number theory
prime numbers
number theory unsolved

Alice and Bob choose each an integer between 1 and 20. They want to know if they chose the same integer, in such a way that if they did not choose the same integer no one would get any information about the number the other one chose. How can they do that?

I posted this under number theory problems since there probably is a solution based on prime numbers and factorization, but I don't have a solution.

I want to see a solution.
Usually, in cryptography "no one would get any information about the number" means that decoding requires too many computations, so it is practically an absurd.

How about the following?  They start with 20 blank sheets of paper, numbered 1 to 20.  While Bob is out of the room, Alice writes a check mark on the back side of the paper corresponding to her number.  Next, while Alice is out of the room, Bob looks to see whether there is a check on his paper.  Bob then lets Alice know the outcome.

This protocol depends on both players being trustworthy.


A right triangle has side lengths of $a+2$, $2a-1$, and $3a-1$. What is the maximum possible value of $a$?

[hide]Substitute $3a-1$ as the hypotenuse.
$(a^{2}+4a+4)+(4a^{2}-4a+1)=(9a^{2}-6a+1)$, which simplifies to $5a^{2}+5=9a^{2}-6a+1$, which on one side is $0=4a^{2}-6a-4$, which divided by 2 is $0=2a^{2}-3a-2$, which factors into $0=(2a+1)(a-2)$, who's solutions are 2 and $\frac{-1}{2}$.  Since sides can't be negative, $a = \boxed{2}$[/hide]

You're both correct, but you should also consider the case where $a+2$ is the hypotenuse, since that [i]could[/i] produce a larger value of $a$. (You don't need to check the case where $2a-1$ is the hypotenuse though, because obviously $2a-1<3a-1$.)

[quote="mad_skillz_aops"]But you also know that when $a > 2$, $3a-1$ will always be greater than $a+2$.[/quote]
I know; that's why your answers are correct. I was just saying that you have to make sure you consider that case in your solutions, because you didn't really do that.  :wink:


limit
calculus
derivative
function
calculus computations

I was trying to solve this problem but its relaly hard

Find lim    { f(8+h)-f(8)}/h
     h-->0                                            for f(x) =x^4/3

now i figured that you can do this

{f(8+h)^4/3-f(8)^4/3 }/ h

but I just dont know how to go on about expanding and how would one go about solving this?
i have asked many peers but they had no idea..please help

The expression
$\lim_{h\to 0} \frac{f(x + h) - f(x)}{h}$
is the formal definition of the derivative of $f(x)$. So you have

$f(x) = x^{4/3}$ (I assume the $/3$ was supposed to be in the exponent) and you're looking at the derivative when $x = 8$.

Do you know how to find $f'(x)$ for a power of $x$? After you've done that, all you have to do is evaluate at $x = 8$.

Has your class discussed the formal definition of the derivative, and how it relates to the "rules" that you learn for taking derivatives of particular functions? I (and plenty of other people here) are probably willing to answer any questions about this that you have. Just ask something specific and we'll be happy to help.

my question was

so in the following

f(8+h) -f (8)
-------------
      h                              for f(X) =x^4/3

so
\lim_{h\to 0} \frac{f(x + h) - f(x)}{h}

Im sorry im not sure i understand

i know that we can find the derivative from nx^n-1

so if f'x  will  be
       (4/3)8 ^ 4/3-1
 ??
Im really confused right now please help..

Also,
$8^{3/4} = (\sqrt[3]{8})^4 = 2^4 = 16$

Ramanujan's way is the formal way to evaluate the derivative. However, it's really much more efficient to see the power rule using the formal definition, and in the future just apply that. Always be aware of what you're actually doing, though. That's important for understanding. :)

\\ \frac{(x+h)^{4/3}-x^{4/3}}{h} = \frac{(x+h)^{4}-x^{4}}{h \cdot [(x+h)^{8/3}+ (x+h)^{4/3}x^{4/3} +x^{8/3}]}

okay well i know that

{(x+h)^{4}-x^{4}}

can be expanded to (X+h)^2(x+h)^2-x^4

but i still dont understand how this will all simply ??
Also.. is there any simpler way to do it.

Xevarion you stated that you can do  8^3/4 gives the answer 16..is that the limit? and if so is that a simpler way to do the question?

Thank you so far ..you all have been kind

[quote="tryingtolearnmath"]i think i kind of understand now
the limit represents the derivative of f(x)= x^4/3  at x=8

so we can find the derivative of f(X)
of x^4/3

and that would be 4/3x^1/3
so f'(8)= 4/3(8)^1/3

= 8/3[/quote]

It is OK!     :)


function
vector
advanced fields
advanced fields unsolved

In an $ n$-dimensional hyperspace given $ k$ hyperplanes such that no two are parallel to each other and no three intersect in the the same $ (n\minus{}1)$-dimensional hyperplane, find the number of distinct (sub)hyperspaces created by these hyperplanes in form of a function $ f(n,k)$. Of course, $ n$ and $ k$ are integers.

You're trying to make conditions that say "in general position," but you haven't got the right ones.  In particular, $ n \minus{} 1$ should probably be $ n \minus{} 2$, but also your "function" isn't a function: for $ n \equal{} k \equal{} 3$, consider the planes $ x \equal{} 0$, $ y \equal{} 0$ and $ x \plus{} y \equal{} 1$ as compared with the planes $ x \equal{} 0$, $ y \equal{} 0$ and $ z \equal{} 0$, for example.

So, we need to strengthen the conditions.  I think that the following will do:
1) given any $ d \leq n$ of the hyperplanes, their normal vectors span a subspace of dimension $ d$, and
2) given any $ d \leq n$ of the hyperplanes, their intersection is of dimension $ n \minus{} d$.
(These may be partially redundant.)

Richard Stanley's [url=http://www-math.mit.edu/~rstan/arrangements/arr.html]Notes on Hyperplane Arrangements[/url] are probably a good place to look for more information about this sort of problem.


Zeige: $ \infty$ ist ein Haufungspunkt von $ \mathbb N$ in $ \bar{\mathbb R}$, aber in $ \mathbb R$ hat $ \mathbb N$ keinen Haufungspunkt!

Fitim

[quote="darij grinberg"]Wie habt ihr Häufungspunkte definiert? Das macht bei so einer Aufgabe einen gewaltigen Unterschied...

  darij[/quote]

Sehr geehrter Darij,

In Universitaet Wien definiren Haufungspunkt so: Sei $ F \subseteq R$. $ x$ heisst Haufungspunkt von [i]F[/i](Hp. von [i]F[/i]), wenn $ \forall \varepsilon > 0$ $ \Rightarrow$ $ U_{\varepsilon} (x) \cap F$ ist unendlich.

Man definiert $ U_{\varepsilon} (x): \equal{} {{y \in R : |y \minus{} x| < \varepsilon}}$

Liebe Gruesse
Fitim Dika

[quote="Fitim"]In Universitaet Wien definiren Haufungspunkt so: Sei $ F \subseteq R$. $ x$ heisst Haufungspunkt von [i]F[/i](Hp. von [i]F[/i]), wenn $ \forall \varepsilon > 0$ $ \Rightarrow$ $ U_{\varepsilon} (x) \cap F$ ist unendlich.

Man definiert $ U_{\varepsilon} (x): \equal{} {{y \in R : |y \minus{} x| < \varepsilon}}$ [/quote]

Die Definition ergibt nur für Teilmengen von $ \mathbb{R}$ Sinn, nicht aber für Teilmengen von $ \overline{\mathbb{R}}$. Du brauchst also für den ersten Teil der Aufgabe eine andere (allgemeinere) Definition von "Häufungspunkt".

Den zweiten Teil ($ \mathbb{N}$ hat in $ \mathbb{R}$ keinen Häufungspunkt) kannst du leicht mit deiner Definition lösen.

  darij


geometry
calculus
integration
function
calculus computations

Find the area of the region bounded by:
[img]http://img82.imageshack.us/img82/5602/boundedbyrc3.jpg[/img]

I got the graph made already, but I am wondering if I am supposed to solve this by writing the integration in the dx (top function - bottom function) form or in the dy (right function - left function) form? :maybe:
[img]http://img239.imageshack.us/img239/9320/graphic8.jpg[/img]

Thanks. :)

Ask this question: Is it easier to integrate $y\,dx$ or $x\,dy$ (the inverse function), given that functional relation?
There's no harm in trying both, except some wasted time.

Agree with jmerry.

Do you really want to solve for $x$ in $y = x\sqrt{x^{2}+5}$?
It would work but it would take you longer than if you tried $ydx$ because just looking at it, you can see that it is simple to integrate.

But both ways work. So try the simpler one and if it becomes hard to integrate try the other.

Thanks, guys. I thought you could only use either one of them to solve each time. :blush:

I am unsure of the indefinite integral of this:[img]http://www.mathlinks.ro/Forum/latexrender/pictures/dd78d918ba899142985dfd99c642c7e1.gif[/img]
Is it: [(x^2 + 5)^3/2]/3 ?

Yes. You can always check by differentiating.


ARML
geometry
3D geometry
probability
LaTeX
search
quadratics

I am hosting a contest, preferably for grades 5-6. If you are younger, feel free to join, but if you are much older, I'd rather not, as you may find these questions too easy.

All right, here are the rules. I will give you from today to Friday, May 30th, Forum Time 3:30 pm to sign up. Then, on the following day in the morning I will post a problem. You have until 3:30 pm Forum Time June 1st to answer it. Please private message your solution and all work to me. On the following Monday, I will post another question. You will have until Wednesday of that week, until 3:30 pm Forum Time to pm it to me. I will choose the top three people who correctly answer both questions and explain them well.

Please sign up! It should be fun!   :)

I'll Sign up!

also sign-up magicsquare please.
[u][b]
Sign up list[/b][/u]

1. BOGTRO
2. magicsquare.

[b]Sign Up List:[/b]

1. BOGTRO
2. magicsquare
3. D3m0n Shad0w

Suggestion: Older members should help out in some cases.

It just is.

I can be the older member that helps out.

Okay, so here are the people who have signed up so far:

1. BOGTRO
2. magicsquare
3. D3m0n Shad0w
4. kookamunga

Please let me know if you signed up and I forgot to put you on the list!

All right, so far we have:

[b]The Sign Up List[/b]

[b]1.[/b] BOGTRO

[b]2.[/b] magicsquare

[b]3.[/b] D3m0n Shad0w

[b]4.[/b] kookamunga

[b]5.[/b] 12markkram34

I will do this for fun (I'll be out of the contest)

[quote="Yongyi781"][quote="D3m0n Shad0w"]Ah, something of my caliber. I'm in grade 5 right now, but moving on to 6 next year  :D I really hope this "marathon" won't get harder. It always leads to that.[/quote]

How are you a moderator of Classroom Math, then? You're also a bit too mature for a 5th grader...[/quote]

Classroom math is for middle schoolers. I am in Sixth Grade Math, even though in fifth grade. As for the maturity, it disgust me how some 13- year olds can act. I try my best to please all.

How do I send my answer by  Private message??? :?:

On the top (Where it says Profile, Private MEssages, Memberlist, Search, etc.),

Click on "Private Messages", and once there, click on "New Private Message".

Type the name of the person you want to send it to in the "Username" Box,

and the Subject in the "Subject Box".

Then Type your message in the "Message Body". When you are done, click "Submit".

Example:

Name: BOGTRO
Subject: Test

Let $ x$ be the desired number. Then, ......

Hope that helped.

Of course, you want to send it to isabella296, not BOGTRO. :)

Sweet! Can we discuss yet?

I got a $ 9$ on my first one and a $ 10$ on the second! OWNED!

lol jk

I got 9 on the first and 7 on the second one  :dry: Trust a fifth grader to know system of equations!

I'm a fifth grader... YOu couldn't solve the system? lol...

Hey now. Don't be making fun of people just because they can't do stuff. A normal 9th grader in TX can barely do systems of equations.

7 means she solved it and explained it ok.

No offense, but TX is really sad then...

Don't insult people. I didn't learn systems of equations until this year, in 6th grade. Last year I didn't know a thing about systems. D3m0n Shad0w is smarter than I was last year, and probably this year too. Okay, here are results!

In first place, we have a tie, between 12markkram34 and BOGTRO, both with scores of 19. I will post the tiebreaker rules below, to find out who's in second and who's our winner!  By the way, in third place we have Canton! Good job, you three!

[b]TIE-BREAKER RULES[/b]

For these problems, I would like some explanation, but you don't have to go into great detail like on the first two problems. For example, you could say, "If we set up a proportion, we find that there are 29 hats."

Each problem below is worth 3 points. Whoever gets the most points, gets first place. If there is another tie, (which would be funny), I will post the "Tie-breaker of the Tie-breaker rules".

1. Sally the sparrow eats 5 worms at every mealtime(3 mealtimes a day). Sammy the sparrow eats 12 worms at every mealtime(3 mealtimes a day). Together, what is the total amount of worms they eat in a year(365 days a year)?

2. I have 34.58 dollars. A pencil costs 0.25 dollars. How many pencils can I buy?

3. For every 5 toys he gets, my dog breaks 3 of them. If I give him 5,000,000 toys, how many will he break?

4. How many arrangements are there of the letters in the following word: TOOTLES

5. Find the root(s) of the following quadratic: $ 4x^2 \minus{} 12x \plus{} 9$

6. Rationalize the following fraction: $ \frac{2\sqrt{8}}{8\sqrt{2}}$

Good luck, BOGTRO and 12markkram34!

By the way, other participants, if you want to try these, feel free!

@BOGTRO

No, TX is not sad by the standard's of U.S. public schools. A lot of people I know from when I went to public school (through last year) can't solve systems well. (I'm a freshman)

[b]WINNERS[/b]

1st Place: BOGTRO and 12markkram34  :winner_first:

2nd Place: Canton  :winner_second:

3rd Place: D3m0n Shad0w  :winner_third:

Since BOGTRO and 12markkram34 both tied in the contest and the tie-breaker, I have decided to simply leave it as a tie.

Good job, you four, and all other participants!  :clap2:  :thumbup:

Thanks, guys.  :winner_third:
That was a fun contest. Maybe it should be longer, with ten problems or so?

Well, take a look at my tournament! There's a 100 problems, total!! YAY!! It'll be even more fun.  :wink:


geometry
geometric transformation
dilation

A $ 10\%$ potassium chloride solution is needed for use in a lab.  How much water (in liters) must be added to dilute 300 liters of a $ 50\%$ potassium chloride solution for the lab's use?

So far $ 0.50 \times 300 \equal{} 150$ liters of the solution is potassium chloride. We want that number to be 10% of something, so $ 150 \div 0.10 \equal{} 1500$. Thus, we need $ \boxed{1200}$ more liters.


linear algebra
matrix
linear algebra unsolved

Let $ A,B \in M_n(\mathbb{R})$, $ n\ge 2$ such that $ A^2\plus{}B^2\equal{}O_n$. Prove that:

a) If $ n\equal{}4k,k\in \mathbb{N}$, then $ det(AB\minus{}BA)\ge 0$

b) If $ n\equal{}4k\plus{}2,k \in \mathbb{N}$, then $ det(AB\minus{}BA)\le 0$

c) If $ n\equal{}4k\plus{}1$ or $ n\equal{}4k\plus{}3$, $ k\in \mathbb{N}$, then $ det(AB\minus{}BA)\equal{}0$

[hide="hint"]Use the fact that if $ T \in \mathcal{M}_{n}(\mathbb{C})$, then $ \det(T)\cdot \overline {\det(T)} \equal{} |\det T|^2 \in \mathbb{R}_{ \plus{} }$ , and notice that $ (A \minus{} iB)(A \plus{} iB) \equal{} i(AB \minus{} BA)$ [/hide]


[hide]Adding the five equations together, we have:
$3x_1+3x_2+3x_3+3x_4+3x_5=186$
$x_1+x_2+x_3+x_4+x_5=62$
Subtract that from the 4th equation, you get $x_4=-14$. Subtract that from the 5th equation, you get $x_5=34$. 3(-14)+2(34)=36.
[/hide]

Somehow, I think you added wrong. Mbe i'm just really out of it. But 1+1+1+1+2 is 6(right????).  Head hurts from AIME...

Ah, I'm the one that's out of it. I need to start paying attention to what I'm writing. No, 2+1+1+1+1 is not equal to 3. :blush:
[hide="On_second_thought"]Okay, so that means $x_1+x_2+x_3+x_4+x_5$ is half of what I thought it was, or 31. So the answer would be 3(17)+2(65)=181...I hope.
[/hide]


[quote="nr1337"]In the middle of doing a problem yesterday, I conjectured this but can't prove it:

For every prime $p$, there exists a positive integer $x$ such that $px - 1$ is prime.

Does anyone know how to prove this or even if it is true or not?  Please move this topic if it is too easy or hard.[/quote]This is a part of Dirchlet's theorem (concerning primes) that states that if $a,b$ are two coprime integers then the sequence $an+b$ contains an infinity of prime numbers. So it's true :P

[quote="nr1337"][quote="Psycho"]seems logical, since there are infinitly many primes, and $x$ can take any value from $1$ till $\infty$  :?[/quote]

Of course it seems logical, but that's not a proof...[/quote]
never said it was a proof ;)

[quote="seamusoboyle"]While were on the topic, the tao/green theorem states that for any n, we can find n primes in arithmetic progression. Not an easy proof, (in fact, a horrible twisted one that takes well over two months of lectures to present), but a lovely theorem!

Tommy Tao went to the IMO (and got gold) at he ripe old age of 11. Wow.[/quote]
 :o  youngest one ever, most likely...


inequalities
LaTeX
inequalities unsolved

$ \text{\LaTeX}$-ed and corrected version:
Let $ a_1,a_2$ be two positive reals and $ \{a_i\}_{i = 1}^{\infty}$ be a sequence satisfying the following condition:
$ a_{n + 2} = \frac {2}{a_n + a_{n + 1}}$.
Also let $ M_n = \max\{a_{n},a_{n + 1},\frac {1}{a_n},\frac {1}{a_{n + 1}}\}$.
Prove that the following inequality holds for all $ n$:
$ M_{n + 3}\leq\frac {3}{4}\cdot M_n + \frac {1}{4}$.

[quote="Allnames"][quote="duj12358"]$ a_1 > 0,a_2 > 0,a_(n \plus{} 2) \equal{} \frac {2}{a_n \plus{} a_{n \plus{} 1}} M_n \equal{} max{a_n,a_(n \plus{} 1),\frac {1}{a_n},\frac {1}{a_{n \plus{} 1}}} Proof: M_(n \plus{} 3) < \equal{} 3/4M_n \plus{} 1/4$[/quote]
$ a_1 > 0,a_2 > 0,a_{n \plus{} 2} \equal{} \frac {2}{a_n \plus{} a_{{n \plus{} 1}}}$
$ M_n \equal{} max {a_n,a_{n \plus{} 1}, \frac {1}{a_n}, \frac {1}{a_{n \plus{} 1}}}$
Proof:$ M_{n \plus{} 3}\leq \frac {3}{4}M_n \plus{} \frac {1}{4}$[/quote]
Thank you Allnames and Erken .


trigonometry
function
integration
calculus
real analysis
real analysis solved

I'm looking for the solution of this problem..
pleaz as soons as possible

f(x)=x(1+cos x)
sf(x):3/2 * sinx + 2 * S
with
S= Sum  n = 2 to infinity of
(-1)^n * sin (nx) / ((n-1)n(n+1))

calculate
Sum n=2 to infinity of
1/((n-1)²(n+1)²).

Sf(x) means the fourrier serie of f(x)..
thanx

i'm terrible at fourier's, so pleaz the full calculation

[quote="Peter Scholze"]$\sum_{n=2}^{\infty} \frac{1}{(n-1)^2(n+1)^2}=\frac{1}{48}(4\pi^2-33)$. you may easily prove this writing it as a sum of fractions.[/quote]
thnx,
but the problem is, the lesson is about fourier.. that is why I would like the answer using fourier series..

First thing about Fourier series: [i]always[/i] give the domain.

You said that $f(x)=x(1+\cos x).$ This is an insufficient description! Fortunately, we can tell what you had to have meant from other clues in the problem. We have a series in $\sin nx$: this means we're dealing with functions periodic of period $2\pi.$ Furthermore, the fact that we have only sines in the series shows us that we are looking at an odd function. Furthermore, the fact that the coefficients are $O(n^{-3})$ suggests that we are looking at a function that is $C^1$ but has a jump discontinuity in the second derivative. Hence our function is:

$f(x)=x(1+\cos x)$ for $x\in[-\pi,\pi],$ continued to be periodic.

This function can be written as a Fourier series in the form

$f(x)=a_0+\sum_{n=1}^{\infty}a_n\cos(nx) + b_n \sin(nx).$

However, since $f$ is odd, $a_n=0\,\forall\,n.$

$b_n=\frac1{\pi}\int_{-\pi}^{\pi}f(x)\sin(nx)\,dx.$

Since $f(x)$ is odd, we can write

$b_n=\frac2{\pi}\int_0^{\pi}x(1+\cos x)\sin(nx)\,dx$

I'll admit that I haven't done these integrals: they can be done by ordinary calculus but the exercise is a bit lengthy. I'll assume that what you have given is correct, namely that

$b_1=\frac32$ and for $n\ge2,\,b_n=\frac{(-1)^n}{(n-1)n(n+1)}.$

Looking ahead to what is asked, we see an extra $n$ in the denominator to get rid of. This can be done by differentiation. (In reasonable cases - which this is - Fourier series can be differentiated term by term.) Hence:

$f'(x)=1+\cos x-x\sin x=\frac32\cos x+\sum_{n=2}^{\infty} \frac  {(-1)^n\cos(nx)}{(n-1)(n+1)},$

valid on $[-\pi,\pi].$

Now we have Parseval's identity (which comes from the trigonometric functions being an ortho-(almost)-normal basis.

If $g(x)=\sum_{n=1}^{\infty}a_n\cos(nx)+b_n\sin(nx),$ then

$\sum_{n=1}^{\infty}\left(a_n^2+b_n^2\right) =\frac1{\pi}\int_{-\pi}^{\pi} g(x)^2dx.$

Apply that in our case to $f'(x)$ and we get

$\frac94+\sum_{n=2}^{\infty}\frac1{(n-1)^2(n+1)^2}= \frac1{\pi}\int_{-\pi}^{\pi}(1+\cos x-x\sin x)^2dx.$

I'll let you do the integrals needed to finish this off.


algebra
polynomial
absolute value
greatest common divisor
algebra proposed

We are given a polynomial $P(x)=x^{n+1}-b_n x^n-......-b_0$ where all the $b_i$ are positive and at least one is nonzero. Prove that $P(x)$ has exactly one positive root, and all other root have modulus less than it.

$f(x) = x^{n+1}(1-\sum_{k=0}^{n}\frac{c_k}{x^k})$ , where all $c_k$ are positive and at least one is nonzero.
But $g(x)=1-\sum_{k=0}^{n}\frac{c_k}{x^k}$ is increasing for $x>0$, and $g(0+)=-\infty , g(+\infty)=1$ , so $g(x), f(x))$ have exactly one positive root, called $p>0$.
Let $u$ a non positive root of $f(x)$. If $u=0$ then $|u|<p$
If $u \neq 0$ then $u$ is a negative real or a complex number.   $f(u)=0 \Longrightarrow g(u)=0 \Longrightarrow \sum_{k=0}^{n}\frac{c_k}{u^k} = 1 \Longrightarrow \sum_{k=0}^{n}\frac{c_k}{|u|^k} \geq 1 $, So $|u| \leq p$ , because $g$ is increasing.
  :cool: q.e.d :cool:

This result is attributed to Cauchy. However, DusT, we may not conclude that every complex root of $P(z)$ has absolute value *strictly* less than the unique positive root $p$. We need an additional condition:

[b]Addition. [/b] (Ostrowski) Let $P(x) = x^n - c_1x^{n-1} - \cdots - c_nx^n$ with $c_i \geq 0$, and assume that the greatest common divisor of the indices $i$ satisfying $c_i > 0$ equals $1$. Then, if $p$ is the unique positive root of $P(x)$, then $|z|<p$ for every other root $z$.

--Vesselin


inequalities
induction
function
Cauchy Inequality
inequalities proposed

Is there a sequence $ a_1,a_2,\ldots$ of positive reals satisfying simoultaneously the following inequalities for all positive integers $ n$:
a) $ a_1\plus{}a_2\plus{}\ldots\plus{}a_n\le n^2$
b) $ \frac1{a_1}\plus{}\frac1{a_2}\plus{}\ldots\plus{}\frac1{a_n}\le2008$?

[edit:] yeah yeah, I misunderstand it...   DON'T READ THIS!!! :)

For $ n\leq 2008$ take $ a_i \equal{} 1$.

When $ 2008k < n \leq 2008(k \plus{} 1)$ take $ a_i \equal{} k \plus{} 1$. Then $ \sum a_i \equal{} n(k \plus{} 1) < n^2$ and $ \sum \frac 1{a_i} \equal{} \frac n{k \plus{} 1} \leq 2008$.
I probably made some mistake  :maybe:
bye

[quote="SpongeBob"]For $ n\leq 2008$ take $ a_i \equal{} 1$.
I probably made some mistake  :maybe: [/quote]It's already gone if you start with this (you've probably misread the question).

It seems that the answer is no, there is no such sequence $ a_n$ (such that the sequence $ h_n \equal{} \sum_{k\equal{}1}^n \frac 1{a_k}$ is bounded), but I don't have a clear proof yet.

Is the inequality $ \sum_{i\equal{}n\plus{}1}^{2n}{\frac{1}{a_i}}\geq 1/4$ clear? Then the solution is clear, too.  :wink:  Ooops, I see I'm posting in the inequalities forum. Sorry....

All right, let's just write it out loud then: $ a_{k\plus{}1}\plus{}\cdots \plus{} a_{2k} < a_1\plus{}\cdots \plus{} a_{2k} \leq 4k^2$, therefore
\[ \frac k{\cfrac 1{a_{k\plus{}1}}\plus{}\cdots\plus{}\cfrac 1{a_{2k}}} \leq \frac {a_{k\plus{}1} \plus{}\cdots \plus{} a_{2k}}{k} < 4k ,\] by AM-GM, which means that $ \frac 1{a_{k\plus{}1}} \plus{} \cdots \plus{} \frac 1{a_{2k}} > \frac 14$.

Obviously this implies that the sequence $ s_n\equal{} \sum_{k\equal{}1}^n \frac 1{a_k}$ is unbounded, so there is no such sequence $ a_n$.

I was actually trying to prove something much harder, and as a matter of fact I'm not even sure that's true: that if $ b_i$ is a sequence such that $ n^2 \geq a_1\plus{}\cdots \plus{}a_i \geq b_1\plus{}\cdots \plus{} b_i$ for all $ i$, then $ \sum_{i\equal{}1}^n \frac 1{a_i} \leq \sum_{i\equal{}1}^n \frac 1{b_i}$ with Abel sums and such, though it seems not to be true ...  :maybe:

I remember that

sum ( k/ (a1+...+ak) ) < 2. (1/a1 +.1/a2 +... )  and 2 is the best constant on the right side

[quote="harazi"]I see I'm posting in the inequalities forum. [/quote]
It's very well! :lol:

[quote="khashi70"]I prove that : $ \sum_{i \equal{} 1}^n \frac {1}{a_{i}} \geq \sum_{i \equal{} 1}^n \frac {1}{2i \minus{} 1}$ by Cauchy Inequality .[/quote]I would like to see that proof, although I doubt very much it's correct.

Ok . Here is my proof :
by Cauchy inequality we have :
$ (\sum _{i \equal{} 1}^n \frac {1}{a_{i}})(\sum_{i \equal{} 1}^n \frac {a_{i}}{(2i \minus{} 1)^2})\geq (\sum _{i \equal{} 1}^n \frac {1}{2i \minus{} 1})^2$
now we need to show that $ \sum_{i \equal{} 1}^n \frac {a_{i}}{(2i \minus{} 1)^2}\leq \sum _{i \equal{} 1}^n \frac {1}{2i \minus{} 1}$.
now we define $ S_{k} \equal{} \sum_{i \equal{} 1}^k a_{i}$ , so we have $ S_{k}\leq k^{2}$ :
$ (\frac {1}{(2n \minus{} 1)^2})S_{n}\leq \frac {n^{2}}{(2n \minus{} 1)^{2}}$
and for all $ 1\leq i\leq n \minus{} 1$ we write :
$ (\frac {1}{(2i \minus{} 1)^{2}} \minus{} \frac {1}{(2i \plus{} 1)^{2}})S_{i}\leq (\frac {1}{(2i \minus{} 1)^{2}} \minus{} \frac {1}{(2i \plus{} 1)^{2}}) i^{2}$
and adding these inequalities :
$ \sum_{i \equal{} 1}^n \frac {a_{i}}{(2i \minus{} 1)^2} \leq \sum_{i \equal{} 1}^{n \minus{} 1} \frac {8i^{3}}{(4i^{2} \minus{} 1)^{2}} \plus{} \frac {n^{2}}{(2n \minus{} 1)^{2}}$
and now by induction u can show that :
$ RH \equal{} \sum _{i \equal{} 1}^n \frac {1}{2i \minus{} 1}$

[quote="d.edgar"][quote="Valentin Vornicu"]I was actually trying to prove something much harder, and as a matter of fact I'm not even sure that's true: that if $ b_i$ is a sequence such that $ n^2 \geq a_1 \plus{} \cdots \plus{} a_i \geq b_1 \plus{} \cdots \plus{} b_i$ for all $ i$, then $ \sum_{i \equal{} 1}^n \frac 1{a_i} \leq \sum_{i \equal{} 1}^n \frac 1{b_i}$ with Abel sums and such, though it seems not to be true ...  :maybe:[/quote]

If the "majorizing" sequence is increasing, then it's true. See here:
http://www.komal.hu/verseny/feladat.cgi?a=feladat&f=A408&l=en[/quote]

Actually if we have that $(a_1,...,a_n) \succ (b_1,...,b_n)$ then applying Karamata inequality to the concave function $f(x)=\frac{1}{x}$ we get $ \sum_{i \equal{} 1}^n \frac 1{a_i} \leq \sum_{i \equal{} 1}^n \frac 1{b_i}$.

But if we don't have the majorizing condition, I don't quite think that it is true :)
Look at this in this way.... We can take the sequences such that $(b_1,...,b_n) \succ (a_1,...,a_n)$, then we have the same inequality, but with the reversed sign !

I don't understand how the solution concludes to the desired result.

@above it shows that the series that should not be greater than $2008$ is actually unbounded, then implying that we can find an $n$ such that the series is strictly greater than $2008$, contradiction.

Let us prove that for any infinite sequence of positive reals satisfying $a_1+a_2+\dots+a_n \le n^2$, and each positive integer $k$ there is an $N\in\mathbb{N}$ for which $\frac{1}{a_1}+\frac{1}{a_2}+\dots+\frac{1}{a_N}>\frac{1}{2}k$. We will induct on $k$.
$k=1$ is trivial, as $a_1\le1 \Rightarrow \frac{1}{a_1}\ge1>\frac{1}{2}$. Assume the induction hypothesis for $k$: $$\frac{1}{a_1}+\frac{1}{a_2}+\dots+\frac{1}{a_N}>\frac{1}{2}k$$
 Now to prove it for $k+1$, assume on the contrary that for any $m\in\mathbb{N}$, $\frac{1}{a_1}+\frac{1}{a_2}+\dots+\frac{1}{a_m}<\frac{1}{2}(k+1)$. Then, $$\frac{1}{2}k<\frac{1}{a_1}+\dots+\frac{1}{a_{N}}<\frac{1}{a_1}+\dots+\frac{1}{a_{N+m}}<\frac{1}{2}(k+1)$$ from where it follows that $$\frac{1}{a_N}+\dots+\frac{1}{a_{N+m}}<\frac{1}{2}$$ for any $m$. But we know from Cauchy that $(\frac{1}{a_N}+\dots+\frac{1}{a_{N+m}})(a_N+\dots+a_{N+m})\ge (m+1)^2>m^2$, which, combined with the above inequality, gives $a_N+\dots+a_{N+m}>2m^2$. So, by the problem condition, we get $(N+m)^2\ge a_1+\dots+a_{N+m}>2m^2$ for any positive integer $m$. But this means $N^2>m(m-2N)$ for any m, which clearly does not hold for sufficiently large $m$. Contradiction. The induction is completed, so we have shown that $\sum \frac{1}{a_i}$ is unbounded.

NOTE: the choice of $\frac{1}{2}$ was arbitrary, and it can in fact be replaced by any real number $0<\lambda<1$, with all the steps remaining analogous.

I will post solution that uses mixing variables and proves stronger fact:

$$ \displaystyle\sum_{i=1}^{c+1} \frac{1}{a_i} > 2008 $$ where $ c = 2^{4014} $.

Assume opposite for the sake of contradiction, that this statement is wrong.

We define parallel sequence $ B = \{ b_1, b_2, \ldots, b_c \} $ such that $ \{ b_1, b_2, \ldots, b_c \} = \{ a_1, a_2, \ldots, a_c \} $ and $  b_1 \le b_2 \le \ldots \le b_c $.
It is obvious that the condition: $ b_1+b_2+\ldots+b_n \le n^2$ is satisfied (we call that $ condition \,1 $) and also that this statement is true: $ \displaystyle\sum_{i=1}^{c} \frac{1}{a_i} \ge \displaystyle\sum_{i=1}^{c} \frac{1}{b_i}$. We can say that $ b_1+b_2+\ldots+b_c = c^2$, because this sum: $ \displaystyle\sum_{i=1}^{n} \frac{1}{b_i} $ decreases when we increase $ b_c $.

[b]Lemma[/b]: $ \frac{1}{x} + \frac{1}{y} \ge \frac{1}{x + d} + \frac{1}{y - d}, x,y > 0, x \le y $ and $ 0 \le d \le \frac{y-x}{2} $.
Proof:
$ \frac{1}{x} + \frac{1}{y} \ge \frac{1}{x + d} + \frac{1}{y - d} \Leftrightarrow \frac{x+y}{xy} \ge \frac{x+y}{(x + d)(y - d)} \Leftrightarrow xy \le (x + d)(y - d) \Leftrightarrow d^2 \le d(y - x) \Leftrightarrow d \le y - x $ which is true.

[b]Transformation[/b]:
$ ({b_t}_i, {b_t}_j) \longmapsto ({b_{t+1}}_i, {b_{t+1}}_j) , {b_{t+1}}_i = {b_t}_i + d, {b_{t+1}}_j = {b_t}_j - d $, where $ d $ is maximum integer such that: $ {b_{t+1}}_i \le {b_{t+1}}_j $. We also apply another condition on choosing $d$: If $ \displaystyle\sum_{i=1}^{i} {b_t}_i \le i^2 $, then $ d \le i^2 - \displaystyle\sum_{i=1}^{i} {b_t}_i $, else $ d = 0 $. We use this transformation on pair $ (i, j), 1 \le i < j \le c $, so $ {b_t}_i \le {b_t}_j $. Because of Lemma: $ \frac{1}{{b_t}_i} + \frac{1}{{b_t}_j} \ge \frac{1}{{b_{t+1}}_i} + \frac{1}{{b_{t+1}}_j} $.

If we use transformations on the these pairs in this order: $ \{ (1,2), (1,3), \ldots, (1,c), (2,3), \ldots, (2,c), \ldots, (c-1,c) \} $ (there are $ c-1 $ trasformation made on each term), we will get sequence: $ B_c= \{ {b_c}_1, {b_c}_2, \ldots, {b_c}_c \} $ (if we start from sequence $ B_1 $).
I will prove that this equality is true: $ b_n = 2n - 1 $.
Assume, for the sake of contradiction, that there exist $ k \in \{2, 3, \ldots, c \} $ such that $ {b_c}_k > 2k - 1 $ and $ {b_c}_i \le 2k - 1 $ for each $ i \in \{1, 2, \ldots, k-1 \} $ ($ k \neq 1 $ because we always choose $ d $ such that $ {b_{t+1}}_1 \le 1 $). After each transformation, $ condition \,1 $ is satisfied, so: $ b_1+b_2+\ldots+b_{k-1} < (k-1)^2$, but that is in contradiction with choosing maximum $ d $ in pairs: $ \{ (1,k), (1,k), \ldots, (k-1,k) \} $. Contradiction, $ b_n = 2n - 1 $, for each $ n \in \{1, 2, \ldots, c \} $.

We are now done because:
$$ \displaystyle\sum_{i=1}^{c+1} \frac{1}{a_i} > \displaystyle\sum_{i=1}^{2^{4014}} \frac{1}{a_i} \ge \displaystyle\sum_{i=1}^{2^{4014}} \frac{1}{b_i} \ge \displaystyle\sum_{i=1}^{2^{4014}} \frac{1}{{b_c}_i} = \displaystyle\sum_{i=1}^{2^{4014}} \frac{1}{i} = \frac{1}{1} + \displaystyle\sum_{i=1}^{4014} \displaystyle\sum_{j=1}^{2^{i-1}} \frac{1}{2^{i-1} + j} \ge \displaystyle\sum_{i=1}^{4014} \displaystyle\sum_{j=1}^{2^{i-1}} \frac{1}{2^{i}} = 1 + \displaystyle\sum_{i=1}^{4014} \frac{1}{2} = 1 + \frac {4014}{2} = 2008 $$

This is contradiction, so answer is no, there is no such a sequence.

Looking at Romanian mathematical competitions 2002 booklet, I just found that the same problem was already posted at "Sixth regional contest 'Alexandru Papiu Ilarian' " that was organized in 2001. There, part b) of the first problem for grade 12 was to prove that under the condition that for positive reals $x_{1}$, $x_{2}$, $\dots$ inequality $x_{1} + x_{2} + \dots x_{n} \leq n ^ {2}$ holds for all positive integers $n$ then we have
\[
\lim_{n\rightarrow \infty} \left(\frac{1}{x_{1}} + \frac{1}{x_{2}} + \dots + \frac{1}{x_{n}} \right) = \infty
\]
so the problem seems to be completely non-original (unless I am completely wrong). Clearly, I do not claim that the author whom this problem was attributed (as far as I remember it was Nikolai Nikolov from Bulgaria) at Balkan M.O. copied it from the other resources. Although the official solutions at both  places are pretty much the same, due to the nature of the problem, it is quite likely that two authors might have come up with it independently. However, it is quite interesting that no one in the jury and problem selection committee noticed this.

No such sequence exists. The key is to observe for each \(k\) that \[\sum_{i=k+1}^{2k}\frac1{a_i}\ge k\cdot\frac1{(a_{k+1}+\cdots+a_{2k})/k}\ge\frac k{4k^2/k}=\frac14.\] It follows that \[\sum_{i=1}^{2^{8032}}\frac1{a_i}\ge a_1+{\underbrace{\frac14+\cdots+\frac14}_{8032}}>2008.\]


Let $n$ be a positive integer number. Prove that:
$\sqrt{n}<1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{n}}$

[quote="hien"]Let $n$ be a positive integer number. Prove that:
$\sqrt{n}< 1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{n}}$[/quote]
Indeed very easy.  :D
Induction. The first step is obvious. We have to show $\sqrt{n+1}< 1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{n+1}}$ when we know that $\sqrt{n}< 1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{n}}$. Of course $\sqrt{n+1}-\sqrt{n}<\frac{1}{\sqrt{n+1}}$ sufficies. It is equivalent to $\sqrt{n+1}-\frac{1}{\sqrt{n+1}}<\sqrt{n}\iff \frac{n}{\sqrt{n+1}}< \sqrt{n}\iff n<\sqrt{n}\cdot\sqrt{n+1}$ which is obviously true since $\sqrt{n}\geqslant \sqrt{n}$ and $\sqrt{n+1}>\sqrt{n}$ and we can multiply these inequalities sidewise to obtain a sharp inequality.

[quote="Rzeszut"]
 Of course $\sqrt{n+1}-\sqrt{n}<\frac{1}{\sqrt{n+1}}$ sufficies.[/quote]
$\sqrt{n+1}-\sqrt{n}=\frac{1}{\sqrt{n+1}+\sqrt{n}}$ and we are done :D


calculus integration probability MATHCOUNTS FTW

How many ways are there to roll 5 dice to get a sum of 10?

[hide="using casework..."]If the first roll gives us a $ 6$, then we can have $ 6 \plus{} 1 \plus{} 1 \plus{} 1 \plus{} 1 \equal{} 10$ which yields $ 5$ ways.

If the first roll gives us a $ 5$, then we can have $ 5 \plus{} 2 \plus{} 1 \plus{} 1 \plus{} 1 \equal{} 10$ which yields $ 10$ ways.

If the first roll gives us a $ 4$, then we can have $ 4 \plus{} 3 \plus{} 1 \plus{} 1 \plus{} 1 \equal{} 10$ for a total of $ 10$ ways.
We can also have $ 4 \plus{} 2 \plus{} 2 \plus{} 1 \plus{} 1$ which gives us $ 30$ ways.

If the first roll gives us a $ 3$, then we can have $ 3 \plus{} 3 \plus{} 2 \plus{} 1 \plus{} 1$ for a total of $ 30$ ways.
We can also have $ 3 \plus{} 2 \plus{} 2 \plus{} 2 \plus{} 1$ for a total of $ 10$ ways.

If the first roll gives us a $ 2$, then we can have $ 2 \plus{} 2 \plus{} 2 \plus{} 2 \plus{} 2 \equal{} 10$ for a total of $ 1$ way.

We have already counted the rest of the ways for 2, and the entire number of ways for 1, so our answer is $ 5 \plus{} 10 \plus{} 10 \plus{} 30 \plus{} 30 \plus{} 10 \plus{} 1$ which gives us $ \boxed {96}$ total ways to get a sum of 10 using 5 dice.[/hide]

[hide="balls and urns way."]Set up an equation: $ a \plus{} b \plus{} c \plus{} d \plus{} e \equal{} 5$ (we have the sum as $ 5$, not $ 10$ because each dice MUST be at minimum a $ 1$, so we get $ 10\minus{}5\equal{}5$)

In general, if you have an equation $ a_1\plus{}a_2\plus{}a_3\plus{} \dots \plus{}a_n\equal{}z$, the number of non-negative integral solutions (which means that $ 0$ is included) is $ \binom{z\plus{}n\minus{}1}{z}$.

So if we have $ a\plus{}b\plus{}c\plus{}d\plus{}e\equal{}5$, by this formula, the number of non-negative integral solutions is $ \binom{5\plus{}5\minus{}1}{5}$, which is equal to $ \binom {9}{5} \equal{} 126$[/hide]

Can anyone tell me what is wrong with my casework???

[quote="athunder"]Personally i think that u should ALWAYS use casework...like it would be out of pure luck you would get a question like this, the balls and urns will not work for like a sum of 15.[/quote]

Your right on the sum of 15 thing (unless you want to split up into looootttts cases to correct for overcounting), but for MC problems and stuff, they always give you these kinds of problems that have minimal, if any, overcounting.

Like one problem on last years state was "whats the probability that of 3 dice rolled, the sum is 9?"

for that, you have x+y+z=6, so there are 8C2 solutions, but we overcounted thrice (Since we already set each dice equal to one, having (x,y,z)=(6,0,0),(0,6,0),(0,0,6) will lead to dice rolls of (7,1,1), (1,7,1), and (1,1,7), which is obviously not possible). So the probability is (28-3)/(6x6x6)=25/216

What Im getting at is that for contests that use this method of "dice equations," like mathcounts, the questions will usually be very straightforward.

@ thunder, nice use of old language english. "Thrice" ftw :D

@ athunder, dang, that was a very stupid mistake....

Yep, analyze every problem separately, and in each situation, find and use the best approach.

In this case, the problem allows you to do balls and urns so do it! If anyone is wondering, we can do it with no complications in this situation, but not most of the time because, when every die is atleast 1, we never accidentally end up with a 7 on a die while trying to achieve a sum of 10.

1+1+1+1+6=10 is the extreme.

http://alt2.artofproblemsolving.com/Wiki/index.php/Distinguishability
You can learn about balls and urns here on this aops site

seriously, he really didn't need that. He knows PERFECTLY how to do balls and urns as shown by his epic pwnage. :)

Yes, that is totally true. The Balls and Urns method was given by Thunder365, who is the true beast. Not me, cuz I am epic failure.


A man makes a trip by automobile at an average speed of $ 50$ mph. He returns over the same route at an average speed of $ 45$ mph. His average speed for the entire trip is:

$ \textbf{(A)}\ 47\frac{7}{19}\qquad
\textbf{(B)}\ 47\frac{1}{4}\qquad
\textbf{(C)}\ 47\frac{1}{2}\qquad
\textbf{(D)}\ 47\frac{11}{19}\qquad
\textbf{(E)}\ \text{none of these}$

[hide="Solution"]
Let $ d$ be half the total distance, and $ r_1,t_1$ and $ r_2,t_2$ be the rate and time to the way there and back respectively, such that:

$ r_1 \equal{} \dfrac{d}{t_1} \equal{} 50$

$ r_2 \equal{} \dfrac{d}{t_2} \equal{} 45$

The average speed in the whole trip is measured by $ \dfrac{2d}{t_1\plus{}t_2}$, the total distance divided by the amount of time.

$ \dfrac{d}{t_1}\cdot \dfrac{9}{10} \equal{} \dfrac{d}{t_2}\implies t_2 \equal{} \dfrac{10 t_1}{9}$.

$ t_2 \plus{} t_1 \equal{} \dfrac{10 t_1}{9} \plus{} t_1 \equal{} \dfrac{19t_1}{9}$

Now we substitute this:

$ \dfrac{2d}{t_1\plus{}t_2} \equal{} \dfrac{2d}{\dfrac{19t_1}{9}} \equal{} \dfrac{18d}{19t_1} \equal{} \dfrac{18}{19}\cdot \dfrac{d}{t_1} \equal{} \dfrac{18}{19}\cdot 50 \equal{} \dfrac{900}{19} \equal{} \boxed{47\plus{}\dfrac{7}{19}}$

So the answer is $ \textbf{(A)}$
[/hide]

[hide="Quick Solution"]
It is known that the average speed of two given speeds is the harmonic mean of those speeds. Therefore, we have the average speed equal to $ \frac{2(50)(45)}{50 \plus{} 45} \equal{} \frac{2(50)(45)}{95} \equal{} \frac{900}{19} \equal{} 47\frac{7}{19}$. Therefore, the answer is $ \fbox{(A)}$.
[/hide]

wow I knew there had to be a faster way, this would have been useful when I used to compete in mathcounts :(


Prove that the orthocentre of the Fuhrmann triangle of triangle $ABC$ is the incentre of triangle $ABC$.

Dear Mathlinkers,
working again on the Fuhrmann triangle, I discovered this problem in the Mathlinks archives.
For a synthetic solution see
Revistaoim 26 (2006) directed by professor Francisco Bellot Rosado or
http://perso.orange.fr/jl.ayme  vol. 1
Sincerely
Jean-Louis

See also [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=101687[/url] for both a trigonometric and synthetic proof.

See the generalization [url=http://www.artofproblemsolving.com/community/c6h624925]here[/url] ([b]Theorem 4[/b]) :)


I met in a Bulgarian book the following theorem:

[quote]If $ k$ is an integer such that there exist an odd $ a$ and an even $ b$ $ \left(b\not\equiv 0 (\mod 3)\right)$, which don't have common divisor of the form $ 4t \plus{} 3$, and $ k \equal{} b^2 \minus{} a^2$, then if $ k\not\equiv \minus{} 1(\mod 8)$, the equation $ x^2 \plus{} k \equal{} y^3$ doesn't have a solution in $ \mathbb{Z}\times\mathbb{Z}$.[/quote]

My question is whether this is really a [b]Mordell[/b]'s theorem. Is it a consequence of the real well-known [b]Mordell [/b]theorem for elliptic curves:

[quote]The group $ E(\mathbb{Q}) \equal{} \{(x,y)\in \mathbb{Q}^2: y^3 \equal{} x^3 \plus{} ax \plus{} b\}$ is finitely generated. [/quote]

 :?:  :?:

No, but obviously both bother with elliptic cuves, which are one of Mordells main research areas.

:maybe:  OK, 10x.
By the way is there a theorem which gives us a sufficient and necessary condition when arbitrary elliptic curve has no solution in integer numbers ? :ninja:

Theoretically, yes. Effective versions of Siegels theorem tell us upper bounds for the size of integer points. But that bounds are very very bad (the original one by Baker was $ e^{{(10^6 \max(|a|,|b|,|c|))}^{10^6}}$ for $ y^2 \equal{} x^3 \plus{} ax^2 \plus{} bx \plus{} c$).


conics
parabola
calculus
function
special factorizations

A farmer sells beans.  His regular price is $1.00 per bean, but after a while, he noticed that people really weren't very likely to buy his beans, so he decided to lower the price by the following formula.  For every bean he has, he subtracts 2 cents from the price per bean.  For example, if he has 25 beans, then they sell for 100 - 2*25 = 50 cents each.  What is the number of beans that will give him the most profit assuming that it costs nothing to get a bean and he sells every bean?

Hint: [hide]Write the profit as an equation with the number of beans as a variable.[/hide]

Another Hint:[hide] Complete the square.[/hide]

BobofBobs wrote:I think I've got it...

Is it: [hide] P=Profit, B=#ofBeans
P=B(100-2B)
  =100B-2(B^2)

The farmer breaks even at 50 beans, so at 25 beans, it must be the peak, which yields a profit of

P=(100*25)-2(25^2)
  =(2500)-2(625)
  =2500-1250
  =1250 cents
  =$12.50[/hide]

I am pretty sure this is right. Please notify me of my mistakes.

It's right, but how did you arive at the anwer of [hide]25[/hide] beans?  How are you sure that is the maximum?

Thanks, nr1337.  I was going to say that :-)

I would [hide]find the vertex of the parabola[/hide] but that doesn't seem very Getting Started-ish. I don't see any more intuitive way, but there probably is one.

This is how I knew the answer was[hide] 25 beans. I worked out the formula for 50 ( I didn' show it). I also worked it out for 1, 2, and 3 as the number of beans (also not shown). It is obvious that if the farmer has 0 beans, then P=0. I then concluded that the profit rose, reached a peak,  and then fell to 0 at the same rate it rose. Yes, this does form a parabola--I didn't notice that. If it falls at the same rate it rises, the number of beans for maximum profit must be the average of the number of beans for 0 profit.

(0+50)/2=50/2=25

You could also see this if you drew a parabola to the equation y=100x-2(x^2)

[hide][/hide][/hide]

Two ways of doing it:

[hide]

p = price

n = number of beans

The profit is the number of beans times the price: np.

p = 100-2n

np = 100n-2n^2

Completing the square on the RHS of the equation above,

np = -2(n-25)^2 - 625.

We want (n-25)^2 to be as small as possible. So n-25 = 0.  n = 25.  Profit = $12.50.

Of course, there's always the obviously not getting started calculus way of doing this problem in which we set dnp/dn equal to 0 and solve for n.  But that's not getting started.

[/hide]

The calc method is essentially the same thing, but it's more general for any function that is convex/concave.


function trigonometry algebra polynomial

Find the ranges of the following functions for $ 0\leq x\leq \pi$.

$ f(x)\equal{}\sqrt{3}\sin x\plus{}3\cos x$.

$ g(x)\equal{}3\sin ^ 2x\plus{}6\sqrt{3}\sin x\cos x\plus{}9\cos ^ 2x\minus{}2\sqrt{3}\sin x\minus{}6\cos x$.

[hide="f(x)"]$ f(x)\equal{}2(\frac{\sqrt{3}}{2}\sin x\plus{}\frac{1}{2}\cos x)\equal{}2(\cos\frac{\pi}{6}\sin x\plus{}\sin\frac{\pi}{6}\cos x)\equal{}2\sin(x\plus{}\frac{\pi}{6})$
We have $ 2\sin\frac{7\pi}{6}\leq f(x)\leq 2\sin\frac{\pi}{2} \iff \minus{}1\leq f(x) \leq 2$.
[/hide]
[hide="g(x)"]$ g(x)\equal{}f(x)^2\minus{}2f(x)$. We see that $ g(x)$ has the range of polynomial $ p(t)\equal{}t^2\minus{}2t$ for $ \minus{}1\leq t \leq 2$. The maximum is when $ t\equal{}\minus{}1$, and the minimum is when $ t\equal{}1$, and we obtain $ \minus{}1 \leq g(x) \leq 3$.[/hide]
Nice problem! :)

hello
[hide="f(x)"]
$ f(x)\equal{}\sqrt{12}sin(\frac{\pi}{3}\plus{}x)$
so range is $ [\minus{}2\sqrt{3},2\sqrt{3}]$

[/hide]
thank u

really i think its correct Kunny.Can u tell me my mistake?I even checked it with my graphical calculator.

hello, i have found $ f(x) \equal{} 2\sqrt {3}\cos(\frac {\pi}{6} \minus{} x)$ and so we have $ \minus{} 3\le f(x) \le2\sqrt {3}$.
Sonnhard.

hello kabi, i think you must have a look at the interval $ 0\le x \le\pi$.
Sonnhard.

@kabi Did you take into consideration of the range $ 0\leq x\leq \pi$?

got it Dr.Thank u :lol: U r simply great.
Sorry Kunny.I haven't seen that. :oops:

 I think Brut3Forc3 just note down the wrong question otherwise his approach is absolutely right.
 :)
so from the method suggested by him we will get this answer.
max will be at $ f(x)\equal{}\minus{}3$
min will be at $ f(x)\equal{}1$
range =$ [\minus{}1,15]$
thank u

Oh, I put 1 cos x instead of 3 cos x. My bad!  :oops:


geometry
circumcircle
incenter
geometric transformation
reflection
ratio
trigonometry

[quote="epitomy01"]I think this problem, in slightly different form was given in Shortlist 99:
http://www.kalva.demon.co.uk/short/soln/sh99g8.html
Strange that it appeared in a Chinese TST then.  :huh:[/quote]

actually it's harder than the one that was given in shortlist.and wants more things.

notice that $ \frac {PN}{PM} \equal{} \frac {NI_1}{MI_2} \equal{} \frac {NI}{MI}$ (cause $ NI \equal{} NI_1,MI \equal{} MI_2$) hence the circumcircle of $ \triangle AIP$ is apollonius circle and devide $ MN$ in a harmonic ratio.
now let DI intersect circumcircle of $ ABC$ at $ P'$ and $ A'$ (Where A' and A are on the same side of MN) cause the line DA is reflection of line DA' wrt MN and D is midpoint of MN hence $ AA'\parallel{}MN$ hence $ \angle NP'A' \equal{} \angle ANM \equal{} \angle IND$ hence $ DN^2 \equal{} DI.DP'$ but $ DN^2$ is power of D wrt circumcircle of AIP hence P' lies on  circumcircle of AIP also hence $ P\equiv P'$

I have used another idea.  :blush:

If $ IP$ passes through the mid-point of $ MN$, then $ \bigtriangleup INP$ and $ \bigtriangleup IMP$ have the same area,

which is equivalent to $ \frac{PB}{PC}\equal{} \frac{sin^2 \frac{C}{2}}{sin^2 \frac{B}{2}}$

But I don't know how to find this ratio easily, so I try trigonometry:

As $ \bigtriangleup PNM$ and $ \bigtriangleup PI_1I_2$ are similar, we have $ \frac{PN}{PM}\equal{} \frac{NI_1}{MI_2}\equal{} \frac{sin \frac{C}{2}}{sin \frac{B}{2}}$

Therefore if we let $ \angle PAB\equal{}\alpha$, $ \angle PAC\equal{}\beta$

We have

$ \frac{sin \frac{C}{2}}{sin \frac{B}{2}}\equal{} \frac{PN}{PM}\equal{} \frac{sin( \frac{C}{2}\plus{} \alpha)}{sin( \frac{B}{2}\plus{} \beta)}$

Also $ \alpha\plus{} \beta \equal{} \angle A$

And $ \frac{PB}{PC}\equal{} \frac{sin \alpha}{sin \beta}$

Thus for $ \alpha\plus{} \beta \equal{} \angle A$ and $ \frac{sin \frac{C}{2}}{sin \frac{B}{2}}\equal{} \frac{sin( \frac{C}{2}\plus{} \alpha)}{sin( \frac{B}{2}\plus{} \beta)}$

We hope to show $ \frac{sin \alpha}{sin \beta} \equal{} \frac{sin^2 \frac{C}{2}}{sin^2 \frac{B}{2}}$

To prove this, I have constructed another right-angled triangle  :wallbash: A very long solution

I have a short solution. It seems that I use similar ideas to Amir.S. I prefer a different notation.

Problem: Let $ \triangle ABC$ have circumcircle $ \omega$ and incenter $ I$. $ BI\cap\omega\equal{}\{B,N\}$; $ CI\cap \omega\equal{}\{C,P\}$; $ M$ is the midpoint of $ NP$. $ G$ is an arbitrary point on arc $ BC$ not containing $ A$. $ I_1;I_2$ are the incenters of $ ABG;ACG$. The circumcircle of $ I_1I_2G$ intersects $ \omega$ again at $ T$ (if they are tangent, $ T\equal{}G$). Prove that $ P,I,T$ are collinear.

Proof:

Lemma 1: $ \triangle TI_1I_2\sim\triangle TPN$ and $ \triangle TPI_1\sim\triangle TNI_2$. Proof: note $ G,I_1,P$; $ G,I_2,N$ are triples of collinear points. From the fact that $ NPTG$ and $ I_1I_2TG$ are cyclic, we get the first similarity. Then we get the second similarity.

Lemma 2: Let $ AI$ intersect the circumcircle of $ ABC$ at $ A'$. Then $ A'I\equal{}A'B\equal{}A'C$. Proof: angle chase.

Lemma 3: $ B,C$ are fixed points. The locus of points, $ P$, such that $ BP: PC\equal{}k\ne 1$ is a circle. Proof: well-known (coordinates? there are other proofs).

Lemma 4: $ ABCD$ with $ AB\parallel CD$. $ E\equal{}AC\cap BD$; $ F\equal{}BC\cap AD$, $ M,N$ are midpoints of $ AB$, $ CD$. Then $ M,N,E,F$ are collinear. Proof: well-known.

Using lemma 1 and 2, $ TP: TN\equal{}PI_1: NI_2\equal{}AP: AN$ which is constant. By lemma 3, there are only two points that lie on $ \omega$ and have that constant ratio since two distinct circles intersect at at most two points. Thus $ T$ is fixed and unique.

Now choose $ G\equal{}T$ and define incenters $ I_1'$ and $ I_2'$. Then $ C(GI_1'I_2')$ must be be tangent to $ \omega$ (if we let there be a second intersection of those circles, we get a contradiction since $ T$ is unique). It follows that $ I_1'I_2'\parallel NP$. Then we are done by lemma 4 using trapezoid $ NPI_1I_2$.

[quote="encyclopedia"]Let me propose another problem  :P

Let $ ABC$ be a triangle inscribed $ (O)$ and $ P$ be a point lies on small arc $ BC$, let $ I_1,I_2$ be A-excenter of triangle $ APB,APC$ prove that circumcircle of $ PI_1I_2$ always pass through a fix point.

[hide="Hint"]This point is intersect of the line connecting $ I_a$(A-excenter of $ ABC$), midpoint $ P$ of $ MN$(with $ M,N$ is midpoints of big arc $ AC,AB$,resp) and $ (O)$[/hide][/quote]
dear encyclopedia
Could you draw a figure?I can't see your result :(

Hi dear plane geometry, maybe I has a typo I denote midpoint of $ MN$ as the same point $ P$, sorry, now I let it be point $ K$.

PROOF:
1.It is easy to prove  A,B,Ia1,Ia are concyclic
 ∠IaBP=90-∠Ia1BIa-1/2∠ABP=90-1/2∠PAC-1/2∠ABP=1/2(∠ACP-∠PAC)
Notice N is the midpoint of big arc APC
so∠NBP=1/2(∠ACP-∠PAC)
this indicates that Ia,N,B are collinear
analogously,Ia,M,C are collinear

2.From 1 we can acquire NIaMQ is a parallelogram
  Denote R,S are midpoints of IaQ,IaA , respectively
  Ia,R,Q are collinear
  From the original problem we know that AQ⊥IA
  And we have MN//AQ  => MN⊥IA => MN is the perpendicular bisector of AIa

3.form 2 we have MN is the perpendicular bisector of AIa
  But we already know that M is on perpendicular bisector of AB
   => M is the circumcenter of ⊙A,B,Ia1,Ia
  analogously, N is the circumcenter of ⊙A,C,Ia2,Ia

4.from 3 , ∠Ia2PM=∠Ia2PC-∠CPM=90-1/2∠B-1/2(∠A-∠B)=90-1/2∠A
  ∠MQN=∠A-1/2(∠A-∠C)-1/2(∠A-∠B)=90-1/2∠A
  N,P,Ia2 are collinear
  analogously, M,P,Ia1 are collinear

5.Denote T=⊙O∩IaQ
MT/MIa1=MT/MA=MT/NQ   NT/Nia2=NT/NA=NT/MQ
  Notice we have   MT/NQ=RT/NR=RT/RM=NT/MQ
 ∠TNP=∠TMP    =>  △NTIa2∽△TMIa1
P,T,Ia1,Ia2 are concyclic

Thus we are done!

[quote="Fang-jh"]Let $ ABC$ be an acute triangle, let $ M,N$ be the midpoints of  minor arcs $ \widehat{CA},\widehat{AB}$ of the circumcircle of triangle $ ABC,$ point $ D$ is the midpoint of segment $ MN,$ point $ G$ lies on minor arc $ \widehat{BC}.$ Denote by $ I,I_{1},I_{2}$ the incenters of triangle $ ABC,ABG,ACG$ respectively.Let $ P$ be the second intersection of the circumcircle of triangle $ GI_{1}I_{2}$ with the circumcircle of triangle $ ABC.$ Prove that three points $ D,I,P$ are collinear.[/quote]
It is not difficult to see that $ NB \equal{} NI_1 \equal{} NI \equal{} NA$ and $ MC \equal{} MI_2 \equal{} MI \equal{} MA$.
Since $ \angle I_1PN \equal{} \angle I_2PM$ and $ \angle NI_1P \equal{} \angle PI_2M$,we have $ \triangle PI_1N$ and $ \triangle PI_2M$ are congruent.
=>$ \frac {PM}{PN} \equal{} \frac {I_2M}{I_2N} \equal{} \frac {AM}{AN} \equal{} \frac {sin\frac {B}{2}}{sin\frac {C}{2}}$
=>$ PM sin \frac {C}{2} \equal{} PNsin\frac {B}{2}$
Since $ \triangle PIN$ and $ \triangle PCM$ are congruent,we have $ \angle IPN \equal{} \angle CPM \equal{} \frac {B}{2}$.Likewise,$ \angle IPM \equal{} \frac {C}{2}$.
$ D,I,P$ are collinear <=> $ [IPN] \equal{} [IPM]$ <=> $ PMsinIPN \equal{} PNsinCPM$ <=> $ PM sin \frac {C}{2} \equal{} PN sin \frac {B}{2}$,which is proven above.$ Q.E.D$

This is too nice a problem!

Basically, after disposing of all the non-sense fluff about the incenter and the mid-point of the arc $BAC$ it is just this

[url]http://www.artofproblemsolving.com/community/u243741h19777p5682901[/url]


analytic geometry
group theory
abstract algebra
arithmetic sequence
number theory unsolved
number theory

Prove: there are no four distinct positive integers a, b, c, d such that a^2, b^2, c^2 and d^2 form an arithmetical progression. I know the way to prove this is by infinite descent (You assume there is a solution and show this implies the existence of a smaller solution, which is impossible since there must be a smallest solution, if a solution exists at all) but I can't prove it.

It give ${a=\frac{k-mn}{2}, b=\frac{k+mn}{2},c=\frac{kn-m}{2},d=\frac{kn+m}2}$, were
(1) ${(k^{2}-m^{2})(n^{2}-1}=4mn(k+1)$,
and $k=m(mod \ 2)$ naturals (n can be rational, but kn,mn - natural).
We can find solutions (1) for n=3. It give
(2) $2k^{2}-3mk-2m^{2}-3m=0.$
Let $13m+3=3x,2k-3m=3y$, then $x^{2}-13y^{2}=1$.
It give infinetely many solutions with $gcm(a,b,c,d)=1$ (if a,b,c,d - solutions, then as,bs,cs,ds - solutions).

I am sorry for my mistake. I prove, that four squares distinct naturals give not arithmetic progression.
For it consider all arithmethic progressions $a^{2},b^{2},c^{2}$. It give $a=m(x^{2}-2x-1),b=m(x^{2}+1),c=m(x^{2}+2x-1)$, were x is natural (x>1), m is natural or $m\in (1/2)N, \ x-\ odd$. Let $b^{2},c^{2},d^{2}$ give arithmetic progression. Then $b=n(y^{2}-2y-1),c=n(y^{2}+1),d=n(y^{2}+2y-1)$. It give elliptic equation: $(x^{2}+1)(y+1)=(x-1)(y^{2}-2y-1).$
We have $y=\frac{x^{2}+2x-1\pm \sqrt{z(x)}}{2(x-1)}, \ z(x)=(x^{2}+4x-7)^{2}+24x-6$. It is easy to check, that if x>20 $(x^{2}+4x-7)^{2}<z(x)<(x^{2}+4x-6)^{2}$, and $z(x)\not =t^{2}$, when $1<x\le 20$.

Hello Rust,
  You say that these are the solutions for $a^{2}+c^{2}=2b^{2}$? I don't think so. Please give me the $m$ and $x$ for $41^{2}+1^{2}+2\cdot 29^{2}$.

[quote="harazi"]Hello Rust,
  You say that these are the solutions for $a^{2}+c^{2}=2b^{2}$? I don't think so. Please give me the $m$ and $x$ for $41^{2}+1^{2}=2\cdot 29^{2}$.[/quote]
I am sorry $x$ are rational. For this case $m=\frac{9}{2},x=\frac{7}{3}$.
Therefore we must find rational solutions of equation:
$(x^{2}+1)(y+1)=(x-1)(y^{2}-2y-1)$.

Elliptic curve in projective coordinates can be read in form:
(1) $\sigma_{1}\sigma_{2}=5\sigma_{3},$
$\sigma_{1}=x+y+z,\sigma_{2}=xy+xz+yz, \sigma_{3}=xyz$.
This curve had 6 points $(1,0,0),(0,1,0),(0,0,1),(1,-1,0),(1,0,-1),(0,1,-1)$. But this pointis give step of arithmetic progression 0 or $\infty$, and give periodic subgroup of group rational points $Z_{2}+Z_{3}$. Your problem had solution if and only if exist integers a,b ($ab\not =0$), suth that all roots of equation
$x^{3}-5ax^{2}+bx-ab$ are integers.
It is equavalent that curve (1) had positive rang, and exist infinetely many numbers a,b,c,d, suth that $(a,b,c,d)=1$ and $a^{2}-b^{2}=b^{2}-c^{2}=c^{2}-d^{2}$.


$a \in R$, solve for $x$: $|x| > \dfrac{6a^2}{a-x}$.

hello, i have got $ a>0$ and $ x<\minus{}2a$  or $ x>a$. In the case $ a\le0$ we have $ x<3a$ or $ x>a$.
Sonnhard.


i want a good higher bounds of the expression

$ \frac{x(a\plus{}b)}{y\plus{}z}\plus{}\frac{y(b\plus{}c)}{z\plus{}x}\plus{}\frac{z(a\plus{}c)}{y\plus{}x}$ :maybe:

really tanx arqady but this page get us the minimum of this expression do you know what the maximum? i just know this maximum is higher than $ x\plus{}y\plus{}z$

[quote="Kamyar1991"]really tanx arqady but this page get us the minimum of this expression do you know what the maximum? i just know this maximum is higher than $ x \plus{} y \plus{} z$[/quote]
But if $ x \equal{} y\rightarrow0^ \plus{}$ then your expression $ \rightarrow \plus{} \infty.$ :wink:
( I think you have meant that $ a,$ $ b,$ $ c,$ $ x,$ $ y$ and $ z$ are positive numbers. )
By the way, the girl in your picture is very pretty.  :lol:


trigonometry
calculus
calculus computations

Find  $\displaystyle \prod_{p\in \mathbb{Q}}\cos(p)$.

I have no idea :blush: . Thats why I posted this problem here. This problem came originaly from $\prod_{n=1}^{\infty}\cos(\frac{1}{n})$ , but then I wondered what would be result if we would consider product on all $\mathbb{Q}$.

we can find a sequence $(r_{n})$ of rational numbers that converges to $\pi/2$. then

$\left| \prod_{r\in\mathbb{Q}}\cos r \right| \leq \left| \cos r_{n}\right|$

for all $n$. you can apply the squeezing theorem to conclude that it converges to 0.

Thanks!I didnt think about that. Could we apply the same argument to $\prod_{n=1}^{\infty}\cos(\frac{1}{n})$ ?  (seems likely to me  :huh: )

The product $\prod_{n=1}^{\infty}\cos\left(\frac1n\right)$ converges to a positive number, not zero.

The general principle: consider $\prod_{n}(1+a_{n})$ (and to make things simple for the moment, assume that $a_{n}$ has constant - or eventually constant - sign). Then the product converges to a nonzero number iff $\sum_{n}a_{n}$ converges.

For the particular case, note that $\cos\left(\frac1n\right)=1-\frac1{2n^{2}}+O(n^{-4}).$

Running up the first thousand or so partial products and employing a Richardson extrapolation suggests that the value of the product is approximately $0.3885362.$

Thanks prof. Marryfield!  Could you direct to a web-page (or a book) so I could find more about Richardson extrapolation.

Richardson extrapolation is a numerical analysis topic and technique. One place to find it would be the Burden and Faires numerical analysis textbook. It is based on certain assumptions about asymptotic behavior and expansion into series. Had I not used that technique, I would simply have said that the 1000th partial product was about 0.3887 and that it seemed to be stabilizing somewhere near there.


Observer A detects two explosions at the same time a distance of $100$ km apart. Observer B detects that the explosions are $160$ km apart. How far apart in time does observe B detect them the two explosions to be? According to a previous post http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=lorentz&t=57472
I should be able to plug in $t_{2}'=\frac{\sqrt{x_{2}^{'2}-x_{2}^{2}}}{c}=\frac{\sqrt{160000^{2}-100000^{2}}}{3\times 10^{8}}=5.3\times 10^{-4}$ s.

But my text has nothing of this sort in it. I did it using Lorentz transforms, we can find $v$ of A with respect to B by $100=160\sqrt{1-v^{2}/c^{2}}\Rightarrow v=.78c$, then since $t=\gamma(t'+\frac{v}{c^{2}}x')=\frac{160}{100}\left(\frac{.78(160000)}{c}\right)=6.6\times 10^{-4}$ s.

Why the discrepancy? I'm assuming it has something to do with the longer way I did.

The invariant interval approach is the correct one, your approach is ok but you made two mistakes. The $x'$ in your solution is $100\text{km}$, and the first numerical result is wrong.


[b] Round 3 due Friday at 9:00 PM.[/b]

If you don't know the rules, search "Half the Average Game 1".

Game 1 was a bit short, so I am going to extend it to 20 people.  Note that with more people, there will be more people who don't submit by the time limit.

[u]SCORES (7/17)[/u]
Alanchou: 67
Life=tennis+math: 15
junggi: 12
BanishedTraitor: 12
Klebian: 9
perfect628: 7
mustafa: 3

/in, but I'm gonna get owned because of ARML and MOSP

I will stretch out the deadlines on the weekends of ARML and MOSP.  I too am going to ARML.  Could somebody tell me when MOSP is so I can stretch the deadline?

Personally, I think "I choose you, Voldermolt" rolls of the tongue better.

Once again, the round ends late.
[size=200][b]ROUND 3 RESULTS[/size][/b]
[hide="Submissions"]Alanchou: No submission (100)
Klebian: No submission (100)
perfect628: No submission (100)
mustafa: No submission (100)
BanishedTraitor: 45
Life=tennis+math: 37
junggi: 32[/hide][hide="Half the Average"]Half the average was about 37[/hide][hide="Below half the average"]junggi[/hide][hide="Scores"]Alanchou: 130
Klebian: 72
perfect628: 70
mustafa: 66
BanishedTraitor: 20
Life=tennis+math: 15 [/hide][hide="Over 100"]AlanChou[/hide][hide="Round 4"]Round 4 due Sunday at 9:00 PM.  Hopefully I wont forget this time.[/hide]

Oops, I forgot to submit :D

Probably a good thing, considering that I would have been far under half the average :roll: .

I'll cautiously move back to 36

[quote="miyomiyo"][quote="junggi"]i lost[/quote]I lost.

[size=9]"The message contains only quote and code tags. Please add some text of your own."  No it doesn't.[/size][/quote]

miyomiyo lost.

Perfectly on time!  :o
[size=200][b]ROUND 4 RESULTS[/size][/b]
[hide="Submissions"]Klebian: No submission (100)
perfect628: No submission (100)
Life=tennis+math: No submission (100)
BanishedTraitor: 39
mustafa: 35[/hide][hide="Half the Average"]Half the average was about 37[/hide][hide="Below half the average"]mustafa[/hide][hide="Scores"]Klebian: 135
perfect628: 133
Life=tennis+math: 78
BanishedTraitor: 22 [/hide][hide="Over 100"]Klebian, perfect628[/hide][hide="Round 5"]Round 5 due Tuesday at 5:00 PM.[/hide]

Okay, I'm ending this game (Future HTA games should have activity checks at the beginning and a maximum of 15 players)
[size=200][b]VERY EARLY ROUND 5 RESULTS[/size][/b]
[hide="Submissions"]Life=tennis+math: No submission (100)
BanishedTraitor: No submission (100)[/hide][hide="Half the Average"]Half the average was 50[/hide][hide="Below half the average"]Nobody[/hide][hide="Scores"]Life=tennis+math: 128
BanishedTraitor: 72 [/hide][hide="Over 100"]Life=tennis+math[/hide][hide="We have a winner!"]Congratulations BanishedTraitor![/hide]

WOOHOO I won for no apparent reason. I reasoned that if I don't submit anything, I should win anyway  :D . My life's not so bad after all.

[quote="mustafa"]Huh. There's a pm in my outbox clearly stating that I wanted to submit 25 for round 3, but you said I didn't submit anything. Not that I mind, because I would have lost, but I just want to say that I had sent a pm to you[/quote]Hm.  I guess I didn't recieve it.
[quote="life=tennis+math"]BUT DUDE WTF
THATS NOT FAIR
i intended to psyche out my opponent by not submitting until tomorrow.[/quote]Sorry, I needed to wrap things up anyway.
[quote="junggi"] i lost

new game?[/quote]I lost.  Somebody else can do the next game; I'm going away for a month, so....


I know this problem is probably really easy but I cant seem to work it out!

Prove that for any two events A and B, the probability that exactly one of the two events will occur is given by the expression P(A) + P(B) - 2P(A ∩ B)

What is the probability of $ (A\text{ and not }B)$? What about $ (B\text{ and not }A)$?

Does this make sence?

P(exactly one of A and B will occur) =
P(A ∪ B - A ∩ B) = P(A ∪ B) - P((A ∪ B) ∩ (A ∩ B)) = P(A ∪ B) - P(A ∩ B) =
P(A) + P(B) - P(A ∩ B) - P(A ∩ B) = P(A) + P(B) - 2 P(A ∩ B)


function
algorithm
abstract algebra
linear algebra
linear algebra unsolved

Ok, just to clarify things, this was a problem from my homework last week, but I want to see if there are any other solutions.
Now, the problem, exactly as it says in the book.
Let $S$ be a set, $\mathbb{F}$ a field, and $V(S,\mathbb{F})$ the space of all functions from $S$ to $\mathbb{F}$
$(f+g)(x)=f(x)+g(x)$
$cf(x)=(cf)(x)$
Let $W$ be any $n$-dimensional subspace of $V(S,\mathbb{F})$. Show that there exists points $x_1,\ldots x_n \in S$ and $f_1,\ldots ,f_n \in W$ such that $f_i(x_j)=\delta_{i,j}$ where $\delta_{i,j}$ is $1$ if $i=j$ and $0$ otherwise.

This problem seems nice to me...
I hope it is not trivial...
Thanks :)

I would prove this by algorithm:
Let $x_1$ be a point at which not all elements of $W$ are zero. This is possible because $W$ contains a nonzero function. Let $g_1$ be an element of $W$ with $g_1(x_1)=1$; take an arbitrary function which is not zero at $x_1$ and scale it.

Now let $W_1=\{f: f\in W\textrm{ and }f(x_1=0)\}$. We have an obvious projection from $W$ to $W_1$ given by $f\mapsto f-f(x_1)g_1$, which has a kernel of dimension 1. By standard theorems, $W_1$ has dimension $n-1$,  and we continue the process.

This gets us points $x_1,x_2,\dots,x_n$ and functions $g_1,g_2,\dots,g_n$ such that $g_k(x_k)=1$ and $g_k(x_j)=0$ if $j<k$. Define $f_n=g_n$, $f_{n-1}=g_{n-1}-g_{n-1}(x_n)f_n$, $f_{n-2}=g_{n-2}-g_{n-2}(x_{n-1})f_{n-1}-g_{n-2}(x_n)f_n$ and so on.

These functions are of course a basis for $W$, and this is basically Gaussian elimination..


Is 2,000,000 minutes closest to 2 years, 3 years, or 4 years?

$ \frac{2000000}{60\times24\times365}\equal{}\frac{2500}{657}\approx3.805\cdots$.
Thus, $ 2,000,000$ minutes closest to $ \boxed{4}$ years.

[b]Remark[/b]
$ 2,000,000 \ \text{minutes} \equal{} 3 \ \text{years} \plus{} 293 \ \text{days} \plus{} 21 \ \text{hours} \plus{} 20 \ \text{minutes}$.

It may be 292
 days, depending on whether there is a leap year

Submitting "2000000 minutes" to wolframalpha.com gives the anwer:
1389 days or 3.803 average Gregorian years

If you've seen the musical "Rent", you would know that 1 year is 525,600 minutes, so 2,000,000 minutes is about 4 years.

How would you do this quickly without a calculator? :o

I mean you proabably would have the best chance guessing, since there are only 3 choices to choose from


function
real analysis
real analysis unsolved

Find all functions f from R+ to R+ which satisfy
            y^(f(x)^3) + [x^(f(y) 2 )]*[y^(f(x))]  =  (y+xy)^[f(x^3+y 2 *x)]

Particular solution f(x)=1 for all x \in R+


geometry
perimeter
conics
ellipse
analytic geometry
algorithm

I am having trouble with a C program. The program is

Write a program that gets as an input:
• A perimeter (boundary) of a “closed” shape, such as an ellipse, circle, etc,.
• Coordinates of an interior point.
The program should fill in a recursive way the area (enclosed by) this shape from the given
interior point without exceeding the boundary of this shape.

Instructions
The input will include:
• m lines, each including n characters. If the character read is not a space, the character belongs
to the boundary of the shape.
• Two numbers that specify the coordinates, namely the row and column indices of the point
from which the program is supposed to start filling the shape.
You can assume that this input point is inside the shape and that the shape is really closed. The
interior point is such that to any direction you “go” from this point, you will eventually reach the
boundary.
Note that the perimeter is represented as a sequence of characters (asterisks).

Which bit are you having trouble with? Think about how you would solve this on paper - that should give you a simple algorithm. The rest is converting the algorithm into code - so it really depends on where you are stuck.

I am having trouble with the recursive algorithm. I can't think of a simple base case.

This is the code of the recursion so far...

[code]
else
{
return
(areafill(array[][],rownum,columnnum+1)&&areafill(array[][],rownum,columnnum-1)&&areafill(array[][],rownum+1,columnnum)&&areafill(array[][],rownum-1,columnnum));
}
[/code]

I am trying to use the [url=http://student.kuleuven.be/~m0216922/CG/floodfill.html#4-Way_Recursive_Method_floodFill4]4-way recursive algorithm[/url]

But I would like to know the way for making the program go over every point in the array once.

If you've already filled in a square, you don't need to fill it in again, right?


Given a traingle $ ABC$with the perimeter $ l$. Determine the shape of the triangle such that $ l\equal{}m\overline{BC}\equal{}n\overline{CA}$ for positive integers $ m,\ n$.

[hide="hint"]$ l\equal{}a\plus{}b\plus{}c\equal{}>l\equal{}l/m\plus{}l/n\plus{}c\equal{}>c\equal{}l[1\minus{}1/m\minus{}1/n]$ clearly this gives $ mn>m\plus{}n$
now $ a\plus{}b>c\equal{}> m\plus{}n\geq mn/2$[/hide]


Support
integration
limit
complex analysis

Let $ \Gamma_n(\alpha): \equal{}\frac{1\minus{}q^{\alpha\minus{}n}}{1\minus{}q^{\minus{}\alpha}}$ where $ q>0$ and $ \alpha\in\mathbb C$.
1) Find residue of $ \Gamma_n$ at $ \alpha\equal{}0$ (with proof).
2) For each $ \phi$ is a intergrable with compact support in $ {\mathbb R}^n$. Show that $ k(\alpha): \equal{}\int \frac{|x|^{\alpha\minus{}n}}{\Gamma_n(\alpha)}\phi(x)dx$ is analytic in $ \alpha\neq0,n$.

For the first one, I'd first check if it's a simple pole by evaluating:

$ \mathop\lim\limits_{\alpha=0}\frac{\alpha q^{\alpha}(q^{\alpha}-q^{2\alpha-n})}{q^{\alpha-1}}$

which if $ n\ne 0$ exist and is not zero.  Thus it's a simple pole and easy to evaluate:

$ \mathop\text{Res}\limits_{\alpha=0}
\{\frac{1-q^{\alpha-n}}{1-q^{-\alpha}}\}=
\frac{1-q^{\alpha-n}}{ln(q)q^{-\alpha}}]_{\alpha=0}$


inequalities
IMO Shortlist
inequalities unsolved

note that $ LHS \ge \dfrac{(a\plus{}b\plus{}c\plus{}d)^2}{4(ab\plus{}ac\plus{}ad\plus{}bc\plus{}bd\plus{}cd)}$ by cauchy schwarz and we need to prove that $ 3(a\plus{}b\plus{}c\plus{}d)^2 \ge 8(ab\plus{}ac\plus{}ad\plus{}bc\plus{}bd\plus{}cd)$ or equivalently $ 3(a^2\plus{}b^2\plus{}c^2\plus{}d^2) \ge 2(ab\plus{}ac\plus{}ad\plus{}bc\plus{}bd)$ which follows from pairing-AM-GM.

[quote="teerawat"]Please write down the complete solution of this problem .Thank you.[/quote]

He posted the complete solution . This problem is from IMO Shortlist 1993


The four partners in a business decide to split the profits of their company in the ratio 2:3:3:5. If the profit one year is $ \$26,000$ what is the largest number of dollars received by any of the four partners?


[color=purple] Prove that :$ m_{a} \plus{} l_{b} \plus{} l_{c}\le\sqrt {3}s\le\sqrt {3s^2 \plus{} \frac {(b \minus{} c)^2}{4}}\le\sqrt {3s^2 \plus{} \frac{3(b \minus{} c)^2}{4}}$
   The first was proposed by Pham Gia Vinh Anh , the 2nd is mine and the 3rd is of a member of a math forum in Viet Nam  :) [/color]

[color=purple]Mod please delete this post [/color]


geometry
3D geometry
tetrahedron
circumcircle
sphere
geometry proposed

Let $O$ be the circumcenter of the face $ABC$ of the tetrahedron $ABCD$. The sphere $\mathcal S$ centered at $O$ through $A,B,C$ cuts $DA,DB,DC$ again at $A',B',C'$ respectively. Show that the tangents through $A',B',C'$ to $\mathcal S$ intersect at the circumcenter of $A'B'C'D$.

Suppose $P$ be the point of concurence of these 3 planes.Prove that projection of $P$ in planes $DAB$, $DAC$,$DBC$ is circumcenter of $DA'B'$,$DA'C'$,$DB'C'$ that is easy. This proves the problem.


Bill and Sam bought a rectangular land plot that is 1 mile long by ½ mile wide. They subdivide the land
by building a fence diagonally across the rectangle. What is the approximate length of the fence?

A) 1584 feet B) 5280 feet C) 5685 feet D) 5903 feet E) 6178 feet

It is necessary to know that $ 1mi.\equal{}5280ft$. The length of the fence, by the Pythagorean Theorem, is then $ \sqrt{2640^2\plus{}5280^2}\equal{}2640\sqrt{5}\approx5903$. So the answer is D.


induction
limit
number theory unsolved
number theory

Let the Let the sequences $(a_n)$ and $(b_n)$ of positive integers satisfy
             $a_{n+1}=na_n+1$ and $b_{n+1}=nb_n-1$ .
Show that these two sequences have only finitely many terms in common

$a_{n+1}=na_n+1=n((n-1)a_{n-1} +1)+1=n(n-1)a_{n-1}+n+1$
$a_{n+1}=n(n-1)(n-2)a_{n-2}+n(n-1)+n+1$

by induction
$a_{n+1}=\frac{n!}{(n-k-1)!} a_{n-k} + \sum_{k=0}^{k-3} \frac{n!}{(n-k)!}$

so for k=n-1 $a_{n+1}=n!a_1 +\sum_{k=0}^{n-4} \frac{n!}{(n-k)!}$
same thing with $b_{n+1}$
$b_{n+1}=n!b_1 -\sum_{k=0}^{n-4} \frac{n!}{(n-k)!}$
it can be seen that a is increasing and b decreases => they have finitely many
common terms,maybe even none.

It is easy problem.
Denote $c_1=a_1,c_n=\frac{a_n}{(n-1)!},d_1=b_1,b_n=\frac{b_n}{(n-1)!},h_n=\sum_{k=1}^n \frac{1}{k!},h_n\to e-1.$
 We have $c_{n+1}=c_n+sum_{k=1}^n \frac{1}{k!}=a_1+h_n,d_{n+1}=d_1-h_n.$ We have $c_n\to a_1-1+e,d_n\to b_1+1-e.$
Let $a_n=b_m$. If n=m $a_n=b_n\Longrightarrow c_n=d_n$. Because 2e is not integer $c_n=d_n$ can be true for a finite numbers n. If $n\not =m,$ then exist K: $|n-m|<K$, and for all l=n-m we have finetely solutions, because $d=\lim_n d_n,c=\lim_n c_n$ are distinct irrational numbers.


geometry
incenter
Euler
geometric transformation
reflection
circumcircle
conics

Internal bisector of $\angle (\ell_b,\ell_c)$ cuts $r$ at the incenter $I.$ Construct incircle $(I)$ centered at $I$ and tangent to $\ell_b,\ell_c$ at $E,F$ and let $D$ be the unknown tangency point of $(I)$ with $\ell_a \equiv BC.$

It's known that $r$ is the Euler line of $\triangle DEF$ containing its unknown orthocenter $T.$ Since the reflection $U$ of $T$ on $EF$ is on $(I),$ then the reflection of $r$ towards $EF$ will intersect $(I)$ at $U$ and the anti-Steiner point of $r$ WRT $\triangle DEF,$ i.e. the Feuerbach point of $\triangle ABC,$ giving rise to 2 solution triangles $\triangle ABC.$ Perpendicular from $U$ to $EF$ cuts $(I)$ again at $D$ and the tangent $\ell_a$ of $(I)$ at $D$ completes $\triangle ABC.$

Variation:
Lb, Lc and HI line (instead of OI ine).

My "natural" (after the original problem) variation was already solved by Luis with projective geometry.
The locus of the orthocenter of the triangle ABC, as BC moves touching the known incircle, is a hyperbola.

http://www.artofproblemsolving.com/Forum/viewtopic.php?f=46&t=224690

I am wondering if there is a construction with no use of conics.

Note: I guess that the locus of the circumcenter is a hyperbola as well.

And now another construction which can be done without conics:

Let Lb, Lc be the lines of AC,AB, resp., Lo, Lh two other lines containing the circumcenter, orthocenter, resp.
To construct the triangle. (As I said, it is possible to construct it without conics).
Note: We can replace Lo with any curve containing O.....  :-)


inequalities trigonometry function logarithms calculus derivative

[hide]
Too hard for "Pre-Olympiad"? If this were posted in "Olympiad", it would be cracked instantly :-)
[/hide]

this is just putting in the law of cosines...and then there is some inequality that finishes it (i don't really know inequalities, so i don't know the name), but it still seems pretty trivial

[quote="fuzzylogic"]Let $ABC$ be an acute-angled triangle with sides $a,b,c$. Prove that
\[ (ab+bc+ca)\biggl(\frac{1}{b^2+c^2-a^2}+\frac{1}{a^2+b^2-c^2}+\frac{1}{a^2+c^2-b^2}\biggr) \geq 9.  \][/quote]

[hide="solution"]
$\cos A = \frac{b^2+c^2-a^2}{2bc} < 1$ since angles are all acute.
Then $b^2+c^2-a^2 < 2bc \Rightarrow \frac{1}{b^2+c^2-a^2} > \frac{1}{2bc}$
$\[ (ab+bc+ca)\biggl(\frac{1}{b^2+c^2-a^2}+\frac{1}{a^2+b^2-c^2}+\frac{1}{a^2+c^2-b^2}\biggr) > \[ (ab+bc+ca)\biggl(\frac{1}{2ab}+\frac{1}{2bc}+\frac{1}{2ca}\biggr)$

$\frac{1}{2}(ab+bc+ca)\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right) \geq \frac{1}{2}(1+1+1)^2 \geq\ \frac{9}{2}$

ugggh...... not 9. help?

[/hide]

[hide]
$\frac{1}{b^2 + c^2 - a^2} = \frac{1}{2bc\cos A}$.  So we just want $(\sum bc)(\sum \frac{1}{bc \cos A} ) \ge 18$.

From Cauchy, $LS \ge ( \sum \sqrt{\frac{1}{cos A} } )^2$.  We have $\sum \sqrt{\frac{1}{cos A}} \ge 3\sqrt[6]{\frac{1}{\prod \cos A}}$, from AMGM.  So we just need $\prod \cos A \le 1/8$.  But the function $\ln \cos x$, with derivative $-\sec^2 x$, is concave.  So by jensen we are done.
[/hide]


The AoPSWiki does that. Start at [[LaTeX:Basics]].

[quote="buzzer11"]I use TeXnicCenter for USAMTS, but now I am starting to use it for school homework. Can someone please teach me how to make a document from scratch? I can't find anywhere online that does... :)[/quote]

Me too. You can also look at the usamts.tex file as an example of a page setting.


We are now taking registrations for Year 19.  If you have already registered online, then all you have to do is log in, go to the Profile page (under My USAMTS) and set your grade.  You can send in your physical entry form with your first round of solutions.  The first round will be released in August.

[quote="kstan013"]So do we set our grade for year 19, or will it automatically adjust?[/quote]

Set your grade.  This will complete your online registration for Year 19.

[quote="ehehheehee"]If we normally e-mail our solutions, does that mean we should fax or mail our entry forms?  I am not sure if USAMTS would accept a scanned image.[/quote]
Last year I e-mailed my entry forms with my first set of solutions.  The only problem that I had was that, at first, they apparently did not realize that my solutions weren't the only attachments, but it worked out in the end.

[quote="ehehheehee"]If we normally e-mail our solutions, does that mean we should fax or mail our entry forms?  I am not sure if USAMTS would accept a scanned image.[/quote]

Fax, mail, or scan - all are fine.


limit
logarithms
calculus
calculus computations

Prove from the definition of the limit of a sequence that $\lim_{n\to\infty}a^{n}= 0$ for $0<a<1$.

[hide="Hint"]Let $\frac{1}{a}=(1+x)$, $x>0$ Then $(1+x)^{n}\le 1+xn, \ n\in\mathbb{N}$. [/hide]

[quote="Jimmy"]Prove from the definition of the limit of a sequence that $\lim_{n\to\infty}a^{n}= 0$ for $0<a<1$.
[/quote]

$(\epsilon>0)\& (0<|a|<1)\&(n>\frac{\ln \epsilon}{\ln |a|}) \Longrightarrow |a|^{n}< \epsilon$  :diablo:
 :cool:

Nice solution, but I forgot to mention that the class from which I got the problem was only allowed to use basic arithmetic, no nth roots or logarithms.

A third approach would be this one:

$\left\{ a^{n}\right\}_{n \geq 1}$ is decreasing and positive, hence has a limit (its $\inf$ that is). $a_{n+1}= a \cdot a_{n}$, so by taking the limit we get that $L = a \cdot L \, \, \Longleftrightarrow \, \, (a-1) \cdot L = 0$. Since $a \in \left( 0, 1 \right)$, it follows that $L = 0$.


If $x=3$ and $y=2$, then what is the value of $\frac{2x^3-3y^2}{6}$?


algebra
polynomial
LaTeX
blogs
complex numbers

Suppose $ r$ is a root of the polynomial
$ x^4\minus{}x^3\plus{}x^2\minus{}x\plus{}1$

Calculate
$ r^20\minus{}r^15\plus{}r^10\minus{}r^5\plus{}1$
Sorry about the LaTex, the powers are supposed to be 20,15,and10 respectively.
Help please!
The answer is [hide]5[/hide]

[hide]If $ r^4\minus{}r^3\plus{}r^2\minus{}r\plus{}1\equal{}0$, then
$ (r^4\minus{}r^3\plus{}r^2\minus{}r\plus{}1)(r\plus{}1)\equal{}r^5\plus{}1\equal{}0$, so $ r^5\equal{}\minus{}1$
It easily follows that the answer is 5.[/hide]

Ah, such an elegant solution and it fell right through my fingers!
Thank you.

Note that the polynomial you gave is the 10th [url=http://mathworld.wolfram.com/CyclotomicPolynomial.html]cyclotomic polynomial[/url], so that all of its roots are the only complex numbers satisfying $ r^5 \equal{} \minus{}1$, with the exception of $ \minus{}1$.

(one of these days I'm going to write that cyclotomic post on my blog so that I can link to that instead :roll: , but [url=http://www.artofproblemsolving.com/Forum/weblog.php?w=984]what I have now[/url] should still be slightly enlightening)


can anyone help and pls pm them to me? (If you have the solutions too, great, but even if only the ans keys it would help me to know what I got right or wrong when practicing. thanks a lot everyone!

uh, sorry if this post was redundant - I just didn't want people thinking I found answer I needed from that posted link ..  :blush:

I suggest you to ask your MC coach or go to official MC sites.

you could ask your coach to buy them
[url=http://www.mathcounts.org/webarticles/anmviewer.asp?a=183&z=6]MC store[/url]


The number of terms in the expansion of $ [(a\plus{}3b)^2(a\minus{}3b)^2]^2$ when simplified is:

$\textbf{(A)}\ 4\qquad
\textbf{(B)}\ 5 \qquad
\textbf{(C)}\ 6 \qquad
\textbf{(D)}\ 7 \qquad
\textbf{(E)}\ 8$

[quote="Smartguy"]The number of terms in the expansion of $ [(a \plus{} 3b)^2(a \minus{} 3b)^2]^2$ when simplified is:
(A) 4
(B) 5
(C) 6
(D) 7
(E) 8[/quote]

we have
\[ (a^2\minus{}9b^2)^4
\]
expand to and we get 5 terms.

He factored out the two squares and put it on the outside to make it to the fourth, then he multiplied the two inside terms.

[quote="Poincare"]How did you get that?[/quote]
\[ \ [(a\plus{}3b)^2(a\minus{}3b)^2]^2 \equal{} [(a\plus{}3b)(a\minus{}3b)]^4 \equal{} (a^2\minus{}9b^2)^4\]
Now it is clear from the binomial theorem that expanding this gives $ 5$ terms (in general, $ (x\plus{}y)^4$ has a term involving each of $ x^4, x^3, x^2, x$, and a constant term with respect to $ x$, which gives a total of $ 5$ terms).


geometry
3D geometry
pyramid
sphere
geometry solved

If it would be possible to receive help on the following questions, it would be great. I really need help, since I have been unable to figure these out yet.

1. A closed cylinder has radius 2 inches and height 6 inches. Thus the are of the base is 4 P square inches (using A=Pr2) The volume is________P  cubic inches.

2.For the same cylinder the total surface area is______ P square inches.

----------------------------------------------------------------------------------------------------
3. A 5 inch high pyramid has a square base with 3 inch sides. It's volume is ____ cubic inches

4. The volume of a sphere with radius (r) may be determined using the formula
(A)  2 Pr
(B) Pr2
(C) 4P2
(D) 4 Pr3
      3

I would really and truly be grateful  for any assitance.

I think this is not a proper section for it. Have you learnt your lesson? ;)  :D
Ok, but if it is needed I write only formulas.
If we have cylinder with area of the base $S$ and height $h$, then it's volume is $Sh$.If the radius of the base is $r$, then the total surface area is $2\pi r(r+h)$.
If we have a pyramid with height $h$ and with base area $S$, then it's volume is $\frac{1}{3}Sh$.
The volume of the sphere is $\frac{4}{3}\pi r^{3}$.


Find all positive integers $ n$ such that there exists a combination $ (k_1,k_2,...,k_n)$ of {$ 1,2,...,n$} satisfying the set {$ k_1,k_1\plus{}k_2,...,k_1\plus{}k_2\plus{}...k_n$} is the complete residue $ mod n$.
I have only found that $ n$ must be even, help me continue please!!

well if you think you've proven that $ n$ must be even.... the following selection shows that all even $ n$ work:

$ (k_1, k_2, ..., k_{2m}) \equal{} (2m, 1, 2m \minus{} 2, 3, 2m \minus{} 4, 5, ..., 2, 2m \minus{} 1)$


modular arithmetic
number theory proposed
number theory

Prove that :
With n be positive integer number , we have :
 $A = (1+2^{n}+3^{n}+4^{n}) \vdots 5$
( :maybe: )
[i]Louis Latin and Vicky[/i]

Checking n=1,2,3,4 suffices since $\text{ord}_{5}k| \phi (5)=4$. All of these cases turn out to be 0 mod 5, and we're done.

It's false if and only if $4|n$ (Fermat theorem: $1+1+1+1 \equiv 4 \pmod 5$)

Anyway you can prove that $p|1^{n}+2^{2}+...+(p-1)^{n}$ iff $p-1 \not | n$

You find a generator $g$ and you change $1,2,3.., p-1$ with all different powers of $g$

Then you have to show that $p|(g^{n})^{0}+(g^{n})^{1}+(g^{n})^{2}...+(g^{n})^{p-2}$

You know that $(g^{n})^{0}+(g^{n})^{1}+(g^{n})^{2}...+(g^{n})^{p-2}=\frac{(g^{n})^{p-1}-1}{g^{n}-1}$

$(g^{n}-1,p)=1$ and by Fermat theorem $p|(g^{n})^{p-1}-1$

[quote="piever"]Anyway you can prove that $p|1^{n}+2^{2}+...+(p-1)^{n}$ iff $p-1 \not | n$
[/quote]
I think I've already seen this somewhere: http://www.mathlinks.ro/Forum/viewtopic.php?t=40171 :)


function
parameterization
algebra
polynomial
integration
combinatorics proposed
combinatorics

Hallo,

I'm trying to evaluate the following sum :
$ n\sum_{i=1}^n \frac{1}{(n-i+1)^2}\binom{n}{i-1}x^{i-1}(1-x)^{n-i+1}$, $0<x<1$.
My guess is that for large n this tends to $\frac{1}{(1-x)^2}$, as $ n \rightarrow +\infty$
I know that
$n\sum_{i=1}^n \frac{1}{n-i+1}\binom{n}{i-1}x^{i-1}(1-x)^{n-i+1}\rightarrow \frac{1}{1-x}$  (1).
as $ n \rightarrow +\infty$.
My first attempt was to solve the differential equation (differentiate with respect to $x$, unknown function is the sum, in the parenthesis on the LHS)
$ \left (n\sum_{i=1}^n \frac{1}{(n-i+1)^2}\binom{n}{i-1}x^{i-1}(1-x)^{n-i+1}\right )'$=
$n\sum_{i=1}^n \frac{i-1}{(n-i+1)^2}\binom{n}{i-1}x^{i-2}(1-x)^{n-i+1}$+$n\sum_{i=1}^n \frac{1}{(n-i+1)}\binom{n}{i-1}x^{i-1}(1-x)^{n-i}$
which after a simple trick (plus and minus n on the numerator of the first sum on the RHS and substitute from the expressions above) becomes
$ \left (n\sum_{i=1}^n \frac{1}{(n-i+1)^2}\binom{n}{i-1}x^{i-1}(1-x)^{n-i+1})\right )'$=
$n^2\sum_{i=1}^n \frac{1}{(n-i+1)^2}\binom{n}{i-1}x^{i-2}(1-x)^{n-i+1}$
- $n\frac{1}{x(1-x)}$ +$n\frac{1}{(1-x)^2}$ as $ n \rightarrow +\infty$.
Unfortunately the solution is rather complex and it won't simplify any further.
Also tried direct computation in Maple v9.5, gave as a result generalized hypergeometric function with parameters [1,1,-n];[2,2];-x which again won't simplify.
Does anybody have any better idea on how I could go about it?
Thanks in advance,
Dimitris

What is your question?   Exact formula, or asymptotics of the sum with respect to $n$?

Asymptotics of the sum with respect to $n$ will be just fine.

If the sum is indeed close to $(1-x)^{-2}$, then you could try to multiply it by $(1-x)^k$ for some $k \geq 2$ (try $k=2$ first, obviously) and compare the result with the Bernstein polynomial approximation of the resulting continuous function on $[0,1]$.

[quote="dimitris11"]I'm trying to evaluate the following sum :
$n\sum_{i=1}^n \frac{1}{(n-i+1)^2}\binom{n}{i-1}x^{i-1}(1-x)^{n-i+1}, 0<x<1$[/quote]
Are you sure it is not $n^2\sum_{i=1}^n \frac{1}{(n-i+1)^2}\binom{n}{i-1}x^{i-1}(1-x)^{n-i+1}$? The latter would make more sense: it is (almost) the Bernstein polynomial for $\frac 1{(1-x)^2}$. Well, the function is not continuous and one term is missing (for a very good reason: it would be infinite), but I believe that the standard proof of convergence of the Bernstein polynomials to the function can be applied here with some modifications, provided that $n$ in front of the sum, indeed, appears with square. Did you try this approach?

Thanks to both fedja and fleeting_guest for bringing to my attention the bernstein polynomial approximation method.
In fact I think that infront of the sum should be $n(n-1)$ (not $n^2$) as with a quick look
it seems nearer the bernstein polynomial formulae I came across. Will work more on it,
but thanks a lot for the suggestions.

uhm..
if i've not become completely idiot, sum can be reduced to some easy sum by substitution and integration...
i mean, say $1-x = y$, and notice that this sum is $\displaystyle\int \sum \dfrac{1}{n+1-i} \binom{n}{i-1}x^i y^{n-i} dy$, and then reduce it to something in the form ${\displaystyle\int \dfrac{\displaystyle\int (x+y)^n dy}y dy}$ (or something similar to it), and then put again $x+y = 1$ and that's it...


A family has 12 children. What is the probability that the family has at least 4 boys, and at least 3 girls? Assume that the probability of a boy and the probability of a girl is equal. Express your answer as a common fraction.

[hide]There's 13 ways we to have boys and girls.The number of ways to have $x$ boys and $12-x$ girls is $\binom{12}{x}$, which adds up to $2^{1}2=4096$ possible cases. To have ate least 4 boys and 3 girls, we want either 4,5,6,7,8, or 9 boys, and 12 minus that number of girls. $\binom{12}{4}+\binom{12}{5}...+\binom{12}{9}=2211$.$\frac{2211}{2096}$[/hide]edited.oops...I thought that was a little easy for a national team :blush:

The cases aren't all weighted the same! Some are more likely than others.

$\frac{2211}{2096}$? That's odd... it seems a little too... big... to me. :rotfl:

Wow, $P(A) \le 1$?

try 'counting the wrong thing.'

I got 3718 when I added up 12C4 + ... + 12C9... so my answer is 3718/4096, whatever that is.

Express your answer as a [b]common[/b] fraction.


inequalities
function
inequalities proposed

If $ x,y,z$ are nonnegative numbers, then

(a) $ \sum\sqrt{x^{2}\plus{}xy\plus{}y^{2}}\ge\sqrt{4(x^{2}\plus{}y^{2}\plus{}z^{2})\plus{}5(xy\plus{}yz\plus{}zx)}$;

(b) $ \sum\sqrt{x^{2}\plus{}xy\plus{}y^{2}}\le\sqrt{5(x^{2}\plus{}y^{2}\plus{}z^{2})\plus{}4(xy\plus{}yz\plus{}zx)}$.

Part (a) is pretty easy. Just squaring and Use Cauchy-Schwarz regularly.

[quote="quangpbc"]
Done.The equality holds if only if $ x \equal{} y \equal{} z$[/quote]

More exactly, for $ x\equal{}y\equal{}z$ and $ y\equal{}z\equal{}0$ and permutation.

Actually, the following nicer inequality holds

$ \sum\sqrt{x^{2}\plus{}xy\plus{}y^{2}}\le 2\sqrt{x^{2}\plus{}y^{2}\plus{}z^{2}}\plus{}\sqrt{xy\plus{}yz\plus{}zx}$.

I think this problem is easy, too. WLOG, assume that $ a\ge b\ge c$ then

$ (a^{2}\plus{}ab\plus{}b^{2},a^{2}\plus{}ac\plus{}c^{2},b^{2}\plus{}bc\plus{}c^{2})$ majorizes $ (a^{2}\plus{}b^{2}\plus{}c^{2},a^{2}\plus{}b^{2}\plus{}c^{2},ab\plus{}bc\plus{}ca).$

The function $ \sqrt{x}$ is concave.

[quote="Vasc"]If $ x,y,z$ are nonnegative numbers, then

(a) $ \sum\sqrt {x^{2} \plus{} xy \plus{} y^{2}}\ge\sqrt {4(x^{2} \plus{} y^{2} \plus{} z^{2}) \plus{} 5(xy \plus{} yz \plus{} zx)}$;

(b) $ \sum\sqrt {x^{2} \plus{} xy \plus{} y^{2}}\le\sqrt {5(x^{2} \plus{} y^{2} \plus{} z^{2}) \plus{} 4(xy \plus{} yz \plus{} zx)}$.[/quote]

[quote="Vasc"]Actually, the following nicer inequality holds

$ \sum\sqrt {x^{2} \plus{} xy \plus{} y^{2}}\le 2\sqrt {x^{2} \plus{} y^{2} \plus{} z^{2}} \plus{} \sqrt {xy \plus{} yz \plus{} zx}$.[/quote]
See here: http://canhang2007.wordpress.com/2009/10/30/inequality-27-v-cirtoaje/


function
vector
real analysis
real analysis theorems

In normal optimizations problems we have a desired function to minimize/maximize with some constraints.

But IF we have 2 things to maximize/minimize (or some combination of both) simultaneously?

Example: The Error of 2 functions is in function of another 2 functions in function of the same dependent variable of the first 2 functions. In this case we don’t want to maximize the error of each case, but maximize the error of the 2 things simultaneously.
For that we can make a lot of process.

Maximize the Sum of the error.
Maximize the generalized mean of the error.
etc.

So the question is. What is the more suitable process to correlate the propagation of uncertainty of this problem?
Of course the problem depend of the statement, But please, be general.
How Can I choose the process without you need to see my problem? Because If you see, the solution wouldn’t be general,

This is called [url=http://www-fp.mcs.anl.gov/otc/Guide/OptWeb/multiobj/]multi-objective optimization[/url] (or vector-valued optimization). The article outlines a few approaches and gives a bunch of references.


If a belongs to gaussian integer and N(a) is a rational prime, then a is not a prime in gaussian integer.
N(a) refers to norm of a.

[quote="t0rajir0u"]Exactly the opposite.  $ a$ must be a Gaussian prime in this case.[/quote]

This is because $ N(wz) \equal{} N(w)N(z)$ for gaussian integers $ w$ and $ z$. So if $ N(a)$ is a rational prime and $ a$ is not a Gaussian prime (i.e. $ a \equal{} pq$, $ p$ and $ q \in \mathbb{Z}[i]$ with neither p nor q of norm 1), then $ N(a) \equal{} N(p)N(q)$ contradicts the fact that $ N(a)$ is a rational prime.

By the way, this should probably be in the Groups, Rings, and Fields Forum.


The product of positive integers $ x$, $ y$ and $ z$ equals 2004. What is the minimum possible value of the sum $ x \plus{} y \plus{} z$?

The minimum possible sum occurs when $ x$, $ y$ and $ z$ are as close to each other as possible.

$ 4\times 3\times 167$

$ 4\plus{}3\plus{}167\equal{}\boxed{174}$


I am not sure whether "Watch this topic for replies" is still working. I haven't received any email notifications since the merge.

Let me know if this persists over the next couple days - I think/hope I've fixed it.


I recently took a test on graphing and got three problems wrong.  I need help figuring out the formulas, as i was able to draw them and it would be impractical on here.  Thanks

1.  Draw a simple disconnected graph for 3 verticies with the maximum number of edges.  Repeat for 4 verticies and for 5 verticies.  If you did this for n verticies how many edges would you expect?  Assume you graph is an n-sided polygon.

Looking back i found: $\frac{(n-2)(n-1)}{2}$  is this right?

2.  Draw a simple connected graph for n verticies and n edges.  In general, how many unique graphs can you construct for a given value of n?  Assume your graph is an n-sided polygon.

3.  Draw a simple connected graph for n verticis and n+1 edges.  In general, how many unique graphs can you construct for a given value of n?  Assume you graph is an n-sided polygon with n>3.

Those are the three problems.  I think i found an answer to the first one but i really do not understand how to approach (2) and (3).  Thanks

number 1 is right: the maximum edge one will have is when you have a full or complete graph with $n-1$ vertices, and then 1 disconnected vertices.  So the number of edges will be $\binom{n-1}{2}$ or the formula you stated


a door of mass $M$ is swinging and has an initial angular velocity of $\omega_{0}$. Air resistance acts on the door, how does this affect its motion? and what is the angular position $\theta$ as  a function of time?

How do you specify air resistance? No coefficients are provided... Do you mean that the air resistance occurs due to gas molecules bombarding the door?

I mean air resistance in this case would be proportional to the linear velocity.  Wouldn't you have to integrate each individual linear velocity of the door to do this?


let: 0<x,y,z,t<1 satisfy:
       xy+xz+xt+yz+yt+xt=1+xyzt
show that: x+y+z+t<2

Not difficult!
We have:
$(1-x)(1-y)(1-z)(1-t) \ge 0$
Then
$1+\sum{xy}+xyzt \ge \sum{x}+\sum{xyz}$
$\leftrightarrow 2(1+xyzt) \ge \sum{x}+\sum{xyz}$
But $0 \le x,y,z,t \le 1$
Then $2xyzt \le xyz+xyt \le \sum{xyz}$
Then $\sum{x} \le 2$
We are done!

[quote="frejer"]Well,I have a different proof:
from the relative xy+yz+zt+tx+xz+yt=1+xyzt
we put x=tg(A/4); y=tg(B/4); z= tg(C/4); t=tg(D/4)
Here:A,B,C,D is angles of a quadrilateral.
The problem is so easy later.[/quote]

hi, can you show a proof of this identity and post more about identities of quadrilteral?


[quote="lucio"]Let triangle $ ABC$, three points $ D,E,F$inside an triangle$ ABC$such that angle
$ DAB$ = angle $ DAE$=angle$ EAC$; angle $ ECA$= angle$ ECF$= angle $ FCB$; angle $ FBC$ = angle$ FBD$= angle $ DBA$.
Prove that : Equilateral triangle $ ABC$[/quote]
I think you mean that we wish to show that $ \triangle DEF$ is equilateral.  This is the well-known [b]Morley's Theorem[/b].  Please consult the following links:
http://www.cut-the-knot.org/triangle/Morley/index.shtml
http://en.wikipedia.org/wiki/Morley%27s_trisector_theorem
http://mathworld.wolfram.com/MorleysTheorem.html

(Sorry, the Wikipedia hyperlink does not work   :(   You will have to copy the URL and paste it into your address bar.)
[moderator edit: you can use %27 instead of ' in hyperlinks.]


algorithm
greatest common divisor
number theory
Euclidean algorithm

Let $c$ be a nonnegative integer, and define $a_n = n^2 + c$ (for $n \geq 1)$. Define $d_n$ as the greatest common divisor of $a_n$ and $a_{n + 1}$.
(a) Suppose that $c = 0$. Show that $d_n = 1,\ \forall n \geq 1$.
(b) Suppose that $c = 1$. Show that $d_n \in \{1,5\},\ \forall n \geq 1$.
(c) Show that $d_n \leq 4c + 1,\ \forall n \geq 1$.

Easy problem... :D

a) We have $ a_n\equal{}n^2$ and $ a_{n\plus{}1}\equal{}(n\plus{}1)^2$, and since $ (n,n\plus{}1)\equal{}1$, the result follows.

b) We have $ a_n\equal{}n^2\plus{}1$ and $ a_{n\plus{}1}\equal{}n^2\plus{}2n\plus{}2$. Now use Euclidean Algorithm.
If $ n$ is even, put $ n\equal{}2k$. So we have $ d_n\equal{}(4k^2\plus{}4k\plus{}2,4k^2\plus{}1)\equal{}(4k^2\plus{}1,4k\plus{}1)\equal{}(4k\plus{}1,k\minus{}1)\equal{}(k\minus{}1,5)$, and the result follows.
If $ n$ is odd, put $ n\equal{}2k\minus{}1$. So we have $ d_n\equal{}(4k^2\plus{}1,4k^2\minus{}4k\plus{}2)\equal{}(4k^2\plus{}1,4k\minus{}1)\equal{}(4k\minus{}1,k\plus{}1)\equal{}(k\plus{}1,5)$, and the result follows.

c) We have $ a_n\equal{}n^2\plus{}c$ and $ a_{n\plus{}1}^2\equal{}n^2\plus{}2n\plus{}c\plus{}1$. Now use Euclidean Algorithm again.
If $ n$ is even, put $ n\equal{}2k$. So we have $ d_n\equal{}(4k^2\plus{}4k\plus{}c\plus{}1,4k^2\plus{}c)\equal{}(4k^2\plus{}c,4k\plus{}1)\equal{}(4k\plus{}1,k\minus{}c)\equal{}(k\minus{}c,4c\plus{}1)$, and the result follows.
If $ n$ is odd, put $ n\equal{}2k\minus{}1$. So we have $ d_n\equal{}(4k^2\plus{}c,4k^2\minus{}4k\plus{}c\plus{}1)\equal{}(4k^2\plus{}c,4k\minus{}1)\equal{}(4k\minus{}1,k\plus{}c)\equal{}(k\plus{}c,4c\plus{}1)$, and the result follows.


A sheet of sixteen stamps is given as shown. Two stamps are connected if they share an edge. (the stamps are actually square, but since I do not know LATEX...)

    SS
    SS
SSSS
SSSS
SS
SS

How many ways are there to choose 3 connected stamps?

I don't know what you mean, but do you mean some arrangement of something like this:

[img]http://img256.imageshack.us/img256/4884/squaresfx4.jpg[/img]

There are 5 ways to get three connected stamps: It could be three vertical stamps in a row, three horizontal stamps in a row, or L-shaped (4 different orientations).

There are 8 ways for selecting 3 stamps in a vertical row and 4 ways to select 3 stamps in a horizontal row. For an proper L configuration of 3 stamps, there are 8 ways to do this, and for a left to right reversed L configuration, there are 7 ways. Each of these 2 L shapes has a mirror image configuration with the number of ways being equal to the original two L shapes due to the symmetry of the figure.

So there are 8 + 4 + 2(7 + 8) = 42 ways.