function
induction
algebra
polynomial
functional equation
complex numbers
Functional Equations

Find all functions $f: \mathbb{N}\to \mathbb{N}$ such that for all $n\in \mathbb{N}$: \[f(f(f(n)))+f(f(n))+f(n)=3n.\]

Answer : $f(n) = n.$

Now, to the proof (by induction) :
At first, taking $n = 1$ in the given relation yields that $f(f(f(1))) = f(f(1)) = f(1) = 1.$
Then, let us assume that, $\forall i \leq n, \ \ f(i) = i.$ Thus $f$ is injective, so that $f(m) > n$ for $m > n.$
So we get that $f(f(f(n+1))) + f(f(n+1)) + f(n+1) \geq 3(n+1).$
But we know that $f(f(f(n+1))) + f(f(n+1)) + f(n+1) = 3(n+1),$ which yields $f(f(f(n+1))) = f(f(n+1)) = f(n+1) = n+1.$
And we're done.

EDIT : I just see that it's also [url=http://www.problem-solving.be/pen/viewtopic.php?t=407]here[/url], and we have the same solution..  :oops:

Just to give another (and typical with those problems involving the iterates of a function - but (imho) less elegant) solution :
for some integer $ x,$ we consider $ a_{n}(x) = f \circ f \ldots \circ f (x).$ (n times)

Then we have : $ a_{n+3}+a_{n+2}+a_{n+1}= 3a_{n},$ whose characteristic polynomial is $ r^{3}+r^{2}+r-3 = (r-1)(r+1+i\sqrt{2})(r+1-i\sqrt{2}),$ hence
$ a_{n}(x) = F(x)+(-1)^{n}(G(x)(1+i\sqrt{2})^{n}+H(x)(1-i\sqrt{2})^{n})$ where $ F(x), G(x)$ and $ H(x)$ are real constants.

Now, letting successively $ n = 0, 1, 2,$ we get that :
$ f(x) = a_{1}(x) = a_{0}(x) = x,$ which is indeed a solution of our functional equation, we're done.

[quote="mathmanman"]Just to give another (and typical with those problems involving the iterates of a function - but (imho) less elegant) solution :
for some integer $ x,$ we consider $ a_{n}(x) \equal{} f \circ f \ldots \circ f (x).$ (n times)

Then we have : $ a_{n \plus{} 3} \plus{} a_{n \plus{} 2} \plus{} a_{n \plus{} 1} \equal{} 3a_{n},$ whose characteristic polynomial is $ r^{3} \plus{} r^{2} \plus{} r \minus{} 3 \equal{} (r \minus{} 1)(r \plus{} 1 \plus{} i\sqrt {2})(r \plus{} 1 \minus{} i\sqrt {2}),$ hence
$ a_{n}(x) \equal{} F(x) \plus{} ( \minus{} 1)^{n}(G(x)(1 \plus{} i\sqrt {2})^{n} \plus{} H(x)(1 \minus{} i\sqrt {2})^{n})$ where $ F(x), G(x)$ and $ H(x)$ are real constants.

Now, letting successively $ n \equal{} 0, 1, 2,$ we get that :
$ f(x) \equal{} a_{1}(x) \equal{} a_{0}(x) \equal{} x,$ which is indeed a solution of our functional equation, we're done.[/quote]

I understand, that $ a_0 (x) \equal{} x,$ but what is the value of $ a_1 (x)$ and $ a_2 (x)$??? We need them to define $ F(x), G(x)$ and $ H(x)$.
Maybe you compare the imaginary part with $ 0$. But then we have:
$ a_0(x) \equal{} F(x) \plus{} G(x) \plus{} H(x) \equal{} x$
$ a_1(x) \equal{} F(x) \minus{} G(x) \minus{} H(x) \plus{} i\sqrt {2}(H(x) \minus{} G(x)) \equal{} f(x)$
From second we get $ G(x) \equal{} H(x)$. Now:
$ a_2(x) \equal{} F(x) \plus{} G(x)(1 \plus{} i\sqrt {2} \minus{} 2 \plus{} 1 \minus{} i\sqrt {2} \minus{} 2) \equal{} f(f(x))$ or
$ a_2(x) \equal{} F(x) \minus{} 2G(x) \equal{} f(f(x))$
We have:
$ F(x) \plus{} 2G(x) \equal{} x$
$ F(x) \minus{} 2G(x) \equal{} f(x)$
$ F(x) \minus{} 2G(x) \equal{} f(f(x))$
What now?

And could we use this method for other equations, for example, for $ f(f(x)) \equal{} x$. In this case we have:
$ r^2 \minus{} 1 \equal{} (r \minus{} 1)(r \plus{} 1)$ and
$ a_{n}(x) \equal{} F(x) \plus{} ( \minus{} 1)^n G(x)$
$ a_0 (x) \equal{} F(x) \plus{} G(x) \equal{} x$
$ a_1 (x) \equal{} F(x) \minus{} G(x) \equal{} f(x)$ and what we should do next? :)

And the last, is $ x$ should be integer (natural) or it may be real?

[hide="Another idea"]If $ f(m)\equal{}f(n),m,n\in\mathbb{N}$, then $ f(f(f(m)))\plus{}f(f(m))\plus{}f(m)\equal{}f(f(f(n)))\plus{}f(f(n))\plus{}f(n)$, so $ m\equal{}n$, i.e., $ f$ is injective. Let $ a$ be the smallest natural number for which $ f(a)\ne a$. Since $ f$ is injective, $ f(a)>a$. Also, $ f(f(a))>a$ and $ f(f(f(a)))>a$, so $ f(f(f(a)))\plus{}f(f(a))\plus{}f(a)>3a$, a contradiction. Therefore $ f(x)\equal{}x,\forall x\in\mathbb{N}$.[/hide]

$f(f(f(1))) \ge 1 ,f(f(1)) \ge 1,f(1) \ge 1$ and $f(f(f(1)))+f(f(1))+f(1)=3 \implies f(1)=1$

and we can easily see that $f$ is one-to-one. now we use induction.

statement:  $\forall i<n, f(i)=i$

the induction step for $n=2$ is trivial so assume it's true for $n-1 \ge 2$ then we shall prove it for $n$, because $f$ in injective then

$f(f(f(n))) \ge n ,f(f(n)) \ge n,f(n) \ge n$ which implies $f(n)=n$ so $\forall n \in N:f(n)=n$

[b]Claim:[/b] $f$ is the identity.
[i]Proof.[/i] We prove by induction. Then the base case follows from $n=1.$ So if $f(k)=k ~\forall n\geq k,$ then $f(n+1)\geq n+1, f(f(n+1))\geq n+1$ and $f(f(f(n+1)))\geq n+1.$ We finish the induction considering $n=n+1$ in the original FE.


I think i have the answer but not sure...could anybody lend a hand?

If \[\frac{4x-1}{x^{2}-5x-14}=\frac{a}{x+2}+\frac{b}{x-7}\] ,

then find \[\frac{ab}{a+b}\].

like i said..i got an answer but im sure of it :maybe:

Just post your answer and your steps and we'll see if you did it right or wrong.

[quote="DaReaper"]
If
\[\frac{4x-1}{x^{2}-5x-14}=\frac{a}{x+2}+\frac{b}{x-7}\]
,

then find
\[\frac{ab}{a+b}\]
.
[/quote]

[hide]
\[\frac{4x-1}{x^{2}-5x-14}=\frac{a(x-7)}{x+2}+\frac{b(x+2)}{x-7}\]
\[\frac{4x-1}{x^{2}-5x-14}=\frac{a(x-7)+b(x+2)}{x^{2}-5x-14}\]

$4x-1=ax-7a+bx+2b$
then i am guessing you would solve for a and b but do, please post your solution because it is really hard to write all of this[/hide]

that answer is what i got..but i wasnt sure if it was right..

and i didnt previuusly post the solution and answer because i want to see if there are different and easier ways of doing the problem.....and the test was multiple choice.. :P


find the smallest number which is divided by 2, 3, 4, 5, 6 gives the remainders respectively 1, 2, 3, 4, 5.

Here's a hint: Use modular arithmetic

Here's the answer: [hide] 59 [/hide]

But could you please show me how you get it

[hide]If x is the number you are trying to find, then x+1 is divisible by 2, 3, 4, 5, and 6, or the LCM of 2, 3, 4, 5, and 6, or 60. Then x is 59.[/hide]

wow his solution is about 300000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000 times better than mine.  But w/e.

[hide]I just said that it had to be congruent to 1 mod 2, 2 mod 3, 3 mod 4, 4 mod 5, 5 mod 6.

Now, it's congruent to 5 mod 6 iff it's congruent to 1 mod 2 and 2 mod 3.
It's congruent to 4 mod 5 and 1 mod 2 iff its congruent to 9 mod 10--i.e. it ends in nine.

Now, the smallest number congruent to 0 mod 4, mod 6, and mod 10 is 60.

60 - 1 = 59.  Don't laugh.  It worked.   :D [/hide]

[quote="seamusoboyle"]Is this not just the chinese reminder theorem?[/quote]

Pretty much.


It comes to my attention that Americans have mispronounced "z" as "zee" instead of the proper "zed."  As Americans do NOT speak American, but English from England, America should follow England's rules for [b][i]their[/i][/b] language.

I do not say this because I am English (I'm not), but because it is a common misconception and all children should be retaught the alphabet.  Good day.

i really don't think it matters-if you have problems with the english language as spoken in the us, i don't think that should be your biggest issue.

I don't really care, I'm just asking what others think.  Is that a crime?

Goodness me, this isn't a very biased poll at all :rotfl:

How is [b]American [/b]English "their" [England] language?  :huh:

Anyway, I think zed sounds cooler, but for all practical purposes, i think zee suits the purpose better. I mean, why would "z" have a d in it? (I know, why would w have a "double" in it, etc., but still)

If we said zed it would ruin the song! Me and zed would not rhyme and it would come out weird.
*sings*ABDCEFGHIJKLMNOPQRSTUVWXY and Zed? Now I know my ABC's next time won't you sing with me?*stops singing*  Lol it would sound kind of weird think of the little children!

Americans do, in fact, speak "American English", another dialect of English.  Hence we pronouce many things differently.  Just as any language has several dialects...

Come on, Ignite168, that analogy is totally irrelevant. No, British English isn't "wrong", and American English isn't "wrong" either. They're just different dialects of the same language.

[quote="Ignite168"][quote="sonny"]It's funny how we think that they (ppl from England) are wrong when they came up with all of it way before we did.[/quote]

The first people thought $ \pi$ was 3. Since they were the first people to come up with it, they are obviously right.[/quote]

If you choose to believe what I wrote and apply it to mathematics sure, keep telling your self that $ \pi$=3.

[quote="Ignite168"][quote="sonny"]It's funny how we think that they (ppl from England) are wrong when they came up with all of it way before we did.[/quote]

The first people thought $ \pi$ was 3. Since they were the first people to come up with it, they are obviously right.[/quote]

Actually, by the time $ \pi$ was generally accepted as the symbol for $ \frac{c}{d}$, it had been calculated to around 100 digits.

[quote="Lazarus"]...by the time $ \pi$ was generally accepted as the symbol for $ \frac{c}{d}$...[/quote]

What? That joke didn't work, at all.
[color=red]
[size=200]The message contains only quote and code tags. Please add some text of your own.[/size][/color]

Does this all really matter? It's not like British ppl will declare war on America for pronouncing the letter "z" wrong or vice versa. Just let everyone pronounce it as they want to...

I think it's fine either way, and that's because British, irish, and Canadian people all understand "Zee", and Americans understand "Zed".

Then there are other options

Zeta(Greek, Spanish)
Zeh(Russian)

apparently, its just because the letter z is "special"

[quote="sapphyre571"]if z is "zed," then why aren't "b" and "v" "bed" and "ved?"[/quote]

LOL... and if it's because "zed" comes from "zeta", "bed" should also come from "beta"...

If the plural of that is those, what's the plural of hat?

the only letter I don't really understand is h

[quote="Ignite168"][quote="SoxFan34"]and Americans understand "Zed".[/quote]

NO. I HAD NEVER HEARD OF ZED AND WILL SLAP ANYONE THAT USES IT[/quote]

ZED

ay, bee, see, dee, eee, ef, jee, aich, eye, jay, kay, el, em, en, oh, pee, cue, ar, es, tee, you, vee, double you, ex, why, ZED

zed zed zed zed zed

Come slap me, I'm not too far away, I live in Massachusetts.

[b]Related[/b]

Who puts bars through the middle of their z's and/or 7's?

[quote="Ignite168"][quote="SoxFan34"]and Americans understand "Zed".[/quote]

NO. I HAD NEVER HEARD OF ZED AND WILL SLAP ANYONE THAT USES IT[/quote]

INTERNATIONAL CAPS LOCK DAY IS OCTOBER 22. Zed zed zed zed zed.

@The Zuton Force: I used to put bars through my 7's for Mathcounts, simply because I write fast and I don't want my 7's and 1's to look alike.

sometimes I have to cross my ZEE's to make sure they don't look like 2's when I do messy algebra.


I am REALLY disappointed that my state does not use CD for Nats. i would've made it. instead, i got 8th written and my mathcounts "season" is over. and yes, i am sorta "venting" :mad:

what did you get in cd?

ours is unofficial

dan got 3 trophies

jorian

1st in IN.  I was paired with the number 1 individual too.

No, ours doesn't count. Not that it would matter for me... :(  I guess I'm done with MATHCOUNTS forever now, never even going to nationals... :(

well my state does count the CD, but thats bad news for me cuz i suck at CD

Indianamath, first off, the poll function is sort of useless because of disproportional # of people from different states.

And our countdown couldn't POSSIBLY be official. It has a freaky setup. If cd was official, it would HAVE to be ladder: 8th v. 7th, 7th v. 6th, etc. Much harder.

Here was ours:

Top 24
Split into four groups of six: first three to answer 1 question advance from each group
Of those remaining 12, split into two groups of six. First three to answer 1 question advance.
Of those remaining six: first four to answer one question advance.
Of those remaining 4: head to head

Yeah that is sort of freaky. Our top four would be me, you, and those two other random kids that didn't even make top 15 but made top 4 countdown. Yeah =/.

lol i feel sorry for joseph. he was stuck with you and me. after we went 1 and 2 on the first round, he probably got nervous and guessed on the third one.

I think i could get to at least fifth. me vs. lyndon if it's a ladder setup

KS is official, and I was so glad last year because of that (went from 7th to 4th).

[quote="Ubemaya"]Dude, our countdown setup is MUCH more freaky.

20 people, 10 doing one ladder and 10 doing another, with each group of 10 doing the countdown 700 miles away from each other.

I suck at countdown.[/quote]

how does the winner get worked out?

is it possible for someone in south cal even to go to nats?

when has it last happened, if it has?

-jorian

Well first, (I forgot to say this), CountDown is unofficial in California, but from my post this should be obvious.

There is a winner in both No. CA and So. CA, I guess. Actually, I don't even know the format for So. CA, I'm just guessing it's t he same as No. CA.

The two competitions are held on the same day, and I think they communicate by e-mail to figure out the top 4 between the two places. Usually at least one person from Southern California makes it, but this year I'm pretty sure that 3-4 people will make it from So. CA, since I already know 3 random people's scores and they are better than the top 3 scores from No. CA.

ok, how about in your posts everyone state your state (lol) and say whether or not cd counts. then someone makes a big list, so that if there are more people from one state, the pole won't get messed up.

jorian

[quote="Fanatic"]
Yeah that is sort of freaky. Our top four would be me, you, and those two other random kids that didn't even make top 15 but made top 4 countdown. Yeah =/.[/quote]

I was one of the random kids.

[quote="jhredsox"]so top 3 in that format automatically make it?

-jorian[/quote]
Yes.
Last year in my state it didn't change the nationals team, the only things were that 10th became 8th and 4th became 3rd.

ok, that kind of shows that Florida is official.

-Jorian

[quote="jhredsox"]list will be edited

COUNTDOWN

Official:
Florida
Kansas
Michigan
New Jersey
Washington
Wisconsin

Unofficial:
California
Connecticut
Kentucky
Illinois
Indiana
Massachusetts
New York
South Carolina[/quote]
States to add:
Official:
Colorado
Deleware
Texas
Utah

Unofficial:
Georgia
Louisiana
Virginia

I think Ohio is unofficial... wait for conformation from someone else before adding it though....

it wouldnt matter to me if countdown mattered in my state or not even though it does not.

[quote="jhredsox"][quote="kstan013"][quote="footballrocks41237"]In Wisconsin, CD counts.  I wish it didn't.  I would have taken 3rd then...[/quote]HA. In 6th grade I moved from 5th to 3rd, 7th grade I moved from 4th to 1st, and this year I stayed in the same spot. But overall in CD, I'm *thinks back* 11-1. only loss was to iris xu in 6th grade.[/quote]

i'm 4-0.

what state are you in cognos?

jorian
______________________________________________________
list will be edited

COUNTDOWN

Official:
Colorado
Deleware
Florida
Kansas
Michigan
New Jersey
Texas
Utah
Washington
Wisconsin

Unofficial:
California
Connecticut
Georgia
Kentucky
Louisiana
Illinois
Indiana
Massachusetts
New York
South Carolina
Virginia[/quote]
Eh, add Alabama to the official list.

Add Arizona to the official list.

My 6th grade year I moved from 7th overall in written to #2 to go to Nats.  Last year I was in 1st for written and held.  Don't know about this year though....

yeah. NY doesn't count(I think). My total wins to total losses is I think 30-11(including practice) but, in chapter or state(haven't made nats yet)2-1. :rotfl:

Official:
Alabama
Arizona
Colorado
Deleware
Florida
Kansas
Michigan
New Jersey
Texas
Utah
Washington
Wisconsin
Oklahoma

Unofficial:
California
Connecticut
Georgia
Kentucky
Louisiana
Illinois
Indiana
Massachusetts
New York
South Carolina
Virginia
OHIO(added)

[quote="mathmatatics"]I'm glad CD doesn't count. I always get crazy during countdown and do stupid things, just because it's the real thing. I do much better on sprint and target since not everyone is watching you[/quote]

Wait, you're in Ohio? Ohio $\neq$ official.


Given that $ x\plus{}y \equal{} 10$ and $ 2x\plus{}y \equal{} 13$, evaluate $ x^2\minus{}y^2$.

Subtract the equations to find $ x \equal{} 3$. Then $ y \equal{} 7$. So $ x^2 \minus{} y^2 \equal{} (x \plus{} y)(x \minus{} y) \equal{} 10\cdot 4 \equal{} 40$.

EDIT: Fail; it's $ 10 \cdot (\minus{}4) \equal{} \minus{}40$.

actually, it is -40, simply 9-40=-40  :wink:

TY on finding my fail error.

I think MCmaniac is being sarcastic about a mistake he made?

EDIT: See my edit on my first post.


The Barringer Meteorite Crater in Winslow, AZ, is a hemisphere of diameter 1.6 km. Pat walks once around the crater on the rim. To the nearest kilometer, how many kilometers does he walk?

Pat walks around the circumference of the hemisphere which has length $ 1.6\pi \approx \boxed{5}$


if x is a positive integer prove that $\frac{81^{x+1}-1}{16^{x+1}}$ is never a integer

[quote="Singular"]Let y = 4x+4.  $2^{y-1} > y$, so we cant have $3^y = 1 \mod{2^y}$.[/quote]

$\phi (2^y)>y$ but $\phi (2^y)$ isnt the smallest for $3^{2^{y-1}}\equiv 1 \mod 2^y$ . Maybe there is another $m<2^{y-1}$ such that $3^m\equiv 1 \mod 2^y$ ?

[hide]

suppose it is an integer then

If $x+1=2k$ for some integer $k$ , then if $16^{2k}|81^{2k}-1 =16^{2k}|(81^k+1)(81^k-1)\implies 16^{2k}|81^k-1$ .

We can keep factorise it until some odd $k$ . So we only need to consider when $x+1$ is odd .

so $16^{x+1}|81^{x+1}-1=80(81^x+81^{x-1}+...+1)\implies 16|(81^x+81^{x-1}+..+1)$

But $81^x+81^{x-1}+..+1\equiv x+1\mod 16$ .

Since $x+1$ is odd , it cannot be divided by 16 hence it cannot be an integer .
[/hide]

Here's what I would say...

$81^{x+1}-1=(9^{x+1}-1)(9^{x+1}+1)$  Those two differ by $2$.  Thus one of them is $2\mod{4}$ and the other $0\mod{4}$.  So the number which is $0\mod{4}$ must be divisible by $\frac{16^{x+1}}{2}$.  However, the number is smaller than $\frac{16^{x+1}}{2}$, and we then have a contradiction.

[quote="Fibfreak"][quote="K81o7"]...So the number which is $0\mod{4}$ must be divisible by $\frac{16^{x+1}}{2}$...[/quote]
i am sorry for to be so stupid, but why is this so?[/quote]

If a number is $2\mod{4}$, then the largest multiple of $2$ it is divisible by is $2$.  If the two numbers, together are divisible by $16^{x+1}$, then the other must be divisible by $\frac{16^{x+1}}{2}$.

[quote] Those two differ by 2.[/quote]
Or more simply,  difference between two powers of two can not be 2 (eg after 1,2 (4,8... it is greater than 3) so each can not be a power of 2.
(every factor of 16^n  has to  be a power of two)


find product of the n nth roots of 1 in terms of n.[/u][/i]

[hide]
$x^{n}= 1 x^{n}-1 = 0$
If n is even then the product is -1, if n is odd then the product is 1, so the answer is $(-1)^{n}$[/hide]

[quote="Ubemaya"][hide]
$x^{n}= 1 x^{n}-1 = 0$
If n is even then the product is 1, if n is odd then the product is -1.[/hide][/quote]

[hide]You might want to check your work again.  Try it if $n=2$.  Also, make sure express it in terms of $n$.[/hide]

[hide] I think my solution is kind of like what Ubemaya was trying to do. The n'th roots of 1 are all, by definition, roots of the equation $x^{n}= 1$, or the polynomial $x^{n}-1$. By vieta's sums, $-1$ is the product of all the solutions, multiplied by $(-1)^{n+1}$. Thus, the product of the solutions is $(-1)^{n+1}$.[/hide]


Prove that the total number of inversions of all even permutations equals the total number of inversions of all odd permutations:
${ \sum_{ \substack {\sigma \in A_n}}m(\sigma) =
\sum_{ \substack {\sigma \in S_n - A_n }} m(\sigma)}=\frac{n!}{4} {{n}\choose{2}}$

A simpler problem:
How many even permutations have the inversion (1,2)?? The answer must be half of them. How do we prove that?
And then prove the same thing for any other inversion.


geometry
geometric transformation
parameterization
calculus
LaTeX
complex analysis
complex analysis unsolved

I have to ...solve these exercises in about 6 hours and haven't got a clue where to start. If anyone could help me with any of them i`ll be grateful.
I`ve attached the image so i wouldn`t have to rewrite the exercises. Below there are the translations for them.
[img]http://www.elves.ro/etc/math.jpg[/img]

(my math english isnt that good so please excuse any errors)
P2.
a) show that f can be expanded in series of powers in the disc |z|<1
b) show that g: C -> C bla bla ...  allows a series expansion in the vecinity of the origin

P3: solve : lim ...

P4: Prove the next inequation

P4: Using limited expansins, solve: ...
Discussion after the parameter p>0

Please next time try to follow the guidelines below

- post each problem in a separate topic
- for easier problems (just like the ones above) use the [url=http://www.mathlinks.ro/Forum/index.php?f=296]Calculus Computations & Tutorial Forum[/url].
- try to learn a bit of $\LaTeX$ to write up the problems nicely (read in my signature about it)
- for Romanian wording there is a [url=http://www.mathlinks.ro/Forum/index.php?f=263]Romanian Forum[/url] to discuss issues in Romanian.

Until then the topic is closed.


[hide="Proof"]Denote the first polynomial by $P(x)$ and the second one by $Q(x).$ It's obvious that $P(x)=Q(x^{1111}).$ The roots of $Q(x)$ are $x_k=\cos\frac{2k\pi}{10}+i\sin\frac{2k\pi}{10}, k=1,2,\dots,9.$ Since \begin{eqnarray*}x_k^{1111}=\cos\frac{2222k\pi}{10}+i\sin\frac{2222k\pi}{10}=\cos(222k\pi+\frac{2k\pi}{10})+i\sin(222k\pi+\frac{2k\pi}{10})=x_k\end{eqnarray*}, it follows that $P(x^k)=Q(x_k^{1111})=Q(x_k)=0$, i.e. all the roots of $Q(x)$ are also the roots of $P(x).$ Therefore, $P(x)$ is divisible by $Q(x).$[/hide]


Does anyone know how to prove the fact that C^2=A^2+B^2 for hyperbolas, where A is half the distance of the transverse axis, B is half the distance of the conjugate axis, and C is the distance from the focus to the center.

Thanks

this might be too hard for this section of the forum, if a admin sees this could he or she please move it to intermediate or higher? my bad.


At what times are the officer hours and the class for the Olympiad Independent Study held?

I don't believe there's a "class". Office hours vary last time I checked.

I'm not entirely sure of how things will work over the summer since the IS is going to be taught in half the time as usual but twice the pace.  I assume they would double their amount of office hours each week.  The office hours generally vary from week to week but the instructors usually try to schedule them in favor of the students' personal schedule as much as they can.

Actually, we decided against folding it into a 7-week class since it would be just too much work.  So, it will be run just like the last one as far as office hours per week and such.

Hmm...is the next available session in the fall?

Yes.  We'll start the fall session in late August or early Sept.

Not yet.  That's something we'll build in the future.

[quote="towersfreak2006"]Do you have an estimate of when it will be ready? (Olympiad B, that is)[/quote]

Because of the books, it's not likely to be a priority anytime soon.  Most students who would take such a course would study at MOP, cutting down on the amount of demand we could possible have.  We might even develop a distinct program that would suit those needs.

We also feel that we can serve more students by building better materials that lead students to such a level.  Most students at that level can study on their own to a significant degree.  The process of getting to that level is more important to students who are not interested in mathematical academia and that includes the majority of AoPSers.  Overall, we try to provide opportunities that suit a broad spectrum of needs/desires.  Hopefully we offer a little or a lot for everyone.

In short, there are an abundance of reasons why we might not offer such a course for quite a while.  However, if it ever seemed like there was significant demand, we would reconsider.

And who knows where the forces of the online world will lead us.  We find ourselves changing our own minds and walking down new paths every few months.  And plans we mention about anything are not set in stone.  We simply look for ways to promote mathematical problem solving education to students who want/need it most.


group theory
abstract algebra
topology
geometry
geometric transformation
invariant
LaTeX

Does there exist an inverse system of finite groups with direct limit $\mathbb{Z}$? Does $\mathbb{Z}$ occur as a Galois group?

I think no: If $Gal(L/K) = \mathbb{Z}$, for every $n$ there is a unique extension $K_{n}/K$ of degree $n$ because every proper subgroup of $\mathbb{Z}$ has finite index and is therefore open (open subgroups <--> finite extensions), hence closed. Because $\cap 2^{n}\mathbb{Z}= 0$, $\cup K_{2^{n}}= K$ ($0$ is a closed subgroup because the Galois group is always Hausdorff), but $\cup K_{2^{n}}$ ha no extensions of odd degree, whereas $K$ has.

Is this correct?

A profinite group can be organized as a compact Hausdorff topological group. However, there are no countably infinite compact Hausdorff groups:

Assume $G=\{g_{n}\ |\ n\in\mathbb N\}$ is such a group. Consider the Haar measure $\mu$ of total mass $1$ on $G$. Because $\mu$ is translation-invariant, all singletons $\{g_{i}\}$ have the same measure. This measure cannot be $0$ because then $1=\mu(G)=\sum_{i}\mu(\{g_{i}\})=0$, and it cannot be positive, because $1=\mu(G)=\sum_{i}\mu(\{g_{i}\})=\infty$. We thus have a contradiction.

Very nice :)

grobber, is my proof also correct?

The grobber's proof is very interesting, but i think that the easiest one is this: [/Z] is countable and infinite, but every profinite group is finite or uncountable. So [/Z] can't be a profinite group!
[/i][/u]

You should at least prove your claim, otherwise it's the same as grobber posted.

Suppose that a profinite group G (so a compact Hausdorff and totally sconnected space) admits a countable family of closed finite subgroups {C_n|n natural} with empty interior such that G is the union of C_i. It follows that D_i = G\C_i is dense in G. It is known that a compact Hausdorff topological space admit a base of clopen sets for its topology. So choose a clopen U_0 of G and a clopen U_1 inside U_0\capD_0. We can so conscruct a   chai of clopen U_n \subseteq U_(n+1). and every U_i is not empty and closed. So, from the compactness of G, is not empty the intersection of the U_i's but it is contained in the intersection of the D_I=G\C_i's which intersection is empty. So G can not be written as union of a countable many closed subsets with empty interior, so G can't be countable (if G is not finite).

I can't use LateX in this forum,  so you could find this proof a bit difficult to read.

And sorry for my english :lol:

@Berri: interesting, you are stating the equivalent of Baire's lemma, i can notice that if the inverse limit is done over a sequence of finite groups ($ G \equal{} lim_{\leftarrow}G_n$), then the profinite group is also metrizable with a metric that induces the profinite topology (just say that the distance between $ x$ and $ y$ is $ \frac{1}{2^n}$, where $ n$ is the smallest index such that $ x$ and $ y$ have a different image in the $ G_n$). But in general i don't know if a profinite group is metrizable (and i'd rather expect counter-examples that are not).

@berti: i think that your proof would be correct if for each profinite group you could find a field that has that group as Galois group, and i don't know if this is true, and if this is a problem that has been studied. Also, a subextension exists only for each [i]open[/i] subgroup, and in our ipothetical profinite topology for $ \mathbb{Z}$ nobody requested finite-index subgroups to be open (and generally in profinite groups subgroups with finite index need not to be open). BTW If each finite-index subgroup is supposed to be open, you don't need to consider fields, because you can just consider the order of the group and apply your argumentation to the quotients with order $ 2^n$ and get an absurd because $ 2$ would be the only prime to appear as divisor of the order (the order of a profinite group is a supernatural number, where a prime divisor can appear at the $ \infty$th power).


trigonometry
limit
calculus
derivative
function
integration
calculus computations

Evaluate

$\lim_{n \to \infty} 4^n\left(1-\cos{\frac{\theta}{2^n}}\right)$.

[quote="t0rajir0u"]Let $u = 2^{-n}$.  We desire

$\lim_{u \to 0^{+} } \frac{1 - \cos u \theta}{u^2} = \frac{u \sin u \theta}{2u} = \boxed{0}$[/quote]

You took the derivative with respect to the wrong variable in the numerator. The derivative of your numerator should be $\theta \sin u \theta$, which will change your answer.

This is an instance, however, where using a Taylor series is probably easier than applying L'Hopital's rule.

[quote="gauss202"][quote="t0rajir0u"]Let $u = 2^{-n}$.  We desire

$\lim_{u \to 0^{+} } \frac{1 - \cos u \theta}{u^2} = \frac{u \sin u \theta}{2u} = \boxed{0}$[/quote]

You took the derivative with respect to the wrong variable in the numerator. The derivative of your numerator should be $\theta \sin u \theta$, which will change your answer.[/quote]

I thought so.  I was wondering why the answer wasn't a function of $\theta$  :blush:

$\lim_{u \to 0^{+}} \frac{\theta \sin u \theta}{2u} = \boxed{ \frac{\theta^2}{2} }$

Edit:  And, of course, this suggests an even better way to do it.

$f(\theta) = \lim_{n \to \infty} 4^n \left( 1 - \cos \frac{\theta}{2^n} \right)$
$f'(\theta) = \lim_{n \to \infty} 2^n \sin \frac{\theta}{2^n} = \theta$
$f(\theta) = \int \theta \, d \theta = \frac{\theta^2}{2} + C$

And $f(0) = 0$.  :)

[quote="t0rajir0u"]
And, of course, this suggests an even better way to do it.

$f(\theta) = \lim_{n \to \infty} 4^n \left( 1 - \cos \frac{\theta}{2^n} \right)$
$f'(\theta) = \lim_{n \to \infty} 2^n \sin \frac{\theta}{2^n} = \theta$
$f(\theta) = \int \theta \, d \theta = \frac{\theta^2}{2} + C$

And $f(0) = 0$.  :)[/quote]

This touches on an interesting question though, is it always true that $\frac{d}{dx} \big(\lim_{n \to \infty} f_n(x)\big) = \lim_{n \to \infty} f'_n(x)$?

And if not, under what conditions is it true?

[quote="gauss202"]
This touches on an interesting question though, is it always true that $\frac{d}{dx} \big(\lim_{n \to \infty} f_n(x)\big) = \lim_{n \to \infty} f'_n(x)$?

And if not, under what conditions is it safe to do so?[/quote]

It's a strange question since we are effectively taking two limits at once.

$\frac{d}{dx} \left( \lim_{n \to \infty} f_n (x) \right) = \lim_{dx \to 0} \left( \lim_{n \to \infty} \frac{ f_n (x + dx) - f_n (x) }{dx} \right)$

Actually, it's probably just easier to do it by Taylor series than to prove that this limit exists.   :|

It does seem like a trivial question. But something very interesting happens! :)

Let $s_{m,n} = \frac{m}{m + n}$

$\lim_{m \to \infty} s_{m,n} = 1$, so $\lim_{n \to \infty} \lim_{m \to \infty} s_{m,n} = 1$

But $\lim_{n \to \infty} s_{m,n} = 0$, so $\lim_{m \to \infty} \lim_{n \to \infty} s_{m,n} = 0$

Therefore $\lim_{n \to \infty} \lim_{m \to \infty} s_{m,n} \neq \lim_{m \to \infty} \lim_{n \to \infty} s_{m,n}$

-----------------------------------------------------------
Another example:
Consider the family of functions $f_n(x) = \frac{\sin nx}{\sqrt n}$

$\lim_{n \to \infty} f_n(x) = 0$ for all x.
So $\frac{d}{dx} \lim_{n \to \infty} f_n(x) = 0$ for all x.

But $f'_n(x) = \sqrt{n} \cos nx$ and $\lim_{n \to \infty} f'_n(0) \neq 0$ for all x!

This is something you will not study until you take Analysis. But the point is to be very careful when you switch limits. They don't always commute!

[quote="Xevarion"][quote]  under what conditions is it safe to do so? [/quote]

Uniform convergence?  :D[/quote]

Yes! :)

I'm still just a student of analysis, and this isn't a classroom either, but I may as well give more than a two word answer for that question.

Basically, the idea is that if you have two variables and you want to take a limit in each, you have to be sure that the convergence of one doesn't depend on the other. That is, you don't want the expression to converge faster or slower in one variable depending on the other variable.

Consider the functions $f_n = x^n$ with $x \in [0, 1]$. Obviously this sequence of functions converges, but it doesn't converge to a continuous function, because the larger is $x$, the slower the sequence $f_n$ converges. Of course, in this case we aren't taking two limits, but it's the same problem, in this case messing up the continuity.

If you have a sequence like gauss202's, $\frac{m}{m+n}$, you can think of the definition of the limit as that you want to be able to draw a huge square, and outside that square (that is, when at least one of $m$ and $n$ are large), all the terms should be close together. But  in this example, that's obviously not true -- for large $n$ and small $m$, you get small, whereas for small $n$ and large $m$, you go to 1. So there is no well-defined limit as $m, n \to \infty$.

There's my attempt at some intuitive explanation of uniform convergence... hope it helps.


Let $a,b,c$ be positive numbers such that $ab+bc+cd+da=1$. Show that
\[ \frac{a^5}{b^3}+\frac{b^5}{c^3}+\frac{c^5}{d^3}+\frac{d^5}{a^3}\geq \left(\frac{a+b+c+d}{2}\right)^8  \]

by hölder's inequality we have that
$\displaystyle (1+1+1+1)^4(ab+bc+cd+da)^3\bigl(\frac{a^5}{b^3}+\frac{b^5}{c^3}+\frac{c^5}{d^3}+\frac{d^5}{a^3}\bigl)\geq \left(a+b+c+d\right)^8 \\\Rightarrow \frac{a^5}{b^3}+\frac{b^5}{c^3}+\frac{c^5}{d^3}+\frac{d^5}{a^3}\geq \frac{(a+b+c+d)^8}{4^4(ab+bc+cd+da)^3}=(\frac{a+b+c+d}{2})^8$


On the AoPS server, why does

[code]
\begin{eqnarray*}
x+2&=&1\\
x&=&-1
\end{eqnarray*}
[/code]

Give the jumbled code below?

\begin{eqnarray*}
x+2&=&1\\
x&=&-1
\end{eqnarray*}

Because it is getting confused. Like tex documents you must either use dollar signs or \ [...\ ] but not both at the same time which makes little sense
[code]\begin{eqnarray*}
x+2&=&1\\
x&=&-1
\end{eqnarray*}[/code]gives
\begin{eqnarray*}
x+2&=&1\\
x&=&-1
\end{eqnarray*}Of course you should not be using eqnarray which is obsolete and replaced by the superior align (see [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=123690]LaTeX FAQ 5[/url])
\begin{align*}
x+2&=1\\
x&=-1
\end{align*}


Simplify $\frac{\sin 10+\sin 20}{\cos 10+\cos 20}$ as much as possible.

[hide]$sin10+sin20=sin(15-5)+sin(15+5)=2sin(15)(cos5)$
$cos10+cos20=cos(15-5)+cos(15+5)=2cos(15)(cos5)$

$\frac{sin10+sin20}{cos10+cos20}=tan(15)$

use half angle formula for tan(15) and done.
[/hide]


How many functions $f(n)$ defined on the positive integers with positive integer values satisfy $f(1) = 1$ and $f(n) f(n+2) = f^{2}(n+1)+1997$ for all $n$?

Denote $f(n)=a_{n}$.
I think the answer is 16 which is the number of divisors of $1998$.
Let $f(2)=t$. Then $f(3)=t^{2}+1997$. Since $f(4)=\frac{f(3)^{2}+1997}{f(2)}$, we must have $(t^{2}+1997)^{2}+1997\vdots t$, or $1997+1997^{2}\vdots t$. We prove that only $t|1998$ satisfies the condition $a_{k}$ to be positive integers. If $1997|t$, then we easily get that $1997|a_{3}$ and that 1997 doesn't divide $a_{4}$. Then $a_{5}=\frac{a_{4}^{2}+1997}{a_{3}}\notin\mathbb{N}$.
Then we prove that for $t|1998$ our sequence takes only positive integer values. Note firstly that for such $t$, no $a_{k}$ will be divisible by 1997, and from the given relation, we get $GCD(a_{n},a_{n+1})=1$.
Now we use induction.

The first 4 terms are positive integers. Suppose $a_{n+2}=\frac{a_{n+1}^{2}+1997}{a_{n}}\in\mathbb{N}$, for some $n\geq2$. We will prove that $a_{n+3}\in\mathbb{N}$. For that we need to have $a_{n+2}^{2}+1997\vdots a_{n+1}$, or $\left(\frac{a_{n+1}^{2}+1997}{a_{n}}\right)^{2}+1997\vdots a_{n+1}$, or $a_{n+1}^{4}+a_{n+1}^{2}\cdot2\cdot1997+1997^{2}+1997a_{n}^{2}\vdots a_{n+1}$, which is true since $a_{n}^{2}+1997\vdots a_{n+1}$ (because $a_{n-1}\in\mathbb{N}$).
The denominator $a_{n}^{2}$ can be ignored since $(a_{n},a_{n+1})=1$
Edited once after seeing Rust's post.

You are right! And your solution is not same as solution where I know, but I no remember. But when $t|1998$ , I think we can prove $a_{n}\in\mathbb{N}$ as follow: From hypothesis $f(n)f(n+2)-f^{2}(n+1)=f(n-1)f(n+1)-f^{2}(n)$  or $f(n)(f(n+2)+f(n))=f(n+1)(f(n+1)+f(n-1))$ or $\frac{f(n+2)+f(n)}{f(n+1)}=\frac{f(n+1)+f(n-1)}{f(n)}$, and obtain $\frac{f(n+2)+f(n)}{f(n+1)}=\frac{f(3)+f(1)}{f(2)}=\frac{t^{2}+1998}{t}\in\mathbb{N}$.


Find all positive integers $n$ such that $-5^4 + 5^5 + 5^n$ is a perfect square. Do the same for $2^4 + 2^7 + 2^n$.

It's been posted several times before!  Here is the post containing your exact question.

http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=-5%5E4++5%5E5++5%5En+perfect+square&t=64535


[quote="i_like_pie"]A certain type of light bulb has a 50% chance of burning out after every 3 months it is in use. Out of 4 such lights, what is the probability that at least 1 light will still work after 12 months of use?[/quote]

[hide="Hint"]Complementary counting, find probability that none of them work after 12 months of use.[/hide]

[hide] The prob. that at least one light is burning =1- the prob that all lights burn out, which is much easier to calculate. The prob that a certain light bulb will burn out during the 12 months = 31/32, so the prob. that all will burn out = $ \left(\frac{31}{32}\right)^{4}$, so answer = $ 1-\left(\frac{31}{32}\right)^{4}$ ??? [/hide]

[hide="solution"]Since the probability of one light bulb continuing to work after 3 months is $ \frac{1}{2}$, the probability of it still working after 12 months is $ \left(\frac{1}{2}\right)^{4}$ or $ \frac{1}{16}$. We use complementary counting to show that the probability of one bulb burning out within the year is $ 1-\frac{1}{16}$ or $ \frac{15}{16}$. So, the probability of all four bulbs burning out is $ \left(\frac{15}{16}\right)^{4}$ or $ \frac{225}{256}$. Finally, complementary counting shows that the the probability of at least one light still functioning after 12 months is $ 1-\frac{225}{256}$ or $ \boxed{\frac{31}{256}}$, as desired.

[/hide]

[quote="emannes"][hide="solution"]Since the probability of one light bulb continuing to work after 3 months is $ \frac{1}{2}$, the probability of it still working after 12 months is $ \left(\frac{1}{2}\right)^{4}$ or $ \frac{1}{16}$. We use complementary counting to show that the probability of one bulb burning out within the year is $ 1-\frac{1}{16}$ or $ \frac{15}{16}$. So, the probability of all four bulbs burning out is $ \left(\frac{15}{16}\right)^{4}$ or $ \frac{225}{256}$. Finally, complementary counting shows that the the probability of at least one light still functioning after 12 months is $ 1-\frac{225}{256}$ or $ \boxed{\frac{31}{256}}$, as desired.[/hide][/quote]
[hide="Slight Error"]$ \tfrac{15}{16}^{4}=\tfrac{50625}{65536}\ne\tfrac{225}{256}=\tfrac{15}{16}^{2}$

$ 1-\tfrac{50625}{65536}=\tfrac{14911}{65536}$[/hide]


How many combinations of coins can you have with 21 cents?

I don't understand how to solve these problems. Please help.

This is essentially the same as woth 20 cents (because other coin must be penny)

2 dimes-1 possibilty
1dime- 3 possibilites (0N,1N,2N)
0dimes-5 possibilities (0N,1N,2N,3N,4N)

9

Oh, I see now. Thanks.

By the way, is there any other way to solve it without having to list them all?

Not really, you can simplify the steps but that pretty much it.


I have no idea how to approach this problem. It's out of Larson's book:

"A well-known theorem asserts that a prome p>2 can be written as a sum of two perfect squares (p=x^2+y^2), with m,n integers) iff p is one more than a multiple of 4. Assuming this, show:

Every prime one more than a multiple of 8 can be written as x^2+16^2

Every prime 5 more than a multiple of 8 can be written as (2x+y)^2 + 4y^2

Thanks so much. Even just a hint would be good, I suppose. But if you want to prove these things, go all out :). Thanks a lot in advance.

for the first one, it should be x^2 + 16y^2, and here is a hint in spoiler

[hide]work mod 8, and show that if p = x^2 + z^2, one of the terms must be divisible by 4[/hide]

for the second one

[hide]

again, work mod 8 and find out that if p = x^2 + y^2, then one of the terms is divisible by 2 but not 4, say that term is y, the show you can solve for k s.t. 2k+y = x, and you would be done.

[/hide]


[quote="mailo"]Παιδια θελω την βοηθεια σας με μια ασκηση. :huh:

Να υπολογισετε με αλγεβρικες μεθοδους την μεγιστη τιμη της παραστασης

$ f(x,y,z) = x(1 - y^2)(1 - z^2) + y(1 - x^2)(1 - z^2) + z(1 - x^2)(1 - y^2)$

αν $ x,y,z$ ειναι θετικοι πραγματικοι αριθμοι τετοιοι ωστε $ xy + yz + zx = 1$[/quote]

Λύση:
Από $ ab < = \frac {a^2 + b^2}{2}$ βγαίνει:
$ f(x,y,z) = x(1 - y^2)(1 - z^2) + y(1 - x^2)(1 - z^2) + z(1 - x^2)(1 - y^2) < =$
$ x\frac {((1 - y^2)^2 + (1 - z^2)^2)}{2} + y\frac {((1 - x^2)^2 + (1 - z^2)^2)}{2} + z\frac {((1 - x^2)^2 + (1 - y^2)^2)}{2} (I)$
Με το ίσον να ισχύει μόνο όταν $ (1 - x^2) = (1 - y^2) = (1 - z^2) < = > x = y = z$
$ < = > 3x^2 = 1 < = > x = \frac {\sqrt{3}}{3}$
και από αντικατάσταση του  $ x = y = z = \frac {\sqrt {3}}{3}$
στο δεύτερο μέλος της $ (Ι)$ βρίσκουμε ότι: $ f(x, y, z) < = \frac {2\sqrt {3}}{9}$ (Η αρχική μου εκτίμηση δεν είχε καμία σχέση κι ας βασιζόταν στην ίδια σκέψη. Αυτά παθαίνει κανείς όταν βιάζεται. Μέσα σε αυτή τη λύση έχω κάνει έμμεση απόδειξη κάποιων εξεζητημένων θεωριών στα κυκλικά πολυώνυμα.)

O τρόπος λύσης που χρησιμοποιείς είναι λάθος . Δεν μπορείς να χρησιμοποιείς οποιαδήποτε ανισότητα και επειδή ισχύει η ισότητα οταν όλα είναι ίσα μετά να κάνεις αντικατάσταση . Εξάλλου γιατί η μέγιστη τιμή να επιτυγχάνεται όταν όλα είναι ίσα ?  :wink:    Δεν μπορεί να έχουμε άλλη περίπτωση επίτευξης του ακροτάτου  ?

Εξάλλου αν αυτό που γράφεις ίσχυε θα είχαμε λύσει όλες τις ανισότητες που η ισότητα ισχύει όταν α=β=γ μόνο με ΑΜ-ΓΜ
Θα δώσω ένα παράδειγμα για αυτό που σου λέω . Με τη λογική σου θα είχαμε και

Έστω ότι θέλουμε να αποδείξουμε τον Nesbitt . $ \sum\frac{a}{b\plus{}c}\geq\frac{3}{2}$

Isxuei apo AM-gm $ \sum\frac{a}{b\plus{}c}\geq 3\sqrt[3]{\frac{abc}{(a\plus{}b)(b\plus{}c)(c\plus{}a)}}$

Η ισότητα ισχύει όταν α=β=γ άρα όταν $ 3\sqrt[3]{\frac{abc}{(a\plus{}b)(b\plus{}c)(c\plus{}a)}}\equal{}\frac{3}{2}$

άρα το ζητούμενο αποδείχτηκε !!!  :o

Α γράφοντας το μήνυμα σκεφτηκα και το εξής άλλο
Χρησιμοποιώντας την ανισότητα παραπάνω βρίσκεις ότι $ f(x,y,z)\leq g(x,y,z)$

και το g είναι μεταβλητό προφανώς άρα δεν είναι μέγιστο  :wink:
Ελπίζω να έγινα κατανοητός .
Αυτά τα έχω εξηγήσει και σε καμιά 50αριά άλλα ποστ απείρου κάλλους

[quote="silouan"]O τρόπος λύσης που χρησιμοποιείς είναι λάθος . Δεν μπορείς να χρησιμοποιείς οποιαδήποτε ανισότητα και επειδή ισχύει η ισότητα οταν όλα είναι ίσα μετά να κάνεις αντικατάσταση . Εξάλλου γιατί η μέγιστη τιμή να επιτυγχάνεται όταν όλα είναι ίσα ?  :wink:    Δεν μπορεί να έχουμε άλλη περίπτωση επίτευξης του ακροτάτου  ?

[...]

Ελπίζω να έγινα κατανοητός .
Αυτά τα έχω εξηγήσει και σε καμιά 50αριά άλλα ποστ απείρου κάλλους[/quote]

Ninja'ed  :ninja:

Hear, hear! Ακριβώς αυτό ήμουν έτοιμος να πω κι εγώ όταν διάβασα την απάντηση του Silouan. Well said sir...

Cheerio,

Durandal 1707

[quote="Durandal"]
Ninja'ed  :ninja:

Hear, hear! Ακριβώς αυτό ήμουν έτοιμος να πω κι εγώ όταν διάβασα την απάντηση του Silouan. Well said sir...

[/quote]

Τι εννοείς ?

Ninja'ed συνήθως σημαίνει ότι κάποιος μόλις έγραψε αυτό που ήθελες να πεις.

Βασικά πάνω που έγραφα μία απάντηση ίδια με τη δική σου για τη λύση του maniopas, πάτησα ενα refresh για να δω μήπως υπήρχε τίποτα καινούριο στο forum και τότε ήταν που είδα τη δική σου απάντηση... οπότε είπα να μη γράψω δύο φορές τα ίδια πράγματα και απλώς έκανα acknowledge την απάντησή σου, επειδή συμφωνούσα απολύτως με αυτά που έγραφες... εξ'ού και το well said.

Hope it makes better sense now :) Cheerio,

Durandal 1707

Η σκέψη που κάνω είναι η εξής: (Θέλω να μου πείτε πού ακριβώς κάνω λάθος)
Έχουμε μια παράσταση f(κ) της πρότασης κ που είναι μικρότερη ή ίση από μία άλλη παράσταση g(κ) με την f(κ) να είναι μέγιστη και την g(κ) ελλάχιστη όταν ισχύει κ = λ.
Άρα g(λ) >= f(λ) >= f(κ)

Aν και ο Σιλουανός είναι ο πλέον αρμόδιος για να σου εξηγήσει θέματα πάνω στις ανισότητες :wink:  θα σου δώσω ένα αντιπαράδειγμα για να καταλάβεις ότι η μέθοδος που χρησιμοποιείς είναι λάθος...
Εστω ότι ψάχνεις την ελάχιστη τιμή του $ A \equal{} x^2 \plus{} 1$. Τότε με τη μέθοδο που λες θα είχαμε:
$ A \equal{} x^2 \plus{} 1\geq2x$ με την ισότητα να λαμβάνεται για $ x \equal{} 1$. Αρα για $ x \equal{} 1$ θα είχαμε το ελάχιστο $ A \equal{} x^2 \plus{} 1 \equal{} 2$ το οποίο είναι φανερά λάθος αφού το ελάχιστο της $ A$ είναι φανερά το 1.
Το ίδιο θα συνέβαινε αν εφάρμοζες αυτήν την μέθοδο και σε άλλα τριώνυμα των οποίων έψαχνες το ελάχιστο.

Ελπίζω να έγινα κατανοητός με αυτό το παράδειγμα...

Κατάλαβα από τον Σιλουανό ότι όντως έχω λάθος στη γενική περίπτωση. Απλά θέλω να βρω το λάθος στο συλλογισμό που ακολούθησα. Μάλλον είναι ότι το μέγιστο της μίας παράστασης και το ελλάχιστο της άλλης ικανοποιούνται για διαφορετικές τιμές των μεταβλητών.

[quote="mailo"]Παιδια θελω την βοηθεια σας με μια ασκηση. :huh:

Να υπολογισετε με αλγεβρικες μεθοδους την μεγιστη τιμη της παραστασης

$ f(x,y,z) \equal{} x(1 \minus{} y^2)(1 \minus{} z^2) \plus{} y(1 \minus{} x^2)(1 \minus{} z^2) \plus{} z(1 \minus{} x^2)(1 \minus{} y^2)$

αν $ x,y,z$ ειναι θετικοι πραγματικοι αριθμοι τετοιοι ωστε $ xy \plus{} yz \plus{} zx \equal{} 1$[/quote]

$ f(x,y,z) \equal{} x(1 \minus{} y^2)(1 \minus{} z^2) \plus{} y(1 \minus{} x^2)(1 \minus{} z^2) \plus{} z(1 \minus{} x^2)(1 \minus{} y^2) \equal{}$
$ x(1 \minus{} y)(1 \plus{} y)(1 \minus{} z)(1 \plus{} z) \plus{} y(1 \minus{} x)(1 \minus{} x)(1 \minus{} z)(1 \plus{} z) \plus{} z(1 \minus{} x)(1 \plus{} x)(1 \minus{} y)(1 \plus{} y)$
$ <\equal{} x(\frac{(1 \minus{} y) \plus{} (1 \plus{} y) \plus{} (1 \minus{} z) \plus{} (1 \plus{} z)}{4})^4 \plus{} y(\frac{(1 \minus{} x) \plus{} (1 \minus{} x) \plus{} (1 \minus{} z) \plus{} (1 \plus{} z)}{4})^4 \plus{} z(\frac{(1 \minus{} x)(1 \plus{} x)(1 \minus{} y)(1 \plus{} y)}{4})$
$ \equal{} x(\frac{4}{4})^4 \plus{} y(\frac{4}{4})^4  \plus{} z(\frac{4}{4})^4 \equal{} x \plus{} y \plus{} z$
Άρα αρκεί να βρούμε τη μέγιστη τιμή του $ x \plus{} y \plus{} z$

Ας δώσω μια λύση για να τελειώσει η υπόθεση  :P
Θα έχουμε

$ f(x,y,z)\equal{}x(xy\plus{}yz\plus{}zx\minus{}y^2)(xy\plus{}yz\plus{}zx\minus{}z^2)\plus{}y(xy\plus{}yz\plus{}zx\minus{}z^2)(xy\plus{}yz\plus{}zx\minus{}x^2)\plus{}z(xy\plus{}yz\plus{}zx\minus{}x^2)(xy\plus{}yz\plus{}zx\minus{}y^2)\equal{}4xyz(xy\plus{}yz\plus{}zx)\equal{}4xyz$

Επιπλέον από ΑΜ-ΓΜ έχουμε $ xy\plus{}yz\plus{}zx\geq 3\sqrt[3]{x^2y^2z^2}$  άρα $ xyz\leq\frac{1}{3\sqrt{3}}$

άρα $ f(x,y,z)\leq \frac{4}{3\sqrt{3}}\equal{}\frac{4\sqrt{3}}{9}$

QED  :D

Ωραια λυση, να σαι καλα silouan!  :clap2:   :)

ΥΓ. εισαι μαθητης ή φοιτητης?

[quote="mailo"]ΥΓ. εισαι μαθητης ή φοιτητης?[/quote]

Για τι σου φαίνομαι ? :P   (μαθητής Γ λυκείου είμαι)

Γιατί όλοι με ρωτάνε αυτό το πράγμα  :roll: ?

Ψέματα λέει  :P   Καθηγητής είναι !!!  Όποτε έχω απορρίες πάντα με έχει βοηθήσει... :lol:


Prove that 111111...1 cannot be a square
prove that 1000000...1 cannot be a square
prove that 1111111...1111101 (n ones, one 0)cannot be a square

2) 10000...1=10000...0+1

10000...0 can be expressed either as a square or as 10*(a square). Since square+1 can't be a square, therefore 1000...1 can't be a square.

I'm not sure if this is the easiest way or even a right way...since I'm a novice when it comes to number theory.  :(

[quote="mandy_pal"]2) 10000...1=10000...0+1

10000...0 can be expressed either as a square [b]or as 10*(a square)[/b]. Since square+1 can't be a square, therefore 1000...1 can't be a square.

I'm not sure if this is the easiest way or even a right way...since I'm a novice when it comes to number theory.  :([/quote]

Well, thats a good start, but what about the second half of your 'or'? (bold in above)


limit
calculus
logarithms
integration
trigonometry
real analysis

A (hopefully not too fallacy filled) calculus solution:

[hide]Let $A = \displaystyle \lim_{n\to \infty}\frac{1}{n^4} \prod_{j=1}^{2n} {(n^2+j^2)}^\frac{1}{n}$

Taking the factor in front and distributing it into the product we get,

$A = \displaystyle \lim_{n\to \infty}\prod_{j=1}^{2n} {\Bigg(1 + \frac{j^2}{n^2}\Bigg)}^\frac{1}{n}$

$\ln {A} = \displaystyle \lim_{n\to \infty}\frac{1}{n} \sum_{j=1}^{2n} {\ln {\Bigg(1 + \frac{j^2}{n^2}\Bigg)}}$

Interpreting this as a Riemann Sum we have

$\ln {A} = \displaystyle \int_{0}^{2} {\ln {(1 + x^2)} \, dx}$

Using integration by parts (details omitted) we find that $\ln {A} = 2\ln (5) + 2\tan^{-1}(2) - 4$

So $A = 25 e^{(2 \tan^{-1}(2) - 4)}$[/hide]


[quote="James Potter"]Let be given a natural number $n\ge 2$ and $n$ positive real numbers $a_1\ge a_2\ge a_3\ge ...\ge a_n$ prove that:
$\frac{a_1}{a_n}+\frac{a_2}{a_{n-1}}+....+\frac{a_{n-1}}{a_2}+\frac{a_n}{a_1}\le n+\frac{(a_1-a_n)^2+(a_2-a_n)^2+...+(a_{n-1}-a_n)^2+(a_n-a_n)^2}{a_n^2}$[/quote]
For all $x,y\geq z>0$ we have $(x-z)(y-z)\geq 0$
\[ \Rightarrow xz+yz\leq xy+z^2 \\\Rightarrow x^2-2xy+y^2\leq x^2-2xz+z^2+y^2-2yz+z^2 \\\Rightarrow (x-y)^2\leq (x-z)^2+(y-z)^2 \]
and $xy\geq z^2>0$. Thus \[ \frac{(x-y)^2}{xy}\leq\frac{(x-z)^2+(y-z)^2}{z^2}. \]
The rest is easy:
\[ 2\cdot LHS=\sum\limits_{i=1}^n\left(\frac{a_i}{a_{n+1-i}}+\frac{a_{n+1-i}}{a_i}\right) \\=\sum\limits_{i=1}^n\frac{a_i^2+a_{n+1-i}^2}{a_ia_{n+1-i}} \\=\sum\limits_{i=1}^n\left(\frac{(a_i-a_{n+1-i})^2}{a_ia_{n+1-i}}+2\right) \\\leq 2n+\sum\limits_{i=1}^n\frac{(a_i-a_n)^2+(a_{n+1-i}-a_n)^2}{a_n^2}=2\cdot RHS \]


Given $ M\equal{}\{1,2,....,n\}$
1) Find the number of 3-subset $ A,B,C$ of $ M$ such that
$ A \cup B \cup C\equal{}M, |B \cap C|>0$
2) Find the number of 4-subset $ A,B,C,D$ of $ M$ such that
$ A \cup B \cup C \cup D\equal{}M , |B \cap C \cap D|>0$

Any ideas ?? Please, give me some hints.
Thanks.

1) Any of the $ n$ elements of $ M$ must belong to exactly one of the $ 7$ sets $ A \setminus (B \cup C)$, $ B \setminus (C \cup A)$, $ C \setminus (A \cup B)$, $ (A \cap B) \setminus C$, $ (A \cap C) \setminus B$, $ (B \cap C) \setminus A$, $ A \cap B \cap C$. There are therefore $ 7^n$ possibilities, but these include cases when $ B \cap C \equal{} \emptyset$, therefore we have to count these. Then the elements of $ M$ may belong to only the first $ 5$ of the sets (exclude $ (B \cap C) \setminus A$ and $ A \cap B \cap C$), with $ 5^n$ possibilities, whence the total number of $ 7^n \minus{} 5^n$ solutions.

2) Similarly, there are $ 15$ sets to consider, after which we must exclude the cases $ B \cap C \cap D \equal{} \emptyset$, where we only consider $ 13$ sets (exclude $ (B \cap C \cap D) \setminus A$ and $ A \cap B \cap C \cap D$). The total number of solutions is now $ 15^n \minus{} 13^n$.

A Venn diagram might help understand this (simple) counting idea. Of course, all this generalizes to more sets.


there are a,b,c real positive number such that
1/a+b+1   +   1/b+c+1     + 1/c+a+1   >=  1

prove that :

a+b+c >= ab+bc+ac

already posted:
from Cauchy:
$ (a\plus{}b\plus{}1)(a\plus{}b\plus{}c^2)\geq (a\plus{}b\plus{}c)^2$
then $ 1\leq \sum \frac{1}{a\plus{}b\plus{}1} \leq \frac{2(a\plus{}b\plus{}c)\plus{}a^2\plus{}b^2\plus{}c^2}{(a\plus{}b\plus{}c)^2}$
$ \Leftrightarrow a\plus{}b\plus{}c\geq ab\plus{}bc\plus{}ca$


I'm going to create The Amazing Race AoPS. There will be 11 teams of 2.

Here's the list to sign up. (please include both [b]usernames[/b])

1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.

Have questions? Ask me!

Could you explain what this is all about? I have seen the show, but I don't understand how this would be pulled off. Also, what do you mean by "both names"?

The rules are similar to the actual show.

Every leg will have a [b]Road Block[/b] where one person completes a task and a [b]Detour[/b] where there are two tasks and one is chosen and completed.

There are [b]Route Info[/b]. Route Information clues instruct the teams where to go next. Along the race, if you come in last at the pit stop, you may be eliminated. There are 1-3 (I can't decide yet) non-elimination legs where the team is spared but they are encountered with a [b]Speed Bump[/b], an extra task only they have to do. There will also be 1-2 [b]Fast Forwards[/b] where a team has to complete a task and if they do, can skip ahead to the pit stop. There are also [b]Yields[/b]. I'm going to change the rule from the actual show. If a team gets yielded, and are still in the next leg, they will leave 30 minutes after their designated time. There is a [b]U-Turn[/b]. This is always put after a detour. If a team is U-Turned, then they must complete the other detour choice. If a team U-Turns a team in front of them, nothing will happen. You will always be aware via you previous clue if a U-Turn or Yield is coming up.

@gogators The "both names" mean a team. First, people converse to be put on a team. Usually, they know each other well. (either from real life of AoPS). After they decide they want to be on a team, they will sign up.

What are the "where's" going to be? And what is the "30 minute delay"?

@Yoshi
I haven't decided yet where the "wheres" will be.
The 30 minute is after a yield. I'll clear it up later, when it comes.

The Amazing Race is mainly about [b]teamwork[/b]. It is important that you and your partner to work together. When choosing your partner, choose well, knowing that you two will embark on this "journey" together. This race is also mainly decided on trust I will post each box after the first team had PMed me the [u]correct[/u] answer. I trust that teams will not open the box unless their name is below the box.

Ex: [img]http://www.caltex.com/amazingraceasia/images/Route-Marker.gif[/img] [hide="Click to open"]The clue[/hide]

People who can open:
(Name) and (Name)
(Name) and (Name)
and so on and so forth.

There will also be penalties. If you disobey a rule, you will encounter a 1 hour penalty. After that, I can check you in, providing you're not the last team. If a task, road block or detour, is too hard for you and you decide to skip it, you will encounter a 3 hour penalty.

If you have any questions, I'll be happy to clear some things up.

I don't know what to do yet, I just waiting for people to sign up. I'm starting the format (kind of).

[quote="Yoshi"]Or you could just PM the box to both people???[/quote]

Is it possible to send one PM to a whole group of people?

If two teams finish at the same time in which you can send the PM...

[quote="oxfz16"]Steven Ban[/quote]

What?
The message contains only quote and code tags. Please add some text of your own.

[quote="spaceguy524"]I want to play.[/quote]

with who??

question.

Does anyone want me to get rid of the "teams" rule?

...if you have any problems with your partner PM me.

Also, something to do for all you people!!

List any phobias you have (fire, heights, water, spiders,...)

PM me the list.

[b]BE TRUTHFUL!![/b]

Just FYI

All times will be in Eastern times...

[quote="andrewjjiang97"]The deadline will be the 21st. The first leg will start promptly at 4:00 p.m. Eastern Time.
[/quote]

Let's move it to 2:00 pm

WTF!!

Let's start now...

Hello everyone!! This is the start of everyone’s Amazing Race!! The winner of the final leg will win 1,000,000 dollars! (jk) The first clue is below. [b] Be careful, this leg has a twist that will be both helpful to one team and horrible to the rest. [/b] Good luck!!

Travel to Giza, Egypt.

There are two flights:
1st: Has 1 stop, leaves at 4:30pm, connects to Spain, and then to Giza. There is 1 hour available for transferring. Arrives 3am. [b] If all goes well [/b] I will give out the clue for the first flight one hour before the 2nd flight.

2nd: Direct flight, leaves at 6pm, arrives 4am next day.

Please PM me telling me what flight you want to be on. If you do not PM me, I will put you on Flight 1. [b] Also, PM me a number between 1 and 10 inclusive. Each team should PM me ONLY one number. It doesn’t matter what the number is. DO NOT tell anyone (besides me) your number. [/b]

PLANE THING DUE BY SUNDAY 11:59 Eastern time

People on 1st flight: zapi2007 and eqjj168 (number is 1)

People on 2nd flight: vahalla and Nachosaurus (number is 3) jjfun1 and meewhee (number is 7)

[quote="eqjj168"][quote="andrewjjiang97"]
[b] DO NOT tell anyone (besides me) your number. [/b][/quote]

You just told everyone the numbers... :ninja:  :|[/quote]

Sorry. Actually, right now it doesn't matter! So, I will continue to post the numbers.

[quote="andrewjjiang97"]PLANE THING DUE BY SUNDAY 11:59 Eastern time [/quote]

ITS MY BIRTHDAY :lol:

Happy Birthday.
Birthday partay.  :P
 :wow:

I'm going to let this die. All my files have been corrupted and half the people aren't playing. Sorry people...


group theory
abstract algebra
superior algebra
superior algebra unsolved

1 (a) Let G be a finitely generated infinite group. Show that G has a normal subgroup N such that G/N is infinite, but all nontrivial normal subgroups of G/N have finite index.

1(b) What happens if we drop the condition that G be finitely generated?

1 (a) Applying Zorn's lemma is a good idea, but I'm afraid you're not using it right.

Let $ S$ be the collection of proper normal subgroups of infinite index in $ G$ (you certainly don't want those of finite index). $ S$ is non-empty since it includes the trivial subgroup $ \{1\}$. Order $ S$ by inclusion, and consider a totally ordered subset $ T$. We need to show that it has an upper bound. This is the heart of the problem. If you can show that, Zorn's lemma gives you a maximal element in $ S$. Such a maximal element is almost by definition what you need.

Proving that $ T$ has a maximal element is where you must use the fact that $ G$ is finitely generated. Guessing what that maximal element is isn't hard: it's just the union of the elements of $ T$ (note that this is indeed a subgroup, as opposed to your definition of $ N$ where you're taking the union over things that don't contain one another). So why is it necessarily of infinite index? Assume the contrary, and see how a finite number of cosets + a finite number of generators implies that one of the members of $ T$ would already need to have finite index, which isn't the case. I'll leave it to you to fill in the details.

1 (b) Whenever you're looking at a non-finitely-generated counterexample to anything, your first thoughts should be $ \mathbb{Q}$ and $ \mathbb{Q} / \mathbb{Z}$. Try these here.

Thank you, random graph. I realized I was not using Zorn's lemma correct.


just curious what do you mean by AoPS student on ESPN? who is going to be on there? and why are they on ESPN?

The Countdown will be on ESPN.  piggipi, Dnas, hexahedron, cutiepi, and big hc will be on TV

Hexahedron is Andrew, is he not? I asked him whether he was an AoPSer or not when he asked me if I was Treething.

[quote="kpavery"]who is big hc?[/quote]

edited

How about someone posts all of the names corresponding to their usernames...

Soem people might consider that a breach of their privacy. Let people post their own names, as seems to be happening in another post right now.

Let me remind you of one paragraph of our [url=http://www.artofproblemsolving.com/Legal/AoPS_L_Privacy.php]privacy policy[/url]:

[quote]This site has forums (message boards) and chat rooms available to its members. Please remember that any information that is disclosed in these areas is published publicly. You should NOT disclose any personal information, such as your real name, address, telephone number, email address, etc. Your user name is the only identity that you need to have, and that is how people should know you on the Internet.[/quote]

In particular, if you happen to know the real name of an AoPS user, keep that information to yourself.  Do [b]not[/b] post another user's real name without his or her explicit permission.  Better yet, let him or her reveal it if he or she is inclined to do so.  Best yet, remain anonymous.


algebra
polynomial
floor function
functional equation
algebra unsolved

Determine the maximum number $ h$ satisfying the following condition: for every $ a\in [0,h]$ and every polynomial $ P(x)$ of degree 99 such that $ P(0)\equal{}P(1)\equal{}0$, there exist $ x_1,x_2\in [0,1]$ such that $ P(x_1)\equal{}P(x_2)$ and $ x_2\minus{}x_1\equal{}a$.

[i]Proposed by F. Petrov, D. Rostovsky, A. Khrabrov[/i]

It seems that this is a tough problem
clearly $ \frac{1}{2}\ge\ h \ge\frac{1}{98}$

Don't you mean $ h \equal{} \frac{1}{98}$?  It obviously can't be more, considering the extremal case $ P(x) \equal{} x(x \minus{} \frac {1}{98})(x \minus{} \frac {2}{98})...(x \minus{} 1)$

cosinator, in your example more than $ \frac {1} {98}$ can be achieved. One can look at larger lengths outside the interval between two consecutive roots!

Let $ p$ be a polynomial of degree $ n$ such that  $ p(0)\equal{}P(1)\equal{}0$.
Suppose there exists a $ r$ such that there exists no $ a,b ,0\le\ a\le\ b \le\ 1$ with $ p(a)\equal{}p(b)$, $ b\minus{}a\equal{}r$
I will prove that the distance between two consecutive roots of $ p(x)$ in $ [0,1]$ is less than or equal to r.
suppose there exists two consecutive roots such that $ a,b$, $ a\minus{}b>r$
consider the polynomial $ g(x)\equal{}p(x)\minus{}p(x\plus{}r)$
                                   $ g(b).g(a\minus{}r)>0$
                          gives  $ p(b\plus{}r)p(a\minus{}r)<0$
so there is a root between $ b\plus{}r$ and $ a\minus{}r$. contradicting the fact that $ a$ and $ b$ are consecutive roots.
now since a polynomial of degree $ n$ can have only atmost $ n$ roots in $ [0,1]$,and since there are two roots in $ [0,1]$
                      $ (n\minus{}1)r\ge 1$ gives $ r\ge\frac{1}{n\minus{}1}$
but clearly $ h>\frac{1}{n\minus{}1}$ for $ n\ge 4$ (Is it? :o)
Also for degree 4 ,$ h\equal{}\frac{1}{2}$

Can anyone give a solution or a hint for this hard problem?

Here is a possible approach....
The general solution will be $\frac{1}{\lfloor \frac{n-1}{2}\rfloor\ +1 }$ where $n$ is the degree of polynomial

step 1:proving that  $h\geq \frac{1}{\lfloor \frac{n-1}{2}\rfloor \ +1}$     (*)

suppose there exists an $r$ such that there exists no $a,b,0\leq a,b\leq 1$ with $p(a)=p(b)$ and $b-a=r$  (1)

i will prove that the distance between two consecutive roots of $p(x)$ in $[0,1]$ is not greater than $r$   (2)

consider the polynomial $g(x)=p(x)-p(x+r)$
since $a,b,b+r,a-r\in\ [0,1]$ , $g(b).g(a-r)>0$ [If it was less than or equal to zero,then $g(x)$ has a root in $[0,1-r]$,contradicting assumption  (1)]
so $p(b+r).p(a-r)<0$ which means $a$ and $b$ are not consecutive roots which is a contradiction.This proves (2)
Now let  $ \frac{1}{k+1}\leq r\leq \frac{1}{k}, k\in \mathbb{N}$
since $g(x)$ has no roots in $[0,1-r],$ assume WLOG $g(x)>0$ for all $x\in \ [0,1-r]$
For $ m\in \{1,2,\text{...}k-1\}$,we have
$\sum _{i=1}^m g((i-1)r)>0\Longrightarrow p(\text{mr})<0$
Now $p(x)$ has a root in $(0,r)$, but $p(mr).p(mr+r)>0$,this means $p(x)$ has atleast two roots in $(mr,mr+r)$ which means $p(x)$ has atleast $2+1+2(k-1)$ zeros $\Longrightarrow $
$2k+1\leq\n$ if $n$ is odd and $2k+2\leq\n$, if $n$ is even
 this proves (*)

Step 2:
proving that the functional equation $Q(x+r)-Q(x)=P(x),Q(0)=0$ where $P$ is a given polynomial, has a polynomial solution for $Q$
Let degree of $P$ be $n$
From the functional equation we can get the allowed values for $Q(r), Q(2r),........$
Now there exists a unique polynomial of degree $n+1$ passing through the points ${(0,Q(0)),(r,Q(r)),..........((n+1)r,Q((n+1)r))}$
I argue that this polynomial (let us call it $R(x)$)satisfies the functional equation for the reason below
Let $R(x+r)-R(x)=T(x)$, now $T(x)$ is a polynomial of degree $n$.
so T(x)-P(x) is also a polynomial of degree atmost $n$ and so have atmost $n$ real zeros or zero everywhere. But we have $T(0)-P(0)=T(r)-P(r)=...........=T(nr)-P(nr)=0$,so as said above $T(x)-P(x)$ is zero everywhere .So $R(x+r)-R(x)=T(x)=P(x)$, $R(x)$ satisfies the functional equation

step 3
Consider the case when $n$ is odd, let $n=2m-1$
let $q(x)$ be the polynomial solution to equation
$q(0)=0$, $q(x+\frac{1}{m})-q(x)=x(x-\frac{m-1}{m})((\prod _{i=1}^{m-2} (x-\frac{i}{m})){}^2)$

the solution has the following properties
The solution as mentioned has degree $n$.
$q(\frac{i}{m})=0,$ for  $i=0,1,2,\text{...}m$
$q'(\frac{i}{m}+\frac{1}{m})=q'(\frac{i}{m})\neq 0$,for $i=0,1,2,\text{...}m-2$. [if it was zero,then $q(x)$ has atleast $2m$ linear factors which is a contradiction]
Also since $q'(x)$ has same sign at $\frac{1}{m}$ and $\frac{2}{m}$ we have a zero for $q(x)$ between $\frac{1}{m}$ and $\frac{2}{m}$
similarly for $\frac{2}{m}$ and $\frac{3}{m}$ and so on till $\frac{m-2}{m}$ and $\frac{m-1}{m}$
so we got all zeroes of $q(x)$.
now since leading coefficent of $q(x)$ is positive,it is increasing at $0$.also $q'(x)$ is negetive at $\frac{1}{m}$.......$\frac{m-1}{m}$     .......(3)

step 4
consider
 $q(x+\frac{1}{m}+\epsilon )-q(x)=x(x-\frac{m-1}{m})((\prod _{i=1}^{m-2} (x-\frac{i}{m})){}^2)+q(x+\frac{1}{m}+\epsilon )-q(x+\frac{1}{m})$,
consider the set closed ${ S:[0,1-(\frac{1}{m}+\epsilon )]-\underset{i=1}{\overset{m-1}{\cup }}[\frac{i}{m}-\lambda ,\frac{i}{m}+\lambda ]}$
where $\lambda$ is a positive number less than half of min[b]{[/b]$|a-b|:q(a)=0,q'(b)=0$[b]}[/b]
It is easy to note that if $\epsilon$ is less than half of min[b]{[/b]$|a-b|:q(a)=0,q'(b)=0$[b]}[/b],then for $i=0,1,.... m-1$ , $q(\frac{1+i}{m}+\epsilon )-q(\frac{1+i}{m})=\epsilon\ q'(c)$, for some $c\in [\q(\frac{1+i}{m}+\epsilon ),q(\frac{1+i}{m})]$ which is less than zero [because of (3)]
which means $q(\frac{1+i}{m}+\epsilon )-q(\frac{i}{m})<0$
So,choose $\lambda$ such that $q(x+\frac{1}{m}+\epsilon )-q(x)$  is less than zero in $\underset{i=1}{\overset{m-1}{\cup }}[\frac{i}{m}-\lambda ,\frac{i}{m}+\lambda ]$
Note that $S$ is independent of $\epsilon$ if $\epsilon\leq\lambda$
Now  $x(x-\frac{m-1}{m})((\prod _{i=1}^{m-2} (x-\frac{i}{m})){}^2)$ is negetive in $S$ and so has a negetive maximum $\alpha$ in that set

Now choose an $\epsilon$ such that $ |q(x+\frac{1}{m}+\epsilon )-q(x+\frac{1}{m})|<-\alpha $ and $\epsilon\leq\lambda$
So in $S$ ,$q(x+\frac{1}{m}+\epsilon )-q(x)<0$

So $q(x+\frac{1}{m}+\epsilon )-q(x)<0$ in $[0,1-(\frac{1}{m}+\epsilon )]$ for all small enough $\epsilon$

Hence $h\leq \frac{1}{\lfloor \frac{n-1}{2}\rfloor \ +1}$
Similarly we can prove it for even $n$s


calculus
integration
logarithms
real analysis
real analysis unsolved

Use the substitution $ y \equal{} (1 \plus{} x) \plus{} \sqrt {\;x\,(1 \plus{} x)\;}$
Then
$ \int_0^\infty\;\;\dfrac{\ln\,\left(\;(1 \plus{} x) \plus{} \sqrt {\;x\,(1 \plus{} x)\;}\;\right)}{(1 \plus{} x)\cdot\sqrt {\;x\,(1 \plus{} x)\;}}\;\;\textbf dx \equal{} \int_1^\infty\;\;\dfrac{2\ln(y)}{y^2}\;\;\textbf dy \equal{} 2$


Another old math league problem.  Give it a try.
What are all real values of $x$ which satisfy:
$\sqrt{x+2\sqrt{x-1}}+\sqrt{x-2\sqrt{x-1}}=2\sqrt{x-1}$?

[hide="solution"]$\displaystyle \sqrt{x+2\sqrt{x-1}}+\sqrt{x-2\sqrt{x-1}}=2\sqrt{x-1}$

Let's denote $\sqrt{x-1} = A$

$\displaystyle \sqrt{x+2\sqrt{x-1}}+\sqrt{x-2\sqrt{x-1}}=2\sqrt{x-1}$
$\displaystyle \sqrt{A^2 + 1 + 2A} + \sqrt{A^2 + 1 - 2A} = 2A$
$\displaystyle (A+1) + (A-1) = 2A$
$\displaystyle 0=0$

thus A = All Real.

since $\sqrt{x-1}$ is all real, $x \in \Re x \ge 1$[/hide]

Plug in 1.  It doesn't work.  Both of you are missing the same thing,
[hide="hint"]$\sqrt{(x-2)^{2}}=\left|x-2\right|$[/hide]

[quote="towersfreak2006"][hide]
Let $y=2\sqrt{x-1}$.
$\sqrt{x+y}+\sqrt{x-y}=y$
$(\sqrt{x+y}+\sqrt{x-y})^2=y^2$
$x+y+2\sqrt{x^2-y^2}+x-y=y^2$
$2x+2\sqrt{x^2-(4x-4)}=4x-4$
$2x+2\sqrt{x^2-4x+4}=4x-4$
$2x+2(x-2)=4x-4$
$0=0$

Thus, it must be true for all $x$ that give valid arguments for square roots.
$x-1\ge0\rightarrow x\ge1$
$x-2\sqrt{x-1}\ge0\rightarrow x^2\ge 4x-4$ since $x\ge1\rightarrow x>0$ and $\sqrt{x-1}\ge0$ for $x\ge1$
$x^2-4x+4\ge0\rightarrow (x-2)^2\ge0$, which is always true by the trivial inequality.

Hence, the equation above is satisfied for all $x\ge1$.
[/hide][/quote]
[hide]
Ah, so instead of $2x+2\sqrt{(x-2)^2}=2x+2(x-2)=4x-4$, it should be $2x+2|x-2|=4x-4$.
Then the equation becomes $|x-2|=x-2$, which is satisfied by $x\ge2$.
[/hide]

Well, my hint was directed towards towersfreak.  But here's the hint for you:
[hide]$\sqrt{A^{2}+1-2A}=\left|A-1\right|$, not just $A-1$. [/hide]

Edit:  I didn't see towersfreak's post while making this one...but he got it.


vector
algebra
polynomial
function
linear algebra
linear algebra unsolved

Hey,

If V being a 3 dimensional vector space of polynomial functions (in the reals) of degree at most 2. Øi belong to V* (Dual vector space) for i=0,1,2. Is
Ø0(f) = f(0) , Ø1(f) = f(1) and Ø2(f) = f(2) a basis for the dual space? Is there a constructive method to find the basis?

Thnx

Interpolating polynomials exist. Consequently, point evaluation at $ n\plus{}1$ distinct points are a basis for the space of polynomials of degree at most $ n$ (over a field of characteristic zero).


linear algebra
matrix
LaTeX
algebra
polynomial

Let A be an nxr mx with orthonormal columns. Show that A(A[size=75]^T[/size]A)^-1A[size=75]^T[/size] is the nxn mx : (I[size=75]r[/size] 0 everywhese else)

Thanks!

$A$ is an $n\times r$ matrix with orthonormal columns.

That forces $n\ge r$, since an orthonormal set must be an independent set. $A$ is at least as tall as it is wide.

The condition that $A$ has orthonormal colums is precisely equivalent to $A^TA=I_r,$ where $I_r$ is the $r\times r$ identity matrix.

I think you wrote $A(A^TA)^{-1}A^T.$ (There would be less ambiguity if you wrote in $\text{\LaTeX},$ or if you used more parenthesis.)

This would simply be $AA^T$, which is an $n\times n$ symmetric matrix of rank $r.$

At this point the argument falters. Certianly, this is similar to the block-structured matrix $\begin{bmatrix}I_r&0\\0&0\end{bmatrix},$ but it doesn't have to be that matrix.

Cheap example: $A=\frac15\begin{bmatrix}3\\ 4\end{bmatrix}.$

Yes, that satisfies the criterion of orthonormal columns. $A^TA=[1].$

But $A(A^TA)^{-1}A^T=AA^T=\frac1{25}\begin{bmatrix}9&12 \\ 12&16\end{bmatrix}.$

This has characteristic polynomial $x^2-x$ so it is certainly similar to $\begin{bmatrix}1&0 \\0&0\end{bmatrix}.$ But it is not equal to that matrix.

So what is the result you wanted?

Your cheap example just show the 2x2 cases. How can I show in nxn case? Thanks!

Show what? If it's not equal to that form for some example, then it's not equal to that form.

On the other hand, consider the matrix $M=AA^T.$ (Leave the $(A^TA)^{-1}$ out since it's just an identity matrix anyway.)

$M^2=(AA^T)(AA^T)=A(A^TA)A^T=AIA^T=AA^T=M.$

If $r<n,$ then the rank of $AA^T$ will be the same as the rank of $A,$ namely $r.$

That means that $M$ is a rank-$r$ matrix whose minimal polynomial is $x^2-x,$ and hence it is similar to $\begin{bmatrix}I_r&0\\0&0\end{bmatrix} .$

Just remember that matrices can be similar without being the same.


I think the following question was asked in a national math. competition:

Find all functions $f: \mathbb{Z}\to \mathbb{Z}$ such that for all integers $a$ and $b$,  $f(a) - b = f(a + f(b))$.

[size=75][color=green]latexed by mod[/color][/size]

[hide="hint"]
Substitute $0\to b$.
[/hide]

[hide]
Substitute 0 for $a$. Then $f(f(b))=f(0)-b$.

Now here's a common thing to do when given things like $f(f(b))$: Look at $f(f(f(b)))$ and evaluate it two different ways (either the outer or inner two $f$'s can be used). Here it is equal to both $f(f(0)-b)$ and $f(0)-f(b)$. Set $b=0$ and set them equal to get $f(f(0))=0$. But $f(f(b))=f(0)-b\Rightarrow f(0)=0$.

Since $f(0)=0$, $f(f(b))=-b$ and $f(-b)=-f(b)$. From the first equation, $f(f(-b))=b$ so let $C=f(-1)$. Then $f(a+1)=f(a)-C$ from the original equation. So $f(a)=-aC$ for all positive integers $a$. Since $f(-b)=-f(b)$ this can be extended to all integers.

So $f(a)$ must have the form $-aC$. But plugging $C$ in for $a$ gives $1=-C^2\Rightarrow C^2=-1$ which is bad. There is $\boxed{\text{no solution}}$ to the functional equation.
[/hide]

Alright, I feel pretty stupid right about now, but what is $\mathbb {Z}$?

[quote="I"]...this can be extended to [b]all integers[/b].

So $f(a)$ must have the form...[/quote]

That should be enough of an explanation of what the set $\mathbb{Z}$ is.

[quote="chess64"][hide]
Substituting $b=0$ we find $f(a)=f(a+c)$, where $c=f(0)$. Thus $f(x)$ must be a constant, or $f(x) = C\forall x\in \mathbb{Z}$. Substituting this in the original equation we find $C-b=C$, which is only true when $b=0$. Hence, there are no solutions for $f(x)$.
[/hide]

Is this right? I'm not sure about the conclusion I made from $f(a)=f(a+c)$.[/quote]

Well, $f(a)=1+(-1)^a$ does satisfy $f(a)=f(a+f(0))$. Of course, it doesn't fit the original equation, but I don't think there is an extremely quick way to solve this.

[hide]f(a)-b = f(a+f(b))
f(a)=f(a+f(b))+b

So the solution is all functions f with f(0)=0.
Is this correct ?[/hide]

From what I remember, doesn't f: Z→Z mean f(z)=z? :huh:
If it does, then the solution is (a,b)=(t,0), t ε R

No.   All Z-Z means is that for all the function maps every integer to another integer.

Well just because our function maps every integer to another integer it doesn't mean that the two integers are the same i.e. $\ f(z)=z$ like you stated.  All it means is that if $\ a$ is an integer $\ f(a)$ is an integer, but not necessarily the same as $\ a$

cooley, can you give the official solution ?


Let $a_{n}$ be the $n$th digit after the decimal point in the repeating decimal expansion of $\frac{28}{111}$.
Compute the value of $\frac{a_{1}}{4}+\frac{a_{2}}{4^{2}}+\cdots+\frac{a^{n}}{4^{n}}$

The decimal expansion is
EDITED:
$0.252\ 252\ 252\ 252$...

Please help

[quote="DiscreetFourierTransform"]

The decimal expansion is $0.25\ 255255255255255255$... so I thought the first two terms had to be treated differently from the repeating 255's.

[/quote]

Its wrong. The correct is $0.25 \ 225225225...$ .

[hide="You might consider looking at the digits this way"] $0.252 \, 252 \, 252 \, 252 \, 252 ...$ [/hide] [hide="Which suggests that you might want to group the summation like this"] $\left( \frac{2}{4}+\frac{5}{4^{2}}+\frac{2}{4^{3}}\right)+\left( \frac{2}{4^{4}}+\frac{5}{4^{5}}+\frac{2}{4^{6}}\right)+...$[/hide] [hide="Which is a"] Regular old geometric series.  Of course, to sum it for finite $n$ you need three cases depending on the value of $n \bmod 3$, but that's not so hard. [/hide]

Ugggh. Thanks for the hints. I'll fix the initial post. Missed this one, but I read the calc screen wrong. Boy, do I feel stupid now.

[hide="And in case you were wondering how to do this without a calculator"] $\frac{28}{111}= \frac{252}{999}= 0.\overline{252}$ [/hide]

Possible solution?
[hide]
Also, we're summing for infinite $n$, I think. The common ratio is less than one, I believe, so the end result should be
$\frac{a}{1-r}$ where a is the first term and r is the ratio, right?

Does this mean that the because the first grouped term is equal to
$0.84375$ and the common ratio is $\frac{1}{64}$,
that the convergence goes to
$\frac{0.84375}{1-\frac{1}{64}}$,
or $\frac{\frac{54}{64}}{\frac{63}{64}}$?
Which works out to
$\frac{54}{63}$?
Is this correct?
[/hide]

Thanks. I completely missed that incorrect decimal expansion thingy. Also, now I know how to do that without a calc. Pretty cool. I'm a newb at this math stuff, just learning the basics of competetion math.
Hoping for a 100+ score on the AMC 12 this year as a freshman.

How about letting $a_{n}$ be the $n$th digit after the decimal point of $\frac{m}{999}$ and computing

$\frac{a_{1}}{b}+\frac{a_{2}}{b^{2}}+\frac{a_{3}}{b^{3}}+\cdots$?

Well, any fraction where A is less than 999 but more than 99 should give a nice repeating decimal of $0.A\ A\ A$...
If $10 \le A\le 99$ then the decimal form is $0.0A \0A\ 0A$...
And finally, if $0 \le A \le 10$ then the decimal form is $0.00A\ 00A\ 00A$...
Thus, for Case 1, where A is a three digit number, we get

((First Digit of A) over b) + ((Second Digit of A) over $b^{2}$) + ((Third Digit of A) over $b^{3}$)

as the initial term, and then find the second term, then find the common ratio, then finally set up the convergence equation. Too lazy to do it now, but it shouldn't be that hard.

[quote="paladin8"][quote="Farenhajt"][quote="paladin8"]How about letting $a_{n}$ be the $n$th digit after the decimal point of $\frac{n}{999}$ and computing

$\frac{a_{1}}{b}+\frac{a_{2}}{b^{2}}+\frac{a_{3}}{b^{3}}+\cdots$?[/quote]

Is $n$ intentionally given in two different usages? (Namely in $a_{n}$ and ${n\over 999}$)[/quote]

No; sorry about that.[/quote]

So, this should be "$a_{k}$ is $k^\mathrm{th}$ digit after the decimal point of ${n\over 999}$ where $n$ is a given natural"?


[quote="nayel"]Let $ x,y,z$ be positive real numbers. Prove that
\[ \sqrt{\frac{x}{x+y}}+\sqrt{\frac{y}{y+z}}+\sqrt{\frac{z}{z+x}}\le \frac{3\sqrt 2}{2}\]
[/quote]
It's very known. See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=17549

i saw your link arqady. and my solution is easier and more elementary than all of those solutions. let's see if anyone can find it. :wink:

[quote="nayel"]Yes, my solution can beat zhaobin's solution. Not actually beat, but my solution is quite similar to zhaobin's solution[/quote]
Not only similar, your proof is [b]the same[/b] as Zhao Bin's: just examine the expression on the other side of $ \le$ in both proofs.

[quote="nayel"][quote="ductrung"]Can you solution beat Zhao Bin's solution posted here: [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=78720[/url] ?
Actually he proved the stronger result of pvthuan.[/quote]

Yes, my solution can beat zhaobin's solution. Not actually beat, but my solution is quite similar to zhaobin's solution of the stronger result, which I proved too. Here's my solution:

$ LHS=\frac{\sqrt{x(y+z)(z+x)}+\sqrt{y(z+x)(x+y)}+\sqrt{z(x+y)(y+z)}}{\sqrt{(x+y)(y+z)(z+x)}}$

$ \le \sqrt{\frac{(x(y+z)+y(z+x)+z(x+y))(x+y+y+z+z+x))}{(x+y)(y+z)(z+x)}}$

$ =\sqrt{4\cdot\frac{(xy+yz+zx)(x+y+z)}{(x+y)(y+z)(z+x)}}$

$ =2\cdot\sqrt{\frac{(x+y)(y+z)(z+x)+xyz}{(x+y)(y+z)(z+x)}}$

$ =2\cdot\sqrt{1+\frac{xyz}{(x+y)(y+z)(z+x)}}$

$ \le 2\cdot\sqrt{1+\frac{1}{8}}$

$ =\frac{3\sqrt 2}{2}$.

so, does that answer your question? :P[/quote]

[hide="If you think this is spam, do not read this"]Someone rated the above post as spam. I would be very happy if you'd tell me the reason, thank you very much. I really care about my rating, that is why I am asking. Is there a fault in my solution? Or are you jealous of me? I see that you are reducing the ratings of most of my posts, why? Was it a mistake of me to find a good solution and posting it? I myself think very carefully before rating a post as spam, and I usually don't rate posts as spams. I even try to increase the rating of a post which is already rated as spam. Because I know how it feels if your rating decreases. Is this what I get in return?[/hide]


modular arithmetic
number theory
prime numbers

Find all triples of positive prime numbers $ p_1,p_2,p_3$ such that

$ 2p_1\minus{}p_2\plus{}7p_3\equal{}1826$ and $ 3p_1\plus{}5p_2\plus{}7p_3\equal{}2007$.

\begin{eqnarray}2p_1 - p_2 + 7p_3 & = & 1826 \\
3p_1 + 5p_2 + 7p_3 & = & 2007\end{eqnarray}

$ (2) - (1)\implies p_1 + 6p_2 = 181$, hence $ p_2 < 30$

Plugging $ p_1 = 181 - 6p_2$ into $ (1)$ we get

$ - 13p_2 + 7p_3 = 1464\implies - 14p_2 + 7p_3 = 1464 - p_2$, hence $ p_2\equiv 1464\equiv 1\pmod{7}$

Hence we're looking for a prime less than $ 30$ which is congruent to $ 1$ modulo $ 7$. In the set $ \{8, 15, 22, 29\}$ only $ 29$ is prime. Therefore $ p_2 = 29\implies p_1 = 7\implies p_3 = 263$, and $ (7, 29, 263)$ is a unique solution.


inequalities
trigonometry
geometry
area of a triangle
Heron\u0027s formula
inequalities unsolved

I know $(a+b+c)^{2}\geq 4(h_{a}h_{b}+h_{b}h_{c}+h_{c}h_{a})$, therefore your problem strong than this inequality.
By $2S=ah_{a}=bh_{b}=ch_{c}=ab\sin{C}=bc\sin{A}=ca\sin{B}=2pr=\frac{abc}{2R}$, where $p=\frac{a+b+c}{2}$, your inequality is equivalent to
\[(\frac{R}{r})^{2}\geq \frac{1}{\sin^{2}{A}}+\frac{1}{\sin^{2}{B}}+\frac{1}{\sin^{2}{C}}. \]
But that is all what i know, at this time :D

Let $K$ be area of the triangle; then $h_{a}=\frac{2K}{a}$.
We must prove that \[(a+b+c)^{2}\geq 16K^{2}(\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}})\]

Note that Heron's Formula yields: \[16K^{2}=(a+b+c)(-a+b+c)(a-b+c)(a+b-c)\]
It remains to show that \[(a+b+c)\geq \left(\prod_{cyc}(a+b-c) \right)\sum_{cyc}\frac{1}{a^{2}}\]

This is tighter than Schur's, since after applying the Schur inequality $abc\geq \prod (a+b-c)$ we must prove: \[a+b+c \geq  abc\sum \frac{1}{a^{2}}\iff a+b+c \geq \sum \frac{bc}{a}\] which in actuality has a reversed inequality sign.

We must rewrite: \begin{eqnarray*}(a^{2}b^{2}c^{2})(\sum a)&\geq &(a^{2}-(b-c)^{2})(b+c-a)(\sum b^{2}c^{2}) \\ (a^{2}b^{2}c^{2})(\sum a)&\geq &\left(\sum a^{3}+3abc-\sum_{sym}a^{2}b \right)(\sum b^{2}c^{2})\end{eqnarray*}

Let $K$ be area of the triangle; then $h_{a}=\frac{2K}{a}$.
We must prove that \[(a+b+c)^{2}\geq 16K^{2}(\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}})\]

Note that Heron's Formula yields: \[16K^{2}=(a+b+c)(-a+b+c)(a-b+c)(a+b-c)\]
It remains to show that \[(a+b+c)\geq \left(\prod_{cyc}(a+b-c) \right)\sum_{cyc}\frac{1}{a^{2}}\]

This is tighter than Schur's, since after applying the Schur inequality $abc\geq \prod (a+b-c)$ we must prove: \[a+b+c \geq  abc\sum \frac{1}{a^{2}}\iff a+b+c \geq \sum \frac{bc}{a}\] which in actuality has a reversed inequality sign.

We must rewrite: \begin{eqnarray*}(a^{2}b^{2}c^{2})(\sum a)&\geq &(a^{2}-(b-c)^{2})(b+c-a)(\sum b^{2}c^{2}) \\ (a^{2}b^{2}c^{2})(\sum a)&\geq &\left(\sum a^{3}+4abc-\sum_{sym}a^{2}b \right)(\sum b^{2}c^{2})\end{eqnarray*}

Now I found a similar one at http://www.mathlinks.ro/Forum/viewtopic.php?t=15558
[quote="tranminhhoang"]Thank you very much fuzzylogic, these nice applications will make my article more interesting.
I also find a nice application

For a,b,c >0 we have :
abc(ab+bc+ca)^2 >= (a+b+c)(a+b-c)(b+c-a)(c+a-b)(a^3+b^3+c^3)

But I think the method we use to prove this inequality also have benefits not only in inequality. We can use Lagrange and Jaccobi Theorem to compute some distance between two points in a triangle.  And the distance also give us an interesting inequality. For example if we compute IH ( I is incenter and H is orthocenter )we have the inequality noted by darij grinberg and we also have
IH <= 2OI   ( O is circumcenter )[/quote]

And I found the following but I think there's a mistake:
since $a,b,c$ are triangle side lengths we just put in for $a+b-c, z$ and the rest also cyclicly over $x,y,z$:
\[{(2\sum x)\geq xyz \sum_{cyc}\frac{1}(2(x+y)^{2}}\iff \sum \frac{1}{2yz}\geq \sum \frac{1}{(x+y)^{2}}\] which follows by AM-GM term by term and using $x,y,z>0$ from Ravi substitution.
I think this might actually be valid, because when I noted the thing about "stronger than Schur" I forgot for a second that these were sides of a triangle and so it is stronger then "semi-Schur" for side lengths which is actually very weak, being the product of three Am-Gm's.

It's easy problem. And here is a stronger inequality:
\[(a+b+c)^{2}\ge 4(m_{a}m_{b}+m_{b}m_{c}+m_{c}m_{a})\]  :wink:

okay, was my solution correct?

Also, I can't determine if it holds also when a,b,c aren't side lengths (the rewritten inequality in my solution)... what do you think?

[quote="nobody1"]In triangle $ABC$, prove that: $(a+b+c)^{2}\ge 4(h_{a}^{2}+h_{b}^{2}+h_{c}^{2})$[/quote]
We have:
\[h_{a}^{2}+h_{b}^{2}+h_{c}^{2}\le l_{a}^{2}+l_{b}^{2}+l_{c}^{2}\\ =\frac{4bc}{(b+c)^{2}}p(p-a)+\frac{4ca}{(c+a)^{2}}p(p-b)+\frac{4ab}{(a+b)^{2}}p(p-c)\\ \le p(p-a)+p(p-b)+p(p-c)=p^{2}\]

[hide="me@home"] I don't know your solution is true or false because it's very long and I very lazy  :blush:  [/hide]
And I'm not sure that my inequality is correct  :blush:
It's should be move to [color=blue]Inequalities Open Question [/color]  :D

[quote="April"]It's easy problem. And here is a stronger inequality:
\[(a+b+c)^{2}\ge 4(m_{a}m_{b}+m_{b}m_{c}+m_{c}m_{a}) \]
:wink:[/quote]

Now i only know an event that is weaker than your conjecture as:
$4(m_{a}m_{b}+m_{b}m_{c}+m_{c}m_{a})\leq 2(a^{2}+b^{2}+c^{2}+ab+bc+ca)$

[quote="hien"]
\[2(m_{a}m_{b}+m_{b}m_{c}+m_{c}m_{a})< a^{2}+b^{2}+c^{2}+ab+bc+ca \]
[/quote]
We have:
\[2\sum m_{b}m_{c}\le 2\sum m_{a}^{2}=\frac{3}{2}\sum a^{2}\]
Then we have to prove that: $\sum a^{2}< 2\sum bc$. But it very simple  :D


If m varies directly as p^2 and q^4. and p doubles while q triples, what happens to m?

Before anyone can answer your question, you have to define what you mean by “varies directly as $ p^{2}$ and $ q^{4}$.”

If, by any chance, you mean $ m(p,q) \equal{} kp^{2}q^{4}$, with $ k$ as a constant, then $ m(2p, 3q) \equal{} k(2p)^{2}(3q)^{4} \equal{} k(4p^{2})(81q^{4}) \equal{} 324kp^{2}q^{4} \equal{} 324 [m(p, q)]$, so we can say that the value of $ m(2p, 3q)$ is $ 324$ times the value of $ m(p, q)$.

However, this is not the only way to interpret your question.

"$ x$ varies directly as $ y$" means that $ x \equal{} ky$, for some constant k.

So... $ m \equal{} k_1p^2\equal{} k_2q^4?$

I think I'm interpreting the problem incorrectly...

He probably means joint variation, I think?


You need to find set of 7 numbers between 2 and 50 which have LCM as maximum of that 7 number

Set of those 7 numbers could be

SET-1 = {2,3,4,6,8,12,48}
SET-2 = {3,4,6,8,12,16,48}
SET-3 = {3,4,6,8,12,24,48}
SET-4 = {3,4,6,8,16,24,48}

Fraction of SET-4 will have sum equal to 1

1/3 + 1/4 + 1/6 + 1/8 + 1/16 + 1/24 + 1/48

Thanks,
Nabin Prasad.
http://www.teacherone.com

---------------------------

[quote="shyong"]Let $S$ be a set $\{\frac{1}{2},\frac{1}{3},\frac{1}{4},\cdots ,\frac{1}{50}\}$ . Choose $7$ distinct fraction in $S$ such that the sum of the $7$ fraction is $1$ .

 :)[/quote] :roll:


geometry
circumcircle
geometric transformation
rotation
similar triangles
complex numbers
geometry proposed

Let  $ABC$  and $AMN$ be two similar triangles with the same orientation, such that  $AB=AC$, $AM=AN$ and having disjoint interiors. Let $O$ be the circumcenter of the triangle $MAB$. Prove that the points $O$, $C$, $N$, $A$ lie on the same circle if and only if the triangle $ABC$ is equilateral.

[i]Valentin Vornicu[/i]

Suppose $AOCN$ is cyclic. This means that $\angle ANC=\pi-\angle AOC$. The triangles $ACN,ABM$ are directly congruent, so $\angle ANC=\angle AMB=\frac{\angle AOB}2$. We thus get $\angle AOC=\pi-\frac{\angle AOB}2$, or, in other words, the line $OC$ is the internal bisector of $\angle AOB$. Since $OA=OB$, this immediately implies $AC=BC$, and we find $ABC$ to be equilateral.

For the converse, we just reverse the reasoning.

In the complex plane let $A=0$ and $B=1$ then $N=CM$.
If $O'$ is the circumcenter of $ACN$ then since $ABM$ and $ACN$ are similar by a factor of $C$ we have $O'=OC$.
Then the assumption becomes that $O'$ has equal distances to $A$ and $O$.
That is
\[ |O'-O|=|O'| \]
Then
\[ |OC-O|=|OC| \]
By canceling $|O|$ we get
\[ |C-1|=|C| \]
that is the length of $BC$ and $AC$ are equal.

Tell me if anything is wrong with the following:

We take a rotation of center $A$ and angle $\angle BAC$. This takes the circumcenter $O$ of $ABM$ into the circumcenter $P$ of $ACN$. Thus, $AO=AP$. The condition that $O,C,N,A$ are cyclic is equivalent to $PA=PO \, \, \Longleftrightarrow \, \, \triangle OAP$ equilateral $\, \, \Longleftrightarrow \, \, \angle BAC = \frac{\pi}{3}$.

[hide="Remark"]If the above is correct, then this problem is one of the most trivial problems given at a Romanian TST, in the last three years, at least.[/hide]

There is nothing wrong with it (In fact it is exactly the same as mine but you expressed it without complex numbers)
And I agree that it is very trivial :lol:

[quote="perfect_radio"]If the above is correct, then this problem is one of the most trivial problems given at a Romanian TST, in the last three years, at least.[/quote]Thanks for the appreciation (the problem is mine :) ).

[b]However[/b], let me point out that for example Micnea Dragos (IMO Gold medalist, JBMO multiple medalist, etc. etc.) didn't solve it. And in fact less than 50% of the contestants solved it. Only $4$ (!!) solved it using the rotations (or directed angles) the rest did either complex numbers or other types of calculatorial arguments. So I really do not know what trivial means ...

For problem 1 I belive it was a good choice.

This is a lovely little problem. Congrats, Valentin! :)

[quote="Valentin Vornicu"]Thanks for the appreciation (the problem is mine :) ).[/quote]
Sorry if I offended you in any way. This was not my intention.

[quote="Valentin Vornicu"]And in fact less than 50% of the contestants solved it.[/quote]
This was "the most solved" problem, along with your problem from the other day. (I didn't manage to solve that one, yet)

[quote="Valentin Vornicu"]So I really do not know what trivial means ...[/quote]
Trivial means here that I could solve a geometry problem in not so much time using synthetic arguments, which is really a rarity. Probably I should have specified that.

PS: The fact that Michnea Dragos didn't solve it is really strange. I knew he was quite good.

[quote="perfect_radio"]The fact that Michnea Dragos didn't solve it is really strange. I knew he was quite good.[/quote]He is, it is just that the problem is not as trivial as you think it is :) Of course, once you see a solution, and/or solve it yourself you might call it easy, but it's not that easy as it seems.

Don't get me wrong: it's not hard, and it's not ment to be hard: it is the first problem. We wanted to give the kids from 9th grade a chance (since all the other problems were not accesible to them).

[color=darkblue]This problem isn't trivial, but is simply and nicely (using a rotation - see the Perfect_Radio's proof). Can solve it without rotation (the level $7^{\mathrm{th}}$ class) and through a chain of equivalencies.[/color]

[color=darkred]Denote the circumcentre $O'$ of the triangle $ACN\ .$ Then, $\triangle ABM\equiv\triangle ACN$ $\Longrightarrow$ $OA=O'A$ and $\widehat {BAO}\equiv\widehat {CAO'}\ ,$ i.e. $\widehat {BAC}\equiv \widehat {OAO'}\ .$ Therefore, the points $O,A,C,N$ are concyclically $\Longleftrightarrow$ $O'A=O'O$ $\Longleftrightarrow$ the triangle $OAO'$ is equilateral $\Longleftrightarrow$ $m(\widehat{OAO'})=60^{\circ}$ $\Longleftrightarrow$ $m(\widehat {BAC})=60^{\circ}\ .$[/color]

"with the same orientation"
what does it mean?


IF  24 H in a D means 24 hours in a day...then figure these out...there are loads more...but these are the only ones I didn't get! :lol:  :D
a)66 B of the B
b)13 S in the U S F
c) 39 B of the O T
d)32 is the T in D F at which W F
e)15 P in a R T
f)3 W on a T
g)100 C in a R
h)11 P in a F (S) T
i)23 P of C in the H B
j)64 S on a C B

[hide="i got a few of them..."]
a. 66 books in the Bible
b. 13 stripes on the US flag
d. 32 is the temperature in degrees fahreinheit at which water freezes
f. 3 wheels on a tricycle
h. 11 players in a football/soccer team
i. 23 pairs of chromosomes in the human body[/hide]
these are fun.  :lol:

[hide="j"]64 squares on a chess board[/hide]

c[hide]39 books of the old testament[/hide]

[quote="mathnerd314"]And here's one:

5 W in the A

[hide="hint"]It's self-referential[/hide][/quote]

5 Words in the.... uhhhhhh  :D  i dunno...have to think....
i thought at fist that it should be 5 V in the A
(5 Vowels in the Alphabet)

Oh..and a few more...
3 B M (S H T R)
15 P in a R T
3 W on a T
100 C in a R
11 P in a F (S) T
13=UFS(this one stumps me the most, so it'll probably be the easiest!!)
8 T on a O
27 B in the N T
hmmm...any ideas?   :D

I love these type of puzzles...anywhere u can get loadz of them?

8 T on an O
[hide="is"]8 tentacles on an octopus[/hide]

11 P on a F(s) t

[hide]11 People on a football(soccer) team has been done already[/hide]

[hide="3 B M (S H T R)"] 3 blind mice (see how they run)[/hide]

882 P of A G

[hide="Hint"] It was in a movie. [/hide]
[hide="Bigger Hint"] Johnny Depp and Orlando Bloom were in it.  :rotfl: [/hide]

[quote="lingomaniac88"]Here are some:
1. 81 N on a F S B
2. 99 B of B on the W
3. 5 T on a F
4. 21 D on a S S D
5. 26 C in the A V of D or N D

Hints:
[hide="1"]Check the Math and Strategy Games Forum.[/hide]
[hide="2"]Think back to long road trips.[/hide]
[hide="3"]Look at yourself.[/hide]
[hide="4"]You can find the answer in a Monopoly set.[/hide]
[hide="5"]It's based on a TV show.[/hide][/quote]

[hide="2"]99 bottles of beer on the wall[/hide]
[hide="3"]5 toes on a foot[/hide]

5

[hide]26 cases in the ?? of Deal or no Deal[/hide]

[quote="lingomaniac88"]Here are some:
1. 81 N on a F S B
2. 99 B of B on the W
3. 5 T on a F
4. 21 D on a S S D
5. 26 C in the A V of D or N D

Hints:
[hide="1"]Check the Math and Strategy Games Forum.[/hide]
[hide="2"]Think back to long road trips.[/hide]
[hide="3"]Look at yourself.[/hide]
[hide="4"]You can find the answer in a Monopoly set.[/hide]
[hide="5"]It's based on a TV show.[/hide][/quote]

4[hide] 21 Dots on a Standard ??(Square) Dice [/hide]

[hide="Some of them"]
101 dalmations
60 seconds in a minute etc.
32 teeth in your mouth?

10 digits in the decimal system
52 cards in a standard deck of cards

[/hide]

366 D I A L Y

all of thse answers are right.  good one...

[quote]882 P of A G [/quote]
[quote="funcia"]That doesn't help!!! Tell me the answer please![/quote]

[hide="882..."]
pieces of Aztec Gold (from Pirates of the Carribean)[/hide]

[quote="nutz_for2.718281828"]101 D
(easy)
60 S in a M. corresponding one: 60 M in an H
(easy)
1000 G in a K
(mild)
32 T in your M
(mild)
12 S on a D
(hard)
10 D in the D S
(hard)
82 S on a D
(Drives you out of your mind)
52 C in a S D of C
(hard)
10 T on a S
(mild)
81 S on a S B
(hard)
6 E in the S W S
(mild)
18 S in E A P's T R
(easy)[/quote]

[hide]101 Dalmatians, 60 Seconds in a Minute, 1000 Grams in a Kilogram, 32 Teeth in your Mouth (unless you got wisdom teeth extracted or something), don't know, don't know, don't know, 52 Cards in a Standard Deck of Cards, don't know, 81 Seats on a School Bus?!?, don't know, 18 Stories in Edgar Allen Poe's something?!?!?[/hide]

2,408 P in M N T F A

[hide="big hint"]I am vain...[/hide]

[quote="mathnerd314"][quote="nutz_for2.718281828"]101 D
(easy)
60 S in a M. corresponding one: 60 M in an H
(easy)
1000 G in a K
(mild)
32 T in your M
(mild)
12 S on a D
(hard)
10 D in the D S
(hard)
82 S on a D
(Drives you out of your mind)
52 C in a S D of C
(hard)
10 T on a S
(mild)
81 S on a S B
(hard)
6 E in the S W S
(mild)
18 S in E A P's T R
(easy)[/quote]

[hide]101 Dalmatians, 60 Seconds in a Minute, 1000 Grams in a Kilogram, 32 Teeth in your Mouth (unless you got wisdom teeth extracted or something), don't know, don't know, don't know, 52 Cards in a Standard Deck of Cards, don't know, 81 Seats on a School Bus?!?, don't know, 18 Stories in Edgar Allen Poe's something?!?!?[/hide]

[/quote]right, right, right, right, wrong, wrong, wrong, right, wrong, OBVIOUSLY wrong, wrong, close to right. Here are all of the answers:

[hide="1."]101 dalmations[/hide]

[hide="2."]60 seconds in a minute and 60 minutes an an hour[/hide]

[hide="3."]1000 grams in a kilogram[/hide]

[hide="4."]32 teeth in your mouth[/hide]

[hide="5."]12 sides on a dodecagon[/hide]

[hide="6."]10 digits in lthe decimal system[/hide]

[hide="7."]82 sections on a dartboard[/hide]

[hide="8."]52 cards in a standard deck of cards[/hide]

[hide="9."]10 tentacles on a squid[/hide]

[hide="10."]81 squares on a Sudoku board[/hide]

[hide="11."]6 episodes in the Star Wars saga[/hide]

[hide="12."]18 stanzas in Edgar Allen Poe's The Raven[/hide]

[quote="pontios"]Here is one more:

23142 U in ML

The above is true at the time I'm writting this[/quote]
23142 users in MathLinks

[quote="Padmavathi"]coool! nice set of problems...
100 C in a R
13=UFS
27 B in the N T
15 P in a R T[/quote]

27 Books in the New Testament

100 cents in a... ruble?!

Has anyone solved 2,408 P in MNTF's A yet? (It's rather contrived...)

[quote="thatin"]6 P in I
42 P in I[/quote]
Aha!... It's
[hide]6 problems in IMO
42 points in IMO
 :D [/hide]

7 years at hogwarts school of w(hm...) and wizardy

4 H at H?

[hide="366 D I A L Y"]
366 days in a leap year
[/hide]
42 is the A to the U Q of L, the U and E  [hide="laughinghead505 & lotrgreengrapes7926"]42 is the Answer to the Ultimate Question of Life, the Universe and Everything - g1, alhough it was really easy :)[/hide]
16 P on a D M C

A B and the 40 T  [hide="mathnerd314"]Ali Baba and the 40 Thieves - n1 mathnerd314[/hide]

[quote="laughinghead505"][hide="42 is the A to the U Q of L, the U and E "]42 is the answer to the U Q of life, the universe and everything[/hide][/quote] ultimate question

[quote="thatin"]7 years at hogwarts school of w(hm...) and wizardy[/quote] witchcraft

:D

[quote="z00ey"][hide="366 D I A L Y"]
366 days in a leap year
[/hide]

42 is the A to the U Q of L, the U and E

16 P on a D M C

A B and the 40 T[/quote]

[hide]
Uhhhhh, 42 is the Apple to the Unidentified Quail of Lobster, the Urchins, and Eel... Just kidding, yes I've read all five books.

16 P on a DMC... don't know.

Ali Baba and the 40 Thieves!!!! :w00tbrows:
[/hide]

[quote="laughinghead505"]23 A on a TI-89 T C
:lol: That one's easy

1. 7 Y at H S of W and W

2. 6 H P B are O

3. 118 R on a D

4. 4 P on M R

5. 104 K on a KB[/quote]

Hints:[hide="2"] Rowling[/hide]
[hide="3"]money[/hide]
[hide="4"]South Dakota[/hide]
[hide="5"]computer[/hide]

[hide]
2. 6 hp books are out
3. don't know
4. 4 presidents on mount rushmore
5. 104 bytes in a kilobyte? thats 1024 so maybe im wrong[/hide]

Oh, I understand now!!! Number three is [hide]118 ridges on a dime[/hide] :D

[quote="laughinghead505"]23 A on a TI-89 T C
:lol: That one's easy

7 Y at H S of W and W

6 H P B are O

118 R on a D

4 P on M R

104 K on a KB[/quote]

[hide]7 Years at Hogwarts School of Witchcraft and Wizardry
6 Harry Potter Books are out
118 ....
4 Presidents on Mount Rushmore
104 keys on a keyboard[/hide]

[quote="mustafa"]Heres a real toughie

7 L in MSN[/quote]

[hide]7 Letters in My Screen Name[/hide]

2,408 P in MNTF's A is similar. Except it will (hopefully) change as soon as I get to 1000 posts. (Rather misleading hint, though...)