taking the infimum won't rescue you :(...  $ d'$ would not be a well-defined metric.[/quote]

Then this is a counterexample. It doesn't really hurt the problem, though admittedly I hadn't thought about all $ \ell$ being non-differentiable.

[quote="n^4+4"]If we look at open sets in $ \mathbb{R}^n$, then $ d'$ corresponds to "distance within the set", and $ d$ corresponds to "distance in $ \mathbb{R}^n$", as I'm sure you know based on your example. In general it seems to me like $ d'$ is the more useful measure of distance, and $ d$ depends too much on the particular embedding of the open set in $ \mathbb{R}^n$.[/quote]

This is precisely why I'm interested - the $ \mathbb{R}^n$ case, which implies that there is an anologue of distance/arclength for metric spaces. But I am not interested in manifolds (you are always try to get me into topology/geometry...). I'm not interested in "wild spaces" either. I'm just interested in general metric spaces that [u]don't[/u] resemble $ \mathbb{R}^n$, where the problem is not particularly interesting. Anyway... perhaps $ b\equal{}b^{\prime}$ when the space is convex and there is some differentiability requirement.

By the way, I know my definition of derivative is standard, but is $ d^{\prime}$ a standard definition?

If you can get JSTOR, look up Vol. 57, No. 2 of Transactions of the American Mathematical Society.  The article is "Arcs and Geodesics in Metric Spaces"

If you cannot access it, tell me and I will summarize the major points of the article.

Ah thanks. I appear to be 60 years behind :roll: .


darkquantum wrote:R and S are the roots of x^2+x+7=0. Compute 2R^2+RS+R-S+4

[hide]R:^2:+R=-7 because it is a root of the given equation. We substitute R:^2:=-7-R into 2R^2+RS+R-S+4 to find that is equal to RS-R-S-10. We know RS=7 and R+S=-1 by Viete, so we substitute that in to get RS-R-S-10=7+1-10=-2

Correct?[/hide]

yup
tho i must admit
brute force would only take 3 min...sigh

[quote]brute force would only take 3 min...sigh[/quote]

That's how I woulda done it.

I notice that "Viete's formula" comes along a lot of times.

a.) Can you post what it is as a sticky so I could keep refering back to it?
b.) When do you learn this?
c.) Is this in AoPS 2? (I have only just acquired 1)

[color=cyan]Hey, whose senior meets are these?[/color]

these are my practice meets, hence i posted them! :D

[color=cyan]Oh, these are the problems Linda is giving out?

How is she as a captain?  How is math team in general?  Actually, don't answer here, e-mail me or something.  How are you doing on them?  Who is the A team going to be?  Yeah, tell me what's going on :)[/color]

hehe
anyway
i've got 7 more practice meets soo

guys, prepare for lotsa questions


Let $MN$ and $PQ$ be two perpendicular diameters in a circle with center $O$ and $A$,$B$ be two points on $MO$ and$ON$ such that $MA=OB$.If line $BQ$ intersect the circle at $C$ then $C\hat{A}Q$ is a right angle.
Prove or disprove.

analytic geometric method :D

suppose $MN$ is $x$-axis and $PQ$ $y$-axis...
the radius of the circle is 1.
Suppose $B(b,0)$ thus $A(x-1,0)$
$Q(-1,0)$
$\ell_{QB}: y=x/b-1$ intersects the circle $x^2+y^2=1$ at $(\frac{2b}{1+b^2},\frac{1-b^2}{1+b^2})$

thus $k_{AC} = \frac{1-b^2}{1+b+b^2-b^3}$
$k_{AQ} = b-1$
the product of the scopes $\neq -1$...


Dear moderator, administrator, or anybody who can help me,

I have sent both my permission form and my entry form but my USAMTS sheet said that I did not. I have already faxed my permission form again, but not the entry form. What should I do? I even qualified for the AIME.

Your USAMTS profile should be up to date now.


There is at least one two-digit number such that when it is added to the two-digit number having the same digits in reverse order, the sum is a perfect square. Find the sum of all such two-digit numbers.

[hide]
Let the digits of the number be $ x$ and $ y$ respectively. Then the number can be written as $ 10x\plus{}y\plus{}10y\plus{}x\equal{}n^2\rightarrow11x\plus{}11y\equal{}n^2$ for some number $ n$. Divide both sides of the equation to get $ x\plus{}y\equal{}\frac{n^2}{11}$. $ \frac{n^2}{11}$ must be a positive integer which implies $ n\equal{}11$. So there are $ 8$ values that $ 10x\plus{}y$ can be: $ 29,38,47,56,65,74,83,92$. This sum is the same as $ 11^2\cdot4\equal{}\boxed{484}$
[/hide]

Why did you multiple $\frac{n^{2}}{11} $ twice?

He wrote n^2/11 then a dot. In the next sentnce, he talked about n^2/11.


I've tried searching for a list of Mock AMC's or AIME's, but I got 256 results, most of which were useless. Could someone take me to a page listing the mock contests created so far, if it exists? Thanks.

Maybe these might help? [url]http://www.artofproblemsolving.com/Wiki/index.php/Mock_AMC#Mock_AMC_.231[/url]

[url]http://www.artofproblemsolving.com/Wiki/index.php/Mock_AIME[/url]

Or try Mildorf's Mock, and a lot harder, Aimes:[url]http://www.tjhsst.edu/~tmildorf/math/Adv.htm[/url]


logarithms
calculus
derivative
limit
function
AoPS Books

Not really enough detail, but possibly:
Could the $e$ mean the base of the natural logarithm? This is one distinction that might be made here.

Otherwise, if $e$ and $b$ are just variables, then the expressions are equal.

The origins of the use of $e^{kx}$ in growth problems concerns a concept known as continuously compounded interest.

http://en.wikipedia.org/wiki/Compound_interest#Continuous_compounding

So I was right:
$e$ is a specific constant.

One cool thing about $e$ (rather randomly)
is that the derivative of $e^{x}$ is $e^{x}$

[quote="jeez123"]$e^{k}$ ... where k is the "continuous growth"
$b^{x}$ ... where b is the annual growth rate

I have trouble understanding what's the difference :([/quote]

To start with $e$ is the base of the natural logarithm.

Just stick that under your bonnet for awhile and consider compound interest.

If you have bank account that offers an Annual Percentage Rate of $r,$ where for purposes of illustration we'll assume that $0 < r \le 1,$ that is $r$ is a decimal fraction, then if you take $100$ and put it on deposit, after $t$ years you'll have a balance in your account of: $100\left(1+r\right)^{t}.$ For example suppose the bank pays 6% and you leave your $100$ on deposit for 5 years then $100\left(1+.06\right)^{5}= 133.82$ which is a little bit more than the $130$ you'd expect to have if they merely paid you $6$ every year for your original $100.$

Thing is, most banks pay interest more often than once per year, usually once per month.
So if you take your same $100$ and put on deposit with a bank that pays monthly then your balance becomes $100\left( 1+\frac{.06}{12}\right)^{12\cdot 5}= 134.88.$

The thing to see is that the more frequently the bank pays interest, the more interest you get paid.

In general the compounding factor then is given by $\left(1+\frac{r}{n}\right)^{nt}.$

In the example I just gave $n = 12, \text{ and }t =5.$ There are 12 months in a year (i.e. 12 compounding periods) and you left your money on deposit for 5 years. Now what happens to the compounding factor if we let $n \to \infty$ ? In other words, were letting interest add up moment by moment rather than adding it only at specific times?

If we take the natural logarithm of the compounding factor we get

$\ln \left(1+\frac{r}{n}\right)^{nt}= nt \ln \left(1+\frac{r}{n}\right)$

Taking the limit of this expression

$\lim_{n \to \infty}\ln \left(1+\frac{r}{n}\right)^{nt}= \lim_{n \to \infty}nt \ln \left(1+\frac{r}{n}\right).$

It is well known that $\ln\left(1+x\right) \to x$ when $x$ is small compared to $1.$

Using that fact, $\lim_{n \to \infty}nt \ln \left(1+\frac{r}{n}\right) =\lim_{n\to\infty}nt\frac{r}{n}= rt.$

Removing the logarithms $\lim_{n\to\infty}\left(1+\frac{r}{n}\right)^{nt}= e^{\ln \left(1+\frac{r}{n}\right)^{nt}}= e^{rt}$

Thus the answer to the question what happens if we compound continuously is

$\boxed{\lim_{n\to\infty}\left(1+\frac{r}{n}\right)^{nt}= e^{rt}.}$

In math, we want to represent some very small or very large quantities sometimes.

We see that for a given function, say $f(x) = \frac{1}{x}$, as the x values get bigger, the fraction and thus the result of the function gets smaller.

The limit form is basically the same as saying in english:
As X grows without bound, the value of the function f(x) decreases to 0.

Thus, mathematically we can say:
$\lim_{x \to \infty}\frac{1}{x}= 0$
Do you see why this is? If we make the denominator smaller and smaller, the value gets closer to zero. If we make the denominator, say, infinitely large, the value gets infinitely closer to zero.
Because we don't like infinitesimal quantities, we say that the limit of $\frac{1}{x}$ as $x \to \infty$ is equal to zero.

Taking the limit of a function is a pretty hefty topic to be covered in calculus class.

In this specific example, we are trying to model continous growth. Because n is the amount of periods, we want there to be as many periods as possible (defn of continous growth).

Thus, if we take the limit at $x \to \infty$, we can see what the function converges to.

If any of this is confusing, I understand completely. That was a crash course intro to limits. If you feel like you simply MUST know more, you can pick up the AOPS books, and read the sections on limits. Quite handy, but they don't often come into play during contests because contests are supposed to be calculus free.

[quote="jeez123"][quote="Dr. No"]
Taking the limit of this expression

$\lim_{n \to \infty}\ln \left(1+\frac{r}{n}\right)^{nt}= \lim_{n \to \infty}nt \ln \left(1+\frac{r}{n}\right).$

It is well known that $\ln\left(1+x\right) \to x$ when $x$ is small compared to $1.$[/quote]

Thanks for the thorough explanation! A few questions please...
What is the "limit" and why are you "taking it"?[/quote]
I believe DiscreetFourierTransform did a good job answering your question, so I'll just add a couple of things to what he said. A limit is the value that a function approaches as the argument of the function approaches some specified value.

For example, $\lim_{x \to 0}y = \lim_{x \to 0}mx+b = b.$ You might think, "Big, fat, hairy deal. Dr. No can evaluate the equation of a line at $x = 0$," but it goes deeper than that.

Suppose you have the graph of $y = x$ but the graph has a hole in it at $x = 2.$ It is incorrect then to say $y\left(2\right) = 2$ because strictly speaking you don't know what $y\left(2\right)$ is. $y\left(2\right)$ is undefined; the graph has a hole in it at $x = 2.$ However, one can say the limit of $y$ as $x$ approaches 2 equals 2, i.e. $\lim_{x \to 2}y = 2.$

Likewise what is $\left. \left(1+\frac{r}{n}\right)^{nt}\right|_{n = \infty}?$ We don't know. Nevertheless, $\lim_{n \to \infty}\left(1+\frac{r}{n}\right)^{nt}= e^{rt}.$ We know that because we proved it.


The sum of the digits of a three-digit number is 26. The number is multiplied by 7, then by 11, and finally by 13. How many times will the digit 9 occur in the final product?

multiplying by 7, then 11, then 13 in effect multiplies by 1001.

i.e. the number is concatenated to itself (334 -> 334334, and etc.)

sum of 26 = 2 9's and 1 8, so the final concatenation gives us 4 9's and 2 8's, so 4.