[quote="mdk"]If a and b are positive integers and b is not the square of an integer, find the relation between a and b so that the sum of $ a+\sqrt{b}$ and its reciprocal is integral.[/quote]

[hide="Possible method"]Let $ x=a+\sqrt{b}$, and $ p$ be the integer.
$ x+1/x = p$
$ x^{2}-px+1 = 0$
$ x = \frac{p}{2}\pm \sqrt{\frac{p^{2}-4}{4}}$

Then, we can match the square root with $ b$, and the integral part with $ a$:
$ a = p/2$
$ b = \frac{p^{2}-4}{4}$

We can eliminate the parameter $ p$:
$ a^{2}= b+1$
I'm not sure about my answer, since I'm assuming that the integer is even...[/hide]

Almost.

[hide="Consider..."] The irrational coefficient of $ a+\sqrt{b}+\frac{1}{a+\sqrt{b}}$ is given simply by $ 1-\frac{1}{a^{2}-b}$, hence $ a^{2}-b = 1$. [/hide]


Everything has been about hard problems on the competition. Looking more on the positive side, which problems were really easy, and why were they easy?

Most of the problems were fairly easy, but the first ten were a cinch.

First few were pretty easy well because....they were the first few!  At states though........the first few might be as hard as the hardest ones at chapters!  :(  :(  :(  :(

[quote="solafidefarms"]Hard as the hardest at chapter. Would that be the age problem? :(  Or #30 :lol:[/quote]
Number 30 was the one I did during the targets... anyways I noticed it worked for all 2 digit numbers that Ended in 9, so it's 19+29+...99, which equals 531 (I think that's the answer I didn't check)

[quote="frost13"]those are the numbers, but it asked for the median(or mean) either way its the same for an arithmetic sequence :)[/quote]
Sorry I was just answering the question from memory...

the first few questions scared me a little bit.... thought i did something majorly wrong because I didn't think it was that easy... this was my first mathcount competition, and I was kinda nervous in the beginning because i thought it would be way harder....

[quote="jasonho2004"]I got 2 wrong! I was probably dreaming. I cannot believe it! Ah well, I still get to go to state. Cheers! lol[/quote]

You put 28%?

[quote="jasonho2004"]I got 2 wrong! I was probably dreaming. I cannot believe it! Ah well, I still get to go to state. Cheers! lol[/quote]

i didn't get what it meant and speant like 5 minutes in it....

some easy questions are scarier than they look, it took me 2 months to finally figure out a simple competition problem!

can someone give me the site where they post the problems? I got the site somewhere for the target problems, but can't figure out the sprint ones

Here you go :

[url=http://www.mathcounts.org/webarticles/anmviewer.asp?a=510&z=47]www.mathcounts.org/webarticles/anmviewer.asp?a=510&z=47[/url]

thanks! oops, my dad is getting mad about the easy problems I missed...

I missed a lot of easy problems because of stupid mistakes, but we're all ordinary humans...... ;)

i never took the chapter (Delaware) but online they were pathetically easy

Trust me... it's so much easier taking them online than actually competing... online there isn't that time pressure.... Looking back on the Chapters, they were pathetically easy... but I made too many stupid mistakes...


[quote="N.T.TUAN"][quote="siuhochung"]what do you mean?
You have registered in this forum when you post this message  ;)
BTW, this is the chinese community..[/quote]

My friend need Register.
Can you help me?[/quote]

why don't you repeat what you have done when you registered here?

i still don't see anything involves the Chinese Community...


Change of topic...

DON'T ACTUALLY **TALK** ABOUT CHAPTER! JUST HOW YOU DID!!!!!!!!!!

I am going to state! Finished 7th Individually (only one from my team) and my team placed 2nd...

But the kids on my team didnt do well at all!

They placed 16th, 17th and 19th!!!! :?

Well, if I were you, I'd wait until the 1st.

Okay.

I came in first in chapter individually and none of the rest of my team was in the top 6, but the team was 1st place too. I was the team captain.

Billy

[quote="Farazy"]Then just delete this topic...Don't want to make the AoPS administrators angry or anything[/quote]
Can't you delete it, since it seems like you wrote it?

I dont think so...
If I can I don't know how.

just pm the moderators or an administrator and ask them to delete this topic...

Have you ever heard of a school being closed unexpectedly?

Yesterday, February 26, was the last [b]scheduled[/b] day for Chapter Competitions.  In the past, there have been extenuating circumstances in which a Chapter was unable to hold their competition.  [Heavy snow would be such an example.  Problems (no heat, no water, no electricity, ...) at the scheduled facility might be another example.]

There are hundreds of Chapters across the nation.  The National MATHCOUNTS organization asks each Chapter to send them notification that their competition has been completed.  Only after the [b]very last[/b] Chapter sends their notification, does the National organization release the competition materials.

If National has not received all of the notifications by Monday, they will attempt to contact the remaining Chapters.  MATHCOUNTS knows that students and coaches and even their own Chapter Coordinators want to talk about the competition.  But it is most important to allow every Chapter to have a fair competition.

In an e-mail from my Chapter's Coordinator, he said National expects to release the competition materials by March 7.  [We all hope it will be sooner.]

DO NOT TALK ABOUT CHAPTER COMPETITIONS UNTIL FURTHER NOTICE.

Ya know, I nearly didn't make it past chapter. It was all thanks to this girl on my math team who helped score an hefty amount of points for us, and then I made it. I still haven't thanked her yet... :?

I think you should run up and kiss her... :P

Well...I am sure we can keep from talking *about* it, yet instead talking about how we did... (Not to forget we CAN'T talk about the test)

So are you going to state or not?

Guess what place I came in at chapters?  Just Guess!
Hint: I didn't get a trophy!!!!!!Grrrrr..... So close but didn't reach it.....grrrr.....lol  :blush:

I got 7th individually =( oh well
1 place away from getting a trophy arrgh...
but our team got 1st =)!
My team members got, 1st, 2nd, and 5th for individual

Oh nice!  In my region, trophies go to top 3 :(
*hint, hint* I just gave another hint about my place in regionals!

[quote="EFuzzy"]I got the lucky number rank in the region, our team got the second odd prime.[/quote]
So you got in 7th and your team on in 5th?  Are you going to states?

Yeah, the top sqrt(625) people go on to states.

Wow...You guys lucked out.

For us it was top 4 teams plus top 4 individuals go to state...

Thats only 20 people.

I was so POed after I got knocked off of countdown! I WROTE THE CORRECT ANSWER BUT CALLED OUT THE WRONG ONE!

Oh well...There is always state.

At my competition only the top 3 teams and top 3 individuals made it to state...

I made it both ways though. Non-prime, non-composite place for individual and first prime number for team.  :lol:

in my chapter top 5 teams and top 15 individuals not on a top team go on to states. I made it with my 9th place individual finish, and 3rd place team finish. my other team members came in 1st, 25th, and 54th.

Team made 4th to 5th (don't know what final standings are because of scoring mistake).

I scored second individual (lost to first in tiebraking procedures), but dropped to third in countdown, at which I am not good at all.

Top 4 teams go to state, and top 5 individuals.

I am!
Good Luck to everyone else going to states!

I am got forth individule. And my two man team didn't do half bad.  :D

I got fourth individual (screwed up the sprint), second team but first countdown... and here in Greater Hartford the top 8 or so teams and top 50 individuals regardless if you're on a top team or not go on...

Post and Logon to AoPS more often. Maybe take a class or two, the MATHCOUNTS Prob Series I highly recommend. I've taken it, and you have no idea how helpful it is

[quote="LadyKn1ght"]ACK!! You guys are lucky! In my chapter only the top team and 2 other top individuals go to states so i got 9th place but am not going :mad: !! (my teammate is, though, because he got 1st  :D . We only had three people on our team so no chance of making to states that way.)[/quote]
But how big is your chapter? Ours has 300-400 people, and over 30 teams competing... plus CT is a really small state and almost all of the population is either here or in the Southwest quadrant...

[color=indigo]This year the Richmond chapter had about 16 teams so thats about 120 people total. So only 1/20 of the people there get to go!!  [/color]

Remember tho that VA isn't exactly the easiest state to win... and because of its size, less people get to advance.


number theory
greatest common divisor
geometry
3D geometry

Find three distinct natural numbers, whose arithmetic mean, geometric mean and harmonic mean are also natural numbers.

Can we find $n \geq 4$ distinct natural numbers that have the same property (the arithmetic mean, the geometric mean and the harmonic mean $\in \mathbb N$)?

[i]Laurenţiu Panaitopol[/i]

If it helps, $H=\frac{G^2}{A}$.
Can $H=3\frac{(abc)^{2/3}}{a+b+c}$ be an integer if $a\neq b\neq c$?
I don't think so, but I'll keep looking.

I believe $H=\frac{G^2}{A}$ is only true for two numbers.

let $\text{gcd}(a,b,c)=d$, and then $a=d*a'$, etc

arithmetic=$d*\frac{a'+b'+c'}{3}$

geometric= $d*\sqrt[3]{a'b'c'}$

harmonic= $d*\frac{3a'b'c'}{a'b'+b'c'+c'a'}$

now we can easily find a solution by making

$a'b'c'$ a perfect cube and $d=3(a'b'+b'c'+c'a')$

there probabily more solutions out there though

and the other problem is also trivial using this method,

$a'b'c'd'$ is a 4th power of an integer

$d=4(a'b'c'+b'c'd'+c'd'a'+d'a'b')$

Wow, that's so...easy.
(And, Jiang, you are right: that equations only works for two numbers.)


[quote="Bojan Basic"]Prove (or disprove) that $x^5+y^5+z^5=2004$ has no integer solutions.[/quote]

You know, I thougth it was easier. But it seems like it doesn't let me solve it totally. It is trivial to prove that it does not have solutions on the positive integers, but that does not solve the problem. Any way, my advances (not many) are that, if such a solution exists:

1) $\gcd(x,y,z)=1$
2) from $x,y,z$, exactly one is even, say, $z$.
3) With the notations above, $x+y\cong4\mod16$, then, the posibilities, mod 16 are (1,3), (5,15), (7, 13) and (9,11).

Best,

Under mod 25, the only possible 5th power residues are $0, 1, 7, 18$, and $24$. The unique way to add three and get 4 is where $x$, $y$, and $z$ are all of the form $5k + 3$. It seems that this would make it possible to argue that the spacing of 5th powers is too sparse if we must increment by 5, but the rationals are dense - we will need more than a few small calculations.

I do think that something like finding all solutions to x 5 + y 5 + z 5 = 2005 could be on some MO, but then this one is much easier... :D


FFind the remainder of 2^87 divided by 3.

[hide] I like to do a couple calculations then find a pattern

2^1 remainder 2
2^2 remainder 1
2^3 remainder 2
2^4 remainder 1

so remainder is 2.

proof of pattern

(x+2/3)*2= 2x+4/3=(2x+1)+1/3
(x+1/3)*2=2x+2/3
[/hide]

[quote="undefined117"][quote="bpms"][hide]
$2 \equiv-1 \mod 3$[/hide][/quote]

so what would one do after that?[/quote]
Uh, that was a hint.
[hide="a little more hintage"]It's trivial to find large integral powers of -1[/hide]


$f(x)$ is a function such that $f(x+y)=f(x) \cdot f(y)$ and $f(1)=\frac12$.

(1) If $n \in N$, find $f(n)$.

(2) Let $a_n=nf(n)$, $n \in N$, find $a_1+a_2+a_3+\cdots+a_n$.

(3) let $b_n=\frac{nf(n+1)}{f(n)}$, $n \in N$, $S_n=b_1+b_2+\cdots+b_n$, find $\frac{1}{S_1}+\frac{1}{S_2}+\frac{1}{S_3}+\cdots+\frac{1}{S_n}$.

P.S. This problem is of the level of the last problem of the Chinese College Entrance Exam. Do you think if it belongs here of Pre-Olympiad?


In response to Ripper1's request, here is another AMC level problem. (Actually, two.)  Both, in fact, are even from this year's AMC12B:

17.  If log(xy^3) = 1 and log(x^2*y) = 1, what is log(xy)?

and

20. The graph of f(x) = ax^3 + bx^2 + cx + d intersects the y-axis at (0,2), and the x-axis at (-1,0), (1,0), and (m,0).  What is b?

I'm sure loads of people have solved these before, so if you have, don't post just yet... anyway, happy solving!

Hmm lets see (I never done AMC so I can post )

[hide]

First one means xy^3 = x^2 y so x=y^2. Now we have log (y^5) = 1,ie log y = 1/5, and we want to find log (y^3) which is then 3/5.

For the second one, just expanding a(x^2 - 1)(x-m) gives am=2 and b=-ma = -2.

[/hide]

I like this method for the first one:

[hide]log(x*y^3)+log(x^2*y)+log(x^2*y)=3=log(x^5*y^5)=5log(xy). Thus log(xy)=3/5.[/hide]

Hmm..I'll be VERY surprised to get either of these right.

[hide]

1) If log (xy^3) = log (x^2y) = 1 Then both of those algebraic expressions equal 10, thus equal each other. We have

xy^3 = x^2y

y^3 = xy     We can divide by x on both sides because we know x is not zero.

Substituting this, we get y^5 = 10, so log y = 1/5

Inserting y into the other equation, we get that log y = 2/5.

When multiplying numbers, we add logs, so log xy = 3/5.

I hope.

2) This one was pretty easy. f(0) = 2, so d must equal 2. Plugging in 1 and 2, we get

-a+b-c=-2

a+b+c=-2

Adding the equations, b = -2

[/hide]

[quote="TheSouthpaw"]Hmm..I'll be VERY surprised to get either of these right.[/quote]

They're both right. You should be more confident in your ability.


calculus
integration
modular arithmetic
number theory
prime factorization

Prove that the number of divisors (including the number and 1) of perfect squares is odd.

Masoud Zargar

Two methods:

Given a number $n=p_1^{k_1}p_2^{k_2}p_3^{k_3}\ldots$ in it's prime factorization form, the number of positive integral factors is $(k_1+1)(k_2+1)(k_3+1)\ldots$ it's perfect square is $n^2=(p_1^{k_1}p_2^{k_2}p_3^{k_3}\ldots)^2=p_1^{2k_1}p_2^{2k_2}p_3^{2k_3}\ldots$. The number of positive integral divisors of $n^2$ is $(2k_1+1)(2k_2+1)(2k_3+1)\ldots$. Since $k_n$ is an element of the set of integers, $2k_n+1\equiv1\pmod2$, and thus $(2k_1+1)(2k_2+1)(2k_3+1)\ldots\equiv1\pmod2$.

This proves the if part.

The only if part can be shown that if there is any even term in the product, it would cause the product to be $0\pmod2$, and thus even. It's corresponding power in the prime factorization of $n^2$ would have to be $2k_n-1$. However, the power must be an even integer because the prime is a factor of a perfect square, and thus taking the square root should result in an integral power, but factoring a two out $2k_n-1$ would leave a non-integral power for the prime, and thus $n$ would not be an integer, which is a contradtion.

There's the only if.

Another way to look at it:

any integral factor $x$ of a number $k$ has a corresponding integral factor $y$ such that $xy=k$ (Since $x$ goes $y$ times into $k$, $y$ goes $x$ times into $k$). In this way, every factor comes with a corresponding pair, and pairs are an even number. You want an odd number, so you want one of these pairs to only contain one distinct number. rewriting the earlier equation, $y=\dfrac{k}{x}$, so if we want only one distinct number, $x$ and $y$ must be equal. Thus, $x=\dfrac{k}{x}\Longrightarrow x^2=k\Longrightarrow x=\sqrt{k}$. Since $x$ is an integer, $\sqrt{k}$ is an integer and thus $k$ must be a perfect square.

Evidently, when we write out the prime factorization of a square number, the exponents of the primes must be divisible by 2.  This is so because a square number is the product of a number multiplied by itslef.  Hence by dividing by 2, we get to equal numbers whose product is a perfect square.

  Next, to find the number of divisors of a certain number, we take the numbers prime factorization, add 1 to each exponent of each prime, and multiply.  Whenever we add 1 to a even number the result is:

  $2n+1$
  and odd number.  Since the product of any series of odd numbers is odd (we can prove this becaause no odd number that have a factor of 2 exist, hence there product will not be divisible by 2)
the number of divisors a perfect square has must alse be odd.

  We can alternatively notice that the factors of non-square numbers can always be paired up to equal our original number.  Since a square number has one number which can be multiplied by itself, we have:

  $2n +1$ factors

  Which again is an odd number of divisors.

i don't see how these proofs are that long, if you have a perfect square, the factorization is $\prod_{i=1}^{m}p_i^{2e_i}$ where m is the number of unique prime factors, (this is because they all must be even otherwise, it will not turn out as an integer when doing the square root of it)
using the divisor formula,
divisiors = $\prod_{i=1}^{m}(2e_i+1)$, and since we want to see if it is even or odd, looking in mod 2,
$\prod_{i=1}^{m}(2e_i+1) \mod{2} \equiv \prod_{i=1}^{m}1 \equiv 1$ therefore it is odd


inequalities
triangle inequality
inequalities unsolved

[quote="The Bogy Man"]a, b, c, complex numbers.

Prove: |a| + |b| + |c| + |a+b+c| \geq |a+b| + |b+c| + |a+c|

Seems like you only need to use the triangle inequality, but I can't figure it out.[/quote]
After squaring use the triangle inequality $ |(a \plus{} b \plus{} c)a| \plus{} |bc|\geq|(a \plus{} b)(a \plus{} c)|.$ :wink:


$n,k$ is natural numbers.find $n$such that $2^{n}+n^{2}=3^{k}$. :)

If $n=1$ then $k=1$.
If $n=2$ then there are no solutions.
Suppose that $n \geq 3$. It's obvious that $n$ is odd. So: $1 \equiv n^{2}\equiv 3^{k}\mod{8}$. It means that $k$ is even, $k=2l$. We have:
$2^{n}= (3^{l}-n)(3^{l}+n)$
so there exist $a, b$ such that:
$2^{a}=3^{l}-n$
$2^{b}=3^{l}+n$
which implies
$2^{a}+2^{b}= 2*3^{l}$
$2^{a-1}+2^{b-1}= 3^{l}$
so $a=1$
Well, now we can solve the equation:
$2^{b-1}+1 = 3^{l}$
but actually I'm to lazy to doing it, we proceed as follows: we check small cases manually, then $\mod{8}$, we prove that $l$ is even and we factor the equation just like we did before. The solutions are $b=2$ and $b=4$.
Finally, we have $n=3$ or $n=5$ but none from this works, so the only solution is $n=1$.

The important thing is that this solution shows that we can replace $n^{2}$ by $m^{2}$ and the equation will still be easily solvable (at least for $n\geq 2$ because I didn't think about $n=1$).


25. What is the greatest possible value of x for the equation

$ \left(\dfrac{4x\minus{}16}{3x\minus{}4}\right)^2$ $ \plus{}$ $ \left(\dfrac{4x\minus{}16}{3x\minus{}4}\right)$ $ \equal{}$ $ 12?$

26. Julian wants to label the five fingers on his left hand with the integers 1 through 5, such that no two
adjacent fingers have consecutive integers, and each integer is used exactly once. In how many ways can Julian label his fingers?

30. Gracie starts at her house and walks one block east. After each block she walks, she flips a fair coin. She walks one block east if the coin lands heads; otherwise one block west. She stops her walk the first time she lands at home. What is the probability she stops her walk after walking exactly eight blocks? Express your answer as a common fraction.
 :|

[hide="hint for #25"] let  $ \frac {4x - 16}{3x - 4}) = y$ This really trivializes the problem[/hide]

30.

We need 3 E's and 4 W's. It is clear that the first must be an E, and the last two must be WW (why?) So, we have $ \binom{4}{2} \equal{} 6$ ways; however, we can't have EEWWEEWW, so we really have 5 ways, out of a total $ 2^7 \equal{} 128$. Therefore, our answer is $ \frac {5}{128}$.

26.

We have 3 cases.

Case 1: The middle finger is labeled 1 (or 5) -

This means the 2nd and 4th fingers are 5 and 4, so the 1st and 5th are 2 and 3, so we have 25143, 24153, 34152, or 35142, a total of 4. Similarly, there are 4 cases if the middle is a 5.

Case 2: The middle finger is labeled 2 (or 4) -

The 2nd and 4th are 4 and 5, and the 1st and 5th are 1 and 3. We have 2 possibilities here, and then another 2 for the 4.

Case 3: The middle finger is 3 -

We have 2 ways.

Therefore, our answer is $ \boxed{14}$.

25.

Let $ \frac{4x \minus{} 16}{3x \minus{} 4} \equal{} y$. We then have $ y^2 \plus{} y \equal{} 12 \implies y^2 \plus{} y \minus{} 12 \equal{} 0$. Factoring gives us $ (y \plus{} 4)(y \minus{} 3) \equal{} 0$. Thus, $ x \equal{} 2$.

[quote="isabella2296"]

26.

We have 3 cases.

Case 1: The middle finger is labeled 1 (or 5) -

This means the 2nd and 4th fingers are 5 and 4, so the 1st and 5th are 2 and 3, so we have 25143, 24153, 34152, or 35142, a total of 4. Similarly, there are 4 cases if the middle is a 5.

Case 2: The middle finger is labeled 2 (or 4) -

The 2nd and 4th are 4 and 5, and the 1st and 5th are 1 and 3. We have 2 possibilities here, and then another 2 for the 4.

Case 3: The middle finger is 3 -

We have 2 ways.

Therefore, our answer is $ \boxed{14}$.
[/quote]

What is your motivation for starting with the middle finger?

If you consider the middle finger, then you get just 3 cases, which doesn't take too long to compute and such.  :)


abstract algebra
Ring Theory
superior algebra
superior algebra solved

Let V be finitely generated R module s.t. MV=V for all maximal ideals M of R.show V=0.

let $ x_1,...,x_n$ be a minimal generating set of $ V$. if $ V \neq 0$, we have $ n \geq 1$. since $ x_1 \neq 0$, $ Ann_R(x_1)$ lies in a maximal ideal $ m$ of $ R$. but then we may write $ x_1 \equal{} a(r_1 x_1 \plus{} ... \plus{} r_n x_n)$ for some $ a \in m$, and $ a x_1$ vanishes on the right hand side, a contradiction to the minimality of $ x_1,...,x_n$.

the proof is nonsense ... a does not have to lie in Ann_R(x1). but I also doubt that the claim is true.

$ V\ne\mathfrak mV\iff\mathfrak m\in\mathrm{supp}\,V\iff\mathrm{Ann}\,V\subseteq\mathfrak m$. owk

You said you know what Nakayama is, so you know that:
$ mV\equal{}V\Rightarrow\exists x\in m$ such that $ (1\plus{}x)\in AnnV$, so $ AnnV\not\subset m$, so $ AnnV\subset m\Rightarrow mV\neq V$
Conversely, we use that the last equivalence is obvious/well known so assume $ V_m\equal{}0$. Then for all $ v\in V$, there exists $ m(v)\not\in m$ such that $ m(v)v\equal{}0$. By multiplying all $ m(v)$'s corresponding to a generating set for $ V$, we get that there exists $ x\not\in m$ such that $ xV\equal{}0$. This $ x$ has a multiple in $ 1\plus{}m$ that we can replace it with, so we can assume $ x\in 1\plus{}m$ (also because $ R/m$ is a field) i.e. $ x\equal{}1\minus{}y$ and $ y\in m$. But then we can write every $ v\in V$ as $ v\equal{}yv\in mV$, so $ V\subset mV$ and since the other inclusion is obvious, we have proved our desired result.


Let $p$ be unique integer and $k(p)$ equals the product of all positive integers divisors of $p$.

Find the smallest $p$ such that following inequality is true.

$k(k(k(p))) + 2 > 5^{10}$

Well, you want an integer that isn't prime. First, start with four, but it comes out to 2^21+2 which is clearly less than 5^10. Next, go to six. k(6)=36, k(36)=36x18x12x9x6x4x3x2= 6^9 which is clearly larger than 5^10.


Prove that for every prime p there exists an infinite set of numbers n such that 2^n-n is divided by p.

2^p-1≡1(mod p) Fermats little theorem.
Let K(n)=(np-1)(p-1) for every n then we have K(n)≡1(mod p)
2^K(n)=(2^p-1)^np-1≡1(mod p)
therefore 2^K(n)-K(n) is divided by p for every n.

So? What is the question?

I think you have done well to post it directly into the solved problems section  :D

Pierre.

Oh yeah, i realised that i have posted it in the unsolved forum only when i loged out.
Well maybe the question would be:"Find what the question is?" :D

Maybe IMO itself will turn someday into a kind of jeopardy  :D

Pierre.


inequalities
calculus
integration
algebra
binomial theorem
inequalities unsolved

Three problems:

1) Prove the following inequality for all integral $ n \ge 2$:
\[ n^{\frac{1}{n-1}} > (n+1)^{\frac{1}{n}}.\]

2) Show that ${ n^{\frac{1}{n}}<1+\left(\frac{2}{n}\right)^{\frac{1}{2}}}$.

3) For integers $ n > 2$ and real numbers $ s > 0$, show that
\[ \left(\prod_{i=0}^{n}(s+i)\right)\left(\sum_{j=0}^{n}\frac{1}{s+j}\right) < (n+1)\prod_{k=1}^{n}\left(s+k-\frac{1}{2}\right).\]

hello, your first inequality i equivalent to
$ \left(1\plus{}\frac{1}{n}\right)^n \le e<n\plus{}1$ for all $ n \ge 2$.
Sonnhard.

[quote="Dr Sonnhard Graubner"]hello, your first inequality i equivalent to
$ \left(1 \plus{} \frac {1}{n}\right)^n \le e < n \plus{} 1$ for all $ n \ge 2$.
Sonnhard.[/quote]

Sorry, but how is it equivalent? I don't see it.

$ \left(\frac{n\plus{}1}{n}\right)^n < n\plus{}1 \Longleftrightarrow (n\plus{}1)^{n\minus{}1} < n^n \Longleftrightarrow (n\plus{}1)^\frac{1}{n} < n^\frac{1}{n\minus{}1}$ :wink:

oh snap... he multiplied by $ n \plus{} 1$. I see :blush:

aha.. thanks!

any ideas on the other 2?

[quote="Dr Sonnhard Graubner"]hello, your first inequality i equivalent to
$ \left(1 \plus{} \frac {1}{n}\right)^n \le e < n \plus{} 1$ for all $ n \ge 2$.
Sonnhard.[/quote]

Sorry for the correction, but I would said that $ \left(1 \plus{} \frac {1}{n}\right)^n\le e$ [b]and[/b] $ e < n \plus{} 1$  [b]imply that [/b] $ \left(1 \plus{} \frac {1}{n}\right)^n < n \plus{} 1$

By the way I remark that the first inequality is a strong one ,but not the final  ! as we can see at TI-83...

That's the power of the implication If $ a < b$ then $ a^n < b^n$ !

By the way, using the same one, 2) is equivalent to $ \left(1 \plus{} \sqrt\frac {2}{n}\right)^n > n$

To prove that, expand $ \left(1 \plus{} \sqrt\frac {2}{n}\right)^n$ with binomial theorem [url]http://en.wikipedia.org/wiki/Binomial_theorem[/url] and you'll be happy ![hide]$ 1 \plus{} n\sqrt\frac {2}{n} \plus{} \frac {n(n \minus{} 1)}{2}\times\frac {2}{n} \plus{} ...$ and we have already much more then $ n$ ![/hide]


prove that for all positive real numbres a,b,c :
\[ (a^2\plus{}2)(b^2\plus{}2)(c^2\plus{}2) \geq 9(ab\plus{}ac\plus{}bc)\]

[quote="stvs_f"]prove that for all positive real numbres a,b,c :
\[ (a^2 \plus{} 2)(b^2 \plus{} 2)(c^2 \plus{} 2) \geq 9(ab \plus{} ac \plus{} bc)
\]
[/quote]

WLOG $ (a \minus{} 1)(b \minus{} 1)\ge 0$.
We have
$ (a^2 \plus{} 1 \plus{} 1)(1 \plus{} b^2 \plus{} 1)\ge (1 \plus{} 1 \plus{} 1)(a^2 \plus{} b^2 \plus{} 1)$(by Tchebychef's Inequality.)
and
$ (a^2 \plus{} b^2 \plus{} 1)(1 \plus{} 1 \plus{} c^2)\ge (a \plus{} b \plus{} c)^2\ge 3(ab \plus{} bc \plus{} ca)$
Done.

step 1:
$ \prod (a^{2}\plus{}2) \geq \prod (b\plus{}c\plus{}1)$
Step 2:
$ \prod (b\plus{}c) \geq \frac{8}{3\sqrt{3}}\sqrt{(\sum bc)^{3}}$
Step 3:
Do by yourself


Is there such a square number where the first 9 digits are

123456789, and the last nine digits are 987654321?

[hide="Yes"] $12345678987654321 = (111111111)^{2}$

This is more obvious if you are familiar with the expansion of $(1+x+x^{2}+...+x^{n})^{2}$. [/hide]
bigboy82892:  Incorrect.  The numbers you named beyond the first one have $123456790$ as their first nine digits.

FYI: i didn't need you to correct me. i just logged off when i remembered that i put a wrong answer here.

[quote="bigboy82892"]i know i just realized my mistake. the only number that does that is 111111111. :!:[/quote]
There are $\infty$ many working numbers (but I agree that it's the only one from your list that works).

PS: do not double-post, especially for such an information...


What is the least whole number value of $ x$ such that $ f(x) \equal{} x^{2} \plus{} x\plus{}11$ is not prime?

Note that $ f(x\plus{}1)\equal{}f(x)\plus{}2(x\plus{}1)$, which makes calculations simpler. Anyway, using this recursion we find that $ x\equal{}\boxed{10}\implies f(x)\equal{}121\equal{}11^2$.

I would assume positive whole numbers, else any negative x divisible by 11 is not prime.

x=0, f(x)=11
x=1, f(x)=13
x=2, f(x)=17
x=3, f(x)=23
x=4, f(x)=31
x=5, f(x)=41
x=6, f(x)=53
x=7, f(x)=67
x=8, f(x)=83
x=9, f(x)=101
x=[b]10[/b], f(x)=121 <- AHA NOT A PRIME!

I don't get the recursion solution. Can somebody help me?

You just keep plugging it back in again and again but each time u add f(1)


Hexagon ABCDEF has the following properties:
i) Diagonals AC, CE and EA are all the same length
ii)angles ABC and CDE are both 90 degrees
iii)all the sides of the hexagon have lengths which are different integers

a)what is the minimum perimeter of hexagon ABCDEF if AC is the square root of 85?

b) What is the smallest length of AC for which ABCDEF has all these properties? What is the minimum perimeter in this case?

This question is given on an on-going math contest; please don't reply to it.


geometry
circumcircle
incenter
geometry unsolved

median point of AB
circumcircle
locus (circle) of incenters of tringles on this circumcircle and with AB side, intersect Li on incenter of the ABC triangle

 :lol:


Prove that if
$ x\sqrt{y^2 \plus{} z^2 \minus{} x^2} \plus{} y\sqrt{z^2 \plus{} x^2 \minus{} y^2} \plus{} z\sqrt{x^2 \plus{} y^2 \minus{} z^2} \equal{} 0$

Then
$ (x \plus{} y \plus{} z)(\minus{}x \plus{} y \plus{} z)(x \plus{} y \minus{} z)(x \minus{} y \plus{} z) \equal{} 0$

hello, i think you mean
$ (x^2\plus{}y^2\minus{}z^2)(\minus{}y^2\minus{}z^2\plus{}x^2)(y^2\plus{}x^2\plus{}z^2)(z^2\plus{}x^2\minus{}y^2)\equal{}0$
this follows from
$ x\sqrt{y^2\plus{}z^2\minus{}x^2}\plus{}y\sqrt{z^2\plus{}x^2\minus{}y^2}\plus{}z\sqrt{x^2\plus{}y^2\minus{}z^2}\equal{}0$
Sonnhard.

No, the problem refers to
$ (x\plus{}y\plus{}z)(\minus{}x\plus{}y\plus{}z)(x\plus{}y\minus{}z)(x\minus{}y\plus{}z) \equal{} 0$

obviously, $ x^2\plus{}y^2\minus{}z^2\ge 0$,$ z^{2} \plus{} x^{2} \minus{} y^{2}\ge 0$,$ y^{2} \plus{} z^{2} \minus{} x^{2}\ge0$
$ x\sqrt {y^{2} \plus{} z^{2} \minus{} x^{2}} \plus{} y\sqrt {z^{2} \plus{} x^{2} \minus{} y^{2}} \plus{} z\sqrt {x^{2} \plus{} y^{2} \minus{} z^{2}} \equal{} 0$
$ x\sqrt {y^{2} \plus{} z^{2} \minus{} x^{2}} \plus{} y\sqrt {z^{2} \plus{} x^{2} \minus{} y^{2}} \equal{} \minus{} z\sqrt {x^{2} \plus{} y^{2} \minus{} z^{2}}$
$ (x\sqrt {y^{2} \plus{} z^{2} \minus{} x^{2}} \plus{} y\sqrt {z^{2} \plus{} x^{2} \minus{} y^{2}})^2 \equal{} (z\sqrt {x^{2} \plus{} y^{2} \minus{} z^{2}})^2$
$ \therefore (z^2 \plus{} y^2 \minus{} x^2)(z^2 \plus{} x^2 \minus{} y^2) \equal{} \minus{} 2xy\sqrt {(y^{2} \plus{} z^{2} \minus{} x^{2})(z^{2} \plus{} x^{2} \minus{} y^{2})}$
$ \therefore 0\ge xy$
similarly, $ 0\ge yz$,$ 0\ge zx$
hence one of $ x,y,z$ equal to zero,
WLOG, let $ x\equal{}0$,
putting $ x\equal{}0$ in the initial equation, we have $ y\equal{}0$,$ z\equal{}0$
hence $ x\equal{}y\equal{}z\equal{}0$


If $ x,y,z$ are integers, solve $ xy(x^2\plus{}y^2)$=$ z^4$.

[hide]First note that either of x,y=0, z=0 is a solution.  We will show that there are no more solutions i.e. no solutions with all three nonzero. We can assume WLOG that x,y are both positive as opposed to both negative.  Since the equation is homogenous, we can assume that gcd(x,y,z)=1.  This also implies x and y are pairwise relatively prime, since if p|x, p|y, then p|$ z^4$ so p|z.  Similarly, x and y are relatively prime to $ x^2\plus{}y^2$.  Thus, all of x, y, and $ x^2\plus{}y^2$ are fourth powers.  Let $ x\equal{}k^4$, $ y\equal{}l^4$.  Then $ k^4l^4(k^8\plus{}l^8)\equal{}z^4$, so $ \frac{z^4}{k^4l^4}$ is an integer, so $ \frac{z}{kl}$ is an integer.  Let $ \frac{z}{kl}\equal{}n$.  Then $ k^8\plus{}l^8\equal{}n^4$, which is a violation of Fermat's Last Theorem for an exponent of 4.[/hide]

Right :!: Now try solving $ xy(x^{2}\plus{}y^{2})$=$ 2z^4$.


function
algebra
functional equation
real analysis
real analysis unsolved

It is well-know that the only continuous function $ f : \mathbb{R} \to \mathbb{R}$ such that
$ (\mathcal P) :$ for all $ x,y$, $ f(x\plus{}y)\equal{}f(x)f(y)$ and $ f(1)\equal{}e$
is $ f : x\mapsto e^x$.
If we replace "continuous function" by "monotonic function", the same conclusion holds.
We can also construct an everywhere-discontinuous function different than $ x\mapsto e^x$ that satisfies $ (\mathcal P )$. So, here is my question:
What happens if $ f$ is only supposed continuous on a smaller subset of $ \mathbb R$ (maybe even one point only)?

more general statement:
if $ f$ is bounded from above or from bellow on $ (a,b)$, and $ f(x\plus{}y)\equal{}f(x)\plus{}f(y)$ then $ f(x)\equal{}ax$.
so if your function is continuous in at least one point then it is exponential.

[quote="anik_andrew"]more general statement:
if $ f$ is bounded from above or from bellow on $ (a,b)$, and $ f(x \plus{} y) \equal{} f(x) \plus{} f(y)$ then $ f(x) \equal{} ax$.[/quote]
Nice, do you have a proof or reference?
I guess the "$ a$" in "$ (a,b)$" and in "$ f(x) \equal{} ax$" are not necessarily the same, right?

A function satisfying $ f(x \plus{} y) \equal{} f(x) f(y)$ continuous at one point must necessarily be continuous everywhere using the functional equation.  This equation is closely related to the [url=http://en.wikipedia.org/wiki/Cauchy%27s_functional_equation]Cauchy functional equation[/url], which describes other hypotheses that also guarantee the "right" solution.


calculus
integration
algebra
function
domain
linear algebra
matrix

[quote="The Captain"]Well I intend to post a question which [i]may[/i] be considered elementary, so [i]everyone's[/i] invited to think on it and give his\her comments. Then I will add another condition on it to make it more advanced.

Find an example of a ring $R$, which does not have a zero divisor and which is not a division ring, or prove that such a ring does not exist.
[hide="Hint"]Can $R$ be finite?.[/hide][/quote]

Take for example $R$ to be any entire ring, then $R[x]$ would also be entire, but not a ring with division.

If you want an example with a finite ring, i guess you can use something like
$\mathbb{Z}_{6}/(2)$ - it's entire, since the ideal $(2)$ is prime in $\mathbb{Z}_{6}$ and not field, since the ideal $(2)$ isn't maximal in $\mathbb{Z}_{6}$, namely $(2) \in (4)$.

[quote="ZetaX"]And $(4) \subset (2)$ ;)
There is no finite counterexample.[/quote]

ups, sorry.

As I'm eagerly waiting for other examples, let me note that why we don't have a finite example (or in your words counterexample). It is easily proved that any finite ring in which cancellation rules hold, is a field. That's why I gave it as a hint.
@eugene: Do you think that any enitire ring works? Give me a definition of an entire ring if neccessary.  :) Thanks.

[quote="The Captain"]
@eugene: Do you think that any enitire ring works? Give me a definition of an entire ring if neccessary.  :) Thanks.[/quote]

Yes, a ring $R$ si called entire if it has no non-zero zero divisiors, it may also be called an integral domain.

No, integral domains are to my knowledge with unity and are commutative, but we really have the stronger fact:

If $a,b \neq 0 \implies ab \neq 0$ and the ring is finite (and not necessary with unit nor commutative), then we have a field.

A proof is quite hard, you have to show that it is a division ring/skew field (easy task) and then to use Wedderburn's theorem (harder task).

[quote="ZetaX"]we really have the stronger fact:

If $a,b \neq 0 \implies ab \neq 0$ and the ring is finite (and not necessary with unit nor commutative), then we have a field.

A proof is quite hard, you have to show that it is a division ring/skew field (easy task) and then to use Wedderburn's theorem (harder task).[/quote]

That is in contradiction with my saying that the proof is easy :lol: I remember proving it this way; $0\neq a\in R,\;Ra=R$ it the yields the

fact that $R$ contains an identity and any nonzero element possesses an inverse, hence $R$ is a field.

Now I want to add being artinian to the conditions above, ie Find an example of an artinian ring $R$, which does not have a zero divisor and which is not a division ring, or prove that such a ring does not exist.

I don' think that eugene can modify his example, in this case. Even if he considers the quotient ring. :)

Yesterday, I asked the same question from my instructor and he proved( :maybe: ) that such a ring does not exist, using several advanced tools, such as Jacobson radical and semi-simple rings. I'm still looking for a simple proof.

Here's a proof using those tools you mentioned:

Denote the Jacobson radical of $R$ by $J$. Since $R$ and $R/J$ "have the same units", in the sense that $a\in R$ projects onto a unit of $R/J$ iff it's a unit in $R$, nothing changes in our problem when we factor through $J$, so we may as well assume that $R$ has trivial Jacobson radical. Being (left, say) Artinian, it's semisimple. The Wedderburn-Artin theorem now says that it must be a finite direct product of matrix rings over division rings, and it's easy to conclude from this and the fact that it has no zero divisors that it must actually be a division ring.

In Lam's "A First Course in Noncommutative Rings" I once saw this exercise:

Prove that if the (unital) ring $R$ is left Artinian and $a\in R$ is not a right zero divisor, then it's a unit.

I remember solving it, but I can't remember my proof, or whether it was "simple", in the sense of not using any of the tools above. I'll try to reconstruct it, and I'll modify this message if I manage to.

[quote="grobber"]Prove that if the (unital) ring $ R$ is left Artinian and $ a\in R$ is not a right zero divisor, then it's a unit.[/quote]
Consider the chain of ideals of the form $ Ra^n$. Now $ Ra^n \equal{} Ra^{n\plus{}1}$, so $ a^n \equal{} ra\cdot a^n$. Since $ a$ is not a right zero divisor, $ 1 \equal{} ra$. The following theorem due to Kaplansky finishes the proof:

If $ 1 \equal{} ra$, then $ (1\minus{}ar)\cdot a \equal{} 0$ and so $ ( a^j\cdot (1\minus{}ar) \plus{} r ) \cdot a \equal{} 1$. Hence if $ r$ is not a two-sided inverse to $ a$, then $ a$ has infinitely many inverses. In particular, in a finite ring or an integral domain, left inverses are right inverses.


In checking the petty cash a clerk counts $ q$ quarters, $ d$ dimes, $ n$ nickels, and $ c$ cents.  Later he discovers that $ x$ of the nickels were counted as quarters and $ x$ of the dimes were counted as cents.  To correct the total obtained the clerk must:

$ \textbf{(A)}\ \text{make no correction} \qquad
\textbf{(B)}\ \text{subtract 11 cents} \qquad
\textbf{(C)}\ \text{subtract 11}x\text{ cents} \\
\textbf{(D)}\ \text{add 11}x\text{ cents} \qquad
\textbf{(E)}\ \text{add }x\text{ cents}$

[quote="brightzhu"][hide]
For the miscounted nickels, there were an extra $ 20x$ cents.
For the miscounted dimes, there were $ 9x$ less cents.
So the correction is adding $ 11x\text { cents} \implies \boxed{D}$.[/hide][/quote]If a nickel was counted as quarter, you'll need to
[hide]subtract 20 cents ... to correct the total because the nickel IS the physical coin and mistreated as a quarter no?[/hide]
or am I just making the wrong interpretation?

If you count a nickel as a quarter, that means that instead of adding 5 cents you added 25 cents, so you would have to substract 20x cents. If you count a dime as a cent you are adding 1 cent instead of ten, so you would have to add 9x cents. -20x + 9x = -11x, so the answer is subtract 11x cents (C).


Hi! I have inequality problem which is actually easy. but i cannot solve it. I don't want a substitution solution please.
Show me AM-GM, Cauchy , or whatever but not substitution.
Anyways the problem is this:

Given
$ x\plus{}y\equal{}3$

Prove that

$ \frac{1}{4\minus{}\sqrt{x}} \plus{} \frac{1}{4\minus{}\sqrt{y}} \leq \frac{8\minus{}\sqrt{3}}{16 \minus{}4\sqrt{3}}$

[quote="shaaam"]
Given
$ x \plus{} y \equal{} 3$

Prove that

$ \frac {1}{4 \minus{} \sqrt {x}} \plus{} \frac {1}{4 \minus{} \sqrt {y}} \leq \frac {8 \minus{} \sqrt {3}}{16 \minus{} 4\sqrt {3}}$[/quote]
Try $ x\equal{}y\equal{}1.5$ :wink:

hello, i think it must be $ \frac{1}{4\minus{}\sqrt{x}}\plus{}\frac{1}{4\minus{}\sqrt{y}}\le \frac{2}{29}(8\plus{}\sqrt{6})$.
Sonnhard.

okay. Arqady is right.. the correct form is this :

$ a\plus{}b\equal{}3$

prove that

$ \frac{1}{4\minus{}\sqrt{x}} \plus{} \frac{1}{4\minus{}\sqrt{y}} \leq \frac{2}{4 \minus{} \sqrt{\frac{3}{2}}}$

please help...

Use Jensen :
$ f(x) \plus{} f(y) \le 2 f(\frac {x \plus{} y}{2}) \equal{} \frac {2}{4 \minus{} \sqrt {\frac {3}{2}}}$

Where $ f(x) \equal{} \frac {1}{4 \minus{} \sqrt {x}}$

hello, $ \frac{2}{29}(8\plus{}\sqrt{6})$ and $ \frac{2}{4\minus{}\sqrt{\frac{3}{2}}}$ are the same terms.
Sonnhard.


What is the sum of all palindromes with 1-9 digits?

wholy **** (u choose the swear :P ) ... is this even possible at mcounts level or any level?

-jorian

Yes, it's actually rather simple.

[hide="hint"]use something similar to what Gauss did to find the sum of the first 100 #s.[/hide]

do you have to add up each case? (the palindromes with 1 digit, the palindromes with two digits, etc).

Yes-there's really only one thing to realize in this problem.

[quote="13375P34K43V312"]What do you mean by 1-9 digits? Do you mean the sum of all one digit, two digit, ..., nine digit palindromes? That would take a long time...although not too hard.[/quote]
I've seen somthing similar to this before.
Here is a hint to all you people out there.
[hide="hint"]Find the average[/hide]

[hide="random guess"]
The smallest is 1 and the biggest is 999,999,999. Their average is 500,000,000. Their are 9 palindromes each for 1 and 2 digits, 90 for 3 and 4 digits, 900 for 5 and 6 digits, 9000 for 7 and 8 digits, and 90000 for 9 digits, so there are a total of 109998 palindromes. Multiply by 500,000,000 to get 54,999,000,000,000.[/hide]

[quote="binonunquineist"][hide]I just did a bunch of arithmetic series sums

First add all the 1 digits, then 2 digits, then 3 digits, so on

I got [b]545045045045040[/b][/hide][/quote]
:gasp: Kevin Chen has returned!

[quote="13375P34K43V312"][quote="binonunquineist"][hide]I just did a bunch of arithmetic series sums

First add all the 1 digits, then 2 digits, then 3 digits, so on

I got [b]545045045045040[/b][/hide][/quote]

How are you sure that they are arithmetic series? I don't think they are...

THE KING IS BACK like Jay-Z[/quote]
lol
[hide="a point"]The sum of the first and last number out of the n digit palindromes is the mean for the n digits.[/hide]


If:

$\displaystyle n^2+1 \geq (t_1+t_2+.....+t_n)(\frac{1}{t_1}+\frac{1}{t_2}+.....+\frac{1}{t_n})$

Prove that $t_i,t_k,t_j$ are sides of a triangle for all $i \ne k \ne j$
$i,k,j \leq n$

Check the IMO forum, it's been solved in many ways there.


Θα σας φορτωθώ λιγάκι παραπάνω απόψε!
Παρακαλώ πολύ αν δεν σας είναι κόπος ρίξτε και μια ματιά σ΄αυτήν.

Να αποδειχθεί ότι η σειρά
$ \sum_{n\equal{}0}^{\infty}{(\minus{}1)}^n \frac{x^2\plus{}n}{n^2}$

συγκλίνει ομοιόμορφα σε κάθε φραγμένο διάστημα, αλλά δεν συγκλίνει απόλυτα για καμία τιμή του χ (δηλαδή για κανένα
πραγματικό αριθμό)

Hehehe, Spinos μου έχεις δώσει τον τέλειο τρόπο για να σκοτώσω λίγο την ώρα μου μέχρι ν'αρχίσει το σινεμαδάκι μου :)

Λοιπόν, έχουμε και λέμε:

[hide]Η σειρά δε συγκλίνει απολύτως εξαιτίας του όρου $ \frac {n}{n^2} \equal{} \frac {1}{n}$ που κάνει το δεύτερο μισό της αρμονικό και, άρα, αποκλίνον.

Τώρα, για την ομοιόμορφη σύγκλιση σε φραγμένα διαστήματα, αρκεί να παρατηρήσεις ότι η σειρά είναι της μορφής $ A_n x^2 \plus{} B_n$ όπου $ A_n, B_n$ συγκλίνουσες ακολουθίες. Αφού μιλάμε για φραγμένα διαστήματα, η $ \sup$-απόσταση τις σειράς από τα μερικά της αθροίσματα θα φράσσεται εύκολα από το max της $ x^2$ στο εν λόγω διάστημα και την ποσότητα $ |A_n \minus{}\lim_{k\to\infty}A_k|$ την οποία και ελέγχεις.
[/hide]

Είμαι λίγο τηλεγράφημα επειδή βιάζομαι, αν θέλεις περισσότερη λεπτομέρεια let me know.

Cheerio,

Durandal 1707

Edit: διόρθωσα τις χαζομάρες που έλεγα περί ομοιόμορφης σύγκλισης επειδή δεν έβγαζαν νόημα (ούτε καν σε εμένα ;) )

Καλά Durandal ,είσαι φοβερός πριν λίγο το έγραψα, πότε πρόλαβες και το έλυσες!
Το πρώτο  με την απόλυτη σύγκλιση το κατάλαβα το έιχα σκεφτει και γω για να μη σου λέω και ψέματα,
αλλά στα επόμενα δυσκολεύτικα να σε παρακολουθήσω.
Γι΄αυτό αν θέλεις όποτε έχεις βρεις λίγο χρόνο κάνε μου τα πιο αναλυτικά,
δεν τα ξέρω και τόσο καλά να τα δουλεύω αυτά ή
πες μου τι θεωρήματα χρησιμοποιείς για να κάτσω να διαβάσω.
Σ΄ευχαριστώ και πάλι για τις συμβουλές σου, καλη διασκέδαση!
καλό βράδυ

Gia na dikseis oti den sigklinei apolita apla pernoume me to kritirio tis sigkriseis oti 1/n = <(x^2+n)/n^2
Twra gia to oti sigklinei omiomorfa i seira , iparxei ena theorima poli kalo to opoio leei oti : An oles oi fn (=(-1)^n*(x^2+n)/n^2 stin periptosi mas) exoun sinexi paragogo , i seira fn sigklinei gia ena toulaxiston x sto diastima [a,b] kai i seira twn paragogwn tis fn sigklinei omiomorfa sto [a,b] tote  i seira fn sigklinei omiomorfa sto [a,b] pros mia sinartisi s , i s einai paragogisimi sto [a,b] kai i paragogos tis s sigklinie omiomorfa stin seira twn paragogwn tis fn

An kaneis mia apli efarmogi aftou mazi me to kritirio tou Wiestrass(tha sou xriastei stin seira tis paragogou tis fn) tha vgaleis to zitoumeno

P.S. sorry alla den kserw na xrisimopoiw latex wste na to grapsw pio omorfa alla pistevw na einai katanoita afta pou egrapsa

Spinos, να σου εξηγήσω τι εννοώ γιατί ήταν γραμμένο εντελώς βιαστικά!

Έστω $ D$ το φραγμένο σου σύνολο κ' έστω $ f_n(x) \equal{} A_n x^2 \plus{} B_n$ όπου $ A_n, B_n$ τα μερικά αθροίσματα των (συγκλίνουσων) σειρών σου. Τότε, αυτό που θέλεις να φράξεις με κάποιο $ \epsilon>0$ είναι το:

\[ \sup_{x\in D}|f_n(x) \minus{} f(x)| \equal{} \sup_{x\in D}|A_n x^2 \plus{} B_n \minus{} Ax^2 \minus{}B| \leq\\
\leq\sup_{x\in D}|(A_n \minus{} A)x^2| \plus{} \sup_{x\in D}|B_n \minus{} B|\]

Ο δεύτερος όρος φράσσεται πολύ εύκολα (δίαλεξε n αρκετά μεγάλο ώστε να είναι μικρότερος από το $ \epsilon /2$) και ο πρώτος φράσσεται αν διαλέξεις n αρκετά μεγάλο ώστε να είναι μικρότερος από το $ \frac{\epsilon}{2L^2}$ όπου $ L^2$ η μέγιστη τιμή της $ x^2$ στο φραγμένο $ D$.

Αυτό εννοούσα, αλλά που να καταλάβεις έτσι όπως το έγραψα :D

Cheerio,

Durandal 1707

PS: Hehehe, πάνω που πήγαινα να γράψω ότι ο Sorrowman θα ξέρει οπωσδήποτε κάποιο θεώρημα που να σκοτώνει την άσκηση γρήγορα και χωρίς να χρειάζεται να λερώσεις τα χέρια σου, είδα την απάντησή του :)

[quote="Durandal"]Spinos, να σου εξηγήσω τι εννοώ γιατί ήταν γραμμένο εντελώς βιαστικά!

Έστω $ D$ το φραγμένο σου σύνολο κ' έστω $ f_n(x) \equal{} A_n x^2 \plus{} B_n$ όπου $ A_n, B_n$ τα μερικά αθροίσματα των (συγκλίνουσων) σειρών σου. Τότε, αυτό που θέλεις να φράξεις με κάποιο $ \epsilon > 0$ είναι το:
\[ \sup_{x\in D}|f_n(x) \minus{} f(x)| \equal{} \sup_{x\in D}|A_n x^2 \plus{} B_n \minus{} Ax^2 \minus{} B| \leq \\
\leq\sup_{x\in D}|(A_n \minus{} A)x^2| \plus{} \sup_{x\in D}|B_n \minus{} B|
\]
Ο δεύτερος όρος φράσσεται πολύ εύκολα (δίαλεξε n αρκετά μεγάλο ώστε να είναι μικρότερος από το $ \epsilon /2$) και ο πρώτος φράσσεται αν διαλέξεις n αρκετά μεγάλο ώστε να είναι μικρότερος από το $ \frac {\epsilon}{2L^2}$ όπου $ L^2$ η μέγιστη τιμή της $ x^2$ στο φραγμένο $ D$.

Αυτό εννοούσα, αλλά που να καταλάβεις έτσι όπως το έγραψα :D

Cheerio,

Durandal 1707

PS: Hehehe, πάνω που πήγαινα να γράψω ότι ο Sorrowman θα ξέρει οπωσδήποτε κάποιο θεώρημα που να σκοτώνει την άσκηση γρήγορα και χωρίς να χρειάζεται να λερώσεις τα χέρια σου, είδα την απάντησή του :)[/quote]

looooooool kala xalarwse den kserw (kai kuriws oxi apeksw ) , oti theorima kai kritirio kikloforei! Wraia lisi sou pantws , apla mia paratirisi.. sto e/2*L^2 giati thelises na to grapseis etsi?katefthian e/2 na teleiwneis wste e/2+e/2 = e kai na vgei i omiomorfi sigklisi.....

Βασικά, χρειάζεσαι το $ L^2$ για να φράξεις τον όρο με το $ x^2$. Αναλυτικά, αυτό που εννοώ είναι ότι:

\[ \sup_{x\in D}|(A_n\minus{}A)x^2| \leq L^2\cdot\sup_{x\in D}|A_n\minus{}A| \equal{} L^2 \cdot |A_n\minus{}A|\]

οπότε χρειάζεσαι ένα $ \epsilon/2L^2$ για να φράξεις την τελευταία προσέγγιση.

Βγάζει καλύτερο νόημα τώρα?

Cheerio,

Durandal 1707

[quote="Durandal"]Βασικά, χρειάζεσαι το $ L^2$ για να φράξεις τον όρο με το $ x^2$. Αναλυτικά, αυτό που εννοώ είναι ότι:
\[ \sup_{x\in D}|(A_n \minus{} A)x^2| \leq L^2\cdot\sup_{x\in D}|A_n \minus{} A| \equal{} L^2 \cdot |A_n \minus{} A|
\]
οπότε χρειάζεσαι ένα $ \epsilon/2L^2$ για να φράξεις την τελευταία προσέγγιση.

Βγάζει καλύτερο νόημα τώρα?

Cheerio,

Durandal 1707[/quote]

egw exw vgalei noima eksarxis  apla gia tixon pareksigisi to eipa to parapano

Παιδιά σας ευχαριστώ πολύ και τους δυό σας για τις συμβουλές σας,
και βασικά για την λύση και των 2 ασκήσεων που έβαλα Durandal!
αλλά πέρα από την λύση τις ασκησης απόλαυσα και την συζήτηση που κάνατε μεταξύ σας,
ήταν αποκαλυπτική στο να καταλάβει κάποιος ότι ένας γνήσιως μαθηματικός έχει ανάγκη να βλέπει την ουσία στα πράγματα,
δεν του αρκεί μια φορμαλιστική απόδειξη.
Σας ευχαριστώ πολύ και πάλι καληνύχτα!


geometry
geometric transformation
rotation
homothety
linear algebra
matrix
trigonometry

Hello,

I am trying to study combined point transformation.

I tried to rotate and then scale a point about origin in a cad package. Here, the sequence does not make any difference. Whether you do scaling first or rotation, the results are correct.

Then I set to get the mathematical representations of this. And here is what I have:

---

To transform the point P to P’ such that distance from origin is increased by S times and the angle the line joining point with origin is increased by t.

Sequence 1: Rotation followed by scaling:

Rotation:

x’’2 = x1 * cos(t) – y1 * sin(t)
y’’2 = x1 * sin(t) + y1 * cos(t)

Scaling

x2 = Sx * x’’2
y2 = Sy * y’’2

x2 = Sx * (x1 * cos(t) – y1 * sin(t))
y2 = Sy * (x1 * sin(t) + y1 * cos(t))

Sequence 2: Scaling followed by rotation:

Scaling

X’’2 = Sx * x1
Y’’2 = Sy * y1

Rotation:

x2 = x’’2 * cos(t) – y’’2 * sin(t)
y2 = x’’2 * sin(t) + y’’2 * cos(t)

x2 = Sx * x1 * cos(t) – Sy * y1 * sin(t)
y2 = Sx * x1 * sin(t) + Sy * y1 * cos(t)

----

The results are affected by sequence of transformations. Is this correct or have I made a mistake somewhere? If this is correct then what is the source of discrepancy?. If this is correct then can I come to the conclusion that scaling should always be applied before translation? Or non-uniform scaling should be avoided as far as possible?

I could not find any convincing solution solution in the couple of books that I have nor on the web.

Please help me.

uhm...
why did you multiply $x$ by $S_x$ and $y$ by $S_y$? in an homothety, $S_x = S_y$...  **
if you set $S_x = S_y = S$, you have that both transformation give the same result.

** in order to prove this, just apply the transformation to points $(0,1)$ and $(1,0)$...

The matrix $\begin{bmatrix}a&0\\0&b\end{bmatrix}$, for $a\neq b$, is not a multiple of the identity, and therefore it doesn't commute with everything- in particular it doesn't commute with the rotation matrix $\begin{bmatrix}\cos\theta&\sin\theta\\-\sin\theta&\cos\theta\end{bmatrix}.$

Obviously, scaling in your CAD package is a multiple of the identity, with one scaling parameter.

I am studying Maths required for Solid Modelling (Transformations is a part of it) and preparing detailed notes for my future reference. Thus I need to investigate all possible aspects.

I am not a mathematician. Can you explain a bit more on the following terms:
multiple of the identity
commute

By the way, are there any general rules / guidelines to handle combined transformation involving scaling so that such pitfalls are avoided?

Thanks for any help.

Have you had linear algebra? That would be the mathematical foundation for what you're doing, since the operations here are linear transformations. If you haven't had linear algebra, do it.

A linear transformation (which fixes the origin) can be represented as multiplication by a matrix: $x=\begin{bmatrix}x_1\\x_2\\\vdots\\x_n\end{bmatrix}$ and $T_A(x)=Ax=\begin{bmatrix}a_{11}&a_{12}&\cdots&a_{1n}\\a_{21}&a_{22}& \cdots&a_{2n}\\\vdots&\vdots&\ddots&\vdots\\a_{n1}&a_{n2}&\cdots&a_{nn} \end{bmatrix}\cdot\begin{bmatrix}x_1\\x_2\\\vdots\\x_n\end{bmatrix}= \begin{bmatrix}\sum_ia_{1i}x_i\\\sum_ia_{2i}x_i\\\vdots\\\sum_ia_{ni}x_i \end{bmatrix}$

Composition of linear transformations is matrix multiplication: $T_A(T_B(x))=A\cdot(Bx)=(AB)\cdot x$. The identity for matrix multiplication is $I$, the matrix with each diagonal element 1 and each off-diagonal element zero. Matrix multiplication is not necessarily commutative, and indeed the only matrices which commute with everything are the constant multiples of the identity.


Solve the following equation:
\[ (z\plus{}i)(z\minus{}1)^2\equal{}4\]

[hide="Hint"]\[ (z\plus{}i)(z\minus{}1)^2\minus{}4 \equal{} (z\minus{}i) \left(z^2\minus{}(2\minus{}2 i) z\minus{}(1\plus{}4 i)\right)\][/hide]


Please explain it more clearier

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=81362

About Megus's solution.

Why
[quote="Megus"]Plug $m=1$ and we get recurrence $f(m+1)=2f(m)+2$ from which we obtain:

$f(n)=2^{n-1}\cdot 5-2$ plugging this back into our original equation we arrive at contradiction.[/quote]

From $f(m+1)=2f(m)+2$ Why we can obtain $f(n)=2^{n-1}\cdot 5-2$

Why we obtain

You should be able to prove it by induction.


linear algebra
matrix
linear algebra unsolved

Prove that a Jordan bloc J(a,n) is a product of a real symmetric matrix and a complex symmetric matrix.

Take $A = \begin{pmatrix}0 & 0 & \ldots & 0 & 1 \\ 0 & 0 & \ldots & 1 & 0 \\ \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & 1 & \ldots & 0 & 0 \\ 1 & 0 & \ldots & 0 & 0 \end{pmatrix}$ and $B = \begin{pmatrix}0 & 0 & \ldots & 0 & \lambda \\ 0 & 0 & \ldots & \lambda & 1 \\ \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & \lambda & \ldots & 0 & 0 \\ \lambda & 1 & \ldots & 0 & 0 \end{pmatrix}$.
Then $A,B$ are symmetric and $AB$ is the desired Jordan block.


geometry
circumcircle
geometric transformation
reflection
trapezoid
homothety
perpendicular bisector

Let $A_b$ and $A_c$ be two points on the sideline BC of triangle ABC, let $B_c$ and $B_a$ be two points on the sideline CA, and let $C_a$ and $C_b$ be two points on the sideline AB. Assume that the six points $A_b$, $A_c$, $B_c$, $B_a$, $C_a$, $C_b$ lie on one circle.

Let X, Y, Z be the circumcenters of triangles $AB_aC_a$, $BC_bA_b$ and $CA_cB_c$, respectively. Prove that the perpendiculars from the points X, Y, Z to the lines BC, CA, AB concur.

  Darij

Excellent! I think it took ten minutes from me.
Let $X'$, $Y'$ and $Z'$ be foot of perpendiculars. It is well known that perpendiculars to sides $BC$, $CA$ and $AB$ at points $X'$, $Y'$ and $Z'$ are concurrent iff
\[BX'^2-X'C^2+CY'^2-Y'A^2+AZ'^2-Z'B^2=0.\quad\quad(*)\]
We know that $XX'$ and bisectors of $AB_a$ and $AC_a$ intersect in common point $X$. Using mentioned criteria we obtain that
\[BX'^2-X'C^2+(AC_a/2)^2-(AB-AC_a/2)^2+(AC-AB_a/2)^2+(AB_a/2)^2=0.\]
It can be rewritten as
\[BX'^2-X'C^2+AB\cdot AC_a-AC\cdot AB_a=0.\quad\quad(**)\]

Since $AB$ and $AC$ are secants to circle $(B_aB_cC_aC_b)$ we conclude $AC_a\cdot AC_b=AB_a\cdot AB_c$, hence, using identities $AC=AB_c+B_cC$ and $AB=AC_b+C_bB$, we conclude from $(**)$
\[BX'^2-X'C^2+BC_b\cdot AC_a-CB_c\cdot AB_a=0.\]
Summing up such three equalities we obtain $(*)$

This solution is similar to the first solution I found. But the really cool thing is that there is a simple solution without any calculation!

In fact, let K, L, M be the reflections of the points A, B, C in the circumcenters X, Y, Z of triangles $AB_aC_a$, $BC_bA_b$, $CA_cB_c$, respectively. Then, of course, the segments AK, BL, CM are diameters in the circumcircles of triangles $AB_aC_a$, $BC_bA_b$, $CA_cB_c$. Hence, for instance, $\measuredangle BA_bL=90^{\circ}$, so that $A_bL\perp BC$. Similarly, $A_cM\perp BC$.

Let K', L', M' be the midpoints of the sides LM, MK, KL of triangle KLM, let $K_1$, $L_1$, $M_1$ be the midpoints of the segments $A_bA_c$, $B_cB_a$ and $C_aC_b$, and let O be the center of the circle through the points $A_b$, $A_c$, $B_c$, $B_a$, $C_a$, $C_b$. Since the center of a circle lies on the perpendicular bisector of any chord, the point O lies on the perpendicular bisector of the segment $A_bA_c$; in other words, $OK_1\perp A_bA_c$, or, equivalently, $OK_1\perp BC$.

But since $A_bL\perp BC$ and $A_cM\perp BC$, the quadrilateral $A_bLMA_c$ is a trapezoid with bases perpendicular to the line BC. Hence, its mid-parallel $K^{\prime}K_1$ is also perpendicular to the line BC. But on the other hand, we know that $OK_1\perp BC$. Hence, the three points K', $K_1$ and O lie on one line perpendicular to the line BC. This yields $OK^{\prime} \perp BC$.

Now, since K', L' and M' are the midpoints of the sides LM, MK, KL of triangle KLM, the triangle K'L'M' is homothetic to the triangle KLM; thus, there exists a homothety which maps the points K', L', M' to the points K, L, M. If U is the image of the point O under this homothety, then UK || OK' (from the properties of a homothety). But $OK^{\prime} \perp BC$; thus, $UK\perp BC$. [Similarly, of course, $UL\perp CA$ and $UM\perp AB$.]

Now, we remember that the point K is the reflection of the point A in the point X. In other words, the point X is the midpoint of the segment AK. On the other hand, if H is the orthocenter of triangle ABC, and T is the midpoint of the segment HU, then the quadrilateral AHUK is a trapezoid with bases AH and UK perpendicular to the line BC, and thus, its mid-parallel XT is also perpendicular to the line BC. In other words, the point T lies on the perpendicular to the line BC through the point X. Similarly, the point T lies on the perpendiculars to the lines CA and AB through the points Y and Z as well, and consequently, the three perpendiculars concur. $\blacksquare$

Sorry for the slightly confusing writeup of the solution; I hope you were able to understand it.  :)

  Darij

Maybe I will read it in next century... :maybe:

Okay. Here is a do-it-yourself version:

[b]1.[/b] Call K, L, M the reflections of A, B, C in X, Y, Z. You can easily see that the problem is equivalent to showing that the perpendiculars to BC, CA, AB through K, L, M concur.

[b]2.[/b] Call K', L', M' the midpoints of LM, MK, KL. Then you can reduce this to proving that the perpendiculars to BC, CA, AB through K', L', M' concur.

[b]3.[/b] But this is easy - they concur at the center of the circle through $A_b$, $A_c$, $B_c$, $B_a$, $C_a$, $C_b$!

  Darij

Here's a cool relation which I had no idea about:

$AC_a^2+BA_b^2+CB_c^2=AB_a^2+CA_c^2+BC_b^2$. It's actually pretty easy to prove, but I'm just saying that I had never seen it before.

[quote="darij grinberg"][b]1.[/b] Call K, L, M the reflections of A, B, C in X, Y, Z. You can easily see that the problem is equivalent to showing that the perpendiculars to BC, CA, AB through K, L, M concur.

[b]2.[/b] Call K', L', M' the midpoints of LM, MK, KL. Then you can reduce this to proving that the perpendiculars to BC, CA, AB through K', L', M' concur.

[b]3.[/b] But this is easy - they concur at the center of the circle through $A_b$, $A_c$, $B_c$, $B_a$, $C_a$, $C_b$![/quote]
I am sure that such kind of solutions is much more useful for olympiad training then extremely long and extremely detailed ones.
Moreover, here I can see beauty of your approach in contrast to your previous message.
I think you should look for compromise between details and lucidity, because long detailed solution doesn't guarantee its lucidity.

probably I misundertud the problem, but it seems to me pretty obvious.
Anyway, here it is because it happens.

We know, as X is the circumcenter of AAbAc, that X lies on the axis of AbAc, and similarly Y and Z lie on the axes of BaBc and CaCb respectively.

But the centers of all circles through AbAc lie on the axis of AbAc, and similarly for Y and Z. Then The concurrence point is the center of circle through the six points AbAcBaBcCaCb.

[quote="sprmnt21"]probably I misundertud the problem,[/quote]

Unfortunately yes  :(

The point X is the circumcenter of triangle $AB_aC_a$, not of triangle $AA_bA_c$.

  Darij


In the expansion of (1 + x)^12 which of the following is correct?

A) ∑(r = 0 to 4) 12C(3r) = 1362
B) ∑(r = 0 to 4) 12C(3r) = 1366
C) ∑(r = 0 to 4) (3r + 1).(12C(3r)) = 9562
D) ∑(r = 0 to 4) 12C(3r) = 1369

More than one options may be correct.
NOTE: 12C(3r) is 12 choose 3r

How can you have an answer that is a number when the binomial you're expanding is in terms of a variable?

The expansion of $ (1\plus{}x)^{12}$ is given by
$ \sum_{i\equal{}0}^{12}\binom{12}{i}(1)^{12\minus{}i}x^{i}\equal{}\sum_{i\equal{}0}^{12}\binom{12}{i}x^{i}$


Prove that $ C_{k}$ is odd iff k is expressible as $ 2^{n}\minus{}1$


Let A and B be two square matrices such that AB=BA.Prove that
(A+B)^n=A^n+nC1*A^(n-1)*B+......+B^n (n=2,3,4......)

Induction on $ n$ same as proof of Newton's identity.


Let a,b,c be positive number such that
 $ a^2 \plus{} b^2 \plus{} c^2 \equal{} \frac {3}{2}$
Prove that
$ \frac {a^3}{b^2 \plus{} c^2} \plus{} \frac {b^3}{c^2 \plus{} a^2} \plus{} \frac {c^3}{a^2 \plus{} b^2} \plus{} 2\sqrt {3(a^6 \plus{} b^6 \plus{} c^6)} \geq 3(a^3 \plus{} b^3 \plus{} c^3)$

$\sum \frac{a^3}{b^2+c^2} =\sum \frac{a^6}{a^3b^2+a^3c^2} \ge \frac{(a^3+b^3+c^3)^2}{\sum a^3(b^2+c^2)}$
Now note that
$2(a^5+b^5+c^5) \ge \sum_{sym} a^3b^2$
Thus,
\[(a^3+b^3+c^3)(a^2+b^2+c^2) \ge \frac32 \sum_{sym} a^3b^2 \\ \implies a^3+b^3+c^3 \ge \sum_{sym} a^3b^2\]
Thus we have
$\sum \frac{a^3}{b^2+c^2} \ge a^3+b^3+c^3$
and by c-s
$2\sqrt{3(a^6+b^6+c^6)} \ge 2(a^3+b^3+c^3)$
Summing up we get the result.
Equality holds when $a=b=c=\frac1{\sqrt2}, \blacksquare$


N {12,18,2,6}
P {1,4,2,3}
If n and p are to be selected at random from sets N and P respectively, what is the probability that n/2p will be a member of set P?

[hide]
Brute force finding only $ 6$ pairs of of $ (n,p)$ work and $ 16$ total pairs so
$ \boxed{\frac{6}{16}= \frac{3}{8}}$.
[/hide]

Yes, I got the Same as AstroPhys...is there a strategy for this....and i don't think this is SAT? Is it?

I've been a SAT teacher for over a year. While there are questions that do test knowledge about sets, I've never seen one similar to this. Set questions usually focus about intersection or union of sets. However, this could have been a question on an experimental section, or just a set question of higher difficulty that I've never seen before.


What is the value of $ \tan{1}\tan{2}\dots\tan{89}$?

If the input is in degrees, we're okay. You can even use cofunctions on this case. Note that the factors are as follows:

$ (\tan 1^\circ)(\tan 2^\circ)\ldots(\tan 45^\circ)\ldots(\tan 88^\circ)(\tan 89^\circ)$

Notice that the right side of $ \tan 45^\circ$ is equivalent to $ (\cot 44^\circ)\ldots...(\cot 2^\circ)(\cot 1^\circ)$

Taking the product of the left side and right side of $ \tan 45^\circ$ is a reciprocal identity, and leaves us with a single expression $ \tan 45^\circ$ which is equal to $ 1$.

Mabuhay!


inequalities
3-variable inequality
cyclic inequality
cauchy schwarz

Expanding and multiply each side by 2, we obtain\[2\sum\frac{a^2}{b^2}+2\sum\frac{b}{a}\ge6+2\sum\frac{a}{b}\]Now by AM-GM, $\frac{a^2}{b^2}+\frac{a^2}{b^2}+\frac{b}{a}\ge3\frac{a}{b}$. Now it suffices to prove $\sum\frac{a}{b}+\sum\frac{b}{a}\ge6$, which is obvious :)
btw, I think you'd better post inequality in the inequality forum, rather than algebra(though $\text{Inequality}\in\text{Algebra}$ ;) )

[color=red][edit]: (I think)post already moved to inequality by mod[/color]

This problem was given in the British Maths Olympiad 2005 as well.

We have from Cauchy Schwarz inequality:

$ \left(\frac ab +\frac bc +\frac ca\right)(ab+bc+ca) \geq  (a+b+c)^2 $  and

$ \left(\frac ab +\frac bc +\frac ca\right)\left(\frac {1}{ab} +\frac {1}{bc} +\frac {1}{ca}\right) \geq  \left(\frac 1b +\frac 1c +\frac 1a\right)^2 $
but
$ (ab+bc+ca)\left(\frac {1}{ab} +\frac {1}{bc} +\frac{1}{ca}\right)=(a+b+c)\left(\frac 1a +\frac 1b +\frac 1c\right)$

Hence multiplying the above two inequalities we get the desired result.

@ British MO: Indeed, see http://www.mathlinks.ro/Forum/viewtopic.php?t=29253 or http://www.mathlinks.ro/Forum/viewtopic.php?t=25355 .

  darij

Here is another solution for this inequality :D

By some obvoius calculations the inequality is equivalent with:

 $(\frac{a}{b}+\frac{b}{c}+\frac{c}{a})^{2}\geq (a+b+c)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}).$
 Anyway, by substitutions $x=\frac{a}{b}, y=\frac{b}{c}, z=\frac{c}{a}$ the inequality becomes

  $(x+y+z)^{2}\geq 3+x+y+z+\frac{1}{x}+\frac{1}{y}+\frac{1}{z}$ plus condition $xyz=1.$

 After all, the inequality is equivalent with this obvious one: $x^{2}+y^{2}+z^{2}+xy+yz+zx\geq 3+x+y+z.$ ;)

Put $x=\frac{a}{b}$ & $y,z$ are defined similarly.now multiplying & expanding we have to show $x^2+y^2+z^2+xy+yz+zx\ge 3+x+y+z$.now $xyz=1$ so am-gm gives $\sum_{cyc}xy\ge 3 (*_1)$ now consider $f(t)=t^2-t$ which is convex.so $ f(x)+f(y)+f(z)\ge3f(\frac{x+y+z}{3})$ again $f$ is increasing for $x\ge 0.5$ now am-gm again gives $x+y+z\ge 3$ so $f(\frac{x+y+z}{3})\ge f(1)=0$ which gives $x^2+y^2+z^2\ge x+y+z(*_2)$.adding $(*_1)$ & $(*_2)$ we have the reqd ineq.
Sorry for reviving this post but your comments will be helpful

[quote="jensen"][quote="Omid Hatami"]Suppose $a,b,c\in \mathbb R^+$. Prove that :\[(\frac ab+\frac bc+\frac ca)^2\geq (a+b+c)(\frac1a+\frac1b+\frac1c)\][/quote]
$(\frac{a}{b}+\frac{b}{c}+\frac{c}{a})^2$=$\frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{c^2}{a^2}+2\frac{b}{a}+2\frac{a}{c}+2\frac{c}{b}$
and $(a+b+c)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})$=$3+\frac{a}{b}+\frac{a}{c}+\frac{b}{a}+\frac{c}{a}+\frac{b}{c}+\frac{c}{b}$
$\rightarrow $ we want to prove that :
$\frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{c^2}{a^2}+2\frac{b}{a}+2\frac{a}{c}+2\frac{c}{b}$ $ \geq$ $3+\frac{a}{b}+\frac{a}{c}+\frac{b}{a}+\frac{c}{a}+\frac{b}{c}+\frac{c}{b}$ $rightarrow $
we want to prove this inequality:
$ \frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{c^2}{a^2}+
\frac{b}{a}+\frac{a}{c}+\frac{c}{b}$ $ \geq$ $3+
\frac{a}{b}+\frac{c}{a}+\frac{b}{c}$
we have:
$\frac{b}{a}+\frac{a}{c}+\frac{c}{b} \geq 3$(?)
so now we should prove that $ \frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{c^2}{a^2}$ $\geq$ $\frac{a}{b}+\frac{c}{a}+\frac{b}{c}$
we have:
$\frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{c^2}{a^2}$ $\geq$ $\frac{(\frac{a}{b}+\frac{c}{a}+\frac{b}{c})^2}{3}$(?) and we have $\frac{a}{b}+\frac{c}{a}+\frac{b}{c}$ $\geq$ $3$(?)
$\rightarrow $ $  \frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{c^2}{a^2}$ $\geq$ $\frac{a}{b}+\frac{c}{a}+\frac{b}{c}$
and now we r done.[/quote]
$  \frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{c^2}{a^2}$ $\geq$ $\frac{a}{b}+\frac{c}{a}+\frac{b}{c}$ please check this step.It is yet not done.

[quote="pirhoosh"][quote="Omid Hatami"]Suppose $a,b,c\in \mathbb R^+$. Prove that :\[ (\frac ab+\frac bc+\frac ca)^2\geq (a+b+c)(\frac1a+\frac1b+\frac1c)  \][/quote]

let's take it easy:
$\frac{a^2}{b^2} + \frac{b^2}{c^2} + \frac{c^2}{a^2}+ \frac{2a}{c} + \frac{2c}{b} + \frac{2b}{a}$ $\geq$ $1+1+1 +$ $\frac ab$ $+ \frac ac$ $+ \frac ba$ $+ \frac{b}{c}+ \frac{c}{b} + \frac{c}{a}$  now:
if we solve this , the problem will solved: $\frac{a^2}{b^2} + \frac{b}{a} - \frac{a}{b} -1 \geq 0;$
$(\frac{a^2}{b^2} + \frac{b}{a} - \frac{a}{b} -1 \geq 0) b^2a \longleftrightarrow a^3 + b^3 \geq a^2b + b^2a$[/quote]
yes the lemma is true and it's a very nice solution

I don't see any error  in Jansen's proof.Please tell me tat, where is the problem ?

hello SMOJ can you show your muirhead solution please?

[b][u][i]My Solution.[/i][/u][/b]

By subtracting $RHS$ from $LHS$ we get $-$
\\$\left(\sum\dfrac{a}{b}\right)^2-\left(\sum a\right)\left(\sum\dfrac{1}{a}\right)$
\\$=\sum\dfrac{a^2}{b^2}+2\left(\sum\dfrac{a}{c}\right)-\left[3+\left(\sum\dfrac{a}{b}+\sum \dfrac{a}{c}\right)\right]~.$
\\$=\sum\left(\dfrac{a^2}{b^2}-\dfrac{a}{b}\right)+\sum\dfrac{a}{c}-3~.$
\\$=\sum\left(\dfrac{a^2}{b^2}-\dfrac{a}{b}+\dfrac{1}{4}\right)+\sum\dfrac{a}{c}-\dfrac{15}{4}~.$
\\$=\sum\left(\dfrac{a}{b}-\dfrac{1}{2}\right)^2+\sum\dfrac{a}{c}-\dfrac{15}{4}.$
\\It follows from $AM-GM$ that $\sum\dfrac{a}{c}\ge 3~.~(1)$
\\And by $CS$ we get $-$
\\$\sum\left(\dfrac{a}{b}-\dfrac{1}{2}\right)^2\geq \dfrac{\left(\sum \dfrac{a}{b}-\dfrac{3}{2}\right)^2}{3}\xrightarrow{A\ge G}\sum\left(\dfrac{a}{b}-\dfrac{1}{2}\right)^2\geq \dfrac{\left(3-\dfrac{3}{2}\right)^2}{3}=\dfrac{3}{4}~.~(2)$
\\By adding $(1),(2)$ we get $\sum\left(\dfrac{a}{b}-\dfrac{1}{2}\right)^2+\sum\dfrac{a}{c}\geq 3+\dfrac{3}{4}=\dfrac{15}{4}~.$
\\Or, $\sum\left(\dfrac{a}{b}-\dfrac{1}{2}\right)^2+\sum\dfrac{a}{c}-\dfrac{15}{4}\geq 0~.$
\\Or, $\left(\sum\dfrac{a}{b}\right)^2-\left(\sum a\right)\left(\sum\dfrac{1}{a}\right)\geq 0$
\\Or, $\boxed{\left(\sum\dfrac{a}{b}\right)^2\geq \left(\sum a\right)\left(\sum\dfrac{1}{a}\right)}~.~\square$

[quote=sqing]http://www.artofproblemsolving.com/Forum/viewtopic.php?f=151&p=512906#p512906:
\[ \left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)^{2}\geq\frac{3}{2}\left(\frac{b+c}{a}+\frac{c+a}{b}+\frac{a+b}{c}\right)\geq\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right). \][/quote]

just  expanding and AM GM

Solution of Iran 2005
$\[\left(\frac ab+\frac bc+\frac ca\right)^2\geq (a+b+c)\left(\frac1a+\frac1b+\frac1c\right)\]\iff\frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{c^2}{a^2}+2\frac{a}{c}+2\frac{b}{a}+2\frac{c}{b}\ge 3+\frac{a+c}{b}+\frac{a+b}{c}+\frac{c+b}{a}\iff \frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{c^2}{a^2}+\frac{a}{c}+\frac{b}{a}+\frac{c}{b}\ge 3+\frac{a}{b}+\frac{b}{c}+\frac{c}{a} $ Which is true because by AM-GM $\frac{a}{c}+\frac{b}{a}+\frac{c}{b}\ge 3$ also we have $\frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{c^2}{a^2}\ge\frac{a}{b}+\frac{b}{c}+\frac{c}{a}$ because of Nordic Math Contest 1987 ;D

[quote=WolfusA]Let $a,b,c$ be positive numbers such that $$(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=\frac{49}{4}$$
Find all possible values of the expression $$E=\frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{c^2}{a^2}$$.
Problem O438 from Mathematical Reflections Issue 1 - 2018
Proposed by Marius Stănean, Zalău, România[/quote]
I think this problem has been released for a while.(I have seen before)

Let $M= \sum_{cyc}\frac{a}{b}$ and $N= \sum_{cyc}\frac{b}{a}$
From the given condition we have: $M+N= \frac{37}{4}$
We want to find the bound of possible values of $E= \sum_{cyc}\frac{a^2}{b^2}=M^2-2N=M^2+2M-\frac{37}{2}$.
Consider cubic polynomial $P(t)$ that have 3 positive real roots:$ x=\frac{a}{b} , y=\frac{b}{c} ,z=\frac{c}{a}$
$P(t)=t^3-Mt^2+Nt-1= t^3-Mt^2+(\frac{37}{4}-M)t-1$
Its discriminant = $((-M)(\frac{37}{4}-M))^2+18(-M)(\frac{37}{4}-M)(-1)-4(\frac{37}{4}-M)^3-4(-M)^3(-1)-27(-1)^2
=\frac{1}{16}(4M-17)(M-5)(4M^2-37M-601)$
must be nonnegative.
Because $0 \le M \le  \frac{37}{4}$ so $4M^2-37M-601=M(4M-37)-601 \le -601 < 0$
and $(4M-17)(M-5) \le 0 , M \in [\frac{17}{4},5]$
We need to check that all $M \in [\frac{17}{4},5]$ cause $P(t)$ to have only positive real roots.
Consider $t \le 0$, we have $t^3,-Mt^2,(\frac{37}{4}-M)t \le 0$ and $-1<0$ so $P(t) < 0$ for all $t \le 0$
Hence all roots of $P(t)$ are positive, so $a,b,c>0$ is  valid.
So $E= M^2+2M-\frac{37}{2} \in [\frac{129}{16},\frac{33}{2}]$
The lowest value $\frac{129}{16}$ is occurred when $(\frac{a}{b},\frac{b}{c},\frac{c}{a})=(\frac{1}{4},2,2)$ or permutations.
The greatest value $\frac{33}{2}$ is occurred when $(\frac{a}{b},\frac{b}{c},\frac{c}{a})=(\frac{1}{2},\frac{1}{2},4)$ or permutations.

[quote=Omid Hatami]Suppose $a,b,c\in \mathbb R^+$. Prove that :\[\left(\frac ab+\frac bc+\frac ca\right)^2\geq (a+b+c)\left(\frac1a+\frac1b+\frac1c\right)\][/quote]
[b]Crux Mathematicorum, Vol. 43(3), March 2017:[/b]
Let $a, b,c \in R^+$  such that $abc = 1. $Prove that
$$a^2b + b^2c + c^2a \geq\sqrt{(a + b + c)(ab + bc + ca)} $$
$$\iff\left(\frac ba+\frac cb+\frac ac\right)^2\geq (a+b+c)\left(\frac1a+\frac1b+\frac1c\right)$$

[quote=Omid Hatami]Suppose $a,b,c\in \mathbb R^+$. Prove that :\[\left(\frac ab+\frac bc+\frac ca\right)^2\geq (a+b+c)\left(\frac1a+\frac1b+\frac1c\right)\][/quote]
$$\left(\frac ab+\frac bc+\frac ca\right)^2= (a+b+c)\left(\frac1a+\frac1b+\frac1c\right)+\left(\frac1{ab}+\frac1{b^2}\right)(a-b)^2+\left(\frac1{bc}+\frac1{c^2}\right)(b-c)^2+\left(\frac1{ca}+\frac1{a^2}\right)(c-a)^2.$$

My solution from WOOT:

Looking at the quantities, we are motivated to expand both sides of the inequality. We have
$$\left(\frac{a}{b} + \frac{b}{c} + \frac{c}{a} \right)^2 \ge (a + b + c)\left(\frac{1}{a} + \frac{1}{b} + \frac{1}{c} \right)$$

$$\iff \frac{a^2}{b^2} + \frac{b^2}{c^2} + \frac{c^2}{a^2} + 2\left(\frac{b}{a} + \frac{a}{c} + \frac{c}{b} \right) \ge 3 + \frac{a}{b} + \frac{b}{a} + \frac{a}{c} + \frac{c}{a} + \frac{b}{c} + \frac{c}{b}$$

$$\iff \frac{a^2}{b^2} + \frac{b^2}{c^2} + \frac{c^2}{a^2} + \frac{b}{a} + \frac{a}{c} + \frac{c}{b} \ge 3 + \frac{a}{b} + \frac{c}{a} + \frac{b}{c}$$
Substituting $x = \frac{a}{b}, y = \frac{b}{c}$ and $z = \frac{c}{a}$, it suffices to prove that
$$x^2 + y^2 + z^2 + \frac{1}{x} + \frac{1}{y} + \frac{1}{z} \ge 3 + x + y + z$$
Now note that
$$x^2 + \frac{1}{x} \ge 1 + x$$
$$\iff (x - 1)^2(x + 1) \ge 0$$
which is true. Adding across $x, y, z$ gives us our earlier result, finishing off the problem. Note that $x, y, z > 0$, with equality occurring when $x = y = z = 1$, or $a = b = c$. Our solution is complete.

Let $a,b,c $ be positive real numbers . [url=https://artofproblemsolving.com/community/c6h535879p3076199]Prove that[/url]
$$\left(\frac ab+\frac bc+\frac ca\right)\left(\frac ba+\frac cb+\frac ac\right)\geq (a+b+c)\left(\frac1a+\frac1b+\frac1c\right)$$

[quote=Omid Hatami]Suppose $a,b,c\in \mathbb R^+$. Prove that :\[\left(\frac ab+\frac bc+\frac ca\right)^2\geq (a+b+c)\left(\frac1a+\frac1b+\frac1c\right)\][/quote]
Let $a,b,c>0.$ [url=https://artofproblemsolving.com/community/c6h3018427p27125358]Prove that[/url]
$$ \left(\frac ab+\frac bc+\frac ca\right)^2\geq \frac{a+2b}{c}+\frac{b+2c}{a}+\frac{c+2a}{b} \geq  (a+b+c)\left(\frac1a+\frac1b+\frac1c\right)$$


calculus
integration
geometry
analytic geometry
calculus computations

I am in AP Calculus BC and one of the things we recently learned was integration applied to force and work problems.  Work equals the integral of force.  One application of this is with pumping water out of objects.  I came upon a problem that we had never done before, and when my math teacher saw it he was not able to do it.  The problem is:

A spherical water tank of radius 4 feet is filled with water up to six feet from the bottom.  How much work is required to pump all the water over the upper rim of the tank? (weight density of water = 62.4 lb/ft^3)

I would really appreciate any help with this problem, since it is really bothering me.  Thank you.

Label a vertical variable with a convenient zero point: say, $y$ is the vertical distance above or below the center of the sphere. Now imagine a thin sheet of water at height $y$; the thickness of this sheet is $dy$. What will this thin sheet look like? It will be a circular disc whose exact radius depends on $y$. I'll let you figure out that formula yourself. So this disc has area $A(y)$ and thus volume $A(y)dy.$ How much does it weigh? If $\sigma$ is the weight density of water (given in your data), the weight is $\sigma A(y)dy.$ Now, lift this sheet of water up to the top of the tank. Work = force * distance, where the force is the weight. The distance lifted depends on $y$ in a very straightforward way, so multiply that in. When you have all that, integrate over all of the $y$'s that matter in the problem.

Go try it.

Let $R = 4$ ft be the tank radius, $W = 6$ ft the initial water level, $h$ the current height from the water level to the tank rim and $r$ the radius of the current water level. It does not matter how deep you suck, moving water from the bottom to the current level requires no work at all - if you turn off the pump, water will rise in the hose up to the current level by itself. In the coordinate system with the vertical axis h and origin at the tank bottom, the vertical cross-section of the tank through its center has an equation

$(h - R)^2 + r^2 = R^2$
$r^2 = 2Rh - h^2$

At any water level, you have to move the mass $dm$ by the distange $h$ against gravity, performing the work $dA$:

$dA = gh\ dm = gh\rho\ dV = gh\rho\ \pi r^2(h)\ dh = g\rho\ \pi (2Rh - h^2)h\ dh$

where $g = 32.2 ft/sec^2$ is the gravitational acceleration, $g \rho = 62.4 lb/ft^3$ the weight density of water and $dV = \pi r^2 dh$ denotes volume of the mass $dm$. Integrating dA from $2R - W$ to $2R$:

$A = \pi g\rho\ \int^{2R}_{2R - W} (2Rh^2 - h^3)\ dh = \pi g\rho\ \left[ \frac {2Rh^3} 3 - \frac {h^4} 4\right]^{2R}_{2R - W}$

$A = \pi g \rho\ (\frac {4096} 3 -\frac {4096} 4 - \frac {64} 3 + \frac {16}4) = 324\ \pi g\rho$

Uh, yetti: if $\rho = 62.4\,\mathrm{lb/ft}^3,$ that's already weight, not mass. You don't need to multiply by $g$ to get weight.

Let think like a physicist: No calculus please!
We must divide the water ideally in odd number of layers. The work for getting out the middle layer is 1/2 r V' h.    h = altitude  = 6 ft and r = specific gravity = 62,4 lb/(ft^3) and V' = volume of layer. Above and below the middle layer there are simmetrical layers and the work  for two layers is:

r V'(1/2 h + d) + rV'(1/2  h - d) =  r V'h

Then we can suppose that all the water is at depth of 1/2 h. The work will be:

1/2 r V h  with V = volume of the water.
The result is 288 pi r.
Is interesting to note tat the numerical factor before r V h for conical tank is 1/4 and for semi sferical tank is 3/8. But this is another story, maybe calculus story.

The tank isn't symmetrical top to bottom because it isn't full.


3 tabs cost the same as 5 taps, while 10 tabs cost the same as 7 tops. If 25 taps cost 16 dollars, find the cost of 105 tops

25 taps are equal to 15 tabs.
15 tabs are equal to 10.5 tops.
10.5 tops cost 16 dollars, so 105 tops cost 160 dollars.


function
limit
algebra
domain
projective geometry
complex analysis
complex analysis unsolved

Let $u: \mathbb{R}^{2}\to\mathbb{R}$ be harmonic in $B(0,1)\setminus\{0\}$ and such that $\lim_{(x,y)\to 0}u(x,y)$ exists.  Let $U$ be the extension of $u$ to all of $B(0,1)$ in the obvious way.  Prove that $U$ is harmonic on $B(0,1)$.

Now, I know a solution to this problem using the Poisson Kernel.  This is nice and fairly natural because it handles all dimensions, but it seems like there should be a simpler solution in the case of $\mathbb{R}^{2}$ writing $f=u+iv$ where $f$ is analytic except at $0$, $v$ is the harmonic conjugate of $u$, and applying the theorem about removable singularities of bounded analytic functions.

Any ideas?

[quote="blahblahblah"] writing $f=u+iv$ where $f$ is analytic except at $0$, $v$ is the harmonic conjugate of $u$[/quote]
There is a catch: since $u$ is harmonic in a not simply-connected domain, $v$ may fail to exist as a single-valued function. I'm not saying that this problem cannot be dealt with, but it surely will make the argument more complicated.


Two numbers are such that their difference, their sum, and their product are in the ratio $ 1: 7: 24$. Determine the product of these two numbers.

[hide="Solution"]Call the two numbers $ a$ and $ b$, and let the ratio between the terms be $ x$. So we have
$ a\minus{}b\equal{}x$, $ a\plus{}b\equal{}7x$, and $ a\cdot b \equal{} 24x$.
$ 2a\equal{}8x\implies a\equal{}4x\implies b\equal{}3x$. $ a\cdot b\equal{}12x^2 \equal{} 24x \implies x\equal{}0,2$. $ x\equal{}0$ doesn't work, so $ x\equal{}2$.
Finally, $ a\cdot b\equal{}24x \equal{} 24\cdot 2 \equal{} \boxed{48}$.[/hide]


Do Liouville numbers have cardinality of the continuum?

Yes. Notice that every real number can be written as the sum of two Liouville numbers.


logarithms
limit
calculus
calculus computations

\[\sum_{x=2}^{\infty}\frac{1}{{(\ln n)}^{3}}\]

The divergence of this I have tried many methods to prove.

But I still can't prove it.

Use Theorem 3.27, page 61, Principles of mathematical analysis, W.Rudin

Use Cauchy condensation criteria, namely the series $\sum a_{n}$ and $\sum 2^{n}a_{2^{n}}$ are both convergent or divergent. (provided that $a_{n}$ is a decreasing sequence.)

More elementary than any of these: we know that as $n\to\infty,\ \ln n$ grows slower than any (positive) power of $n.$

${\lim_{n\to\infty}\frac{\frac1{(\ln n)^{3}}}{\frac1n}}= \lim_{n\to\infty}\left(\frac{n^{1/3}}{\ln n}\right)^{3}= \infty.$

Thus the terms of this series are eventually larger than $\frac1n$ and it diverges by ("simple" or one-sided) comparison to the harmonic series.


Points $ A \equal{} (3,9), B \equal{} (1,1), C \equal{} (5,3),$ and $ D \equal{} (a,b)$ lie in the first quadrant and are the vertices of quadrilateral $ ABCD$. The quadrilateral formed by joining the midpoints of $ \overline{AB}, \overline{BC}, \overline{CD},$ and $ \overline{DA}$ is a square. What is the sum of the coordinates of point $ D$?

$ \textbf{(A)} \ 7 \qquad \textbf{(B)} \ 9 \qquad \textbf{(C)} \ 10 \qquad \textbf{(D)} \ 12 \qquad \textbf{(E)} \ 16$

This is really easy for a #20. :|

[hide="Answer"]Midpoint of $\overline{AB}$ is $(2,5)$ and midpoint of $\overline{BC}$ is $(3,2)$. Between the two midpoints with given coordinates, we move one space to the right and three down, so for the perpendicular line we must move three spaces over and one space up. This gives us the point $(6,3)$, which is the midpoint of $CD$, and we know $C(5,3)$, so $D(7,3)$, and $7+3=10\Rightarrow \boxed{C}$.[/hide]

How would you solve this with complex numbers?

You would assign each of the points numbers, then find two of the points, then rotate it -90 and then 90 degrees to get the other two points of the square. Then just find D since those two points are the midpoints. Fancier imo :P


calculus
integration
trigonometry
function
calculus computations

find a [b][u]closed-form [/u][/b] expression for the integral:

$ \int_{0}^\infty\;\left( \;\prod_{k=1}^\infty\;\cos \left( \frac{y}{k}\right) \;\right) \;\textbf dy$

[ PS: i spent almost 2 hrs on this one last night and
am no-where even close...  :P  :rotfl: ]

i think we cannot evaluate this. just a idea

Forget the integral.

take the limit of the product itself in $ y \to  oo$

this limit does not exist, is oscilanting in the interval $ (-1,1)$

So, the integral will oscillate too! and the "accurance" of the oscilation depende in how we make the product and the integral goes to infinity.

we have a double limit, and a indefinied function in infinity.

so is more common think that this integral does not converge and oscillate


algorithm algebra polynomial calculus integration quadratics function

I would like everybody to check out the new features of the CAS Maple.

New Features of Maple 9.5

  Mathematical Computations
Differential Equations

Exact Solutions
 New algorithms for solving Riccati type ODEs, second order linear ODEs of Mathieu type, polynomial solutions for nonlinear ODEs and systems of ODEs, linear and nonlinear PDEs and systems of PDEs
 New methods for initial value ODE problems, and piecewise ODEs
 Hypergeometric solutions free of integrals for linear ODEs
 Efficiency improvements for difficult first order Abel type ODE problems

Numeric Solutions
 Three new numerical solution methods for Differential Algebraic ODE initial value problems (DAEs), both stiff and non-stiff
 Maplet assistant extended to allow interactive solving of DAE problems
 Optimization option for large or complex systems can increase calculation speed by up to 30 times
 New implicit option for the stiff IVP solver is useful for large, dense ODE systems

Differential Equation Tools
 Seven new commands, some based on original algorithms, were added to DEtools
 PDEtools now includes a command for computing traveling wave solutions
 Method for using algebraic triangular techniques were added to diffalg package

New Packages

Optimization
 Numeric methods for solutions of optimization problems
 Easy-to-use interactive Maplet assistant
 Arbitrary-precision solutions
 Solvers for linear, quadratic and nonlinear programs, including constrained and unconstrained problems
 Solvers for linear and nonlinear least-squares problems

Logic
 The Logic package is a collection of commands for manipulating and transforming expressions using two-valued Boolean logic
 Simplify logical expressions, test two expressions for equivalence, convert logical expressions to algebraic expressions modulo 2, and perform a variety of other logical operations

RootFinding
 The RootFinding package enhances Maple's ability to compute and locate roots numerically
 Computes zeros of analytic functions and solutions of two or more bivariate polynomials

Enhanced Packages

Groebner
 Now includes two algorithms to compute the reduced Groebner basis of a toric ideal

LREtools
 Determine necessary conditions for the solution of a linear recurrence equation to be analytic, in terms of the initial values
 Finds all d'Alembertian solutions of a linear recurrence equation
 Determine if it is possible to construct a desingularizable operator with
polynomial coefficients for the linear recurrence, and compute it if it exists

QDifferenceEquations
 Sum the solutions of a q-shift operator using the methodof accurate q-summations
 Compute a series solution of a linear q-difference equation
 Extend a series solutions of a linear q-difference equation to a higher degree
 Finds all q-hypergeometric solutions of a given linear q-difference equation

SolveTools
 Extended to include functions to solve inequalities
 Linear inequalities with respect to one variable
 Linear univariate and multivariate systems of inequalities

SumTools
 Functionality involving sums of hypergeometric type sums has been extended

General

Symbolic Calculations
 Enhancements have been made to many top-level routines for symbolic calculations and manipulations, including the functions for:

     Integration
     Summation
     Simplification
     Combinations
     Series expansions
     Conversions

 Function decomposition routines performs rational function decomposition on overdetermined systems in the presence of parameters
 The FunctionAdvisor command has been enhanced regarding the computation of non-trivial specializations of mathematical functions for particular values of their parameters

Efficiency Improvements
 Significant improvements have been made in the efficiency and accuracy of the routines that calculate elliptic and Jacobi functions
 Maple 9.5 has more efficient computation of the quotient of sparse polynomials
 Conversions between D and diff notation for derivatives are now much faster
 Numeric linear algebra calculations on Windows, Macintosh and Linux use state-of-the art routines optimized for each platform

  Education

Dictionary

 Over 5000 mathematical and engineering terms and definitions are fully incorporated into the Help system
 Includes over 300 figures to clearly explain the concepts
 Definitions can be accessed as help pages using an alphabetical listing in the Maple Help system, or through links in a worksheet or help page
 Short definitions are conveniently displayed as pop-ups

Student[MultivariateCalculus]

 The Student[MultivariateCalculus] package will assist with the teaching and learning of multivariate calculus
 Interactive routines use Maplet technology to assist you to work through the standard problems of multivariate calculus in a visually directed manner
 Visualization routines are provide to aid in the understanding of concepts including Taylor approximation, change of variables, center of mass, gradient, Jacobian, surface area, and more

Student Packages

 Student package tutors are now easily accessible from the Tools menu
 Most interactive tutors now include an option to see the equivalent Maple command
 Student[Precalculus] now includes more functions for visual explorations of topics in greater depth, including a routine to explore lines

  Graphical User Interface
Usability

 Maple 9.5 adds the ability to dock your palettes to the edges of your workspace
 Improved palettes use meaningful placeholder names. Expressions can be dragged from the palette into the workspace
 Improved command completion facilities automatically display the full command when a unique completion is detected
 The Tools menu provides quick access to the Maple Assistants, such as the Unit Converter, Plot Builder and ODE Analyzer, and to the Maple Tutors associated with the Student packages, including Linear Algebra and Multivariate Calculus
 The Options dialog has been redesigned to allow easier and more consistent access to the settings, and control over more items

Document Formatting

 You can emphasize information in your worksheets by using the Text Highlight feature
 Maple 9.5 allows you to include superscript and subscript characters in your worksheets
 You can now easily control the number of displayed terms of an expression and digits of a number. A threshold can be set at which elision is invoked as well as the number of leading and trailing terms or digits
 Two new types of hyperlinks are supported: a link to a definition contained in the math dictionary and a link to a Maplet

Graphics

 The Interactive Plot Builder has been expanded to allow easier creation of a greater variety of plots
 The Plot Builder now supports plotting multiple expressions simultaneously, and allowing interactive adding of expressions to be plotted
 Animations can be created using the Plot Builder
 Plots created with slider controls for parameters permit a plot to be adjusted interactively
 Maple 9.5 provides slider control for viewing individual frames of an animated plot
 Oscillate control allows you to play animations backward and forward
 Point-probe cursor makes is easy to identify the coordinates of a particular point on a 2-D plot
 Scale and pan 2-D and 3-D plots and animations

  Programming and Connectivity

OpenMaple

 OpenMaple allows access to Maple using C and now Java and Visual Basic programs

Maple Library Archive

 All components of a Maple library can now be saved in a single file
 Other types of files can be saved inside the archive for convenient sharing of related materials

Debugging

 A program that is already running can be debugged by clicking on a Debug button inside the worksheet
 If a computation cannot be interrupted, you are now given the option of restarting the math engine

MapleNet

 Maplet applications created in Maple 9.5 can easily be deployed to MapleNet

  New Packages and Functions

Mathematica Conversion Tools

 Automatically convert Mathematica notebooks to Maple worksheets
 Translate commands from Mathematicas syntax to Maple syntax
 Maplet assistant makes translating commands and documents easy

Context Menus

 The new context menu package provides new easy-to-use tools to control and customize Maple context-sensitive menus
 Replace the default context menu, add new entries to the existing context menu, or alter the criteria under which entries are displayed in a menu

Cache

 This package contains function to manipulate a new cache data structure. A cache table allows both permanent and temporary storage inside the same data structure
 Used as a remember table, the important items can be stored permanently while temporary items can be replaced to avoid wasting memory

Package Unloading

 The unwith command allows you to reverse the effects of a with for module-based packages

Modules and Operators

 Modules can now be applied to objects
 Modules can control how they are printed
 Routines can be defined that will run when a module is loaded or unloaded
 The list of overloadable operators has been extended to include list construction, set construction, function application, and index application
 Overloaded operators and functions can be controlled by the argument type

  Enhanced Packages and Functions

Code Generation

 The new CodeGeneration[Save] command allows user-contributed language definitions to be saved in a Maple archive more easily

Inert Representation

 The ToInert command for converting to an inert representation of a Maple object has been extended to allow for finer control over which objects to convert

Partial Fraction Decomposition

 Partial fraction decomposition now has a convenient programmer interface

Polynomial Tools

 Four more tools have been added for calculating Taylor shift distances, shiftless decompositions, greatest factorial factorizations and gcd-free bases

String Tools

 Includes a function for seeding the random string generator

I agree with Simon. Mathematica all the way!

[quote="ComplexZeta"]What's the point of buying Maple when Mathematica is the industry standard and is more powerful and costs the same?[/quote]
There are three leading products: Maple, Mathematica, and Mupad.
Its not my fault that the university here has Maple as a standard. So any class would have Maple stuff included.

List of some places that use Maple.

Maple at Universities

    * University of Waterloo. (Engineering).
    * Universit Polytechnique de Montral.
    * MIT (Massachusetts Institute of Technology)
    * . Arizona State University. Foundation Coalition.
    * Brown University.
    * Denison University.
    * Indiana University.
    * Monmouth University. Department of Mathematics.
    * North Carolina State University.
    * Saint Louis University. Department of Mathematics and Computer Science.
    * Texas A&M University. Calclab - Department of Mathematics.
    * Villanova University. Department of Mathematical Sciences.
    * University at Buffalo. State University of New York.
    * University of Illinois at Chicago. Mathematics, Statistics & Computer Science.
    * University of Michigan.
    * University of Pennsilvania. (see also.)
    * University of South Florida.
    * University of Texas.
    * University of Utah. (see also).
    * Universitt Essen.
    * University of Oldenburg.
    * University of Regensburg.
    * Universitaet Karlsruhe.
    * Technische Universitt Chemnitz.
    * Isolde-Kurz-Gymnasium Reutlingen.
    * Royal Melbourne Institute of Technology (RMIT).
    * Universidad Nacional Autnoma de Mexico (UNAM).
    * Universidad de los Andes. CeCalCULA.

Research Groups

    * Symbolic Computation Group, University of Waterloo.
    * Institute of Scientific Computing (ETH).
    * Maple V Symbolic Manipulation Group, MIT.
    * Centre for Experimental and Constructive Mathematics (CECM), Simon Fraser University (SFU). Burnaby, Canada.
    * Formal Geometry and Mathematical Physics. (FG&MP) Utah State University.
    * Relativity & Symbolic Computing, School of Computing and Mathematics, Deakin University. Geelong, Victoria, Australia.
    * Projet EURECA, INRIA Lorraine, Nancy.
    * Project Algorithms, INRIA Rocquencourt.
    * SAFIR (Systmes Algbriques Formels pour l'Industrie et la Recherche), INRIA Sophia-Antipolis.
    * Groupe de Calcul Formel, LACO (Laboratoire d'Arithmtique, de Calcul Formel et d'Optimisation), Universit de Limoges.
    * Equipe CALCUL FORMEL, LMC (Laboratoire de Modlisation et Calcul), IMAG (Institut d'Informatique et de Mathmatiques Appliques de Grenoble).
    * Groupe Calcul Formel, IREM (Institut de Recherche sur l'Enseignement des Mathmatiques). Universit Joseph Fourier - Acadmie de Grenoble.
    * Groupe Calcul Formel, IREM, Universit de Picardie Jules Verne.
    * quipe CalFor. LIFL (Laboratoire d'Informatique Fondamentale de Lille).
    * CSO (Computer Algebra division). Universit Libre de Bruxelles.
    * CAG (Computational Algebra and Geometry). Department of Mathematics, Statistics and Computation, Universidad de Cantabria.

Plus Maple offers a suite of apps to help students. There is Linear Algebra, Precalculus, Calculus, Multivariable Calculus, and some more. They are really helpfull. My favorite is the Integration Tutor. It shows you step by step on how to integrate stuff.

I aggree with you COmplexzeta that Mathematica has more users. Just look at China.

    * Beijing University
    * Tunghai University
    * China University of Mining and Technology
    * Fudan University
    * Hunan Normal University
    * Nankai University
    * Southwest University
    * Tsinghua University
    * Academia Sinica
    * Feng Chia University
    * National Taiwan University
    * Soochow University


AMC
USA(J)MO
USAMO
MATHCOUNTS
geometry
trigonometry
analytic geometry

[quote="ra5249"]Just a quick question: do you have to be a freshman to make Red MOP, or can you make it if you're in like 7th or 8th grade?[/quote]
Freshman, 8th graders and below aren't allowed.

I think its the Akamai foundation that has something that says it only sponsers freshman

To be clear, 8th graders and have qualified for blue/black MOP in the past.  It's not common, but it happens.

but only freshmen can qualify for red? ok sure. thanks.

Isn't Red MOP pretty new as it is? like 2005 was the first time they had it?

They did something like it, only even larger, in 2002.

I think they took all the freshmen into red MOP in 2004, but there were less than 30 then.

How many middle schoolers have qualified for Blue/Black MOP in the past and what are their names?

There were 2 in 1999; Dani Kane (7th grade) and Po-Ru Loh (8th grade). With Po-Shen on the IMO team, it was a good year for Madison.

There's been more

If you look at the AMC website, there's also Tianki Liu, Allison Miller, and then I remember hearing about Dani Kane and Gabe Carroll.

Looking here - [url]http://www.unl.edu/amc/a-activities/a6-mosp/a6-1-mosparchives/1999-ma/moplist-0599.html[/url]

You can kind of guess by looking at the dates they attended MOP, but I don't recognize any of the names.

Although Gabriel Carroll made the USAMO in 1997 as an 8th grader, he didn't make MOP.

The lists at the bottom of the MOSP archive page are a good place to look- they include grade.

[quote="jmerry"]There were 2 in 1999; Danny Kane (7th grade) and Po-Ru Loh (8th grade). With Po-Shen on the IMO team, it was a good year for Madison.[/quote]yeah...yet again proving how great Wisconsin really is :P but yeah going into MOP at such a young age helped them. I know that Po-Ru had 2 golds and a silver, making the IMO team as early as a freshman. Dan Kane got 2 golds of his own...and then there's Reid Barton, who makes MOP as a 7th grader and goes on to win 4 golds at imo...yeah so getting that jump start in middle school REALLY helps

[quote="ra5249"]i think there have only been two - eric larson and reid (dont know last name)[/quote]
eric larson... I haven't heard from him in a while but back in 8th grade before I knew what the amc was he was on the state mathcounts team with me and kept busting out analytic geometry on these mathcounts problems and pwning them but nobody knew what he was doing...

Also Sherry Gong made MOP as an 8th grader in 2003.

[quote="noneoftheabove"]isnt analytic geometry overkill for mathcounts?[/quote]
lol yeah
i know this one kid who uses trig instead of the pythagorean theorem
he seems to think its easier...  :huh:

So, middle schoolers have to get into blue or black mop?


function
integration
trigonometry
real analysis
real analysis unsolved

let $f: R^2 -> R^2$ be an harmonic function.
$K(r)=\int_0^{2\pi} f(r.cos(t),r.sin(t)) dt$
Prove that $K$ is constant.

$K(r)=2\pi f(0).$

More generally, $f(x,y)=\frac1{2\pi}\int_0^{2\pi}f(x+r\cos\theta, y+r\sin\theta)\,d\theta.$

It's called the mean value property of harmonic functions. There are several proofs; I don't have time to write one up just now.