Which kingdoms of organisms are motile?

Are plants or fungi motile?

I already checked wikipedia but couldn't find it...

EDIT: Incorrect, sorry. (t0rajir0u's response is correct)

Plants are generally nonmotile.

(Adult) plants and fungi are sessile as a general rule.  They can produce motile sperm/spores.


[hide]This is like one of the BAMAs....:

the first line starts as 1 (nothing u can do about that)

There's only one 1, so we have 11.

1
11
21 (2 1's)
1211(one 2 and one 1)
111221 (one 1 one 2 and two 1's)
312211 (three 1's, two 2's, and one 1)
i think you get the pattern now
the next two lines are
[b]13112221
1113213211 There's your answer.[/b][/hide]


let p is an odd prime.
As we know 2^(p-1) - 1 is divided by p
And when [2^(p-1) - 1 ]/p is a perfect square number?

This has been posted before, twice if I remember correctly. I'll edit this post and put the link here when/if I find one of those two topics.

Edit: [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=1055]Here it is[/url]

oh,so quick...
I am new here.
Sorry for my repeats.


calculus
function
analytic geometry
graphing lines
slope
derivative

When ur given the function and then it is graphed, there is a point P on the slope of the tangent line which you get from the derivative of the function, but how do you solve for the point?

i don't quite understand  your question; also this topic should be moved

I may have understood this question incorrectly due to bad wording. Find the derivative of the function, set it equal to the given, I'm assuming, slope, and then solve.  I can't explain it any clearer than this if that is what you asked..


Easy question, but nice enough.

We start with the following obvious identity:
\[(a+b+c)^{2}=(-d)^{2}=a^{2}+b^{2}+c^{2}+2(ab+bc+ac)\]
This identity can be written in the following form:
\[d^{2}-(a^{2}+b^{2}+c^{2})+(ac+bc+ab)=3(ab+bc+ac)\]

Multiplying both sides by $d$ gives us that
\[d^{3}-d(a^{2}+b^{2}+c^{2}-ac-bc-ab)=3(abd+bcd+acd)\]

We know that \[a^{3}+b^{3}+c^{3}-3abc=(a+b+c)(a^{2}+b^{2}+c^{2}-ac-bc-ab)\]
\[=-d(a^{2}+b^{2}+c^{2}-ac-bc-ab)\]

So we can rewrite this last identity as \[d^{3}+a^{3}+b^{3}+c^{3}-3abc=3(abd+bcd+acd)\]
Which completes the proof. $\blacksquare$

[quote="kunny"]We can also prove that $a^{3}+b^{3}+c^{3}+d^{3}=3(a+d)(b+d)(c+d)$. :D[/quote]
Please post your prove!  :huh:

[quote="N.T.TUAN"]Let $a,b,c,d$ be reals such that $a+b+c+d=0$. Prove that
\[a^{3}+b^{3}+c^{3}+d^{3}=3(abc+bcd+cda+dab). \]
[/quote]
Another solution (in a book :P ).
We assume that $abcd\not =0$. Consider the equation \[x^{4}-(\sum a)x^{3}+(\sum ab)x^{2}-(\sum abc) x+abcd=0.\] with roots $a,b,c,d$. Substituting $x$ with $a,b,c,d$ and simplifying by $abcd\not =0$, after summing up we obtain \[\sum a^{3}-\sum a\sum a^{2}+\sum ab\sum a-3\sum abc=0.\]
By $\sum a=0$ we're done.

If $abcd=0$: trivial! :D


pigeonhole principle
number theory
greatest common divisor
algorithm
limit
integration
algebra

Hello guys, this is my first post here. I wasn't quite sure where this should fit, so I put it here, but feel free to move it elsewhere.

While working through another problem, I stumbled upon another question. Given two irrational numbers, can you pick two natural numbers such that the difference of each irrational number multiplied by the respective natural number is arbitrarily small? In a more formal way, is the following statement true:

Given any two irrational numbers, $ r_1$ and $ r_2$:
$ \forall \epsilon > 0$
$ \exists n_1, n_2 \in \{ 1, 2, 3, 4, ... \}$ such that
$ | n_1*r_1 \minus{} n_2*r_2 | < \epsilon$

I've done some work on this which I can post later, but so far I haven't been able to prove this one way or the other. Still, intuitively I feel like it should be true. If anyone could shed any insight, I would be appreciative.

Note that it is enough to prove this statement for $ r_1 \equal{} 1$.  There are two standard ways to go about this:  one is to use pigeonhole (consider the sequence $ \{ r_2 \}, \{ 2r_2 \}, ... \{ n r_2 \}$ where $ \{ x \}$ denotes fractional part), and the other is to remark that $ r_2$ has an infinite [url=http://en.wikipedia.org/wiki/Continued_fraction]continued fraction[/url] expansion and take its convergents.

The other comment is that this is true for any [i]real[/i] numbers $ r_1$ and $ r_2,$ but that to make it true, you have to allow for the possibility that the only way to make that linear combination be smaller than $ \epsilon$ may be to make it equal to zero.

There are two cases: $ \frac{r_1}{r_2}$ rational and $ \frac{r_1}{r_2}$ irrational. In the former case, there is a positive gcd of $ r_1$ and $ r_2$ and the continued fraction that t0rajir0u mentioned terminates. That's the case in which we have to make the linear combination be zero in order to get it to be arbitrarily small. (Note that this case could occur with both $ r_1$ and $ r_2$ irrational; the issue is whether they're rational multiples of each other.)

In the second case, with $ \frac{r_1}{r_2}$ irrational, we have an infinitely continued continued fraction, and we can make the linear combination arbitrarily small but never zero. In fact, the set of all integer linear combinations of $ r_1$ and $ r_2$ is dense in $ \mathbb{R}.$

To put any computational muscle behind this, I would use a four-column version on a spreadsheet of the Euclidean algorithm. Have I ever posted anything here with that algorithm? I can't remember whether I have.

[quote="t0rajir0u"]Right, I forgot about that case.  I should also mention that it is possible to prove the much stronger result that if $ \alpha$ is irrational then the sequence $ \{ n \alpha \}$ is [i]equidistributed[/i] in $ [0, 1]$ by Fourier-analytic methods.[/quote]

I have been working on this problem and similar ones but was unable to make any progress. Could you elaborate on the Fourier-analytic methods proof?

Here's an outline.

Suppose $ \alpha$ is irrational.

Step 1. If $ k\in \mathbb{Z}\setminus\{0\},$ then $ \lim_{N\to\infty}\frac1N\sum_{n\equal{}1}^Ne^{2\pi ikn\alpha}\equal{}0.$

Step 2. For all $ k\in\mathbb{Z},$ $ \lim_{N\to\infty}\frac1N\sum_{n\equal{}1}^Ne^{2\pi ikn\alpha}\equal{}\int_0^1e^{2\pi ikx}\,dx.$ (Note that $ k\equal{}0$ gets handled separately.)

Step 3: If $ f(x)$ is a trigonometric polynomial, then $ \lim_{N\to\infty}\frac1Nf(n\alpha)\equal{}\int_0^1f(x)\,dx.$

Step 4: Use the trigonometric Weierstrass theorem -- the trig polynomials are dense in the metric space $ C(\mathbb{T})$ --  to show that the identity in Step 3 holds for all periodic continuous functions.

Note for contrast that is $ \alpha$ is rational, then Step 1 fails for some values of $ k.$


geometry
incenter
circumcircle
angle bisector
geometry proposed

In a triangle $ABC$ , $B_{1}$ and $C_{1}$ are the midpoints of $AC$ and $AB$ respectively, and $I$ is the incenter. The lines $B_{1}I$ and $C_{1}I$ meet $AB$ and $AC$ respectively at $C_{2}$ and $B_{2}$ . If the areas of $\Delta ABC$ and $\Delta AB_{2}C_{2}$ are equal, ﬁnd $\angle{BAC}$ .

Dear All My Friends,
We solve this problem by using cosin and angle bisector theorems:
Suppose $ a$, $ b$, $ c$ are side lengths of $ ABC$.
We calculate $ AB_{2}$ and $ AC_{2}$. They are:
$ AB_{2}\equal{}\frac{b \cdot c}{(a\minus{}b\plus{}c)}$ and $ AC_{2}\equal{}\frac{b \cdot c}{(a\plus{}b\minus{}c)}$
Two given triangle areas are equal if and only if:
$ b \cdot c\equal{}\frac{b\cdot c}{(a\plus{}b\minus{}c)}\cdot \frac{b\cdot c}{(a\minus{}b\plus{}c)}$
From this: $ b^{2}\minus{}b\cdot c\plus{}c^{2}\minus{}a^{2}\equal{}0$
Moreover by cosin theorem: $ a^{2}\equal{}b^{2}\plus{}c^{2}\minus{}2\cdot b\cdot c\cdot cos(A)$
Therefore: $ cos(A)\equal{}\frac{1}{2}$ and we are done.

Calculation: $ AC_{2}\equal{}\frac{b \cdot c}{a\plus{}b\minus{}c}$
Denote $ AC_{2}\equal{}x$, $ r$ is radius of incircle, $ h_{c}$ is length of altitude from $ C$,  $ N$ is touch point of incircle with $ AC$
We use areas of some triangles. Denote $ S(ABC)$ as area of $ \triangle ABC$
$ S(AB_{1}C_{2})\equal{}S(AB_{1}I)\plus{}S(AC_{2}I)\equal{}S(AB_{1}I)\plus{}S(AB_{1}I)\cdot \frac{AC_{2}}{AB_{1}}$
$ S(AB_{1}C_{2})\equal{}S(AB_{1}I)\plus{}S(AB_{1}I)\cdot \frac{2 \cdot x}{b}\equal{}S(AB_{1}I)\cdot \frac{b\plus{}2 \cdot x}{b}$
$ x\cdot \frac{h_{c}}{4}\equal{}S(AB_{1}I)\cdot \frac{b\plus{}2 \cdot x}{b}$
$ x\cdot \frac{2\cdot S(ABC)/c}{4}\equal{}\frac{r\cdot b}{4}\cdot \frac{b\plus{}2\cdot x}{b}$
$ x\cdot\frac{S(ABC)}{2\cdot c}\equal{}\frac{2\cdot S(ABC)/(a\plus{}b\plus{}c)}{4}\cdot (b\plus{}2\cdot x)$
$ \frac{x}{c}\equal{}\frac{(b\plus{}2\cdot x)}{(a\plus{}b\plus{}c)}$
$ x\cdot (a\plus{}b\plus{}c)\equal{}c\cdot (b\plus{}2\cdot x)$
$ AC_{2}\equal{}x\equal{}\frac{b\cdot c}{a\plus{}b\minus{}c}$

Best regards,
Bui Quang Tuan

Dear All My Friends,
I suggest another problem. It is exactly as original but only one slightly changed: we replace Incenter $ I$ with Circumcenter $ O$.
How about the solution now?
Best regards,
Bui Quang Tuan


Circle $c_{1}$ is concentric with and smaller than $c_{2}$.  Circle $w_{1}$ is concentric with and smaller than $w_{2}$.  Circles $c_{1}$ and $w_{2}$ are externally tangent, as are $w_{1}$ and $c_{2}$.  If $|Circle|$ is the area of a circle and $|c_{1}| : |c_{2}| = |w_{1}| : |w_{2}|$, prove that $|c_{2}| = |w_{2}|$ and $|c_{1}| = |w_{1}|$

[quote="Inquisitor"]Are you absolutely sure that it is not:
Prove $|c_{1}|=|w_{1}|$ and $|c_{2}|=|w_{2}|$? I think I can find a counterexample to the stated problem, if my condition above is followed...[/quote]

Sorry, that's right.  I'll edit it.


Hello AoPS,
I am a high school student living in Colorado, I have a strong interest in pretty much three things, games, web design and math.  This led me to an idea, why not start an MMORPG based on math, for those of you who don't know, MMORPG stands for Massive Multplayer Online Role Playing Game.  Basically it would be a game where users answer math questions to raise their "army size," each question answered correctly would add 1 to their army.  The user has a Turn Based Gold, which is affected by the army size, the larger the army, the more gold they get.  They can buy weapons (or the equivalent of these in math terms) and attack other users to steal gold.   Anyone interested in helping or anyone with idea or anyone who thinks this is stupid, email me at virtus@comcast.net.  This is not limited to people with coding experience, to start a game like this we would definitely need a plethora (i like that word) of problems to use, so anyone interested email me.

I apologize if this is the wrong thing to post on these forums and if it is an admin may delete it at their will.

Thanks,
Eric Conner

This is a cool idea.  With a tad of tweaking it could be done well.  You could have "parties" a la Diablo II, where people can work together to solve stuff.

Wow.  That's a really cool idea.  Regarding the points-earning system, would you have a question-of-the-day (or several-questions-of-the-day) system, or a large database from which one could solve problems forever?  I think the first option is better, since otherwise people could write down their answers and keep feeding them in.  Not to mention the repetitiveness factor.

I can help with writing/garnering problems, although I don't have much experience with coding.

I think you've suggested a nice idea, but it will not be easy to put together.  Most significantly, you'd need thousands and thousands of problems in order to make the experience really worthwhile -- ideally, not everyone would always get the same problem and so on.  I don't know what kind of resources yo would need to collect enough to make it worthwhile.  Additionally, you'd have to design some system for checking if the problems are well-posed.  The simplest solution to this is multiple-guess questions, but that's not too much fun :)

Yea, I dunno prolly a few problems a day although it would be hard to update problems every day.  Maybe a database that allocated problems each day but then the game would have limits but i guess a huge database would be hard to put together.

Dunno... if it was put together, I'd probably join...

Seriously, this is not a good idea.  Consider the following.

The whole concept of any turn based MMORPG in the long run is to keep players in by forcing them to log in each day to use their daily turns, else they are forfeit and that player does not advance.  This builds a daily revenue from advertising that keeps the game in existence.  The problem is, people get bored from doing the same repetitive things everyday to keep their player/character intact and eventually leave the game.

Solving math problems is fine.  But why do you introduce the turn based aspect?  This is a often used _gimmick_ that will only hurt your game, by detracting the focus of the game from solving math problems into "making sure I log in to get my daily credits."  And believe me, people get tired of that fast.

If you want further proof, look closely at games like "Kings of Chaos", "Outwar", etc.

This is an awsome idea man!
I'll see if my programing friends will try to make one.
It would be cool though, because the treee things i'm interested in is also math, computers, and gaming.

My friend Claudiu and I are working on an isometric-2d network game system. When it's done it should be extensible enough to use for what you are suggesting. http://cs.stuy.edu/cswiki/Multiplayer_20Thing

Turn-based MMORPG?
Try
[url=http://www.the-reincarnation.com]Magewars[/url]
It's a very very mathematically based game. Fighting requires some hard calculations (well, only the good players do it)


Need help!

Prove that

C(7,5) = C(4,4) + C(5,1) + C(6,2)

thanks

Or...

$ \binom{4}{4}+\binom{5}{1}+\binom{6}{2}= \binom{4}{0}+\binom{5}{1}+\binom{6}{2}= \binom{7}{5}$ by the Hockeystick Identity.


This thread is for game moderators and prospective absent players. Post here, if your game needs some replacements. If you want to fill an open slot, you can respond to the game moderators here, too.

Please always specify the game and give a link to the corresponding thread.

Note: it's up to the moderators/players if they accept your wish for participation, depending on the game it may not be possible. You also should be prepared to acquaint with the game before joining a game.

[quote="AIME15"]I'll replace in any game that is not totally pointless.[/quote]
This.
Apparently this is only quote and code tags. Oops.

I thought that you were only supposed to post stuff like this after someone has posted that they need replacements.

Indeed, as long as you don't want to promiss that you will join anything at any time  :wink:

Possible replacement needed for a player in my Survivor game, message me if interested.

I need a replacement for Mediocrity, PM if interested

I will need one replacement for [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=248199]Beat the Mod! II[/url]. PM if you're interested.

I need another replacement for Mediocrity.

I need one replacement for The Race to Survival, possibly more. PM me if interested. Note: You must be fairly active for this game, i.e. check the thread at least once a day.

EDIT: Make that three replacements. Anyone?

I'll replace for Race To Survival if I'm not too late.

I need one replacement for The Legend of the Thirteenth Moon.

One replacement needed for [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=271422&sid=9e1d95dc4d34dc06507bf3ea0550eb14]The Mole 5[/url]

No application required.

https://artofproblemsolving.com/community/c1202119h2147751_welcome_to_the_hobbies_club

Would you like to join my forum?

[b]Looking for a replacement in Mafia Game 237.  Please message me if you're interested.[/b]

First post since literally my $14^\text{th}$ birthday!!!

[quote=SpeedCuber7]lock nobodys doing games anymore[/quote]

yes people are

[quote=SpeedCuber7]still lock this thread[/quote]

noo

[quote=SpeedCuber7]still lock this thread[/quote]

that's all your saying

You know the FF announcement is locked, right?

I need people for my Wof roleplay message me :)


LaTeX
parameterization
real analysis
real analysis unsolved

I am new to this forum so if this is posted in the wrong section just let me know!

I have an idea for a website but I am having difficulty working out the mathematical aspect of it and deciding if it will be profitable or not. Here is a short overview of how the site will work:

It is a website that users must pay USD 25 a year to join. On the website, members engage in one on one battles with eachother. If they win a battle they receive 1 point. If they lose a battle they are deducted 1 point. Members can engage in a maximum of 730 battles per year. Members can turn in their points in at a store on the website in exchange for merchandise. The US dollar value of the merchandise ranges from USD 25 - 4000. What I need to figure out is how many US dollars each point should be worth so that I can associate a point value with the merchandise in the store. The dollar value of merchandise purchased at the store per year should be less than USD 25 x the number of members so that the website makes a profit.

I think this might have to do with statistics because I would assume that the graph of member's wining percentages would be a normal curve. Maybe somehow you could use that to figure out on average how many points should be redeemed in the store in order for the website to make a profit.

I expect the website to have a relatively small amount of members at first but grow to about 4000 members eventually. It would really help if you could show the profit for varying amounts of members between 50 - 4000.

Any help would be greatly appreciated!

I edited your post a bit (replacing dollar signs that are used for bounding $ \text{\LaTeX}$ formulae here by USD).

Let's clarify some details first:
1) Do you keep negative scores or the scores can never go below $ 0$?
2) Is it realistic to assume that the outcome of each battle is random or it depends heavily on the skills of the battlers? (I mean, if you have a group of skilled users constantly playing against weaklings to gain points, you will see quite a different picture from the one when they just play against each other. It can be so different that I would even suggest creating some ranking system and awarding/removing points according to the strength of the opponent, so the strong person playing against a weak one will have no chance to gain an easy point. It will make the thing much more stable, allow you to avoid heavy losses, and will also sound fair to the users).

Sorry, but personally I am not really supporting using a mathematical battle for money.

fedja-

To answer your questions:

1) I was planning on having negative scores.  Would it be better to have the scores just stop at 0?
2) The players will be skilled and it will not be random who wins.  Do you suggest I do somethig along the lines of  awarding 4 points for battling someone above you, 2 points for battling somene within 75% - 100% of your points, 1 point for battling someone within 50% - 75% of your points, and not allow you to battle other members who have less than 50% of your points?

Thanks for the help so far!

@hungkhtn Well, since nobody said that battles will be mathematical, you probably meant "I don't like using mathematics to make money". This basically reduces your possible career choices to a pure researcher in a non-teaching institution. True, there are a few places there, but to get one of them is not at all easy so I predict that you'll end up using mathematics to make money for somebody else anyway. :P

@looking4answers I just want to show you the difference between a pure ranking system and a pure point system in a game of skill.

Suppose that the ranking system is the following: each person initially has rank 10. The loser pays the winner a certain portion (say 1/10) of his rank (so rank 20 versus rank 5 game results in 18,7 if the top guy loses and in 20.5,4.5 if he wins). The total rank will be redeemed in the end of the year. Here you do not need to care about anything because you know [b]exactly[/b] what the sum of all ranks is, so as long as each rank point is below USD 2.5, you make profit.

Assume now that you have your original scheme (each game is worth 1 point) Suppose that you have 100 players in 2 groups: 50 strong and 50 weak and they play randomly. Suppose that the strong player always beats the weak one and in other cases the outcome is random. Suppose that each of them plays 300 games. Then each strong person is expected to accumulate 150 points. This should be less than USD 50 to make profit, so you need your points to be worth about 33c or less. With 700 possible games this value drops to 16c.

So, I hope you see the logic. I would strongly recommend something closer to the first scheme (it is safer for you; much more advantageous for the strongest player; everybody has some noticeable amount to redeem initially; the point value is much higher; if you are skilled, you do not need to play like crazy just to get even: all you need is to beat a few opponents who didn't lose their initial rank yet (if you can find them :P); the strong persons will seek to play against each other; etc.)  but it is up to you to make the decision. In the first case you basically run a loosely organized tournament with an entry fee and the total prize amount equal to a certain portion of collected money. In the second case you just give free money to the stronger part of the group according to how many weaklings they managed to beat.

Once you figure out the exact scheme you want to use, the mathematics shouldn't be that hard except if you choose to rely on probabilities, the answer may depend heavily on a few unknown parameters involved. So I suggest that at this stage you just think a bit of what exactly you want to do.

Thanks a lot fedja! All of your info has been extremely helpful!  I think I am going to go with the first scheme you described and allow players to chose what percentage of points they would like to wager in each battle.  Also if a player loses all of their points they will be able to buy 10 more points for USD 25.  This seems like the most reasonable way to set up the website.


Apply Huygens inequality for (1+abcd) and then break into inequalities in one variable.

Manlio, you must have meant

$2\left( a^{3}+1\right) \left( b^{3}+1\right) \left( c^{3}+1\right) \left( d^{3}+1\right) \geq \left( a^{2}+1\right) \left( b^{2}+1\right) \left( c^{2}+1\right) \left( d^{2}+1\right) \left( abcd+1\right) $.

This is problem 97 (by Vasile Cîrtoaje) of "Old and new Inequalities". I would be glad to try this but I don't have the time...  :(

  Darij


algorithm
induction
combinatorics proposed
combinatorics

Let $m$ circles intersect in points $A$ and $B$. We write numbers using the following algorithm: we write $1$ in points $A$ and $B$, in every midpoint of the open arc $AB$ we write $2$, then between every two numbers written in the midpoint we write their sum and so on repeating $n$ times. Let $r(n,m)$
be the number of appearances of the number $n$ writing all of them on our $m$ circles.

a) Determine $r(n,m)$;

b) For $n=2006$, find the smallest $m$ for which $r(n,m)$ is a perfect square.

Example for half arc:    $1-1$;
$1-2-1$;
$1-3-2-3-1$;
$1-4-3-5-2-5-3-4-1$;
$1-5-4-7-3-8-5-7-2-7-5-8-3-7-4-5-1$...

Consider all different half arc sequences:
$s_1: 1,1$
$s_2: 1,2,1$
$s_3: 1,3,2,3,1$ and so on.
The following assertions are true:
1. If $a,b$ is an ordered pair of consecutive numbers from one of this sequences. Then $(a,b)=1$. (obvious, follows from Euclid's algorithm).
2. All ordered pairs $a,b$ of positive integers are found exactly once in exatly one of the sequences.
We'll prove it using induction on $a$.
For $a=1$ it's obvious ($1$s are found only in the beginning and in the end of each sequence, second element of $s_i$ is $i$).
Let it be true for all $a'<a$, $a\ge2$. Consider ordered pairs $i,a$ with $i<a, (a,i)=1$. Each of them is found exactly once. But each sequence is symmetrical (a palyndrome), hence each ordered pair $a,i$ is found exatcly once.
But $a,i$ is found exactly once in $s_j$ iff $a,i+a$ is found exactly once in $s_{j+1}$. Hence $a,i$ is found exactly once in $s_j$ iff $a,i+ka$ is found exactly once in $s_{j+k}$.
Therefore the assertion #2 is true for $a$.

It can be easily proved using induction on $j$ that:
1. If $a,b$ is in $s_j$ then $j\le max(a,b)$.
2. If $a,b$ and $a,c$ are in $s_j$ then $b\neq c (mod a)$.

For $n\ge2$ consider all numbers $x$ less than and coprime with $n$. Hence for each such $x$ we can find exactly one ordered pair $n,y$ from $s_n$ with $x=y (mod n)$. Therefore there are exactly $\phi(n)$ such pairs in $s_n$.

The result follows easily.


Prove that for each set $S=\{1,2,...,2^{n}\}$ there exist $A\subset S$ and $B\subset S$ such that $A\cup B=S$ and \[\sum_{x\in A}x^{k}=\sum_{x\in B}x^{k}\] for $k=0,...,n-1.$

maybe http://www.mathlinks.ro/Forum/viewtopic.php?t=98965  :wink:


algebra unsolved
algebra
Sums and Products

given the two conditions :
-   $\displaystyle \sum_{k=1}^{1999} a_k=2$

 -   $a_1.a_2+\cdots+a_{1998}.a_{1999}+a_{1999}.a_1=1$

$\displaystyle \mathcal{S}=\sum_{k=1}^{1999}a_k^2$.
     find $\min \mathcal{S}$ and $\max \mathcal{S}$

l don't know the nature of the $a_i$'s (l think they $\in \mathbb{Z}$ or $\mathbb{R}$). this problem has already been given, but l can find it

this is probably not a solution, but l found that :
l began with 3 instead of 1999.
$a_1^2+a_2^2+a_3^2=(a_1+a_2+a_3)^2-2(a_1a_2+a_2a_3+a_3a_1)$,  so
$\displaystyle\sum_{k=1}^{1999}a_k^2=a_1^2+\cdots+a_{1999}^2=\left( \sum_{k=1}^{1999}a_k \right)^2-2(a_1a_2+\cdots+a_{1998}a_{1999}+a_{1999}a_1),$ and $\mathcal{S}=2^2-2\times1=2$

l am surprised that nobody found it, that's why l wonder lf l haven't miss something...

arthur! i begin to wonder...
we actually have $(\sum a_i)^2=\sum a_i^2+2a_1a_2+2a_1a_3+2a_1a_4+...+2a_1a_{1999}+2a_2a_3+2a_2a_4+...+2a_2a_{1999}+...+2a_{1998}a_{1999}$, so much more terms than we know(all combinations of a's)! just in the case of three variables that's the same.

Peter

umm... anyway, this doesn't solve the minimum and maximum problem.
and peter, l think you are false. actually
$(a_1+\cdots+a_{1999})(a_1+\cdots+a_{1999})=a_1(a_1+\cdots+a_{1999})+a_2(a_1+\cdots+a_{1999})+\cdots+a_{1999}(a_1+\cdots+a_{1999})$

sure and now compare with my formula. we have $(\sum a_i)^2=\sum a_i^2+2\sum_{i<j}a_ia_j$. that's the way one normally writes it down, but i was not sure if you understand the second sum...it's over all pairs (i,j) of indices such that i<j.

Peter

yes, l have already meet that kind of sum like \[    \left( \sum_{i,j=1}^n|x_i-x_j| \right)^2\leq \frac{2(n-1)}{3} \sum_{i,j=1}^n(x_i-x_j)  ,  \] (remember) where there is always a distance between the $x_i$ and the $x_j$


[quote="shyong"]A vessel of volume $0.02m^{3}$ contains an ideal gas at pressure $1.5\times 10^{5}$ Pa. a tap attached to the vessel is opened and the gas expands adiabatically until its pressure equals the atmospheric pressure of $1.05\times 10^{5}$ Pa. The tap is then closed. The gas is allowed to return to its original temperature. in the final equilibrium state, the pressure of the gas that remains in the vessel is $1.13\times 10^{5}$ Pa .

Find the volume of the gas remains in the vessel .

I try to use PV/T = constant to find but feel that it is not correct because the V here is volume of vessel and it is constant . Any help ?[/quote]
shyong,

The key equation for an adiabatic expansion is $pV^{\gamma}= \text{ const.}$

Since the temperature of the gas is returned to $T_{1}$ from $T_{2}$ via a constant volume process, by manipulating the ideal gas law one finds $\frac{p_{1}}{p_{3}}= \frac{V_{2}}{V_{1}}.$

By taking the logarithm of the key equation one finds,

$\log \left(\frac{p_{1}}{p_{2}}\right)+\gamma \log \left(\frac{V_{1}}{V_{2}}\right) = 0.$

$\implies \log \left(\frac{p_{1}}{p_{2}}\right)-\gamma \log \left(\frac{p_{1}}{p_{3}}\right) = 0.$

$\implies \gamma = \frac{\log \left(\frac{p_{1}}{p_{2}}\right)}{\log \left(\frac{p_{1}}{p_{3}}\right)}.$

Once you know $\gamma$ you can plug it into the key equation and find the volume you want.

[quote="shyong"]Thanks for replying Dr.No .

Actually ,finding $\gamma$ is the second part of the question and fiding volume that remains is the first part .

I follow your first step

$\frac{p_{1}}{p_{3}}=\frac{V_{2}}{V_{1}}$

so $\frac{1.5}{1.13}=\frac{V_{2}}{0.02}$

which yields $V_{2}=0.02655m^{-3}$

Now I get this value of $V_{2}$ which it is obviously greater than original volume $V_{1}=0.02$ . But isn't that when we open the tap,some of the gases have run out the vessel ? Shouldn't the volume be something else smaller than $V_{1}$ ? Can you clear my doubt ? Thanks .[/quote]

Edit: I got the sign on the exponent wrong. Correcting it:

From the conditions given in the problem I calculate $\gamma = 1.259.$ This value gives me a ratio $\frac{V_{3}}{V_{1}}= \left(\frac{p_{3}}{p_{1}}\right)^{-\frac{1}{\gamma}}= 1.252.$

I assume it's a trick question, the volume of gas in the container remains the same but at a lower pressure.

Okay let's see if I can do this right this time.

In an adiabatic process some volume of gas, raised to an exponent greater than 1, is inversely proportional to pressure. That means that if pressure drops, the volume will increase. In the problem at hand the pressure does drop; therefore the volume of gas increases. The volume of the container however is fixed, therefore in order for the adiabatic process to take place some volume of gas must escape the container.

An adiabatic expansion takes place in two stages: 1) The gas expands isentropicly from state one to state two (this is the adiabatic portion of the process), and 2) the gas returns from temperature $T_{2}$ to temperature $T_{1}$ isochoricly (along a line of constant volume).

$T_{3}= T_{1}\implies nR T_{3}= nR T_{1}\implies P_{3}V_{3}= P_{1}V_{1}.$

$V_{3}= \frac{P_{1}}{P_{3}}V_{1}.$

$V_{2}= V_{3}$

The term [b]adiabatic[/b] means there is no heat added; this implies

$dQ =\frac{\partial Q}{\partial V}dV+\frac{\partial Q}{\partial P}dP = 0.$

From the chain rule,

$\frac{\partial Q}{\partial V}= (\frac{\partial Q}{\partial T})_{P}\frac{\partial T}{\partial V}) \text{ and}$

$\frac{\partial Q}{\partial P}= (\frac{\partial Q}{\partial T})_{V}\frac{\partial T}{\partial P}).$

$(\frac{\partial Q}{\partial T})_{P}\equiv C_{P}\text{ and }(\frac{\partial Q}{\partial T})_{V}\equiv C_{V}.$

From the ideal gas law:

$\frac{V}{nR}= \frac{\partial T}{\partial P}\text{ and }\frac{P}{nR}= \frac{\partial T}{\partial V}.$

Putting the pieces together: $dQ = C_{P}\frac{P}{nR}dV+C_{V}\frac{V}{nR}dP = 0$

$\implies C_{P}\frac{dV}{V}+C_{V}\frac{dP}{P}= 0$

$\implies \frac{C_{P}}{C_{V}}\frac{dV}{V}+\frac{dP}{P}= 0.$

$\gamma \equiv \frac{C_{P}}{C_{V}}$

$\therefore \gamma \frac{dV}{V}+\frac{dP}{P}= 0.$

$\int \gamma \frac{dV}{V}= \int-\frac{dP}{P}$

$\ln V^{\gamma}=-\ln P+\ln k$

$\therefore V^{\gamma}= k/P \Rightarrow \boxed{PV^{\gamma}= k.}$

We have $\frac{V_{3}}{V_{1}}= \frac{P_{1}}{P_{3}}= \frac{1.5\times 10^{5}}{1.13 \times 10^{5}}= 1.327.$

We also have that $P_{1}V_{1}^{\gamma}= P_{2}V_{2}^{\gamma}\Rightarrow \ln P_{1}+\gamma \ln V_{1}= \ln P_{2}+\gamma \ln V_{2}$

$\gamma = \frac{\ln P_{1}-\ln P_{2}}{\ln V_{2}-\ln V_{1}}= \frac{\ln P_{1}-\ln P_{2}}{\ln V_{3}-\ln V_{1}}= \frac{\ln P_{1}-\ln P_{2}}{\ln (1.327 V_{1})-\ln V_{1}}= \frac{\ln P_{1}-\ln P_{2}}{\ln (1.327)+\ln V_{1}-\ln V_{1}}.$

$\gamma = \frac{\ln (\frac{1.5 \times 10^{5}}{ 1.05 \times 10^{5}})}{\ln (1.327)}= 1.259.$


[quote="anirudh"]Two numbers have a difference of 3 and a sum of 31. What is the larger of the two numbers?[/quote]
[hide]$17$[/hide]

[hide]
call the two numbers x and y

x - y = 3
x + y = 31

2x = 33
x = 33/2
x = 16.5

16.5 + y = 31
y = 14.5

now these are not actual values for this but it gives a good estimate of what the two terms are. now it is guess and check. you know the numbers must add up to 31 and you know one of them has to be 3 digits greater than the other.

the only two i can think of are
x = 17
y = 14

x is the larger number having a value of 17[/hide]

[hide="We can"]add 31 and 3 to get twice the larger number. We divide by two to get 17.[/hide]

[hide][quote="anirudh"]Two numbers have a difference of 3 and a sum of 31. What is the larger of the two numbers?[/quote]

x-y=3
x+y=31
2x=34
x=17
y=14[/hide]

I've learnt that in the second grade.
[hide=":showoff:"]
$x-y=3\implies x=y+3$
Knowing that $x+y=31$ we substitute and get  $2y+3=31\implies2y=28\implies y=14$ we will get that $x=\boxed{17}$[/hide]

[hide]The average is 15.5 and they are 3 apart, so each is 1.5 from the mean, the larger number in the positive direction, and 15.5+1.5=[b]17[/b][/hide]

Have fun, think of it this way:
   x-y=3
+ x+y=31
----------- (equal sign)
2x=34
x=17
[/quote]

[quote="#H34N1"]Two numbers have a difference of 3 and a sum of 31. What is the larger of the two numbers?[/quote]

[hide]LOL

$ x-y=3$
$ x+y=31$
$ 2x=34$
$ x=17$
$ 17-y=3$
$ 17+y=31$
$ y=14$
larger is (drumroll)
$ 14!!$
Just kidding
$ 17$[/hide]

this is from the 2006 countdown round state or chapter i cant remeber for mathcounts.

add the 2 numbers together and divide by 2. [b]17[/b]

[hide="how does this work?"]let a and b be the 2 numbers.
we have
a+b=31
a-b=3
since a-b is greater than zero, a is the larger number. so we get rid b. add them together. thatss where we get adding 31 and 3
2a=34
then solve for a: divide by 2.[/hide]


trigonometry
probability
geometry
3D geometry

I want to see how many odd perception questions there are out there.

This will be a marathon!!! Please continue it as you please. :D

Thanks, and please follow the rules of regular marathons.

:D:D:D

What people call "trick questions". Do you have any?

[quote="nonie"][quote="zeb"]Is it legal for a man born in Utah to marry his widow's sister?[/quote]

how does a dead man marry? :lol:

a very old one:

thers this 1 story house every thing in it is purple the walls are purple the  charis the couches the carpet EVERYTHING what colors are the stairs?[hide=" answer"][hide=" sure you want to know?"][hide="fine have it ur way.."]hahaha its a one story house[/hide][/hide][/hide][/quote]

It's a one story house. :D

Someone else post next question.

Do not post answer :D (if you posted the problem...)

you are the driver of a bus

you pick up 3 people, you drop off 0

then, you pick up 45, drop off 3

then, you pick up 37, drop off 34

then, you pick up 1289, drop off 46

whats the eye color of the driver

my eyes are reallyyy dark brown!

If Mr Smith's peacock lays an egg in Mr Jones' yard, who owns the egg?

the persons eye color is unknown cuz he is blind

[quote="zeb"]Can peacock's lay eggs?[/quote]

Actually no. Peacocks are the male of the species. The peahens lay eggs. so neither owns the egg sine it's non-existant.

[quote="236factorial"]I'm not sure if this is a trick, but I want ot know,...

If a baby is born, half on one side of the international date line, half on the other side. It is 12:15AM on one side, but 11:15PM on the other. When is the baby's birthday?[/quote]

the probablility an event like being born/crashing etc. on a line is sooo close to zero. actually, for all practical purposes it is 0 and thus you can't be born on the line. you're on one side or the other, by some amount, no matter how small.
and i would say that you are on the side of the line that the majority of your body is on.

This one is quite an odd perception question. Can anyone guess the answer?
[color=white] 5 [/color]

What was the president's name in 1960?  :P

[quote="zeb"]Bread. :rotfl: Good one.

A lady is walking down the street, and she sees a beggar. The woman is the beggar's sister, but the beggar is not the woman's brother. How is this possible?[/quote]
The beggar is the woman's sister.

What do cows drink?

[quote="IntrepidMath"][quote="ZennyK"]
What do cows drink?
[/quote]

Its best to ask this question after making the person repeat the word silk about 20 times. :D[/quote]

I think it's even better after milk.

Cows do drink milk. At least baby cows do. Like all mammals...

. . . but a baby cow is called a calf. ;)

If a red house is made of red bricks and a blue house is made of blue bricks, what is a green house made of?
(don't answer if you've heard this one before)

[quote="ZennyK"]If a red house is made of red bricks and a blue house is made of blue bricks, what is a green house made of?
(don't answer if you've heard this one before)[/quote]
[hide=" :)  :(  :?  :D  :D  :lol:  ;)  :blush:  :P  :o  :mad:  :oops:  :roll:  :  :|  :rotfl:  :huh:  :ninja:"]um, i haven't heard this one but my answer is bricks.[/hide]click on a smiley

Glass! :lol:

I don't have a question/problem. Someone else should go.

[quote="aznmonkey1992"]do you know what a greenhouse is?
just google it

say this quickly:
i slit a sheet,
a sheet i slit,
and on a slitted sheet i sit[/quote]

w00t I did it without cursing

@greenhouse...

On a related note,

If Mr. Red lives in the red house and Mr. Blue lives in the blue house, who lives in the white house?

the president.

someone post a new one. I don't have any :)


geometry
analytic geometry
symmetry
3D geometry
prism
AMC
AIME

Points $A$, $B$, $C$, $D$ and $E$ are located in 3-dimensional space with $AB = BC = CD = DE = EA = 2$ and $\angle ABC = \angle CDE = \angle DEA = 90^{o}$. The plane of $\Delta ABC$ is parallel to DE. What is the area of $\Delta BDE$?

Well this implies that A, C, D, E are on the same plane. However, we do not know the angle between the planes ABC and the one above, so we are not able to calculate the area of the triangle, unless there is some technique to do this which I do not know of.

I didn't take this test or look at the answer,
[hide="but I will try"]
We are given that line DE is parallel to the plane containing A, B, C.  Let D' and E' be points in the plane containing A, B, C, such that DD'=EE', and both segments are as short as possible. Because of this, we get D'E'=2, angleD'E'A=angleE'D'C=90, and since EA=DC, E'A=D'C=1. Since ABC is a right triangle of length 2, either AB or CB is parallel to D'E', meaning either BD' or BE' is equal to 1. By symmetry, these possibilities are equivalent, so let's say BD'=1. Since CD=2 and CD'=1 and angleDD'C=90, by congruent triangles, we now have the information to tell that BD'E' is right with right angle at D', so angleBDE is right and BD=CD=2. So the area is (2)(2)(1/2)=2.
[/hide]

Could someone explain why it is a prism?

I didn't solve it during the test because I kept thinking that angles CDE and DEF were coplanar.  When I got past that and figured out that it was a prism formed from an equilateral triangle, I wanted to hit myself. :huh:

[quote="paladin8"][quote="samath"]Ummm I sorta cheated  :D

[hide="Cheap but effective solution"]
I put the points on a 3-D coordinate system, with B at (0,0,0).

I came up with:
A=(2,0,0)
B=(0,0,0)
C=(0,2,0)
D=(1,2,$\sqrt{3}$)
E=(1,0,$\sqrt{3}$)

Then BE=ED=2 and $BD=2\sqrt{2}$, so BDE is a right isosceles triangle and it's area is $\frac{2\times 2}{2}=2$.[/hide][/quote]

How is that cheating? :huh: The only hard part was visualizing it...[/quote]

Well, compared to their solution, it is :P
They have a working solution, I just found some stuff that works

 :D well it gave me the right answer, so it's all good

[quote="Altheman"]coordinate brute force...come on...it is a prism with equilateral triangle bases...this is easily found by thinking about loci...[/quote]
Yeah, I'm going to have to agree.
This may sound ridiculous coming from me (for those of you who know me), but analytic geo should only be used as a last resort.

Analytic geo is really fun though  :roll:

Seriously, if you're good at anal. geo you can brute force a hella lot of things, especially in longer competitions like AIME and USAMO ... though the graders might get pissed when you turn in a 6 page proof, 5 pages of algebra.


group theory
abstract algebra
induction
superior algebra
superior algebra unsolved

I'm self-studying some algebra now, and I have several problems:

1. (a) Let G be a group and H be a subgroup of finite index. Show that there exists a normal subgroup N of G contained in H and also of finite index.
(b) If G is a group and $ H_{1}, H_{2}$ are subgroups of finite index, show that $ H_{1}\cap H_{2}$ has finite index.

2. Let X and Y be finite sets and let C be a subset of $ X\times Y$. For $ x\in X$, let $ f(x)$ be the number of elements $ y\in Y$ such that $ (x,y)\in C$. Prove:

$ |C| \equal{}\sum_{x\in X}f(x)$.

3. Let P be a p-group. Let A be a normal subgroup of order p. Prove that A is contained in the center of P.

1b) Decompose the (product) set $ H_{1}H_{2}\subseteq G$ into a disjoint set of left cosets of $ H_{2}$, i.e. $ H_{1}H_{2}\equal{}\bigcup_{h\in I}hH_{2}$, where we can choose a finite set of representatives $ I\subseteq H_{1}$, since $ H_{1}$ has finite index. Then $ H_{1}\equal{}H_{1}\cap H_{1}H_{2}\equal{}H_{1}\cap\bigcup_{h\in I}hH_{2}\equal{}\bigcup_{h\in I}H_{1}\cap hH_{2}\equal{}\bigcup_{h\in I}h(H_{1}\cap H_{2})$. This shows that $ H_{1}\cap H_{2}$ has finite index in $ H_{1}$ and hence in $ G$.

1a) Let $ G\equal{}\bigcup_{x\in I}Hx$ be a disjoint, finite union of right cosets of $ H$ and denote $ N\equal{}\bigcap_{g\in G}H^{g}$. Clearly, $ N$ is a subgroup of $ H$ and normal in $ G$. The transformation $ N\equal{}\bigcap_{g\in G}H^{g}\equal{}\bigcap_{h\in H,x\in I}H^{hx}\equal{}\bigcap_{x\in I}H^{x}$ demonstrates that $ N$ is the intersection of finitely many subgroups each of finite index in $ G$. Using an induction argument plus 1b) we conclude that $ N$ has finite index in $ G$ as well.

2) Let $ F(x)\equal{}\{y\in Y|(x,y)\in C\}$ and $ \Omega\equal{}\{x\in X|\exists y\in Y: (x,y)\in C\}$, then $ x\in\Omega\Longleftrightarrow F(x)\not\equal{}\emptyset\Longleftrightarrow f(x)\not\equal{}0$, i.e. by definition
$ |C|\equal{}\sum_{x\in\Omega}|F(x)|\equal{}\sum_{x\in X}f(x)$.

3) Let $ P$ be a minimal counterexample with regard to the group order, i.e. we assume $ A\cap Z(P)\equal{}1$ and clearly $ P$ cannot be abelian. By minimality the group $ AZ(P)/Z(P)$ lies in the center of $ P/Z(P)$. This implies for the commutator group $ [A,P]\subseteq Z(P)$ and since $ A$ is normal, we have $ [A,P]\subseteq A\Longrightarrow[A,P]\subseteq A\cap Z(P)\equal{}1$. This means $ A$ commutes with any element of $ P$ and therefore, must lie in the center, a contradiction.


algebra
polynomial
linear algebra
linear algebra unsolved

Let $ A,B\in  M_{3}(\mathbb R)$ such that
$ A^2\plus{}B^2\equal{}A\plus{}B\equal{}AB$  $ (*)$

a)Prove that there are infinitly many matrices $ A,B$ satisfying this property
b)Find $ det(A)$ and $ det(B)$ ,for $ A,B \in M_{3}(\mathbb R)$ satisfying  property. $ (*)$

for a) consider nilpotent matrices

A=-B=
(0 0 x)
(0 0 0)
(0 0 0 )   with x any real

b) with *
A,B commute

and $ (2A\minus{}B)^2\equal{}\minus{}3B^2$ so $ det(B)\equal{}0$

similarly $ det(A)\equal{}0$

We prove that $ \forall\lambda\in\mathbb{R},det(A\plus{}\lambda{B})\equal{}0$; in particular $ det(A)\equal{}det(B)\equal{}0$.
1) $ trace(AB\minus{}BA)\equal{}trace((AB\minus{}BA)^2)\equal{}0$ (cf. http://www.artofproblemsolving.com/Forum/viewtopic.php?t=144008). $ AB\minus{}BA$ has at least one real eigenvalue $ \lambda$. Then $ spectrum(AB\minus{}BA)\equal{}\{\lambda,j\lambda,j^2\lambda\}$ where $ j\equal{}exp(2i\pi/3)$. Let $ AB\equal{}U$; $ U^2\equal{}A^2\plus{}B^2\plus{}AB\plus{}BA$ thus $ BA\equal{}U^2\minus{}2U$. $ spectrum(AB)\equal{}spectrum(BA)\equal{}\{\alpha,\beta,\gamma\}\equal{}\{\alpha^2\minus{}2\alpha,\beta^2\minus{}2\beta,\gamma^2\minus{}2\gamma\}$. Then the sequence $ u_0\equal{}\alpha,u_{n\plus{}1}\equal{}u_n^2\minus{}2u_n$ is 1,2 or 3-periodic. Therefore $ \alpha,\beta,\gamma$ are roots of the polynomial: $ x(x\minus{}3)(x^2\minus{}x\minus{}1)(x^3\minus{}3x^2\plus{}3)(x^3\minus{}2x^2\minus{}x\plus{}1)$. In particular $ \alpha,\beta,\gamma$ are real; thus $ BA\minus{}AB\equal{}U^2\minus{}3U$ has a real spectrum, $ \lambda\equal{}0$ and $ \alpha^2\minus{}3\alpha\equal{}0,\cdots$. Therefore $ spectrum(U)\subset\{0,3\}$.
2)$ (A\minus{}B)^2\equal{}\minus{}U^2\plus{}2U$, $ spectrum((A\minus{}B)^2)\equal{}\{\minus{}\alpha^2\plus{}2\alpha,\cdots\}\subset\{0,\minus{}3\}$. $ (A\minus{}B)^2$ has at least one non negative real eigenvalue then $ 0\in{spectrum(U)}$ and $ det(U)\equal{}det(A\plus{}B)\equal{}det(A\minus{}B)\equal{}0$.
3) $ det((A\plus{}iB)(A\minus{}iB))\equal{}|det(A\plus{}iB)|^2\equal{}det((1\minus{}3i)U\plus{}iU^2)\equal{}0$.
$ \phi(\lambda)\equal{}det(A\plus{}\lambda{B})$ is a complex polynomial of degree 3 which is $ 0$ for $ \lambda\equal{}1,\minus{}1,i,\minus{}i$ then it's $ 0$ for each real $ \lambda$.                         QED

Hi loup blanc

A+B=AB

(A-I)(B-I)=I

the inverse commute

(B-I)(A-I)=I

BA=A+B

A,B commute

OK Moubi; a biblical simplicity.
Moreover your proof is valid in $ \mathcal{M}_n(\mathbb{R})$ where $ n$ is odd.


There's no way I'm forgetting the solution for this one :D:

A quadrilateral ABCD is given in the plane. Circles ACD and BCD cut BC and AC at F and E respectively. Show that AD*DE/AE2 = BD*DF/BF2

AD*DE/AE^2 = sin AED sin DAE/ sin^2 ADE=sin BFD sinDBF/sin^2 BDF=  BD*DF/BF2

this is one of the way to prove this fact.

Another solution is:

Very easy fact: Let S and S' two circles intersected in P and Q. Consider the rotational gomotetion H witn center P such that H(S)=S'. Let M\in S and H(M)=M'. Then M,M' ,Q are collinear.

I thought it was tough at first, but then I noticed that triangles ADE and BDF are similar, and that certainly does the trick... :D

Thanks for the solutions.


calculus trigonometry limit integration function LaTeX search

I mean like how does a computer calculate pi to like a trillion digits. How is it possible that it is that accurate? In order for it to calculate it, wouldn't it need an equation of some sort as its input? What is that equation?
Maybe this is in the wrong forum i dunno... :?

Take a circle with infinite radius and find its circumference. Divide that by its diameter and you have $\pi$.

Just kidding. I was wondering about this too. They probably have some super-complex algorithms. :P

there's limits and stuff that approaches $\pi$, so they probably use that

[url]http://en.wikipedia.org/wiki/Pi[/url] and [url]http://mathworld.wolfram.com/PiFormulas.html[/url] have some approximation formulas...

[quote="frt"]There's also a traditional geometric way to approximate $\pi$.
Basically, consider a regular polygon inscribed in a circle and compare their perimeters.
That gives you a lower bound for the value of $\pi$.
Similarly, a regular polygon circumscribed about a circle gives an upper bound.

[hide="Example"]
Consider a regular octagon inscribed in a circle.
Its side lengths is $2r\sin{22.5^\circ}$.
We have 8 sides, so the perimeter is $16r\sin{22.5^\circ}$.
This has to be smaller than the circumference, so $16r\sin{22.5^\circ} < 2\pi r$
This gives $\pi >8\sin{22.5^\circ} =8\sqrt{\frac{1-\frac{\sqrt{2}}{2}}{2}}\approx 3.061$

Watch what happens if you calculate $100000\sin \left(\frac{\pi}{100000}\right)$ using calculator ;)
[/hide]

Although this is not a good way to calculate $\pi$ to trillion digits, this is how $\pi$ was calculated in old days.[/quote]
Give it it's proper name, frt :)
This algorithm is known as Archimedes' method, although many suspect that it was known even before that, maybe Babylonians...

[quote="chess64"]Take a circle with infinite radius and find its circumference. Divide that by its diameter and you have $\pi$.[/quote]

Nice...the problem is that it usually takes a lot of time to perform an infinite task...

-Using power series and calculus, it can be shown that
$\arctan {1} = 1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\cdots=\frac{\pi}{4}$ (Hence $\pi=4\displaystyle\sum_{n=1}^{\infty} (-1)^{n-1}\frac{1}{2n-1}$).  "However, this series (developed by James Gregory in 1671) does not give us a practical way of approximating $\pi$ because it converges so slowly that hundreds of terms wold have to be used to obtain reasonable accuracy." (so says my calculus book)

A much better and faster converging series (by John Machin, 1706) is
$4\arctan {\frac{1}{5}-\arctan{\frac{1}{239}}=\frac{\pi}{4}}$
"By using only five terms from each of the series for [the arctangents], you obtain" an approximation "which agrees with the decimal representation of $\pi$ with an error of less than .0000001" (from the same book)

I actually tried to make my own way of representing pi as a limit.  Here's what I came up with:
[hide]${\lim_{n\to\infty}n\sqrt{2-2\cos\frac{2\pi}n}}=2\pi$
What I'm doing is taking the perimeter of a regular n-gon, in which the distance from a vertex to the center (a point equidistant from all vertices) is 1.  I split the n-gon into $n$ congruent isosceles triangles.  Then I used the law of cosines to determine the length of the base of each individual triangle.  Multiply that by $n$ and you should get $2\pi$.  But when I plugged in high values for $n$, I get larger than $\pi$.  999,999 is okay, but not 1,000,000.  Oh well...[/hide]

[quote="lingomaniac88"]I actually tried to make my own way of representing pi as a limit.  Here's what I came up with:
[hide]${\lim_{n\to\infty}n\sqrt{2-2\cos\frac{2\pi}n}}=2\pi$
What I'm doing is taking the perimeter of a regular n-gon, in which the distance from a vertex to the center (a point equidistant from all vertices) is 1.  I split the n-gon into $n$ congruent isosceles triangles.  Then I used the law of cosines to determine the length of the base of each individual triangle.  Multiply that by $n$ and you should get $2\pi$.  But when I tried graphing it on my TI-84+, it got a bit "fuzzy" as I approached 7-digit or 8-digit numbers... Oh well.[/hide][/quote]

It kind of defeats the purpose if your $\pi$ approximating formula requires the use of $\pi$. After all, if the formula has an approximation for $\pi$, wouldn't it return even less accurate approximations?

I say just use a bunch or inverse tangents that add up to equal pi, then apply the taylor series. :D

Ramanujan's formula is the best one I know of. It gives ~7 digits per iteration.

Look it up if on wikipedia/google.

[quote="0714446459923"]by dividing a regular polygon into isocles triangles you cna see that a n sided polygon has an area of
(n/2)*sin(360/ndegrees) times the radius squared or something like that.
could you use that to approximate pi?[/quote]
Yeah, that works, too.

BTW, I remember that one of the problems for the entrance exam for Tokyo University (in the year of 2003 I believe) was "Prove that $\pi>3.05$". (No calculator!) :roll:

I remember that I saw

$\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...=\frac{\pi^2}{6}$.

I don't know how to write that in $\sum$ notation with latex, but you get the idea of the series.

I like pointing out 236!'s mistakes :D

The sum should be $\sum_{j=1}^\infty\frac1{j^2}=\frac{\pi^2}6$

yeah yeah $\pi^{\underline{2}}$  :roll:

(and thanks for the sigma notation. I [hope] I can figure out the way to put things in sigma notation).

Well, here are a few methods i know of:

[hide="1"] It is easy to prove that $2\cos \frac{\pi}{2^{n+1}}$(denoted by $r_n$) $=\underbrace{\sqrt{2+\sqrt{2+\sqrt{2+\cdots }}}}_n$ using half-angle formulas and induction. This means that $2-r_n=4\sin^2 \frac{\pi}{2^{n+2}}$. Using $\lim_{\theta \rightarrow 0} \frac{\sin \theta}{\theta}=1$,
\[ \lim_{n\rightarrow \infty} 4^n(2-r_n)=\frac{\pi^2}{4}  \]
Here any finite sequence is less than $\frac{\pi^2}{4}$ because $\sin \theta <\theta$ for $0<\theta \leq \frac{\pi}{4}$.[/hide]

[hide="2"]More trig based on an explicit formula plus a limit: by using the identity $\frac{\sin x}{x}=\prod_{n=1}^\infty \cos \frac{x}{2^n}$ and plugging in $\frac{\pi}{2}$,
\[{ \frac{2}{\pi}=\prod_{n=1}^\infty \frac{r_n}{2}}  \]
Any finite sequence is clearly greater than $\frac{2}{\pi}$ because each term is less than 1.[/hide]

[hide="3"] Calculus time: the integral $\int_0^\frac{\pi}{2} \cos^n x\ dx$ gives different formulas based on whether $n$ is odd or even, but always decreases as $n$ increases. This gives \[ \frac{1}{\sqrt{\pi}}=\lim_{n\rightarrow \infty} \sqrt{n}\prod_{k=1}^n \frac{2k-1}{2k}  \] (which can also be expressed as an infinite product). An error formula is given by
\[ \frac{1}{\sqrt{\pi (n+1/2)}}<\prod_{k=1}^n \frac{2k-1}{2k} <\frac{1}{\sqrt{\pi n}}  \] [/hide]

[hide="4"] Use trigonometric identities plus the arctangent series: a well known one is
\[ \frac{\pi}{4}=\sum_{n=0}^\infty \frac{(-1)^n}{2n+1}  \]
but this converges very slowly compared to exponential series such as
\[ \frac{\pi}{4}=\arctan \frac 12 +\arctan \frac 13 =\sum_{n=0}^\infty \frac{(-1)^n}{2n+1} \left( \frac{1}{2^{2n+1}}+\frac{1}{3^{2n+1}} \right)  \] [/hide]

[hide="5"] Values of the function $\zeta(s)=\sum_{n=1}^\infty n^{-s}$ are known for positive even integers $s$ and are expressed in terms of $\pi^s$ and rational numbers. It is easy to show that
\[ \sum_{n=1}^\infty (-1)^{n+1} n^{-s}=(1-2^{1-s})\zeta(s)  \]
and this can be used to calculate $\pi^s$ with an error of $\frac{1}{(n+1)^s}$. For large $s$, this formula gets REALLY complicated, while for $s=2$ and $s=4$ they give
\[ \frac{\pi^2}{12}=\sum_{n=1}^\infty (-1)^{n+1}n^{-2}  \]
\[ \frac{7\pi^4}{720}=\sum_{n=1}^\infty (-1)^{n+1}n^{-4}  \]
[/hide]

woooooowwww :wow:  *runs off and tries to calculate pi using these way, gives up after 5 minutes*

[quote="4everwise"]I'm not usually the moderator who says this, but try using the search function.  ;)

I won't lock the topic though. :)[/quote]

lol sorry XD

A very simple metod:

$\large{\pi\,=\,\lim_{x\,\,\rightarrow\,\,\infty}\,\,\,x.sen(\frac{180}{x})}$

Sorry, i`m brazilian and don`t know how this function is write. My English is very bad. :|

Well, it's easy to guess : sen -> sin. :)
(At least I think so..)

I'm a fan of the isosceles triangle method that was already discussed...  Here's it in great detail:

Take a regular polygon with $n$ sides and split it up into $n$ isosceles triangles.  The heights of these triangles will be called $r$ (when $n\to\infty$ this height becomes the radius of a circle).  Because the polygon is regular, however, it's angles can be found.  The formula for the angle measure of each angle in a regular polygon with $n$ sides is: $\frac{180(n-2)}{n}$

[hide="Explanation of formula"]
An $n$-sided polygon can be divided into $n-2$ triangles, each of whose angles add up to $180$ degrees, and this total is divided by $n$ because there are $n$ angles in an $n$-sided regular polygon.[/hide]

Dividing this angle by $2$ yields the angle measure of each of the base angles of the isosceles triangles, and splitting the triangles into two right triangles yields the ratio $\frac{r}{\frac{1}{2}}$, where the length of each side is $1$, equal to the tangent of the half-angle:

$\tan{\frac{180(n-2)}{2n}^{\circ}}=\frac{r}{\frac{1}{2}}$
$\frac{\tan{\frac{180(n-2)}{2n}^{\circ}}}{2}=r$

From here $\pi$ can be determined using areas or distances:

[hide="Area"]
The area of each triangle is $\frac{1}{2}r$, and the area of the whole polygon is $n*\frac{1}{2}r$ because there are $n$ sides and therefore $n$ isosceles triangles.  Substituting $\frac{\tan{\frac{180(n-2)}{2n}^{\circ}}}{2}$ for $r$ gives this equation for the area of the polygon:

$n*\frac{1}{2}*\frac{\tan{\frac{180(n-2)}{2n}^{\circ}}}{2}=$
$\frac{n}{4}*\tan{\frac{180(n-2)}{2n}^{\circ}}$

And since $\pi$ is the ratio between the area and $r^2$ in a regular polygon with infinitely many sides, we divide this expression by $r^2$:

$\frac{n}{4}*\tan{\frac{180(n-2)}{2n}^{\circ}}*\frac{1}{r^2}$

Substituting the tangent expression for $r$:

$\frac{n}{4}*\tan{\frac{180(n-2)}{2n}^{\circ}}*\frac{1}{(\frac{\tan{\frac{180(n-2)}{2n}^{\circ}}}{2})^2}=$

$\frac{n}{\tan{\frac{180(n-2)}{2n}^{\circ}}}$

Not knowing the mathematical way to go from here, plug this into the $y=$ of your calculator and use the table to plug in large values for $x$ and find a close approximation of the limit as $x\to\infty$[/hide]

[hide="Perimeter"]
The perimeter of the polygon is $n$, and since $\pi$ is the ratio between the circumference twice the radius in a regular polygon with infinitely many sides, use the expression

$\frac{n}{2*r}$

$\frac{n}{2*\frac{\tan{\frac{180(n-2)}{2n}^{\circ}}}{2}}$

$\frac{n}{\tan{\frac{180(n-2)}{2n}^{\circ}}}$

and plug it into the $y=$ of your calculator and use the table to plug in large values for $x$ and find a close approximation of the limit as $x\to\infty$[/hide]

Sorry if I was over explanatory... I'm still in USAMO mode :(

Just for fun, you could try using $e^{i\pi}=-1$, by somehow finding out how to do $\ln{-1}$...

[quote="quantum leap"][quote="chess64"]Take a circle with infinite radius and find its circumference. Divide that by its diameter and you have $\pi$.[/quote]

Nice...the problem is that it usually takes a lot of time to perform an infinite task...

-Using power series and calculus, it can be shown that
$\arctan {1} = 1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\cdots=\frac{\pi}{4}$ (Hence $\pi=4\displaystyle\sum_{n=1}^{\infty} (-1)^{n-1}\frac{1}{2n-1}$).  "However, this series (developed by[b] James Gregory in 1671) [/b]does not give us a practical way of approximating $\pi$ because it converges so slowly that hundreds of terms wold have to be used to obtain reasonable accuracy." (so says my calculus book)

A much better and faster converging series (by John Machin, 1706) is
$4\arctan {\frac{1}{5}-\arctan{\frac{1}{239}}=\frac{\pi}{4}}$
"By using only five terms from each of the series for [the arctangents], you obtain" an approximation "which agrees with the decimal representation of $\pi$ with an error of less than .0000001" (from the same book)[/quote]

Yes, the second part is nice form..

Just one comment, the  first series you posted should actually be credited to Madhava of Kerala  Inida who puplished it almost 300 years before James Gregory!

 see the link:
http://www-groups.dcs.st-and.ac.uk/~history/Miscellaneous/Pearce/Lectures/Ch9_3.html
for example.

[quote="me@home"]still talking about the arctan formula (with 239):
ever since I saw it first, I thought, "well how did the 239 get in there?"
can anyone explain?[/quote]

Well  it is fairly standard trick...   it is pretty easy to see that  $\arctan (1/5)$  is pretty close to  $\pi /16$

Let $\alpha = \tan ^{-1} \frac 1 5$
then $\tan 2\alpha = \frac{(2/5)} {1 - (1/25)} = \frac 5 {12}$
so $\tan 4\alpha = \frac{120} {119}$ !!   which is pretty close to  $1$  (slightly greater than 1)
So let us assume  $4\alpha = \pi / 4 + \beta$ now if you take the tangent of both sides, you get

$\frac{120} {119} = \frac{1 + \tan \beta }{1 - tan \beta} \rightarrow \tan \beta = \frac 1 {239}$

Hence you get $\frac{\pi} 4 = 4 \tan ^ {-1} \frac 1 5 - \tan ^ {-1} \frac 1 {239}$

You can get many similar results ...

[quote]would someone come up with that without knowing it already, each step seems random
[/quote]

It's not random .. and that's what I was trying to show.. basically you know  tan(pi/4) = 1 but 1  converges too slowly so you want to start with smaller number.. you can start with  $x \ tan ^{-1} (1/y)$  where x and y are  small numbers but the result is pretty close tp $45^o$ (If nothing else you use a graph paper and measure the angle) and then calculate the "correcting" factor.

For example if you thought (1/5) is not small enough..  you start with the previous equation, break (1/5) into two (1/10) and have
(Assuming I have not made any mistake in calculation below ...:)

  $\tan ^{-1} (1/5) = 2 \tan ^{-1} (1/10) - \tan^{-1} (1/k)$
and simple calculation as the  previous post and you will see $k=515$
and you have a "better" (well a little faster converging) series for pi:
 $\pi / 4 = 8 \tan ^{-1} (1/10) - \tan ^{-1} (1/239)- 4 \tan^{-1} (1/515)$

Hope that helps ...

BTW some of the other nice such series  I like are: (Probly google/reference would give many more)  using these fractions ...

(1/2) (1/3)
(1/2) (1/5) (1/8)
4*(1/5) - (1/70) +(1/99)
2*(1/3) (1/7)

And  1985 was a good year so:  3*(1/4) (1/20)  (1/1985)

Of course these number means , add  arctan before these, for example first line is short for:
$\pi / 4 = \arctan (1/2) + \arctan (1/3)$

Hope this helps.

Read Wikipedia. The main way to compute $\pi$ is either a variant of the convergence Ramanujan gave: $\frac{1}{\pi} = \frac{2\sqrt{2}}{9801} \sum^\infty_{k=0} \frac{(4k)!(1103+26390k)}{(k!)^4 396^{4k}}$ which nowadays has been replaced by the quicker converging $\frac{1}{\pi} = 12 \sum^\infty_{k=0} \frac{(-1)^k (6k)! (13591409 + 545140134k)}{(3k)!(k!)^3 640320^{3k + 3/2}}$. How Ramunujan arrvied at that formula is still a mystery.

But don't forget computers work in binary or hexadecimal and so it would be nicer if we could find the hexadecimal expansion of $\pi$ which currently is currently best given by: $\pi = \frac{1}{2^6} \sum_{n=0}^{\infty} \frac{{(-1)}^n}{2^{10n}} \left( - \frac{2^5}{4n+1} - \frac{1}{4n+3} + \frac{2^8}{10n+1} - \frac{2^6}{10n+3} - \frac{2^2}{10n+5} - \frac{2^2}{10n+7} + \frac{1}{10n+9} \right)$.

[quote="me@home"]does that mean that theoretically there exist faster converging series with larger numbers?
btw, how would someone come up with that without knowing it already, each step seems random[/quote]

For anyone interested in the Machin formula given above,

http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=%5C%5Cpi+239&t=69571 has a nice list of Machin-like formulae.   :)

The steps are far from random, but the math behind the general theory is a little complicated.  I once saw a brilliant derivation from Gaussian integers, but have forgotten it.

[quote="me@home"]still talking about the arctan formula (with 239):
ever since I saw it first, I thought, "well how did the 239 get in there?"
can anyone explain?[/quote]

$(12 + 5i)^2 = 119 + 120i$
$\frac{120}{119} \approx 1$
$120 + 119 = 239$

See also : http://www.mathlinks.ro/Forum/viewtopic.php?t=4262 . :)


inequalities
LaTeX
algebra unsolved
algebra

Prove that any positive real number A1, A2, A3, ... ,An, (where n is a nature number) the following inequality is true:
 sigma(A[i]i[/i] / (A[i]i+1[/i] + A[i]i+2[/i])  >= n/2
 where {n|n<= 12 or n >= 27} or {n| 14<= n <= 26 & n is even}

sorry about not using LaTeX since my computer's sort of in short of GB

This is Shapiro's inequality.
I don't think someone will solve it here because the general case is very long and difficult.

Pierre.

yes, and i saw Vasc proved this in a paper if i'm not mistaken. :?

Well, you should take a look at that (yeah, Pierre's involved in this! :D) :

http://www.math.niu.edu/~rusin/known-math/99/shapiro

Oh, and check this too : http://www.artofproblemsolving.com/Forum/topic-1874.html :)

actually the rigth side of the inequality could be n/4, which is correct for all A[i]i[/i]

frankthegoblintinker

The results were are known are
i)the ineq $\sum\frac{a_i}{a_{i+1}+a_{i+2}}\geq\frac{n}{2}$ is true for odd

numbers $n\leq 11$ and even numbers $n\leq 12$.

ii) But Dringlet , who was one of the winners of the Olympiad of 1969 found an impreesive result: He founds the best $c\frac{n}{2}$  and this $c$ is very close to

$0,989$ and his ordinate is the point where the common tangent of the fuctions

$y=e^{-x}$ and $y=\frac{2}{e^{x/2}+e^x}$ intersect the axis $Oy$.

I hope my English are written.

[quote="silouan"]

ii) But Dringlet , who was one of the winners of the Olympiad of 1969 found an impreesive result[/quote]

Hmm... Drinfeljd...and better than IMO, he won a fields medal  (sure not for that)...

Pierre.

By the way, for a proof of Drinfield's result about Shapiro's inequality, one could try to solve the problem given in 6 hours at ENS 1997. It's not an easy task, but it really worths.

That's why I said nobody will prove it here...I do not imagine someone who solve a 6 ours ENS problem in some post... :D

Pierre.

yes but if someone has this result  proven he could write it in to pdf file and post it here. It would be great :)


function
symmetry
calculus
derivative
algebra proposed
algebra

Find all functions $f: \mathbb R \rightarrow \mathbb R^{+}$ that for each $x,y\in \mathbb R,x\neq y$ we have: \[ f(x)f(y)\leq (x-y)^2  \]

By symmetry, we have $|f(x)-f(y)|<(x-y)^2$. Dividing by $x-y$ and taking limits, $f$ is differentiable everywhere with derivative zero. By the Mean Value theorem, $f$ is constant.

[quote="Omid Hatami"]Sorry I edited my problem and it's a more difficult problem :(  :([/quote]
Are you sure that you copied the problem correctly? The one you wrote has a lot of non-zero solutions, for example $ f\left(\frac {k}{2^n}\right) \equal{} \frac {1}{2^{2n}}, \ \gcd(k,2^n) \equal{} 1,$ defined by continuity in other points.

I think the problem is correct, and I think the function can not be extended continuously to $ \mathbb R$.
But here is the solution:
If the function exists, for each $ x\in R$ draw circles in $ \mathbb R^2$, with center $ (x,f(x))$ and radius $ \frac{f(x)}2$. Then for every two of  these circles don't intersect. But it is impossible to have uncountable circles in $ \mathbb R^2$ which their interior do not intersect.


How to solve q in inequality which have squareroot.
1) Solve for value of x :?:
  (x+61)^1/2<x+5

2)(x+3)<(x+33)^1/2

hello, to 1), because of the range of definition we have $ x>\minus{}5$ after squaring both sides of your equation and simplifying we get $ 0<x^2\plus{}9x\minus{}36$, together with $ x>\minus{}5$ we have the solution $ 3<x<\plus{}\infty$.
Sonnhard.

[hide="2"]Let us square the equation to get: \[ (x\plus{}3)<(x\plus{}33)^{\frac{1}{2}}\equal{}x^2\plus{}6x\plus{}9<x\plus{}33\] Now, we will simplify the equation: \[ x^2\plus{}5x<27\] Let us plug in numbers to find the range. We get $ \minus{}9<x<4$.[/hide]

hello, are you sure? I have found $ \minus{}33<x<3$ and $ x \in \mathbb{R}$.
Sonnhard.

Yes, there seems to be a mistake in my solution.

Hi
Thanks Doc and akhil for sol..
But Dr i need solution for second Q.....
Because as in first u have taken a range as x>-5 what is the range for the second  q
(x+3)<(x+33)^1/2

[quote="Nidhi"]When it |x|<a then it is expanded as -a<|x|<a
but how to solve following :?:

(x^2-7|x|+10)/(x^2-6x+9)  < 0

and

(|x+2|-x)/x  < 2[/quote]

Inequality [b]with modulus[/b], why? :?:

For the first inequality we have $ x^2\minus{}6x\plus{}9\equal{}(x\minus{}3)^2>0$ for $ x \neq 3$, so $ x \neq 3$ and $ (|x|\minus{}2)(|x|\minus{}5)<0$
gives $ \minus{}5<x<\minus{}2$ or $ 2<x<5$, with $ x \neq 3$, as solutions.

For the second one we have if $ x$ is negative, then $ \frac{|x\plus{}2|\minus{}x}{x}<0<2$
If $ x$ is positive, then the question $ \Leftrightarrow x\plus{}2<3x \Leftrightarrow x>1$
So $ x<0$ or $ x>1$ are the solutions.

Hi
Thanks Doc and akhil for sol..
But Dr i need solution for second Q.....  for root one...
Because as in first u have taken a range as x>-5 what is the range for the second q
(x+3)<(x+33)^1/2

hello, to solve the inequality $ x\plus{}3<\sqrt{x\plus{}33}$ we denote the condition $ x\geq\minus{}33$ because of the range of the square root. If we have $ \minus{}33\le x \le\minus{}3$ then our inequality is fulfilled for all real $ x$ with
$ \minus{}33\le x \le\minus{}3$. In the other case $ x>\minus{}3$ we have a contradiction to the condition $ x\le\minus{}33$, so the solution is $ \minus{}3\le x\le \minus{}33$ and $ x \in \mathbb{R}$.
Sonnhard.


Can $ 1$ be written as a sum of $ 2005$ different terms of the form $ \frac{1}{3n-1}$, where $ n\in N$?


What is the product of the smallest one-digit prime number times the smallest two-digit prime number?

The smallest 1-digit prime is 2.
The smallest 2-digit prime is 11.
2*11=22.


complex analysis
complex analysis unsolved

Let $ f$ be analytic on $ G\equal{}\{ z: Re(z)>0\}$ , $ f$ is one-to -one, and $ Re(f(z))>0\,\,\forall z\in G$, and $ f(a)\equal{}a$ for some $ a\in\mathbb{R}$. Prove that $ |f'(a)|\leq 1$.

thanks

Find a fractional linear transformation that sends your half plane to the open unit disk, then use the Schwartz lemma.

thanks, but i cannot see it, could you please show me?
Note : $ G\equal{}\{ z: Re(z)>0\}$ is not the upper-half plane

which is why you can not use the Cayley map  :lol: . But still the right Half-plane (and any other half-plane) is conformal to the unit disk.
As the map $ \phi: G\to D(0,1)$ defined by $ \phi (z)\equal{}\frac{z\minus{}1}{z\plus{}1}$ shows.
Now if you look at $ \phi f \phi^{\minus{}1}$ it has a fixed point at $ \phi(a)$ and your claim follows from the Schwarz-Pick lemma.

thanks Albanian Eagle, I got the problem
By the way, I have another question :

================================================
let $ G_{1}, G_{2}$ be simply connect regions neither of which is the whole complex plane. Let $ f$ be analytic, one-to-one, onto mapping from $ G_{1}$ to $ G_{2}$, let $ a\in G_{1}$  and $ \alpha\equal{}f(a)$. Prove that for any one-to-one analytic map $ h$ from $ G_{1}$ into $ G_{2}$ with $ h(a)\equal{}\alpha$ , prove that $ |h'(a)|\leq |f'(a)|$?

thanks


${\frac{G}{\left(t^{2}+y^{2}\right)^{2}}=\frac{\left(\frac{d^{2}y}{dt^{2}}\right)^{2}}{\left(\frac{dy}{dt}\right)^{2}+1}}$

How can I Solve this?

mathematica 5.2 did not solve this.

but, the solution exist in polar coordinates.

It's a physics problem.

I cannot say what problem is, becuase, if i do so, the answer is simple.

but, without know the answer, how can i solve?

hello, i don't think that an explicit solution exists.
Sonnhard.


Find an appropriate expression for the n-th term of the following sequence and also the sum of first $n$ terms:
$4,2,10,58,146,388,1010...$

wow...the online encyclopedia of integer sequences couldnt get this one

Two random things that I have come up with looking at this so far...

[hide]If you add then all up, you get 1618, which is phi-ish

If you find the largest prime number other than two for each number, then you get 1, 1, 5, 29, 73, 97, 101. The difference between these are 0, 4, 24, 44, 24, 4. If you take the other numbers, you get 4, 2, 2, 2, 2, 4, 10, which doesn't really make much sense, so my previous sequence might have been a coincidence.[/hide]

Mustafas first idea of largest primes seems [b]really[/b] promising.
However, his later ideas are just junk (brainteaser won't have long solution)

I am stumped for now though...

[quote="mustafa"]Where did you get this from?[/quote]
A teacher from this math enrichment thing I go to gave it to me. He told me that he does not remember where he took the problem from and he does not know the solution either.

Is it really tricky like "days per month" or like "number of letters of certain words" (Its obviously not either but you understand what I mean)
or is it pure mathematical?

[quote="rem"]It was included in a set of problems on recurrence relations, so it should be purely mathematical.[/quote]

The way the ratio jumps around, it looks to me like one or more terms are missing, or that it has typos.  (If it's supposed to be any kind of simple, nice recurrence relation.)

Hah, I win!

[url=http://www.research.att.com/~njas/sequences/A069921]This sequence[/url] begins 1, 5, 10, 29, 73, 194, 505.  Twice these terms is 2, 10, 20, 58, 146, 388, 1010, or exactly your sequence except it begins one term later and has an additional term, just about where I would have predicted :-)


trigonometry
algebra
polynomial
algebra unsolved

Prove, that $\cos\frac{5\Pi}{7}$ is a root of the equation :
$8x^{3}-4x^{2}-4x+1=0$ and that it's irrational.

For the latter, consider $x=y-1$, the resulting polynomial is $8y^{3}-28y^{2}+28y-7$, thus irreducible by Eisenstein.
To see it's a root, you can write the number in terms of roots of unity.

Let $z\in \mathbb C$ ,$z=cos\frac{5\pi}{7}+i\cdot\frac{5\pi}{7}$ $\rightarrow$ $z^{7}=-1$ and $cos\frac{5\pi}{7}=\frac{z^{2}+1}{2z}$ and it's easy from here.

[quote="mathmanman"]Analogously, prove that $\cos\left(\frac{2 \pi}7\right)$ is irrational. (first you'll have to find that cubic polynomial whose root that number is)[/quote]
If $\cos{\frac{2\pi}{7}}\in\mathbb{Q}$ then $\cos{\frac{5\pi}{7}}\in\mathbb{Q}$. Done! :D

We can solve the problem by another simple approach as:
Note that $\frac{5\pi}{7}$ is a solution of equation:
$7x=\pi+k.2\pi\Leftrightarrow 3x=(\pi-4x)+k.2\pi\Leftrightarrow \cos3x=-\cos4x$
$\Leftrightarrow \cos4x+\cos3x=0$
alternatively, we have: $\cos4x=8\cos^{4}x-8\cos^{2}x+1$,  $\cos3x=4\cos^{3}x-3\cosh$
Hence: $\frac{5\pi}{7}$ is a root of equation $8\cos^{4}x+4\cos^{3}x-8\cos^{2}x-3\cosh+1=0$
Put $t=\cos x\Rightarrow \cos\frac{5\pi}{7}$ is a root of equation:
$8t^{4}+4t^{3}-8t^{2}-3t+1=0\Leftrightarrow(t+1)(8t^{3}-4t^{2}-4t+1)=0$
Since $\cos\frac{5\pi}{7}\neq-1$ then:
$\cos\frac{5\pi}{7}$ is a root ofeq: $8t^{3}-4t^{2}-4t+1=0$

bacause if $\frac{p}{q}\in\mathbb{Q}(\gcd (p,q)=1)$ is root of $a_{n}x^{x}+a_{n-1}x^{n-1}+...+a_{0}\in\mathbb{Z}[x]$ then $p|a_{0}$ and $q|a_{n}$.

From my solution we also have: $\cos\frac{\pi}{7}$,  $\cos\frac{3\pi}{7}$ are solutions of that equation, too.
Alternative, i also have the equality: $\frac{1}{\cos\frac{\pi}{7}}+\frac{1}{\cos\frac{3\pi}{7}}+\frac{1}{\cos\frac{5\pi}{7}}=4$

[quote="mathmanman"]Analogously, prove that $\cos\left(\frac{2 \pi}7\right)$ is irrational. (first you'll have to find that cubic polynomial whose root that number is)[/quote]

But i have a question: How do you deduce it?

I'm not sure I fully got your question, so if my reply is way off, let me know.

Basically, the approach indeeds consists in finding a cubic polynomial with integer coefficients whose roots are $\cos \frac{2\pi}7, \ \ \cos \frac{4\pi}7 , \ \ \cos \frac{6\pi}7$, so we of course begin with $(X-\cos \frac{2\pi}7)(X-\cos \frac{4\pi}7)(X-\cos \frac{6\pi}7) = 0$, or
$X^{3}-(\cos \frac{2\pi}7+\cos \frac{4\pi}7+\cos \frac{6\pi}7)X^{2}+(\cos \frac{2\pi}7 \cos \frac{4\pi}7+\cos \frac{2\pi}7 \cos \frac{6\pi}7+\cos \frac{4\pi}7 \cos \frac{6\pi}7)X-\cos \frac{2\pi}7 \cos \frac{4\pi}7 \cos \frac{6\pi}7 = 0.$
Let us then compute these three coefficients.
Let $\omega = e^{2 i \pi/7}.$ We have :
$\cos \frac{2\pi}7+\cos \frac{4\pi}7+\cos \frac{6\pi}7 = \frac{1}{2}(\omega+\omega^{6}+\omega^{2}+\omega^{5}+\omega^{3}+\omega^{4}) =-\frac{1}{2},$
then,
$\cos \frac{2\pi}7 \cos \frac{4\pi}7+\cos \frac{2\pi}7 \cos \frac{6\pi}7+\cos \frac{4\pi}7 \cos \frac{6\pi}7 = \frac{1}{4}((\omega+\omega^{6})(\omega^{2}+\omega^{5})+(\omega+\omega^{6})(\omega^{3}+\omega^{4})+(\omega^{2}+\omega^{5})(\omega^{3}+\omega^{4}))$
$= \frac{1}{4}((\omega^{3}+1+\omega+\omega^{4})+(\omega^{4}+\omega^{5}+\omega^{2}+\omega^{3})+(\omega^{5}+\omega^{6}+\omega+\omega^{2}))$
$=-\frac{1}{4},$
and eventually,
$\cos \frac{2\pi}7 \cos \frac{4\pi}7 \cos \frac{6\pi}7 = \frac{1}{8}(\omega+\omega^{6})(\omega^{2}+\omega^{5})(\omega^{3}+\omega^{4})$
$=\frac{1}{8}.$
Hence we get that those three numbers are roots of the equation $X^{3}-\frac{1}{2}X^{2}-\frac{1}{4}X+\frac{1}{8}= 0$ or of the equation
\[8X^{3}-4X^{2}-4X+1 = 0. \]
Now, to show that this solution has no rational roots, show that if there were any, they would belong to that set given by [b]N.T.TUAN[/b] above, and check that none of these is a root of our equation, so we're done.

[quote="mathmanman"]I'm not sure I fully got your question, so if my reply is way off, let me know.

Basically, the approach indeeds consists in finding a cubic polynomial with integer coefficients whose roots are $\cos \frac{2\pi}7, \ \ \cos \frac{4\pi}7 , \ \ \cos \frac{6\pi}7$, so we of course begin with $(X-\cos \frac{2\pi}7)(X-\cos \frac{4\pi}7)(X-\cos \frac{6\pi}7) = 0$, or
$X^{3}-(\cos \frac{2\pi}7+\cos \frac{4\pi}7+\cos \frac{6\pi}7)X^{2}+(\cos \frac{2\pi}7 \cos \frac{4\pi}7+\cos \frac{2\pi}7 \cos \frac{6\pi}7+\cos \frac{4\pi}7 \cos \frac{6\pi}7)X-\cos \frac{2\pi}7 \cos \frac{4\pi}7 \cos \frac{6\pi}7 = 0.$
Let us then compute these three coefficients.
Let $\omega = e^{2 i \pi/7}.$ We have :
$\cos \frac{2\pi}7+\cos \frac{4\pi}7+\cos \frac{6\pi}7 = \frac{1}{2}(\omega+\omega^{6}+\omega^{2}+\omega^{5}+\omega^{3}+\omega^{4}) =-\frac{1}{2},$
then,
$\cos \frac{2\pi}7 \cos \frac{4\pi}7+\cos \frac{2\pi}7 \cos \frac{6\pi}7+\cos \frac{4\pi}7 \cos \frac{6\pi}7 = \frac{1}{4}((\omega+\omega^{6})(\omega^{2}+\omega^{5})+(\omega+\omega^{6})(\omega^{3}+\omega^{4})+(\omega^{2}+\omega^{5})(\omega^{3}+\omega^{4}))$
$= \frac{1}{4}((\omega^{3}+1+\omega+\omega^{4})+(\omega^{4}+\omega^{5}+\omega^{2}+\omega^{3})+(\omega^{5}+\omega^{6}+\omega+\omega^{2}))$
$=-\frac{1}{4},$
and eventually,
$\cos \frac{2\pi}7 \cos \frac{4\pi}7 \cos \frac{6\pi}7 = \frac{1}{8}(\omega+\omega^{6})(\omega^{2}+\omega^{5})(\omega^{3}+\omega^{4})$
$=\frac{1}{8}.$
Hence we get that those three numbers are roots of the equation $X^{3}-\frac{1}{2}X^{2}-\frac{1}{4}X+\frac{1}{8}= 0$ or of the equation
\[8X^{3}-4X^{2}-4X+1 = 0. \]
[/quote]

I think that's a general approach for the type problem


calculus
integration
calculus computations

[url=http://img3.imageshack.us/my.php?image=47586628.jpg][img]http://img3.imageshack.us/img3/7558/47586628.th.jpg[/img][/url]

how can I start to solve this?

Since it looks like you're integrating in a hollowed out cylinder, use cylindrical coordinates.


The product of two consecutive even whole numbers is 2 more than their sum. What is the sum of these two whole numbers?

If the smaller one is X,

X(X+2)=X+X+2+2
X^2+2X=2X+4
X^2=4
X=2,

So the 2 numbers are 2 and 4, which adds up to 6.


function algebra domain calculus integration polynomial Princeton

This is a question puzzled me whenever I begin to learn math. What is on earth a formule? For example if a series $ a(n) \equal{} x^n$, $ x^n$ is called a formule for it. But in someme "counting question" such as partition of numbers, number of n-order group, $ \sum \frac {1}{n}$  …… they are thought to have no explicit formule or simply no formule solution. But, this seem to be our prejudice that we are more familiar with some functions (such as $ x^n$) and less familiar with someothers (such as algorithm). Under this consideration, I came up with a "definition" of "formule": suppose $ S$ to be a basic set of operations (calculation), a formule is a function, 1. generated by $ S$.(in another word $ f$'s calculation consists of operations from $ S$ by composing). 2. the steps of operation for each value in $ f$'s domain is bounded and the order of operations is unique.

Is this "definition" "right"? And what is or is there a common agreement on this "definition"? I'd like to read your suggestions.

And in our customs $ S$ is a particular set of Oper that we are familiar with. However this "definition" still "failed" in many cases. For example, $ \sum \frac {1}{n^3}$ is thought to be unsolved, but if one accosiate it with $ h_3(x) \equal{} \sum \frac {1}{n^3}e^{ \minus{} nx}$ and solve inductively the eqs $ h_3'(x) \equal{} h_2(x)\cdots$ one gets an integral solution, which is not regard as a solution. So, what on earth is a solution to this question. What people want to know, actually, are some usual properties about the solution such as whether it's algebraic ect. But what propert will finally satisfy us. The answer to $ \sum \frac {1}{n^{2k}}$ satisfy us because it's the polynomial of the $ \pi$, since both "polynomial" and $ \pi$ are "renown". But the solution to $ \sum \frac {1}{n^3}$ might be a new comer, then to find a solution is eq to find the wanted property for this solution, but what kinds of properties?? People say they don't know the answer so it shall means they know what kinds of properties it'll have or what kind of form it shall be represented in. What's you opinion??

As is the same with formule, by any counting problem we get straighaway a "function" and our goal is to find the relation between this function with others. Is it?? I really want to hear your opinion. (Thanks for reading till here.)

Your post is really hard to understand, but what I *think* you are asking about is not quite what people call a "formula" (which is extremely general) but rather a "closed form solution."  The phrase "closed form solution" is not really a technical term.  Also, one can refer to a formula as being "written in terms of elementary functions," which might also be what you are thinking of.

So, this is a fairly big question you're asking, and it's more meta-mathematical than mathematical.  Doron Zeilberger has some thoughts on the question of what "an answer" is in the context of enumerative combinatorics in his [url=http://www.math.rutgers.edu/~zeilberg/mamarim/mamarimhtml/enu.html]entry[/url] in the [url=http://press.princeton.edu/titles/8350.html]Princeton Companion to Mathematics[/url], and I had some thoughts on this question [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=253408&start=20]here[/url] (you have to go to the previous page to read what I'm commenting on).  The consensus of those two sources is that really the notion of "a formula" reduces to the notion of "an algorithm," though I disagree with Wilf (via Zeilberger, who also doesn't fully agree) that polynomiality is necessarily part of the definition.  There are two good reasons to like "an algorithm" as the definition of "a formula." the first is that it means that you can (with sufficient patience, in principle) actually write down the answer for any case you care about.  The second is that (if you believe the Church-Turing thesis) the notion of "algorithm" is so robust that it really does seem to capture the notion of "everything we can compute."  The biggest reason to [i]dis[/i]like "algorithm" as the meaning of "formula" comes from Wilf's point (and the parenthetical comment two sentences preceding this one): without some requirement of efficiency, the existence of an algorithm doesn't do us much good.  This is, for example, the problem with asking for the value of the Ramsey number $ R(6, 6)$.

So to tackle your particular examples:  we say that there is no closed form for $ H_n$ because there is not (to my knowledge) no algorithm to compute it significantly faster than adding all $ n$ terms.  (If there were, there might be a way to compute $ n!$ efficiently, which might make Wilson's theorem a practical primality testing.)  We say that we don't know what Apery's constant is [b]specifically[/b] because we have not expressed it in terms of other constants that we already understand.  If we had such an expression, it would deepen our understanding of Apery's constant (hence probably the zeta function, etc.), but we don't.

I don't think mathematicians have the "prejudice" you think they have.  There are very good reasons to prefer the function $ x^n$, for example, over other types of functions:  it counts the number of functions from an $ n$-element set to an $ x$-element set, and this is a general enough framework to encompass a great deal of other counting problems.  For fixed $ x$ and varying $ n$, for example, you can obtain sequences related to walks on graphs, which count functions from $ n$ to the vertices of a graph $ G$ satisfying certain requirements.  For fixed $ n$ and varying $ x$, you can obtain polynomials like the chromatic polynomial, which counts functions from the vertices of a graph $ G$ to a set of colors satisfying certain requirements.

So, simply put:  we prefer certain answers over others because familiar answers to an unfamiliar question place them in a framework we understand, and this deepens our understanding of the unfamiliar question.

Thanks a lot. After reading, I make sure of something. 1, this is not a stupid question and has been considered.    2, It is, very likely, our prejudice and ignorance on "other" functions that leads to our refusal to "other" functions. (If yenlee's comment "……only elementary func……" is right, and if t0rajir0u's comment "……constant we are familiar with……" is right.) (and t0rajir0u: I will discuss about your comment in paregraph 2 latter) When we are familiar with a set of func, (knowing their alge relationship, analytic property and computation aspect [how to compute quickly] ect) we may also allow this set of func to appear in a "closed form solution". And as I said $ \sum \frac{1}{n^{2k\plus{}1}}$ may be a "new number" that is not related, in ways we are familiar with, with constants we are familiar with. That is to say, such judgement as "$ \sum \frac{1}{n^{2k\plus{}1}}$ hasn't been solved" is not "very appropriate", or even "an answer to $ \sum \frac{1}{n^{2k\plus{}1}}$" is not appropriate. Since Zeta func and other methods have produced integral solution, and since some properties have been found.    3, I'm now more confident about my "definition". (Still, I need to hear more advice)

    Actually, my ques come up from two ways.  What is a closed form solution, and what is an appropriate formule (algorithm).
    According to previous discussion, a closed form solution is a solution that 1. yeilds (easily) all properties "we want". (or some particular properties in some cases) 2. "easy" to compute. So this definition depend on what we need (in some cases we may not even need to compute it then the "2" condition is not necessary), and by "find an answer" to "blabla" means to find properties (as much as we can or at least those we need) to "blabla".
    An appropriate formule is required to be easy to compute. At first sight this require seem to be of no problem. However, "easy to compute" till now means easy to compute with Turing model (or models that can be transformed in relative short time to this model.). But in some cases we might not need the "numerical" answer but only some properties that is we might need some relation between the answer to other "things". In these cases, we can or have to change the model to other models fit to the problem. As t0rajir0u said, "There are very good reasons to prefer the function $ x^n$" since it fit many questions well. However we've already seen, in many counting problem, such function no longer fit. So, we may have to find some other set of "appropriate" functions (operations) to fit those class of "counting problems". That's why I add a "man-choosed" basic operations set $ S$ in the definition. Even the set $ S$ can not be easily computed on Turing Model, it may still reveal the relations between questions, if many questions can be "easily computed by this set of operations". It's just like different CPU use different set of instructions. (Still, I need more advice.)

A nice formula is a "nontrivial" / "useful" / "fruitful" / "clever" / "tricky" / "unexpected" / "beautiful" method to compute some quantity.

One may just define the quantity to be a constant, or write down some trivial formulas for it, but that will NOT be very nice.

The aesthetic taste is very important.

I think you might misunderstood my question. Any way, "nontrivil" "beautiful" are ambiguos and "unexpected" can't be defined.


[youtube]Ooa8nHKPZ5k[/youtube]

This is so funny...
 :rotfl:  :rotfl:  :rotfl:

this is retarded...
i dont get y the last 40 seconds are just black screen
...
kinda dumb but ya...
w/e


ceiling function
floor function
calculus
calculus computations

I'm trying to find a second order recursive relation for the edge set of a directed graph.
The exact problem is listed [url=http://math.berkeley.edu/~strain/55.F09/final.sample.pdf]here.[/url], if anyone's interested in seeing the whole problem.

Anyway, I've figured that the sequence looks like this:
$ 0,1,2,4,6,9,12,16,20,25,30,...$, as in add 1 twice, add 2 twice, add 3 twice, etc.

Can anyone see a second-order recursive relation for this sequence? Thanks.

I haven't looked at the original problem, but if we define $ a_0 \equal{} 0,a_1 \equal{} 1$, a second-order recursive relation for your sequence is $ a_{n} \minus{} a_{n \minus{} 2} \equal{} \begin{cases}n & n \text{ is even} \\
n \minus{} 1 & n \text{ is odd}\end{cases}$, or equivalently $ a_n \equal{} \left \lceil \frac{n}{2} \right \rceil \left \lfloor \frac{n\plus{}2}{2}\right \rfloor$.

Thanks for the response JoeBlow. Your answer tipped me off to something, and I think another nice recursive relation is
$ a_n \equal{} a_{n \minus{} 2} \plus{} n$, when we have the initial conditions $ a_0\equal{}0$ and $ a_1\equal{}1$ as you stated. Again, thanks.


vector
linear algebra
linear algebra theorems

I have my final coming up on Tuesday, and I've been practicing problems from my textbook...one of them lists several types of vectors and we have to choose which form subspaces for R3...anyways, I'd like to know why any vectors of the form (a, b, 0) do NOT form a subspace for R3...while vectors of the form (a, 0, 0) do; it would seem to me that vectors of the form (a, b, 0) also work with the two subspace axioms (maybe it's the textbook's error?). Thanks a lot.

Of course they form a subspace . :)

Edit: a pressed submit instead of preview by mistake sorry....:)
Anyway there is nothing spesial here.

If $V = \{(a,b,0) \ \, a,b \in \mathbb{R} \}$ then you only have to show
that  $x,y \in V \rightarrow x+y \in V$ and  $x \in V \, ,\lambda \in \mathbb{R} \rightarrow \lambda x \in V$.


A floor is covered by 120 eight-inch, non-overlapping, square tiles.  How many new two-inch square tiles must be bought in order to cover the same area covered by the old tiles?

for each 8*8, we need 16 2*2's, that means 120*16=1920 tiles.

each 8 by 8 is 64 square inches and there are 120 of them, so 64*120=7680 square inches of floor, then divide by the area of a 2 by 2, which is 4 and get 1920.


calculus
derivative
function
induction
calculus computations

$f$ function n-times differentiable on an interval $I$
Show that for  $1/x \in I$

$\frac{1}{x^{n+1}}f^{(n)}(\frac{1}{x})=(-1)^{n}D^{n}(x^{n-1}f(\frac{1}{x}))$

where $D$ is derivative operator

Suppose that $f$ function n-times differentiable on an interval $I$, and that $1/x \in I$. Prove that $\forall n\in\mathbb{N},$
\[\frac{1}{x^{n+1}}f^{(n)}\left(\frac{1}{x}\right) = (-1)^{n}\frac{d^{n}}{dx^{n}}\left[ x^{n-1}f\left(\frac{1}{x}\right)\right] . \]
Proof: Proceed by induction on $n$:

i) For $n=0$ the formula certianly holds;

ii) assume that for some fixed $n\in\mathbb{N}$, the formula holds, then
\[(-1)^{n+1}\frac{d^{n+1}}{dx^{n+1}}\left[ x^{n}f\left(\frac{1}{x}\right)\right] = (-1)^{n+1}\frac{d^{n}}{dx^{n}}\left[ nx^{n-1}f\left(\frac{1}{x}\right)-x^{n-2}f^{\prime}\left(\frac{1}{x}\right)\right] \]

\[=-n\cdot (-1)^{n}\frac{d^{n}}{dx^{n}}\left[ x^{n-1}f\left(\frac{1}{x}\right)\right]+(-1)^{n}\frac{d^{n}}{dx^{n}}\left[ x^{n-2}f^{\prime}\left(\frac{1}{x}\right)\right] \]

\[=-\frac{n}{x^{n+1}}f^{(n)}\left(\frac{1}{x}\right)+(-1)^{n}\frac{d^{n}}{dx^{n}}\left[ x^{n-2}f^{\prime}\left(\frac{1}{x}\right)\right] \]
by the inductive hypothesis... [color=darkred]got stuck here, why don't one of you mathlinkers finish this post?[/color]


I have posted (not spam of course) in the Classroom Math, MATHCOUNTS, High School basics and my post count still stayed at 84...what's happening???

read's VV's last post [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=118586]here[/url]


A semicircle $(O)$ is given, with diameter $AB$ and let $P$ be, a variable point on $(O).$

We draw the circle $(A),$ centered at $A,$ with radius $AP$ which intersects the segment $AB$ at a point, so be it $M.$

Determine the point $P,$ in order to be the maximum of the segment $PM.$

Kostas Vittas.

Nice problem vittasko.
Let $PA=PM=x,AB=a\Rightarrow 0\le x\le a$ by law of cosine for triangle $PAM$ we have $PM^{2}=AP^{2}+AM^{2}-2AP\cdot AM\cos\angle{PAB}=2x^{2}-2x^{2}\cos\angle{PAB}$
note that triangle $APB$ right at $P$ therefore $\cos\angle{PAB}=\frac{PA}{AB}=\frac{x}{a}$ thus we get
$PM^{2}=2x^{2}-2\frac{x^{3}}{a}\Rightarrow PM\ , \forall x\in [0,a]$ max iff $f(x)=-\frac{x^{3}}{a}+x^{2}$ max on $[0,a]$ we have
$f'(x)=-\frac{3x^{2}}{a}+2x=0\Leftrightarrow x=0 \vee x=\frac{2a}{3}$ easily seen $\max_{x\in [0,a]}=f(\frac{2a}{3})$ equality when $x=\frac{2a}{3}\Leftrightarrow PA=\frac{2AB}{3}$.

Thank you dear [b][size=100]gemath[/size][/b] for your interest and your solution, sorter than the one I have in mind.

I forgot to note an interesting result in this configuration.

When the segment $PM$ becomes maximum $($ $maxPM = \frac{4R\sqrt{2}}{3\sqrt{3}},$ where $R,$ is the radius of $(O)$ $),$ then it is intersected from the midperpendicular of $AB,$ in two segments which the one is triple of the other.

Best regards, Kostas vittas.

Sorry there is typo in my solution it must be
Let $AP=AM=x,AB=a\Rightarrow 0\le x\le a$ and thank you vittasko for your interest, oh but your result $\max PM = \frac{4R\sqrt{2}}{3\sqrt{3}}$ thus mine is wrong?  :maybe: , I must check my solution.


geometry
3D geometry
sphere
tetrahedron
calculus
integration
combinatorics proposed

1. Consider $4$ points chosen at random in an unit cube. These points form a tetrahedron.
Determine the average value for:
1a. the total area of this tetrahedron.
1b. the volume of this tetrahedron.

2. Consider $4$ points chosen at random in a sphere with diameter equal with $1$. These points form a tetrahedron.
Determine the average value for:
2a. the total area of this tetrahedron.
2b. the volume of this tetrahedron.

Well, the average surface area of the tetrahedron is 4 times the average area of a triangle.  Is the a nice way to compute this without lots of integrals?


[quote="hurdler"]I signed up for both WOOT and the AMC  12 class, but now I think the AMC 12 class would be too easy for me. However, I paid for those two classes at once and used a 50 dollar AoPS gift certificate in the process. If I drop the AMC 12 class, will I get a full refund or 50 dollars less than the full refund (since I am not sure which of the two classes the gift certificate got used on)?

Thanks.[/quote]

After you click Drop Class on the AMC 12 class, follow up with an email explaining this situation and we'll sort it out properly (apply the 50 dollar discount to the class you keep, and refund the full 150 on the AMC 12).

How long does it usually take before the refund shows up on the credit card statement?

That's out of our control -- depends on your credit card processor (and these days, I bet credit card companies are dragging their feet as much as possible).  We submit the refund to the credit card company within a day or two.  If it's been more than 6 weeks, send us an email.


inequalities
function
inequalities proposed

Let $ x_{1},x_{2},...,x_{n}$ be nonnegative numbers, only one from which is greater than or equal to 1. If $ x_{1}\plus{}x_{2}\plus{}...\plus{}x_{n}\equal{} n$, then

$ (x_{1}\minus{}1)^{3}\plus{}(x_{2}\minus{}1)^{3}\plus{}...\plus{}(x_{n}\minus{}1)^{3}\ge 0$.

WLOG $ x_{1}\geq 1$.Let $ x_{1}\minus{}1\equal{}a_{1},1\minus{}x_{2}\equal{}a_{2},1\minus{}x_{3}\equal{}a_{3}$ and others similar.
We have to prove
$ a^{3}_{1}\geq a^{3}_{2}\plus{}...\plus{}a^{3}_{n}$,while we have
$ a_{1}\equal{}a_{2}\plus{}a_{3}\plus{}...\plus{}a_{n}$
Thus by substituting $ a_{1}\equal{}a_{2}\plus{}a_{3}\plus{}...\plus{}a_{n}$,we are done. :)

The inequality below is harder.

If $ a,b,c,d$ are nonnegative numbers such that $ a\ge b\ge 1\ge c\ge d$ and $ a\plus{}b\plus{}c\plus{}d \equal{} 4$, then

$ 2(a^{3}\plus{}b^{3}\plus{}c^{3}\plus{}d^{3})\plus{}12\ge 5(a^{2}\plus{}b^{2}\plus{}c^{2}\plus{}d^{2})$.

We can go ahead and homogenize...notice for the following function, that doesn't matter...

$ f(a,b,c,d)\minus{}f\left(a,b,\frac{c\plus{}d}{2},\frac{c\plus{}d}{2}\right)$
$ \equal{}2*\left(c^{3}\plus{}d^{3}\minus{}2*\left(\frac{c\plus{}d}{2}\right)^{3}\right)\plus{}\frac{(a\plus{}b\plus{}c\plus{}d)}{4}*\left(c^{2}\plus{}d^{2}\minus{}2*\left(\frac{c\plus{}d}{2}\right)^{2}\right)$
$ \equal{}(c\minus{}d)^{2}*\left(\frac{3}{2}*(c\plus{}d)\plus{}\frac{a\plus{}b\plus{}c\plus{}d}{4}\right)\ge 0$

We are done by the theory of mixing variables.

[quote="Altheman"]We can go ahead and homogenize...notice for the following function, that doesn't matter...

$ f(a,b,c,d)\minus{}f\left(a,b,\frac{c\plus{}d}{2},\frac{c\plus{}d}{2}\right)$
$ \equal{} 2*\left(c^{3}\plus{}d^{3}\minus{}2*\left(\frac{c\plus{}d}{2}\right)^{3}\right)\plus{}\frac{(a\plus{}b\plus{}c\plus{}d)}{4}*\left(c^{2}\plus{}d^{2}\minus{}2*\left(\frac{c\plus{}d}{2}\right)^{2}\right)$
$ \equal{} (c\minus{}d)^{2}*\left(\frac{3}{2}*(c\plus{}d)\plus{}\frac{a\plus{}b\plus{}c\plus{}d}{4}\right)\ge 0$.[/quote]
After me,
$ f(a,b,c,d)\minus{}f\left(a,b,\frac{c\plus{}d}{2},\frac{c\plus{}d}{2}\right)$
$ \equal{} (c\minus{}d)^{2}*\left(\frac{3}{2}*(c\plus{}d)\minus{}\frac {5}{2}*\frac{a\plus{}b\plus{}c\plus{}d}{4}\right)$.

@Altheman,$ a,b,c,d$ not equal so you can't use Mixing Variable Method  :(

ack wow...i thought that after normalizing, that would remove the condition...but it is very weird...i.e.

if $ a\plus{}b\plus{}c\plus{}d\equal{}4t$, then two of the variables are greater than $ t$ and two are less than $ t$...very strange condition to deal with...? all I can think of is either an ugly substitution or bashing the problem with lagrange multipliers [which isn't particuarly hard... $ (6a^{2}\minus{}10a,...)\equal{}k*(1,1,1,1)$ then we check all the boundaries...]


If Alex gives Bob a penny, Bob will have three times as many pennies as Alex has. If Bob give Alex
a penny, Bob will have twice as many pennies as Alex has. How many pennies does Bob currently have?

Setting up a system of equations, we have $ 3(a\minus{}1) \equal{} b \plus{} 1$, and $ 2(a \plus{} 1) \equal{} b\minus{}1$. Solving these two equations for $ b$, we get $ b \equal{} 17$. Thus Bob has $ 17$ pennies.


Do you how else we can obtain previous USNCO exams? [see the broken link as of Feb 11]
The local exams open just fine, but the national ones do not. i.e. http://portal.acs.org/portal/PublicWebSite/education/students/highschool/olympiad/us/competition/archives/CTP_005487

http://portal.acs.org/portal/acs/corg/content?_nfpb=true&_pageLabel=PP_SUPERARTICLE&node_id=1508&use_sec=false&sec_url_var=region1&__uuid=85ef9e53-bbab-4311-9aea-b38d09afadb2

not sure if i can upload them, but i do have them saved...

Could you please upload them?  I don't have any USNCO tests saved.

I think theyre back up... you can check again.


inequalities
calculus
integration
real analysis
real analysis unsolved

Let $ f: [0,1]\to [0,\infty)$ and it is continuous on $ [0,1]$. Setting $ g(x)=1+2\int_{0}^{x}f(t)dt$, assume that $ g(x)\geq f^{2}(x)\forall x\in [0,1]$. Prove that $ g(x)\leq (1+x)^{2}\forall x\in [0,1].$

SO we have $ g'(x)=2f(x) \leq 2\sqrt{g(x)}\Rightarrow \frac{g'(x)}{2\sqrt{g(x)}}\leq 1$.So $ (\sqrt{g(x)})' \leq 1$, $ \int_{0}^{x}(\sqrt{g(x)})'dx \leq x \Rightarrow \sqrt{g(x)}\leq x+1,(g(0)=1)$.So $ g(x) \leq (x+1)^{2}$