[quote="antony"] France hosted the imo in 1988 or something like this,[/quote]

IMO1988 was in Australia, but there was an IMO in Paris in 1983. I am not sure at the moment, maybe there also were IMOs in France before.

You are right Orlando, it was in Paris in 1983.
And, as Antony said, it seems that the IMO will not be hosted by France until probably many years...It's a pity, maybe a shame for some of us. Yet, the president of the IMO jury for the last 10 years was Claude Deschamps, leader of the french team.

Pierre.

[quote="antony"]
And PLEASE notice that many, many French really hate eating frogs and snails... :D[/quote]

i often heard that Frenchmen likes snails but I never heard that they also like frogs... and it is said that Chirac (french president) does like snails.... isn't it?

I agree with you on France's beauty (although I wouldn't affirm it is really the most beautiful country).
  I'm sorry you guys won't host the IMO anytime soon, but it was to be expected from country whose political members come mostly from the same school in Paris and who has never had (correct me if I'm wrong) a prime minister (or a president) from outside of Paris. Except Raffarin, of course...by the way is he still prime minister?...
  My point is that your sistem is too centralized to be effective. And of course not only education has to suffer...I heard that agriculture is also affected by it...that would explain the funds France gets from the European Union for agriculture...but at least you people have a system... :D

as a Romanian, I would not dare talk about centralized system, and other political issues :)

But if no one talks about them, how are they going to change even in our country?...because they have to change...

yes you are right, but I was saying that we should not talk about France's politics as long as ours really sux :)

Well, Raffarin is still prime minister, and I don't know if he comes from the Paris ENA school. About centralization... I think it must change !!!

[quote="antony"]I live in France, the most  :D  beautiful country of the world... France hosted the imo in 1988 or something like this, and I don't think that it will host it a new time before long, as our ministry of education doesn't seems really interested in math (eg, there was a math competition last march, and the winners still haven't got the prizes).
(I still haven't been to imo. Athens' my last chance...)
The French are the ones who led the revolution against the G.W.Bush, and we're proud of it (just kidding...) !!!

And PLEASE notice that many, many French really hate eating frogs and snails... :D[/quote]

Just so you know - France was also home to the world's first capitalist revolution and first commnuist revolution.

[quote="bubka"]
Just so you know - France was also home to the world's first capitalist revolution and first commnuist revolution.[/quote]
Communist revolution? When?

What I know is that the terms "left" and "right originated during the French Revolution.

[quote="bubka"]That's dumb. Paris commune of course.[/quote]
Ouch Bubka  :(

About the Paris commune : apart from the relatively short time they were in power, how much of France did they really control?

France sucks at wars. They're the only people to have lost 2 wars to Italians. Lol. Italians.

Never mind how much of France PC controlled: it was the world's first ever communist revolution.

[quote="fredbel6"][quote="Ignite168"]France sucks at wars.[/quote]
  :| The fact that that dumb myth was only invented to quell the bitching about a proposterous war seems..... "less relevant".

Here we go again :

1066: William the Conqueror, from France, goes to Britain to claim his throne and wins the battle of Hastings (the current British queen is still a direct descendant of Willam)

1337-1453 : France (barely) wins the Hundred Years war,.

American revolution : only through French help can the colonies obtain independence

1804-1814 : Napoleonic empire : France controls parts of the Italian coastline, all of Belgium, Luxemburg, the Netherlands and part of Germany

1853-1856 : Crimean war, France allies itself with the Ottoman empire and Britain against Russia to protect western christianity in the Ottoman empire.  France is on the victorious side, and provides more troops than all other allies together.

World War I : France is the biggest participant in the "Second Battle of the Marne" and wins under Foch

1923 : Belgium and France invade Germany to enforce the Versailles Treaty
[quote] They're the only people to have lost 2 wars to Italians. Lol. Italians[/quote]
About the Italians, they did beat Carthago three times in the Punic wars... and established a Roman empire that hanged around for about a millennium :ninja:[/quote]

http://www.albinoblacksheep.com/text/france.html

I'm well aware of that file.  I've read it when it was somehow meant to defend the war in Iraq.

Don't get me wrong, I don't support the French in everything they do or have done either (conquering our country, institutionalizing the French language, bombing the Rainbow Warrior, harming innocent people with their nuclear tests, thinking Algeria is theirs ...)
But I am wary when a country with a timeline starting in 1776 starts talking about history...

Can we not play this the other way :

1812-1814 : USA goes to war with United Kingdom and Canada.  One of the goals : getting their hands on more British colonies in Canada.  In 1813, William Hull surrenders Detroit to the British out of fear for Indian tribes.  In 1812 they managed to conquer Toronto, which was actually tiny back then.  They dumbly walk into a trap when the British blow up their own fort while retreating.  The British got back at them for that when they managed to burn Washington DC in 1814, including the White House.
Apparently the French were interesting enough to side with.
The magnificent victory for the USA.... a return to the original situation.

1864-1865 :USA wins a war!  Maybe because the other side was USA too!

1950-1953 : the United States leads a UN coalition against North Korea.  The result.... a tie.

1954 : USA is attacked by Israeli agents during the Lavon affair through their Egyptian institutions.
USA does.... nothing.  And keeps helping Israel.

1961 : USA decides to sneak up on Castro in the "Bay of Pigs".  They get kicked out by Castro's troops

1964-1973: USA decides to help south Vietnam in the struggle against north Vietnam.  The result : a retreat, leaving Laos and Cambodia communist too.

1983 : USA loses more than 200 troops sent to intervene in the Lebanese Civil War.  Reagan says the USA presence will be maintained.
Less than half a year later....USA leaves.

1980-1988 : USA supports Islamic resistance against the Soviet occupation.  Eventually, the Soviets withdraw.  Side effect : the same people who were trained and paid by the USA bite back at them in 2001

1974-2002 : USA decides to support rebel groups fighting against the communist MPLA in the civil war in Angola.  The result : MPLA victory.

I guess USA history books look like this :

[img]http://upload.wikimedia.org/wikipedia/en/thumb/a/a1/WW2_Iwo_Jima_flag_raising.jpg/300px-WW2_Iwo_Jima_flag_raising.jpg[/img]

This is reality :

[img]http://www.fbesp.org/pix/cards.jpg[/img]

[quote="fredbel6"][quote="Ignite168"]France sucks at wars.[/quote]
  :| The fact that that dumb myth was only invented to quell the bitching about a proposterous war seems..... "less relevant".

Here we go again :

1066: William the Conqueror, from France, goes to Britain to claim his throne and wins the battle of Hastings (the current British queen is still a direct descendant of Willam)

1337-1453 : France (barely) wins the Hundred Years war,.

American revolution : only through French help can the colonies obtain independence

1804-1814 : Napoleonic empire : France controls parts of the Italian coastline, all of Belgium, Luxemburg, the Netherlands and part of Germany

[/quote]

they ruled in other parts of europe too.

[quote]1853-1856 : Crimean war, France allies itself with the Ottoman empire and Britain against Russia [b]to protect western christianity[/b] in the Ottoman empire.  France is on the victorious side, and provides more troops than all other allies together.
[/quote]

come on, frederik. Britain and France didn't want that bosphorus strait fell into russian hands. and true, they wanted to solve the problem with the holy land.

[/quote]
[quote]
World War I : France is the biggest participant in the "Second Battle of the [b]Marne[/b]" and wins under Foch
[/quote]
isn't that in france?
[quote]
1923 : Belgium and France invade Germany to enforce the Versailles Treaty
[/quote]
france invade germany to make sure that they would pay her the reparations.
[quote] They're the only people to have lost 2 wars to Italians. Lol. Italians[/quote]
[quote]About the Italians, they did beat Carthago three times in the Punic wars... and established a Roman empire that hanged around for about a millennium :ninja:[/quote]

romans defeated carthago, not italians.

you have forgotten the french military role in laos, vietnam etc. before USA and war in vietnam

second post:
[quote="fredbel6"]I'm well aware of that file.  I've read it when it was somehow meant to defend the war in Iraq.

Don't get me wrong, I don't support the French in everything they do or have done either (conquering our country, institutionalizing the French language, bombing the Rainbow Warrior, harming innocent people with their nuclear tests, thinking Algeria is theirs ...)
But I am wary when a country with a timeline starting in 1776 starts talking about history...

Can we not play this the other way :

1812-1814 : USA goes to war with United Kingdom and Canada.  One of the goals : getting their hands on more British colonies in Canada.  In 1813, William Hull surrenders Detroit to the British out of fear for Indian tribes.  In 1812 they managed to conquer Toronto, which was actually tiny back then.  They dumbly walk into a trap when the British blow up their own fort while retreating.  The British got back at them for that when they managed to burn Washington DC in 1814, including the White House.
Apparently the French were interesting enough to side with.
The magnificent victory for the USA.... a return to the original situation.

1864-1865 :USA wins a war!  Maybe because the other side was USA too!

1950-1953 : the United States leads a UN coalition against North Korea.  The result.... a tie.

1954 : USA is attacked by Israeli agents during the Lavon affair through their Egyptian institutions.
USA does.... nothing.  And keeps helping Israel.

1961 : USA decides to sneak up on Castro in the "Bay of Pigs".  They get kicked out by Castro's troops

1964-1973: USA decides to help south Vietnam in the struggle against north Vietnam.  The result : a retreat, leaving Laos and Cambodia communist too.

1983 : USA loses more than 200 troops sent to intervene in the Lebanese Civil War.  Reagan says the USA presence will be maintained.
Less than half a year later....USA leaves.

1980-1988 : USA supports Islamic resistance against the Soviet occupation.  Eventually, the Soviets withdraw.  Side effect : the same people who were trained and paid by the USA bite back at them in 2001

1974-2002 : USA decides to support rebel groups fighting against the communist MPLA in the civil war in Angola.  The result : MPLA victory.

[/quote]

how useless wasting of money (tax money from federal funds).

If France under Napoleon controlled even more territory, that proves my point even more.

Victory was obtained during the Crimean war, true intentions in war are never that obvious.  It still proves my point.

Yes, Marne is in France, the battle was not far from Paris.

I checked my history atlas and it shows the Roman Empire in 262 before Christ as about all of the Italians peninsula south of the Arno river.  Considering the Romans later even had emperors who were from Spain, I think there is a big chance that, at least during the last Punic War, there were quite a few non Roman soldiers from the Empire fighting too.  And also : they did have a considerable influence on all of Italy (look at the language), so I don't think we can only attribute the Roman values etc.. to the city Rome

About French Indochina, I decided not to add it, because they were gonna see another French defeat in the eventual loss of control.(Even though USA didn't manage to control it either, when even allied  with South Vietnam)

 :roll:  I just found out that this topic is three years old and that it's just Bubka who decided to push it up. :wink:

fredbel, what you said is incorrect, as you blatantly didn't mention the wars the US DID win. Also, your argument about the US was an ad hominem.

[quote="Ignite168"]
fredbel, what you said is incorrect, as you blatantly didn't mention the wars the US DID win.[/quote]
 :roll:  That's the point  :roll:  : as I wrote earlier , France has victories too, and USA has defeats too, anyone can play that game....

[quote]Also, your argument about the US was an ad hominem[/quote]
"ad hominem" means "against the person", which person?  Which argument are we speaking of??
EDIT : Oh I get it, about the timeline I guess?  Well that is how I introduced my 'non ad hominem' arguments.


A regular octagon is formed by cutting an isosceles right triangle from each of the corners of a square with sides of length 2000. What is the length of each side of the octagon?

Let the base of each isosceles triangle be x. Then 2000 - 2x = x :rt2: . 2000 = 2x + x :rt2:  = x(2+ :rt2: ). So 2000 / 2+ :rt2:  = x and you rationalize the denominator to get 2000(2- :rt2: )/2 = x and so x=1000(2- :rt2: ). since you want x :rt2: ( also 2000 - 2x) , then [hide]x :rt2:  = 2000 :rt2: - 2000 .[/hide]


Suppose that $a,b,c > 0$ and $a+b+c=1$.  Prove that

\[\sqrt{\frac{ab}{c}+1}+\sqrt{\frac{bc}{a}+1}+\sqrt{\frac{ca}{b}+1} \geq 2(\sqrt{a}+\sqrt{b}+\sqrt{c}).\]

Haha yeah, I wrote the wrong condition--it's edited now.  :)

If I made no mistake, it's minkowski + cauchy.

I don't think it needs the equals sign.  AM-GM seems to give the result... (two variable)

[quote="siuhochung"][quote="blahblahblah"]Can you elaborate?  I'm not seeing it.[/quote]
I forgot my solution too... anyway, i find this simpler
\[
\sqrt {\frac{{ab}}
{c} + 1}  = \sqrt {\frac{{(c + a)(c + b)}}
{c}}  \geqslant \sqrt {\frac{{c\left( {\sqrt a  + \sqrt b } \right)^2 }}
{c}}  = \sqrt a  + \sqrt b
\]
by cauchy.
The result follows.[/quote]

same result here...

And I find this even easier:

$\frac {ab}{c}+a+b+c\geq a+b+2\sqrt{ab}=(\sqrt{a}+\sqrt{b})^2$

I was just asking because I couldn't see the Minkowski (but then again, I never see when Minkowski can be applied, unless it just slaps me in the face)


How do you show that the average number of representations of a natural number as a sum of two squares is $ \pi$?


LaTeX
Ring Theory
superior algebra
superior algebra solved

Let P be a Prime Ideal of the commutative ring R. Prove that if I and J are Ideals of R and I $intersection$ J $subset$ P, then either I $subset$ P or J $subset$ P

P.S. I tried unsing Latex but I guess it didn't work out. Sorry :blush:

hint: try showing that $IJ\subset I\cap J$ . once u've showed this, use the definition of a prime ideal (namely that if P is a prime ideal and whenever $ab\in P$ then one of a or b is in P) to get what you wanted. i hope this is correct  :|

greets

Assume $I \cap J \subseteq P$, but neither $I \subseteq P$ nor $J \subseteq P$. Then there exist $x \in I \backslash P,~y \in J \backslash P$. But $xy \in I \cap J \subseteq P$ implies $x \in P$ or $y \in P$, a contradiction.

[quote="mikejones"]
P.S. I tried using Latex but I guess it didn't work out. Sorry :blush:[/quote]

Usually latex just comes down to putting between dollar signs.
But for symbols like intersection you have put a \ before those words(intersection doesn't exist, cap does however) :

Let P be a Prime Ideal of the commutative ring R. Prove that if I and J are Ideals of R and I $\cap$ J $\subset$ P, then either I $\subset$ P or J $\subset$ P

You can 'steal'  :wink:  other people's code by clicking the quote button on their messages

Maybe this list can help you too :

[url]http://www.mathlinks.ro/LaTeX/AoPS_L_GuideSym.php[/url]


Let $a$ and $b$ be positive integers greater than $2$. Prove that there exists a positive integer $k$ and a finite sequence $n_1$, $\cdots$, $n_k$ of positive integers such that $n_1 =a$, $n_k =b$, and $n_i n_{i+1}$ is divisible by $n_{i}+n_{i+1}$ for each $i$ $(1 \le i \le k)$.

[b]Definition[/b] We will denote by $ a\rightarrow b$ the fact that there exists a positive integer $ k$ and a finite sequence $ n_{1},\cdots, n_{k}$ of positive integers such that $ n_{1}\equal{} a$, $ n_{k}\equal{} b$, and $ n_{i}n_{i\plus{}1}$ is divisible by $ n_{i}\plus{}n_{i\plus{}1}$ for each $ i$ ($ 1\le i\le k$).
[b]Observation 1[/b] If we reverse the sequence of $ k$ positive integers such that $ n_{1}\equal{} a$, $ n_{k}\equal{} b$, and $ n_{i}n_{i\plus{}1}$ is divisible by $ n_{i}\plus{}n_{i\plus{}1}$ for each $ i$ ($ 1\le i\le k$), we get a sequence of $ k$ positive integers such that $ n_{1}\equal{} b$, $ n_{k}\equal{} a$, and $ n_{i}n_{i\plus{}1}$ is divisible by $ n_{i}\plus{}n_{i\plus{}1}$ for each $ i$ ($ 1\le i\le k$). Therefore, $ a\rightarrow b\Leftrightarrow b\rightarrow a$.
Also, note that if $ (x\plus{}y)|xy$ then $ x\rightarrow y$ and $ y\rightarrow x$.
[b]Lemma 1[/b] $ a\rightarrow a(a\minus{}1)$.
[b]Proof[/b] $ a\plus{}a(a\minus{}1) \equal{} a^{2}|a^{2}(a\minus{}1) \equal{} (a)(a(a\minus{}1))\Rightarrow (a\plus{}a(a\minus{}1))|(a)(a(a\minus{}1))$ so $ a\rightarrow a(a\minus{}1)$ and the lemma is proved.
[b]Lemma 2[/b] If $ a$ is odd, then $ a(a\minus{}1)\rightarrow a(a\plus{}1)$.
[b]Proof[/b] $ a(a\minus{}1)\plus{}a(a\plus{}1) \equal{} 2a^{2}|a^{2}(a\minus{}1)(a\plus{}1) \equal{} (a(a\minus{}1))(a(a\plus{}1))$ because $ a\minus{}1$ is even as $ a$ is odd. So, $ (a(a\minus{}1)\plus{}a(a\plus{}1))|(a(a\minus{}1))(a(a\plus{}1))$ and the result of lemma 2 follows.
[b]Lemma 3[/b] $ a\rightarrow b, b\rightarrow c\Rightarrow a\rightarrow c$.
[b]Proof[/b] We can just take the finite sequence $ n_{1},\cdots, n_{k}$ of positive integers such that $ n_{1}\equal{} a$, $ n_{k}\equal{} b$, and $ n_{i}n_{i\plus{}1}$ is divisible by $ n_{i}\plus{}n_{i\plus{}1}$ for each $ i$ ($ 1\le i\le k$) and the a finite sequence $ m_{1},\cdots, m_{l}$ of positive integers such that $ m_{1}\equal{} b$, $ m_{l}\equal{} c$, and $ m_{i}m_{i\plus{}1}$ is divisible by $ m_{i}\plus{}m_{i\plus{}1}$ for each $ i$ ($ 1\le i\le l$), and the sequence $ a \equal{} n_{1}, n_{2},...,n_{k}\equal{} b \equal{} m_{1}, m_{2},...,m_{l}\equal{} c$ will satisfy the fact that the sum of very two consecutive terms in the sequence will divide their product, so $ a\rightarrow c$ and lemma 3is proved.
[b]Lemma 4[/b] For any positive integer $ a\geq 3$, $ 3\rightarrow a$.
[b]Proof[/b] We will prove the result using mathematical induction.
[b]Base Step[/b] If $ a \equal{} 3$ then $ a\rightarrow 3\Rightarrow 3\rightarrow 3$ which is true.
[b]Induction step[/b] Assume, for $ i \equal{} 3,4,...,k$ we have $ 3\rightarrow i$. Let us prove $ 3\rightarrow k\plus{}1$.
If $ k\plus{}1$ is even, $ k$ is odd, so by lemmas 1 and 2, have $ k\plus{}1\rightarrow k(k\plus{}1)\rightarrow k(k\minus{}1)\rightarrow k$ ($ k(k\minus{}1)\rightarrow k$ because of observation 1 and $ k\rightarrow k(k\minus{}1)$ by lemma 1). Hence, by lemma 3 $ k\plus{}1\rightarrow k$, so $ k\rightarrow k\plus{}1$. But $ 3\rightarrow k, k\rightarrow k\plus{}1$ so by lemma 3 $ 3\rightarrow k\plus{}1$ and the result is true for $ a \equal{} k\plus{}1$.
If $ k\plus{}1$ is odd, by lemmas 1 and 2 have $ k\plus{}1\rightarrow k(k\plus{}1)\rightarrow (k\plus{}2)(k\plus{}1)\rightarrow k\plus{}2$ so by lemma 3, $ k\plus{}1\rightarrow k\plus{}2$ so $ k\plus{}2\rightarrow k\plus{}1$. Because $ k\plus{}1$ is odd, $ k\plus{}2$ is even, let $ k\plus{}2 \equal{} 2l$, then $ k \equal{} 2l\minus{}2$. Now, $ (2l\plus{}2l(l\minus{}1))|(2l*2l(l\minus{}1))\Rightarrow 2l(l\minus{}1)\rightarrow 2l \equal{} k\plus{}2$. Also, $ (2l(l\minus{}1)\plus{}2(l\minus{}1)(l\minus{}2)) \equal{} 4(l\minus{}1)^{2}|(2l(l\minus{}1))(2(l\minus{}1)(l\minus{}2))\Rightarrow 2(l\minus{}1)(l\minus{}2)\rightarrow 2l(l\minus{}1)$. Finally, $ (2(l\minus{}1)\plus{}2(l\minus{}2)(l\minus{}1)) \equal{} 2(l\minus{}1)^{2}|(2(l\minus{}1)*2(l\minus{}2)(l\minus{}1))\Rightarrow 2(l\minus{}1)\rightarrow 2(l\minus{}2)(l\minus{}1)$.
Because $ k\plus{}2 \equal{} 2l$, then $ k \equal{} 2l\minus{}2 \equal{} 2(l\minus{}1)$.
So, $ k \equal{} 2l(l\minus{}1)\rightarrow 2(l\minus{}2)(l\minus{}1)\rightarrow 2l(l\minus{}1)\rightarrow 2l \equal{} k\plus{}2\rightarrow k\plus{}1\Rightarrow k\rightarrow k\plus{}1$, but because $ 3\rightarrow k$ by lemma 3 it follows that $ 3\rightarrow k\plus{}1$ and the result is true for $ a \equal{} k\plus{}1$.
Hence, in both cases we have proved the result for $ a \equal{} k\plus{}1$ and by mathematical induction it follows that $ 3\rightarrow a$ for any positive integer $ a\geq 3$ and lemma 4 is proved.

Now, for any two positive integers $ a,b$ greater than $ 2$, by lemma 4, $ 3\rightarrow a\Leftrightarrow a\rightarrow 3$ and $ 3\rightarrow b$. So, $ a\rightarrow 3$ and $ 3\rightarrow b$ so by lemma 3, $ a\rightarrow b$ and the result of the problem follows.

First, we can see that if there is a finite sequence satisfying the problem requirements such that $n_{1} = a$ and $n_{k} = b$, then we can make a sequence that satisfies the problem requirements such that $m_{1} = b$ and $m_{k} = a$ with the simple mapping of $m_{j} = n_{k-j+1}$. Also, note that if we have a term X in our sequence and D is a divisor of X then (D-1)X may be the next term in the sequence because $\frac{(D-1)X^2}{DX}$ is an integer. Thus if X is a term in our sequence, then we can make X! a term in our sequence as well. If we want to construct a sequence that takes a to b, we can assume that a is less than b. Now, we can construct a sequence that takes a to a! and we now want to construct a sequence that takes a! to b. Let us consider the case in which b is odd. We can see that b+1 is now even and thus divisible by 2. Also, let us note that we can take a! to (2a)! by simply taking a! to 2*a! and noting that this is divisible by 2a which lets us take it to (2a)!. If $a < \frac{b+1}{2} \rightarrow 2a < b+1$. We can now take a! to $(2^k*a)!$ such that $\frac{b+1}{2} \le 2^k*a < b+1$. We can now note that 2 and $\frac{b+1}{2}$ both divide $(2^k*a)!$ thus we can take $(2^k*a)!$  to $b*(2^k*a)!$ and take this to b!. However, we can take b! to b because we can take b to b!. Now let us consider the case in which b is even. We can take b to b(b-1) to (b-2)(b-1) to b-1 which is odd and thus we can construct a path from a to b-1 to b. Thus we can always construct a path from a to b for any pair of integers (a, b).


function
geometry
analytic geometry
algebra
domain
functional equation
algebra proposed

Find all continuous functions $ f: \mathbb{R}\rightarrow\mathbb{R}$ satisfying
$ f(f(x)y \minus{} xf(y)) \equal{} yf(f(x)) \minus{} f(x)f(y)$.

I don't know whether the condition continuous is necessary.
My interest is not finding the solution but how such functions satisfy this equation.. After you solve this, I think you would feel same as me...

[quote="TTsphn"]Take $ x \equal{} y$ then $ f(x)^{2} \equal{} xf(x)$[/quote]
Why?? We just get $ f(0) \equal{} xf(f(x)) \minus{} f(x)^{2}$ so $ f(x)^{2} \equal{} xf(f(x))$...
Because $ f(x) \equal{} ax$ is one of its roots, your equation can't be hold...Try more.

I have proved that $ f(0)\equal{}0$ . Take $ x\equal{}y$ then $ xf(f(x))\equal{}f(x)^2$

What do you mean? You wrote "$ f(x)^{2}\equal{}xf(x)$".
This problem is not that easy I think.....

I'm sorry ,it is only a typo . I just have repair its .

[quote="Little Gauss"][quote="TTsphn"]
Take $ y \equal{} x_0$ then $ f(x_0f(y)) \equal{} x_0f(f(x))$ (1)[/quote]
It should be $ f(x_{0}f(x)) \equal{} x_{0}f(f(x))$ and you can't do anything more.
about Case 2 : $ x \minus{} 1$ is solution, too. So your argument is false.....
Please check your whole solution and edit it...[/quote]
I will complete solution ,it is case 2 of my problem .
$ f(1)\equal{}0,f(0)\equal{}\minus{}1$

You still have a problem in Case1.
For the final step of Case 2, you can't say like that. try $ f(x)\equal{}2x\minus{}\sqrt{3x^{2}\plus{}1}$.

[hide][quote="TTsphn"]Take $ x \equal{} y \equal{} 0$ then $ f(0)^2 \plus{} f(0) \equal{} 0$
Therefore $ f(0) \equal{} 0$ or $ f(0) \equal{} \minus{} 1$
Case 1 : $ f(0) \equal{} 0$
Take $ x \equal{} y$ then $ f(x)^2 \equal{} xff((x))$
We will prove that if exist a real number $ x_0$ different from 0 and $ f(x_0) \equal{} 0$ then $ f(x) \equal{} \equiv 0$
Take $ y \equal{} x_0$ then $ f(x_0f(y)) \equal{} x_0f(f(y))$ (1)
Take $ x \equal{} x_0$ then $ f( \minus{} x_0f(y)) \equal{} 0$
Take $ x \equal{} \minus{} x_0f(y)$ then $ f(x_0f(y)^2) \equal{} 0$
Take $ y \equal{} 1$ then we have $ f(x_0f(f(1)) \equal{} 0$
Take $ x \equal{} f(1)$ in (1) then $ f_3(1) \equal{} 0$
Easy to check that $ f_3(1) \equal{} \frac {f(1)^3}{1^2}$ so $ f(1) \equal{} 0$
Take $ x \equal{} 1$ then $ f( \minus{} f(x)) \equal{} 0$ therefore $ f(f(x)f(y)) \equal{} 0$
If exist a real a such that $ f(f(a))$ is different from 0 then we have :
$ f(f(a)y \minus{} af(y)) \plus{} f(a)f(y) \equal{} xf(f(a))$
So there exist at least of $ f(f(a)y \minus{} af(y)),f(y) > 1$ or exist a real r such that $ f(r) > 1$
Consider $ g(x) \equal{} f(x) \minus{} 1$
$ G(0) \equal{} \minus{} 1 ,f(r) > 0$ and because g continuous so exist $ r'$ such that $ f(r') \equal{} 1$
Take $ x \equal{} r'$ then $ f(f(x)) \equal{} 0,\forall x$
But from $ xf(f(x)) \equal{} f(x)^2$ so $ f(x)\equiv 0 ,\forall x$ ,it gives contradiction .
Now consider problem when has no real different from 0 such that $ f(x_0) \equal{} 0$
From condition $ xf(f(x)) \equal{} f^2(x)$ it shows that f is an inject function .
$ xf(f(x)y \minus{} yf(x)) \equal{} f(x)(yf(x) \minus{} xf(y))$
Take $ y\to x,x\to y$ then
$ xf(yf(x) \minus{} xf(y)) \plus{} yf(xf(y) \minus{} yf(x)) \equal{} 0$
Take $ y\to \minus{} x$ and from f is inject then
$ f(x) \plus{} f( \minus{} x) \equal{} 0$ so f is an odd function .
Therefore $ (x \minus{} y)(f(f(x)y \minus{} xf(y)) \equal{} 0$
It gives that for all $ (x,y)$ then $ f(x)y \minus{} yf(x) \equal{} 0$ or $ f(x)\equiv ax$
Case 2 :$ f(0) \equal{} \minus{} 1$
Take $ y \equal{} 1,x \equal{} 0$ then $ f(0)f(1) \equal{} 0$ and because $ f(0) \equal{} \minus{} 1$ so $ f(1) \equal{} 0$
Take $ x\to y$ then $ xf(f(x)) \equal{} f(x)^2 \minus{} 1$
Take $ x \to 1 then$ then $ f( \minus{} f(y)) \equal{} \minus{} y$
Therefore take $ x\to \minus{} f(x)$ then
$ f(x)f( \minus{} x) \equal{} 1 \minus{} x^2$
More than take $ x\to 0$ then $ f( \minus{} y) \equal{} yf( \minus{} 1) \plus{} f(y)$
Then we have recursion $ f(x)f( \minus{} x) \equal{} 1 \minus{} x^2,f(x) \minus{} f( \minus{} x) \equal{} xf( \minus{} 1)$
Then easy to check that $ f(x) \equal{} x \minus{} 1$ is solution
1.$ f(x)\equiv 0$
2.$ f(x)\equiv ax$
3. $ f(x)\equiv x \minus{} 1$

I think we don't need condition ''continuous '' ,but now i can not find out a complete solution  :)[/quote][/hide]I don't see how "Take $ y \equal{} x_0$ then $ f(x_0f(y)) \equal{} x_0f(f(y))$ (1)."  Perhaps I'm just stupid.

It is true ,because $ f(x_0)\equal{}0$
Take $ y\to x_0$ then $ f(x_0f(x))\equal{}x_0f(f(x))$
It is also true for any real y  .

Oh yeah, I forgot you stated that $ f(x_0) \equal{} 0$.  I always make this kind of stupid mistakes of forgetting the previously stated conditions.

[quote="TTsphn"]
$ xf(f(x)y \minus{} yf(x)) \equal{} f(x)(yf(x) \minus{} xf(y))$
Take $ y\to x,x\to y$ then
$ xf(yf(x) \minus{} xf(y)) \plus{} yf(xf(y) \minus{} yf(x)) \equal{} 0$[/quote]
I can't understand this part... It can be deduced for only case $ f(x)\equal{}f(y)$... isn't it?

Actually you used the fact "f is continuous"!

Case 1: $ f(0) \equal{} \minus{}1$.

Then, we have $ f(x) \equal{} kx\minus{}\sqrt{(k^2\minus{}1)x^2\plus{}1}$ for some $ k \geq 1$.

Case 2: $ f(0) \equal{} 0$.

2.1 $ f(t) \equal{} 0$ for some $ t \neq 0$.
   Then, for some $ k \geq 0$,  $ f(x) \equal{} 0$ if $ x \geq 0$ and $ f(x) \equal{} kx$ if $ x<0$.

2.2 $ f(t) \equal{} 0$ if and only if $ t\equal{}0$.
   I didn't complete this case, but I believe the solutions are in the form $ f(x) \equal{} kx$ for some $ k \neq 0$.

I did finish Cases 1 and 2.1, but my solution is lengthy (and brutally messy).  Do you have any short solution for this problem?

My solution is also long, messy. It seems to be impossible to show that above root of your Case 1 is really root without substituting and calculating both side.
I have solution for $ f(0) \equal{} 0$ using area of triangle in coordinate plane.. how did you do 2.1?

Anyway, It is very curious. How such irregular function satisfy this tidy(?) equation?
I think there are deeper reasons of it.

Case 2.1: $ f(0) \equal{} 0$ and $ f(t) \equal{} 0$ for some $ t \neq 0$

I borrowed a lot of ideas from TTsphn.

I will use his results that
(1) $ f(1) \equal{} 0$,
(2) $ xf\left(f(x)\right) \equal{} \left(f(x)\right)^2$ for all $ x$,
(3) $ f\left( \minus{} f(y)\right) \equal{} 0$ for all $ y$, and
(4) $ f\left(f(x)f(y)\right) \equal{} 0$ for all $ x$ and $ y$.

Let $ S: \equal{} f^{ \minus{} 1}\left(\mathbb{R} \minus{} \{0\}\right)$.  If $ S \equal{} \emptyset$, then $ f$ is the zero function.  Suppose $ S$ is nonempty and let $ F$ be the restriction of $ f$ on $ S$.  With slight effort, we get $ F$ is injective.

Suppose that there exists $ a \neq 0$ such that $ f(a) > 0$.  Thus, $ f\left(f(a)\right) \equal{} \frac {\left(f(a)\right)^2}{a} \neq 0$.  TTsphn suggests that for all $ y$,
\[ yf\left(f(a)\right) \equal{} f\left(f(a)y \minus{} af(y)\right) \plus{} f(a)f(y)\,.
\]
Hence, we can choose $ y$ positively large or negatively large enough that $ yf\left(f(a)\right) \gg \max\left\{f(a),1\right\}$.  Thus, the above equation implies that either
\[ f\left(f(a)y \minus{} af(y)\right) \gg 1
\]
or
\[ f(y) \gg 1\,.
\]
In any case, using the continuity of $ f$, there must be a number $ u \neq 0$ such that $ f(u) \equal{} 1$.  Consequently,
\[ 0 \equal{} uf(1) \equal{} uf\left(f(u)\right) \equal{} \left(f(u)\right)^2 \equal{} 1
\]
is a blatant contradiction.  Therefore, $ f(x) \leq 0$ for all $ x$.

Now, if $ f(p) < 0$ and $ f(q) < 0$ for some $ p > 0$ and $ q < 0$, then the injectivity of $ F$ is violated (argued by continuity of $ f$).  Thus, either
(i) $ f(x) \equal{} 0$ for all $ x \geq 0$  and $ f(x) \leq 0$ for $ x < 0$, or
(ii) $ f(x) \equal{} 0$ for all $ x \leq 0$ and $ f(x) \leq 0$ for $ x > 0$.
The latter case cannot happen; otherwise, there would exist $ s > 0$ such that $ f(s) < 0$ and $ f\left(f(s)\right) \equal{} \frac {\left(f(s)\right)^2}{s}$ would be positive.

Hence, $ f(x) \equal{} 0$ if $ x \geq 0$ and $ f(x) \leq 0$ for $ x < 0$.  Now, assume that $ f(r) < 0$ for some $ r < 0$.  Thus, for all $ y \geq 0$,
\[ f\left(f(r)y\right) \equal{} f\left(f(r)y \minus{} rf(y)\right) \equal{} yf\left(f(r)\right) \minus{} f(r)f(y) \equal{} y\frac {\left(f(r)\right)^2}{r} \,,
\]
or $ f\left(f(r)y\right) \equal{} k\left(yf(r)\right)$, where $ k: \equal{} \frac {f(r)}{r} > 0$.  Since $ y$ is arbitrary, we then prove that $ f(x) \equal{} kx$ for all negative $ x$.

We can easily check that both the zero function and the functions in the form discussed above satisfy the condition.  Therefore, for any $ k \geq 0$,  $ f(x) \equal{} 0$ if $ x \geq 0$ and $ f(x) \equal{} kx$ for $ x < 0$ is a solution to the functional equation.

PS:
Can anyone show me how to deal with the case where $ f(t) \equal{} 0$ if and only if $ t \equal{} 0$, please?

Case 1: $ f(0) \equal{} \minus{} 1$

Let $ k : \equal{} \minus{} \frac {f( \minus{} 1)}{2}$.  We know that
(1) $ xf\left(f(x)\right) \equal{} \left(f(x)\right)^2 \minus{} 1$ for all $ x$,
(2) $ f( \minus{} y) \equal{} \minus{} 2ky \plus{} f(y)$ for all $ y$ (replacing $ x$ by $ 0$ in the original equation),
(3) $ f(1) \equal{} 0$ (replacing $ y$ by $ 1$ in (2)), and
(4) $ f\left( \minus{} f(y)\right) \equal{} \minus{} y$ (replacing $ x$ by $ 1$ in the original equation).
Replace $ y$ by $ \minus{} f(y)$ in (2) and use (4) to get
\[ f\left(f(y)\right) \equal{} 2kf(y) \minus{} y\,.
\]
From (1) and the above,
\[ \left(f(x)\right)^2 \minus{} 1 \equal{} xf\left(f(x)\right) \equal{} x\left(2kf(x) \minus{} x\right)\,.
\]
This shows that $ f(x) \equal{} kx \pm \sqrt {\left(k^2 \minus{} 1\right)x^2 \plus{} 1}$ for each $ x$.  By continuity of $ f$ and the fact that $ f(0) \equal{} \minus{} 1$, we conclude that
\[ f(x) \equal{} kx \minus{} \sqrt {\left(k^2 \minus{} 1\right)x^2 \plus{} 1}\,.
\]
Now, $ f(1) \equal{} 0$ implies that $ k$ is nonnegative.  Since the domain of $ f$ is $ \mathbb{R}$, then $ k \geq 1$.

You should check that any $ f$ in the above form satisfies the condition.  (This is not an easy task, but I confirm it is true.)  Hence, all functions $ f$ satisfying the condition with $ f(0) \equal{} \minus{} 1$ are
\[ f(x) \equal{} kx \minus{} \sqrt {\left(k^2 \minus{} 1\right)x^2 \plus{} 1}\,,
\]
where $ k \geq 1$.  (For $ k \equal{} 1$, we get $ f(x) \equal{} x \minus{} 1$, for example.)

Case 2.2: $ f(t) = 0$ iff $ t = 0$.

Borrowed ideas from TTsphn are
(1) $ f$ is injective and
(2) $ xf(f(x)y - xf(y)) = f(x)(f(x)y - xf(y))$ for all $ x$ and $ y$.

For each $ x$, define $ S_x : = \{f(x)y - xf(y): y \in \mathbb{R}\}$ (of course, $ 0 \in S_x$ for each $ x$).  Moreover, if $ x \neq 0$, $ m_x : = \frac {f(x)}{x}$.  Also, let $ T$ denote the set $ \{x \in \mathbb{R}: S_x \neq \{0\}\}$.  Thus, for all $ x \in \mathbb{R}$ and $ y \in S_x$,
\[ f(y) = m_xy\,.
\]
If $ T = \emptyset$, then there exits $ k\in\mathbb{R}$ such that $ f(x) = kx$ for all $ x$.

Assume that $ T \neq \emptyset$ (but keep in mind that $ 0 \notin T$).  Hence, whenever $ a$ and $ b$ are in $ T$, $ S_a$ and $ S_b$ contain a common nonzero element or else, one of them contains only nonnegative elements and the other contains only nonpositive elements.  Thus,
\[ T = T^ + \cup T^ - \,,
\]
where $ T^ + = \{x \in T: S_x \subseteq [0, + \infty)\}$ and $ T^ - = \{x \in T: S_x \subseteq ( - \infty,0]\}$.  Hence, $ m_x$ are equal for all $ x \in T^ +$; let say that they are all $ m_ +$ (if $ T^ + = \emptyset$, let just say $ m_ + = 0$).  The number $ m_{ - }$ is defined similarly.

Now, if $ x \in \mathbb{R} \setminus T$ and $ y \in T$, we have $ S_x = \emptyset$, or
\[ f(x)y - xf(y) = 0\,.
\]
Since $ f(y) = m_{\pm}y$, we then have
\[ f(x) = m_{\pm}x
\]
as well.  This shows that $ f(x) = m_{\pm}x$ for all $ x \in \mathbb{R}$.

By continuity of $ f$, for all $ x \geq 0$, $ f(x) = k_{ + }x$ for some $ k_ + \in \{m_{\pm}\}$.  Similarly, for all $ x \leq 0$, $ f(x) = k_{ - }x$ for some $ k_ - \in \{m_{\pm}\}$.

(i) If $ k_ + > 0$ and $ k_ - > 0$, then for $ x > 0$ and $ y < 0$,
\[ f((k_ + - k_ - )xy) = f(f(x)y - f(x)y) = yf(f(x)) - f(x)f(y)\,.
\]
This shows that $ f((k_ + - k_ - )xy) = k_ + ((k_ + - k_ - )xy)$ or $ 0 < k_ + < k_ -$.

(ii) If $ k_ + > 0$ and $ k_ - < 0$, then for $ x > 0$ and $ y < 0$,
\[ f((k_ + - k_ - )xy) = f(f(x)y - f(x)y) = yf(f(x)) - f(x)f(y)\,.
\]
Hence, $ f((k_ + - k_ - )xy) = k_ + ((k_ + - k_ - )xy)$.  However, since $ (k_ + - k_ - )xy < 0$, $ f((k_ + - k_ - )xy) = k_ - ((k_ + - k_ - )xy)$.  Therefore, $ k_ + = k_ -$, which is absurd.

(iii) If $ k_ + < 0$, then for $ x > 0$ and $ y < 0$,
\[ f((k_ + - k_ - )xy) = f(f(x)y - f(x)y) = yf(f(x)) - f(x)f(y)
\]
implies $ f((k_ + - k_ - )xy) = 0$.  Thus, $ k_ + = k_ -$.  This contradicts the fact that $ T$ is nonempty.

Therefore, the solutions are
(1) $ f(x) = kx$ for all $ x$, and
(2) $ f(x) = k_ + x$ if $ x\geq0$ and $ f(x) = k_ - x$ if $ x\leq0$, where $ 0 < k_ + < k_ -$.
Combining all Cases I have writen, the solutions to this functional equations are
(1) $ f(x) = kx$ for all $ x$, where $ k$ is an arbitrary constant,
(2) $ f(x) = k_ + x$ if $ x\geq0$ and $ f(x) = k_ - x$ if $ x\leq0$, where $ 0 \leq k_ + < k_ -$, and
(3) $ f(x) = kx - \sqrt {(k^2 - 1)x^2 + 1}$, where $ k \geq 1$.

PS: This is the weirdest (and toughest) functional equation I have ever seen.  By the way, there is no need to deal with Case 2.1 at all.  I think Case 2.2 can be made to cover 2.1.


vector
topology
linear algebra
linear algebra unsolved

Let X be a real vector space and let M:X->X be a map such that 2M( (x+y)/2 ) =M(x)+M(y) for any x,y in X. Deduce that M is linear.

(May need to assume that M is continuous).

[quote="chungyc"]
(May need to assume that M is continuous).[/quote]

What is X? A subset of $\mathbb{C}$?

X is just a vector space over $\mathbb{R}$. However, I suppose we need a topology on X to talk about continuity. Let's assume that there is a norm defined on X.


function
real analysis
real analysis unsolved

If $ A\subseteq\mathbb{R}$, and $ g: \mathbb{A}\to\mathbb{R}$ is a lower semi-continuous function, prove that for each $ f: \mathbb{A}\to\mathbb{R}$ with $ f(x)\geq g(x)$ for every x in A, we have $ \liminf_{x\rightarrow y}{f(x)}\geq g(y)$ for every y in A.

Split it into two parts: $ \liminf_{x\rightarrow y}{g(x)}\geq g(y)$ and  $ \liminf_{x\rightarrow y}{f(x)} \ge \liminf_{x\rightarrow y}{g(x)}$.


Prove that the inradius of a right angled triangle with integer sides is an integer.

We have a right triangle $\Delta = s \cdot r = a \cdot b / 2$

$a \cdot b = (a+b+c)(a+b-c)$

So $r = (a+b-c)/2$

So we have to prove that $2 \mid a+b-c$ but since $a^2 + b^2 = c^2$, $a+b$ has same parity as $c$.

[quote="Valentin Vornicu"]Use \cdot instead of \times or "X" next time, ok? ;)[/quote]

OK ......... [hide]but why ?[/hide]

Because \times is used either for vectorial product, or for sizes of matrices, arrays and tabulars. \cdot (which means "central dot") is for multiplication.


From a circular piece of paper with radius BC, Jeff removes a sector of the circimference. Using the large sector left, he joins edge BC to edge BA(no overlap) to form a cone of radius 12 centimeters and of volume 432$\pi$ cubic centimeters. What is the number of degrees in the measure of angle ABC of the sector that is not used?
Enjoy... :D

[hide]You find the height to be 9 if you use the volumn and the radius.
 You can then easily tell that the slant height or the radius of circle B is 15 using the Pythagorean Theorem.
Therefore the circumference of circle B is 30$\pi$
You find the circumference of the cone is 24$\pi$

$360(\frac{30-24}{30})=\boxed{72}$[/hide]

This question is actually number 29, by the way.

[hide]h=height of cone, r=radius of circle=slant height of cone, x=degree of arc

$\frac{12^2\times h}{3}=432\pi \implies h=9$.
So $r=15$. Then $x=360-360\times\frac{2\times 12\pi}{2\times 15\pi}=72$.[/hide]


algorithm
number theory
relatively prime

[hide]Let  be the smaller of the two integers, and  be the larger.  Thus, to see if they are relatively prime, a fraction containing them must not reduce.  This fraction is  or its inverse.  Expanding this fraction, it gives .  Since x is an integer,  is always not an integer, making the two numbers relatively prime.

There are 2 special cases to this:

, in which case 1 and 2 are relatively prime

 or , so that x or x+1 is 0, therefore making relative primality trivial.[/hide]

mrblueskies wrote:prove that any two consecutive integers are relativley prime.

[hide]we could use euclids algorithm which completely trivializes it...[/hide]

Oh really?  I was thinking of somthing along the lines of towbomb.  What is Euclid's algorithim and how do you use it?

Suppose the integer [i]n[/i] is divisible by any number of facors {a,b,c...}.

n must be congruent to all of these integers mod 0.

n + 1 must be congruent to all of these integers mod 1 . Consequently, n and n  + 1 have no factors in common (except 1) and are therefore relatively prime. [/i]


Posted here:
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=3048[/url]

And please choose better titles for your posts; the title "Inequalities" is too generic.


topology
function
advanced fields
advanced fields unsolved

$E$ the set of function from $[0;1]$ to $\mathbb{R}$. Does there exist a topology on $E$ that generates the simple convergence.

It's the topological space $\mathbb R^{[0,1]}$, i.e. the topological product of $c$ copies of $\mathbb R$ ($c$ being the continuum cardinality).

in fact this question was a particular case of the following question. On a topological space $E$, if one define a kind of convergence, (i.e. if we know all convergent sequence) does there always exists a topology that generates this convergence.

At the beginning we wanted to define the set of closed sets of $E$ as follows:
Let $C$ the set of all convergent sequences of $E$. To each $U \in C$, let $L(U)$ the limit of $U$.
One says that $F$ is a closed subset if any sequence $U$ of element of $F$ that belongs to $C$ satisfies $L(U) \in F$.
But it does not work because maybe there is more convergent sequence wiht this new topology. So, is the general statement true ? but maybe this is a very stupid question ..

dear Grobber, I think i've not perfectly understand your example. Could you precise it ? because it seems it doesn't work isn't it ?

The functions from $[0,1]$ to $\mathbb R$ can be regarded simply as elements of $\mathbb R^{[0,1]}$. A net (generalized sequence) of functions $(f_{\alpha})_{\alpha}$ from $[0,1]$ to $\mathbb R$ converges pointwisely to a function $f: [0,1]\to\mathbb R$ iff the corresponding elements form a net converging to the element corresponding to $f$. There really isn't anything to prove. This is how you define the product topology.

I thought the product topology on $\mathbb{R}^{[0;1]}$ was the topology generated by th sets of the form :
$\Pi_{x \in F} w_x$ where $F$ is a finite set of $[0;1]$ and $w_x$ an open set of $\mathbb{R}$ for any $x \in F$. Is it ?

And why would you think that this is different from the topology of pointwise convergence? :)

consider the  topology generated by the sets of the form :
$\Pi_{x \in S} w_x$ where $S$ is an arbitrary set of $[0;1]$ and $w_x$ an open set of $\mathbb{R}$ for any $x \in S$. What is now the convergence ? it is much stronger than uniform convergence ..

There are several topologies you can put on an (infinite) cartesian product.
what you need for simple convergence is the smallest topology for which the following functions are continuous:
$ev_x: E\to\mathbb R$, $ev_x(f)=f(x)$, for $x\in [0,1]$.
In other words a basis of open sets is
$\bigcap_{i=1}^nev_{x_i}^{-1}(G_i)$
where $x_1,\dots, x_n$ are finitely many points and $G_i$ are open in $\mathbb R$.
Your example is not of this kind.

yep, I got this (this is exactly the usual product topology). But I still wonder what kind of convergence we obtain with the new topology ?

$f_n\to 0$ if for any positive function $\phi: [0,1]\to (0,\infty)$, there exists $N(\phi)$ such that
$|f_n(x)|\leq\phi(x)$, $\forall x\in [0,1]$, $\forall n\geq N(\phi)$.

example:  if $f_n(x)=0$ for $x\neq 1/2$ and $f_n(1/2)=1/n$, then $f_n\to 0$.


[hide][quote="anirudh"]What common fraction is equal to half of 80%?[/quote]

80%=4/5
half of that is 2/5[/hide]

A good way to do this problem, or any problem which has a % sign, is to think of % as a variable which stands for 1/100. Then just work with the resulting fraction.

[hide]80% is 4/5, half of which is 2/5[/hide]

Percent = percentage/100, therefore 80/100 * 1/2

[quote="bpms"]Wasn't this a state countdown problem from this year?[/quote]

Yes it was (2006 State Countdown #12). If you want to see all of the problems, click [url=http://www.mathcounts.org/webarticles/articlefiles/510-06SCountdown.pdf]here[/url].


Let $\large a;b;c > 0$.
$\large \sum \sqrt{\frac{a}{a^2+bc}} \leq\ \sum \frac{1}{\sqrt{a+b}}$

Yes it is true. Put $x=\frac1a$ etc., rename and see
http://www.mathlinks.ro/Forum/viewtopic.php?t=67457
It is the $n=\frac12$ case of (2). (posts #4 and #5 prove it for all positive $n$) :D


Sooo, who got a perfect?  There have got to be some on the forum.

Thazn or whatever his username is. :blush:

That 's so wonderful. Congratulations, ThAnz1!


calculus integration limit trigonometry algebra function domain

I'm trying to solve some problems from the Measure & Integral class and I'm having difficulties with the following problems:

1. Calculate

$ \lim_{n\to\infty}\sum_{m \equal{} 0}^{\infty}\left(\frac {1}{2}\right)^{nm}\cos{(nm)}$

and justify your calculations.

2. Let $ f_n(x) \equal{} \left(1 \minus{} \left(1 \minus{} \frac {1}{x^2}\right)^2\right)^n$ be a sequence of functions. Calculate

$ \lim_{n\to\infty}\int_{2}^{ \plus{} \infty}f_n(x)dx$

and justify your calculations.

For the first problem, I don't know how to approach it. I've solved some problems where an integral sign is placed instead of the sum by using the LDCT (Lebesgue's Dominant Convergence Teorem) but I don't know if something similar applies when there is a sum.

For the second problem, if I take the domain of $ f_n$ to be $ X \equal{} \left[1, \plus{} \infty\right\rangle$, than the limit $ \lim_{n\to\infty}f_n(x)$ equals the indicator function $ \textbf{1}_{\{1\}}$ so to apply the LDCT I need to find an integrable function $ g\colon X \to \left[0, \plus{} \infty\right]$ such that $ \forall n\in\mathbb{N}\quad |f_n(x)|\leq g(x) \quad (\mu \minus{} a.e.)$ on $ X$, but I'm having difficulties with finding $ g$.

Thanks for your help. :)

For the first one, I'm thinking Weierstrass M-test.  Fix $ m$ so that we get $ \frac {\cos(mn)}{2^{mn}}\leq\frac {1}{2^{mn}} \equal{} \frac {1}{(2^{m})^{n}} \equal{} M_{n}$.  Then $ \sum_{n \equal{} 1}^{\infty}M_{n} \equal{} \sum_{n \equal{} 1}^{\infty}(2^{ \minus{} m})^{n} \equal{} \frac {1}{1 \minus{} 2^{ \minus{} m}} \equal{} \frac {1}{1 \minus{} \frac {1}{2^{m}}} \equal{} \frac {2^{m}}{2^{m} \minus{} 1}$, which converges to 1.  This means that $ f_{m}(n) \equal{} \frac {\cos(mn)}{2^{mn}}$ converges uniformly.  So you can swap limits and bring $ \lim_{n\rightarrow\infty}$ inside the summation.  That is, $ \sum_{m \equal{} 0}^{\infty}\lim_{n\rightarrow\infty}\frac {\cos(mn)}{2^{mn}} \equal{}1\plus{} \sum_{m \equal{} 1}^{\infty}0 \equal{}1\plus{} 0\equal{}1$.  I'm not sure if this works.

For the second one, as far as I can tell, that function is monotone on your interval.

The DCT does apply to sums. A sum [i]is[/i] an integral over the integers (or some subset of the integers) with counting measure - that is, the measure of each integer point is 1. You could argue that the in the first problem, the terms tend pointwise to the the function $ (1,0,0,\dots)$ - that is, the $ m\equal{}0$ term tends to $ 1$ while every other term tends to $ 0.$ All terms are bounded above by $ 2^{\minus{}m},$ which is summable (that is, integrable), so $ 2^{\minus{}m}$ serves as the dominating function. So the limit is $ 1\plus{}0\plus{}0\plus{}\cdots \equal{}1.$

But it's not a problem that showcases the need for the DCT. You can justify the interchange of sum and limit by more elementary means, as JRav showed. (Even if he did overlook the fact that the $ m\equal{}0$ term is different.)

The second problem is also not a particularly convincing display of the need for the DCT or MCT. Note, by the way, that having a decreasing positive monotone sequence of functions doesn't guarantee anything - just consider $ \int_0^{\infty}\frac{x}{n}\,dx$ as $ n\to\infty.$ To get anything out of that, you have to have some function be integrable.

How I'd argue that: start with $ 0\le1\minus{}\left(1\minus{}\frac1{x^2}\right)^2\le\frac{2}{x^2}$ just by direct computation.

Then $ \int_2^{\infty}f_n(x)\,dx\le\int_2^{\infty}\frac{2^n}{x^{2n}}\equal{}\left.\minus{}\frac{2^n}{(2n\minus{}1)x^{2n\minus{}1}}\right|_2^{\infty}\equal{}\frac1{(2n\minus{}1)2^{n\minus{}1}}.$

And that goes to zero. The argument is just calculus estimation and didn't need any big theorems. Yes, you could call it DCT using $ f_1$ as the dominating function - but that seems like overkill.

Or is the in integral for the second problem actually from $ 1$ to $ \infty$ rather than starting at $ 2?$ Because then we need a little more. If we are integrating on $ [1,\infty),$ we see that $ f_n(x)\to 0$ for $ x\in (0,\infty)$ but $ f_n(1)\equal{}1.$ That's still going to zero almost everywhere. Then we can use that $ f_n(x)\le f_1(x)\le\frac2{x^2}$ and that's an integrable function. So the limit is zero by the DCT.

Yes, I forgot to include that $ m\equal{}0$ term in my sum :).  I've edited it.


inequalities
trigonometry
inequalities open

I am recalling trigonometric identities. (For inequalities)

I know that if I have $ a\plus{}b\plus{}c\equal{}abc$ then $ a\equal{}\tan{\alpha},b\equal{}\tan{\beta},c\equal{}\tan{\gamma}$ with $ \alpha\plus{}\beta\plus{}\gamma\equal{}\pi$ may work. (there are more, but my main question is:)

Is there any similar trig substitution for $ ab\plus{}bc\plus{}ca\equal{}abc$?

In such situation you can always change variables to $ x\equal{}\frac{1}{a}$, $ y\equal{}\frac{1}{b}$, $ z\equal{}\frac{1}{c}$ and then $ x\plus{}y\plus{}z\equal{}1$.

You can try :

$ a\equal{} tan X tan Y$
$ b\equal{} tan Y tan Z$
$ c\equal{} tan Z tan X$

For example, prove that :

$ \frac{3}{2}\sqrt{x\plus{}y\plus{}z\minus{}1} \ge \sqrt{x\minus{}1}\plus{}\sqrt{y\minus{}1}\plus{}\sqrt{z\minus{}1}$

where $ xy\plus{}yz\plus{}zx\equal{}xyz$ and $ x,y,z>1$

[quote="Nemion"] Is there any similar trig substitution for $ ab \plus{} bc \plus{} ca \equal{} abc$?[/quote]

This is not trigonometic, but it works in many inequalities. It's easy to check that for any $ x$, $ y$ and $ z$:

\[ {\frac{3}{1\plus{}x\minus{}y} \cdot \frac{3}{1\plus{}y\minus{}z} \plus{} \frac{3}{1\plus{}y\minus{}z} \cdot \frac{3}{1\plus{}z\minus{}x} \plus{} \frac{3}{1\plus{}z\minus{}x} \cdot \frac{3}{1\plus{}x\minus{}y} \equal{} \frac{3}{1\plus{}x\minus{}y} \cdot \frac{3}{1\plus{}y\minus{}z} \cdot \frac{3}{1\plus{}z\minus{}x}}\]

So we can substitute $ a\equal{}\frac{3}{1\plus{}x\minus{}y}$, $ b\equal{}\frac{3}{1\plus{}y\minus{}z}$ and $ c\equal{}\frac{3}{1\plus{}z\minus{}x}$.

Sergey Markelov


Using Riemann sums, find

$
\lim_{n\to\infty}\sum_{k=1}^{n}\sin(k/n)\cdot\sin(k/n^2)
$

hmmm, if you are right please explain why..

$\lim \sum k/n\sin(k/n)=\int_0^1 x\sin x\mathrm{d}x$

if your result is true, then then we should prove

$\lim \sum (k/n-\sin(k/n^2)\cdot\sin(k/n)=0$

why?

[quote="probability1.01"]Well as k/n^2 goes to 0 we have sin(k/n^2) = k/n^2...[/quote]that is not a valid proof for it's summation being zero ;)

[quote="sebas"]... please explain

[/quote]

Use

$x-\frac{x^3}{6}\leq \sin x \leq x$ for $x\geq 0$

this is what I tried

$
\sin(k/n)\cdot\sin(k/n^2)\leq k/n^2\cdot\sin(k/n)
$

then

$
S_n=\sum_{k=1}^n \sin(k/n)\cdot\sin(k/n^2)\leq
\sum_{k=1}^n k/n^2\cdot\sin(k/n)
\to\int_0^1 t\sin t\mathrm{d}t
$

So we get the bounds

$
0\leq S_n\leq \int_0^1 t\sin t\mathrm{d}t
$

Similarly

$
\sin(k/n)\cdot\sin(k/n^2)\geq \bigl(\sin(k/n^2)\bigr)=
\frac{1-\cos(2k/n^2)}{2}
$

then

$
S_n\geq \sum_{k=1}^n \frac{1-\cos(2k/n^2)}{2}
$

I've found the sum of the right side, made $n\to\infty$ and the limit is different from  $\int_0^1 t\sin t\mathrm{d}t$

$\sin(k/n)\cdot\sin(k/n^2)\geq (k/n^2-k^3/(6n^6))\sin(k/n)$

$S_n\geq 1/n\sum_{k=1}^{n}k/n\sin(k/n) - \sum_{k=1}^{n}k^3/(6n^6)\sin(k/n)$

The second sum in RHS tends to 0


[quote="cincodemayo5590"]Given a rectangle $ABCD$ with $AB > BC$, pick an arbitrary point $P$ in the plane. Prove that for any $P$, one of the two expressions:
\[m= | [APD]+[BPC] | , \ \ \ n = |[APD]-[BPC] | \]
remains constant, and find that constant.

$( [XYZ]$ denotes the area of triangle $XYZ$.$)$[/quote]

Um...
[hide]
If we draw the altitudes from P to AD and BC...[APD]+[BPC] is $\frac{BC}{2}(PE+PF)$ if E and F are the points of intersection of the altitude...but PE+PF is constant? and BC is constant as well...or is that not it... :maybe: [/hide]

[quote="paladin8"][hide="hint"] Consider the two cases: (1) $P$ is between lines $AD$ and $BC$, and (2) $P$ is not between them. Figure out which case corresponds to which expression being constant. [/hide][/quote]

Whoops.
[hide]
So then if P is between AD and BC then it's the addition one, and wouldn't it be basically the same if P isn't between AD and BC, except you subtract it instead...since if P were outside AD and P was closer to AD than BC you'd get PF-PE as a constant where F is the altitude to BC and E the altitude to AD...and same if P were closer to BC?[/hide]


group theory
abstract algebra
superior algebra
superior algebra unsolved

What is the order of the group $ G$ with presentation
$ \left < a,b|a^{16} \equal{} b^6 \equal{} 1, bab^{ \minus{} 1} \equal{} a^3 \right >$

No hint given  :(

Also, can someone reccomend me a book that discusses free groups and finitely presented groups in a "teaching" way (i.e. not for researchers, but for students). I find the material we have is not enough.

$ a \equal{} b^6 a b^{ \minus{} 6} \equal{} a^{3^6} \equal{} a^9$, I think you can do the rest.

[hide="solution"]Thus it's $ \langle a,b : a^8 \equal{} b^6 \equal{} 1 ,  b^{\minus{}1} a b \equal{} a^3 \rangle  \cong \mathbb{Z}/6 \rtimes \mathbb{Z}/8$ (remark $ gcd(3,8)\equal{}1$ and $ 3^6 \equiv 1$ mod $ 8$).[/hide]

Ok, thanks! Did you write something else at first? Because from $ a^8 \equal{} 1$ I have also found the order to be $ 48$, but I thought you wrote $ Z_3 \rtimes Z_8$ so I have been thinking about if perhaps I made a mistake in my reasoning.

Another question: You write $ \langle a,b : a^8 \equal{} b^6 \equal{} 1 , b^{ \minus{} 1} a b \equal{} a^3 \rangle \cong \mathbb{Z}/6 \rtimes \mathbb{Z}/8$. Can you really write isomorphism here? I thought the second expression doesn't uniquely determine a group unless both subgroups are normal, although it does give the order and many other properties.

What about book advice?

yes I edited the post a couple of times. it's virtually impossible to determine the exact order if you don't give a more concrete representation of this group. so you should review your own proof. from the relations alone you just get an upper bound.

in the context it is clear that the semidirect product is given by $ \mathbb{Z}/6 \to Aut(\mathbb{Z}/8), 1 \mapsto (z \mapsto 3z)$.

I did give a concrete operation to the group by defining $ (a^xb^y)(a^ib^j)$ by looking at $ ba \equal{} a^3b$, and then checking that the operation works ($ 3^6 \equal{} 1$ mod 8 comes into play as you wrote). I'm not too sure about what the semi-direct product stuff means, but I'll try to read a little about it and think.

ok you probably worked out the semidirect product in this special case. you may have a look at http://en.wikipedia.org/wiki/Semidirect_product. but one important thing is missing there (maybe I'll add it): if $ \alpha : G \to Aut(N)$ is  given, $ G \rtimes_{\alpha} N$ (in the article $ N \rtimes_{\alpha} G$) is the free group with generators of $ G$ and $ N$, relations of $ G$ and $ N$ as well as $ g^{ \minus{} 1} n g \equal{} \alpha(g)(n)$ for all $ g \in G, n \in N$.


summer program
PROMYS
Ross Mathematics Program
number theory

[quote="DavidL"]Hallo! I am wavering between CTY Number Theory and Promys. Which one would I learn more at?

Thanks a muffinload,

05.1 Logic
David/Johnny[/quote]

As a PROMYS alumni, I would have to say that I'd be surprised if a high school student could learn that much NT in one summer anywhere else... I haven't heard much about CTY's NT course, but I'd bet PROMYS is much more rigorous and challenging, not to mention more selective (I think).

You would learn much much much much ... much more at PROMYS. CTY NT is pretty basic. ;)

Errr yeah CTY NT is pretty trivial. They're just like hmmm let's learn how to count I mean induct. This is true of CTY courses in general, I think. Of course, PROMYS may require a certain amount of mathematical experience (correct me if I'm wrong), but even so, you probably satisfy the requirements. It's always a good idea to do something challenging. You should ask beta, who went to PROMYS. OK, now I'm expecting that muffin: http://www.artofproblemsolving.com/Forum/weblog.php?w=46

This thread would certainly be on topic in, and might get more replies in, the [url=http://www.artofproblemsolving.com/Forum/index.php?f=136]Other U.S. Contests and Programs Forum[/url].

I have read in online discussion from one alumnus of Ross Program (similar to PROMYS) and the CTY Number Theory that two weeks of Ross >>>>> three weeks of CTY.


$p_{1}$,$p_{2}$,……$p_{n}$ are primes,and $p_{i}<>p_{j},i<>j$
prove :sqrt: $p_{1}$+ :sqrt:$ p_{2}$+……+ :sqrt:$ p_{n}$can't be an integer

Have a look here :
http://www.mathlinks.ro/Forum/viewtopic.php?highlight=rational&t=4496

Pierre.


How many equilateral triangles in the plane have two vertices in the set $ \{(0, 0), (0, 1), (1, 0), (1, 1)\}$?

The four points make up a square. There are 4 eq. triangle that share a side of the square and lies outside the square. Similarly, there are 4 eq. triangles that share a side and is inside the square. There are also 4 eq. triangles that use the diagonals as a side. Thus, there are a total of 12 different eq. triangles.

Choose 2 points from the given 4 in 4C2 ways = 6
From 2 points, 2 different triangles are possible. Hence 12.


inequalities
function
inequalities unsolved

[quote="vardanch"]Prove that

\[ (x + y + z)^4 \ge 27(x^2 y^2 + y^2 z^2 + x^2 z^2 )  \][/quote]

This is wrong; take x = 2, y = 2, z = 1.

Beat, you applied Chebyshev the wrong way round...

  Darij

[quote="pvthuan"]It must be
\[ (x+y+z)^4\geq16(x^2y^2+y^2z^2+z^2x^2).  \]
The proof involves AG, and derivative.[/quote]

 I just curious to see the nice proof for this one.
I have done it by substitute $y=x+a$, $z=x+a+b$ with $x,a,b \geq 0$. After expanding,  :blush: I get all of terms are positive. So this inequality is perfectly correct.

Problem. Prove that if $a,b,c\geq 0$, then $(a+b+c)^4\geq16(a^2b^2+b^2c^2+c^2a^2)$.

Solution. Without loss of generality, we can suppose that $a+b+c=1$, and $a\geq b\geq c$. It follows that $0\leq c\leq\frac13$. Let $f=a^2b^2+b^2c^2+c^2a^2$, by AG inequaliy
\[ f\leq \frac{(a+b)^4}{4^2}+c^2(a+b)^2=\frac{(1-c)^4}{16}+c^2(1-c)^2.  \]
Consider function $f(c)=\tfrac1{16}(1-c)^4+c^2(1-c)^2$ over $[0,\tfrac13]$, differentiating $f(c)$ gives
\[ f'(c)=-\frac14(1-c)(17c^2-10c+1).  \]
We have $f'(c)=0$ when $c=1$, or $c=\tfrac{5-\sqrt8}{17}$. Fuction $f(c)$ is decreasing on $[1,\tfrac{5-\sqrt8}{17}]$, increasing on $[\tfrac{5-\sqrt8}{17},\tfrac13]$. It follows tha
\[ f(c)\leq\max\{f(0),f(\tfrac13)\}=\frac1{16}.  \]
Equality holds when $(a,b,c)$ is some permutation of $(0,\tfrac12,\tfrac12)$.
 :lol:

@darij: All I could find is http://www.mathlinks.ro/Forum/viewtopic.php?t=2848 . Is this what you were looking for?

BTW, is your avatar supposed to represent a version broken-line version of Pappus' Theorem?

Lovasz, it is nice too. Have you got this kind of inequality in four numbers?

Thanks, pvthuan.
I think this inequality is true, it maybe trivial.

Let $x,y,z,t \ge 0$. Show that:
\[ (x+y+z+t)^4 \ge 16\left(x^2y^2+y^2z^2+z^2t^2+t^2x^2 \right)  \]

It's easy. Right?
I had another one but I'm not sure it's true.

[quote="Lovasz"]Can you post it?
I wonder if our solution is same or different.[/quote]
Ok, let's wait me one day.

[quote="zhaobin"]my solution:
for the orginal problem,it is equal to:$\sum_{cyc}x^4+4\sum_{sym}x^3y+6\sum_{cyc}x^2y^2+12\sum_{cyc}x^2yz \ge 16\sum_{cyc}x^2y^2$
just use schur's:$\sum_{cyc}x^4+\sum_{cyc}x^2yz \ge \sum_{sym}x^3y$
and $\sum_{sym}x^3y \ge 2\sum_{cyc}x^2y^2$

for the n variables:assume x be the minimal one:
we can prove:$(z+t)^2x^2+y^2(z+t)^2 \ge y^2z^2+z^2 t^2+t^2x^2$
it reduce to $n-1$ variables..[/quote]

Oh... master of expanding messy expression. I'm not brave enough like you...
Indeed, I don't really like yours, zb.

[quote="Lovasz"]I will prove the inequality by using Mixing Variables.

[quote] Let $a,b,c$ be nonnegative reals. Prove that:
\[ (x+y+z)^4 \ge 16(x^2y^2+y^2z^2+z^2x^2)  \] [/quote]
Let $f(x,y,z)=(x+y+z)^4-16(x^2y^2+y^2z^2+z^2x^2)$.
Wlog, assume $0 \le z \le y \le x$.
\begin{eqnarray*} f(x,y,z)-f\left(x,\frac{y+z}{2},\frac{y+z}{2}\right) &=& 16\left[x^2\left(\frac{(y+z)^2}{2}-y^2-z^2\right)+\left(\left(\frac{y+z}{2}\right)^2-yz\right) \left(\left(\frac{y+z}{2}\right)^2+yz\right) \right] \\ &=& 4(y-z)^2\left(\left(\frac{y+z}{2}\right)^2+yz-2x^2\right) \\ & \ge & 0 \\ \implies f(x,y,z) \ge f\left(x,\frac{y+z}{2},\frac{y+z}{2}\right) \end{eqnarray*}
Normalize choice y letting $x+y+z=2$,that follows $x \in \left[0,\frac{2}{3}\right]$.
Now:
\begin{eqnarray*} f(x,y,z) \ge f\left(x,\frac{y+z}{2},\frac{y+z}{2}\right) \\ &=& 16-16\left(2x^2\left(\frac{y+z}{2}\right)^2+\left(\frac{y+z}{2}\right)^4\right) \\ &=& 16-16\left(2x^2\left(\frac{2-x}{2}\right)^2+\left(\frac{2-x}{2}\right)^4 \right) \\ &\ge& 0 \end{eqnarray*}
This is obviously true, because:
\[ 2x^2\left(\frac{2-x}{2}\right)^2+\left(\frac{2-x}{2}\right)^4 \le 1 \\ \iff x\left(9x^3-40x^2+56x-32\right) \le 0\ \text{for all}\ x\in\left[0,\frac{2}{3}\right]  \]
Done.[/quote]
sorry,I prefer zhaobin's solution to yours.

[quote="Lovasz"]But I don't think you want to expand the 4-variable inequality?[/quote]
but I wrote a method to reduce 4-varialbes to 3-varialbes :)

Thanks,Hawk Tiger

Problem: Let $a,b,c,d\geq0$, prove that $(a+b+c+d)^4\geq16(a^2b^2+b^2c^2+c^2d^2+d^2a^2)$.

Solution: WLOG, we suppose that $0\leq a\leq b\leq d\leq c$ and $a+b+c+d=1$. The right hand reads
\[ f=(a^2+c^2)(b^2+d^2).  \]
We need to show that $f\leq 1/16$. Note that
\begin{eqnarray*} f&=&\{(a+c)^2-2ac\}\{(b+d)^2-2bd\}\\ &\leq& (a+c)^2\{(b+d)^2-2bd\}\\ &=&(1-s)^2(s^2-2p). \end{eqnarray*}
where $s=b+d$, $p=bd$. Consider $f(p)=-2p(s-1)^2+s^4-2s^3+s^2$ over $0\leq p\leq s^2/4$. We have $f(p)$ is decreasing over its interval of definition. So that $f(p)\leq f(0)$. Moreover, $f(0)=s^2(1-s)^2$.

By AG inequality, we have
\[ s^2(1-s)^2\leq\Bigl(\frac{s+1-s}2\Bigl)^4=\frac1{16}.  \]

pvthuan

Problem: Let $a,b,c,d \ge 0$, prove that $(a+b+c+d)^4 \ge 16 (a^2b^2+b^2c^2+c^2d^2+d^2a^2)$.

There is another straightforward proof:

$(a+b+c+d)^4 \ge 16(a+c)^2(b+d)^2 \ge 16(a^2 b^2 + b^2c^2 + c^2d^2 + d^2a^2)$.

The first inequality is by AM-GM, the second one is obvious.  :)

[quote="darij grinberg"]See Theorem 1 in [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=316023#p316023]http://www.mathlinks.ro/Forum/viewtopic.php?t=43904 post #13[/url] (if anyone can find the topic Harazi is speaking about, I would be grateful).

  Darij[/quote]
http://www.mathlinks.ro/Forum/viewtopic.php?highlight=%28ab%29%5En&t=2934
http://www.mathlinks.ro/Forum/viewtopic.php?highlight=%28ab%29%5En&t=2070

So, I think that the more general case also holds:

If $a,b,c\geq 0$ and $n \in \mathbb N$ with $n\geq 2$ then

$(a+b+c)^{2n}\geq 4^n(a^nb^n+b^nc^n+c^na^n)$

Stelios...

What about this one?
[hide="Click to reveal hidden content"]
[quote]
Let $a,b,c,d$ be positive reals. Prove that:
\[ (a+b+c+d)^5 \ge 256(a^2b^2c+b^2c^2d+c^2d^2a+d^2a^2b)  \]
[/quote] [/hide]
It seems to be quite ugly.

[quote="pvthuan"]Even the following is true
\[ (a+b+c)^4\geq16(a^2b^2+b^2c^2+c^2a^2)+11abc(a+b+c).  \][/quote]
Yes, it is true for all non-negative real $a,$ $b$ and $c.$

[quote="Lovasz"]What about this one?
Let $a,b,c,d$ be positive reals. Prove that:
\[ (a+b+c+d)^5 \ge 256(a^2b^2c+b^2c^2d+c^2d^2a+d^2a^2b)  \]
It seems to be quite ugly.[/quote]
No, it is very nice, but wrong. Tty $a=b=c=1$ and $d\rightarrow0^{+}.$ ;)


Didn't know where to post it, move it if needed.

So basicly you have to prove the equations [b]using only "Algebra" of sets, dont use the term x[/b], choose one side of the eauation and develop it to the other side while using known and proved sentences.

[b]In the second equation, A' or B' means [u]Complement[/u], the wierd plus with the circle around him in  means [u](A-B)U(B-A)[/u], the unit of the sets without there cut.[/b]

[img]http://img80.imageshack.us/img80/7541/qaaub7.jpg[/img]

THANKS

Let me try the second: $ A \oplus B \equal{} A' \oplus B'$.

The operation $ A \oplus B$ (which was also represented as $ A \Delta B$) is (or was) called [i]symetric difference[/i], and equals $ A \oplus B \equal{} (A \cup B) \minus{} (A \cap B)$. So we are to prove:

$ (A \cup B) \minus{} (A \cap B) \equal{} (A' \cup B') \minus{} (A' \cap B')$.

But we have:

$ A' \oplus B' \equal{} (A' \cup B') \minus{} (A' \cap B') \equal{} (A' \cup B') \cap (A' \cap B')'$

$ \equal{} (A' \cup B') \cap (A \cup B)$

$ \equal{} (A \cap B)' \cap (A \cup B)$

$ \equal{} (A \cup B) \cap (A \cap B)'$

$ \equal{} (A \cup B) \minus{} (A \cap B) \equal{} A \oplus B$

and so our equality is proved.

Great solution, i managed to solve it in another way.
for each A,B, A-B= A$ \cap$B' and then we get A+B= [(A') $ \cap$ (B')'] $ \cup$ [(A')' $ \cap$ (B')]= A'+B'
But i can't solve the first, please help!

For the first equality, let's deal with the second member:

$ (A_1 \minus{} B_1) \cup (A_1 \minus{} B_2) \equal{} (A_1 \cap B_1') \cup (A_1 \cap B_2') \equal{} A_1 \cap (B_1' \cup B_2') \equal{} A_1 \cap (B_1 \cap B_2)'$

where I have used the definition of difference, the distributive law, and Morgan's law. In a similar way:

$ (A_2 \minus{} B_1) \cup (A_2 \minus{} B_2) \equal{} (A_2 \cap B_1') \cup (A_2 \cap B_2') \equal{} A_2 \cap (B_1' \cup B_2') \equal{} A_2 \cap (B_1 \cap B_2)'$

and so

$ (A_1 \minus{} B_1) \cup (A_1 \minus{} B_2) \cup (A_2 \minus{} B_1) \cup (A_2 \minus{} B_2) \equal{} [A_1 \cap (B_1 \cap B_2)'] \cup [A_2 \cap (B_1 \cap B_2)']$

$ \equal{} (A_1 \cup A_2) \cap (B_1 \cap B_2)' \equal{} (A_1 \cup A_2) \minus{} (B_1 \cap B_2)$

where I have used the distributive law and the definition of difference. So, the equality is proved.


logarithms
integration
inequalities proposed
inequalities

Let $H_n=\sum\limits_{k=1}^{n}\frac{1}{k}$  and  $A_n=H_n-\frac{1}{2^n}\sum\limits_{k=1}^{n}{n \choose k}H_k\; .$ [color=green][b] Prove   [/b][/color]\[ \begin{array}{|c|} \hline \\ \displaystyle \frac{1}{(n+2)2^{n}} < \displaystyle \ln{2} -A_n < \displaystyle\frac{1}{(n+1)2^{n}}\\ \\ \hline \end{array} \; , \; (n\in {\mathbb N}^* ).  \]

[quote="flip2004"]Let $H_n=\sum\limits_{k=1}^{n}\frac{1}{k}$  and  $A_n=H_n-\frac{1}{2^n}\sum\limits_{k=1}^{n}{n \choose k}H_k\; .$  Prove that  $\frac{1}{(n+2)2^{n}} < \ln{2} -A_n < \frac{1}{(n+1)2^{n}}\; \; ,\; \; (n\in {\mathbb N}^* ).$[/quote]
Some hints for possible complete solutions:
[i]First solution:[/i]   $\ln{2}-A_n=\frac{1}{2^n}\sum\limits_{k=0}^{\infty} \frac{(-1)^k}{(k+1){k+n+1\choose n}}\; .$ Further use properties of alternating series.
[i]Second solution:[/i]   If $S_n: =\sum\limits_{k=1}^n {n \choose k}H_k ,$ then  $S_n=2S_{n-1}+\frac{1}{n}\sum\limits_{k=1}^n {n\choose k} .$ Thus $S_n= 2^n\left(H_n-\sum\limits_{k=1}^n \frac{1}{k2^k}\right)$ ,   $A_n=\sum\limits_{k=1}^{n} \frac{1}{k2^k}$  and $\ln{2}- A_n= \frac{1}{2^n}\int\limits_{0}^{1}\frac{x^n}{2-x}\; dx .$


i dont understand proof of $Inn(G) \unlhd Aut(G)$ where $Inn(G)$ interior automorphisms of G

Let $\sigma \in \mbox{Aut} ( G )$. And $c_{x} \in \mbox{Inn} ( G )$ defined by $c_{x} ( a ) = xax^{- 1}$ for all $a \in G$ and an $x \in G$.
Then for all $a \in G$,
\begin{eqnarray*} \left( \sigma c_{x} \sigma^{- 1} \right) ( a ) & = & \left. \sigma \left( c_{x} ( \sigma^{- 1} ( a ) \right) \right)\\ & = & \left. \sigma ( x \sigma^{- 1} ( a ) x^{- 1} \right)\\ & = & \sigma ( x ) \sigma \left( \sigma^{- 1} ( a ) \right) \sigma ( x )^{- 1}\\ & = & \sigma ( x ) a \sigma ( x )^{- 1}\\ & = & c_{\sigma ( x )} \end{eqnarray*}
Since $\sigma$ is an automorphism, as $x$ goes through $G$, $\sigma ( x )$ goes through $G$.
Hence $\sigma \mbox{Inn} ( G ) \sigma^{- 1} = \left\{ \sigma c_{x} \sigma^{- 1} | x \in G \right\} = \left\{ c_{\sigma ( x )} | x \in G \right\} = \left\{ c_{x} | x \in G \right\} = \mbox{Inn} ( G )$ for all $\sigma \in \mbox{Aut}(G)$.
If anything is not clear, just tell me.

Well isn't that the definition of normality ?
Let $H$ be a subgroup of $G$.
$H$ is normal in $G$ if and only for all $x \in G$, we have $xHx^{-1}=H$.
Correct ?

[quote="spx2"]so how does the last thing
$\sigma \mbox{Inn} ( G ) \sigma^{- 1}=\mbox{Inn} ( G )$
proove that innG is normal subgroup of G ?[/quote]
You of course ment normal subgroup of $AutG$

ok...i understood.
could i get some bibliography that i can download from the internet ?
i found something wrote by beachy http://www.math.niu.edu/~beachy/abstract_algebra/study_guide/contents.html
but i dont seem to have all proofs there.
i need also a book with allot of exercices and examples

amfulger thank you for the resource.
now comes a problem about that course of j.s. milne.
i started reading it,i find that some of the theorems used are said
to have been taught in a different course.
where might that be?
also i understand that it is math594 - the name of that course
what does this exactly mean ? that number 594 corresponds to some branch of
mathematics and how advanced is the course.
i think i read somewhere that courses (outside of romania ofcourse) have
these numbers attached to them...


during a super collision ,as in the case of neutrons what is the external force that provides an increase in velocity after collision?

What? You mean like in fission? In fission the neutrons that hit the nuclei disturb the strong nuclear force holding the particles, which liberates the potential energy in the protons held together.

serialk11r what is the generic term of that force was my question .your answer was for a specific case


[quote="Interval"](1) What is the best way to master, if you will, solving word problems?

(Please, don't answer the obvious like PRACTICE, PRACTICE, PRACTICE, etc.)

(2) Why do college textbooks call word problems APPLICATIONS?

(3) Why do elementary school teachers call word problems MATH STORIES?

Thanks[/quote]

(1) PRACTICE, PRACTICE, PRACTICE, etc.
Seriously, though, learn to look for the key information in a problem. Translate some of the relationships into mathematical expressions. Etc.

(2) Because college students are mainly focusing on how math can be "applied" to whatever their major is. So the word problems are "applications" of the math concepts to the real world.

(3) Because the teachers don't want to scare the little kids away from math by calling them "applications".

(1)
Learn the lingo. You already know 'of' means '$*$' and stuff, but word problems are written in words, and you need to be able to translate any words you find into math.

(2)
mathnerd314 said it

(3)
because at the elementary school level they are stories. Bob the clown has 7 balloons.... that's a story!

For #3, the kids most likely don't know a lot of vocabulary but they probably know the words "math" and "story".


AMC USA(J)MO USAMO inequalities geometry function algebra

What general subjects do you hope each problem will cover?

My dream USAMO ( :P ):

1.  Inequality
2.  Geometry
3.  Number theory/divisibility

4.  Game theory
5.  Number theory/general
6.  Algebra/polynomial

1. Number theory
2. Inequality or something bashable
3. Don't care

4. Geometry
5. Functional equation
6. Don't care

[quote="t0rajir0u"]Last year's USAMO was as close to my dream as it'll get; I think the main reason I did well is that my solution to #4 was more rigorous than most.  I seem to be better at those kinds of problems than at problems that require large theorems, for example.[/quote]

Last year's day 1 was pretty sweet; I got all 3.  Then day 2 pwn'd me; I only got #5 and the answer (un-rigorous proof) to #4.

Of course, last year was the year I didn't qualify for the USAMO :(

[quote="mathcrazed"]1. Geometry
2. Geometry
3. Geometry

4. Geometry
5. Geometry
6. Geometry

Notice a trend here? They're the only things I can do xD[/quote]

I'd say that's my best too, but I tried to be more realistic :D

Mine are

1.Logic
2. Algebra

4. Number Theory
5. Inequality

Don't care about others...

1. combo
2. geo
3. ...

4. alg/ineq
5. geo
6. ...

I don't really plan on working on more than 2 problem per day.

NOT inequalities. I epic fail those.

1. Geo
2. Number Theory
3.

4. Combo
5. Geo
6.

I have [i]never[/i] solved a 3/6, so I don't really care what goes there. I probably have a best shot if it's number theory or geo...

and as I was writing this, I realized I really have no idea what my strengths/weaknesses are...

[quote="nr1337"]Last year's day 1 was pretty sweet; I got all 3.  Then day 2 pwn'd me; I only got #5 and the answer (un-rigorous proof) to #4.

Of course, last year was the year I didn't qualify for the USAMO :([/quote]

Wow, that really stinks

[quote="nr1337"][quote="tjhance"]NOT inequalities. I epic fail those.[/quote]

My theory is that if there's an inequality for #1, it will be a relatively straightforward application of AMGM/Cauchy/Jensen/etc.  I'm not all that great with inequalities, but I'm sure I could do a #1.[/quote]

True, but if there is an inequality, it's probably either going to be geometric, or just one of those freaky weird ones that's asymmetric and involve weird functions, and not a #1.

I hope I'm wrong.

[quote="Hamster1800"]Here's my hopes:

1. Not Geometry
2. Not Geometry
3. Not Geometry

4. Not Geometry
5. Not Geometry
6. Not Geometry

Sorry, mathcrazed <_<[/quote]
QFAAAAAAAAT

Anyway,
1. Combo
2. Algebra
3. Game Theory
4. Combo
5. Algebra
6. Game Theory

I think all inequalities on the USAMO are meant to be applications of those basic 3 inequalities (plus rearrangement), just probably not always straight forward (though the last one we had was a #1, and it was really straight forward...)

I just hope that

Please let geo on the test is purely synthetic, and doesn't involved much length and/or angle measurement.  I can't stand lengthy equations in geometry problems.
Not more than one inequality.  And a nice one, please.
More problems like #3 last year, pl0x.  Gimme good combo.
NT: Meh, as long as it's not a factorization like last year's #5.
Algebra: Please go easy on the polynomial stuff, it seems to suddenly be a weakness of mine
Yay game theory.

In general, nothing bash, because that's something I've avoided in problems ever since MOP, and I think it would be hard to go back to it.

Hmmm I'm not really sure...

Best Performance:

1. nothing weird
2. geometry
3. combinatorics that lends itself to analyzing small cases

4. nothing weird
5. combinatoricsy NT
6. geometry

Most Exciting:

1. ...
2. any combinatorics
3. polynomials

4. ...
5. something unconventional
6. combinatorial geometry

Most Annoying:

1. casework
2. a "does there exist" problem where the answer is yes
3. 3 variable inequality

4. casework
5. diophantine equation
6. functional equation with lots of variables

1.number theory
2.graph theory
3.geometry(something like USAMO 2005 #3)

4.inequality
5.sequence
6.geometry(something like USAMO 2006 #6)
 :D

1/4 anything besides geometry
2/5 game theory, number theory, inequalities
3/6 geometry. eww.

As you can see, I don't like geometry very much. :P

Hmm I know I can't get in on pure skill...

1. Game theory
2. random creativity problem
3. a collection of 100 mathcounts state sprint problems

4. Game theory
5. random creativity problem
6. a collection of 300 mathcounts national countdown problems

ok realistically.

1. Geometry
2. NT involving prime numbers
3. Dont care won't get it

4. Game theory / random creativity
5. Algebra
6. Dont care wont get it.

hopefully number 3/6 is a geo, cuz those are the only 3/6s i have any remote chance at getting

if at least three of the remaining four problems are bashable, i'll be VERY happy (and i'll be seeing you all at MOP ;p )

[quote="Anaxerzia"][quote="K81o7"][quote="cyberspace"]Game Theory is pretty much all im good at and thats just because i get bored in 4.5 hrs and i dont have anything else to do except play the game :)[/quote]

Ergh...I lost...

Sure Anaxerzia...but you can't possibly suck as much as I do.  I'm going to end up getting a negative zero.

Argh...can't stand the tension waiting for the problems!  No matter what problems come up, there's some luck involved...[/quote]

Yeah you suck so abysmally you shot be shot... -_-

Seriously this self-depreciation helps us feel better about ourselves, and your just mocking it by being a USAMO winner an all...

bah.[/quote]

In case anyone didn't quite catch it, the negative zero is a joke.

[quote="Hamster1800"][/quote][quote="Anaxerzia"][quote="K81o7"][quote="cyberspace"]Game Theory is pretty much all im good at and thats just because i get bored in 4.5 hrs and i dont have anything else to do except play the game :)[/quote]

Ergh...I lost...

Sure Anaxerzia...but you can't possibly suck as much as I do.  I'm going to end up getting a negative zero.

Argh...can't stand the tension waiting for the problems!  No matter what problems come up, there's some luck involved...[/quote]

Yeah you suck so abysmally you shot be shot... -_-

Seriously this self-depreciation helps us feel better about ourselves, and your just mocking it by being a USAMO winner an all...

bah.[/quote][quote="Hamster1800"]

In case anyone didn't quite catch it, the negative zero is a joke.[/quote]

In case anyone didn't quite catch it, mine was only 3/17ths serious :P

1. Functions
2. Algebra/Sets
3. Geometry (no inequalies/equalities in it)
4. Number theory
5. Game theory/logic
6. Polynomials (dont really care since ill get a 0 on it)

the only 3/6 i can get are geometries and only with like the ones with angle chasing/concyclicity

[quote="Anaxerzia"]
Yeah you suck so abysmally you shot be shot... -_-

Seriously this self-depreciation helps us feel better about ourselves, and your just mocking it by being a USAMO winner an all...

bah.[/quote]

I'm trying to make a point here.  Everyone says they will do so badly...

I was taking the USAMO for the first time last year, and said the same thing.  And look at what happened.

As someone who has experienced the [b]Korean Mathmatical Olympiads.....[/b]

I daresay the USAMO is about the same as the Final Round of the KMO series...Hmm.

As both tests select the 12-member national team~

My ideal USAMO:

1 - Something like Easy Algebra or Easy Combinatorics (But Alas, expecting an EASY on the USAMO is stupid :| )

2 - A "sensible" Geometry/ Number Theory

3 - I'd really like some Algebra here...something I can tackle  :P

4 - Number Theory/Geometry

5 - Algebra

6 - Don't care that much...I just wish I'd be able to UNDERSTAND the question and write a few lines... :P

"Hey, it's TODAY! Just a few more hours and we'll see for OURSELVES  :lol: "

[quote="Hamster1800"]Here's my hopes:

1. Not Geometry
2. Not Geometry
3. Not Geometry

4. Not Geometry
5. Not Geometry
6. Not Geometry

Sorry, mathcrazed <_<[/quote]
Agreed!
Except no inequalities either.

[quote="DonkeyKingKong"]I got a 15... probably 7/1/0/7/0/0[/quote]

You got a 3 on the AIME per your post http://www.artofproblemsolving.com/Forum/viewtopic.php?t=203500 :wink:

obviously donkeykingkong is full of crap :D

[quote="DonkeyKingKong"]I got a 15... probably 7/1/0/7/0/0[/quote]

That's bullshit
You're bragging about your 3 on AIME and you're trying to skip to MIT


Let $ S \equal{}{99}!^{\frac{1}{99}}$ and $ T \equal{}{100!}^{\frac{1}{100}}$
a) without using a calculator, which number is larger, $ S$ or $ T$?
b) Which is larger: $ \frac{S}{99}$ or $ \frac{T}{100}$?
c) Limit as $ n$ tends to infinity of  $ \frac{{n!}^{\frac{1}{n}}}{n}$

[quote="lordWings"]Your question b) is exactly equivalent to a).
Looking at c), I think that question b) should be:
Which is larger: $ \frac{S}{99}$ or $ \frac{T}{100}$?
c) The limit is $ \frac{1}{e}$, using Stirling's approximation:
[url]http://en.wikipedia.org/wiki/Stirling's_approximation[/url][/quote]

Any takers for (a) or (b)?

I might be wrong, but can we say that $ \frac{1}{e}\approx\frac{99!^{\frac{1}{99}}}{99}$
and therefore $ \frac{99}{e}\approx 99!^{\frac{1}{99}}$?

Same process for 100

And we have that $ \frac{100}{e}>\frac{99}{e}$

I'm not sure how to find the error term on that expression though.

a) is much simpler than that.  [hide="All you have to say is..."] $ S$ is the geometric mean of the set $ \{ 1, 2, ... 99\}$.  The geometric mean will stay unchanged if we toss in the mean itself, which is clearly less than $ 99$.  Hence tossing in the extra value $ 100$ increases the value of the new geometric mean $ T$. [/hide]

Or you can just clear the exponents and substitute $ n \equal{} 99!$ to compare

$ n^{100}\quad\rho\quad (100n)^{99}$

where $ \rho$ is the desired relation. This simplifies to

$ n\quad\rho\quad 100^{99}$

and $ \rho$ is obviously $ <$, since all the factors on LHS are less than $ 100$.

Another solution for $ c)$

Let $ a_{n}\equal{}\frac{n!}{n^{n}}$.
We have to compute $ \lim_{n\to\infty}\sqrt[n]{a_{n}}$

From Cauchy-d'Alembert criteria we know that $ \lim_{n\to\infty}\sqrt[n]{a_{n}}\equal{}\lim_{n\to\infty}\frac{a_{n\plus{}1}}{a_{n}}\equal{}\lim_{n\to\infty}\left(1\minus{}\frac{1}{n\plus{}1}\right)^{n}\equal{}e^{\lim\minus{}\frac{n}{n\plus{}1}}\equal{}e^{\minus{}1}$


hi evryone i am new and i wich that i became more good in math

[quote="Qruni"]Can the moderator please add the subforum for the Linear Algebra, Complex Analysis, Calculus and Real Analysis[/quote]

...but they already exist, under college playground...

Can I know how to be a moderator here ??? :roll:


function
inequalities
limit
real analysis
real analysis unsolved

Prove that there's no function $f: \mathbb{R}^+\rightarrow\mathbb{R}^+$ such that $f(x)^2\ge f(x+y)\left(f(x)+y\right)$ for all $x,y>0$.

Note that the inequality can be rewritten as $ f(x)(f(x) \minus{} f(x\plus{}y)) \geq f(x\plus{}y)f(y)$ so it is clear that $ f$ is strictly decreasing.
There are two possibilities: $ \lim_{x\to 0^{\plus{}}} f(x) \equal{} M$ where $ M$ is positive real number or $ \lim_{x\to 0^{\plus{}}} f(x) \equal{} \plus{}\infty$.

Suppose that the first holds. Then consider the sequence $ (x_n)$ given by $ x_{n\plus{}1}\equal{}\frac{x_n}{2}$ and $ x_0\equal{}x$. Put $ x\equal{}y\equal{}x_{n\plus{}1}$ in the original equation. We get
$ (f(x_{n\plus{}1}))^2 \geq f(x_n)(f(x_{n\plus{}1}) \plus{} f(x_{n\plus{}1}))$
Taking the limit with $ n \to \infty$ we obtain that $ M^2 \geq 2M^2$ which is a contradiction since $ M > 0$.

So let's assume that $ \lim_{x\to 0^{\plus{}}} f(x) \equal{} \plus{}\infty$. If $ y<x$ then $ f(x\plus{}y)f(x) \plus{} f(x\plus{}y)f(y) > f(2x)f(x) \plus{} f(2x)f(y)$, thus
$ (f(x))^2 > f(2x)f(y)$
taking limit with $ y \to 0^{\plus{}}$ we obtain a contradiction.

TomciO, you solved $ f(x)^2\ge f(x \plus{} y)\left(f(x) \plus{} f(y)\right)$ but the problem asked for $ f(x)^2\ge f(x \plus{} y)\left(f(x) \plus{} y\right)$   :P .

Here is a solution:

Substituting $ g(x) \equal{} \frac {f(x)}{x}$ we get: $ g(x)^2\geq g(x \plus{} y)\left( 1 \plus{} \frac {y}{x}\right) \left( g(x) \plus{} \frac {y}{x} \right) > g(x)g(x \plus{} y)$ so $ g$ is strictly decreasing and $ l \equal{} \lim_{x\to 0}g(x)$ exists.
Putting $ x \equal{} y$ we get $ g(x)^2\geq 2g(2x)\left( g(x) \plus{} 1 \right)$. For $ x\to 0$ we get $ l^2\geq 2l(l \plus{} 1)$, a contradiction because $ l > 0$.

Why would g decreasing imply that the limit exists? 1/x?

In the case limit does not exist, we can follow the earlier suggestion. y > x means g(x) > g(y) so take limit as y tends to +0


limit
algebra
polynomial
calculus
calculus computations

If $f$ is $C^{2}$ on an interval prove that

$\lim_{h\to 0}\frac{f(x+h)-2f(x)+fx-h)}{h^{2}}=f''(x).$

yeah... why didn't i think of l'hopital...

thanks for the help.

Alternatively, since $f\in C^{2},$ we can say:

$f(x+h)=f(x)+f'(x)h+\frac12f''(x)h^{2}+o(h^{2})$

$f(x-h)=f(x)-f'(x)h+\frac12f''(x)h^{2}+o(h^{2})$

Add those together, subtract $2f(x),$ and divide by $h^{2}$ to get

$\frac{f(x+h)-2f(x)+f(x-h)}{h^{2}}=f''(x)+o(1).$

[The next line is of interest to numerical analysts.]

If we assume more smoothness, say $f\in C^{4},$ then we can include the $h^{3}$ and $h^{4}$ terms of the Taylor polynomials, the $h^{3}$ terms cancel, and the estimate becomes

$\frac{f(x+h)-2f(x)+f(x-h)}{h^{2}}=f''(x)+O(h^{2}).$

----

The L'Hôpital computation works, but for robustness, flexibility, and the possible devising of other similar formulas (hence my reference to numerical analysis), I would prefer to work with the Taylor polynomials, exactly as above. As an example: can you find $a,b,$ and $c$ such that

$\frac{af(x)+bf(x+h)+cf(x+2h)}{h}=f'(x)+o(h^{2})\ ?$


let   a1,a2,......,an     n ordered positive integers
prove that there exists i and j where 1<=i<j<=n  such that  ai+aj   divides one of the integers
3a1,3a2,3a3,.....,3an

does someone has the solution please??

There is a missing hypothesis.
The result does not hold for $ n\equal{}3$ and $ a_1 \equal{} 1, a_2 \equal{} 3, a_3 \equal{} 5$.

Pierre.


Suppose $ X$ and $ Y$ are two sets and $ f: X\rightarrow Y$ is a function. For a subset $ A$ of $ X$ ,define $ f(A)$ to be the subset {$ {f(a): a\in A}$} of $ Y$ .For a subset $ B$ of $ Y$ ,define $ f^{ \minus{} 1}(B)$ to be the subset {$ x\in X$:$ f(x)\in B$} of $ X$.Then which of the following statements is true?

(A)$ f^{ \minus{} 1}(f(A)) \equal{} A$ for every $ A\subset X$

(B)$ f^{ \minus{} 1}(f(A)) \equal{} A$ for every $ A\subset X$ if and only if $ f(X) \equal{} Y$

(C)$ f(f^{ \minus{} 1}(B)) \equal{} B$ for every $ B\subset Y$

(D)$ f(f^{ \minus{} 1}(B)) \equal{} B$ for every $ B\subset Y$ if and only if $ f(X) \equal{} Y$

To me it appears that every option is correct.But then I think conditions for inverse are also to be applied .Why are (A),(B),(C) wrong and (D) correct.I am not being able to put the concepts of ONE-ONE and ONTO functions in place

Example for A: Let $ f: \mathbb{R}\to\mathbb{R}$ be given by $ f(x)\equal{}x^2.$ Then $ f([0,1])\equal{}[0,1],$ but $ f^{\minus{}1}(f([0,1])\equal{}f^{\minus{}1}([0,1])\equal{}[\minus{}1,1].$

Although this function was not onto, we couldn't save it by making it onto (part B) - the primary issue is about $ f$ not being one-to-one.

As an example for C, just let $ f$ be a constant function - say, $ f(x)\equal{}0$ for all $ x.$ Then $ f^{\minus{}1}([\minus{}1,1])\equal{}\mathbb{R}$ but $ f(f^{\minus{}1}([\minus{}1,1]))\equal{}f(\mathbb{R})\equal{}\{0\}.$


How many positive integers less than 1000 are multiples of 7 and have a units digit of 8?

So we must have $ x\equiv0\equiv \minus{} 42\pmod7$ and $ x\equiv8\equiv \minus{} 42\pmod{10}$, so
$ x\equiv \minus{} 42\pmod{\gcd(10,7)}\implies x\equiv \minus{} 42\equiv28\pmod{70}$. Thus,
$ 1\le70k \plus{} 28\le999\implies \minus{} 27\le70k\le971,k\in\mathbb Z\ge0\implies0\le k\le13$
$ \implies13 \minus{} 0 \plus{} 1 \equal{} \boxed{14}$.

I'm not sure how to express my solution in mathematical notation, but there are 142 integers > 0 and < 1000 that are divisible by 7: 1000 / 7 = 142 Remainder 6. Every 4 + 10i, i=0,1,2,3,... of those end in digit 8. 142 / 10 = 14 remainder 2. Perhaps that is what the previous reply said, more elegantly. I will try to learn proper notation for future replies.


calculus
algebra
polynomial
geometry solved
geometry

Is there a (finite) formula for sin(x^2) and is it possible to construct the square of a given angle?

[quote="clinkmath"]Is there a (finite) formula for sin(x^2)?[/quote]
Yes, indeed. For example: $\sin(x^{2})=\sqrt{1-cos(x^{2})^{2}}$ :D

[quote="clinkmath"]is it possible to construct the square of a given angle?[/quote]

Squares often have perpendicular sides... :rotfl:

Disregarding the post above me  :roll:

No, there isn't any nice synthetic geometrical method of finding squares of an angle as such, since there is no operation which multiplies angles (this would be absurd, also, due to units).

Of course using calculus we can solve problems involving polynomials of angles although it is messy with Taylor polynomials.

Furthermore, there might be interesting non standard geometries which involve angle multiplication (maybe in a distance formula or something)


So, any of you guys going to state ciphering at the Univ of AL at Tuscaloosa?

its the Alabama statewide ciphering event. Your team had to go to the alabama state math test, and the comprehensive team had to place top 16 in each division. (so if you went, you pretty much qualify for ciphering).   Its this weekend on saturday at tuscaloosa (univ of al).

yay it does not coincide with my piano competition. Thus I may go. I hope the questions are aime level as usual (as in i always hope the questions are hard), but that is a ridiculous hope.

But it does coincide with State science olympiad... oh well I suck at math anyway

Psh, xinke, dont get your hopes up too high.... :wink:

State ciphering never has been difficult in the past, at least since I went to it in 8th grade. The questions are hardly AMC 12a level, nevertheless AIME. its funny, 99% of the people who go to ciphering cant make higher than a 120 on the AMC 12.

Plus, only 4 ppl get to cipher, so taking an entire team is useless. (but there is a team ciphering, so who knows :P )

no, it starts in 9th grade, but our high school wanted to take me b/c i was also taking geometry courses in high school, while in middle school. And, i was on the math team and was pretty good at it at the time  :P  so i went
 :ninja:

so, no YOU cant go in 8th grade, but if youre straight up baller like i am, then you can go  :wink: (jk) :rotfl:


consider the sequence $ S_1 $, $ S_2 $, ...$ S_r $, ... where each $ S_i $ is itself a sequence of the natural numbers 1, 2, 3, 4, ...

let the $ n $th sequence $ S_n $ be denoted by $ u_1 $, $ u_2 $, $ u_3 $, ... and $ S_{n+1} $ = $ u'_1 $, $ u'_2 $, $ u'_3 $, ...

For each $ n $, the terms of the sequence $ S_{n+1} $ are derived from those of the sequence $ S_n $ as follows:

$ u'_i $ = $ u_i $ + 1 if $ n $ is a divisor of $ u_i $
              $ u_i $, if otherwise.

Determine those positive integers $ k $ for which the first $ k-1 $ terms of $ S_k $ are all equal to the number $ k $, but for which the $ k $th term is not.

Source: AMOC extension programme 4 Set 1 problem 6

test the first 10 cases (i know..  but it isn't too hard) and it should be really very apparent.

The proof itself was quite annoying... proving the k-1th term isn't too much effort.. trying to prove the rest.. well...


geometry
3D geometry
analytic geometry
geometry unsolved

$ABCDA_{1}B_{1}C_{1}D_{1}$ is a cube with side length $4a$. Points $E$ and $F$ are taken on $(AA_{1})$ and $(BB_{1})$ such that $AE=B_{1}F=a$. $G$ and $H$ are midpoints of $(A_{1}B_{1})$ and $(C_{1}D_{1})$, respectively.
Find the minimum value of the $CP+PQ$, where $P\in[GH]$ and $Q\in[EF]$.

I really hate when I see a post with you do that and that and rest
is easy but in fact, the rest is what makes the problem hard. This
is such post, but really, the rest of it is as not so important as
the first part. Maybe I had made a mistake in calculation somewhere,
but problem can be solved like this for sure.

Put the origin of Cartesian system in $A$ such that
$A(0,0,0),B(4,0,0),C(4,4,0)...$ (we put $a=1$). Then it's not hard
to se that $P(2,y,4)$ and $Q(x,0,\frac{x+2}2)$ where $x,y\in [0,4]$.
So $CP+PQ=\sqrt{20+(y-4)^{2}}+\sqrt{(x-2)^{2}+(\frac{x+2}2-4)^{2}+y^{2}}$. We want to calculate the minimum of this expression when
$x,y\in [0,4]$. The minimum of $(x-2)^{2}+(\frac{x+2}2-4)^{2}$ can
easily be calculated, and if I am not wrong it is $\frac65$ and it
is obtained for $x=\frac{16}5\in [0,4]$ (you see that only this part
depends on $x$, so we can calculate its minimum and use it in
further solving). Now we should calculate minimum of $\sqrt{(y-4)^{2}+20}+\sqrt{y^{2}+\frac{6}5}$. If you put $M(4,\sqrt{20})$ and
$N(0,\sqrt{\frac{6}{5}})$, then $\sqrt{(y-4)^{2}+20}+\sqrt{y^{2}+\frac{6}5}$ is $YM+YN$ where $Y(y,0)$. This is well known problem,
and solution is (again if I am not wrong, i really calculated all
this in maximum speed) is $y=\frac{4\sqrt{6}}{10+4\sqrt{5}}$. This
is all for $a=1$, but only multiply everything whit $a$, and that's
it.

Well, I like coordinate geometry, and this is the first time I use it on 3D that's why I was so exciting, and didn't look for other "easier" solutions :).
I usually use it in 2D like in http://www.mathlinks.ro/Forum/viewtopic.php?t=136680v
Bye

[quote=freemind]I didn't even think about solving this with coordinates! I guess you made some mistaken calculations... The answer is $\sqrt{\frac{276}{5}}a$. I took it "easier". I expressed $CP+PQ$ by the right triangles $CHP$ and $QGP$. Setting $HP=x$, we get $CP+PQ=\sqrt{CH^{2}+x^{2}}+\sqrt{QG^{2}+(4a-x)^{2}}$ and we use Minkovski. So, to attain the minimum, $QG$ must be firstly minimal, that is $Q$ is the projection of $G$ on $EF$.
Then we apply Minkovski to "get rid" of $x$. So minimum is $\sqrt{(CH+QG)^{2}+(4a)^{2}}$. Well, I don't make all the calculations. It is easy to get $CH$ from right triangle $CHP$, and $QG$ as being altitude in $EGF$.[/quote]

I think apply Minkowski is wrong because your equation never attain $\sqrt{\frac{276}{5}}a$. I think to solve your equation need derivative equalize to zero and $x$ will be nearly $2.38$ and answer is nearly $7.47938$