I personally am a fan of all forms of card games, and I recently learned about a new card game, Mao,  which is a parody of Communist China :D . So, I got to wondering how many card games are there. So, please, post any weird card games you may know and a brief discription!

For example,
1)Mao-a card game learned by experience and is a parody of Communist China

I have played Mao, and it is sooo sadistic and fun, but you can only play it with people who've never played it before.

I also like poker, usually we get a group together and do dealer's choice.
Also, there's this new game that our family loves-- Apples to apples.

Yeah, I enjoy playing poker, but never seem to get anything, but Apples to Apples? Hmmm.. Never Heard of it before. What is it about?
I also know:

1)Double 7-Card game where people go around pick a card from the person on their right's hand and eliminating all the pairs and who ever is left with the last joker is eliminated. (Played with one joker obviously  :P )

Apples to Apples is played like this: everyone gets 7 red cards, each one has a person, place, thing or event on them. Then one person is selected to be the judge, who picks a green card which contains an adjective. When the judge lays down the green card, everyone else must choose the red card which is best described by the green card. The judge collects all the red cards (without knowing whose is whose), reads each one aloud, and chooses the card which they think is best described by the green card. Whoever's red card gets chosen gets to keep the green card. Then the next player to the left becomes the judge, everyone draws a new red card, and another round begins.
The object is to collect a certain number of green cards (depending on the number of players.)

Simple but very entertaining. :D

Mafia card game is also very fun.

King is storyteller

Ace of Hearts is Doctor

Queens are mafia

Jacks are werewolves

8's of clubs are inspectors

2's are magicians

3's are masons

anything else is a villager.

That's for if you going on a really long trip...not really any discussion going on random lynching lol.

It's still fun though if you put role claiming in.

[quote="Dreamer"][quote="fireforce"]
For example,
1)Mao-a card game learned by experience and is a parody of Communist China[/quote]

Sorry,are you going to have a further explanation?[/quote]

by all means, NO.

and mafia isn't really a "card game"; it just so happens that cards prove to be useful

unless this is some other variant about which I do not know

[quote="jmerry"][quote="Hexaditidom"]I have played Mao, and it is sooo sadistic and fun, but you can only play it with people who've never played it before.[/quote]
That's not exactly true- the version I learned at MOP is a bit friendlier, and can be played with people who already know the game.

The essence of the game: The winner of each round makes up a rule for all subsequent rounds, and doesn't tell anybody else what that rule is. After enough rounds, the game ends and everybody reveals their rules. There are a few basic rules (which everyone knows) to get the game started, and some metarules to keep things fair.

I suspect the evil version you know starts with a bunch of rules that the newcomers don't know. It's more fun to start small and build up the rules gradually.[/quote]

eh, I learned mine at MOP too and we also started with only the basic rules - it was still enough to confuse the newcomers though (especially the ------ rule) (darn I wanted to say it but then I realized I couldn't)

WOOHOO the really sadistic one is where you can't talk :). But anyways no one said anything about rock paper scissors the collectible card game. Another good card game: throwing the cards.

[quote="probability1.01"]Another good card game: throwing the cards.[/quote]

I would like that except that I'm really bad at it :(

[quote="probability1.01"]Another good card game: throwing the cards.[/quote]

hahaha that's my favorite game!

[b]For the record:
Mao is a game that SHOULD only be learned by experience$\dagger$. For those who play the game know what I mean. For those who do not I suggest you play with someone who has played before.[/b]

$\dagger$ Personally, I have played with the version of no talking with a bad moderator/dealer/judge who half the time would screw us over by tricking us into breaking the rules and getting punished. :diablo: Hopefully u will have a better experience.:jump:

Also, I agree with MysticTerminator ( :o  :o ) in Mafia even though u use cards to select the player u really don't play with the cards soooooooooooo... I don't think it should be considered a card game. What do u think :?:

[quote]hearts: a game wher you either avoid hearts and the Q of spades, or you try to get all of them.[/quote]
Personally, I dislike hearts because (I play on the computer) no matter what I try to do the CPU opponents always anhilate me!

Here is my family's personal favorite:
1)Egyptian Rat Screw (ERS) or Egyptian Rhapsody, Egyptian Rattrap, Slap, Bloodystump, Egyptian Rat Race, or Egyptian War- A awsome crazy slapping card game that basically only requires:
 A)fast slapping hand
 B)rapid reflexes
No matter how bad your luck is, these 2 things will always make you win!!

[quote="fireforce"] [quote]hearts: a game wher you either avoid hearts and the Q of spades, or you try to get all of them.[/quote]
Personally, I dislike hearts because (I play on the computer) no matter what I try to do the CPU opponents always anhilate me!
[/quote]

Haha on computer I just shoot every time because the computers are really bad at stopping people from shooting.

I don't know, when I play and try to shoot, I'm always stopped right near the end, but when the other comp players try to shoot, they ALWAYS seem to succeed unless I stop them...

My friends and I have been obsessed about hearts since the beginning of school and we played it almost everyday so now I'm like a master at that game...or not.

But now we're into Kemps.
2 players are on a team and there are normally 4 or 6 people in a game. The goal of the game is to get 4 of a kind and your partner has to say "kemps". Before you play you and your partner agree on a signal so if you ever get 4 of a kind your partner will know. You start with 4 cards and every round there are 4 cards on the table where you can switch on of your cards for another card. You mucst always have 4 cards in your hand. After no one else wants any of the cards you put 4 new cards out and so on. If you suspect an opposing team has 4 of a kind you can say "stop kemps" and the opposing team loses. Fun game.

Other fun card games are Rummy and Texas Hold'em

I'm surprised that no one mentioned Bridge yet, I think it's a cool game although I almost never play it.

Good ol' Texas hold em poker!  Though, I watch WPT and WSOP more than I actually play it :( (and the game has really been hyped up a lot because of those shows...).

[quote]But now we're into Kemps. [/quote]

I have also heard this game called Camps, Bob, and Kims.
 :ninja:

[quote]Good ol' Texas hold em poker![/quote]
Also, there are multiple variants of Poker, though I too like Texas Hold'em the best.

[quote]Other fun card games are Rummy[/quote]
Rummy is also alot of fun:
1)Rummy-Each person gets either 10 or 7 cards, (depending on the number of people playing, if more then 3 people are playing then 7,). The object of the game is to make one series of 4 cards (i.e. \[ 2\clubsuit\ 3\clubsuit\ 4\clubsuit\ 5\clubsuit\ \] ) and make a series or all of a kind (i.e. jacks, queens, etc.) of 3 cards each. In, the 7 card version it has the same rules except its only one set of 3. Then, the person right of the dealer draws a card and places one card that they do not want on the discard pile. Then the next person goes (moves counterclockwise) and either takes the card on the [b]top[/b] of the discard pile or draws a card feom the deck. They play their unwanted on top of the previous card. When you are done you place your last unwanted card face down, thus winning the game

 :!: [size=200]What card games do you know of that originate from your country?[/size] :!:

[quote="fireforce"][quote]But now we're into Kemps. [/quote]

I have also heard this game called Camps, Bob, and Kims.
 :ninja:

[/quote]

Round here, we call it Peanut butter and Jelly... XD

I forgot what this game was called, but here's how it was played (I learned this at ARML 2005 San Jose).

Each person has their own deck of cards. You sit in a circle (or whatever convex polygon you prefer), and you basically play mass Solitaire.

These aren't all the rules (I don't remember them all), so if anyone knows this game, please tell me the name and any additional rules there are.

My favorite card game is definently Poker, prefebly Hold 'em.


I'm not the only mafia freak here, right?

cause in that case i'm not bringing cards


number theory
prime numbers
number theory proposed

I have the following problem(sorry if it was posted before):
Both grobber and harazi loves Valiowk :D :D  :D  .They find away choose a person who will with draw. They play a below game, the loser will leave without returning.
  They both choose a natural N , being as big as they want. Then grobber secret chooses a natural M>N. After that, each time harazi gives a natural n, grobber must says  the number M+n is a prime number or not. Prove that harazi has a method that after no more than N times asking, he can say the number M. And of course then he win.  :lol:
I hope grobber, harazi and Valiowk aren't angry with me  :P  .
The truth is out there ;)

[quote]I hope grobber, harazi and Valiowk aren't angry with me.[/quote]
I am angry, because I love Valiowk too!  :D

Anyway, I am sure that you mean $M<N$. I will assume it.
Let $p_1$, $p_2$, ..., $p_{N-1}$ are pairwise distinct prime numbers greater than $N$. Define $D_s$ a natural number s.t. $p_k\,|\,D_s+k$ for all $k=1,...,N-1$ except $k=s$ (here we use Chinese residue theorem); $L_s=p_1p_2...p_{N-1}/p_s$. It is clear $\gcd(L_s,D_s+s)=1$.
Due to Dirichlet theorem for each $s\in\{1,2,...,N-1\}$ there is a prime number $q_s=kL_s+D_s+s$, $k\in\mathbb{N}$.
Now harazi's strategy is very simple: he should try $n=q_1-1,\ q_2-2,\ ...,\ q_{N-1}-(N-1)$. It is clear that he will find unique index $r$ s.t. $M+(q_r-r)$ is a prime number. It follows that $M=r$.

I'm angry too  :(  I love Valiowk too but I have not succeeded to solve the problem before Myth... :(  :(  :(
Sad day....

Pierre.

[quote="pbornsztein"]I love Valiowk too[/quote]
Duel!

Hmmm...russians  :maybe:  ;)
A french guy knows waiting....

Pierre.

:blush: I'm flattered. ;)

I think Myth forced both grobber and harazi to withdraw. :D ;)

[quote="Myth"][quote="grobber"]Yup.. He frightened me :) :D[/quote]
Prepare for battle! ;) :D[/quote]

Wow. Valiowk could have a huge harem.  :)  Everything a girl always dreamt of...   ;)

I'm younger, I will just wait... I have all the time... :D . By the way, there is indeed a problem, since grobber is even younger.

In order to seduce Valiowk, I think we may use a more attractive solution than Myth's one  :P  namely :

It is really well-known (without Dirichlet's theorem nor Chinese remainders lemma) that we must find a prime $p>N$ such that each of the numbers $p+k$, where $k \in \{1, \cdots, N \}$, are composite.
Therefore, Pierre's strategy is clear :
He/I first asks for $p-1$. If the answer is 'prime' then $M = 1$. Otherwise, the answer is composite and $M > 1$.
Thus, he/I asks for $p-2$ to test $M=2$, and so on.

The worse case is to ask for $p-(N-2)$ because if the answer is composite too, then $M > N-2$. And since we know that $M<N$ this forces $M=N-1$.
Thus Pierre can find $M$ in at most $N-2$ questions (but so does Myth...).

Pierre.


Has anyone read this book?

For my biology class a couple years ago, it was required reading. I really enjoyed it. It takes 23 chromosomes, breaks them down into characteristics of human, and gives great insight. I thought it was a really fun read.

That year, I also read a fiction thriller by Richard Preston called The Cobra Event. It was a book about the release of biological warfare...a page turner.

Does anyone have any good bio fction or nonfiction books that are good reads? Not textbooks but leisure reading (which may be textbooks again, lol)

hmmmmm ive never heard of those books...i should get into reading stuff about bio thats fiction. itd probably be pretty cool and a break from textbook stuff. lol

[u]The Seven Daughters of Eve[/u], by Bryan Sykes is very good.

Ridley's [i]Red Queen[/i] is also very interesting. . .

Another interesting, bioesque book that you might want to check out is [i]The Selfish Gene[/i] by Richard Dawkins.


ARML geometry calculus integration trigonometry Vieta circumcircle

So..anyone want to try working the problems?

[quote="MCrawford"]
[hide="The Statements"]
1. In right triangle $ABC$, $\angle C = 90^\circ$, $c=3$, and $a+b=\sqrt{17}$.  Compute the area of the triangle.

2. If $(a+b) : (b+c) : (c+a) = 6:7:8$, and $a+b+c=14$, compute the value of $c$.

3. Let $z$ be a root of $z^5-1=0$, with $z \not= 1$.  Compute the value of
\[z^{15} + z^{16} + z^{17} + \cdots + z^{50}. \]

4. A regular polygon of $2n$ sides is inscribed in a circle of radius 5.  The sum of the squares of the distances from any point on the circle to the vertices of the polygon is 600.  Compute the value of the integer $n$.

5. The circles, each of radius 5, lie in perpendicular planes.  They intersect in two points, $X$ and $Y$, thus determining a common chord $\overline{XY}$.  If $XY = 8$, compute the distance between the centers of the circles.

6. How many positive integers less than 54 are equal to the product of their proper divisors?

7. Find all positive integers $n$, less than 17, for which $n! + (n+1)! + (n+2)!$ is an integral multiple of 49.

8. In triangle $ABC$, $a \ge b \ge c$.  If $\displaystyle \frac{a^3+b^3+c^3}{\sin^3{A} + \sin^3{B} + \sin^3{C}} = 7$, compute the maximum possible value of $a$.[/hide][/quote]

Is there some moderator here that can edit the post. It seems like the only problem is that there should be a space between each parenthesis and the colon.

On the other hand, here you have some suggestions:

[hide="Some Suggestions"]1. Pithagoras, of course

2. $c$ is the only integer in the solution of the equations.

3. $z^5-1=0\Rightarrow\left\{\begin{array}{l}z^{5k}=1\forall k\in\mathbb{N}\\z^4+z^3+z^2+z+1=0\end{array}\right.$

6. Do you know what does square free means.

7. Wilson's theorem (well used) gives you the answer almost inmediately.

8. Did you know that $\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{b}=\frac{c}{d}=\frac{\alpha a+\beta c}{\alpha b+\beta d}$ for any real numbers $\alpha$ and $\beta$ such that non of the ractions is undefined?
[/hide]

Let's see if you can solve them.

Best,

I'm sorry, I'm not too sure about what you want me to do?

I took care of it.

wow lol, i actually worked some of these out and got answers...i was gonna ask if anyone had the answers to these, but since no one but me actually pays attention to those probs, nvm lol. well, if u happen to be readin this and know the answers, please tell me so i can see if mine r right. b/c some of them im not too sure, b/c i did some weird stuff, but thats what makes math competitions fun, the weird stuff

[hide="1"]\begin{eqnarray*}c^2=a^2+b^2&=&9\\ a+b&=&\sqrt{17}\\ a^2+2ab+b^2&=&17\\ 2ab+9&=&17\\ \frac{1}{2}ab&=&2\\ &=&[ABC]\end{eqnarray*}[/hide]
[hide="2"]\begin{eqnarray*}\frac{14-c}{14-a}&=&\frac{6}{7}\\ 14+6a&=&7c\\ \frac{14-c}{14-b}&=&\frac{3}{4}\\ 14+3b&=&4c\\ \frac{7c-14}{6}+\frac{4c-14}{3}+c&=&14\\ \frac{21c-42}{6}&=&14\\ c&=&6\\\end{eqnarray*}
We can verify subsequently that $a=\frac{14}{3}$ and $b=\frac{10}{3}$.

Not exactly elegant, but...[/hide]
[hide="3"]Since $z$ is a fifth root of unity, this can be rewritten as $7(z^5+z^1+z^2+z^3+z^4)+z^5$, or seven times the sum of the roots of $x^5-1$, plus any one of these roots to the fifth, which is, of course, 1.  We use Vieta's formulas to see that the sum of the roots must be zero.  Therefore the answer is $7(0)+1=1$.[/hide]

I have to go now, so I'll work on the others later.

#5

[hide]We draw out the circles, and draw radii meeting the two points of intersection for both circles. Let's say that the circles lie on the xy and xz planes. An isoceles triangle is formed on the xz circle, and the distance from the center of the circle to the chord formed is $3$, because of Pythagorous's Theorem. Likewise, the distance is $3$ for the xy circle due to symmetry. Thus, the distance from the centers is $\boxed{3 \sqrt{2}}$.[/hide]

#6

[hide]This is completely a wild guess... but here it goes :)

All numbers in the form $p^3$ or $p_1 p_2$ work where $p$ is a prime number. For the first case, the proper divisors of $p^3$ are $1, p, p^2$, which multiply to equal $p^3$. For $p_1 p_2$, the only proper divisors are $1, p_1, p_2$, which multiply to $p_1 p_2$. This doesn't work with any other combination, as $p^4$ would yield $p p^2 p^3$ which doesn't equal $p^4$. So obviously, no value $p^x$ where $x > 3$ works. This doesn't work with $p^2$ either, as the only proper divisors are $1, p$, which don't multiply to $p^2$. For the second case, $p_1 p_2 p_3$'s proper divisors are $p_2, p_2, p_3, p_1p_2, etc.$, which obviously don't multiply out to $p_1 p_2 p_3$. Thus, our only answers are:

$8, 27$, for the first case
$6, 14, 22, 26, 34, 38, 46, 21, 11, 13, 17, 35$

So, we have $\boxed{14}$ numbers.
[/hide]

#7

[hide]We have:

$n! + (n+1)n! + (n+1)(n+2)n! =$
$n!(1 + n + 1 + (n+1)(n+2)) =$
$n!((n+2)^2)$

So, either $(n+2)^2$ has to be divisible by 49, or $n!$ must be. By the first case, $n = 7$. By $n!$, $n = 14, 15, 16$, because $14! = 1 * 2 * ... * 7 * 8 * ... * 13 * 14$, which has two factors of 49: 7 and 14. Likewise, since $15! = 15 * 14!$ and since $16! = 16 * 15 * 14!$, 15 and 16 are solutions as well.

So, the answer is $\boxed{n = 7, 14, 15, 16}$.[/hide]

#8
[hide]
Since $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R$ by the extended Law of Sines, we replace, getting

$\frac{a^3 + b^3 + c^3}{\frac{a^3}{8R^3}+\frac{b^3}{8R^3}+\frac{c^3}{8R^3}} = 7$.

Simplyfying, we have

$8R^3 = 7$, or $R = \frac{7^{1/3}}{8}$. The maximum value of a side given the circumradius is $2R$, when the triangle is inscribed in a circle and the hypoteneuse is the circumradius. Thus, our answer is $\boxed{2*\frac{7^{1/3}}{8}}$.[/hide]


geometry
trapezoid
perpendicular bisector
similar triangles

[quote]
The distance $AB$ is $12$. The circle center $A$ radius $8$ and the circle center $B$ radius $6$ meet at $P$ (and another point). A line through $P$ meets the circles again at $Q$ and $R$ (with $Q$ on the larger circle), so that $QP = PR$. Find $QP^{2}$
[img]http://www.kalva.demon.co.uk/aime/soln/diag8314.jpg[/img]
[/quote]

So basicly, what I do is draw an altitude from A and B to QR, such that the one from A bisects QP and the one from B bisects PR. I call the altitude from A b, and the altitude from B, c. Line QR is now broken up into 4 parts which I'll call $a$.
I then draw a line from A to Q, and from B to R. And also a line parallel to QR passing through B at ending at the perpendicular bisector from A.

This gives 3 triangles and 3 variables to consider

$8^{2}= b^{2}+a^{2}$
$6^{2}= c^{2}+a^{2}$
$12^{2}= (2a)^{2}+(b-c)^{2}= 4a^{2}+b^{2}-2bc+c^{2}$

Now I should be able to solve for a... but I keep getting really ugly equations I cant simplify due to the $(b-c)^{2}$.. isolating for b and c gives root values and then they dont square very neatly.. >_<

I just end up getting
$2a^{2}= 44+2bc$

and if we solve for b or c we get
$b = \sqrt{64-a^{2}}$
$c = \sqrt{36-a^{2}}$

am I not seeing something right?

:rotfl: eghgh im slow sometimes

thanks chickendude :D

[hide="Just another solution... "]
$\begin{array}{c}8^{2}= b^{2}+a^{2}\\ 6^{2}= c^{2}+a^{2}\end{array}\}\Rightarrow b^{2}-c^{2}= 28 \ \ (1)$

Let $M$ be the midpoint of $AB$

$AA'=b, \ BB'=c$

[img]8541[/img]
From the trapezoid $AA'B'B$ we get that $PM\parallel a \parallel b \Rightarrow PM\perp PR$

So, $MR$ is a diameter of the second circle, or else , $R \in AB$.
Then from the similar triangles $RAA', RBB'$ we get $b=3c \Rightarrow b^{2}= 9c^{2}$.

Now from $(1)\Rightarrow 8c^{2}=28 \Rightarrow c^{2}= \frac{7}{2}\Rightarrow$
$a^{2}= 6^{2}-c^{2}= \frac{65}{2}\Rightarrow (2a)^{2}= 130$[/hide]

I like that solution, but stating that $PM\parallel a \parallel b \Rightarrow PM\perp PR$ seems kind of pointless because it is never used, but none the less it seems a much simpler solution.

Also, what program did you use to draw that diagram?


floor function
number theory unsolved
number theory

Find all integer number n such that $S(9^{n})=9$
where S(x) is sum of digit of in base 10.

$9^{n}= (10-1)^{n}= 10^{n}-{n\choose1}10^{n-1}+\cdots+(-1)^{n}$

So, $S(9^{n}) = \sum_{i=0}^{n}{(-1^{i}){n\choose{i}}}+10\left\lfloor\frac{n+1}{2}\right\rfloor$

That last term accounts for the fact that $9^{2}$ = 81 and not
-20+1...

Anyway, when n is odd and greater than 1, the first sum is 0.

When 4|n, the sum is also 0.

ARGH WHAT DID I DO WRONG

I have confused myself.

However, the answers are 1 and 2...

See [url=https://math.stackexchange.com/questions/1911009/digit-sum-of-9n]here[/url] for an "elementary" proof.


integration
real analysis
real analysis theorems

We know that for 0<f1<f2<...<fn,and fn->f,n->00 that
int_A fn dm = int_A f dm (LMC)
and also
for sets A1 subset A2 subset An .... and An->A
lim int_An f dm = int_A f dm (LMC)

Can I state that lim int_An fn dm = int_A f dm?
My claim is yes and I prove it as
lim_m lim_n int_Am fn dm = lim_m int_Am f dm = int_A f dm
so since lim_m lim_n (int_Am fn dm) exists it must be equal to lim_m lim_n for m=n
(because this is a subcase when m=n ie convergence in the line m=n)
hence equal to lim_n int_An fn dm
thus lim_n int_An fn dm

Do you find any mistakes,insufficiencies in that?
Thnx again..

I think you are correct.

In fact,

$\int_{A_n} f_n dm = \int_A f_n\cdot\chi_{A_n} dm$

and $\{f_n\cdot\chi_{A_n}\}$ is monotonically increasing.


The largest whole number such that seven times the number is less than 100 is

A. 12
B. 13
C. 14
D. 15
E. 16

[hide]
$\\7\cdot 15=105\\7\cdot 14=98$
So the answer is $14\implies\boxed{C}$.
[/hide]

Masoud Zargar

[hide]
15*7=105
14*7=98
So C is the answer.[/hide]

[hide="faster"]100/7 give you 14.blahblahblah so the answer is 14[/hide]

[hide]
$7 \cdot 13=91$
$7 \cdot 14=98$
98 =7 is over 100 so the anwer is $14$[/hide]

[quote="apusey"]sometimes guess and check is good but on this problem it requires a calculator to do it fast[/quote]

You require a calc to do 100/7???  :o  :o

i think he meant to do it fast....
i bet he can do 100/7 in 10 seconds but with a calc, it would take him 5.


Can someone post a lot of algebra problems???
I really need some practice.

what kind of problems.... like a+3=5? lol... well ur posting it in classroom.. but i havent really gotten the jist of whats 'too easy' and whats 'too hard' for classroom.. post an example of a prob ur looking for.

Maybe some factoring, systems of equations, quadratic equations, graphs.  But i want it to be difficult.

doesn't belong in this forum...

and I have a post about a request for some quadratic equation problems in Basics Forum...

http://artofproblemsolving.com/Forum/viewtopic.php?t=42121

can someone move this to where this belongs?

[quote="IntrepidMath"][quote="math92"]Thanks guys!!!!![/quote]
Although what you really should do is work through the AoPS book's algebra. There are some very nice problems. You could also take intermediate algebra.[/quote]
yeah, i'm doing that too.

i don't care...he can do it if he wants...

it's not too easy.  In fact it's a little difficult.(for me)

c'mon you can get it... dont go deep.. the answer lies in the shallow.

[quote="riddler"]Solve for the values of $i$,$j$,$k$
$(LT^{-1})^i(L^3T^{-2}M^{-1})^j(ML^2T^{-1})^k=T$
Sorry if this is too easy[/quote]

Well, let's see...

[hide="Answer"]$(L^iT^{-i})(L^{3j}T^{-2j}M^{-j})(M^kL^{2k}T^{-k})=T$
$L^i\cdot L^{3j}\cdot L^{2k}=L^0 \Rightarrow i+3j+2k=0$
$M^{-j}\cdot M^k=M^0 \Rightarrow -j+k=0 \Rightarrow j=k$
$i+3j+2k=0 \Rightarrow i+5k=0$
$T^{-i}\cdot T^{-2j}\cdot T^{-k}=T^1 \Rightarrow -i-2j-k=1 \Rightarrow i+3k=-1$
$(i+3k=-1)-(i+5k=0)=(-2k=-1) \Rightarrow k=\frac{1}{2}$
$i+\frac{5}{2}=0 \Rightarrow i=-\frac{5}{2}$
$(i,j,k)=\displaystyle\left(-\frac{5}{2}, \frac{1}{2}, \frac{1}{2}\right)$[/hide]

Hmmm... I'll get some harder problems for the equation solviong contest rounds so math92 can practice  :)


A right triangle has legs with lengths of 5 units and 12 units. How many units long is the hypotenuse?

You want to fix what you wrote up? It looks like a mess to me.

[quote="BOOYA"][hide]umm.... use the Pythagorean Theorem.

a2 + b2 = c2
when it is a right triangle and c is the hypotenuse.
52 + 122 = 132[/hide][/quote]
what????
i got [hide]13, by using the pythagoran threom which is
$a^2+b^2=c^2$
just plug in the numbers and you get 13 :) [/hide]

[hide]$c=\sqrt (a^2+b^2)$, where $c$ is the Hypotenuse, and $a$ and $b$ are the legs of the Right triangle.

Therefore, $c=\sqrt (5^2+12^2)=\sqrt 169=13$[/hide]

a^2 + b^2 = c^2

Sorry, it got messed up, i used < s u p > 2 < s u p > (as in superscript)
I don't know any other way of using "square" or "to the power of"

put a^2+b^2=c^2 within $, and you'll get exponents

[hide]
Use Mr. Pythagorean to get:
$5^2+12^2=x^2$
There is a pythagorean triple, so x=$13$

[/hide]

[hide]If you memorize Pythagoreon Triples, then you can easily get that the hypotenuse is 13[/hide]


$ \frac{11!}{9! \plus{} 2\cdot 8!} \implies \frac{9\cdot 10\cdot 11}{9 \plus{} 2} \implies 9 \cdot 10 \implies \boxed{90}$


What is the difference between the largest and the smallest values of $ x$ satisfying the following system of inequalities?
Express your answer as a mixed number.

\begin{eqnarray*} 2x - 6 \le 6 \\
4x + 3 \ge 13  \end{eqnarray*}

[hide]

Just solve the inequalities.

so x is less than or equal to 6

so x is greater than or equal to 2.5

So 2.5 is smallest 6 is largest.

6-2.5=3.5 As a mixed number, 3 1/2[/hide]


Find the pH of a saturated solution of zinc hydroxide ($ K_s \equal{} 3.0 \times 10^{\minus{}23}$).

$ K_s \equal{}x \times (2x)^2$ where $ x\equal{}$Molar conc. of $ Zn(OH)_2$ that is dissolved
That gives $ x\equal{} \sqrt[3]{\frac{30}{4}}\times{10^{\minus{}8}}$

$ [OH]^\minus{}$(due to dissolution)$ \equal{} 2\times x$

What's that??
$ [OH]^\minus{}$ is very less....

By the way $ K_s$ is solubility product??
Then the value is not correct

check http://en.wikipedia.org/wiki/Zinc_hydroxide :blush:  :maybe:

It doesnt appear that you have made right approach as dissociation of water is another variable.
let x=Zn2+ and y=H+.
then from charge conservation 2x+y=OH-
thus x(2x+y)=Ksp
and y(2x+y)=Kw
thus (2x+y)2=2Ksp+Kw is roughly Kw as ksp is so small
thus ph=7
if someone is very particular ph is 7+1.3*10^-9

Savy using data from wiki ph= 7.33
Come on Even Class Xi ncert has this sort of trash.


Is it true that any afirmation about something that does not exist is considered to be true? For example: the afirmation "The lions which are not mammals live in Africa" is considered to be true?

to my knowlege, it is definitely considered true.
basically, you are saying, "is it true that the null set obeys every property?" which I think is true, the same way that zero is even (multiple of two) therefore multiple of all numbers.

1) The sentence of the lions is logically true. You have to realize that real-life examples like this sound strange, because we transport more information with speech than just the logical content. ( As in: Why are you talking about lions which are not mammals if you know that they don't exist?)

2) The null set does not have every property. Elements of the null set fulfill every property (as you will not be able to provide a counterexample).

3) Zero is even by definition, not by some counterintuitive logical construction. This has nothing to do with the lions sentence. $2\cdot 0$ is a straightforward multiple of 2, therefore even.


[quote="sharkman"]Give the rectangular equation of r=2+sin(theta)[/quote]
[hide]Just substitute: (Remember that $r= \sqrt{x^{2}+y^{2}}$, $x=r \cos \theta$, and $y=r \sin \theta$.)

$r=2+\sin \theta$
$r^{2}= 2r+r \sin \theta$
$x^{2}+y^{2}=2 \sqrt{x^{2}+y^{2}}+y$

You can simplify or solve how you like.[/hide]


Okay this is top country on AoPS, I know, it looks like it will last very long.
In this game hurt= -2 and heal=+1
U.S.: 31
Canada: 30
Mexico: 30
Brazil: 30
Argentina: 30
Chile: 30
Venezuela: 30
Peru: 30
Cuba: 30
Bolivia: 30
Columbia: 30
Uruguay: 30
Ecuador: 30
Finland: 30
Iceland: 30
Greenland: 30
Denmark: 30
Lithuania: 30
U.K.: 30
Ireland: 30
Germany: 30
Poland: 30
Belgium: 30
France: 30
Slovakia: 30
Hungary: 30
Turkey: 30
Albania: 30
Afghanistan: 30
Armenia: 30
Azerbaijan: 30
Bangladesh: 30
China: 30
Georgia: 30
Hong Kong: 30
India: 30
Indonesia: 30
Iran: 30
Iraq: 30
Israel: 30
Japan: 30
Jordan: 30
Korea: 30
Mongolia: 30
Russia: 30
Saudi Arabia: 30
Thailand: 30
Taiwan: 30
Vietnam: 30
Ukraine: 30
Algeria: 28
Nigeria: 30
Chad: 30
Egypt: 30
Sudan: 30
Ethiopia: 30
Luxembourg:30
New Zealand: 30
South Africa: 30
Madagascar: 30
Spain: 30
Austria: 30
Australia: 30
Antarctica: 30
Jamaica: 30

I don't have anything against Algeria, but I jsut randomly chose to take away from it.

P.S. - there are 65 countries so we better start now.

Wickedestjr
Poincare Conjecture

Offline
Joined: 06 Apr 2008
Posts: 179
Location: Antarctica

#1top country on AoPS
click on me

--------------------------------------------------------------------------------
Okay this is top country on AoPS, I know, it looks like it will last very long.

U.S.: 31
Canada: 30
Mexico: 30
Brazil: 30
Argentina: 30
Chile: 30
Venezuela: 30
Peru: 30
Cuba: 30
Bolivia: 30
Columbia: 30
Uruguay: 30
Ecuador: 30
Finland: 30
Iceland: 30
Greenland: 30
Denmark: 30
Lithuania: 30
U.K.: 30
Ireland: 30
Germany: 30
Poland: 30
Belgium: 30
France: 30
Slovakia: 30
Hungary: 30
Turkey: 30
Albania: 30
Afghanistan: 30
Armenia: 30
Azerbaijan: 30
Bangladesh: 30
China: 30
Georgia: 30
Hong Kong: 30
India: 30
Indonesia: 30
Iran: 30
Iraq: 30
Israel: 30
Japan: 30
Jordan: 30
Korea: 30
Mongolia: 30
Russia: 30
Saudi Arabia: 30
Thailand: 30
Taiwan: 30
Vietnam: 30
Ukraine: 30
Algeria: 28
Nigeria: 30
Chad: 30
Egypt: 30
Sudan: 29
Ethiopia: 30
Luxembourg:30
New Zealand: 31
South Africa: 30
Madagascar: 30
Spain: 30
Austria: 30
Australia: 30
Antarctica: 30
Jamaica: 30

I heard New Zealand is really cool.
Never heard of Sudan.

U.S.: 31
Canada: 30
Mexico: 30
Brazil: 30
Argentina: 30
Chile: 30
Venezuela: 30
Peru: 30
Cuba: 30
Bolivia: 30
Columbia: 30
Uruguay: 28
Ecuador: 30
Finland: 30
Iceland: 30
Greenland: 30
Denmark: 30
Lithuania: 30
U.K.: 30
Ireland: 31
Germany: 30
Poland: 30
Belgium: 30
France: 30
Slovakia: 30
Hungary: 30
Turkey: 30
Albania: 30
Afghanistan: 30
Armenia: 30
Azerbaijan: 30
Bangladesh: 30
China: 30
Georgia: 30
Hong Kong: 30
India: 30
Indonesia: 30
Iran: 30
Iraq: 30
Israel: 30
Japan: 30
Jordan: 30
Korea: 30
Mongolia: 30
Russia: 30
Saudi Arabia: 30
Thailand: 30
Taiwan: 30
Vietnam: 30
Ukraine: 30
Algeria: 28
Nigeria: 30
Chad: 30
Egypt: 30
Sudan: 29
Ethiopia: 30
Luxembourg:30
New Zealand: 31
South Africa: 30
Madagascar: 30
Spain: 30
Austria: 30
Australia: 30
Antarctica: 30
Jamaica: 30
Italy: 30

There's actually 66 countries in the list.

U.S.: 31
Canada: 30
Mexico: 30
Brazil: 30
Argentina: 30
Chile: 30
Venezuela: 30
Peru: 30
Cuba: 30
Bolivia: 30
Columbia: 30
Uruguay: 28
Ecuador: 30
Finland: 30
Iceland: 30
Greenland: 30
Denmark: 30
Lithuania: 30
U.K.: 30
Ireland: 31
Germany: 30
Poland: 30
Belgium: 30
France: 28
Slovakia: 30
Hungary: 30
Turkey: 30
Albania: 30
Afghanistan: 30
Armenia: 30
Azerbaijan: 30
Bangladesh: 30
China: 30
Georgia: 30
Hong Kong: 30
India: 30
Indonesia: 30
Iran: 30
Iraq: 30
Israel: 30
Japan: 30
Jordan: 30
Korea: 30
Mongolia: 30
Russia: 31
Saudi Arabia: 30
Thailand: 30
Taiwan: 30
Vietnam: 30
Ukraine: 30
Algeria: 28
Nigeria: 30
Chad: 30
Egypt: 30
Sudan: 29
Ethiopia: 30
Luxembourg:30
New Zealand: 31
South Africa: 30
Madagascar: 30
Spain: 30
Austria: 30
Australia: 30
Antarctica: 30
Jamaica: 30
Italy: 30

My best friend is Russian, and my enemy is french.

#4U.S.: 31
Canada: 30
Mexico: 30
Brazil: 30
Argentina: 30
Chile: 30
Venezuela: 30
Peru: 30
Cuba: 30
Bolivia: 30
Columbia: 30
Uruguay: 28
Ecuador: 30
Finland: 30
Iceland: 30
Greenland: 30
Denmark: 30
Lithuania: 30
U.K.: 30
Ireland: 31
Germany: 30
Poland: 30
Belgium: 30
France: 28
Slovakia: 30
Hungary: 30
Turkey: 30
Albania: 30
Afghanistan: 30
Armenia: 30
Azerbaijan: 30
Bangladesh: 30
China: 30
Georgia: 30
Hong Kong: 30
India: 30
Indonesia: 30
Iran: 30
Iraq: 30
Israel: 30
Japan: 31
Jordan: 30
Korea: 30
Mongolia: 30
Russia: 31
Saudi Arabia: 30
Thailand: 30
Taiwan: 30
Vietnam: 28
Ukraine: 30
Algeria: 28
Nigeria: 30
Chad: 30
Egypt: 30
Sudan: 29
Ethiopia: 30
Luxembourg:30
New Zealand: 31
South Africa: 30
Madagascar: 30
Spain: 30
Austria: 30
Australia: 30
Antarctica: 30
Jamaica: 30
Italy: 30

Waaaaaaaaaaaaaaaaaaaaay... too many countries

U.S.: 31
Canada: 30
Mexico: 29
Brazil: 30
Argentina: 30
Chile: 30
Venezuela: 30
Peru: 30
Cuba: 30
Bolivia: 30
Columbia: 30
Uruguay: 28
Ecuador: 30
Finland: 30
Iceland: 30
Greenland: 30
Denmark: 30
Lithuania: 30
U.K.: 30
Ireland: 31
Germany: 30
Poland: 30
Belgium: 30
France: 28
Slovakia: 30
Hungary: 30
Turkey: 30
Albania: 30
Afghanistan: 30
Armenia: 30
Azerbaijan: 30
Bangladesh: 30
China: 30
Georgia: 30
Hong Kong: 30
India: 31
Indonesia: 30
Iran: 30
Iraq: 30
Israel: 30
Japan: 31
Jordan: 30
Korea: 30
Mongolia: 30
Russia: 31
Saudi Arabia: 30
Thailand: 30
Taiwan: 30
Vietnam: 28
Ukraine: 30
Algeria: 28
Nigeria: 30
Chad: 30
Egypt: 30
Sudan: 29
Ethiopia: 30
Luxembourg:30
New Zealand: 31
South Africa: 30
Madagascar: 30
Spain: 30
Austria: 30
Australia: 30
Antarctica: 30
Jamaica: 30
Italy: 30[/quote]

Idea: HIDE the countries! :D

[hide="Like this! :D"]U.S.: 32
Canada: 30
Mexico: 29
Brazil: 30
Argentina: 30
Chile: 30
Venezuela: 30
Peru: 30
Cuba: 30
Bolivia: 30
Columbia: 30
Uruguay: 28
Ecuador: 30
Finland: 30
Iceland: 30
Greenland: 30
Denmark: 30
Lithuania: 30
U.K.: 30
Ireland: 31
Germany: 30
Poland: 30
Belgium: 30
France: 28
Slovakia: 30
Hungary: 30
Turkey: 30
Albania: 30
Afghanistan: 30
Armenia: 30
Azerbaijan: 30
Bangladesh: 30
China: 30
Georgia: 30
Hong Kong: 30
India: 31
Indonesia: 30
Iran: 30
Iraq: 30
Israel: 30
Japan: 31
Jordan: 30
Korea: 30
Mongolia: 30
Russia: 31
Saudi Arabia: 30
Thailand: 30
Taiwan: 30
Vietnam: 28
Ukraine: 30
Algeria: 28
Nigeria: 30
Chad: 29
Egypt: 30
Sudan: 29
Ethiopia: 30
Luxembourg:30
New Zealand: 31
South Africa: 30
Madagascar: 30
Spain: 30
Austria: 30
Australia: 30
Antarctica: 30
Jamaica: 30
Italy: 30[/hide]

U.S.: 32
Canada: 30
Mexico: 29
Brazil: 30
Argentina: 30
Chile: 30
Venezuela: 30
Peru: 30
Cuba: 30
Bolivia: 30
Columbia: 30
Uruguay: 28
Ecuador: 30
Finland: 30
Iceland: 30
Greenland: 30
Denmark: 30
Lithuania: 30
U.K.: 31
Ireland: 31
Germany: 30
Poland: 30
Belgium: 30
France: 28
Slovakia: 30
Hungary: 30
Turkey: 30
Albania: 28
Afghanistan: 30
Armenia: 30
Azerbaijan: 30
Bangladesh: 30
China: 30
Georgia: 30
Hong Kong: 30
India: 31
Indonesia: 30
Iran: 30
Iraq: 30
Israel: 30
Japan: 31
Jordan: 30
Korea: 30
Mongolia: 30
Russia: 31
Saudi Arabia: 30
Thailand: 30
Taiwan: 30
Vietnam: 28
Ukraine: 30
Algeria: 28
Nigeria: 30
Chad: 29
Egypt: 30
Sudan: 29
Ethiopia: 30
Luxembourg:30
New Zealand: 31
South Africa: 30
Madagascar: 30
Spain: 30
Austria: 30
Australia: 30
Antarctica: 30
Jamaica: 30
Italy: 30

U.S.: 31
Canada: 30
Mexico: 29
Brazil: 30
Argentina: 30
Chile: 30
Venezuela: 30
Peru: 30
Cuba: 30
Bolivia: 30
Columbia: 30
Uruguay: 29
Ecuador: 30
Finland: 30
Iceland: 30
Greenland: 30
Denmark: 30
Lithuania: 30
U.K.: 31
Ireland: 31
Germany: 30
Poland: 30
Belgium: 30
France: 28
Slovakia: 30
Hungary: 30
Turkey: 30
Albania: 28
Afghanistan: 30
Armenia: 30
Azerbaijan: 30
Bangladesh: 30
China: 30
Georgia: 30
Hong Kong: 30
India: 31
Indonesia: 30
Iran: 30
Iraq: 30
Israel: 30
Japan: 31
Jordan: 30
Korea: 30
Mongolia: 30
Russia: 31
Saudi Arabia: 30
Thailand: 30
Taiwan: 30
Vietnam: 28
Ukraine: 30
Algeria: 28
Nigeria: 30
Chad: 29
Egypt: 30
Sudan: 29
Ethiopia: 30
Luxembourg:30
New Zealand: 31
South Africa: 30
Madagascar: 30
Spain: 30
Austria: 30
Australia: 30
Antarctica: 30
Jamaica: 30
Italy: 30

It's Colombia, not Columbia

U.S.: 31
Canada: 30
Mexico: 29
Brazil: 30
Argentina: 30
Chile: 30
Venezuela: 30
Peru: 30
Cuba: 30
Bolivia: 30
Colombia: 31
Uruguay: 29
Ecuador: 30
Finland: 30
Iceland: 30
Greenland: 29 (not a country, I think)
Denmark: 30
Lithuania: 30
U.K.: 31
Ireland: 31
Germany: 30
Poland: 30
Belgium: 30
France: 28
Slovakia: 30
Hungary: 30
Turkey: 30
Albania: 28
Afghanistan: 30
Armenia: 30
Azerbaijan: 30
Bangladesh: 30
China: 30
Georgia: 30
Hong Kong: 30
India: 31
Indonesia: 30
Iran: 30
Iraq: 30
Israel: 30
Japan: 31
Jordan: 30
Korea: 30
Mongolia: 30
Russia: 31
Saudi Arabia: 30
Thailand: 30
Taiwan: 30
Vietnam: 28
Ukraine: 30
Algeria: 28
Nigeria: 30
Chad: 29
Egypt: 30
Sudan: 29
Ethiopia: 30
Luxembourg:30
New Zealand: 31
South Africa: 30
Madagascar: 30
Spain: 30
Austria: 30
Australia: 30
Antarctica: 30
Jamaica: 30
Italy: 30

[size=200]Guys, hurt=-2 and heal=+1 (this makes the game go faster)[/size]

U.S.: 31
Canada: 30
Mexico: 29
Brazil: 30
Argentina: 30
Chile: 30
Venezuela: 30
Peru: 30
Cuba: 30
Bolivia: 30
Colombia: 31
Uruguay: 29
Ecuador: 30
Finland: 30
Iceland: 30
Greenland: 29 (not a country, I think)
Denmark: 30
Lithuania: 30
U.K.: 31
Ireland: 31
Germany: 30
Poland: 30
Belgium: 30
France: 28
Slovakia: 30
Hungary: 30
Turkey: 30
Albania: 28
Afghanistan: 30
Armenia: 30
Azerbaijan: 30
Bangladesh: 30
China: 30
Georgia: 30
Hong Kong: 30
India: 31
Indonesia: 30
Iran: 30
Iraq: 30
Israel: 30
Japan: 31
Jordan: 30
Korea: 30
Mongolia: 30
Russia: 31
Saudi Arabia: 30
Thailand: 30
Taiwan: 30
Vietnam: 28
Ukraine: 30
Algeria: 28
Nigeria: 30
Chad: 29
Egypt: 30
Sudan: 29
Ethiopia: 30
Luxembourg:30
New Zealand: 31
South Africa: 30
Madagascar: 30
Spain: 30
Austria: 28
Australia: 30
Antarctica: 31
Jamaica: 30
Italy: 30

U.S.: 31
Canada: 30
Mexico: 29
Brazil: 31
Argentina: 30
Chile: 30
Venezuela: 30
Peru: 30
Cuba: 30
Bolivia: 30
Colombia: 31
Uruguay: 29
Ecuador: 30
Finland: 30
Iceland: 30
Greenland: 29 (not a country, I think)
Denmark: 30
Lithuania: 30
U.K.: 31
Ireland: 31
Germany: 30
Poland: 30
Belgium: 30
France: 28
Slovakia: 30
Hungary: 30
Turkey: 30
Albania: 28
Afghanistan: 30
Armenia: 30
Azerbaijan: 30
Bangladesh: 30
China: 30
Georgia: 30
Hong Kong: 30
India: 31
Indonesia: 30
Iran: 30
Iraq: 30
Israel: 30
Japan: 31
Jordan: 30
Korea: 30
Mongolia: 30
Russia: 31
Saudi Arabia: 30
Thailand: 30
Taiwan: 30
Vietnam: 28
Ukraine: 30
Algeria: 28
Nigeria: 28
Chad: 29
Egypt: 30
Sudan: 29
Ethiopia: 30
Luxembourg:30
New Zealand: 31
South Africa: 30
Madagascar: 30
Spain: 30
Austria: 28
Australia: 30
Antarctica: 31
Jamaica: 30
Italy: 30

[quote="7h3.D3m0n.117"]This topic's locked because Wickedestjr and alingaling are the same account.

Wickedestjr, stop creating new accounts and impersonating them or else I'll take this to the admins and have your accounts banned.

Plus, this thread is really annoying.
[/quote]

From http://www.artofproblemsolving.com/Forum/viewtopic.php?t=224772


find $a\in\mathbb{Q},b\in\mathbb{Q},c\in\mathbb{Q}$, where

$\sqrt[6]{7\sqrt[3]{20}-19}=\sqrt[6]{a}-\sqrt[6]{b}+\sqrt[6]{c}$

[quote="seamusoboyle"]As a matter of interest, is this problem more than endless number crunching?[/quote]
Are you fed up? OK. I don't going to delect more. :)

Can those numbers be found in a "smart" way? I can't see any...

[quote="perfect_radio"]Can those numbers be found in a "smart" way?  [/quote]
Yes, there is a "smart" way!
Try it:
Find $a\in\mathbb{Q},b\in\mathbb{Q},c\in\mathbb{Q},$ where $\sqrt[3]{21\sqrt[3]{6}-37}=\sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c}.$ ;)

[quote="Klio"]
Can it be done?[/quote]
Certainly! And very easy, already. :)

[quote="Klio"][quote="Yury"]\[ \sqrt[6]{7\sqrt[3]{20}-19} = \sqrt[3]{5/3} - \sqrt[3]{2/3}.  \][/quote]
That is short , clean and correct answer!
 :lol:
But ,If I got the author correctly,he requires  answer where [u][i]three[/i][/u] different fractions are put under
6-th roots on the right  side of the equality. .
Right? :)
Can it be done?[/quote]
\[ \sqrt[6]{7\sqrt[3]{20}-19} = \sqrt[6]{25/9} - \sqrt[6]{4/9} + \sqrt[6]{0}.  \]

:rotfl:

[quote="Klio"][quote="Yury"]\[ \sqrt[6]{7\sqrt[3]{20}-19} = \sqrt[3]{5/3} - \sqrt[3]{2/3}.  \][/quote]
That is short , clean and correct answer!
 :lol:
But ,If I got the author correctly,he requires  answer where [u][i]three[/i][/u] different fractions are put under
6-th roots on the right  side of the equality. .
Right? :)
Can it be done?[/quote]

For example:

$\sqrt[6]{7\sqrt[3]{20}-19}=\sqrt[6]{\frac{81}{625}}-\sqrt[6]{\frac{4}{9}}+\sqrt[6]{\frac{64}{5625}}$


algebra
polynomial
Vieta
number theory theorems
number theory

I know some problems involving vieta-jumping (e.g. IMO 1988 Q6, IMO 2007 Q5, .. ).

I think I can grasp what it is, but can anybody give me the 'definition' of vieta-jumping?

Not sure whether this is what you are looking for?

[url]http://reflections.awesomemath.org/2007_5/vieta_jumping.pdf[/url]


ratio calculus logarithms function integration

solve
$\displaystyle\sum_{n=1}^\infty\frac{1}{n3^n}$

i know that this sum converges (by using the ratio-test) but have no idea how to begin evaluating it.

I approached this one using calculus:

Note first that

$\frac{1}{1-x} = 1 + x + x^2 + ...$

Now, integrating both sides and multiplying by -1, we have

$- \ln (1 - x) = x + \frac{x^2}{2} + \frac{x^3}{3} + ...$

Now let $x = 1/3$, and we have

$- \ln (1 - 1/3) = \ln \left(\frac{3}{2}\right) = \frac{1}{3} + \frac{1}{2 \cdot 3^2} + \frac{1}{3 \cdot 3^3} + ... = \sum_{n=1}^{\infty} \frac{1}{n \cdot 3^n}$.

We can expand this result to find that, in general for $-1 \leq \frac{1}{x} < 1 \Rightarrow | x | > 1$ or $x= - 1$, we have

$\sum_{n=1}^{\infty} \frac{1}{n \cdot x^n} = \ln \left(\frac{x}{x-1}\right)$.

wow.
that is amazing...i would never have thought of that.
thank you very much for the help.

[quote="Farenhajt"]Just another thing, though:

Since an undetermined integral generally requires the integration constant, we must write

$-\ln(1-x)=C+x+\frac{x^2}{2}+\dots$,

and we must find some way to determine the value of $C$. Therefore, so far we can say only that the given series is $\ln\frac{3}{2}+C$[/quote]
The constant of integration is trivial: let $x = 0$ and we find that we must have $C = 0$.

[quote="pkerichang"]just a comment: generating function rocks![/quote]
I tend to think of them as Taylor or McLaurin series because I was introduced to them before I had learned about the concept of a generating series, but yes, they do rock!  :lol:

[quote="zanttrang"]The constant of integration is trivial: let $x = 0$ and we find that we must have $C = 0$.[/quote]

Agreed on the triviality part, but it must be mentioned in the solution.


1. An Ampere is a unit of current, equivalent to Coulombs/Second, so (Current)(Time)=Charge.

No clue about the second one.

2. "grounded" is used the describe the place at which the voltage is 0.

[quote="qweretyq"]Ok this won't take you very long to answer so please do

1) Car batteries are rated in Ampere-hours (A*h). What quantity is being measured?
       I'm not sure if this means couloumbs??

2) In a car, one terminal of the battery is said to be connected to "ground." Since it is not really connected to the ground, explain what is meant by this expression?

THANKS[/quote]

1) Amperes have dimensions of $\text{charge}\cdot\text{time}^{-1}$

Hours have dimensions of time. $\\
\\
\therefore A\cdot h = \text{charge}\cdot\text{time}^{-1}\cdot\text{time} =\text{charge}$

2) In electrical jargon any point in a circuit having a potential of zero volts is referred to as ground.  Since you're dealing with a circuit, however, a point having a potential of zero volts may be defined arbitrarily; then the potential (voltage) of any other point in the circuit can be given with referrence to this defined "ground."


Can someone post a proof of why angles extending from the center of a circle, ending on the endpoints of a chord are twice the measure of an angle extending from a point on the circumfrence of the circle ending on the same endpoints of the same chord?

Consider the altitudes from the circumcenter of a triangle.  :)

Here's what I came up with:

Imagine a circle is encompassed by another circle of radius twice itself, the center tangent to the point on the circumfrence of the smaller circle where the half angle would radiate from. Smaller circle's radius is $2\pi r$ while the large one's is $4\pi r$. So basically:

$2\pi r \cdot \frac{\theta}{360} = 4\pi r \cdot \frac{\phi}{360}$
$\frac{\theta}{\phi} = 2$

Also I think this proof shows that the angle ratio is actually proportional depending on how far away the other angle's point of origin is.


$1.$Punem inegalitatea sub forma:
$\frac{a}{1-a^{2}}+\frac{b}{1-b^{2}}+\frac{c}{1-c^{2}}\geq \frac{3\sqrt{3}}{2}\ (a^{2}+b^{2}+c^{2})$
Spargem inegalitatea in:
$\frac{a}{1-a^{2}}\geq\frac{3\sqrt{3}}{2}\ a^{2}$, unde $a\in\ [0,1]$
Se ajunge la $\frac{2}{1-a^{2}}\geq\ 3\sqrt{3}a$
Pentru o descompunere in factori mai usor de gasit notam $x=a\sqrt{3}$, deci $x\in\ [0,\sqrt{3}]$
Se ajunge la $x^{3}-3x+2\geq\ 0$, echivalent cu $(x-1)^{2}\ (x+2)\geq\ 0$, care este evident pentru ca $x\geq\ 0$.

$2.$Analog cu $1.$, spargem inegalitatea in $\frac{a^{2}}{1-a^{3}}\geq\frac{4\sqrt[3]{4}}{3}\ a^{3}$, ajungandu-se, dupa notatia $x=a\sqrt[3]{4}$, la $x^{4}-4x+3\geq\ 0$, care este echivalent cu $(x-1)^{2}\ (x^{2}+2x+3)\geq\ 0$, lucru evident adevarat.

$3.$Notam cu $K$ valoarea minima a expresiei. Spargem inegalitatea si il cautam pe K minim pentru care $\frac{a^{3}}{1-a^{8}}\geq\ K\cdot\ a^{4}\Leftrightarrow\ a-a^{9}\leq \frac{1}{K}$
Asadar $\frac{1}{K}=max(a-a^{9})$.Fie $f(a)=a-a^{9}$, Derivata $f\prime(a)=1-9a^{8}$ si se anuleaza pentru $a=\frac{1}{\sqrt[8]{9}}$ in intervalul $[0,1]$. Intre $0$ si $\frac{1}{\sqrt[8]{9}}$. $f(a)$ creste, iar intre $\frac{1}{\sqrt[8]{9}}$ si $1$ $f(a)$ este descrescatoare. Rezulta ca maximul lui $f(a)$ este $f(\frac{1}{\sqrt[8]{9})})=\frac{8}{9}\cdot \frac{1}{\sqrt[8]{9}}=\frac{1}{K}$.Rezulta ca minimul cautat este $K=\frac{9\sqrt[8]{9}}{8}$.

Felicitari! Ai descoperit ideea comuna de rezolvare a problemelor!


[quote][color=green]There are given three circles $ \omega_1,\omega_2,\omega_3$ such as $ \{\begin{array}{cccc} A, D \in \omega_1 \cap \omega_2\\
\\
B,E \in \omega_1 \cap \omega_3\\
\\
C,F\in \omega_2 \cap \omega_3\end{array}$   so that $ ABCDEF$ is a hexagon(posibly non-convex).Prove that:

\[ AB\cdot CD\cdot EF=BC\cdot DE\cdot FA\][/color] [/quote]


How many digits are in the square root of the perfect square 152,399,025?

Since that number has $ 9$ digits, its square root can have at most $ 5$ digits.
Now we have to see that its square root can't have $ 4$ digits.
Since we can "take off" the last $ 6$ digits (for ease of approximation), also take off the last $ 3$ digits of the square root. Thus, we see that the remaining number has to be two digit, because $ 152$ is the square of a two digit number.

That means the square root has $ 2\plus{}3\equal{}\boxed{5}$ digits.

If you actually wanted to find the square root of that number, have fun :D (it's [hide="some number"]$ 12345$[/hide])


Assume that s,t >= 1 and 1/s + 1/t = log 3/ log 2, prove that 1 + x + x^2 \geq (1 + x^s)^(1/s)(1 + x^t)^(1/t) for all x \in [0,1] if and only if (s + t) <= 8/3.

Don't kill me please.

Is this one too ugly? I have a few other similar to this.

I don't think so.
It is strange but not ugly.
Is it difficult?
Have you a solution?

I have a solution, but it's not mine. The solution was pretty comprehensive and used cantor measures to prove the results. I'm sure there's some more elementary way of showing this. Also, I belive the given bounds are quite sharp.

Anyone would care to try finding a more elementary solution, I've failed.


quadratics
search
number theory proposed
number theory

Prove that if p is a prime number of the form $ p\equal{}4k\plus{}1$ then $ p|k^k\minus{}1$

It suffices to show $ k$ is the fourth power of something modulo $ p$.  Then it suffices to show $ 16k$ is a fourth power.  Then it suffices to show $ \minus{} 4$ is a fourth power.  This is true if and only if $ \left(\frac {2}{p}\right) \equal{} \left(\frac {\sqrt { \minus{} 1}}{p}\right)$.  Now $ \sqrt { \minus{} 1}$ is a quadratic residue if and only if $ p \equiv 1\mod{8}$, while $ 2$ is a quadratic residue if and only if $ p \equiv \pm1 \mod{8}$.  But $ p$ cannot be $ \minus{} 1\mod{8}$, because $ p\equal{}4k\plus{}1$.  Therefore, $ \left(\frac {2}{p}\right) \equal{} \left(\frac {\sqrt { \minus{} 1}}{p}\right)$, as desired.

Is there a more elementary way?

It is C 37 at Pen but I cannot find any discussion for this one . :maybe:
Could you give me a link ?

There is no C37, you use some old list or something else. It's definitely there.

I cannot find it sorry . Anyway here is my solution . It suffices to prove that there exist a such that (as K81o7 said)
$ a^4\equiv \minus{}4(\mod p)$ or  $ ((a\minus{}1)^2\plus{}1)((a\plus{}1)^2\plus{}1)\equiv 0(\mod p)$

But we know that there exist x such that $ p|x^2\plus{}1$ so we are done . :)


Can you give a "Euclid-style" proof that there are infinitely many primes $ p\equiv 1 \, \text{mod} \, n$ for every $ n>0$ .

See http://www.mathlinks.ro/viewtopic.php?t=126566 .

By the way, in http://www.mathlinks.ro/viewtopic.php?t=100567 you can find the $ \minus{}1 \mod n$ case.

Let $ \Phi_n(x)$ cyclotomic polinom. Then, from $ p|\Phi_n(x)$ get $ n|p\minus{}1$ or $ p\equal{}1\mod n$.
Let $ p_i$ is primes form $ p_i\equal{}1\mod n$ (or $ p_i\equal{}\minus{}1\mod n$), let $ P\equal{}\prod_i p_i$, then any prime divisor $ p|\Phi_n (P)$ give new prime divisor form $ p\equal{}1\mod n$, and $ p|\Phi_{2n}(P)\to p\equal{}\minus{}1\mod n$.

[quote="Rust"]and $ p|\Phi_{2n}(P)\to p \equal{} \minus{} 1\mod n$.[/quote]
It's not that easy  :wink:
In fact, $ p|\Phi_{2n}(P)\to p \equal{} 1\mod n$.


[quote="jli"][quote="Ignite168"]A regular tetrahedron has all edge lengths of 5 cm. The midpoints of two opposite edges are connected together. What is the length of the segment connecting these two midpoints?[/quote]

yay! more non-interval posts!

[hide="solution"]just look at the face containing the line segment.  It should be an equilateral triangle with side 5.  because the endpoints of the line segment are midpoints of the triangle, hte length should be half of 5 which is 2.5[/hide][/quote]

Sorry, what I mean by opposite edges is ones that don't intersect at all.

[quote="Hychu"][hide]Let ABCD be a tetrahedron. Let the midpoint of AB be M. Let the midpoint of CD be N. Draw NA. It is on the side ACD and is perpedicular to CD. Use Pythagorean (spelled correctly?) Theorem and get AN = 5(3)^(1/2)/2. AMN is a right triangle. Use the same theorem again and get MN = 5/(2)^(1/2).[/hide][/quote]
$\frac{5}{\sqrt{2}}=\frac{5\sqrt{2}}{2}$.


[hide="My answer"]7080[/hide]

I found a really long solution but I don't have any time right now so I'll post it later.

EDIT: Oh yeah, I messed up. THe correct answer is [hide="here"]3540[/hide]

[hide="My solution"]

2x3y=(24(1/2)+(1/3)+(1/4)+...+(1/60))1 (24(1/3)+(1/4)+...+(1/60))2 (24(1/4)+(1/5)+...+(1/60))3 ...(24(1/60))59

This is the same as $(24^{1/2}*24^{1/3}*24^{1/4}...)^1...$ which is equal to $24^{1/2}*24^{1/3}*24^{1/4}...*24^{2/3}*24^{2/4}...*24^{3/4}...$ or $24^{1/2}*24^{3/3}*24^{6/4}$.

We immediately notice that the numerators of the fractions $1/2,3/3,6/4...$ are triangle numbers, or numbers that are the sum of all numbers from 1 to n.

Furthermore, where the denominator is n, the numerator is the (n-1)th triangle number, or the sum of all numbers from 1 to (n-1). Using the formula $\frac {n(a(1)+a(n))}{2}$ we find that the (n-1)th triangle number is equal to $\frac {n(n-1)}{2}$ Multiplying by $1/n$ we get $ \frac {n-1}{2}$ !

Therefore the sequence is equal to $1/2+2/2+3/2+4/3...$ or the sum of all numbers from 1 to 59, divided by 2. therefore we get $\frac {59*60} {2*2}$ or $59*15=885$.

Since $24^x=2^{3x}3^x$, the answer would be $885*4 = 3540$.

[/hide]

I did it the way Bob did it, and I like yours, and yes that is the answer


linear algebra
matrix
group theory
abstract algebra
quadratics
algebra
polynomial

Hey guys, I have what I think might be an interesting question:

What is the maximal order of an element in PGL(2,F_31). What is the maximal order among the elements in PGL(2,F_31) with an order divisible by 31?

(Note that PGL = GL/(center of GL)).

I'd like to know how one approaches this question exactly. My professor has been asking similar questions to this all the semester, but his explanations are totally incoherent and no one in the class really gets how to do it. Some sort of methodology or way of motivating an answer to this question or similar ones would be really great, perhaps even better than the answer itself.

Thanks guys.

$GL_{2}(F_{31})$ has $(31^{2}-1)(31^{2}-31)=30^{2}*31*32$ elements, so $PGL_{2}(F_{31})$ has $30*31*32$ elements.
A typical element of order 31 is $\begin{pmatrix}1&1\\0&1\end{pmatrix}$; all others are conjugates of this element.
The elements of the form $\begin{pmatrix}a&0\\0&1\end{pmatrix}$ form a subgroup of order 30, isomorphic to the multiplicative group of the field and thus cyclic.
If $c$ is not a quadratic residue mod 31, matrices of the form $\begin{pmatrix}a&cb\\b&a\end{pmatrix}$ can be identified with the field with $31^{2}$ elements. Taking the invertible elements ($a$ and $b$ not both zero) and a projective quotient, we get a cyclic subgroup of order 32.

It makes sense to look at the characteristic polynomial $p$, eigenvalues, and eigenvectors of a matrix $A\in GL_{2}(F_{31})$. We have three cases:
Case 1: $p(x)=(x-a)^{2}$. Then either $A=aI$ or $A=aI+N$, with $N$ nilpotent. $aI$ is the identity in $PGL_{2}(F_{31})$, so $A$ has order either 1 or 31 in $PGL_{2}(F_{31})$.
Case 2: $p(x)=(x-a)(x-b)$. Let $v$ and $w$ be the eigenvectors for $a$ and $b$ respectively. We have $A^{30}(cv+dw)=a^{30}cv+b^{30}dw=cv+dw$, so $A^{30}=I$ and the order of $A$ divides 30.
Case 3: $p(x)=x^{2}-2ax+a^{2}-cb^{2}=(x-(a-b\sqrt{c}))(x-(a+b\sqrt{c}))$. The map $y\to y^{31}$ is a nontrivial automorphism of $F_{31^{2}}$, so it exchanges $a-b\sqrt{c}$ and $a+b\sqrt{c}$. Consequently, $(a-b\sqrt{c})^{32}=a^{2}-cb^{2}=(a+b\sqrt{c})^{32}$, and this element is in the base field $F_{31}$. Therefore $A^{32}=(a^{2}-cb^{2})I$, and $A^{32}$ is the identity in the quotient $PGL_{2}(F_{31})$.

A summary: The largest possible order is 32, achieved by many matrices with eigenvalues outside $F_{31}$. The only possible orders are divisors of $30,31,$ and $32$.


The second round is over, but discussion has not yet been opened.

Any standard programming language could be used for what you're looking at, but you still need some insight- you can't just test all positive integers, since that would take an infinite amount of time..

I've never done much with programming languages, but I do like to use a spreadsheet for numerical experimentation.

You could just use maple, mathematica, pari ... ?

ok thanks...y eah I'll check them all out.. I was just wondering what computer programming to use to solve equations..

you can use any language really, it doesn't matter

you could even use php

[quote="g1sk8terz"]I'm a newbie in computer programming..

Does anyone know what computer programming program to use to solve math equations, like find all real solutions to or find all integer solutions to...

I came across problem 1 in the USAMTS round 2, it was find all integers such that the squares of its digits equal 2006 less than it.  What computer programming (Java, C+, stuff like that) can I use to solve this?

Thanks![/quote]

try BASIC on ur TI calc? lol

not TI calc that's too obvious..I mean for computers..

Just use anything (I like c++ and pascal, but that depends very much on the user), using some multiprecision library if you want to work with integers.


inequalities
complex numbers
geometry proposed
geometry

Let $ ABC$ be a triangle of sidelengths $ a,b,c$ and $ X$ a point in plane. We denote the distances of $ X$ from the vertices $ A,B,C$ with $ x,y,z$. Prove that

$ ayz\plus{}bzx\plus{}cxy \geq abc$

WITHOUT USING COMPLEX NUMBERS OR POLAR INVERSION !!!!! :wink:

Does it exist such a solution?  I cannot ind it. :(

[hide="A hint"]We proceed from the inequality $ mXA^2\plus{}nXB^2\plus{}pXC^2\ge \frac{mnc^2\plus{}npa^2\plus{}pmb^2}{m\plus{}n\plus{}p}$ for every triples m,n,p of real numbers.
Apply that ineq with $ m\equal{}\frac{a}{XA},n\equal{}\frac{b}{XB},p\equal{}\frac{c}{XC}$, hence result.[/hide]


function
combinatorics unsolved
combinatorics

A.Let A(n) be the number of partitions n in which no part occurs more than twice.  Find the generating function.

My Answer:
$ (1 \plus{} x)(x \plus{} x^{2})(x^{2} \plus{} x^{3})...........(x^{i} \plus{} x^{i \plus{} 1})$
$ \prod_{i \equal{} 0}^{\infty}$$ (x^{i} \plus{} x^{1 \plus{} i})$

B.  let B(n) be the number of partitions of n in which no part is a multiple of 3.  Find the Generating function.

$ \dfrac{1}{1 \minus{} x}$ - $ \dfrac{1}{1 \minus{} x^{3}}$
I'm still new with these, I'm sure I messed up, can anyone help me out?

The partition generating function with no restrictions is given by $ (1 \plus{} x \plus{} x^2 \plus{} ...)(1 \plus{} x^2 \plus{} x^4 \plus{} ...)(1 \plus{} x^3 \plus{} x^6 \plus{} ...)...$.  That's because each term in each factor is a number of times a part occurs; the $ x^2$ term in the first factor means $ 1$ appears twice, the $ x^6$ term in the third factor means $ 3$ appears twice, and so on.

Start from this function and mess around with it.

Ok so I'm still a little confused then.  What does the $ x^{6}$ factor in the first factor mean 3 appears twice?

If so wouldn't it just be then

$ (1 \plus{} x \plus{} x^{2} \plus{} x^{6} \plus{} x^{8}.....)$ = $ \dfrac{1}{1 \minus{} x^{2}}$

or would it just be

$ (1 \plus{} x \plus{} x^{2})(1 \plus{} x^{2} \plus{} x^{4})(1 \plus{} x^{3} \plus{} x^{6}).....$

or more simpily $ \prod_{i \equal{} 1}^{\infty}$$ (1 \plus{} x^{i} \plus{} x^{2i})$
My guess is the second one, I just want to make sure though.

And for the one that contains no multiple of 3 should be

$ \prod_{i \equal{} 1}^{\infty}\dfrac{(1 \minus{} x^{3i})}{1 \minus{} x^{i}}$

Can anyone confirm this?

[quote="Jrb599"]$ (1 \plus{} x \plus{} x^{2})(1 \plus{} x^{2} \plus{} x^{4})(1 \plus{} x^{3} \plus{} x^{6}).....$[/quote]

That's correct.

To explain my first comment, let's talk about how we can organize a partition in a unique way.  Generally, a partition is sorted from greatest to least:  for example, $ 9 \equal{} 4 \plus{} 2 \plus{} 1 \plus{} 1 \plus{} 1$ uniquely describes a partition.  We could also sort from least to greatest:  $ 9 \equal{} 1 \plus{} 1 \plus{} 1 \plus{} 2 \plus{} 4$.

To start writing a generating function, imagine that to create a partition we have an infinite number of boxes, one for every positive integer:  the first box contains $ 1$s, the second contains $ 2$s, and so forth.  To create a partition, we take some number of symbols from each box and add them all up.

In this way, we see that to create the partition above we take three $ 1$s, one $ 2$, and one $ 4$, which is a third way to sort them:  first we have [i]how many we took from a given box[/i], and then [i]what box we took it from.[/i]

The generating function for the first box is $ (1 \plus{} x \plus{} x^2 \plus{} x^3 \plus{} ...)$.  That's because each term $ x^k$ represents the number of ways we can take $ k$ $ 1$s from the box (exactly one way, since they're all the same).  The generating function for the second box is $ (1 \plus{} x^2 \plus{} x^4 \plus{} x^6 \plus{} ...)$.  That's because each term $ x^{2k}$ represents the number of ways we can take $ k$ $ 2$s from the box (again, exactly one way).  We multiply each exponent by $ 2$ because each $ 2$ we take contributes $ 2$ to the sum.  And so forth.

To answer the first question, we restrict our boxes to only two symbols each.

To answer the second question, we ignore the boxes that contain multiples of $ 3$.  That makes the answer

$ (1 \plus{} x \plus{} x^2 \plus{} ...)(1 \plus{} x^2 \plus{} x^4 \plus{} ...)(1 \plus{} x^4 \plus{} x^8 \plus{} ...)(1 \plus{} x^5 \plus{} x^{10} \plus{} ...)(1 \plus{} x^7 \plus{} x^{14} \plus{} ...)...$

or

$ \prod_{i \equal{} 0}^{\infty} \frac {1}{(1 \minus{} x^{3i \plus{} 1})(1 \minus{} x^{3i \plus{} 2})}$.

If the following two generating functions, how might one check to see if these are equivalent to each other?

$ \prod_{i \equal{} 0}^{\infty} (1 \plus{} x^{i} \plus{} x^{2i})$  ==  $ \prod_{i \equal{} 0}^{\infty} \frac {1}{(1 \minus{} x^{3i \plus{} 1})(1 \minus{} x^{3i \plus{} 2})}$

They're not.  Those are the answers to two different questions.

Actually, they are, except that the first one should begin with $ i \equal{} 1$.  (No parts of size $ 0$ allowed.)  Use the equality $ 1 \plus{} x^i \plus{} x^{2i} \equal{} \frac{1 \minus{} x^{3i}}{1 \minus{} x^i}$ and then cancel when you take the product of all such terms.

Oh, that was the third part of the question.  I see.  My apologies  :oops:


trigonometry
complex numbers
calculus
calculus computations

I have a question about a step in lagranges trig identity,
so i want to prove this identity and I understand everything except a single step

We have this
$ [1\minus{}Exp(i(n\plus{}1)\theta)]/(1\minus{}Exp(i \theta))$

I am interested in the real part, but in the book they simplify the expression a bit more
but first they multiply both the numerator and the denominator with

$ Exp(\minus{}i\theta /2)$

But why do they do that?
So why dont they use +(i/theta) or something

thanks guys

[quote="Abelois"][color=darkred] I have a question about a step in Lagrange's trig identity ...[/color] [/quote]

What is [b]Lagrange's trig identity[/b] ? Thank you !

[quote="Abelois"]
We have this
$ [1 \minus{} Exp(i(n \plus{} 1)\theta)]/(1 \minus{} Exp(i \theta))$

I am interested in the real part, but in the book they simplify the expression a bit more
but first they multiply both the numerator and the denominator with

$ Exp( \minus{} i\theta /2)$

But why do they do that?
So why dont they use +(i/theta) or something
[/quote]

Think of
$ (Exp(\minus{} i\theta /2) \minus{} Exp(i \theta /2)) \equal{} \minus{}2*Sin(\theta /2)$
and split the numerator using
$ Re(Exp(i \alpha)) \equal{} Cos(\alpha)$
and you are done.

[quote="Virgil Nicula"]What is Lagrange's trig identity ?[/quote]

Lagrange's identity is the name given to the following sum: $ \sum_{k\equal{}0}^n \cos(k \theta) \equal{} \frac12 \plus{} \frac{\sin\left[\left(n \plus{} \frac12\right) \theta \right]}{2 \sin \frac{\theta}{2}}$

This can also be proved by a slightly easier way.
Since $ \displaystyle\sin(nt\plus{}\frac{t}{2})\equal{}\sin nt\cos\frac{t}{2}\plus{}\cos nt\sin\frac{t}{2}$, and $ \displaystyle\sin(nt\minus{}\frac{t}{2})\equal{}\sin nt\cos\frac{t}{2}\minus{}\cos nt\sin\frac{t}{2}$
so $ \displaystyle\sin(nt\plus{}\frac{t}{2})\minus{}\sin(nt\minus{}\frac{t}{2})\equal{}2\cos nt\sin\frac{t}{2}$
Therefore, $ \displaystyle\frac{1}{2}\left[\sin(n\plus{}\frac{1}{2})t\minus{}\sin(n\minus{}\frac{1}{2})t\right]\equal{}\cos nt\sin\frac{t}{2}$
The sum can be written as: $ \displaystyle\sin\frac{t}{2}\sum_{n\equal{}1}^{M}\cos (nt)\equal{}\frac{1}{2}\sum_{n\equal{}1}^M\left[\sin(n\plus{}\frac{1}{2})t\minus{}\sin(n\minus{}\frac{1}{2})t\right]$ which telescopes.
The result is: $ \displaystyle\sin\frac{t}{2}\sum_{n\equal{}1}^{M}\cos (nt)\equal{}\frac{1}{2}\left[\sin(M\plus{}\frac{1}{2})t\minus{}\sin\frac{t}{2}\right]$.
After dividing both sides by $ \sin(t/2)$ and adding both sides by $ 1$, we have "Lagrange's Trig Identity" $ \displaystyle 1\plus{}\sum_{n\equal{}1}^{M}\cos (nt)\equal{}\sum_{n\equal{}0}^{M}\cos(nt)\equal{}\frac{1}{2}\plus{}\frac{\sin\left(M\plus{}\frac{1}{2}\right)t}{2\sin\frac{t}{2}}$.

Consider the real sequence $ (a_n)_{n\in\mathbb N^*}$ for which exists $ r\in \mathbb R^*$ so that $ a_{n + 1} = a_n + r$ for any $ n\in\mathbb N^*$ .

Then there is the identities ${ \sum_{k = 1}^n\sin a_k} = \frac {\sin\frac {a_1 + a_n}{2}\cdot\sin\frac {nr}{2}}{\sin\frac r2}$ and $ \sum_{k = 1}^n\cos a_k = \frac {\cos\frac {a_1 + a_n}{2}\cdot\sin\frac {nr}{2}}{\sin\frac r2}$ .

[b]Abelois[/b], what is the name of the upper trigonometrical identities ?!

Observe that ${{ \sum_{k = 1}^n\sin a_k}\ =\ \tan\frac {a_1+a_n}{2}\cdot\sum_{k = 1}^n\cos a_k}$

[b]Particular case.[/b] $ \sum_{k=1}^n\sin kx=\frac {\sin \frac {(n+1)x}{2}\cdot\sin\frac {nx}{2}}{\sin\frac x2}$ and $ \sum_{k=1}^n\cos kx=\frac {\cos \frac {(n+1)x}{2}\cdot\sin\frac {nx}{2}}{\sin\frac x2}$.


superior algebra
superior algebra unsolved

Describe all homomorphisms $ \phi: \mathbb{Z}^{\plus{}}\rightarrow\mathbb{Z}^{\plus{}}$. Determine which are injective, which are surjective, and which are isomorphisms.

Smells a bit like homework, but...

[hide="hint"]
Assuming by $ \mathbb{Z}^\plus{}$ you mean the integers under addition, homomorphisms $ \varphi$ are completely determined by $ \varphi(1)$.
[/hide]


consider the sequence of number {$a_n$} defined by
$a_0=0 ;a_1=1; a_{n+1}=4a_n+a_{n-1}$.prove that:
$a_{10k}$ is divibie by $20^{k}$ ; with $k\in N$

$a_n = \frac{(2+\sqrt{5})^n - (2-\sqrt{5})^n}{2\sqrt{5}}$ by induction

$a_{10k} = \frac{(2+\sqrt{5})^{10k} - (2-\sqrt{5})^{10k}}{2\sqrt{5}} = \frac{\sum_{i=0}^{10k} \binom{10k}{i} 2^i \left( \sqrt{5} \right)^{10k - i} - \sum_{i=0}^{10k} \binom{10k}{i} 2^i \left( -\sqrt{5} \right)^{10k - i}}{2\sqrt{5}}$ $= \frac{\sum_{n=1}^{5k}  \binom{10k}{2n-1} 2^{2n} \cdot 5^{5k-n}}{2}$. Every term in the expansion is a multiple of $40$, so upon division by $2$, it'll be a multiple of $20$. I'm not sure that it'll be a multiple of $20^k$, though; this seems false for $k = 2, 3$. I might be wrong.


The incircle (I) of triangle ABC touches BC,CA,AB. AD,BE,CF are the bisectors and AM,BN,CP are the medians of triangle ABC. DA',EB',FC' are three tangents (other than BC,CA,AB) of (I) (where A',B',C' are tangency points). Prove that A'M,B'N,C'P are concurrent.


Here is the problem Which is general form of many old inequalities posted in this forum. Please try to post a complete solution...

If $a,b,c,d$ are non-neagtive reals and $a+b+c+d=4$ then :

\[ (abc)^k+(bcd)^k+(cda)^k+(dab)^k \leqslant \max \{ 4 ; (\frac{4}{3})^{3k} \}  \]

The special case is one problem of Harazi :

if $a^2+b^2+c^2+d^2=4$ then:
$(abc)^3+(bcd)^3+(cda)^3+(dab)^3 \leqslant 4$

(I don't remamber exactly...)

 :)

nice inequality! It can be more general if we put a=4x/x+y+z+t, b=4y/x+y+z+t etc.
I want to see and a beautiful solution! :)

[quote="hungkhtn"]I see that noone has right idea for it? Please try again because It's the general form of many old problem!

 :([/quote]

Hung, reserve this hard problem for our book.

[quote="hungkhtn"]
If $a,b,c,d$ are non-neagtive reals and $a+b+c+d=4$ then :

\[ (abc)^k+(bcd)^k+(cda)^k+(dab)^k \leqslant \max \{ 4 ; (\frac{4}{3})^{3k} \}  \]

 :)[/quote]
Here is my some little ideas:
1)Substituting:$x=\frac34*a$,etc...we will prove that:
$\sum(xyz)^k\leq 1$ where $x+y+z+t=3$
Supposing that:$x\leq y\leq z\leq t$
By cauchy inequal,we got:$\sum(xyz)^k\leq x^ky^k(z^k+t^k)+kx^ky^k(z^{k-1}t+zt^{k-1})\leq x^k.y^k.(z+t)^k\leq 1$ for all natural number $k\geq2.$
2)Substituting $a=x^{\frac{1}{4k}}$,etc....Now we need prove that:
$\sum(xyz)^{\frac{1}{4})\leq 4}$ where $\sum x^{\frac{1}{4k}}=4$
But we have:
$\sum(xyz.1)^{\frac14}\leq \sum\frac{x+y+z+1}{4}=\frac{3(x+y+z+t)}{4}+1\leq 4$
It's only true for $k\leq \frac14$
They are not enough to solve the problem but anyway they are only ideas.. :)

Of course, It's nice and a good idea.  :)

[quote="hungkhtn"]One month Unsolved, also Unsolved in 2005 .

This problem I call " Secrect Inequality" because It's in my incomplete book. But 2 month later, I'll attack the full solutions of all unsolved inequality of mine in this forum. Well, there're quite alot problems of mine unsolved here. For all, it won't be secrect in 2006 :lol:. I promise.  :)[/quote]

Hung, you forgot my words to you? See me before you post the proofs. Thuan

[quote="hungkhtn"]One month Unsolved, also Unsolved in 2005 .

This problem I call " Secrect Inequality" because It's in my incomplete book. But 2 month later, I'll attack the full solutions of all unsolved inequality of mine in this forum. Well, there're quite alot problems of mine unsolved here. For all, it won't be secrect in 2006 :lol:. I promise.  :)[/quote]
I have said the inequality can use largrange multiply.
That's because I have seen the topic [url=http://www.mathlinks.ro/Forum/viewtopic.php?highlight=n-1+principle&t=33456]here[/url]
Or that is n-1 Equal Variable Principle :)

Here is the summarized solution:

WLOG, we suppose that $a\geq b\geq c\geq d\geq0$

If $k\geq2$, let $f$ be the left hand side, we have
\[ f\leq (abc)^k+3(abd)^k=(ab)^k(c^k+3d^k). \]
Note that for $k\geq2$
\[ \Bigl(\frac{c}{c+d}\Bigl)^k+3\Bigl(\frac{d}{c+d}\Bigl)^k\leq \Bigl(\frac{c}{c+d}\Bigl)^2+3\Bigl(\frac{d}{c+d}\Bigl)^2=\frac{c^2+3d^2}{(c+d)^2}. \]
Since $a\geq b\geq c\geq d$, $(c+d)^2=c^2+2cd+d^2\geq c^2+3d^2$. Then we have
\[ (c^k+3d^k)\leq (c+d)^k . \]
Now, we have by AM-GM
\[ f\leq (ab)^k(c+d)^k\leq (4/3)^k. \]

If, $0\leq k\leq1$, first we prove that
\[ abc+bcd+cda+dab\leq4. \]
We have by AG inequality
\[ ab(c+d)+cd(a+b)\leq\frac14(a+b)^2(c+d)+\frac14(c+d)^2(a+b)\leq4. \]
Then, the problem reduces to proving that
\[ x^k+y^k+z^k+t^k\leq4, \]
with $x+y+z+t\leq4$.

We begin with proving that $x^k+y^k\leq 2$ if $x+y=2$. Then, we prove that $x^k+y^k+z^k\leq 3$ with $x+y+z\leq4$ and finally, we come to the
\[ x^k+y^k+z^k+t^k\leq4, \]
with $x+y+z+t\leq4$.

Case $1\le k \le 2$  is the most difficult case. If $k \ge 2$ or $k \le 1$, the problem is not hard at all.

Hmm, how hard it is  :!:

Give me a hand! ;)

Ok, I only use mixing variable method, but need a stronger manner. I used SMV theorem, which is very strong for 4-variable problems.

Hung, can you introduce the UMV theorem, or at least solve the inequality using this method?


hi folks. it is possible to make it so that the e-mail you receive when a replay is made to one of your watched topics contains the new message posted. would you like this feature embellished? please vote.


Normal british / indian system of expressing dates is dd.mm.yy.

tell me if you are told that all the dates between 1st jan, 2001 and 31st dec 2100 ( both dates inclusive ) are expressed either in dd.mm.yy or in the reverse yy.mm.dd style and you do not know which date is written in which format, what percentage of the dates in the present century will confuse you.

for example if the date is wrtten as 06.03.12, it can be either 6th march, 2012 in dd.mm.yy format or 12th march, 2006 in yy.mm.dd format. so you're confused.

how many such confusions might occur in this century ?

[hide="hint"]

since we can determine whether it is a day or year when the number goes above 31 that limits the confusions to up to 2031.  And since we always know the month we can also figure out how many days are in each, except for Feb due to leap years.
As a result it will occur as combinations of of the number of possible daysper month.  I suggest making 12 cases, with an additional case for Feb of leap years.

Also we must subtract one for every time we get dd=yy.
[/hide]

[hide="solution"]

$\boxed{5206}$
[/hide]


A regular polygon has interior angles of 120 degrees. How many
sides does the polygon have?

The exterior angles are $ 180\minus{}120\equal{}60$, so the answer is $ \frac{360}{60}\equal{}\boxed6$.


Find all possible solutions to the equation $x^3+y^3+z^3=2008$, where x,y and z are positive integers.

Given $12^3 < 2008 < 13^3$, it's not too laborious just to check all the cases for positive integers $x, y, z$.

For a slightly more tidy argument, note $n^3 \equiv 0, 1, 5, 8, 12 \pmod{13}$ and you can go from there. Note $2008 > 13^3$ and given we are looking for positive integers we can discount $n^3 \cong 0 \pmod{13}$. Therefore as $2008 \equiv 6 \pmod{13}$, $x, y, z \cong 8, 12, 12 \pmod {13}$.

Then you can just case check to find a solution is $(x, y, z) = (10, 10, 2)$ and all permutations.

(Edit...:oops: forgot a case...)

Another solution is $(x, y, z) = (4, 6, 12)$ and all permutations.

That are not all the solutions.
Hint
[hide]x,y,z can't be odd.[/hide]


[quote="James Potter"]Find the least value of the expression:
$A=\frac{a^4}{(b-1)^3}+\frac{b^4}{(a-1)^3}$
where $a,b$ are numbers greater than $1$,satisfying the condition $a+b\le 4$[/quote]
Let $a-1=x,b-1=y.$ Then $x>0,y>0,x+y\leq2.$
Hence, $A=\frac{(x+1)^4}{y^3}+\frac{(y+1)^4}{x^3}\geq\frac{(2\sqrt x)^4}{y^3}+\frac{(2\sqrt y)^4}{x^3}=16(\frac{x^2}{y^3}+\frac{y^2}{x^3})\geq$
$\geq8(x+y)(\frac{x^2}{y^3}+\frac{y^2}{x^3})\geq8\cdot4=32.$
$A=32$ for $a=b=2.$ :)

[quote="tranthanhnam"]i think that you aren't English ,
you are Vietnamese
and this problem is this issue of the youth and maths at Vietnam
i don't like you make it , you should make it by yourselve [/quote]

Oh sorry if it's true because I received this problems and some others from one of my friend in Vietnam.I also know the magazine that you have talked about.another friend of mine scans it to me every months.But I haven't received the magazine 1-2006(may be this problem in this issue).
By any way sorry sory everybody :( ,thank you [i]tranthanhnam[/i] for discovering this mistake.
P.S 1:You are right [i]tranthanhnam[/i] I'm a Vietnamese but now I am studying in England.(I left Vietnam 3 years ago.)
P.S 2:[quote="tranthanhnam"]i don't like you make it , you should make it by yourselve[/quote]
Your English is a little bad.I know what you mean and the right sentence is "I don't like you doing like that , you should do it by yourselve" :lol:(just a joke)


Let $(a_{n})_{n\geq 1}$ and $(b_{n})_{n\geq 1}$ be the sequence of positive integers defined by  $a_{n+1}=na_{n}+1$ and $b_{n+1}=nb_{n}-1$ for $n\geq 1$. Show that the two sequence cannot have infinitely many common terms.

[i]Laurentiu Panaitopol[/i]

The idea is that $\frac{a_{n+1}}{n!}$ tends to $a_1-1+e$, while $\frac{b_{n+1}}{n!}$ tends to $b_1+1-e$ (or something like that :)). For sufficiently large $n$, the only common terms must be of the form $a_n=b_n$, because otherwise, if, for example, $a_n=b_m$ with $n<m$, this would mean that for some small $\varepsilon,\varepsilon'$ we would have $n!(a_1-1+e-\varepsilon)=m!(b_1+1-e-\varepsilon')$, and this is impossible for large $n<m$. On the other hand, $a_n=b_n$ for infinitely many $n$ leads to $a_1-1+e=b_1+1-e$, which is, again, impossible because $e$ is irrational.


1)We've known that $ \sin18^0 \equal{} \frac {\sqrt {5} \minus{} 1}{4}$.Prove that:
$ \sin2^o\sin18^o\sin22^o\sin38^o\sin42^o\sin58^o\sin62^o\sin78^o\sin82^o \equal{} \frac {\sqrt {5} \minus{} 1}{1024}$
2) We've known that $ \cos36^o \equal{} \frac {\sqrt {5} \plus{} 1}{4}$ and with $ 0^o < \alpha < 45^o$ we have $ \cot\alpha \equal{} \cot2\alpha \plus{} \sqrt {1 \plus{} \cot^22\alpha}$. Prove that:
$ 4\cos36^o \plus{} \cos7^o30' \equal{} \sqrt {1} \plus{} \sqrt {2} \plus{} \sqrt {3} \plus{} \sqrt {4} \plus{} \sqrt {5} \plus{} \sqrt {6}$

[color=green]LaTeX fixed by mod.[/color]

[quote="ngoclam1088"]1)We've known that $ sin18^0 \equal{} \frac {\sqrt {5} \minus{} 1}{4}$.Prove that:
$ sin2^osin18^osin22^osin38^osin42^osin58^osin62^osin78^osin82^o \equal{} \frac {\sqrt {5} \minus{} 1}{1024}$
2) We've known that $ \cos36^o \equal{} \frac {\sqrt {5} \plus{} 1}{4}$ and with $ 0^o < \alpha < 45^o$ we have $ \cot\alpha \equal{} \cot2\alpha \plus{} \sqrt {1 \plus{} \cot^22\alpha}$. Prove that:
$ 4\cos36^o \plus{} \cos7^o30' \equal{} \sqrt {1} \plus{} \sqrt {2} \plus{} \sqrt {3} \plus{} \sqrt {4} \plus{} \sqrt {5} \plus{} \sqrt {6}$[/quote]

[color=darkblue]Oh! You are wrong at problem 2. The true is $ 4\cos36^o \plus{} \cot7^o30' \equal{} \sqrt {1} \plus{} \sqrt {2} \plus{} \sqrt {3} \plus{} \sqrt {4} \plus{} \sqrt {5} \plus{} \sqrt {6}$

This is solution:

$ \bullet$ $ \cot15^0 \equal{} \cot30^0 \plus{} \sqrt {1 \plus{} \cot^230^0} \equal{} \sqrt {3} \plus{} \sqrt {4}$

$ \bullet$ $ \cot7^030' \equal{} \cot15^0 \plus{} \sqrt {1 \plus{} \cot^215^0} \equal{} \sqrt {3} \plus{} \sqrt {4} \plus{} \sqrt {1 \plus{} (\sqrt {3} \plus{} \sqrt {4})^2}$

$ \equal{} \sqrt {3} \plus{} \sqrt {4} \plus{} \sqrt{(\sqrt {2} \plus{} \sqrt {6})^2} \equal{} \sqrt {2} \plus{} \sqrt {3} \plus{} \sqrt {4} \plus{} \sqrt {6}$

$ \bullet$ $ 4\cos36^0 \equal{} \sqrt {5} \plus{} 1$

Hene $ 4\cos36^o \plus{} \cot7^030' \equal{} \sqrt {1} \plus{} \sqrt {2} \plus{} \sqrt {3} \plus{} \sqrt {4} \plus{} \sqrt {5} \plus{} \sqrt {6}$[/color]

No, I can't. Because Mr ngoclam1088 was wrong.

If $ \sin{18^0}\equal{}\frac{\sqrt{5}\minus{}1}{4} \Longrightarrow \frac {1}{4^3}.\frac {\sin18^0}{\sin62^0}\equal{}\frac{\sqrt{5}\minus{}1}{256\sin{62^0}} \neq \frac{\sqrt{5}\minus{}1}{1024}$

[quote="thanhnam2902"]No, I can't. Because Mr ngoclam1088 was wrong.

If $ \sin{18^0} \equal{} \frac {\sqrt {5} \minus{} 1}{4} \Longrightarrow \frac {1}{4^3}.\frac {\sin18^0}{\sin62^0} \equal{} \frac {\sqrt {5} \minus{} 1}{256\sin{62^0}} \neq \frac {\sqrt {5} \minus{} 1}{1024}$[/quote]

Sorry , Mr ngoclam1088 's correct, because

(1) You missed the term $ \sin62^\circ$ in the origional form ;

$ \boxed{\sin2^0.\sin18^0.\sin22^0.\sin38^0.\sin42^0.\sin58^0.\sin62^0.\sin78^0.\sin82^0}$.........(9 terms)

$ \equal{} \boxed{(\sin18^0.\sin42^0.\sin78^0).(\sin22^0.\sin38^0.\sin82^0).\sin2^0.\sin58^0}$.......(8  terms)

(2)you got illogically $ \frac {1}{4^3}\cdot \frac {\sin18^\circ}{sin62^\circ}$at the end.In my mind ,it should be $ \frac {1}{4^4}\cdot \frac {\sin18^\circ}{sin62^\circ}$

Do you agree?

Hi vinskman! I think mr_Ngoclam1088 have change his post. Yesterday it's $ \boxed{\sin2^0.\sin18^0.\sin22^0.\sin38^0.\sin42^0.\sin58^0.\sin78^0.\sin82^0}$


If I have $ y^x$, how can I make the $ x$ show up as large as the $ y$?

Use y^{\displaystyle x} in a document, though it won't work here as \displaystyle is stripped out. On the forum you can use  $ y^{\textstyle x}$ which is almost as good.
If you are using e^{complicated expression} then it's better to use exp(compicated expression). See also [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=232863]Mix of tall and short equations in a line[/url]


We say that a function $f: \mathbb{N}\rightarrow\mathbb{N}$ has the $(\mathcal{P})$ property if, for any $y\in\mathbb{N}$, the equation $f(x)=y$ has exactly 3 solutions.
a) Prove that there exist an infinity of functions with the $(\mathcal{P})$ property ;
b) Find all monotonously functions with the $(\mathcal{P})$ property ;
c) Do there exist monotonously functions $f: \mathbb{Q}\rightarrow\mathbb{Q}$ satisfying the $(\mathcal{P})$ property ?

c) No. Assume the contrary.
Then there is $f(a)=f(b)=f(c)=1$ then $f(x) \neq 1$ for $x \neq a,b,c$

$WLOG$, $a<b<c$

Since $a,b \in \mathbb{Q}$, $\frac{a+b}{2}\in \mathbb{Q}$.

Since $a<\frac{a+b}{2}<b$ and f is monotonous, $1=f(a) \leq f(\frac{a+b}{2}) \leq f(b)=1 \Rightarrow f(\frac{a+b}{2})=1$ or $1=f(b) \leq f(\frac{a+b}{2}) \leq f(a)=1 \Rightarrow f(\frac{a+b}{2})=1$

Contradiction.

[quote="N.T.TUAN"][quote="felixmann"]
a) Prove that there exist an infinity of functions with the $(\mathcal{P})$ property ;
[/quote]
If exist a function with $(\mathcal{P})$ property then there exist an infinity of functions with the that property. We have function follow has $(\mathcal{P})$ property
$f(\{1,2,3\})=\{1\},f(\{3,4,5\})=\{2\},...$ etc.
[quote]
b) Find all monotonously functions with the $(\mathcal{P})$ property ;[/quote]
Now, I can't.
[quote]
c) Do there exist monotonously functions $f: \mathbb{Q}\rightarrow\mathbb{Q}$ satisfying the $(\mathcal{P})$ property ?[/quote]
I don't understand  :huh:[/quote]
You have to determine if there exist monotonously functions $f: \mathbb{Q}\rightarrow\mathbb{Q}$ satisfying: for each $y\in\mathbb{Q}$ the equation $f(x)=y$ has exactly 3 solutions. [this is the $(\mathcal{P})$ property for $\mathbb{Q}$]


geometry
3D geometry
algebra
factorization
difference of cubes
special factorizations

[hide]Let A=1000 and B=999

The given expressoin equals:
$\frac{(A+B)^3-A^3-B^3}{(A+B)AB}$
Expanding the numerator gives:
$\frac{A^3+3A^2B+3AB^2+B^3-A^3-B^3}{(A+B)AB} = \frac{3AB(A+B)}{(A+B)AB}=3$[/hide]

wow tiredepartment that is a beautiful answer!!!

[quote="10000th User"][quote="236factorial"]That is a very nice answer! In reality though, it is just Farenhajt's answer, somehow, in terms of A and B.[/quote]But you missed one point, tiredepartment's answer is more general. For any real numbers $A\ne 0,B\ne 0, A+B\ne 0$, the expression is always 3. So you can have $A=92387498344737$ and $B=\pi$, you still get the same answer.[/quote]

Thanks, I didn't see that until you told me  :blush:

Can't it be simplified also with the difference of cubes?

[quote="236factorial"]Can't it be simplified also with the difference of cubes?[/quote]Sure why not?

Let $A=1999$, $B=1000$ and $A-B=999$, then you can apply [b]the cube of difference[/b] and you still get the same answer :)


MATHCOUNTS
complex analysis
linear algebra

of all the topics u have started, how many pages did the longest one end up being???

edit: and include what the topic was

9pages... but cant remember which one.... i just know that by the next time i looked at it, it was already at 7 pages ;)

The Google your name topic maybe.  I know I've started other topics that might have been longer, but I forgot what they were.

Note: Some people have never made a new topic before (Im not one of them, but Im suggesting an addition to the poll- 0pgs)

[quote="djimenez"]The one of Never Ending Story on the Fun & Games forum, it has 8 pages so far, but it turned a little different of what I spected (I spected to have a more structured story!) Any way is the one with more replays from the ones I have originally posted![/quote]
I think the only reason the story is not structured is becasue of its length.  I don't think many people actually read the first few pages before posting since it would take too long.

maybe the Your April Fools Day in the Games and Fun Factory...

[quote="djimenez"][quote="Peter VDD"]It wasn't on mathlinks, but the longest one I started had 12000 posts. :D[/quote]I suppose that here we should vote for the thread in MathLinks![/quote]Some sites have statistics on longests topics etc, but apparently Valentin didn't want that :)

Well, my last attempt got 14 pages in 2 days :) not bad, but apparently Valentin rather sees me working in the Complex Analysis and Linear Algebra forums on this site.

[size=200](HINT VALENTIN HINT)[/size] ;)

NBA-16 pages and counting...

edit: 13, the pages went down

Hey, did you noticed that the number of pages in all those topics have gone lower...... Now every page has 20 posts instead of 15! Cool!

Mine was 10 pages :D.  "What we really need to know" in the Mathcounts Forum.  It used to be the one with the greatest number of replies on the Mathcounts Forum, but it got locked (not sure why, probably cause it went off topic), and now its the one with the second greatest number of replies. :(
Still a good length though :D

[quote="xdl615x"]Makes me think... I should make one.[/quote]The idea is to start posts that are really interesting and with some "real" purpose, otherwise it would be considered spamming, and probably you will get some warning from an administrator (I think that in the last week or so there has been at least two or three attempts). Now, for example, Silverfalcon had a great idea when he thought about the [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=33505]Problem Solving Marathon 2[/url]. May be it is not the longest ever, but I would say that it should between the 10 more replyed ever in the forum... And is 100% related with the forum topic. I think the topic is great, and goes fast! They have 116 problems in less than a month, and 369 replys (18 pages so far).

Think about something similar, it could be funny and productive for the group :P

[quote="sheepwarrior"]well then, would this be considered spam or is it ok cuz this is the Games&Fun Factory section
cuz this topic is pointlesss i guess lol[/quote]What I meant is that it shouldn't be one of those [i]The last post wins[/i], or something like this specifically done to spam the site. This one, well, is sort of, but not totally spam. Any way, as someone already said (I think it was MithApprendice) G&F is the forum for all those posts that do not fit anywhere else in the forum.

[quote="furious"]sad but the most i've gotten was two pages. anything i start, basically dies almost instantly. i have no skills.[/quote]

The point is not to start the longest topic ever.  The point of AoPS forums is to meet new people, have fun, get some answers, and help some people.  It doesn't matter who has the longest topic.  I had to learn it the hard way :(


lets talk about actors and actresses.(because at last i think we can't talk about a movie becasue few of us had seen the same movie and ...)
i think they have a hard but lovely job and ...
here is a list of actors who i think they are good:
1)RUSSEL CROWE
2)ROBERT DENIRU
3)TOM HANKS
4)BRAD PITT
5)GEORGE CLOONEY
6)AL PACINO
...
ok join now and lets start with anyone you want. :D

Its spelled Robert De Niro, I actually just watched a movie in SS about him today.

thanks.so do you want to talk about him?

[quote="Kalimdor"]I think we can kick Mr. Pitt out of the list, and move Mr. Tom Hanks up one seat(Have you ever seen "Philadelphia"? Tom hanks is so awesome in it!

Frank Morgan[/quote]
yes Tom Hanks is a very good actor.i didn't see Philadelphia but i will see it.but i think Brad Pitt is good,may be he is not very very good but good.what do you think about him?

[quote="Kalimdor"]Mr. Pitt is more likely to be popular among actors depending on lookings, though he might need more practice on his acting skill. And I have a idea about all actors and actresses in general: The ones who are good at acting are not always the most popular and ( :roll: morally) the best actor
And if any of we mathlinkers are good at acting, I'd love to donate my lunch today(a huge size hamburger) to support him/her to goto Hollywood :lol:[/quote]
yes you are right about Pitt and i am agree with you.but today beauty is more important than acting for an actor (specialy actresses),so what should we do?

[quote="Kalimdor"]I guess we might need a purification of humanity itself, about the "modern" moral aspects of the socialty, especially america. :D
Maybe it's impossible, so we may ask the Acadamy to establish a new award for those who are not so Goooooood-Looking, but really good at acting. :)
And for the donation idea, we may suggest Valentin to make up a new topic with poll, see if we really get someone good at acting among us.  :roll: :ninja:[/quote]
yes you are right.we have a lot of people who are not good looking but they are really good actors.i agree with you. :D


function calculus derivative algebra polynomial trigonometry integration

Well, I am studying Classical Mechanics and I am currently on the chapter of Linear Oscillation. The Mathematics are getting to me a bit. I have a few questions to ask.

1. When solving the SHM Equation the solution says 'we seek solutions of the form $ x \equal{} e^{\lambda t}$ for $ \frac {d^{2}x}{dt^{2}} \plus{} \Omega ^{2} x \equal{} 0$'. Why are we seeking solutions of that form? Also solving the thing we get $ x \equal{} e^{\pm i \Omega t}$ and how does this form the basis of the real solutions?

2. Considering Damped Simple Harmonic Motion $ \frac {d^{2}x}{dt^{2}} \plus{} 2K \frac {dx}{dt} \plus{} \Omega ^{2} x \equal{} 0$. Can someone show me how to derive the equations for subcritical, supercritical and critical damping using the above method?

3. How do we solve $ \frac{d^{2}x}{dt^{2}} \plus{} 2K \frac {dx}{dt} \plus{} \Omega ^{2} x \equal{} F(t)$? What is Time Harmonic?

Thanks in advance.

For the first part we seek solutions of that kind $ e^{\lambda t}$ because it is an eigenfunction of the differential operator. The differential operator is now $ L(D)=D^2 +\Omega^2$ So we need to find functions, on which we "put" the operator.
So:
$ Dx=\lambda x$ that is we need to find a function that is equal to its derivative, times something this function is $ e^{\lambda t}$
similarly: $ D^2x=\lambda^2 x$...
So the differential operator for this system acts, on a function which is $ e^{\lambda t}$ and since the operator is a "polyinomial" of the differential operators, we get an $ L(\lambda)$  polynomial multiplied by $ e^{\lambda t}$
And this has to be zero in our case.
 $ L(\lambda)=\lambda^2+\Omega^2 =0 \rightarrow \lambda_{1,2}=\pm i\Omega$
now $ e^{\lambda_1 t}$ is clearly a solution of the DE. $ e^{\lambda_2 t}$ is also a solution, Hence their linear combination must be a solution too. So the general solution is:
$ x(t)=A_1e^{i\Omega t}+A_2e^{-i\Omega t}$ Here $ A_1$ and $ A_2$ are constants and depend on the initial conditions.
From this by e manipulation we can get the usual result, of $ x(t)=A\sin(\Omega t +\phi)$

Now lets look at the second question:
our differential operator is now: $ L(D)=D^2+2KD+\Omega^2$, hence the polynomial: $ L(\lambda)=\lambda^2+2K\lambda +\Omega^2=0$
From here:
$ \lambda_{1,2}=-K\pm \sqrt{K^2-\Omega^2}$
Now we see that we have something under the root. so we do case work:
a.) $ K>\Omega$
So $ K^2-\Omega^2>0$
Hence the solution, as before:
$ x(t)=A_1 e^{(-K+\sqrt{K^2-\Omega^2})t}+A_2e^{(-K+\sqrt{K^2-\Omega^2)t}
= e^{-Kt}(A_1e^{\sqrt{K^2-\Omega^2}t}+A_2e^{-\sqrt{K^2-\Omega^2}t})}$
We see that we will get hyperbolic functions if we manipulate it.
b.)$ K<\Omega$
Now$ K^2-\Omega^2<0$
Just as before:
$ x(t)=A_1 e^{(-K+i\sqrt{K^2-\Omega^2})t}+A_2e^{(-K+i\sqrt{K^2-\Omega^2)t}
= e^{-Kt}(A_1e^{i\sqrt{K^2-\Omega^2}t}+A_2e^{-i\sqrt{K^2-\Omega^2}t})}$
Now we will get $ x(t)=Ae^{-Kt}\sin{\omega t+\phi}$ Here: $ \omega^2=\Omega^2-K^2$
c.)$ K=\Omega$
I think you can do this now.

3. Now there are two famous methods to solve this one. One is using Green functions, the other is Fouriers method.
The basic idea is that we "break" up the $ F(t)$ function into many "little" funtions, we see the response function the system gives us, and we "add" those together then we get the wanted "full" response $ x(t)$ of the system. So symbolicaly:
1.) Let $ F(t)=\sum_k c_k \varphi_k(t)$ Here $ \varphi_k$ are the "little" funtions.
2.) We put the differential operator on this: $ L(D)\xi_k=\varphi_k$ so the response functions are $ \xi_k$
3.) we sum up, so the wanted response of the system: $ x(t)=\sum_k c_k \xi_k$
Now we now that functions can be "broken" up with the dirac-delta function. That is :
$ F(t)=\int_{-\infty}^{\infty} d\tau f(\tau)\delta(t-\tau)$ this is the first step.
We know that the response function of the dirac-delta is the Green function:
$ L(D)G(t-\tau)=\delta(t-\tau)$ second step.
so the response:
$ x(t)=\int_{-\infty}^{\infty}d\tau F(t)G(t-\tau)$
The Green function for this system is:
$ G(t)=\begin{cases}
0 & \mbox{if} \_ t<0 \\
e^{-Kt} \frac{\sin\omega t}{\omega} &  \mbox{if} \_  t>0
\end{cases}$
Now you can solve the DE.
The Fourier method is much longer so I wont write that down now.

[quote="BanishedTraitor"]1. When solving the SHM Equation the solution says 'we seek solutions of the form $ x \equal{} e^{\lambda t}$ for $ \frac {d^{2}x}{dt^{2}} \plus{} \Omega ^{2} x \equal{} 0$'. Why are we seeking solutions of that form?[/quote]

If we have any $ n^{th}$-degree linear homogeneous ODE with constant coefficients, say

\[ a_n \frac {d^{n}x}{dt^{n}} \plus{}a_{n\minus{}1} \frac {d^{n\minus{}1}x}{dt^{n\minus{}1}}\plus{}\cdots \plus{} a_1 \frac {dx}{dt}\plus{}a_0 x\equal{}0\]

then the substitution $ x(t)\equal{}e^{\lambda t}$ produces

\[ a_n \lambda^{n}e^{\lambda t} \plus{}a_{n\minus{}1} \lambda^{n\minus{}1}e^{\lambda t}\plus{}\cdots \plus{} a_1 \lambda e^{\lambda t}\plus{}a_0 e^{\lambda t}\equal{}0\]

\[ e^{\lambda t} \big(a_n \lambda^{n}\plus{}a_{n\minus{}1} \lambda^{n\minus{}1}\plus{}\cdots \plus{} a_1 \lambda \plus{}a_0 \big)\equal{}0\]

and since $ e^{\lambda t}\neq 0,$ dividing it out of the equation is ok, and we are left with a polynomial of degree $ n$ in $ \lambda$, namely

\[ a_n \lambda^{n}\plus{}a_{n\minus{}1} \lambda^{n\minus{}1}\plus{}\cdots \plus{} a_1 \lambda \plus{}a_0 \equal{}0\]

which is the so-called characteristic equation (of the ODE). The fundamental theorem of algebra tells us that there will be $ n$ roots of this equation, say $ \lambda_{1},\lambda_{2},\ldots ,\lambda_{n}$. So that $ \left\{e^{\lambda_{1}t},e^{\lambda_{2}t},\ldots ,e^{\lambda_{n} t}\right\}$ form a fundamental set of solutions to the original ODE, and the general solution is then

\[ x_H\equal{}c_1 e^{\lambda_{1}t}\plus{}c_2e^{\lambda_{2}t}\plus{}\cdots \plus{}c_ne^{\lambda_{n} t}\]

[quote="BanishedTraitor"]Also solving the thing we get $ x \equal{} e^{\pm i \Omega t}$ and how does this form the basis of the real solutions?[/quote]

If $ x\equal{}u\plus{}iv$ is a complex-valued solution to an ODE, where $ u$ and $ v$ are real-valued functions, then $ x\equal{}u$ and $ x\equal{}v$ are both real-valued solutions to the same ODE. Since Euler's equation gives $ e^{i\alpha}\equal{}\cos \alpha \plus{}i\sin\alpha$, we have that $ x \equal{} e^{\pm i \Omega t} \equal{} \cos\Omega t\pm i\sin\Omega t$ and hence $ x\equal{}\cos\Omega t$ and $ x\equal{}\sin\Omega t$ are also both real-valued solutions to the given ODE.

[quote="Thaakisfox"]3. Now there are two famous methods to solve this one. One is using Green functions, the other is Fouriers method.[/quote]

I think it would be more appropriate here to solve that differential equation by the usual method: first solve the associated homogeneous equation to get the solution $ x_h$; then find a particular solution $ x_p$ for the given dif. eq. by using the known methods if F(t) is of the form $ e^{at}[P_n(t) \cos(bt) \plus{} Q_m(t) \sin(bt)]$; finally, the general solution of the given dif. eq. will be $ x \equal{} x_h \plus{} x_p$. If F(t) is not of the form given above, use Lagrange's method.

Yes thats right carcul, but sometimes most of the time its much easier to solve with the Green method. Plus the Green method is also very usefull in solving wave equations, and electrostatic, optical problems etc. So this is the reason I wrote down this method since it can be used very widely, and is simple.


Hello :roll: Are there any methods for solving equation:
$ \frac{1}{\frac{1}{\frac{1}{\frac{1}{x}\minus{}x}\minus{}(\frac{1}{x}\minus{}x)}\minus{}(\frac{1}{\frac{1}{x}\minus{}x}\minus{}(\frac{1}{x}\minus{}x)}\minus{}(\frac{1}{\frac{1}{\frac{1}{x}\minus{}x}\minus{}(\frac{1}{x}\minus{}x)}\minus{}(\frac{1}{\frac{1}{x}\minus{}x}\minus{}(\frac{1}{x}\minus{}x)))\equal{}x$
(Denoting $ f(x)\equal{}\frac{1}{x}\minus{}x,$ we get $ f(f(f(f(x))))\equal{}x$ )
Some ideas? :wink:

Note that $ f^2 (a) \equal{} a \implies f^4 (a) \equal{} f^2 ( f^2 (a)) \equal{} f^2 (a) \equal{} a$, so solutions $ x \equal{} a$ to the equation $ f^2 (x) \equal{} x$ are also solutions to $ f^4 (x) \equal{} x$ (the reverse is not true).

$ f^2 (x) \equal{} x \implies \frac {1}{f} \minus{} f \equal{} x \implies 1 \equal{} f \cdot (f \plus{} x) \equal{} \left( \frac {1}{x} \minus{} x \right) \cdot \frac {1}{x} \equal{} \frac {1}{x^2} \minus{} 1$

Thus, $ x \equal{} \pm \frac {1}{\sqrt {2}}$ are two of the solutions.

We can say that $ f^2 (x) \equal{} x$ is a [i]root equation[/i] of $ f^4 (x) \equal{} x$. Notice that $ f(x) \equal{} x$ is a root equation of both $ f^2 (x) \equal{} x$ and $ f^4 (x) \equal{} x$.

[b]What about the other roots?[/b]

There are other solutions as well that you won't find using the root equation. Combining fractions and simplifying:

$ \frac {(2 x^2 \minus{} 1) (x^6 \minus{} 3 x^4 \plus{} 1)}{ x (x \minus{} 1)(x \plus{} 1) (x^2 \minus{} x \minus{} 1) (x^2 \plus{} x \minus{} 1)} \equal{} 0$

Since the numerator and denominator have no common roots, we have:
$ (2 x^2 \minus{} 1) (x^6 \minus{} 3 x^4 \plus{} 1) \equal{} 0$

We already discovered the solutions of the first factor from the root equation, which was much simpler. The second factor has six roots, four of which are real and none of which are nicely representable algebraically. See [b]math154[/b]'s post below.

Well, they are nicely representable... substituting $ x^2 \equal{} 1/(2\cos{u})$ we find that $ \cos{3u} \equal{} \minus{} 1/2$, and then that the roots are $ x \equal{} \pm\sqrt {1/(2\cos(k\pi/9))}$ for $ k \equal{} 2,4,8$. The roots of $ 2x^2\minus{}1$ correspond to $ k\equal{}0$.

TZF, did you mean that $ f(f(x))$ and $ f^2(x)$ are the same number? :maybe:

Right, by $ f^n$ I mean $ f$ applied $ n$ times. It is one convention for how to write several applications of the same operator. This is the same reason why $ f^{ \minus{} 1}$ means inverse of $ f$.

One exception to this notation is that a [i]positive[/i] superscript to a trigonometric function usually means exponentiating the result.


I drew circle arc from point A to point B like this:
[code]draw (arc (A, O, B));[/code]

Now I'm interested in drawing complementary circle arc - any help?

What do you mean by complementary arc? If you mean arc(B,O,A), you could program a new command to produce it.

[code]
path carc ( arc(a,b,c) ) {return arc(c,b,a) ;}
[/code]


complex numbers
geometry proposed
geometry

Given a circle $C$ and an exterior point $A$. For every point $P$ on the circle construct the square $APQR$ (in counterclock order). Determine the locus of the point $Q$ when $P$ moves on the circle $C$.

Note that if $a$ and $p$ are complex numbers representing points $A$ and $P$, then $a+(1+i)(p-a)=-ia+(1+i)p$ represents $Q$.

By centring $C$ at the origin, it is easy to see that the locus of $Q$ must be a circle whose centre is obtained by rotating $A$ $90^o$ clockwise about the centre of $C$, and whose radius is $\sqrt{2}$ times the radius of $C$.


1.) A horizontal beam has a pinned end at point A (that can sustain forces in the x and y directions but not a moment) and an additional brace CD is added with a roller (only has perpendicular force) end at point C and a clamped end at point D.
(my problem is i dont understand what [i]clamped [/i]is)

Treating ACB as a free body, write equation for Sum Y_forces=0. Then write equation for moment about any point on ACB. Then use those two equations to find Fc_y (y force on C)

[img]http://ronney.usc.edu/AME101F07/AME101-F07-PS3_files/image004.gif[/img]

edit: just one confusing problem =)

[quote="Qweretyq"](...) a pinned end at point A (that can sustain forces in the x and y directions but not a moment) (...) with a roller (only has perpendicular force)[/quote]

That information is not needed: a pinned end has always associated a force which in general has x and y components (other examples are rough surfaces), and rollers have always associated only a perpendicular force (other examples are rockers, frictionless surfaces, short links and cables, collars on frictionless rods, and frictionless pins in slots).

The "clamped end" is also known as a fixed support and has associated with it not only a force with unknown direction and magnitude (in general, there are components in the x and y directions) but also a couple whose moment has unknown direction and magnitude.