I am a 7th grader that did not qualify for USAMO. my scores are  AMC 10B:123, AIME:5. what should be my goal for next year? i do plan on qualifying for USAMO. ignore whats in my sig.

what ever you think you can get.

aim for an IMO Gold Medal.  :P

You must get IMO gold in 8th grade or you are NOTHING.

No srsly, getting to the USAMO in 8th grade sounds like a good enough goal.  I wouldn't expect to get more than one problem right on it (or really any at all) on your first try.  But if you study enough then by the next year you should be good enough to at least make Red MOP.

I would suggest aiming to max out the AMC10 (150) and aiming for an 8+ on the AIME (whichever you feel more comfortable with).

Most likely, I'd suggest increasing your goal to 11 next, and then 13, 15. Of course, you might work to it much faster :wink:


A Tetris piece is one of the following seven shapes consisting of four unit squares:

[img]http://www.touchgenerations.com/_media/enGB/content/tetris_ds/tetris_blocks_all.gif[/img]

Two Tetris pieces are said to be identical if and only if the shape of one piece can be achieved by rotating (but not reflecting) the other.  For example, the purple piece and the yellow piece are considered distinct.

Define a region $ R$ made of unit squares to be "large" if and only if for all Tetris pieces $ T$, two $ T$s can fit inside $ R$ without overlapping.  For example, a 4x4 square is "large."  This can be achieved by rotation.

The question:
Compute the smallest possible area of a "large" region.

[hide="Solution"]Assume the height is at least the width

The height is at least 4, since the maximum maximum size (in any dimension) of a block is 4 (the stick), and the width is at least two, since the maximum minimum size (in any dimension) of a block is 2 (for example, the L).

Hence, the area is at least 8. However, we cannot fit two zigzags or two T's in an area of 8.

The next value up is 9, but this must be either 9x1, which fails W >= 2, or 3x3, which fails H >= 4, so this is not valid.

The next value up is 10, and indeed a 5x2 rectangle works:
[code]
oo  oo  ox  xo  +x  oo  oo
x+  ++  xx  xx  xx  ++  ++
x+  ++  x+  +x  +x  +x  x+
x+  xx  ++  ++  ++  +x  x+
x+  xx  +o  o+  +o  xx  xx
[/code]

and we are done.[/hide]

To the above post: I don't believe the region must be rectangular. That doesn't change the answer, but it means you have to strengthen the proof.

[hide]Area $ 10$ works: $ 2 \times 5$ rectangle. We will now show a region of area $ 9$ is not possible.

First of wall, we need to be able to fit two long pieces. Observe that if their fit has them perpendicular, using $ 8$ separate squares, then no square pieces can fit. Since we are using an area of at most $ 9$, we only can add one additional square, so we can only fit one square piece at most.

So the long piece's fit has to be parallel, and one of the following up to symmetry:

[code]OOOO _ .OOOO _ ..OOOO _ ...OOOO
OOOO _ OOOO. _ OOOO.. _ OOOO...
1... _ 2.... _ 3..... _ 4......[/code]
In fits $ 3$ and $ 4$, adding one square can't allow a fit of two square pieces, so only $ 1$ and $ 2$ are possible.

In case $ 2$, consider the following diagram:

[code].1234
5OOOO6
OOOO.[/code]
We can add the 9th square to any of the 6 designated spaces, and this is exhaustive by symmetry. But no matter which space we add it to, we can't fit two green $ S$ pieces.

In case $ 1$, we have the following:

[code].12
3OOOO
.OOOO[/code]

We have 3 choices here. But we can't fit two blue $ S$ pieces no matter where it is added.

Therefore area $ 9$ is not possible. The following shows how to fit each piece into a $ 2 \times 5$.

[code]1111O 1112O 1122O 1112O
2222O O1222 1122O 1222O

1222O O1122 1122O
1112O 1122O O1122[/code][/hide]

[quote="MellowMelon"]I don't believe the region must be rectangular.[/quote]
I did not specify that the region had to be rectangular.  Nevertheless, the answer is the same for each of them, as MellowMelon pointed out.


Find all positive integers n and x such that
$5^n- 5^4 + 5^5 = x^2.$

Easy to be on a national olympiad. :lol:

[hide]
Checking $n = 1,2,3,4$ we find no solutions. So assume $n > 4$.

Then $5^n-5^4+5^5 = 5^4(5^{n-4}+4) = x^2$. Let $x = 25y$.

So $5^{n-4}+4 = y^2 \Rightarrow 5^{n-4} = (y-2)(y+2)$ but only one of $y-2$ and $y+2$ can be divisible by $5$ so $n-4 = 1$. Checking, we find the only solution is $n = 5$, $x = 75$.[/hide]

Nice solution, paladin. :)
[hide]The easiest way, in my opinion, to do this would be to rewrite as
$5^n = x^2 - 2500$
$5^n = (x- 50)(x+50)$
The factors on the LHS differ by $100$. The only powers of 5 that differ by $100$ are $25$ and $125$
So $5^n = 25 * 125$
$n = 5, x = 75$[/hide]


Post all the tips and tricks you have collected about runescape!

Alright,
I like to
1) mine copper and tin (it goes really fast) over and over and then sell all of it at the grand exachange or the general store. You can make bout 80 gp each.
2)Also, cowhide is really great. Get a LOT of it. Like a 100. Then, if you sell it you'll probably get like 40 gp each, but if you go to al kharid and get it tanned you can get about 90 gp each.
3)Raw meat. I prefer beef (it really easy to get, just go to the cow fields and pick it up, you don't even have to kill anything) or chicken (same as beef, just go to the chicken coops).

Yeah, I have a level 95 account, but I gave it to a friend. I like to log in once in a while to see how things are going.

Having a Runescape account automatically entitles your grades to a 10% discount, with an additional 5% for every hour you play per day.

Then again, I do have a level 120 something that I don't use anymore... The majority of the RS community is simply ridiculous, so I quit. I recommend that you stop before you become like those poor people in Asia dying from too much gaming.

[quote="n0vad3m0n"]Woodcutting. Yews bring you at least 300 gp each (I haven't logged in in over a year so who knows what the rate is now).

(In any case, RuneScape is ridiculous; the only useful application is to sell the accounts you have for real cash).[/quote]

Is this your new account Henry?

[quote="jonathanchou711"]Isn't Runescape failing? I think that WoW is kind of pwning it.[/quote]

I think all MMOs are detrimental to the betterment of mankind.

(Actually, never mind, people addicted to them just get wiped from the gene pool anyways.)

[quote="Aristocrat"]I think all MMOs are detrimental to the betterment of mankind.

(Actually, never mind, people addicted to them just get wiped from the gene pool anyways.)[/quote]

Hahaha. Touche. First person shooters always win. At least there are the fan girls.

And Poincare, I just got my username changed, that's all.

RuneScape's community is full of whiny 12 year olds who think they're cool, leaving Jagex to force the over-strict policies (language filters, trade restrictions, etc.) - which is why I quit.

finally started playing Runescape again after taking a 2 year break...great game
_________________
[url=http://gamefriends.com/game_videos][color=red]Game Videos[/color][/url], [url=http://gamefriends.com/ask][color=blue]Ask a Gamer[/color][/url], [url=http://gamefriends.com/game_news][color=purple]Game News[/color][/url]

[quote="CircleSquared"][quote="Aristocrat"][quote="CircleSquared"]In my opinion, Runescape is not as good as MapleStory.  And that says something, I think.[/quote]

Maplestory boss runs at high levels have a countdown timer telling you how long you have to kill the boss.

It is measured in days.

This tells you how unhealthy the game is.[/quote]

It is NOT measured in days.[/quote]

HOURS, not days

Meh. I used to play. Not anymore, gave the account to a friend. Still know the password though.

[quote="Aristocrat"]...

You are given 24 hours to kill Horntail; granted, no one uses that much, but the fact is that you are given 24 hours... of consecutive play time. What does beating him get you? A necklace that upgrades your character. And you need to beat it four times to get a full necklace. That means staying logged in for a total of more than 96 hours (Since apparently you need to stay logged in for the whole duration to get the reward).

This cannot be casual gaming.[/quote]

people can log out- as long as there are at least 2 people you can still get the award


Find all increasing functions $ f: \mathbb R^{\plus{}} \rightarrow \mathbb R$ such that $ f(x\plus{}1)\equal{}f(x)\plus{}2^{\minus{}x}$ for every $ x\in \mathbb R^{\plus{}}$. ($ \mathbb R$ is the set of real numbers, $ \mathbb R^{\plus{}}$ is the set of positive real numbers)

We have $ (f(x\plus{}1)\plus{}2^{\minus{}x})\equal{}(f(x)\plus{}2^{\minus{}(x\minus{}1)})$, so $ g(x) \equal{} f(x)\plus{}2^{\minus{}(x\minus{}1)}$ is periodic, of period $ 1$. Assume $ g(x)$ is not constant; then there exist $ x_1 < x_2$ with $ g(x_1) > g(x_2)$, so $ g(x_1 \plus{} n) > g(x_2 \plus{} n)$ for all positive integers $ n$. But then $ f(x_2 \plus{} n) \minus{} f(x_1 \plus{} n) \equal{} \minus{}(g(x_1\plus{}n) \minus{} g(x_2\plus{}n)) \plus{} 2^{\minus{}(n\minus{}1)}(2^{\minus{}x_1} \minus{} 2^{\minus{}x_2}) < 0$ for large enough $ n$, and that contradicts $ f$ being increasing. So $ g(x) \equal{} c$ is constant, and $ f(x) \equal{} c  \minus{} 2^{\minus{}(x\minus{}1)}$.


If $p$ is a prime number, prove that $p^2+26$ is always composite.

Thanks jli! Now I'll leave it for other people  :P

[hide]
primes in mod 4 are 1, 2, or 3.

1^2+2=0(mod4)
2^2+2=2(mod4)
3^2+2=2(mod3)

I am not sure how to do that last one(case)...
[/hide]

yes...mod 6  might work

try mod 9, many problems are ably to solve in mod 9

$p^2+26\bmod 9=p^2-1\bmod 9 = (p-1)(p+1)\bmod 9$

[hide]

Letting $p=2,3$ we can see that it is not a prime . For  $p>3$  , we know that $p^2\equiv 1\mod 3$ . And this follows immediately that $p^2+26\equiv 27\equiv 0 \mod 3$ . [/hide]


function
algorithm
algebra unsolved
algebra

Find all natural number k and real number a such that there exists a function f continuous from R to R satisfying
$f_k(x) = a.x +1$   $\forall x \in R$

where $f_0(x) = x, f_{n+1}(x) = f_n(x)$  $\forall n \in N$
-------------------------
Sorry, there's a mistake in my problems.

Find all natural number k and real number a such that there exists a function f continuous from R to R satisfying
$f_k(x) = a.x +1$   $\forall x \in R$

where $f_0(x) = x, f_{n+1}(x) = f_n(x)$  $\forall n \in N$
For which k and a, find all function f

for a=0 take f(x)=1.
for k=1, take f(x)=ax+1.
Now suppose that a!=1 and that k>1.
clearly f(x) is one to one and since it is continous it should be either increasing
or decreasing.
let g(x) denote inverse of f(x).
suppose k=2.
since f(x) is monotonic, we have that f(f(x))>0 for  x>N for some N.
hence a>0.
clearly for a>0 we can take f(x)=  (:sqrt: a)x+ b, where b=1/(1+ :sqrt: a)
Now suppose that fk(x)=ax+1 where k is odd.
then take v^m = a and (v^m-1+...+v+1)u = 1
then f(x)=vx+u satisfies the relation.
by the same argument a function exists for even values of k iff a>0.
if a>=0 any k works. if a<0 only odd values of k.

As stated, the problem is rather simple. But I see that the subject is " Find all functions" . I think this is a very hard task. Do you happen to have a complete solution for this problem? I would be very interested. I tried several times to solve it, but I failed each time.

I think that I have a complete solution in the case of k=2, which is as follows:
I give the general solution of functions f:[a1,a2] -> R such that
f(f(x))=ax+1 and that f(x) is continous.

clearly f(x) is monotonic.
case 1. f(x) is increasing. we know that a>0 .  clearly since f(x) is increasing:
f(f(x))>x iff f(x)>x.
so f(x)>x iff (a-1)x+1>0  so we can divide [a1,a2] into two subintervals, such that
in one of them f(u)>u and in the other f(u)<u.
wlog, let f(x)>x for all x in [a1,a2].
Now there is a  \beta >0 such that if x,y in [a1,a2] then abs(x-y) <=  \beta
implies that abs(f(x)-f(y))< \varepsilon  for some  \varepsilon >0.
also there is an M>0 such that f(x)-x>m.
take d to be min( \beta , abs( \varepsilon - m) ).
partition the set [a1,a2] to [a1,a1+d,a1+2d,...,a1+td,a2] where abs(a2-a1-td)<d.
Now note that for any x,y in any interval [u-d,u+d]. we have that:
abs(f(x)-u)=abs(f(x)-f(u)+f(u)-u)>=abs(m- \varepsilon )>d.
This means that we can arbitrarily assign a continous and increasing function g(x) to the interval [a1,a1+d] the only constraints are:
i) g should be increasing.
ii) g(x)>x.
iii) g(a1+d)<= f(f(a1+d))=a(a1+d)+1.
after assigning g(x) to [a1,a1+d] the interval [f(a1),f(a1+d)] can be uniquely determined. (these two are disjoint) Now we will follow the same algorithm for the remaining segments [a1+d,a1+2d],... .
this does the problem if f(x) be increasing.

case ii) f(x) decreasing. consider the same interval [a1,a2] the equation f(x)=x has a finite number of solutions in interval [a1,a2].( I'm having problem with this statement!) isolate each of these special points and construct the function as before. at the end you can take care of these finite number of points.

I forgot to handle f(x)=1 as a seperate case.
in this case the continouty of f(x) yields that f(x)=1.
I think that the method presented in my last post can be generelized to all values of k.
Also there might be a problem with the argument of extending [a1,a2] to the whole R.
For this I present the following argument:
suppose f(x) is increasing. then a>0 implies that f(x)-x is unbounded in whole R.
we can infact choose T such that f(x)-x>M for all x>T and that enables us to extend [a1,a2] to the whole R.

I think this Question doesn't have a complete solution


graph theory
combinatorics unsolved
combinatorics

An $n\times n$ is fill with 0 and 1 so that if we chose randomly n cell(that is no two of them is in the same row or column) then there atre at least one contain 1.Prove that we can find i row and j column so that $i+j\geq\ n+1$ and their intersection contain  only 1

This is just another way of looking at the marriage theorem:

Consider a bipartite graph with bipartition $A,B,\ |A|=|B|=n$ encoded by this $n\times n$ table as follows: the rows represent the vertices in $A$, the columns represent the vertices in $B$, and $i\in A,j\in B$ are connected iff the entry in the $i$'th row and $j$'th column is $0$. The hypothesis simply says that this graph has no perfect matching, so, by the marriage theorem, we can find some $i$ vertices in $A$ which have $<i$ neighbours in $B$. This means that there are $<i$ columns which contain a $0$ at the intersection with one of the $i$ mentioned rows, or, in other words, there are $>n-i$ columns whose intersections with the $i$ rows are all $1$'s. This is exactly what we wanted to prove.


http://www.mathlinks.ro/Forum/memberlist.php

Why somebody with more than ten million posts?  :D

that was for a different user. now there's two people with over 16M posts.

If you try to find all their posts, it says "nothing matches your search criteria." Apparently, they have 0 posts but for some reason it shows 16777215 and 16777214.

Did these people like randomly hack into the system? :huh:

They might have somehow got into the admin panel... :huh:  :|

[quote="nebula42"]Actually, I'm thinking that they might not have hacked the database. Notice that the number $16777215 = 2^{24}-1$, so it looks more like some weird computer error, since computers generally deal with these type of numbers.[/quote]

Notice that theccy and hoavokhuyet have the title Moderator. :wink:

Just an offhand question, but is theccy and ccy the same person?

[quote="joml88"]Hm, I really wonder how they did this  :?: It seems like they somehow hacked the database.[/quote]It's amazing where a small "-" can lead people to think :) Problem (finally) fixed. It was no hack, just an extra minus that made it so that each time a post was deleted a person's post count drops by 2 (instead of 1).

[quote="13375P34K43V312"]Now what's the deal with micheal?[/quote] :bomb:  :wallbash_red: There must be another "-" somewhere it should be.

[quote="i_like_pie"]Maybe the "-" wasn't the real problem? :ninja:[/quote]It was. I know where the problem is and I will fix it, it's just that it implies a big update of the site which I have been preparing for a couple of days now (and it will be ready soon).

Don't worry it takes only a couple of seconds to resync everyone's post count after the update is done.

[quote="i_like_pie"]Three things I noticed:

1) Micheal has 16M+ posts.
2) Everybody lost a post (V.V. had 5080, now 5079).
3) Post count isn't going up anymore.

Maybe the "-" wasn't the real problem? :ninja:[/quote]

Ohh, that's why people have 0 posts now. Will our post counts get fixed once Valentin fixes the problem?

[quote="13375P34K43V312"]Will our post counts get fixed once Valentin fixes the problem?[/quote]
[quote="Valentin Vornicu"]Don't worry it takes only a couple of seconds to resync everyone's post count after the update is done.[/quote]
Already answered. :)

yeah I've posted like 20 times today. I just noticed that it didn't go up at all. But posting still is fun. Is there some computer guru just trying to get attention on this website by doing this?

Can't wait till this site is updated. It would probably look a lot better , new and improved  :lol:

You should have seen the format they had in like april, it's much better now.

They also have moderator status, something that should be fixed...

[quote="Ihatepie"][quote="bpms"][quote="nebula42"]Actually, I'm thinking that they might not have hacked the database. Notice that the number $16777215 = 2^{24}-1$, so it looks more like some weird computer error, since computers generally deal with these type of numbers.[/quote]
Okay, how the heck did you figure that out?[/quote]

Its not that hard, I can do that in my head. Btw I used to do exponents of 2 in my head after I finished my math tests (2,4,8,16,32,...) before evry1 else  :lol:[/quote]
Okay, but did you have it memorized up to 24?

Well, everything important has been said here and the five or so other threads with that topic...
Thus such a small question isn't really offtopic.

But I find it very interesting how many people posted, especially those whose post count doesn't rise anymore :D

[quote="bpms"][quote="nebula42"]Actually, I'm thinking that they might not have hacked the database. Notice that the number $16777215 = 2^{24}-1$, so it looks more like some weird computer error, since computers generally deal with these type of numbers.[/quote]
Okay, how the heck did you figure that out?[/quote]

It isn't that hard to figure out...

Anyways, now I don't think anyone's post count is increasing...since I remember my post count being 2517, and I definitely posted and it is still 2517.

same, ive posted like a lot since it stopped going up. My post count should be 330+.

[quote="Valentin Vornicu"]Don't worry it takes only a couple of seconds to resync everyone's post count after the update is done.[/quote]

I think you should read this.  :wink:

[quote="bpms"][quote="i_like_pie"]Dude... Way of topic. :|[/quote]
Have you read this thread? It isn't offtopic.
To ZetaX no one's post count is going up, read the thread.[/quote]
 :|
Read my post again. I did not ask why etc., I just made a remark about how many people complain about non-increasing post count (since these people possibly post for the counter, not for the content).

lol before my post was 84...now it's 78???

I can make it 0 if you don't like 79 :)
Remember post count is [b]purely informational[/b]. You should not post in order to increase your post count.

I concur...

But yes, thanks for the fixing of the problem...

yay, last night my post count said 304.  :lol:


I don't know if this has already been addressed but:

How often do the classes meet and at what times, usually?

Would it be difficult/beneficial to take Intermediate algebra and AIME problem series at the same time?

About which parts of AoPS V2 are covered in Intermediate algebra? (because some chapters I don't understand that well)

Thanks!

All of our summer classes meet once a week at 7:30 Eastern / 4:30 Pacific and last 90 minutes.

It is perfectly reasonable to take both the Intermediate Algebra and the AIME Problem Series class simultaneously.  Many of our students take more than one class at a time.

As far as which chapters in AoPS Vol 2 are covered in the Intermediate Algebra class: I'm not really sure how to answer that, since the online classes don't directly follow the AoPS books.  I think that most of the material in the Intermediate Algebra class is at a level beyond the AoPS books.  MCrawford (who is the instructor for both of the classes that you mentioned) might be able to give you more detailed information -- he's out of the office today though.

You can get a pretty good idea about what's in the class by looking at the "Are You Ready?" and "Do You Need This?" problems listed for each class on the classes page at [url]http://www.artofproblemsolving.com/Classes/AoPS_C_ClassesS.php[/url].

From what I've seen you done, I would say you're more than well-prepared for Intermediate Algebra. In fact, you might want to make sure you can't do all of the problems first. ;)

I took three classes (including AIME Problem Series) at the same time, so you'll definitely be fine with taking those two over the summer.

Ok, thanks for the information.  I think I am going to take both of them, and it's summer too so I'll have time.  Thanks, towersfreak, but I can't do all the problems. ;)

Towersfreak2006, did you enjoy the intermediate algebra class/did you find it useful?

Thanks again.

I haven't taken it yet, and I don't know that I will this upcoming semester due to other commitments.
But from what I've heard, it's one of the easier intermediate classes.

[quote="towersfreak2006"]But from what I've heard, it's one of the easier intermediate classes.[/quote]

That's because you own algebra. -.-

Daniel, Intermediate Algebra is a great class to take if you're trying to brush up your algebra skills and cover any holes that may have been neglected from previous classes/books/etc.

[quote="towersfreak2006"]
But from what I've heard, it's one of the easier intermediate classes.[/quote]

We generally recommend this be the first of the Intermediate classes a student takes, though students who are very proficient with algebra can feel comfortable taking one of the other intermediate classes first.


AMC
USA(J)MO
USAMO
number theory
greatest common divisor
number theory unsolved

I got a solution to USAMO 2004 #2 but nobody else I saw had it so I was wondering if somebody could check it for me.  Thanks so much.

Suppose a1...an are integers whose greatest common divisor is 1. Let S be a set of integers with the following properties:

(a) For i=1...n , ai is in S.
(b) For i,j = 1...n (not necessarily distinct), ai - aj is in S.
(c) For any integers x and y, if x + y is in S, then x - y is in S.  (Edit: x and y must be IN S)

Prove that S must be equal to the set of all integers.

[hide="Solution"]Noticing that 0 must be in the set, we find that if x = 0+x is in the set, so 0-x = -x is in the set.

Also, due to this, if x-y is in S then x+y is in S.

Also, we can easily prove that any linear combination of two of the ai's is in the set.

Now, I shall prove that any linear combination of n variables with at least n-2 even coefficients is in the set.

The 2 case is true.
For the three case, suppose we want an arbitrary combination $ c_1a_1 \plus{} c_2a_2 \plus{} 2c_3a_3$ (this has at least one even).  Since $ c_1a_1 \plus{} c_3a_3$ is in the set and $ \minus{} c_3a_3 \minus{} c_2a_2$ is in the set, and their sum is in the set, so their difference, the desired combination is in the set.

Now, inducting, we suppose we have a linear combination of k of them that has at least k-2 evens is in the set.
$ c_1a_1 \plus{} c_2a_2 \plus{} 2c_3a_3 \plus{} \cdots \plus{} 2c_ka_k \in S$.

Now, since $ \minus{} c_2a_2 \minus{} c_{k \plus{} 1}a_{k \plus{} 1}$ is in the set, and the sum of this an our linear combination of k is in the set, so any linear combination of k+1 of the ai with at least k-1 evens is in the set.

More generally, we have proved that any linear combination of n with at least n-2 evens is in the set.

Now, we know that there is some combination $ d_1a_1 \plus{} d_2a_2 \plus{} ... \plus{} d_na_n \equal{} 1$ (because they have GCD of 1).

However, in order to use our lemma, we need that at most two of the $ d_i$ are odd.
We know that at least one of the $ a_i$ is odd; if not, then the GCD is at least 2.  Hence, suppose this is $ a_1$.  Now, suppose $ d_j$ is odd, but we want to make it even.

$ d_1a_1 \plus{} d_ja_j \equal{} (d_1 \plus{} a_j)a_1 \plus{} (d_j \minus{} a_1)a_j$.

Using this shift, we have changed the parity of the coefficient of $ a_j$.  It doesn't really matter what happens to the coefficient of $ a_1$.  At the end, we can just have it be odd if the rest are even.

Thus, applying this to all the other odd coefficients in our linear combination, we have found a linear combination with at least n-2 evens that equals one.  For all the rest of the integers, we can just multiply the combination by that integer and it wouldn't change the parities from even to odd in any of the coefficients.  Thus, the set is all integers as desired.  [/hide]

$ \exists i\{1,2,...,n\}: \ a_i\ odd$
we have $ \frac {a_i - 1}{2} + \frac {a_i + 1}{2} = a_i\in S$ so $ 1 = \frac {a_i + 1}{2} - \frac {a_i + 1}{2}\in S$
and $ \forall k\in\mathbb{Z}: \ (k + 1) + ( - k) = 1\in S$ so $ 2k + 1 = (k + 1) - ( - k)\in S$

if $ \forall i\in\{1,2,...,n\}: \ a_i \ odd$ then $ a_1 - a_2 = even\in S$
if $ \exists i\in\{1,2,...,n\}: \ a_i\ even$ so $ a_i%Error. "even" is a bad command.
\in S$

then $ \exists h\in\mathbb{Z}: \ 2h\in S$
and we have $ h + h\in S$ gives $ 0 = h - h\in S$
and $ \forall k\in\mathbb{Z}: \ (k) + ( - k) = 1\in S$ so $ 2k= (k) - ( - k)\in S$

so $ S = \mathbb{Z}$

I'm sorry I copied the problem a little bit wrong :( (but you have a nice solution to my mistaken problem).

The property (c) should say "for any integers x and y [i]in S[/i]"  I will edit that.

Here is the original LaTeX.
http://www.artofproblemsolving.com/Forum/resources.php?c=182&cid=27&year=2004

I would say that it's clean.  I don't really know how to score it, but I'd put it at least in the 5-7 range, rather than 0-2.

Maybe be a little more specific on "any linear combination" of two $ a_i$'s is in S.

[quote="dasherm"]Maybe be a little more specific on "any linear combination" of two $ a_i$'s is in S.[/quote]This is obvious given his first two statements ($ 0\in S$, $ x\in S$ implies $ \minus{}x \in S$ and $ x\minus{}y \in S$ implies $ x\plus{}y \in S$).


I want to add a foot note but normally they're numbered and I don't want to reader to confuse it as an exponent.

How do I change the footnote symbol to like *? I don't understand the command they gave in the TexnicCenter Help. Can someone please help me.

In the preamble of your document (i.e. before \begin{document}) put:

[code]\renewcommand{\thefootnote}{\fnsymbol{footnote}}[/code]

Then, where you want the footnote to appear put

[code]\footnote{[Put footnote text here.]}[/code]


Find all functions $ f: \mathbb R\rightarrow \mathbb R$ satisfying $ (x \plus{} y)(f(x) \minus{} f(y)) \equal{} (x \minus{} y)f(x \plus{} y)$ for every $ x,y\in \mathbb R$

This problem is really hard, I hope you will post solutions soon.

It is not hard  :wink: .
Take $ y\to \minus{}x$ then $ f(0)\equal{}0$
Take $ y\to 1$ and let $ a\equal{}f(1)$
\[ f(x\plus{}1)\equal{}f(x)\plus{}a
\]
Take $ y\to y\plus{}1$ then
\[ (x\plus{}y\plus{}1)(f(x)\minus{}f(y)\minus{}a)\equal{}(x\plus{}y\plus{}1)(f(x\plus{}y)\plus{}a
\]
Therefore :
$ f(x)\plus{}f(x\plus{}y)\minus{}f(y)\equal{}2ax$
Take $ y\to 0$ then  we have :
\[ f(x)\equal{}ax,\forall x\in \mathbb{R}
\]
This is solution of this function equation .

I agree with you TTsphn,simplely,you can choose $ y\equal{}\minus{}x$ instead of $ y\to \minus{}x$ :wink:
[hide="to TTsphn"]Anh là nguyễn thọ Tùng phải ko ạ[/hide]

[quote="thaithuan_GC"][quote="TTsphn"]
Take $ y\to 1$ and let $ a \equal{} f(1)$
\[ f(x \plus{} 1) \equal{} f(x) \plus{} a
\]
Take $ y\to y \plus{} 1$ then
\[ (x \plus{} y \plus{} 1)(f(x) \minus{} f(y) \minus{} a) \equal{} (x \plus{} y \plus{} 1)(f(x \plus{} y) \plus{} a
\]
[/quote]
Your post made a follish mistake: my equation is [b](x+y)[/b](f(x)-f(y))=[b](x-y)[/b]f(x+y)   :rotfl:
It is really hard, i hope to receive your solutions.[/quote]
OK .it is stiil not hard
Let $ f(1) \equal{} a,f( \minus{} 1) \equal{} b$

Then$ take y \equal{} 1$ we have $ f(x \plus{} 1) \equal{} (x \plus{} 1)\frac {f(x) \minus{} a}{x \minus{} 1}$
        $ y \equal{} \minus{} 1,x\plus{}1$ we have        $ f(x \plus{} 1) \equal{} (x \plus{} 2)\frac {f(x)}{x} \plus{} b$
The last is only solve this equation with variable is $ f(x)$
and with some small comment ,we can finish
Is it right

Thanks guys for your interests,
But to prove f(x)=xf(1), we must prove that $ f( \minus{} 1) \equal{} \minus{} f(1)$, it will finish the problem. Any ideas?

I solved it, thanks Allnames very much  :lol:


LaTeX
geometry
geometric transformation
reflection
\/closed
noooowhybumped

When I type a post in the message board with spoiler anything inside the
dollar signs shows up without being highlighted.How do you spoiler tex?

Can't spoiler TeX - the TeX gets rendered as images, not text, so you can't spoiler it.

Can we use backgrounds here? If so, we could make black backgrounds around $\text{\TeX}$ as a sort of ad hoc spoiler system.

[quote="tandjsnell"]I don't think we can create backgrounds.
So what do I do?Just not use TeX?[/quote]
No, please, use LaTeX so it's easy to read, and if it isn't sufficiently hidden because of the LaTeX... oh well. Perhaps there will be a work-around in the future.

[quote="Valentin Vornicu"][quote="tandjsnell"]I don't think we can create backgrounds.
So what do I do?Just not use TeX?[/quote]it's very simple: don't use spoliers :P :) .. ;)[/quote]

Valentin, as has been discussed in the other post, can you please stop telling people not to use spoilers. If you do theres going to be big problems, as spoilers have been widely used at AoPS.

tandjsnell: if what you are writing is unreadable with normal text, then sure use latex and just use spoiler on any text you write. Not ideal, but its the best that can be done. If its just simple things like squaring, writing ^2 is probably easier, that will allow spoiler.

[quote="TripleM"][quote="Valentin Vornicu"][quote="tandjsnell"]I don't think we can create backgrounds.
So what do I do?Just not use TeX?[/quote]it's very simple: don't use spoliers :P :) .. ;)[/quote]

Valentin, as has been discussed in the other post, can you please stop telling people not to use spoilers. If you do theres going to be big problems, as spoilers have been widely used at AoPS. [/quote] Dear TripleM, I am very aware of the contents of the other post you are refering to.

However, as you can see above tandjsnell asked a question. I gave him a suggestion. If I cannot do that anymore, please tell me so. My (personal) opinion is against using spoilers. However as you can easily see I am conformating myself to the big majority of the people, ie spoilers can be used in all forums.

I do not belive I need to be told what to or not to say, especially if it reflects a personal opinion (which was also underlined by the emoticons :) :P and ;)!) ...

[quote="TripleM"]If its just simple things like squaring, writing ^2 is probably easier, that will allow spoiler.[/quote]
For superscripts, you can also use the HTML tags [sup] and [/sup].  For subscripts, you can use [sub] and [/sub].  Both get hidden under spoiler.

[quote="TripleM"]tandjsnell: if what you are writing is unreadable with normal text...[/quote]
Or hard to read, at least. Sorry MMM, just clarifying a bit.

[quote="Valentin Vornicu"][/quote][quote="TripleM"][quote="Valentin Vornicu"][quote="tandjsnell"]I don't think we can create backgrounds.
So what do I do?Just not use TeX?[/quote]it's very simple: don't use spoliers :P :) .. ;)[/quote]

Valentin, as has been discussed in the other post, can you please stop telling people not to use spoilers. If you do theres going to be big problems, as spoilers have been widely used at AoPS. [/quote][quote="Valentin Vornicu"] Dear TripleM, I am very aware of the contents of the other post you are refering to.

However, as you can see above tandjsnell asked a question. I gave him a suggestion. If I cannot do that anymore, please tell me so. My (personal) opinion is against using spoilers. However as you can easily see I am conformating myself to the big majority of the people, ie spoilers can be used in all forums.

I do not belive I need to be told what to or not to say, especially if it reflects a personal opinion (which was also underlined by the emoticons :) :P and ;)!) ...[/quote]

I'm sorry if you took my post the wrong way - to me it seemed the topic was about how best to use spoiler, not whether to use spoiler or not. If I misinterpreted your post, then other people may also do so, and I just want to make sure people get the right idea.

[quote=akliu]Luckily we have hide tags for hiding latex now. Quick question, when was this added to AoPS?[/quote]

Old AoPS is different from AoPS now.  It might have had a different, less reliable hide feature called "spoiler"

[quote=pieMax2713][quote=akliu]Luckily we have hide tags for hiding latex now. Quick question, when was this added to AoPS?[/quote]

please dont bump 17 year old threads and how did you find this

well idk when this was added to aops
there arent a lot of ppl who joined when hide tags werent a thing and still use aops now[/quote]

Search function, and the first AoPS user. Some people say something about the "new aops", was the whole AoPS site remade?

And also, why did Richard Rusczyk join after Valentin Vornicu??? isn't richard rusczyk the creator of aops???

[quote=pith0n][quote=akliu][quote=Critical]Valentin (my instructor!) made the site, and then later joined with Aops to make this site.[/quote]

OHHHH that makes sense now...[/quote]

He made mathlinks.ro which combined with AoPS (the previous link redirects to the college and olympiad portal).

And it really wasn't necessary to bump this.[/quote]

[s]The oldest topic on SS, finally destroyed... :evilgrin: [/s]

Thanks for telling me about that @all above

AKLIU WHY DID YOU BUMP THIS? Now it will get locked. At least there is one more thread.

[quote=jmiao]WOW this is old[/quote]

PM me if you want to know another old one.

I got postbanned from site support for bumping g old threads


When General Han counts the soldiers in his army, he uses the following method. He orders them to line up in rows of 11, then in rows of 13, and finally in rows of 17, and each time he counts the number of soldiers not in a row. One morning, he finds that there are 3 soldiers left when the rest are in rows of 11, 4 soldiers left when the rest are in rows of 13, and 9 soldiers left when the rest are in rows of 17. He knows that there are 1000 soldiers in his army. How many of the soldiers are present this morning.

I don't know how to solve this problem other than guess and check :oops: ....Can anyone help explain? Thanks

Learn mods if you haven't already, they're a very powerful tool. And if you have, learn when to use them. :)

For this problem, say $n$ is the number of soldiers. $n\equiv 3\bmod{11}$, $\equiv 4\bmod{13}$, and $\equiv 9\bmod{17}$. From there, we say that $(n-3)\equiv 0\bmod{11}$, $\equiv 1\bmod{13}$, and $\equiv 6\bmod{17}$. Therefore, we have a multiple of $11$ that is one more than a multiple of $13$. We have this at $66+143k$, where $k$ is an integer. Next, we see that $66+143k$ is six more than a multiple of $17$. You only need to check a few numbers at that point.

[hide="Only number under 1000 where this occurs"]$924$[/hide]

So, add $3$ to that to get the answer.

What are mods? :huh: I have heard of that word a couple of times but have no idea what it is.  Also, what does the 3 bars stand?

Thanks

$\equiv$ means equivalent to. It's the remainder when $n$ is divided by the $\bmod$ number. For example, $73\equiv 3\bmod{7}$, since $3$ is the remainder when $73$ is divided by $7$.


vector
analytic geometry
graphing lines
slope
calculus
calculus computations

Everything I have been doing so far with 3D stuff has been tangent planes. Then I got to this:

Find the parametric equations for the tangent [b]line[/b] to the curve of intersection of the paraboloid $z=x^{2}+y^{2}$ and the ellipsoid $4x^{2}+y^{2}+z^{2}=9$ at the point (-1,1,2).

I'm confused.... :?:

Just take the cross product of the gradient vectors to the two surfaces at that point. That will give you the "direction" vector.

Okay, so the cross products of the gradients.
$\nabla P=\left<2x, 2y, -1\right>$ $\nabla E=\left<8x, 2y, 2z\right>$
$\left<2x, 2y, -1\right>\times\left<8x, 2y, 2z\right> =\left<y(4z+2),x(-z-8),-12xy\right>$
Okay, for point (-1,1,2) = $\left<10,16,12\right>$

Um...so what is the point slope form for 3D? Is it related to the linear approximation?


function
modular arithmetic
algebra solved
algebra

Let $ S \equal{} \{0,1,2,\ldots,1999\}$ and $ T \equal{} \{0,1,2,\ldots \}.$ Find all functions $ f: T \mapsto S$ such that

[b](i)[/b] $ f(s) \equal{} s \quad \forall s \in S.$
[b](ii)[/b] $ f(m\plus{}n) \equal{} f(f(m)\plus{}f(n)) \quad \forall m,n \in T.$

I think the function is determined if we know f(2000). We have no restrictions on f(2000), so we can take it to be any number from 0 to 1999. By looking at some cases we observe a pattern: f(2000)=k => the function increases with 1 unit until it reaches 1999 and then it starts from k again.

For example, assume f(2000)=1997. Then f(2001)=f(2000+1)=f(1997+1)=1998; f(2002)=f(2001+1)=f(1998+1)=1999; f(2003)=f(2002+1)=f(1999+1)=f(2000)=1997 and so on (in this case k=1997).

There are 2000 such functions, corresponding to the cases f(2000)=k, with k from 0 to 1999.

We have
\[f(m)=f(m+0)=f(f(m)+f(0))=f(f(m)\\
\Rightarrow f(m+n)=f(f(m)+f(n))=f(f(m)+f(f(n)))=f(m+f(n))\]
Now suppose that $f(2000)=a$, for some $0\leq a\leq 1999$. Then \[f(m+2000)=f(m+f(2000))=f(2000+a)\]
and so $f(m)=f(m+a-2000)$ for all $m\geq 2000$. Thus the function determine uniquely by definition of $f(2000)$. Therefore there are 2000 functions satisfy the problem:
\[f(n)=n,\qquad \forall n,\quad 0\leq n\leq 1999,\\
f(n)=(n-a)\pmod {(2000-a)}+a,\qquad \forall n,\quad n\geq2000\]


Find the force applied to the dielectric in the electric field .Value of electric field is E and dielectric constant is e.Surface of the dielectric perpendicular to the electric field is A.

Here's one approach:

[hide]The (potential) energy stored in the electric field is given by:

$W_{e}=\int Q\mathrm{d}V$

Force is equal to minus the gradient of potential.

It follows that $\vec{F}=-\nabla W_{e}=-\nabla\left(\int Q\mathrm{d}V\right)$

$\Rightarrow \vec{F}=Q\left(-\nabla V\right) = Q\left|\vec{E}\right| \mathbf{\hat{n}}.$

According to Gauss' Law the electric flux across a surface is equal to the charge enclosed by said surface, i.e. $\vec{D}\cdot \mathbf{\hat{n}}A = Q.$

With that in mind, making the appropriate replacement:

$Q\left|\vec{E}\right| \mathbf{\hat{n}}=\vec{D}\cdot \mathbf{\hat{n}}A \left|\vec{E}\right| \mathbf{\hat{n}}.$

Also we know that $\vec{D}= \epsilon \vec{E}.$

Making that replacement:

$\vec{D}\cdot \mathbf{\hat{n}}A \left|\vec{E}\right| \mathbf{\hat{n}}= \epsilon \vec{E}\cdot \mathbf{\hat{n}}A \left|\vec{E}\right| \mathbf{\hat{n}}$
$=\epsilon \left|\vec{E}\right|^{2}A\mathbf{\hat{n}}.$

Conclusion:

$\boxed{\vec{F}= \epsilon \left|\vec{E}\right|^{2}A\mathbf{\hat{n}}.}$
[/hide]


geometry
3D geometry
tetrahedron
rotation

We have a cube. From every vertex, draw the plane formed by the three adjacent vertices. There are 8 such planes. How  many regions do they divide the cube into?

this problem probably is very trivial but i cant c how to solve it and cant visualise  :(  :( obvioulsy, for each plane we have a parellel plane - and the problem is just equivalant to: how many regions do a regular tetrahedron, and the same regular tetrahedron rotated by 180 degs ( i think) intersect? i think the answer is 17 but am not sure.

 How would you generalize to n dimensions?

it are actually 21 if i'm not wrong. take 4 vertices of the cube which form a regular tetrahedron. the faces of this tetrahedron are planes of the task. look at the region out of the tetrahedron but inside the cube. there are four congruent areas behin every face of the tetrahedron. three planes go through the vertex behind that face dividing that area into 4 pieces, no other planes go through there. thus we get 16 areas at all outside the tetrahedron. the tetrahedron itself is cut by all the four other planes dividing it into 5 pieces(these planes don't intersect inside the tetrahedron). thus we get 21 areas i think.

You can also see it this way: there are 12 pieces such that each piece contains 1 edge of the cube, 8 pieces such that each piece contains a vertex of the cube (the other 12 pieces contain 2 vertices of the cube each, but anyway the edges are more obvious), and one central piece, to make a total of 21.

This question cost me quite a bit of money once upon a time :( .


calculus
integration
derivative
LaTeX
complex analysis
complex analysis unsolved

Let a > 0. Prove:

integral from 0 to + infinity of (e^(-ax)dx) = 1/a

Diﬀerentiate with respect to a, n-times and prove

integral from 0 to + infinity of ( [x^n]*[ e^(-ax)] dx) = (n!) / a^(n+1)

Considering the integral:

integral from n to + infinity of ( [x^n]*[ e^(-ax)] dx)

prove that

for any a > 0.

(n^n)(a^n)(e^−an) ≤ n!

Just note $ n^n\int_n^{\infty} e^{\minus{}ax} dx \leq \int_n^{\infty} x^n e^{\minus{}ax} dx < \int_0^{\infty} x^n e^{\minus{}ax} dx$.

Also, learn some LaTex to post messages.


Hi!
I have a couple of questions on chemical bonding. I would be grateful if someone could answer them.

1) Compare the C-C distance (distance between 2 carbon atoms) in Ethane and  1,1,1-Trifluoroethane.
One of the reason which I can think is that fluorine being highly electronegetive would induce a partial positive charge on carbon. Due to this the carbon attached to flourines would pull the bond pair electrons between the two carbons towards itself. This would strengthen the bond and inturn decrease the bond length. Is it correct? However it isnt completely convining.

2)Why does the most electronegetive atom occupy axial positions on Trigonal Bipyramidal hybridisation(sp3d). I tried searching the net but dint get a convincing answer.
Again one of the answer which I can think of is that the electronegetive atoms would pull the elctron from the centre atom and thus reduce the electron desity.
That would reduce the repulsion in the equatorial plane. But that could be done even if the atom is in equatorial plane. So why axial?
Could u guys plz answer?

Thanks!

2. the most electronegative atom attracts electrons most strongly. the closer the electrons are to the atom, the less space the bonding pair can take up.

The less electronegative atoms then pull less on electrons and so their corresponding bonding pairs take up more space. They tend to occupy equatorial positions then because they have roughly 2 120 degree and 2 90 degree repulsions, instead of the more repulsive axial position with 3 90 degree repulsions. As less electronegative atoms occupy equatorial positions to minimize repulsions, more electronegative atoms then occupy axial positions.


We know the space diagonal of a cube is $ s\sqrt{3}$ where $ s$ is the side length. Since $ \sqrt{27}\equal{}3\sqrt{3}, s\equal{}3$, and the volume is $ 3^3\equal{}\boxed{27}$


conics
ellipse
function
calculus
calculus computations

Knowing that any polygon is a combination of finite triangles, is there a similar relationship with curved shapes? What I mean, is any curved shape a finite combination of another curved shape?

So far, I haven't found any such relationship. However, I recently thought that perhaps a curved shape can be defined as a polygon with an infinite number of sides. If this were true, then any curved shape could be defined as an infinite number of triangles merged together.

Any ideas?

If by "curved shape" you mean any simple closed curve, then this is clearly absurd.  Curves can take on arbitrary continuous shapes, whereas polygons are discrete, that is, any polygon can be finitely described, but an arbitrary curve comprises an infinite amount of information.

[quote="t0rajir0u"]  Curves can take on arbitrary continuous shapes.[/quote]

I'm not exactly sure what you mean by this.

One thing that distinguishes curves from straight lines is that there are many different types of curves (ex/ curves forming an ellipse versus the curves forming a logarithmic function), whereas there is only one type of straight line segment i.e. the shortest path passing through two points. I've even read that a straight line is a special type of curve. Is this what you meant?

Some curves can be thought of, not as polygons with infinitely many sides but more properly as limits of polygons.  However, there is no nice categorization of curves equivalent to, say, the categorization of polygons by number of sides.  Certainly, if you start listing types of curves ("elipse, logarithmic curve, ..."), you will spend all of your life and never make a dent in the family of all smooth curves.  We can deform a given curve essentially arbitrarily to give massive families of curves.


... to MC alums Zach Abel, Zhou Fan, Sherry Gong, Adam Hesterberg, Albert Ni, and Tim Pollio, USAMO winners/honorable mentions this year!

[quote="MithsApprentice"]Any of them coming back this year?[/quote]

Two of them, I think, at this point... unless, of course, they make IMO!

As listed on the alumni forms, the probability that I will return to math camp is now 1-P(I will die or become mentally disabled by July 3, 2005), changed from [1-P(I will die or become mentally disabled by July 3, 2005)][1-P(I make it to the IMO)P(The IMO goes through August 2)] ~= 1 because I was two points short of the TST.


abstract algebra
group theory
superior algebra
superior algebra unsolved

Let G be a finite Abelian group whose order is not divisible by $ p$, where $ p$ is a prime. Determine the number of solutions to the equation $ g^p \equal{} a$, where $ a$ is any element in $ G$.

This is indeed a homework question. I am not asking for a solution (but help if mine is wrong =))
I have that for each $ a \in G$ there is a [i]unique[/i] (ie 1) element/solution $ g$.

[b]1.[/b] For any $ a \in G$ there is a $ g \in G$ such that $ a \equal{} g^{p}$.

[b][color=orange]Proof.[/color][/b] Since $ p$ and $ |G|$ are coprime we have that $ 1 \equal{} m \cdot p \plus{} n \cdot |G|$ for some $ m, n \in \mathbb{Z}$. Hence,

$ a \equal{} a^{1} \equal{} a^{m \cdot p \plus{} n \cdot |G|} \equal{} a^{m \cdot p} \cdot a^{n \cdot |G|} \equal{} (a^{m})^{p}(a^{|G|})^{n} \equal{} (a^{m})^{p}.$

The result follows once we let $ a^{m} \equal{} g.$

[b]2.[/b] For any $ a \in G$ there is exactly one $ g \in G$ such that $ a \equal{} g^{p}$.

[b][color=orange]Proof.[/color][/b] In view of [b]1[/b] all that remains to be shown is unicity. We proceed by [i]reductio ad absurdum[/i] here. Let us suppose that there exist $ g_{1}, g_{2} \in G$ such that

$ g_{1}^{p} \equal{} a \equal{} g_{2}^{p}.$

From the equalities in the previous line we get that

$ e \equal{} g_{1}^{p}(g_{2}^{p})^{ \minus{} 1} \equal{} g_{1}^{p}g_{2}^{ \minus{} p} \equal{} (g_{1}g_{2}^{ \minus{} 1})^{p}.$

Hence, the group $ G$ possesses an element of order $ p$ ([b][color=red]contradiction![/color][/b]) and the proof terminates.

[b]Conclusion:[/b] Little Turtle's guess is right!  :D

Best regards,

José.


Find all rational numbers a, such that [tex]\left(\frac{a^{m+n}+1}{a^{n}+1}\right)^{\frac{1}{m}}\in\mathbb{Q}[/tex], where m and n are positive integers, with m>1.

[quote="pcalin"]Find all rational numbers $a$, such that $\displaystyle \left(\frac{a^{m+n}+1}{a^{n}+1}\right)^{\frac{1}{m}}\in\mathbb{Q}$, where $m$ and $n$ are positive integers, with $m>1$.[/quote]

Anyone? :)


algebra
polynomial
arithmetic sequence
algebra unsolved

Find all positive integers $ n\geq 3 $ such that there exists an arithmetic progression $ a_0 , a_1, \ldots, a_n $ such that the equation $ a_nx^n + a_{n-1}x^{n-1} + \cdots+ a_1x+a_0 = 0 $ has $n$ roots setting an arithmetic progression.

[quote="Myth"]Let $a_n=a$, $a_{n-1}=a+b$, ..., $a_0=a+nb$, then initial equation can be rewritten as
\[ax^{n+2}-(a-b)x^{n+1}-(a+b(n+1))x+(a+bn)=0.\]
This equation has additional root $1$ of order 2, so totally it has $n+2$ reall roots. But we see that there are $n\geq 3$ consecutive zero coefficients. So from Descartes rule we conclude there is an imaginary root. Contradiction.

Hmm... It works for $n=2$ too  :?[/quote]
Why roots of the polynomials setting arithmetic progression??  :P  :P  :D

???
What do you mean? :?
I have showed that such polynomials cannot have $n$ real roots.


algebra
system of equations
algebra unsolved

[quote="Przemo"]Let n be positive integer. Solve system of equations
$ x_1 \plus{} x_2^2 \plus{} ... \plus{} x_n^n \equal{} n$
$ x_1 \plus{} 2x_2 \plus{} ... \plus{} nx_n \equal{} \frac {n(n \plus{} 1)}{2}$
for n-tuples $ (x_1,...,x_n)$ of nonnegative real numbers.[/quote]
Add $ 1 \plus{} 2 \plus{} 3 \plus{} ... \plus{} (n \minus{} 2) \plus{} (n \minus{} 1)$ in both side of the first equation, we have:
$ \frac{n(n \plus{} 1)}{2} \equal{} x_1 \plus{} (x_2^2 \plus{} 1) \plus{} (x_3^3 \plus{} 2) \plus{} ... (x_n^n \plus{} n \minus{} 1) \geq x_1 \plus{} 2x_2 \plus{} 3x_3 \plus{} ...nx_n$
$ \Rightarrow x_1 \equal{} x_2 \equal{} x_3 \equal{} ...x_n \equal{} 1$


vector
LaTeX
linear algebra
matrix
inequalities

This should be a very basic problem, but I have some doubts.

Suppose that U and V are finite-dimensional vector spaces and that S is a linear map from V to W, T is a linear map from U to V.
Prove that $ dimnullST$ is less than or equal to $ dimnullS\plus{}dimnullT$.

Also, can anyone give an example when it is strictly less than?

$ \text{\LaTeX}$ tip: first see if the word you want is encoded by putting \ in front of it:

\dim gives $ \dim$, \nul gives an error message.

In that case, use \text{nul} to get what you want: $ \dim\text{nul}S.$

Very simple example: $ S = T = \begin{bmatrix}1 & 0 \\
0 & 0\end{bmatrix}.$ Then $ ST = S$ and $ \dim\text{nul}S = \dim\text{nul}T = \dim\text{nul}ST = 1.$

I'll channel my current colleague from whom I learned all of my linear algebra to say this: why are you calling these "linear maps on finite dimensional vector spaces"? These are [b]matrices[/b], and the sooner you get to thinking of them as matrices, the better.

Sorry, I've just begun linear algebra for several days, I only read three chapters of a linear algebra book that did not even mention matrix for many times.

So what about the prove for the inequality?


Let's expect results of the IMO 2008 Spain.

1.China
2.Russia
3.South Korea
4.USA
5.Vietnam
6.Bulgaria
7.Romania
8.Taiwan
9.Iran
10.Ukraine

[quote="edriv"]Isn't it too early?  :D[/quote]

Of course not :P
By the way, the Hungarian team will be:
...
:lol:

[quote="Valentin Vornicu"]Will North Korea participate?[/quote]

That is a touchy question.. I had a conversation with Pelikán about that, but i think i'm not at liberty to talk about that in a public forum...
Anyway, it's not known yet.

Well, it's important for next poll, as they are in the top10.

[quote="ringos"][quote="Valentin Vornicu"]Will North Korea participate?[/quote]

Why did not North Korea participate from 1993 to 2005?[/quote]It did not participate in 2006 either. It most likely was because the Olympiad wasn't held in a communist country anymore so the political leadership didn't want them to send a team. It was disqualified in 1991 in Sweden for (suspicion of) cheating. I doubt they will come to Spain next year as well, so that means that probably in the top 10 poll I will exclude it and put the next country in the list.

Does anyone know which country will be strong next year?
(For example, who will be high school students next year among this year's contestants in your country?)

It's too early now but I think Georgian Team will be in 15 best country(at least).

[quote="pephuc_93"]I think Viet Nam will win because next year chienthan will 16 years old! :wink:[/quote]
It is really strong reason :rotfl: I'm kidding!

[quote="pephuc_93"]I think Viet Nam will win because next year chienthan will 16 years old! :wink:[/quote]

[color=green]hnm maybe.  :D [/color]

i'm sure $ INDIA$ has the potential to win
i will surely expect a good performance from $ INDIA$

(although for some last years the performance is degrading)


A curve has the equation $ y\equal{}f(x)$ where

$ f(x) \equal{} \cos (2x\plus{}\frac{\pi}{3}) \plus{} \sin (\frac{3x}{2} \minus{} \frac{\pi}{4})$

Find the period of $ f(x)$.

Determine all values of $ x$ in the interval $ \minus{}\pi \le x \le \pi$ for which $ f(x)\equal{}0$. Find a value of $ x$ in this interval at which the curve touches the $ x\minus{}$axis without crossing it.

Find the value or values of $ x$ in the interval $ 0 \le x \le 2\pi$ for which $ f(x)\equal{}2$.

\[ f(x)\equal{}0 \implies 0\equal{}\cos(2x\plus{}\frac{\pi}{3})\plus{}\sin(\frac{3x}{2}\minus{}\frac{\pi}{4}) \implies \sin(\frac{\pi}{6}\minus{}2x)\equal{}\sin(\frac{\pi}{4}\minus{}\frac{3x}{2})\]

therefore \[ \frac{\pi}{6}\minus{}2x\equal{}\frac{\pi}{4}\minus{}\frac{3x}{2} \implies x\equal{}\minus{} \frac{\pi}{6}\]


geometry
Triangles
rotation
IMO 1986
Fixed point
Equilateral Triangle
IMO

Write $x= A_1-A_2 , y= A_2-A_3 , z= A_3-A_1$.
The mappings are (with $\zeta$ being the 'first' third root of unity):
$P \mapsto \zeta (P-A_1)+A_1$
$P \mapsto \zeta (P-A_2)+A_2$
$P \mapsto \zeta (P-A_3)+A_3$
and using them in a row gives (easy to calculate out):
$P \mapsto P + \zeta^2 x + \zeta y + z$
Using the chain of mappings $1986= 3 \cdot 662$ times is the same as using that mapping $662$ times, thus the mapping
$P \mapsto P+ 662(\zeta^2 x + \zeta y + z)$
Now the last shall equal $P$, so $\zeta^2 x + \zeta y + z = 0$, which means (by subtracting $x+y+z=0$) that $(\zeta^2-1) x +(\zeta-1) y =0 \iff x= \zeta y$.
Similary, $z=\zeta x$, thus $|x|=|y|=|z|$, thus the triangle is equilateral.

Does anyone have a synthetic solution or know if a nice one exists?

[quote="FOURRIER"]There's one here, but in french
p.20

http://www.animath.fr/cours/deho_geo/deho_geo.pdf[/quote]
Thank you very much!  Luckily "Math French" is a little similar to "math English."  So I think I have understood the solution, and this is not a literal translation from that file.
[hide="Synthetic"]A composition of three $ 120$ rotations is a rotation of $ 120 \plus{} 120 \plus{} 120 \equal{} 360$, i.e. a translation.  Thus, $ \vec{P_0 P_3} \equal{} \vec{P_3 P_6} \equal{} \dots \equal{} \vec {P_{1983} P_{1986}}$.  But $ P_0 \equal{} P_{1986}$, so the vector is null and $ P_0 \equal{} P_3 \equal{} \dots \equal{} P_{1986}$.  Since $ P_0$ had no restrictions, we can say that any point in the plane gets mapped to itself after the three rotations.  In particular, let's examine the behavior of $ A_0$.  After the first rotation, $ A_0$ remains $ A_0$.  After the second, it gets mapped to some point $ B$.  Finally, by our previous result, the third rotation takes $ B$ to $ A_0$ again.  Now noting that $ \angle A_0A_2B \equal{} \angle B A_1 A_0 \equal{} 120$, and that $ BA_2 \equal{} A_2 A_0$ and $ BA_1 \equal{} A_1 A_0$, it is easy to deduce that $ A_0A_1A_2$ is equilateral.[/hide]

I think the solution found by Zetax is shorter and better (I also did like him).

[quote=orl]Given a point $P_0$ in the plane of the triangle $A_1A_2A_3$. Define  $A_s=A_{s-3}$ for all $s\ge4$. Construct a set of points $P_1,P_2,P_3,\ldots$ such that $P_{k+1}$ is the image of $P_k$ under a rotation center $A_{k+1}$ through an angle $120^o$ clockwise for $k=0,1,2,\ldots$. Prove that if $P_{1986}=P_0$, then the triangle $A_1A_2A_3$ is equilateral.[/quote]
Professor Chang Gengzhe, one of the propositional person of this problem, died in Beijing on November 18, 2018.

[quote=FOURRIER]There's one here, but in french
p.20

http://www.animath.fr/cours/deho_geo/deho_geo.pdf[/quote]

Sir can you please repost this link because this link is dead when you open this there is no solution of above problem.kindly give me other link for this solution

Let $a_1, a_2, a_3, p_0$ be the complex numbers corresponding to their points in the complex plane, and set $\omega = e^{2i \pi/3}$. We can compute via a series of rotations that$$p_3 = p_0 - a_1 + \omega^2 a_1 + \omega a_2 - \omega a_3 - \omega^2 a_2 + a_3.$$In other words, we must have$$a_3-a_1+\omega^2(a_1-a_2) + \omega(a_2-a_3) = 0$$in order to have $P_{1986} = P_0$ because $3 \mid 1986$. Now, consider the polynomial$$x^2(a_1-a_2) + x(a_2 - a_3) + (a_3 - a_1) = 0.$$It has roots $\omega$ and $1$, so by Vieta's,$$\frac{a_3 - a_1}{a_1 - a_2} = \omega,$$which implies that $a_1, a_2, a_3$ are the vertices of an equilateral triangle, as required.

Observe that by the theory of rotation composition, we have that $P_{k+3}$ is a translation of $P_k \forall k \ge 0$. But since $P_{1986}$ is reached by performing $662$ such translations, and $P_0 = P_{1986}$, we have that $P_k = P_{k+3} \forall k \ge 0$.
Now construct an equilateral triangle $A_1A_2B$ such that $B$ is on the same side of line $A_1A_2$ as $A_3$. Then clearly if $P_0 = B$, we also have $P_2  = B$.Therefore $P_2 = P_3 = B$, but this is only possible if $B = A_3$. Since $\Delta A_1A_2B$ is equilateral by definition, we are done. $\square$


MATHCOUNTS
summer program
MathPath
AMC 8

how do  i prepare for mathcounts chapters?

Just do a bunch of problems. The mathcounts website has a bunch of problems from previous years and it might be a good idea to get hold of warmups and workouts from your teacher.

[quote="Walk Around The River"]what i do is the night before the competition, stay up all night.  make sure to bring lots of mt dew.  this isnt a joke.  then your body will keep you up long enough for the competition, but you will be so tired that your body just works on instinct so i make less mistakes.[/quote] That approach only works if you're REALLY, REALLY confident with the material. Instinct can trick you, and if you are tired, you're more likely to misread the question or make stupid, mistakes. I'd recommend getting a lot of sleep and trying to be as alert and ready for the competition as possible. That way you can pay attention to detail and still work quickly.

what i'd do
______________________________________________________
write a summary of small things like easy formulas, how to do stuff like permutations, pythag., etc. and just look over it the day before. try not to do any problems

ull get too used to those, u just want to LOOK OVER the general concept.

-jorian

read aops v.1.
then practice by doing past problems, and understand the solutions and concepts, so that when you see it again, you won't miss it.
work on making as few careless mistakes as possible.

"i have a MC competition today??"

 :rotfl:

at one time i almost forgot until it was 9 pm friday and my dad reminded me

You might check out this web site for more info.

[url]http://www.geocities.com/danb80/PrepForContest.html[/url]

I really don't know how to make myself have a "good day" or a "bad day". I wish I did. Sometimes on practices I just do horrible, but sometimes I do really well. The most noticeable difference is on countdown. I think sleep and exercise are factors but I'm not sure what other factors there are.

For preparing, do lots of problems. But test makers are making problems that have never been seen before, so doing a lot of problems will make you good at those old problems but not necessarily on the real test, because you've never seen the concepts of those problems. I doubt it will matter much at chapters though so just do old tests.

Ubemaya, stop lying
the worst you have ever gotten on a practice was a 43.
and any way, i agree that just doing past rounds won't always help.
its good to just do tough problems that really open up your mind. By doing so, your mind will be trained to solve tough problems you have never seen before

[quote="laughinghead505"]Are you allowed to bring a sheet with a bunch of formulas on it?  :P Because I was planning to write out all the primes under 100 because I always forget them.[/quote]

I wish...

Well, they would never know if you just take a piece of scratch paper and write them out right before it starts, would they? :P

[quote="kyyuanmathcount"]I STRONGLY discourage you do the mountain dew thing.

Strategy during the prep time months before competition:  Use Red Bull or caffeinated energy drink of choice and madly do problems.

Stategy right before competition:  If you haven't prepared enough, that's too bad.  You should relax anyways though, because cramming in an extra formula is not worth as much as having a good state of mind so you can think of a good method to do a few problems.[/quote]

what if you can't get your hands on an energy drink :P ?

-jorian

[quote="laughinghead505"]Then you eat sugar.  :D Although I don't think I should do that this year because when I do, I usually get so phsyced up and hyper that I get a headache and waste some of the team time talking. :([/quote]
And you end up tired at the end. I recommend Propel if you need something to keep you up-it isn't entirely energy drink and people need water.
It's like those gummy vitamins!
Never eat sugar you get at hotels. That is my only warning if you will eat sugar.

so, any other solutions to my energy drink crisis?

-jorian

EDIT: I THINK I CAN GET ONE (I ASKED MY MOM :P )

-jorian (again)

[quote="Rafaelloaa"]i go to http://www.saab.org[/quote]
THe problem with saab is that the problems get really repetative after a while.

I do no problems the day of the comp. until the comp. the week b4, I do TONS of problems. my team, as i believe i've stated, wrote a song last year during the team. it wasn't bad. i've got it written somewhere. i'll post i if i find it.

If we have extra time this year, I'm going to make my team memorize the pi song (copyrighted to SHBoB, MathPath 2006 :lol: )

To the tune of Yankee Doodle:

3.1415926535897932384626433
 :D

during a practice chapter team, we wrote a song with 16 minutes left :P

we got screwed, we ended up with a 9 :blush: i mean, i did better on the amc 8!!!!!!!!!!!!!!!!!!!

-jorian

so your team did all ten atleast 2 times in four minutes?

sort of. :P

except for one which we purposely held off til the last minute as a joke. i think we got THAT one right

-jorian


In a right triangle a = 30 yards and tan A = 2. Find b and c

does the pythagero theorm play a role in this question??

[hide]$\tan A=2=\frac ab$
$2=\frac{30}{b}$
$b=15$
Use Pythagorean to get $c=15\sqrt5$[/hide]

[hide="Solution"]

Draw a right triangle and label one of the acute vertices A.

Its tangent is 2. Since you know that the length of its opposite is 30, you know that the adjacent side is $15$.

The hypotenuse is then $15\sqrt{5}$.

So, you know all the sides.[/hide]


geometry
perpendicular bisector
geometry proposed

Let $ ABCD$ be a quadrilateral inscribed in a circle $ (O)$ and let $ x,y$ be the perpendicular bisector of the of the line segments $ OC,OD$ respectively. $ S$ is a point on the circle $ (O)$. The line  $ SA$ meets $ y$ at $ I$, the line  $ SB$ meets $ a$ at $ J$.Prove that the line $ IJ$ is always tangent to a fixed circle when $ S$ moves on the circle $ (O)$
P/s:I have just edited. Thank arqady

[quote="Brown"]Prove that the line $ IJ$ is always tangent to a fixed circle when $ M$ moves on the circle $ (O)$[/quote]
$ M???$ What is this?
Do you mean  the following?
Prove that the line $ IJ$ is always tangent to a  circle $ (M)$ when $ M$ moves on the circle $ (O)$ :wink:
If so it's obvious.

Dear Mathlinkers,
This problem is from Mathematics and Youth magazine- a Mathematic magazine in Viet Nam, the latest number, so I think that there should be no discussions about this problem until the next 2 months. Hope that we will all respect the fair in this magazine's contest.
MathVNpro


advanced fields
advanced fields unsolved

Find the cardinality of $ X\equal{}\{f | f: \mathbb R \rightarrow \mathbb R, bijective\}$.

$ |Sym(M)| \equal{} 2^{|M|}$ for infinite sets $ M$.


Given a convex quadrilateral $ABCD$, in which:

$\angle BAC = 30^\circ$
$\angle CAD = 20^\circ$
$\angle ABD = 50^\circ$
$\angle DBC = 30^\circ$

If $P = AC \cap BD$, show that $PC = PD$

Can you find a [i]synthetic[/i] proof?

(A popular trigonometric proof can be found here: [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=79800[/url])

It's been almost a week, and unfortunately there are no answers. :( I know of 2 synthetic proofs for this nice problem; I will wait a little longer and post them (hopefully) in a few days.


Fix the series $a_{n}=\sum_{k=1}^{n}\frac{(-1)^{k+1}}{k}=\frac{1}{1}-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+.....$
Show that for every $x \in \mathbf{R}$ there is a bijection $\pi: \mathbf{N}\to \mathbf{N}$ such that the series $\sum_{k=1}^{\infty}\frac{(-1)^{\pi(k)+1}}{\pi(k)}=x$.

A simple application of the classic theorem on conditionally convergent series. The construction:
Take positive terms starting at the beginning until their sum is greater than $x$. Take negative terms starting at the beginning so that the sum is less than $x$. Take positive terms starting from the first excluded until the sum is greater than $x$. Repeat indefinitely.
There's some work to do to prove that the construction works and the constructed series converges to $x$. It's not very hard.

For $x=1.1$, the series produced this way is
$1+\frac13-\frac12+\frac15+\frac17-\frac14+\frac19+\frac1{11}-\frac16+\frac1{13}+\frac1{15}-\frac18+\frac1{17}+\frac1{19}+\frac1{21}-\frac1{10}+\cdots$


Prove it.   :lol: (This is significantly harder than [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=26559]this[/url] ... ).

As I mentioned there, I believe the solution in post number $3$ of [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=166331#p166331]this topic[/url] can easily be extended to solve this problem by working with $I$ and $\{\frac 1x|x\in I\}$ for intervals $I$ which do not contain $0$ instead of $I$ and $-I$, but I didn't check it thoroughly. I might be wrong.


Prove by induction the following:

Suppose you are in a country that has bank notes of value 3 dollars and 5 dollars only. Prove that it is possible to pay (without requiring change) for any item which is worth a whole number of dollars greater than 7 dollars.

*The one thing I do know is n >= 8.

Thank you for your help in advance.

Chicken McNugget Theorem.

Go look it up or ask somebody else.

Chicken McNugget, great name  :rotfl: ! (OK, I just googled it, it DOES exist, I'd just never heard of it before).

Anyway, here is my (pretty detailed) proof:

$8 = 5+3$
$9 = 3+3+3$
Now, every number is 0, 1 or 2 (mod 3), which means they are all 0, -2 or -1 (mod 3).
Since
$5 \equiv-1$ (mod 3)
every number x can be written as a sum of five and three
$x \equiv 0$ (mod 3) then you don't need 5
$x \equiv-1$ (mod 3) then you need one 5
$x \equiv-2$ (mod 3) then you need two 5
Since every number left is greater than or equal to 10, this is not a problem.

[quote="scorpius119"][quote="MathStudent2"]I need to solve the problem using a base case and then assuming k + 1 is true.  I am being taught to do it this way and in steps like below.

Prove P(n): n = 3*a + 5*b or n >= 8

Step 1:  Let n = 8
	8 = 3*1 + 5*1
	8 + 8

Step 2:  Assume P(k) is true for n = k.
	Then, k = 3*a + 5*b

Step 3:  Must show P(k+1) is true.
	That is n = k +1

From here I have trouble.  I have been looking at it and trying a few things, but I don’t get anywhere.  I need it broke down step by step if I could.  Also, why do you show cases 9 and 10?

Thank You![/quote]
[hide="using induction"]
The base case $n=8=3+5$ is clearly true. For the inductive step (assume true for $n$, prove for $n+1$), write $n=3a+5b$ for nonnegative integers $a,b$. If $b\neq 0$, then $n+1=3(a+2)+5(b-1)$. Otherwise, $n=3a$, so $a\geq 3$ and $n+1=3(a-3)+5\cdot 2$.
[/hide][/quote]

Where does (a+2) and (b-1) come from?  Thank you!

[quote="MathStudent2"]Okay, I add one to both sides, then we get:
k + 1 = 3a + 6 + 5b - 5, another way to write
k + 1 = 3a + 5b + 1, then next line is just factoring out a 3 and 5.

So, I am still not sure where to go from here:
k + 1 = 3(a + 2) + 5(b - 1)?

Thanks,[/quote]

Basically, he was just showing that if k can expressed as the sum of 3's and 5's, then k+1 can also be expressed as the sum of 3's and 5's.

Thank you tjhance.  Where do I go from here: k + 1 = 3(a + 2) + 5(b - 1) to complete the proof.  Thanks!

I don't quite see what's left to prove, except the case where $b=0$. If so, then $k=3a+5(0)=3a$. $k\ge 9$ so $a\ge 3$, so $k+1=3(a-3)+5(2)$ would then work, as shown by scorpius119.

So I need to do induction on three cases: k, k+1, or k+2? And, the fact that it is k + 1 and k + 2 is because 8 +1 = 9 and 8 + 2 = 10? Just trying to clear up my thinking process. Thanks!

we proved that if it is true for $k$ then it is true for $k+1$. Then, since it's true for $k+1$, we automatically know it is true for $k+2$. And then $k+3$. and so on, to infinite.

[quote="MathStudent2"]Okay, I am understanding the problem a little better.  I am still looking at it though.  Why do you set b = 0 and solve?  Why not a = 0?

When b = 0, I see we get n = 3a.  So, just to clarify my thoughts we have a >= 3 because n has to be >= 8 giving us n = 9?  Am I correct on my thoughts here?

Thanks again for your help![/quote]

Well, $3a=n\geq 8\Rightarrow a\geq \frac{8}{3}$. And $a$ is an integer, that gives $\boxed{a\geq 3}$ ($n$ doesn't necessarily equal 9 though; you can't have $3a+5b=12$ unless $b=0$)

I have a question in regards to the second case below:

2) Assume that d is zero.
Then, 3*a + 5*b = k  (Ind. Hyp.)
3*a = k
Because k >= 8, then x >= 3
Then, 3*a + 5*b + 1 = k + 1
3*a – 9 +5*b + 10 = k + 1
3(a – 2) + 5(b + 2) = k + 1

Choose c = a – 2 and d = b + 2 and we have:
3*c + 5*d = k + 1, with both in the nonnegative integers.

OR

3) Assume that d is zero.
Then, 3*a + 5*b = k  (Ind. Hyp.)
3*a = k
Because k >= 8, then x >= 3
Then, 3*a + 1 = k + 1
3*a – 9 + 10 = k + 1
3(a – 3) + 5*2 = k

Since I am trying to show the form 3*c + 5*d = k + 1 I am not sure which of the two above would be the better one to show.  The first one, I am assuming d is zero and then deriving down to 3(a – 2) + 5(b + 2) = k + 1 so that c = a – 2 and d = b + 2.

On the second one, I only end up with c = a – 3 and no d, but I did assume is zero.

Not sure which I should use?  Any suggestions?

Thank you for all of your help throughout this problem!


Let $ a,b,c\geq 0$ s.t. $ abc\equal{}1$.Prove that:
$ (\dfrac{a\plus{}b\plus{}c}{3})^5\geq\dfrac{a^2\plus{}b^2\plus{}c^2}{3}$

[quote="Inequalities Master"]Let $ a,b,c\geq 0$ s.t. $ abc \equal{} 1$.Prove that:
$ (\dfrac{a \plus{} b \plus{} c}{3})^5\geq\dfrac{a^2 \plus{} b^2 \plus{} c^2}{3}$[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=87919


Find the positive difference between the largest and the second largest elements of the set $ \left\{\frac13,\frac3{10},\frac15,\frac14\right\}$. Express your answer as a common fraction.

It's not immediately obvious which ones are largest, smallest, etc.. So put everything over a common denominator. The set becomes $ \{\frac{20}{60}, \frac{18}{60}, \frac{12}{60}, \frac{15}{60}\}$. Now it becomes easy to see. The largest is $ \frac{20}{60}$ and the second-largest is $ \frac{18}{60}$. Taking their difference, we have $ \frac{20}{60} \minus{} \frac{18}{60} \equal{} \frac{2}{60} \equal{} \boxed{\frac{1}{30}}$


search
combinatorics theorems
combinatorics

Why $ \binom{2n} {n}$  is divisible by n+1?

Thank you very much

$ \binom{2n}{n}\equal{}\frac{(2n)!}{n!(n!)}\equal{}\frac{2n\cdot(2n\minus{}1)\cdots(n\plus{}1)}{n!}$, which as you see, is divisible by $ n\plus{}1$.

The question has been asked several times here.

So, you want to prove that Catalan numbers are integers.

Take $ A \equal{} {2n \choose n}$ and $ B \equal{} {2n \choose n \plus{} 1}$

Now, it's easy to show that $ C \equal{} A \minus{} B \equal{} \frac {(2n)!}{n!(n \plus{} 1)!} \equal{} \frac {{2n \choose n}}{{n \plus{} 1}}$.

$ A$ and $ B$ are integers, so $ C$ is an integer too, qed.

$ \binom{2n}{n}\equal{}\frac{2n\cdot(2n\minus{}1)\cdots(n\plus{}1)}{n!}$ that is divisible by $ n\plus{}1$

BTW, here are two more proofs (Search button is your friend  :lol: ):

http://www.mathlinks.ro/viewtopic.php?t=18917

http://www.mathlinks.ro/viewtopic.php?t=167490

P. S. Desmo90, haven't you noticed that your one line proof is exactly what JRav already wrote?  :)

[quote="hsiljak"] P. S. Desmo90, haven't you noticed that your one line proof is exactly what JRav already wrote?  :)[/quote]

no, i'm sorry. :oops:

Well, no problems. But if you like, you could show why does the fact you wrote holds. It isn't completely trivial, and the proof could be useful for the O. P. to understand the logic of binomial divisibility and so ;)

[quote="JRav"]$ \binom{2n}{n} \equal{} \frac {(2n)!}{n!(n!)} \equal{} \frac {2n\cdot(2n \minus{} 1)\cdots(n \plus{} 1)}{n!}$, which as you see, is divisible by $ n \plus{} 1$.[/quote]
This isn't a proof.  Once you remove that term from the numerator, you have no guarantee that $ n!$ divides what's left except in the case that $ n \plus{} 1$ is prime.

One straightforward way to do this is to define the [url=http://en.wikipedia.org/wiki/Catalan_number#Proof_of_the_formula]Catalan numbers[/url] in a combinatorial fashion and then show that they are given by this formula, from which it immediately follows that the formula must return integer values.


I need:
     Complete handbook solutions for 2001-02, 2002-03, 2003-04, 2005-06.

I have:
            National 1984-2008, missing some countdowns
            State 1991-2008, missing 95,96,97,98 team
            Chapter 2000-08
            school 2006-08
            Handbook 02-08 with answer keys

PM me if you want to trade.

Thanks!

Hi, I am a new member to AOPS and basically new to this whole trading thing. What do I do if I don't have ANYTHING. How do I go about aquiring new tests from Nationals and Older States and the handbooks. I am deperate but not really sure how to start. How did some of you guys start aquiring tons of stuff?

The way I see it, once I actually have some stuff to trade, I can aquire more and actually trade. If I don't have anything, I can't trade so I am stuck.

Can someone PLEASE HELP!!

Please post all trading requests in this [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=133189]thread[/url]

What I have:

2000-2005 AMC 10s
1996-2005 AMC 12s
2001-2005 AIMEs

2000-2008 State Rounds
2005-2008 Chapter Rounds
08-09 Mathcounts Handbook
1999-2005 USAMOs

I have solutions for all of them

I AM DESPERATE FOR MC HANDBOKS, PM IF INTRESTED< I AM A NEWB SO I NEED HELP!!!!!!!!!!!!!!!!!!!!11

First of all, don't capitalize all you words and add so many exclamation points; it won't persuade others to help you.

Your best bet is to post on forums like these and see if anyone else wants to trade with you. Make your own topic; people will notice it more.

[quote="akhil0422"] Make your own topic; people will notice it more.[/quote]

But that is also spam when we have a topic intended for it.

Sorry, I didn't see that. I don't have anything to trade so I [b]don't[/b] trade (great logic isn't it).


group theory
abstract algebra
invariant
superior algebra
superior algebra unsolved

Let $ G$ be a finite simple group of even order. Prove that $ \mbox{Inn}(G) \stackrel{char}{\trianglelefteq} \mbox{Aut}(G)$.

What is $ \text{Inn}(G)$ ? And what is the $ char$?...

Oh, I'm sorry. $ \mbox{Inn}(G)$ is the group of inner automorphisms of $ G$, while $ \mbox{Aut}(G)$ is the group of all automorphisms of $ G$.

Well, the definition of a normal subgroup $ N$ is a subgroup which is invariant under all inner automorphisms, i.e. all elements of $ \mbox{Inn}(G)$.
We denote $ N \trianglelefteq G$.
A characteristic subgroup is something stronger, since it is a subgroup $ N$ which is invariant under all automorphisms, i.e. all elements of $ \mbox{Aut}(G)$.
We denote $ N \stackrel{char}{\trianglelefteq} G$.

Note that it is well known (or otherwise quickly proved) that $ \mbox{Inn}(G) \trianglelefteq \mbox{Aut}(G)$. The problem asks to prove that for finite simple groups of even order, it is even a characteristic subgroup.

Here's a hint: First prove that $ G$ is generated by it's involutions. Then prove that $ C_{\mbox{Aut}(G)}(\mbox{Inn}(G))$ is trivial.

Ok, since nobody has solved this problem, I will post the solution myself. This question is from the $ 2005$ exam of Group Theory here at Ghent University (UGent)

We can easiley prove that $ C_{\mbox{Aut}(G)}(\mbox{Inn}(G))$ is trivial. Prove this by contradiction, and you will get that $ Z(G)$ contains a non-trivial element, which is impossible since $ G$ is simple and can obviously not be abelian (unless $ G \cong C_2$ but we can check this case)

It is well known that $ \mbox{Inn}(G) \trianglelefteq \mbox{Aut}(G)$.

Assume that there exists an element in $ \mbox{Aut}(\mbox{Aut}(G))$ that does not fix $ \mbox{Inn}(G)$, and maps it to $ H$. Obviously, \[ H \trianglelefteq \mbox{Aut}(G),\hspace{1,5 cm} \mbox{Inn}(G) \cap H \trianglelefteq \mbox{Inn}(G).\]

Because $ Z(G)$ has to be trivial, we know that \[ G \equal{} G \slash Z(G) \cong \mbox{Inn}(G).\]
It follows that $ \mbox{Inn}(G) \cap H \equal{} \left\{e\right\}$. Now, $ H$ and $ \mbox{Inn}(G)$ are two normal subgroups of $ \mbox{Aut}(G)$, and they have trivial intersection.

This implies that $ H$ and $ \mbox{Inn}(G)$ commute, which is impossible since $ C_{\mbox{Aut}(G)}(\mbox{Inn}(G))$ is trivial.


How hard would it be to program in a multi-option poll (ie you can choose more than one option if you want). Would it be easy enough to just have a option to replace the radio button (I think they're called radio buttons anyway) by checkboxes? You'd want both options, of course.


A group of people went to a restaurant for lunch. They agreed to split the luncheon bill equally. Later, when the bill arrived, three of them discovered that they had forgotten their money. The others in the group agreed to make up the difference by each paying an extra $ 1.80. If the total bill was$78.00,
how many people were in the group?

[hide] Let $ x$= the number of people, and $ y$= the price each person was to originally pay. Then,

$ (x \minus{} 3)(y \plus{} 1.8) \equal{} xy \implies 1.8x \minus{} 3y \equal{} 5.4$

Since $ xy \equal{} 78$, we can say that $ y \equal{} \frac {78}{x}$.

So,

$ 1.8x \minus{} 3(\frac {78}{x}) \equal{} 5.4$

$ \implies x^2 \minus{} 3x \minus{} 130 \equal{} 0$

$ \implies (x \minus{} 13)(x \plus{} 10) \equal{} 0 \implies x \equal{} \boxed{13}$.[/hide]


1) When you play [url=http://blackjack.how2gamble.com]blackjack[/url],Which dealer's up-card(s) will cause him to bust more than 50% of the time? ? do you know?
the 6
the 5 and 6
the 4, 5 and 6
there aren't any

Can a moderator please delete this topic?  This is like the 10th time that Kotsba has been advertising for online gambling sites.


That's right, but you can find $ a_{n}$ by solving this reccursive directly. :P

Hmmm I have no clue how to do that. Demonstrate?


inequalities
superior algebra
superior algebra theorems

Could someone please write or give a reference what is Schur-Zassenhaus Theorem(in group theory)

From google:
[quote] ...And I came across something pretty nifty in a book called "Noncommutative Algebra" by Farb and Dennis: the Schur-Zassenhaus theorem, which says that if |H| is relatively prime to |G/H| then G is a semidirect product H X| G/H. Apparently there is a short simple proof using group cohomology. I don't know if there is a short proof with straight group theory. So cohomology of groups seems interesting since you can prove things easily...[/quote]


algebra
polynomial
function
integration
calculus
calculus computations

Part a)

Compute the 4th degree Taylor polynomial for the function $ f\equal{}e^{x^2}$ at 0.

Part b)

Approximate $ \int e^{x^2} dx$ and determine the accuracy of the approximation.

I have done part a, and I have done the first half of part b. I am not sure about the second half. I have computed the Lagrange Remainder, but since it came out to be 0, I don't know if that means there is no error?

for values of x sufficiently close to 0, the error is approximately equal to the first left out term

in this case, the $ \frac {x^7}{42}$ term because the x^6 term is 0.


In a triangle $ ABC$ $ AD,BE,CF$ are concurrent .also let $ P \in EF, R \in ED, Q \in FD$ such that $ DP,FQ,ER$ are concurrent .Prove that $ AP,BQ,CR$ are concurrent.

Check out the Cevian Nests theorem in my post at http://www.mathlinks.ro/Forum/viewtopic.php?t=3536 .

  darij


geometry
power of a point
radical axis
geometry proposed

Let BB' and CC' the altitudes of the acute triangle ABC and H its orthocenter. Prove that the projection of J on the median of the angle <BAC lies on the line segment joininig the midpoints of BC and B'C'.

Define A' similar. M is a midpoint of BC.
Let projection of H on the median is P.
Then $AB'HPC'$, $HPMA'$, $A'B'C'M$ are cyclic.
so radical axis $B'C'$, $BC$, $HP$ concour at one point.

Ups...it was not the median but the bisector of BAC :blush:

For the correct problem, see http://www.mathlinks.ro/Forum/viewtopic.php?t=23523 post #3.

  darij


In $ \triangle ABC$ , $ AB \equal{} 13$, $ AC \equal{} 5$, and $ BC \equal{} 12$. Points $ M$ and $ N$ lie on $ \overline{AC}$ and $ \overline{BC}$, respectively, with $ CM \equal{} CN \equal{} 4$. Points $ J$ and $ K$ are on $ \overline{AB}$ so that $ \overline{MJ}$ and $ \overline{NK}$ are perpendicular to $ \overline{AB}$. What is the area of pentagon $ CMJKN$?
[asy]unitsize(5mm);
defaultpen(linewidth(.8pt)+fontsize(10pt));

pair C=(0,0), B=(12,0), A=(0,5), M=(0,4), Np=(4,0);
pair K=foot(Np,A,B), J=foot(M,A,B);

draw(A--B--C--cycle);
draw(M--J);
draw(Np--K);

label("$C$",C,SW);
label("$A$",A,NW);
label("$B$",B,SE);
label("$N$",Np,S);
label("$M$",M,W);
label("$J$",J,NE);
label("$K$",K,NE);[/asy]$ \textbf{(A)}\ 15 \qquad
\textbf{(B)}\ \frac{81}{5} \qquad
\textbf{(C)}\ \frac{205}{12} \qquad
\textbf{(D)}\ \frac{240}{13} \qquad
\textbf{(E)}\ 20$

[hide]

We drop an altitude from $ C$ to $ AB$. Let the foot of the altitude be $ H$. Hence, $ \triangle{AJM}\sim\triangle{AHC}$, so $ MJ/CH\equal{}\frac{MJ}{60/13}\equal{}1/5$. Hence, $ MJ\equal{}\frac{12}{13}$. By the Pythagorean Theorem, $ AJ\equal{}\frac{5}{13}$. By the same similarity reasoning, we have $ NK\equal{}\frac{40}{13}$, and with the Pythagorean Theorem, $ BK\equal{}\frac{96}{13}$. From these side lengths, $ [NKB]\equal{}1920/169$ and $ [MJA]\equal{}30/169$. Finally,

$ [CMJKN]\equal{}[ABC]\minus{}([NKB]\plus{}[MJA])\equal{}30\minus{}(\frac{30}{169}\plus{}\frac{1920}{169})\equal{}\frac{240}{13}\Longrightarrow\mathrm{ (D) \ }$

[/hide]


For any positive integer $ n$, let $ f(n)$ be the sum of the digits in the odd digits of $ n$, and let $ g(n)$ be that of the digits in the even digits of $ n$. (We count the digits from right to left.)

Let $ h(n) \equal{} f(n) \minus{} g(n)$

Let $ A \equal{} \begin{bmatrix} a \plus{} c \plus{} 2b & b & c \plus{} 2b \\
c \plus{} b \plus{} 2a & a & b \plus{} 2a \\
c \minus{} a & c \minus{} a & 2c \minus{} a \minus{} b \end{bmatrix}$

Let $ N$ be the number of triples $ (a,b,c)$ of integers such that $ \minus{} 2007\leq a < b < c\leq 2007$ and $ \det (A) > 0$.

Find $ h(N)$

[hide="Brawler Style Start"]$ \text{det} \left( \begin{bmatrix} a \plus{} c \plus{} 2b & b & c \plus{} 2b \\ c \plus{} b \plus{} 2a & a & b \plus{} 2a \\ c \minus{} a & c \minus{} a & 2c \minus{} a \minus{} b \end{bmatrix} \right)$

$ \equal{} \text{det} \left( \begin{bmatrix} a \plus{} b\plus{}c & b & c \plus{} 2b \\ a\plus{} b\plus{}c & a & b \plus{} 2a \\ 0 & c \minus{} a & 2c \minus{} a \minus{} b \end{bmatrix} \right)$

$ \equal{} \text{det} \left( \begin{bmatrix} a \plus{} b\plus{}c & b & c \plus{} 2b \\ 0 & a\minus{}b & 2a\minus{}b\minus{}c \\ 0 & c \minus{} a & 2c \minus{} a \minus{} b \end{bmatrix} \right)$

$ \equal{} (a\plus{}b\plus{}c) \text{det} \left( \begin{bmatrix} a\minus{}b & 2a\minus{}b\minus{}c \\ c \minus{} a & 2c \minus{} a \minus{} b \end{bmatrix} \right)$

$ \equal{} (a\plus{}b\plus{}c)((a\minus{}b)(2c\minus{}a\minus{}b) \minus{} (2a\minus{}b\minus{}c)(c\minus{}a))$ $ \equal{} (a\plus{}b\plus{}c)((a\plus{}b\plus{}c)^2 \minus{} 3(ab\plus{}bc\plus{}ca))$ $ \equal{} (a\plus{}b\plus{}c)^3 \minus{} 3(a\plus{}b\plus{}c)(ab\plus{}bc\plus{}ca)$

Thus, $ (a\plus{}b\plus{}c)^3 > 3(a\plus{}b\plus{}c)(ab\plus{}bc\plus{}ca)$[/hide]


algebra
polynomial
integration
calculus
function
domain
real analysis

I wonder if the following statement is true or false.
Let $n\geq 1$ be an integer, $P(x_{1},x_{2},...,x_{n})$ and $Q(x_{1},x_{2},...,x_{n})$ be two real polynomials. For example, if $n=3$, $P(x,y,z)=x^{2}y+z^{3}+xz+1$ is such a polynomial. Denote by $d(P)$ the degree of $P$, i.e. the highest power of $P$. In the example above, $d(P)=3$, or if $Q(x,y,z)=x^{2}+yz^{2}$, $d(Q)=3$.
[b]Statement :[/b] $\frac{P(x_{1},x_{2},...,x_{n})}{Q(x_{1},x_{2},...,x_{n})}$ is (L-?)integrable on ${[0,1]}^{n}$ if and only if $d(Q)-d(P)\leq n$.

It's true for $n=1$, it seems true for $n=2$...
Can someone correct this, complete this, prove this or give a counterexample ?

Counterexample, consider the case when the two polynomials are polynomials in two variables.

$\int_{0}^{1}\int_{0}^{1}\frac{dxdy}{x^{np}}=\int_{0}^{1}\frac{dx}{x^{np}}=\infty$
hence $P/Q=\frac{1}{x^{n}}$ is not $L^{p}$ integrable for all $p\geq 1$.
$degree(P)-degree(Q)=0-n=-n<2$.

The same works with $P=y$ $Q=yx^{n}$ where $n$ is a natural number.

I'm sorry, it should be d(Q)-d(P) and not d(P) - d(Q). I've edited and thanks for interest

It is still false, since if $P=1$ and $Q=x$ then
$deg(Q)-deg(P)=1<2$ and the double integral is $\infty$ hence the function $P/Q$ is not $L^{p}$ integrable for all $p>1$.

You want the one in the denominator to have no zeros in $R^{n}$ either. Anyone have counterexamples now? :)

Xevarion,
I do not understand what you are saying, could you write down the statement of the problem?

Okay so, I didn't verify this carefully, but here is a modified version of what OP was asking:

Consider $f(x) = \frac{P(x_{1}, \ldots, x_{n})}{Q(x_{1}, \ldots, x_{n})}$

Assume that

1) The degree of $P$ in variable $x_{i}$ is at least 2 smaller than the degree of $Q$ in the variable $x_{i}$ for all $i$.

2) If $Q(y_{1}, \ldots, y_{n}) = 0$ for some point $(y_{1}, \ldots, y_{n})$, then $P(y_{1}, \ldots, y_{n}) = 0$ also and the degree of the root of $P$ is at least as large as the degree of the root of $Q$. In other words, $f$ has no singularities in the domain of integration.

Then $f(x)$ is in $L^{p}$ all $p$ on all of $R^{n}$.


With $ a,b,c$ are three lenghs of sides of a triangle such that $ abc\equal{}2$ Prove that
  $ \sqrt{a\plus{}b\minus{}c}\plus{} \sqrt{a\plus{}c\minus{}b}\plus{} \sqrt{b\plus{}c\minus{}a} \geq 2 \sqrt{2}$

Substitute $ a\equal{}x\plus{}y$ ,$ b\equal{}y\plus{}z$ and $ c\equal{}z\plus{}x$.
The inequality becomes $ \sqrt{2x} \plus{}\sqrt{2y}\plus{}\sqrt{2z} \geq 2\sqrt{2}$ $ \implies$ $ \sqrt{x} \plus{}\sqrt{y}\plus{}\sqrt{z} \geq 2$

$ \sqrt{x} \plus{}\sqrt{y} \plus{}\sqrt {z} \geq (x\plus{}y)(y\plus{}z)(z\plus{}x)$

I can not continue  :blush:

[quote="Mathias_DK"][quote="2424"]With $ a,b,c$ are three lenghs of sides of a triangle such that $ abc \equal{} 2$ Prove that
  $ \sqrt {a \plus{} b \minus{} c} \plus{} \sqrt {a \plus{} c \minus{} b} \plus{} \sqrt {b \plus{} c \minus{} a} \geq 2 \sqrt {2}$[/quote]
$ a \equal{} x^2 \plus{} y^2, b \equal{} y^2 \plus{} z^2, c \equal{} z^2 \plus{} x^2$ for some $ x,y,z \ge 0$ since $ a,b,c$ is length of triangles.
Then we should prove:
$ x \plus{} y \plus{} z \ge 2$ when $ (x^2 \plus{} y^2)(y^2 \plus{} z^2)(z^2 \plus{} x^2) \equal{} 2$
Assume that $ x \plus{} y \plus{} z < 2$ when $ (x^2 \plus{} y^2)(y^2 \plus{} z^2)(z^2 \plus{} x^2) \equal{} 2$. Then by increasing $ x,y,z$ we can get $ x \plus{} y \plus{} z \equal{} 2$ and $ (x^2 \plus{} y^2)(y^2 \plus{} z^2)(z^2 \plus{} x^2) > 2$. So it is enough to prove that:
$ (x^2 \plus{} y^2)(y^2 \plus{} z^2)(z^2 \plus{} x^2) \le 2$ when $ x \plus{} y \plus{} z \equal{} 2$.(This is btw. easily solved with the uvw-method)
Wlog assume $ x \le y,z$. Let $ Y \equal{} y \plus{} \frac {x}{2}$ and $ Z \equal{} z \plus{} \frac {x}{2}$. Then $ Z^2 \equal{} z^2 \plus{} xz \plus{} \frac {x^2}{4} \ge z^2 \plus{} zx \ge z^2 \plus{} x^2$ and likewise $ Y^2 \ge y^2 \plus{} x^2$. Also $ Z^2 \plus{} Y^2 \ge z^2 \plus{} y^2$.
So we should only prove:
$ Z^2Y^2(Z^2 \plus{} Y^2) \le 2$ when $ Z \plus{} Y \equal{} 2$. That is:
$ 8Z^2Y^2(Z^2 \plus{} Y^2) \le (Z \plus{} Y)^4 \equal{} Z^4 \plus{} 4Z^3Y \plus{} 6Z^2Y^2 \plus{} 4ZY^3 \plus{} Y^4 \iff$
$ (Z \minus{} Y)^4 \ge 0$. QED :)[/quote]
Or you can prove $ (x^2y^2\plus{}y^2z^2\plus{}z^2x^2)(x^2\plus{}y^2\plus{}z^2) \leq 2$


[quote="e.lopes"]one correction:

the numerator of the fraction in the rio-plate problem isn't this![/quote]
Your solution?

Could anyone solve it for me,please. :blush:

If $ 3|b$, then $ 3|a$ and $ a^b\plus{}b\equal{}3\mod 9$ - not solutions.
For any b, suth that $ 3\not |b$, we can find infinetely many a, suth that $ \frac{a^b\plus{}b}{b^2\plus{}9}\in N$.
For example $ b\equal{}1, a\equal{}10k\minus{}1$, $ b\equal{}2,a\equal{}\pm 4 \plus{}13k$,...

No.The problem from Rio-Plate isn't that (because tmbtw type wrong). :oops:
This is the real problem:
Find all (a,b) $ \in N*$ such that $ \frac{a^b\plus{}b}{ab^2\plus{}9} \in N*$


[quote="Arne"]Find all real solutions of the following equation: \[ (x + 1995)(x + 1997)(x + 1999)(x + 2001) + 16 = 0. \][/quote]

Or, just note $(x+1995)(x+2001)=a, (x+1997)(x+1999)=a+8$, and we have $(a+4)^2=0,$so $a=-4$, and we get ( using second degree function) $x=\sqrt{5}-1998, x=-\sqrt{5}-1998$


Hi have been selected for kvpy 09-10 interview.but am yet to get the interview letter.:(  :(
Should i hav got it by now? or Indian post has lost it??   :mad:


geometry
inequalities
inequalities solved

Here is a nice problem :D

If a,b,c,d are the lenghts of the sides of a convex quadriteral and if S is its area then prove or disprove the inequality:

3*sqrt(S)<=p+(a*b*c*d)^(1/4).   :?

Just an idea the area of a quadrilateral ABCD  in term of a,b,c,d is

S = sqrt( (p-a)(p-b)(p-c)(p-d) - abcd.cos^2(w/2) ) with

w = 1/2(angleB + angle D) = half of oposite angle in the
convex quadrilateral

There is another formula for the area for an quadrilateral ABCD
AB=a,BC=b,CD=c, DA=d

S = 1/2(ab.sinB + cd.sin(D))

Did you try use these formula for your inequality ?

cheers = che 2 rs

[quote="Lagrangia"]If $ a,b,c,d$ are the lenghts of the sides of a convex quadrilateral and if $ F$ is its area, then prove or disprove that

$ 3\sqrt {F}\leq s \plus{} \sqrt [4]{abcd}$,

where $ s$ is the semi-perimeter.[/quote]A counterexample: $ a \equal{} 1,b \equal{} c \equal{} d \equal{} 12\Longrightarrow$

$ 3\sqrt {F_{max}} \equal{} \frac {3\sqrt [4]{76895}}{2} \equal{} 24.978\cdots$;

$ s \plus{} \sqrt [4]{abcd} \equal{} \frac {37}{2} \plus{} 2\sqrt [4]{108} \equal{} 24.947\cdots$.

By the way, we have the following inequality

$ 3\sqrt {3}F\leq s^2 \plus{} \left(3\sqrt {3} \minus{} 4\right)\sqrt {abcd}$,

with equality if and only if the convex quadrangle  is a square


trigonometry
calculus
derivative
function
linear algebra
matrix
topology

Is it true that all the solutions of the equation

$ x''\equal{}1\plus{}2\sin x$

can be defined on the whole real axis?

if the second derivative of a function is defined at a point, wouldn't that imply the function is definined at that point too?

No, that misses the point that this is a differential equation, not a simple statement about derivatives.

As an example, the differential equation $ x'\equal{}x^2$ has no obvious singularities, but nonetheless, other than the zero function, it has no solutions that are defined on the whole real axis.

None of which answers the question here, one way or another.

Yes, it is true. In general, you have the following theorem. If $ F: \mathbb R^n\to\mathbb R^n$ is Lipschitz, then the solutions of the system $ X' \equal{} F(X)$ exist on the entire real line. You second order equation is equivalent to the system $ x_1' \equal{} x_2$, $ x_2' \equal{} 1 \plus{} 2\sin x_1$.

I'm sorry, but what does it actually mean by $ F: \mathbb{R}^n\to\mathbb{R}^n$ is Lipschitz?

A function $ f$ from a metric space $ X$ to another metric space $ y$ is Lipschitz when $ d(f(x),f(y))\le C\cdot d(x,y)$ for some real $ C$ and all $ x,y$. This implies continuity, of course.

In this case, the derivative of $ F$ is $ \begin{bmatrix}0&1\\2\cos x_1&0\end{bmatrix}$. The 2-norm of this matrix is at most 2, so $ |F(x)\minus{}F(y)|\le 2|x\minus{}y|$ in the Euclidean metric.
We could use other norms in $ \mathbb{R}^n$- this might result in different Lipschitz constants, but it wouldn't change whether the functions are Lipschitz.


How to find $ sup|x^{n}\minus{}x^{n\plus{}1}| (x\in[0,1])$ when $ n\rightarrow\infty$, or how to prove that
 if $ f_{n}(x)\equal{}x^{n}\minus{}x^{n\plus{}1}$ then $ f_{n}$ is uniformly convergent when $ n\rightarrow\infty$?

Consider $ f\left( x \right) \equal{} x^n  \minus{} x^{n \plus{} 1}$. Then
\[ f'\left( x \right) \equal{} nx^{n \minus{} 1}  \minus{} \left( {n \plus{} 1} \right)x^n  \equal{} x^{n \minus{} 1} \left( {n \minus{} \left( {n \plus{} 1} \right)x} \right).
\]
This means that $ f(x)\equal{}0$ occurs if either $ x_1\equal{}0$ or
\[ x_2 \equal{} \frac{n}{{n \plus{} 1}}.
\]
In both cases, $ f$ attains its maximum value at $ x_2$. Of course, it is also its sup.

In the last question, it is known that $ f_n \to f$ uniformly in $ D$ iff
\[ \mathop {\lim }\limits_{n \to \infty } \mathop {\sup }\limits_{x \in D} \left| {f_n \left( x \right) \minus{} f\left( x \right)} \right| \equal{} 0.
\]
We observe that if $ x \in (0,1)$ then $ f_n$ converges to 0 pointwise. This mean that we just claim
\[ \mathop {\lim }\limits_{n \to \infty } \mathop {\sup }\limits_{x \in \left[ {0,1} \right]} \left| {f_n \left( x \right)} \right| \equal{} 0.
\]

if you understand right then this will be good example for you. :)
$ f_{n}(x)\equal{}x^{n}\minus{}x^{2n}$ when  $ 0 \leq x \leq 1$

OK.
It is obvious that $ f_{n}(x)\rightarrow f(x)\equal{}0$ for $ x\in [0,1]$ when $ n\rightarrow\infty$.

Considernig $ f_{n}(x)\equal{}x^{n}\minus{}x^{2n}$ we have $ f'_{n}(x)\equal{}nx^{n\minus{}1}\minus{}2nx^{2n\minus{}1}\equal{}nx^{n\minus{}1}(1\minus{}2x^{n})$.
So $ f'_{n}(x)\equal{}0$ in points $ x\equal{}0$ or $ x_{n}\equal{}\frac{1}{\sqrt[n]{2}}$.
It means that $ f_{n}(x)$ attains its maximum in point $ x_{n}\equal{}\frac{1}{\sqrt[n]{2}}$, and it's also its supremum.
Hence $ \mathop{\sup}\limits_{x\in\left [ {0,1}\right ]}|f_{n}(x)\minus{}f(x)|\equal{}\mathop{\sup}\limits_{x\in\left [ {0,1}\right ]}|f_{n}(x)|\equal{}(\frac{1}{\sqrt[n]{2}})^n\minus{}(\frac{1}{\sqrt[n]{2}})^{2n}\equal{}\frac{1}{4}$.

So $ f_{n}(x)$ is not uniformly convergent.
Is this correct?


Lets consider $ f(x,k,m) \equal{} x(k \minus{} x^m)$  a positive value function, where $ m$ is a positive integer, $ x$ and $ k$ are positive.

A] Find the maximun value $ f$ takes in this conditions, for given $ k$ and $ m$; in other words express the maximun in therms of(the given) $ k$ and $ m$

B] Find the value $ x$ must have, in order to take this maximun, also with given $ k$ and $ m$

C] Find the triple $ (x,k,m)$ wich maximize this maximun, if there is one, if not explain it.

 :)

Edited____
            Sorry I hope it is clear now.

I'm confused. Let $ x \equal{} 1$, then set $ k$ to be an arbitrarily high number. There doesn't seem to be a maximum.

Are $ k$ and $ m$ supposed to be fixed or something?

[hide="Solution to A"]By inspection, $ f \geq 0$ for $ 0 \leq x \leq \sqrt [m]{k}$, and is otherwise negative. Clearly, the negative does not produce a maximum, so we will ignore those cases. Then, $ x$ and $ (k \minus{} x^m)$ are positive terms.

Rewrite as $ f \equal{} \frac {1}{\sqrt [m]{m}} \left( (\sqrt [m]{m} x) (k \minus{} x^m) \right)$.

It suffices to maximize $ (\sqrt [m]{m} x) (k \minus{} x^m)$, and thus to maximize $ (m x^m) (k \minus{} x^m)^m$.

By AM-GM, $ (m x^m) (k \minus{} x^m)^m \leq \left( \frac {mx^m \plus{} m(k \minus{} x^m) }{m \plus{} 1} \right) ^{m \plus{} 1} \equal{} \left( \frac {mk}{m \plus{} 1} \right) ^{m \plus{} 1}$

Replacing this value into the original function:

$ f \leq \frac {1}{\sqrt [m]{m}} \left( \left( \frac {mk}{m \plus{} 1} \right) ^{m \plus{} 1} \right)^{\frac {1}{m}} \iff \boxed{ f \leq \frac {mk}{m \plus{} 1} \sqrt [m]{\frac {k}{m \plus{} 1}}}$

[i]I hope I didn't do anything dumb in there![/i]
[/hide]

[hide="Solution to B"]
The equality case occurs when all terms in AM-GM are equal. We require $ mx^m \equal{} (k\minus{}x^m) \iff (m\plus{}1) x^m \equal{} k \iff x \equal{} \sqrt[m]{\frac{k}{m\plus{}1}}$
[/hide]


inequalities
trigonometry
geometry proposed
geometry

In the ABC triangle    ∠C=90°  and ∠A=α    the following inequality proof.
(4+sin3α-cos3α)/(3√2)>cos(α-45°)

hello, do you mean
$ \frac{(4\plus{}\sin(3\alpha)\minus{}cos(3\alpha))}{3\sqrt{2}}>\cos(\alpha\minus{}\frac{\pi}{4})$?
Sonnhard.